{"question_id": "P-20-Q7", "question": "The current $I_{1}$ (in A) flowing through $1\\Omega$ resistor in\nthe following circuit is", "question_images": ["images/image17.png"], "option_1": "0.2", "option_2": "0.4", "option_3": "0.5", "option_4": "0.25", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] Resistance of branch PQ = 2.5 [IMAGE] Apply Ohm's law:\nI = = 0.4A \\\\\ni = = 0.2A", "solution_images": ["images/image18.png", "images/image19.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-24-Q5", "question": "A small spherical droplet of density d is floating exactly half\nimmerse in a liquid of density ρ and surface tension T. The radius of\nthe droplet is (take note that the surface tension applies an upward\nforce on the droplet)", "question_images": [], "option_1": "$r = \\sqrt{\\frac{T}{(d + \\rho)g}}$", "option_2": "$r = \\sqrt{\\frac{3T}{(2d + \\rho)g}}$", "option_3": "$r = \\sqrt{\\frac{T}{(d + \\rho)g}}$", "option_4": "$r = \\sqrt{\\frac{2T}{3(d + \\rho)g}}$", "correct_option": 2, "numerical_answer": null, "solution": "$\\Rightarrow d \\cdot \\frac{4}{3}\\pi r^{3}g = \\rho \\cdot \\frac{2}{3}\\pi r^{3}g + 2\\pi rT$\n$\\Rightarrow \\frac{2}{3}r^{2}g(2d - \\rho) = 2T$\n$\\Rightarrow r = \\sqrt{\\frac{3T}{(2d - \\rho)}}$", "solution_images": ["images/image29.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-27-Q31", "question": "A container of a large uniform cross-sectional area A resting on\na horizontal surface holds two immiscible, non-viscous and\nincompressible liquids of densities and each of height\n(1/2)H as shown. The smaller density liquid is open to atmosphere. A\nhomogeneous solid cylinder of length\n[IMAGE] cross-sectional area (1/5) A is immersed such that it floats with its\naxis vertical to the liquid-liquid interface with length (1/4) L in\ndenser liquid. If D is the density of the solid cylinder then D=5d/n.\nFind n.,d yEck,d leku vuqizLFkdkV\ngSa,d leku ?kuRo\njgk gS fd bldh dqy yEckbZ dk ?kuRo okys nzo A;fn D csyu", "question_images": ["images/image202.png", "images/image203.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "", "solution_images": ["images/image204.png", "images/image205.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Hydrostatic pressure", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-07-Q14", "question": "One end of a uniform rod of mass m and length\n[IMAGE] is clamped. The rod lies on a smooth\nhorizontalsurface and rotates on it about the clamped end at a uniform\nangular velocity$\\omega$. The force exerted bythe clamp on the rod has a\nhorizontal component", "question_images": ["images/image28.png"], "option_1": "$m\\omega^{2}\\mathcal{l}$", "option_2": "zero", "option_3": "mg", "option_4": "$\\frac{1}{2}m\\omega^{2}\\mathcal{l}$", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image29.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Reaction from the hinge", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-24-Q7", "question": "The vessel shown in the figure has two sections of areas of\ncross-section [IMAGE] and\n[IMAGE]. A liquid of density ρ fills both the\nsections, up to a height h in each. Neglect atmospheric pressure", "question_images": ["images/image34.png", "images/image35.png", "images/image36.png"], "option_1": "the pressure at the base of the vessel is $2\\ h\\ \\rho\\ g$", "option_2": "the force exerted by the liquid on the base of the vessel is\n$\\ 2\\ h\\rho gA_{2}$", "option_3": "the walls of the vessel at the level X exert a downward force\n$h\\rho g\\left( A_{2} - A_{1} \\right)$ on the liquid.$\\ $", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$0 + \\rho g(2h) =$Press at base\n$F = A_{2}(\\rho g2h)$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Hydrostatic pressure", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-07-Q20", "question": "A uniform rod AB of mass m and length I at rest on a smooth\nhorizontal surface. An impulse P is applied to the end B. The time taken\nby the rod to turn through a right angle is", "question_images": [], "option_1": "$\\frac{2\\pi ml}{P}$", "option_2": "$\\frac{\\pi mI}{3P}$", "option_3": "$\\frac{\\pi ml}{12P}$", "option_4": "$\\frac{2\\pi ml}{3P}$", "correct_option": 3, "numerical_answer": null, "solution": "Impulse = change in momentum\n$\\therefore\\ \\ \\ \\ \\ \\ \\ P \\cdot \\frac{\\mathcal{l}}{2} = \\frac{mc^{2}}{12} \\cdot \\omega$\n(about centre of AB)\n$\\Rightarrow \\omega = \\frac{6P}{m\\mathcal{l}}$\n$\\frac{\\pi}{2} = \\omega t \\Rightarrow t = \\frac{\\pi}{2\\omega} = \\frac{\\pi m\\mathcal{l}}{2 \\times 6p}$\n$\\Rightarrow t = \\frac{\\pi m\\mathcal{l}}{12p}$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Angular impulse", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "Ph-25-Q17", "question": "The figure shows the P-V plot of an ideal gas taken through a\ncycle DCBAD then\ndks pØ DCBAD", "question_images": ["images/image126.png"], "option_1": "maximum temperature will occur during the process C → B → A.", "option_2": "negative work is done by the gas in the cycle DCBAD.", "option_3": "heat flows out of the gas during C → B → A.", "option_4": "All of the above.", "correct_option": 4, "numerical_answer": null, "solution": "Since temperature at C and A are same so maximum temperature\noccurs at a state between them during process CBA.\nSince cycle in PV graph is anticlockwise so work done by gas in\nnegative.\nDuring CBA work done by gas is negative and [IMAGE] so\n[IMAGE]. So heat is released by gas.\nD;ksafdC rFkk A ij rkieku leku gS rks eku bu izØe CBA\nds nkSjku gksxkA D;ksfd PV vkjs[k okekorZ", "solution_images": ["images/image127.png", "images/image128.png", "images/image127.png", "images/image128.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Cyclic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-03-Q9", "question": "A boy sits in a chair connected to a rope that passes over a\nfrictionless pulley. He pulls the loose end of the string with such a\nforce that the spring scale reads 250 N. The boy's weight is 320 N, and\nthe chair weighs 160 N. The force that the boy exerts on the chair is\n(Take g = 10 m/s2)", "question_images": ["images/image51.png"], "option_1": "250 N", "option_2": "320 N", "option_3": "160 N", "option_4": "[IMAGE] \n 250 N\n 320 N 160 N \n ( g = 10 m/s2)", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] TOPIC: NLM\nSUB TOPIC: EQUILIBRIUM", "solution_images": ["images/image53.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "Ph-26-Q4", "question": "If $\\rho$ is density of the material of a uniform rope of length\n$\\mathcal{l}$ and $\\sigma$ is the breaking stress. It is rotated in\nhorizontal circular path then maximum angular velocity it can have for\ncircular path:\n$\\mathcal{l}$yEckbZ dh le:i jLlh ds inkFkZ dk ?kuRo $\\rho$ gS rFkk\njLlh dh ruu lkeF;Z(breaking stress)$\\sigma$gSA bl ry esa\no`Ùkh; iFk ij ? \nosx Kkr djks", "question_images": [], "option_1": "$\\frac{1}{\\mathcal{l}}\\sqrt{\\frac{2\\sigma}{\\rho}}$", "option_2": "$\\frac{1}{\\mathcal{l}}\\sqrt{\\frac{\\sigma}{2p}}$", "option_3": "$\\frac{2}{\\mathcal{l}}\\sqrt{\\frac{2\\sigma}{\\rho}}$", "option_4": "$\\frac{2}{\\mathcal{l}}\\sqrt{\\frac{\\sigma}{\\rho}}$", "correct_option": 1, "numerical_answer": null, "solution": "$\\omega^{2} = \\frac{2\\sigma}{\\mathcal{l}^{2}\\rho} \\Rightarrow \\omega = \\frac{1}{\\mathcal{l}}\\sqrt{\\frac{2\\sigma}{\\rho}}$", "solution_images": ["images/image3.png"], "subject": "Physics", "topic": "Mechanical properties of matter", "subtopic": "Breaking stress of rod", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-09-Q17", "question": "A particle is executing SHM between points\n-[IMAGE] and [IMAGE], as shown in\nfigure-I. The velocity V(t) of the particle is partially graphed and\nshown in figure-II. Two points A and B corresponding to time\nare marked on the V(t) curve", "question_images": ["images/image99.png", "images/image100.png", "images/image101.png", "images/image102.png", "images/image103.png"], "option_1": "At time [IMAGE], it is going towards", "option_2": "At time [IMAGE], its speed is increasing.", "option_3": "At time [IMAGE], its position lies in between\n[IMAGE] and O", "option_4": "The phase difference ∆φ between points A and B must be expressed\nas 90°\n<∆φ< 180°.", "correct_option": 3, "numerical_answer": null, "solution": "At time [IMAGE], velocity of the particle is\nnegative i.e. going towar. From the graph, at\ntime [IMAGE], its speed is decreasing. Therefore\nparticle lies in between [IMAGE] and", "solution_images": ["images/image104.png", "images/image106.png", "images/image104.png", "images/image106.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Position of particle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-18-Q15", "question": "A body is given a velocity $\\sqrt{2gr}$ at the highest point of\na half circular smooth track which is joined by arough horizontal track\nwhose co-efficient of friction is $\\mu = 0.5$. Then the distance\ntravelled by body before it stops on horizontal track is: (r=1m)", "question_images": ["images/image23.png"], "option_1": "1 m", "option_2": "4 m", "option_3": "6 m", "option_4": "none of these", "correct_option": 4, "numerical_answer": null, "solution": "$W_{g} + W_{t} = 0 - \\frac{1}{2}m(2gr)$\n$mg2r - \\mu mgx = mgr\\ $ $x = \\frac{r}{\\mu} = \\frac{1}{0.5} = 2m$.", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work energy theorem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-13-Q11", "question": "The dimensions of$\\frac{B^{2}}{2\\mu_{0}}$, where B is magnetic\nfield and $\\mu_{0}$ is the magnetic permeability of vacuum, is", "question_images": [], "option_1": "MLT^-2", "option_2": "ML^-1T^-2", "option_3": "ML^2T^-1", "option_4": "ML^2T^-2", "correct_option": 2, "numerical_answer": null, "solution": "Energy density in magnetic field$= \\frac{B^{2}}{2\\mu_{0}}$\n$= \\frac{\\ Force\\ \\times \\ displacement\\ }{(\\ displacement\\ )^{3}} = \\frac{{MLT}^{- 2}L}{L^{3}} = {ML}^{- 1}T^{- 2}$", "solution_images": [], "subject": "Physics", "topic": "Magnetism", "subtopic": "Dimension of energy density in magnetic", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-19-Q9", "question": "Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of\nradii$R_{1}$and $R_{2}$, respectively. The electric fields $E_{1}$(on\n$S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that\n$E_{1}/E_{2} = R_{1}/R_{2}$. Then the\nratio$V_{1}(onS_{1})/V_{2}(onS_{2})$ of the electrostatic potentials on\neach sphere is", "question_images": [], "option_1": "${(\\frac{R_{1}}{R_{2}})}^{3}$", "option_2": "$\\frac{R_{1}}{R_{2}}$", "option_3": "$(\\frac{R_{2}}{R_{1}})$", "option_4": "${(\\frac{R_{1}}{R_{2}})}^{2}$", "correct_option": 4, "numerical_answer": null, "solution": "$\\frac{E_{1}}{E_{2}} = \\frac{r_{1}}{r_{2}}$\n$\\frac{V_{1}}{V_{2}} = \\frac{E_{1}r_{1}}{E_{2}r_{2}} = \\frac{r_{1}}{r_{2}} \\times \\frac{r_{1}}{r_{2}} = {(\\frac{r_{1}}{r_{2}})}^{2}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric potential", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-10-Q2", "question": "A bead of mass m is kept at a distance from the axis of\nrotation of a smooth massless tube inside it. The tube is rotating on a\nsmooth horizontal surface about a vertical axis perpendicular to the\nlength L of the tube and passing through one of its end with\nconstant angular velocity[IMAGE].", "question_images": ["images/image7.png", "images/image8.png"], "option_1": "The speed of the bead relative to tube when it is at a distance r\nfrom axis of rotation is2[IMAGE]", "option_2": "The speed of the bead relative to tube when it is at a distance\n'r' from axis of rotation is[IMAGE]", "option_3": "The work done by the centrifugal force in the frame of tube when\nthe bead reaches at a distance the axis of rotation starting", "option_4": "The work done by the centrifugal force in the frame of the tube\nwhen the bead reaches at a from the axis of rotation", "correct_option": 3, "numerical_answer": null, "solution": "$\\frac{v^{2}}{2} = \\omega^{2}\\frac{\\left( r^{2} - a^{2} \\right)}{2}$\n$V = \\omega\\sqrt{r^{2} - a^{2}}$\nApplying WET equation\n$W_{(Centrifugal\\ force\\ )} = \\frac{1}{2}{mV}^{2}$\n$W = \\frac{1}{2}m\\omega^{2}\\left( r^{2} - a^{2} \\right)$", "solution_images": ["images/image13.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Work energy theorem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-16-Q25", "question": "Consider a particle initially moving with a velocity 7 m/s\nstarts decelerating at a constant rate of$2m/s^{2}$.\nThe distance travelled in the fourth second is$\\frac{x}{10}m$. Find\nthe value of x.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "Here particle is decelerating so velocity of particle becomes\n after t=3.5 sec. the particle has a turning point at t=3.5 sec.\n$0 = 7 - 2t$\n$t = 3.5\\sec$\n$x_{1} = x_{3} - \\frac{1}{2} \\times 2 \\times 3^{2} = 12m$\n$y_{2} = 7 \\times 3.5 - \\frac{1}{2} \\times 2 \\times (3.5)^{2} = \\frac{49}{4}m$\n$\\Rightarrow y_{2} - x_{1} = \\frac{49}{4} - 12 = \\frac{1}{4}m$.\nDue to symmetry displacement of the particle at t=3 and t=4sec. is same.\nso total distance traveled in the fourth sec\nis$= \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}m$", "solution_images": ["images/image45.png"], "subject": "Physics", "topic": "Kinematics", "subtopic": "Uniform accelerated motion", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-21-Q12", "question": "Both the diodes used in the circuit shown are assumed to be\nideal and have negligible resistance whenthese are forward biased. built\nin potential in each diode is 0.7 V. For the input voltages shown in\nthefigure, the voltage (in volts) at point A is..........\nekusa rFkk vxzfnf'kd ¼QkjoMZ½ bu varjfufeZr foHkokUrj(built in potential), 0.7 VgSAfp=k\n (input) oksYVrk fcUnq A ij oksYVrk", "question_images": ["images/image31.png"], "option_1": "3", "option_2": "2", "option_3": "4", "option_4": "5", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] Let [IMAGE] Right diode is reversed biased and left diode is forward biased\n + Ik'p", "solution_images": ["images/image32.png", "images/image33.png", "images/image33.png", "images/image34.png"], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Diodes", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-13-Q17", "question": "An electron gun is placed inside a long solenoid of radius R on\nits axis. The solenoid has n turns/lengthand carries a current I. The\nelectron gun shoots an electron along the radius of the solenoid with\nspeedv. If the electron does not hit the surface of the solenoid,\nmaximum possible value of v is (all symbolshave their standard meaning)", "question_images": ["images/image19.png"], "option_1": "$\\frac{e\\mu_{0}nIR}{2m}$", "option_2": "$\\frac{e\\mu_{0}nIR}{m}$", "option_3": "$\\frac{2e\\mu_{0}nIR}{m}$", "option_4": "$\\frac{e\\mu_{0}nIR}{4m}$", "correct_option": 1, "numerical_answer": null, "solution": "$R_{\\max} = \\frac{R}{2} = \\frac{{mv}_{\\max}}{e\\mu_{0}in}$\n$V_{\\max} = \\frac{{Re}{\\mu_{0}in}}{2m}$", "solution_images": [], "subject": "Physics", "topic": "Magnetic effect of current", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-16-Q22", "question": "An elevator of height ascends with constant acceleration\n When it crosses a platform bolt drops from the top of the\nelevator. If the time for the bolt to hit the floor of the elevator\nis$\\sqrt{\\frac{\\lambda h}{g + a}}$ then find $\\lambda$.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "Choose the floor of the elevator as reference. The observer is\ninside the elevator take the downwarddirection as positive.\nAcceleration of bolt relative to elevators is\n$a^{'} = g - ( - a) = g + a$\n$h = \\frac{1}{2}a^{'}t^{2} = \\frac{1}{2}(g + a)t^{2}$\n$t = \\sqrt{\\frac{2h}{g + a}}$", "solution_images": [], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Collision with wall", "difficulty": "Tough", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-15-Q13", "question": "When photon of energy [IMAGE] strikes the surface\nof a metal A, the ejected photoelectrons have maximum kinetic energy Ta\neV and de-Broglie wavelength[IMAGE]. The maximum kinetic\nenergy of photoelectrons liberated from another metal B by photon of\nenergy [IMAGE] is [IMAGE] If the de-Broglie\nwavelength of these photoelectrons[IMAGE], then the work\nfunction of metal B is", "question_images": ["images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png", "images/image97.png"], "option_1": "2 eV", "option_2": "3 eV", "option_3": "1.5 eV", "option_4": "4 eV", "correct_option": 4, "numerical_answer": null, "solution": "Relation between De-Broglie wavelength and K. E. is", "solution_images": ["images/image98.png", "images/image99.png", "images/image100.png", "images/image101.png", "images/image102.png", "images/image103.png", "images/image104.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "Einstein's photoelectric equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-19-Q12", "question": "A long, straight wire of radius a carries a current distributed\nuniformly over its cross-section. The ratioof the magnetic fields due to\nthe wire at distance $\\frac{a}{3}$ and 2a, respectively from the axis of\nthe wire is", "question_images": [], "option_1": "$\\frac{1}{2}$", "option_2": "2", "option_3": "$\\frac{3}{2}$", "option_4": "$\\frac{2}{3}$", "correct_option": 4, "numerical_answer": null, "solution": "$B_{A} = \\frac{\\mu_{0}ir}{2\\pi a^{2}} = \\frac{\\mu_{0}i\\frac{a}{3}}{2\\pi a^{2}} = \\frac{\\mu_{0}i}{\\pi a^{2}}\\frac{a}{6} = \\frac{\\mu_{0}i}{6\\pi a}$\n$B_{B} = \\frac{\\mu_{0}i}{2\\pi(2a)}$\n$\\frac{B_{A}}{B_{B}} = \\frac{4}{6} = \\frac{2}{3}$", "solution_images": ["images/image17.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Ampere's circuital law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "Ph-25-Q3", "question": "Pressure versus temperature graph of the ideal gas at constant\nvolume V of an ideal gas is shown by a straight line A. Now, pressure P\nof the gas is doubled and the volume is halved: then the corresponding\npressure versus temperature graph will be shown by the line:-,d vkn\"kZ xSLk ds fLFkj xSl dk ncko cuke rkieku xzkQ\nrkieku xzkQ ykbu tk,xkA", "question_images": ["images/image4.png"], "option_1": "A", "option_2": "B", "option_3": "C", "option_4": "none of these buesa", "correct_option": 2, "numerical_answer": null, "solution": "$P = \\frac{nRT}{V} = \\frac{mRT}{MV}$\n(Where[IMAGE] number of moles)\nSo, at constant volume pressure versus temperature graph is a straight\nline passing through origin with slope mR/MV. As volume is halved,\nslope becomes two times. Therefore, pressure versus temperature graph\nwill be shown by the line B.\n eku ds fo:) fLFkj 0\n$\\therefore$Comets will never returns to the solar system because total\nenergy is +ve.", "solution_images": [], "subject": "Physics", "topic": "Gravitation", "subtopic": "Binding energy", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "Ph-26-Q12", "question": "For a fluid which is flowing steadily, the level in the vertical\ntubes is best represented by\nLFkk;h:i ls izokfgr nzo", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "From continuity equation, velocity at cross-section (1) is more\nthan that at cross-section (2).\nlkarR; lehdj.k ls vuqizLFk dkV(1) ij osx ] vuqizLFk dkV(2) ij osx", "solution_images": [], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Buoyancy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-11-Q8", "question": "A particle of mass m is dropped from a height h above the ground.\nAt the same time another particle of the same mass is thrown vertically\nupwards from the ground with a speed of[IMAGE]. If they\ncollide head-on completely in elastically, the time taken for the\ncombined mass to reach the ground, in units of [IMAGE] is", "question_images": ["images/image71.png", "images/image72.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] time for collision [IMAGE] After [IMAGE] And [IMAGE] at the time of collision", "solution_images": ["images/image77.jpeg", "images/image78.png", "images/image79.png", "images/image80.png", "images/image81.png", "images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png"], "subject": "Physics", "topic": "Collision", "subtopic": "elastic collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-21-Q22", "question": "The circuit shown below is working as a 8V dc regulated voltage\nsource. When 12V is used as input,the power dissipated (in mW) in each\ndiode is 8n. Find n. (considering both zener diodes are identical)\n__________.\n A tc blesa 12V oksYVrk yxk;h tkrh gS rks + esa gksus okyh mW esa 8n gksxh (nksuks T+khuj\nMk;ksM,d leku gSa) __________.", "question_images": ["images/image78.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "$i = \\frac{(12 - 8)}{(200 + 200)}A = \\frac{4}{400} = 10^{- 2}A$\nPower loss in each diode gkfu = (4)(10^-2 ) W\n= 40 mW", "solution_images": [], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Zener diode", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-12-Q8", "question": "The current i in the network is", "question_images": ["images/image9.png"], "option_1": "0.2A", "option_2": "0.3A", "option_3": "0A", "option_4": "0.6A", "correct_option": 2, "numerical_answer": null, "solution": "Both diodes are in reverse biased", "solution_images": ["images/image10.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-19-Q8", "question": "In the given circuit, value of Y is", "question_images": ["images/image11.jpeg"], "option_1": "toggles between 0 and 1", "option_2": "0", "option_3": "will not execute", "option_4": "1", "correct_option": 2, "numerical_answer": null, "solution": "$$\n[IMAGE] = 0", "solution_images": ["images/image13.png", "images/image14.png", "images/image15.png"], "subject": "Physics", "topic": "Digital electronics", "subtopic": "Lopic pates", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-20-Q24", "question": "A beam of electromagnetic radiation of intensity\n[IMAGE] is comprised of wavelength,\n[IMAGE] It falls normally on a metal (work function\n[IMAGE] ) of surface area of [IMAGE]. If\none in [IMAGE] photons ejects an electron, total number\nof electrons ejected in[IMAGE] is\n[IMAGE] then x is....", "question_images": ["images/image87.png", "images/image88.png", "images/image89.png", "images/image90.png", "images/image91.png", "images/image92.png", "images/image93.png", "images/image94.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "Energy of photon. [IMAGE] (so photoelectric\neffect will take place)\n[IMAGE] No. of photons falling per second\n[IMAGE] No. of photoelectron emitted per second", "solution_images": ["images/image95.png", "images/image96.png", "images/image97.png", "images/image98.png"], "subject": "Physics", "topic": "Modern physics", "subtopic": "Photoelectric emission", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-01-Q8", "question": "An electron starting from rest and accelerated through V =\n20,000 V in an x-ray tube loses half the energy it acquired as it\napproaches an atom of the target material. The remaining KE is lost in\nsubsequent interactions with other target atoms. Find the wavelength of\nthe radiation emitted in the first interaction.", "question_images": [], "option_1": "6.63", "option_2": "4.14", "option_3": "1.24", "option_4": "12.4", "correct_option": 3, "numerical_answer": null, "solution": "∆E = 1 × 104 eV = [IMAGE] λ = [IMAGE] nm = 0.124 nm", "solution_images": ["images/image29.png", "images/image30.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-07-Q4", "question": "A circular platform is free to rotate in a horizontal plane about\na vertical axis passing through its centre. A tortoise is sitting at the\nedge of the platform. Now the platform is given an angular velocity\n$\\omega_{0}$. When the tortoise moves along a chord of the platform with\na constant velocity (with respect to the platform) the angular velocity\nof the platform $\\omega(t)$ will vary with time t as", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "since, there is no external torque, angular momentum will\nremain conserved. The moment of inertia will first decrease till the\ntortoise moves from A to C and then increase as it moves from C and D.\nThe re fore $\\omega$ will initially increase and then decrease. Let R be\nthe radius of platform m the mass of disc and M is the mass of platform.\nMoment of inertia when the tortoise is at A\nand moment of inertia when the tortoise is at B\n$I_{2} = {mr}^{2} + \\frac{{MR}^{2}}{2}$\nhere $r^{2} = a^{2} + {\\lbrack\\sqrt{R^{2} - a^{2}} - vt\\rbrack}^{2}$\nFrom conservation of angular momentum\n$\\omega_{0}I_{1} = \\omega(t)I_{2}$\nsubstituting the values we can see that variation of $\\omega(t)$ is\nnonlinear.", "solution_images": ["images/image10.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "Ph-27-Q28", "question": "A vessel contains two immiscible liquids of densities\n[IMAGE]. A solid block of volume\n[IMAGE] and density [IMAGE] is tied to one end of a string and other\nis tied to the bottom of the vessel as shown in figure. The block is\nimmersed with [IMAGE] of its volume in the liquid of higher density and\n[IMAGE] in the liquid of lower density. The entire system is kept in an elevator\nwhich is moving upwards with an acceleration of a = g/2. Find the\ntension in the string. (Take g = 10 m\n[IMAGE] ) (in N),d ik=k esa nks ρ_1 = 1000 kg\nm^-3rFkk ρ_2 = 1500 kg m^-3 Hkjs gSaV = 10^-3 m^3vk;ru rFkk d\n= 800 kg m^-3?kuRo dk,d Bksl CykWd jLlh ca\ndk nqljk fljk fp=kkuqlkj ik=k ds ry ij ca\n $\\frac{2^{th}}{5}$Hkkx mPp ?kuRo nzo esa rFkk$\\frac{3^{th}}{5}$Hkkx\nfuEu ?kuRo ds nzo esa Mqck gSA lEiw.kZ Roj.k", "question_images": ["images/image180.png", "images/image181.png", "images/image182.png", "images/image183.png", "images/image184.png", "images/image185.png", "images/image186.png", "images/image187.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "[IMAGE] We analysed this problem From the reference frame of Elevator. Total\nbuoyant force On the block,\nCykWd ij dqy mRIykou cy\n[IMAGE] For equilibrium,", "solution_images": ["images/image188.png", "images/image189.png", "images/image190.png", "images/image191.png", "images/image192.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Archimedes principle", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-06-Q14", "question": "A particle of mass m is released from point A and it stops at B\nfind out the coefficient of friction", "question_images": ["images/image24.png"], "option_1": "$\\mu = \\frac{1}{4}$", "option_2": "$\\mu = \\frac{1}{3}$", "option_3": "$\\mu = \\frac{2}{3}$", "option_4": "None of these", "correct_option": 1, "numerical_answer": null, "solution": "$W_{g} + W_{f} = K_{f} + K_{i}$\n$mg\\left( \\frac{3h}{4} \\right) - \\mu mg3h = 0 - 0 \\Rightarrow \\mu = \\frac{1}{4}$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work energy theorem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-01-Q2", "question": "Four functions given below describe motion of a particle. (I) y\n= sin ωt - cos ωt, (II) y = sin3 wt, (III) y = 5 cos\n[IMAGE], (IV) y = 1 + ωt + ω2t2. Therefore, simple\nharmonic motion is represented by", "question_images": ["images/image6.png"], "option_1": "only I", "option_2": "I, II, III and IV", "option_3": "I and III", "option_4": "I and II", "correct_option": 3, "numerical_answer": null, "solution": "I and III are showing S.H.M. on the basis of superposition\nprinciple.", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-07-Q13", "question": "Find the moment of inertia of a uniform rectangular plate of\nmass M, edges of length\n and about itsaxis passing through\ncentre and perpendicular to", "question_images": ["images/image25.png", "images/image26.png"], "option_1": "$\\frac{M\\left( \\mathcal{l}^{2} + b^{2} \\right)}{6}$", "option_2": "$\\frac{M\\left( \\mathcal{l}^{2} + b^{2} \\right)}{12}$", "option_3": "$\\frac{M\\left( \\mathcal{l}^{2} + b^{2} \\right)}{3}$", "option_4": "None", "correct_option": 2, "numerical_answer": null, "solution": "Using perpendicular axis theorem l_3 = I_1 + I_2\n$I_{2} = \\frac{M\\mathcal{l}^{2}}{12}$\n$I_{3} = \\frac{M\\left( \\mathcal{l}^{2} + b^{2} \\right)}{12}$\n$I_{3} = \\frac{M\\left( \\mathcal{l}^{2} + b^{2} \\right)}{12}$", "solution_images": ["images/image27.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-23-Q30", "question": "The radius of a coil of wire with 21 turns is 0.22 m, and i_2\ncurrent flows clockwise in the coil as shown. A long straight wire\ncarrying a current is toward the left is located 0.05 m from the edge of\nthe coil. Themagnetic field at the centre of the coil is zero tesla. The\nratio $\\frac{i_{1}}{9i_{2}}$ is (Use$\\pi = 22/7$)", "question_images": ["images/image106.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "9", "solution": "$\\frac{\\mu_{0} \\cdot i_{1}}{2\\pi(d + r)} = \\frac{N \\times \\mu_{0}i_{2}}{2r}$\n$\\Rightarrow N = \\frac{i_{1}}{i_{2}} \\cdot \\frac{r}{\\pi(d + r)}$\n$\\frac{i_{1}}{i_{2}} = 81$", "solution_images": [], "subject": "Physics", "topic": "Moving charges and magnetism", "subtopic": "Magnetic field due to current carrying conducting", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-22-Q21", "question": "A uniform solid sphere of radius R and mass M purely rolls down\nan inclined plane. The coefficient of friction between the sphere and\nthe inclined plane is[IMAGE]. If the maximum value of\nangle of inclination for this to be possible is\n[IMAGE] then n is: (Take[IMAGE] )", "question_images": ["images/image55.png", "images/image56.png", "images/image57.png", "images/image58.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "7", "solution": "[IMAGE] put in (1)\n[IMAGE] Now [IMAGE] TOPIC:Rotaional dynamics\nSUB TOPIC: Rolling down the incline\nLEVEL: Moderate", "solution_images": ["images/image59.png", "images/image60.png", "images/image61.png", "images/image62.png", "images/image63.png", "images/image64.png", "images/image65.png", "images/image66.png", "images/image67.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-07-Q17", "question": "A small sphere of mass 1 kg is moving with a velocity\n$(6\\widehat{i} + \\widehat{j})ms^{- 1}$. It hits a fixed smooth wall and\nrebound with velocity $(4\\widehat{i} + \\widehat{j})ms^{- 1}$. The\ncoefficient of restitution between the sphere and the wall is", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "Impulse = Change in momentum\n$= 1(4\\widehat{i} + \\widehat{j}) - 1(6\\widehat{i} + \\widehat{j}) = - 2\\widehat{i}$\n \nWhich is perpendicular to the wall. Componentof initial velocity along\n$\\widehat{i} = 6\\widehat{i}$\n$\\Rightarrow$ Speed of approach =6 \nTherefore $e = \\frac{4}{6} = \\frac{2}{3}$", "solution_images": [], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-20-Q9", "question": "The distance between two slits in a Young's double slit\nexperiment is 3 mm. The distance of the screenfrom the slits is 1 m.\nWaves of wavelength 1 mm are incident on the plane of the slits\nnormally.The distance of the first maxima on the screen from the central\nmaxima will be: (in cm)", "question_images": [], "option_1": "36.6", "option_2": "35.4", "option_3": "34.4", "option_4": "35.4", "correct_option": 4, "numerical_answer": null, "solution": "At first maxima d sinθ = λ\n$3sin\\theta = 1 \\Rightarrow sin\\theta = \\frac{1}{3}$\n$\\tan\\theta = \\frac{y}{D} \\Rightarrow y = D \\times \\tan\\theta = 1 \\times \\frac{1}{\\sqrt{8}}m = \\frac{100}{\\sqrt{8}}cm = 35.4cm$", "solution_images": [], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-12-Q16", "question": "Two infinite planes each with uniform surface charge density\n$+ \\sigma$ are kept in such a way that the anglebetween them is 30°. The\nelectric field in the region shown between them is given by", "question_images": ["images/image18.png"], "option_1": "$\\frac{\\sigma}{2\\epsilon_{0}}\\left\\lbrack (1 + \\sqrt{3})\\overset{\\hat{}}{y} - \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "option_2": "$\\frac{\\sigma}{2\\epsilon_{0}}\\left\\lbrack \\left( 1 - \\frac{\\sqrt{3}}{2} \\right)\\overset{\\hat{}}{y} - \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "option_3": "$\\frac{\\sigma}{2\\epsilon_{0}}\\left\\lbrack (1 + \\sqrt{3})\\overset{\\hat{}}{y} + \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "option_4": "$\\frac{\\sigma}{\\epsilon_{0}}\\left\\lbrack \\left( 1 + \\frac{\\sqrt{3}}{2} \\right)\\overset{\\hat{}}{y} + \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "correct_option": 2, "numerical_answer": null, "solution": "$\\overset{\\rightarrow}{E} = \\frac{\\sigma}{2\\varepsilon_{0}}\\left\\lbrack \\left( 1 - \\frac{\\sqrt{3}}{2} \\right)\\overset{\\hat{}}{y} - \\frac{1}{2}\\overset{\\hat{}}{x} \\right\\rbrack$.", "solution_images": ["images/image19.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field due to infinite sheet of charge", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "Ph-27-Q8", "question": "The densities of wood and benzene at 0 ºC are 880 kg m^-3 and\n900 kg m^-3, respectively. The coefficientsof volume expansion of wood\nis 1.2 × 10^-3 ºC^-1.and that of benzene is\n1.5 × 10^-3 ºC^-1. Value oftemperature (in ºC) at which a piece of\nthis wood would just sink in benzene at the same temperature isT. Then\nfind value of 3×10^-4T (where T is tempture in ºC).\nydM+h rFkk csathu dk ?kuRo 0ºC ij Øe'k 880 kg m^-3rFkk 900 kg\nm^-3gSA ydM+h rFkk csathu izlkj xq.kkad Øe'k 1.2 × 10^-3\nC^-1rFkk 1.5 × 10^-3 ºC^-1gSA ftl rkieku ij ydM+h csathu esa\niw.kZ:Ik ls Mqcrh gS rks og rkieku T (ºC esa) gSA rc 3×10^-4T", "question_images": [], "option_1": "40", "option_2": "20", "option_3": "60", "option_4": "80", "correct_option": 2, "numerical_answer": null, "solution": "For just sink condition\nMqcus", "solution_images": ["images/image39.png", "images/image40.png", "images/image41.png", "images/image42.png", "images/image43.png", "images/image44.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Archimedes principle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-03-Q23", "question": "Two blocks of masses 2 kg and 4 kg are connected through a massless\ninextensible string. The\nco-efficient of friction between 2 kg block and ground is 0.4 and the\ncoefficient of friction between 4 kg block and ground is 0.6. Two forces\nF1 = 10 N and F2 = 20 N are applied on the blocks as shown in the\nfigure. Calculate the frictional force ( in N )between 4 kg block and\nground (Assume initially the tension in the string was just zero before\nforces F1 and F2 were applied)\n[IMAGE] 2 kg 4 kg 2 kg\n 0.4 4 kg \n0.6 F1 = 10 N F2 = 20 N \n4 kg (, F1 F2 )", "question_images": ["images/image141.png", "images/image141.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "18", "solution": "System will not move (Equilibrium)\nFor equilibrium of 2 kg block\n[IMAGE] 10 = 8 + T\n⇒ T = 2N\nFor equilibrium of 4 kg block\n[IMAGE] T + f2 = 20\n⇒ f2 = 18 N\nTOPIC: NLM\nSUB TOPIC: FRICTION", "solution_images": ["images/image142.png", "images/image143.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-03-Q5", "question": "A ball is thrown horizontal from the top of a tower 40 m high. The\nball strikes the ground at a point 80 m from the bottom of the tower.\nFind the angle that the velocity vector makes with the horizontal just\nbefore the ball hits the ground.", "question_images": [], "option_1": "45°", "option_2": "90°", "option_3": "37°", "option_4": "53°", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-08-Q5", "question": "If the length of steel rod is 5cm longer than copper rod at all\ntemperatures. ($\\alpha$ for copper $= 1.7 \\times 10^{- 5}ºC^{- 1}$ and\n$\\alpha$ for steel $\\left. \\ = 1.1 \\times 10^{- 5}C^{- 1} \\right)$", "question_images": [], "option_1": "Length of steel rod is 14.17cm and length of copper rod is 9.17cm", "option_2": "Length of steel rod is 15.17cm and length of copper rod is 6.17cm", "option_3": "Length of steel rod is 14.17cm and length of copper rod is 10.17cm", "option_4": "Length of steel rod is 15.17cm and length of copper rod is 9.17cm", "correct_option": 1, "numerical_answer": null, "solution": "$\\mathcal{l}_{1}\\left( 1 + \\alpha_{1}\\Delta T \\right) - \\mathcal{l}_{2}\\left( 1 + \\alpha_{2}\\Delta T \\right) = \\mathcal{l}_{1} - \\mathcal{l}_{2}$\n$\\mathcal{l}_{1}\\alpha_{1}\\Delta T = \\mathcal{l}_{2}\\alpha_{2}\\Delta T$\n$\\mathcal{l}_{1}\\alpha_{1} = \\mathcal{l}_{2}\\alpha_{2}$\n$\\Rightarrow \\mathcal{l}_{1} \\times 1.7 = \\mathcal{l}_{2} \\times 1.1$\n$17\\mathcal{l}_{1} = 11\\mathcal{l}_{2}$\nAs $\\left( \\mathcal{l}_{2} - \\mathcal{l}_{1} = 5 \\right)$\n$\\Rightarrow 17\\mathcal{l}_{2} - 11\\mathcal{l}_{2} = 5 \\times 17$\n$6\\mathcal{l}_{2} = 85$\n$\\Rightarrow \\mathcal{l}_{2} = \\frac{85}{6} = 14.17cm$\n$\\mathcal{l}_{1} = 9.17cm$", "solution_images": [], "subject": "Physics", "topic": "Heat", "subtopic": "Thermal expansion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-24-Q29", "question": "The surface tension of a liquid is 5 Newton per meter. If a film\nof this liquid is held on a ring of area $0.02m^{2}$, if its surface\nenergy is about $\\left( n \\times 10^{- 1} \\right)$ Joule then n is", "question_images": [], "option_1": "$\n$= 5frac times 2 times 0.02m_ = 0.2J$", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "We know that surface energy\n$U_{s} = T \\times \\ Area\\ $\nHere, as 2 films are formed because of ring.\nSo $U_{s} = T \\times 2 \\times (A)$\n$= 5\\frac{N}{m} \\times 2 \\times 0.02m_{2} = 0.2J$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-25-Q20", "question": "32 g of [IMAGE] is contained in a cubical\ncontainer of side 1 m and maintained at a temperature of\n[IMAGE]. The isothermal bulk modulus of elasticity of\nthe gas is [IMAGE]. Then the value of K is:[IMAGE] 32 s 1m Hkqtk ds ?kukdkj crZu esa 127°C A rks K", "question_images": ["images/image145.png", "images/image146.png", "images/image147.png", "images/image148.png", "images/image145.png", "images/image147.png", "images/image148.png"], "option_1": "32.30", "option_2": "33.40", "option_3": "33.20", "option_4": "33.00", "correct_option": 3, "numerical_answer": null, "solution": "Bulk modulus of elasticity at constant temperature ij", "solution_images": ["images/image149.png", "images/image150.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Bulk modulus", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-02-Q37", "question": "If the dipole moment vector[IMAGE] of a dipole\nand the electric field[IMAGE] (uniform) are oriented as\nshown then which of the following statement is incorrect?", "question_images": ["images/image28.png", "images/image29.png", "images/image30.png"], "option_1": "potential energy in I = potential energy in II", "option_2": "potential energy in I = negative of potential energy in II", "option_3": "in I rotates anticlockwise", "option_4": "in II rotates clockwise", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] where α is always the shorter angle\nbetween[IMAGE] and [IMAGE] ∴ in both cases I and II, α = π − θ ⇒ (cos α)_I = (cos α)_II\n∴ U_I = U_II ⇒ (B) is wrong.\n(C) and (D) are correct as direction of [IMAGE]", "solution_images": ["images/image31.png", "images/image32.png", "images/image33.png", "images/image34.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-09-Q2", "question": "The minimum time in which the potential energy of a particle\nexecuting SHM changes from maximum to minimum is 5 s. Then the time\nperiod of SHM is", "question_images": [], "option_1": "5 s", "option_2": "10 s", "option_3": "15 s", "option_4": "20 s", "correct_option": 4, "numerical_answer": null, "solution": "P.E. is maximum at extreme position and minimum at mean\nposition.\nTime to go from extreme position to mean position is,\n where T is time period of SHM.", "solution_images": ["images/image10.png", "images/image11.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Time period of oscillation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-18-Q21", "question": "The speed of a projectile at the highest point is $v_{1}$ and at\nthe point half the maximum height is $v_{2}$. If\n$\\frac{v_{1}}{v_{2}} = \\sqrt{\\frac{2}{5}}$, then the angle of projection\nis π/n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "$0 = {(v_{2}sin\\beta)}^{2} - 2g\\frac{H}{2}$\n$v_{1}^{2} = v_{1}^{2} - gh$\nAlso $\\frac{v_{1}}{v_{2}} = \\sqrt{\\frac{2}{5}}$\nFrom above\n$v_{1} = \\sqrt{\\frac{2gH}{3}}$\n$v_{2} = \\sqrt{\\frac{5gH}{3}}$\n$H = \\frac{u^{2}\\sin^{2}\\theta}{2g} \\Rightarrow sin2\\theta = \\frac{2gH\\cos^{2}\\theta}{v_{1}^{2}}$\n$\\tan^{2}\\theta = \\frac{2gH}{v_{1}^{2}}{or\\ tan}^{2}\\theta = \\frac{2gH \\times 3}{2gH}$\n$tan\\theta = \\sqrt{3}$\n$\\theta = 60^{\\circ}$", "solution_images": ["images/image28.png"], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Angle of projection", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-16-Q19", "question": "A block of mass 60 kg is released from rest when compression in\nthe spring is 2 m (natural length of spring is 8 m). Surface AB is\nsmooth while surface BC is rough. Block travels x distance before coming\nto complete rest. Value of x is: $\\lbrack g = 10m/s^{2}\\rbrack$", "question_images": ["images/image36.png"], "option_1": "2 m", "option_2": "3 m", "option_3": "1 m", "option_4": "4 m", "correct_option": 2, "numerical_answer": null, "solution": "By Work Energy\nTheorem$W_{net} = \\Delta K.E = 0 \\Rightarrow \\frac{1}{2}k(x_{0}^{2} - x^{2}) - \\mu mgx = 0$\n$\\Rightarrow \\frac{1}{2} \\times 200(2^{2} - x^{2}) = \\frac{1}{2} \\times 60 \\times 10x$\n$\\Rightarrow x = 1m$\nAlso at this moment $f_{\\max} > kx$\nSo, block will not move so total distance travelled =2+1=3 m.", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work -energy theorem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-01-Q17", "question": "The diagram below shows the lowest four energy levels for an\nelectron in a hypothetical atom. The electron is excited to the -1eV\nlevel of the atom and transitions to the lowest energy state by emitting\nonly two photons. Which of the following energies could not belong to\neither of the photons ?", "question_images": ["images/image86.png"], "option_1": "2 eV", "option_2": "4 eV", "option_3": "5 eV", "option_4": "6 eV", "correct_option": 2, "numerical_answer": null, "solution": "3eV to -7eV is not possible.", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-15-Q9", "question": "An electron (mass m) with initial velocity [IMAGE] is in an electric field [IMAGE] If [IMAGE] is initial de-Broglie wavelength of electron, its de-Broglie wave length\nat time t is given by", "question_images": ["images/image68.png", "images/image69.png", "images/image70.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "Initially [IMAGE] Velocity as a function of time [IMAGE] so wavelength [IMAGE]", "solution_images": ["images/image75.png", "images/image76.png", "images/image77.png", "images/image78.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "De Broglie wavelength", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-22-Q17", "question": "An impulsive force F acts horizontally on a solid sphere of\nradius R placed on a horizontal surface. Theline of action of the\nimpulsive force is at a height h above the centre of the sphere. If the\nrotational andtranslational kinetic energies of the sphere just after\nthe impulse are equal, then h is equal to", "question_images": ["images/image44.png"], "option_1": "$\\frac{R}{2}$", "option_2": "$\\frac{2}{5}R$", "option_3": "$\\frac{R}{\\sqrt{2}}$", "option_4": "$\\sqrt{\\frac{2}{5}}R$", "correct_option": 4, "numerical_answer": null, "solution": "Transitional k.e.= rotational k.e.\n$\\frac{1}{2}{mv}^{2} = \\frac{1}{2}I\\omega^{2}$\n$V^{2} = \\frac{2}{5}\\omega^{2}R^{2}$...(1)\nImpulse eq. Fdt = mV...(2)\n$\\ Fhdt\\ = I\\omega = \\frac{2}{5}mR^{2}\\omega$...(3)\nDividing (2) and (3)$h = \\frac{2}{5}\\frac{R^{2}\\omega}{v}$...(4)\nSolving (1) and (4)$h = \\sqrt{\\frac{2}{5}}R$", "solution_images": [], "subject": "Physics", "topic": "Rotational motion", "subtopic": "Angularimpulse", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-01-Q10", "question": "Two liquids of densities 1 are mixed in equal\nvolumes. A cube of external dimension 20 cm has an inner cubical portion\nof side 10 cm whose density is twice that of the outer portion. If this\ncube is just floating in the liquid mixture, find the density of the\ninner portion.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "ρ = [IMAGE] = 2 mg = 2ρ1 × 103 + ρ1 × (203 - 103)\n= 9ρ1 × 103 g = ρ × 8 × 103 g", "solution_images": ["images/image44.png", "images/image45.png", "images/image46.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "Ph-25-Q21", "question": "Consider a mixture of n moles of helium gas and 2 n moles of\noxygen gas (molecules taken to be rigid) as an ideal gas. Its\n[IMAGE] value will be:\nds 2n eksYl dh feJ.k xSl eku s rks bl feJ.k", "question_images": ["images/image151.png", "images/image152.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image157.png", "images/image158.png", "images/image159.png"], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Heat capacity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-06-Q2", "question": "Find out velocity of block B in the given figure: (all the\npulleys are ideal and string are inextensible massless)", "question_images": ["images/image3.png"], "option_1": "$80\\ m/s\\ \\widehat{j}$", "option_2": "$- 40\\ m/s$", "option_3": "$0\\ m/s\\ \\widehat{j}$", "option_4": "$- 80\\ m/s\\ \\widehat{j}$", "correct_option": 3, "numerical_answer": null, "solution": "$\\mathcal{l}_{1}^{'} + \\mathcal{l}_{2}^{'} = 0$\n$x = 4$``0\n$\\mathcal{l}_{3}^{'} + \\mathcal{l}_{4}^{'} = 0$\n$40 - y + 40 - 80 = 0$\n$y = 0$", "solution_images": ["images/image4.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Constraint motion", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "Ph-28-Q34", "question": "In the figure shown switch [IMAGE] remains\nconnected and switch S2 remains open for a long time. Now\n[IMAGE] is also closed. Assuming\n[IMAGE] and [IMAGE]. The magnitude of\nrate of change of current is 2n A/s in inductor just after the switch S2\nis closed. Find n", "question_images": ["images/image299.png", "images/image300.png", "images/image301.png", "images/image302.png", "images/image303.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "[IMAGE] Using [IMAGE] TOPIC: Electromagnetic induction\nSUB TOPIC: Self induction\nLEVEL: Moderate", "solution_images": ["images/image304.png", "images/image305.png", "images/image306.png", "images/image307.png", "images/image308.png", "images/image309.png", "images/image310.png", "images/image311.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-23-Q5", "question": "A uniform elastic rod of cross-section area A, natural length L\nand Young's modulus Y is placed on asmooth horizontal surface. Now two\nhorizontal forces (of magnitude F and 3F) directed along the lengthof\nrod and in opposite direction act at two of its ends as shown. After the\nrod has acquired steady state,the extension of the rod will be.", "question_images": ["images/image7.png"], "option_1": "$\\frac{2F}{YA}L$", "option_2": "$\\frac{4F}{YA}L$", "option_3": "$\\frac{F}{YA}L$", "option_4": "$\\frac{3F}{2YA}L$", "correct_option": 1, "numerical_answer": null, "solution": "Tension in rod at a distance x from right edge is\n$T = F\\left( 3 - 2\\frac{x}{L} \\right)$\n[IMAGE] $\\therefore$net extension in\nrod $= \\int_{0}^{L}\\mspace{2mu}\\frac{T}{YA}dx = \\frac{2F}{YA}L$", "solution_images": ["images/image8.png"], "subject": "Physics", "topic": "Elasticity", "subtopic": "Elongation in rod due to unbalanced force", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-18-Q10", "question": "In the fig. shown a cart moves on a smooth horizontal surface\ndue to an external constant force of magnitude F. The initial mass of\nthe cart is $M_{0}$and velocity is zero. Sand falls on to the cart with\nnegligible velocity at constant rate$\\mu\\ kg/s$ and sticks to the cart.\nThe velocity of the cart at time t is", "question_images": ["images/image20.png"], "option_1": "$\\frac{Ft}{M_{0} + \\mu t}$", "option_2": "$\\frac{F}{\\mu}\\mathcal{l}n\\frac{m_{0} + \\mu t}{m_{0}}$", "option_3": "$\\frac{Ft}{M_{0}}$", "option_4": "$\\frac{Ft}{M_{0} + \\mu t}e^{\\mu t}$", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] Formula $F = m\\frac{dv}{dt} + (V - u)\\frac{dm}{dt}$\nHere u=velocity of sand =0\n$m = M_{o} - \\mu t = mass\\ at\\ time\\ t$\nand$\\frac{dm}{dt} = \\mu$\n$\\therefore F = (M_{0} + \\mu t)\\frac{dm}{dt} + v\\mu$\n$(F - \\mu v)dt = (M_{o} + \\mu t)dv$\n$\\int_{0}^{t}\\frac{dt}{M_{0} + \\mu t} = \\int_{0}^{v}\\frac{dv}{F - \\mu v}$\n${\\lbrack log(M_{0} + \\mu t)\\rbrack}_{0}^{t} = \\frac{1}{\\mu}\\lbrack log(F - \\mu v)\\rbrack_{0}^{v}$\n$log\\frac{(M_{0} + \\mu t)}{M_{0}} = log(\\frac{F}{F - \\mu v})$\n$F - \\mu v = \\frac{M_{o}F}{M_{0} + \\mu t} \\Rightarrow v = \\frac{Ft}{M_{0} + \\mu t}$", "solution_images": ["images/image21.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Variable mass", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-04-Q21", "question": "Power delivered to a particle varies with time as P = (3t2 - 2t)\nwatt, where is in seconds, then change in its kinetic energy\nbetween time t = 1 to t = 3s.\n P = (3t2 - 2t),, t = 1 t = 3s", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "18", "solution": "[IMAGE] W =[IMAGE] = ∆ K.E.\nTOPIC: WPE\nSUB TOPIC: POWER", "solution_images": ["images/image117.png", "images/image118.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "Ph-26-Q34", "question": "A simple open U-tube contains mercury. Now water is poured\nslowly upto height 27.2 cm in the left arm. How high (in cm) does the\nmercury rise in the right arm from its initial level in equilibrium\nstate?\n(Take density of mercury $= 13600\\ kg/m^{3}$ and that of water\n$= 1000\\ kg/m^{3}$and $g\\ = 10\\ m/s^{2}$)\n 27.2cm rd Hkjk fLFkfr \nHkqtk es a blds çkjfEHkd Lrj ls ikjk fdruh (cm esa) rd tk;sxkA\nikjs dk ?kuRo $= 13600\\frac{kg}{m^{3}}$ rFkk ikuh", "question_images": ["images/image149.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "[IMAGE] Pressure at A that is at interface of water and mercury and B is same as\nboth are at same horizontal level in same liquid. Thus\nAij vFkkZr~ ikuh dh vkUrfjd lrg rFkk ikjs ij rFkkBij nkc leku gksxk\nD;ksafd nksauks leku nzo esa leku kjk çkIr \n$= \\frac{x}{2} = 1\\ cm$", "solution_images": ["images/image150.png"], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Hydrostatic pressure", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-26-Q11", "question": "The efficiency $\\eta$ of the cyclic process\n$\\left( \\gamma = \\frac{5}{3} \\right)$ is given by $\\frac{50n}{23}$ then\nfind out the value of n.\npØh; izØe $\\eta\\frac{50n}{23}$gS rksn", "question_images": ["images/image31.png"], "option_1": "12.5", "option_2": "13.5", "option_3": "18.5", "option_4": "11.5", "correct_option": 1, "numerical_answer": null, "solution": "$\\eta = \\frac{W}{+ Q}$\nTemperature at A is T_0 then$P_{0}V_{0} = {nRT}_{0}$\n T_0gS rks$P_{0}V_{0} = {nRT}_{0}$\nat$T_{A} = T_{0},T_{B} = 2T_{0},T_{C} = 6T_{0},T_{D} = 3T_{0}$ \n$W = 2P_{0}V_{0}$\n$+ Q = Q_{AB} + Q_{BC}$\n$= n \\cdot C_{v}dT + nC_{p}dT$\n$= \\frac{3}{2}{nRT}_{0} + \\frac{5}{2}nR\\left( 4T_{0} \\right)$\n$= \\frac{23}{2}{nRT}_{0} = \\frac{23}{2}P_{0}V_{0}$\n$\\eta = \\frac{2P_{0}V_{0}}{\\frac{23}{2}P_{0}V_{0}} = \\frac{4}{23}$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Efficiency of a cyclic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-08-Q8", "question": "32 g of O_2 is contained in a cubical container of side 1 m and\nmaintained at a temperature of 127ºC. The isothermal bulk modulus of\nelasticity of the gas in terms of universal gas constant R is", "question_images": [], "option_1": "127 R", "option_2": "400 R", "option_3": "200 R", "option_4": "560 R", "correct_option": 2, "numerical_answer": null, "solution": "Bulk modulus of elasticity at constant temperature\n$B = \\frac{dP}{dV/V} - = - V\\left( \\frac{dP}{dV} \\right)_{T} = + P = \\frac{nRT}{v}$\n$= \\frac{1 \\times R \\times 400}{1^{3}} = 400R$", "solution_images": [], "subject": "Physics", "topic": "Heat", "subtopic": "Bulk modulus", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-17-Q15", "question": "A fiber sheet with relative permittivity 5 is inserted between\ntwo parallel metal plates 0.25 cm apart. A potential difference of 2500\nv is applied between the plates. If dielectric strength of air is\n$3 \\times 10^{6}v/m$ than", "question_images": ["images/image13.png"], "option_1": "Air will not breakdown", "option_2": "Air will breakdown", "option_3": "Potential gradient strength in fiber sheet exceeds dielectric\nstrength of air", "option_4": "All the above are correct", "correct_option": 2, "numerical_answer": null, "solution": "As $V = V_{1} + V_{2}$\n$2500 = 0.0002E_{1} + 0.0023E_{2}$\n$2E_{1} + 23E_{2} = 25 \\times 10^{6}$\nAlso $E_{1} = 5E_{2}$\nSolving $E_{2} = 0757 \\times 10^{8}V/m$\nAs $E_{1} > 3.6 \\times 10^{6}$ Hence Air will breakdown", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Dielectric", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-09-Q19", "question": "Two particle of mass m each are fixed to a massless rod of\nlength [IMAGE]. The rod is smoothly hinged at one end to\na ceiling. It performs oscillation of small angle in vertical plane. The\nlength of the equivalent simple pendulum is", "question_images": ["images/image115.png", "images/image116.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image121.png", "images/image122.png", "images/image123.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Time period of pendulum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-04-Q8", "question": "A shot is fired at an angle θ to the horizontal such that it strikes\nthe hill while moving horizontally Find initial angle of projection θ.", "question_images": ["images/image45.png"], "option_1": "tan θ =[IMAGE]", "option_2": "tan θ =[IMAGE]", "option_3": "tan θ =[IMAGE]", "option_4": "none of these\n (shot) θ \n ", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] sin θ =[IMAGE] 4 sin θ = 8 sin θ - 6 cos θ\n4 sin θ = 6 cos θ ⇒ tan θ = \nTOPIC:KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image49.png", "images/image50.png", "images/image51.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-17-Q9", "question": "An electric field\n$\\overset{\\rightarrow}{E} = 4 \\times \\overset{\\hat{}}{i} - (y^{2} + 1)jN/C$\npasses through the box shown in figure. The flux of the electric field\nthrough surface ABCD and BCGF and marked as $\\phi_{I}$ and $\\phi_{\\Pi}$\nrespectively. The difference between\n$(\\phi_{I} - \\phi_{\\Pi})is\\ (in{Nm}^{2/C})$", "question_images": ["images/image9.jpeg"], "option_1": "47", "option_2": "46", "option_3": "42", "option_4": "48", "correct_option": 4, "numerical_answer": null, "solution": "Flux via ABCD\n$\\phi_{1} = \\int\\overset{\\rightarrow}{E} \\cdot d\\overset{\\rightarrow}{A} = 0$\nFlux via BCEF\n$\\phi_{2} = \\int\\overset{\\rightarrow}{E} \\cdot d\\overset{\\rightarrow}{A} = 0$\n$\\phi_{2} = \\overset{\\rightarrow}{E} \\cdot \\overset{\\rightarrow}{A} = (4 \\times \\overset{\\hat{}}{i} - (y^{2} + 1)\\overset{\\hat{}}{j}) \\cdot 4\\overset{\\hat{}}{i}$\n$= 16x,x = 3$\n$\\phi_{2} = 48\\frac{N - m^{2}}{C};\\phi_{1} - \\phi_{2} = - 48\\frac{N - m^{2}}{C}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Gauss\"s law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-04-Q3", "question": "A point moves according to the law, x = at and y = at (1- αt),\nwhere a and α are positive constants and t is time. Find the moment at\nwhich angle between velocity vector and acceleration vector", "question_images": ["images/image7.png"], "option_1": "α sec", "option_2": "", "option_3": "", "option_4": "Not possible\n x = at y = at (1- αt) a α\n t \n [IMAGE]", "correct_option": 2, "numerical_answer": null, "solution": "x = at ⇒ vx = a ⇒ ax = 0\ny = at - aαt2 ⇒ vy =[IMAGE] = a - 2αat\nay =[IMAGE] = - 2αa\n[IMAGE] = av cos θ\n= (2αa) [IMAGE] Solve to get, [IMAGE] TOPIC:KINEMATIC\nSUB TOPIC: ACCERATION", "solution_images": ["images/image10.png", "images/image11.png", "images/image12.png", "images/image13.png", "images/image14.png", "images/image15.png", "images/image16.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-06-Q15", "question": "The force exerted by a compression device is given by\n$F(x) = kx(x - \\lambda)$ for $0 \\leq x \\leq \\lambda$, where $\\lambda$\nis the maximum possible compression, x is the compression and k is a\nconstant. The work required to compress the device by a distance d will\nbe maximum when", "question_images": [], "option_1": "$d = \\lambda/4$", "option_2": "$d = \\lambda/\\sqrt{2}$", "option_3": "$d = \\lambda/2$", "option_4": "$d = \\lambda$", "correct_option": 4, "numerical_answer": null, "solution": "For W to be maximum \ni.e. $F(x) = 0 \\Rightarrow x = \\lambda,x = 0$\nClearly for $d = \\lambda$,the work done is maximum.", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work done by a variable force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-09-Q21", "question": "A sinusoidal travelling wave in a string has a velocity of\npropagation 300 m/sec. The time period of oscillations of the particles\nof the string is 0.04 sec. The phase difference between the oscillations\nof two points at distances 10 m and 16 m respectively from source of\noscillation is nπ. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "", "solution_images": ["images/image135.png", "images/image136.png"], "subject": "Physics", "topic": "Wave on string", "subtopic": "Phase difference", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "Ph-27-Q3", "question": "Three bars having lengths [IMAGE] and\n[IMAGE] and area of cross section A, A/2, 2A are joined\nrigidly end toend. The compound rod is subjected to a stretching force\nF. It young's modules of the material of eachrod is Y then the increase\nin length of the rod is.\n yEckbZ dh o vuqizLFk dkV\n{ks=kQyA,[IMAGE], 2Adh rhu NM+s n`<+:i ls fljs ls fljs\nNM+ xq.kkadY", "question_images": ["images/image8.png", "images/image9.png", "images/image8.png", "images/image9.png", "images/image10.png"], "option_1": "$\\frac{13F\\mathcal{l}}{2AY}$", "option_2": "$\\frac{F\\mathcal{l}}{AY}$", "option_3": "$\\frac{9F\\mathcal{l}}{5AY}$", "option_4": "$\\frac{19F\\mathcal{l}}{2AY}$", "correct_option": 1, "numerical_answer": null, "solution": "$\\ \\Delta l = \\Delta l_{1} + \\Delta l_{2} + \\Delta l_{3}$\n$\\frac{F}{\\frac{YA}{\\mathcal{l}}} + \\frac{F}{\\frac{YA}{4\\mathcal{l}}} + \\frac{F}{\\frac{2YA}{3\\mathcal{l}}} = \\frac{F\\mathcal{l}}{YA}\\left( 1 + 4 + \\frac{3}{2} \\right) = \\frac{13F\\mathcal{l}}{2YA}$", "solution_images": [], "subject": "Physics", "topic": "Elasticity", "subtopic": "Extension in compound rod", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-24-Q26", "question": "A non-viscous ideal fluid is flowing vertically downwards in a\npipe. The area of cross-section at section is double that at\nsection Vertical distance between section A and section B is h\nand the height of water column in tube (2) is more than that in tube", "question_images": [], "option_1": "by $\\frac{h}{2}$ distance. If the velocity(m/s) of the fluid at section\nA is x, then find the value of x. (if h = 2.7 m)", "option_2": "is more than that in tube", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "Applying Bernoulli's equation between section A and B\n$P_{A} + \\frac{1}{2}\\rho v_{A}^{2} + \\rho{gh}_{A} = P_{B} + \\frac{1}{2}\\rho v_{B}^{2} + \\rho{gh}_{B}$\n$P_{A} + \\frac{1}{2}\\rho v^{2} + \\rho gh = P_{B} + \\frac{1}{2}\\rho(2v)^{2} + 0$\n$P_{B} - P_{A} = \\rho gh - \\frac{3}{2}\\rho v^{2}\\ $\nand $P_{B} - P_{A} = \\rho g\\frac{h}{2}$\n$\\rho g\\frac{h}{2} = \\rho gh - \\frac{3}{2}\\rho v^{2}$\n$\\Rightarrow v = \\sqrt{\\frac{gh}{3}} = 3$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-20-Q25", "question": "In the arrangement shown in figure, wavelength of light used is\n[IMAGE]. The distance between slits\n[IMAGE] The distance between [IMAGE] and\n[IMAGE] is [IMAGE] If the ratio of\nmaximum to minimum intensity received on screen P is K. Then K is", "question_images": ["images/image99.png", "images/image100.png", "images/image101.png", "images/image102.png", "images/image103.png", "images/image104.png", "images/image105.png", "images/image106.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "9", "solution": "If intensity of light incident on slits [IMAGE] and\nThe intensity through [IMAGE] will be\n[IMAGE] The intensity through [IMAGE] will be\n[IMAGE] TOPIC:Wave Optics\nSUB TOPIC:YDSE\nLEVEL:Moderate", "solution_images": ["images/image100.png", "images/image101.png", "images/image107.png", "images/image104.png", "images/image108.png", "images/image109.png", "images/image110.png", "images/image111.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-12-Q5", "question": "The emf of cell E_2 is", "question_images": [], "option_1": "0.5 V", "option_2": "1 V", "option_3": "1.5 V", "option_4": "2 V", "correct_option": 1, "numerical_answer": null, "solution": "$i = \\frac{2}{15 + 10}$\n$V_{AB} = i10 = \\frac{20}{25}$\n$E_{2} = \\frac{V_{AB}}{50} \\times 31.25$\n$= \\frac{20}{25} \\times \\frac{31.25}{50}$\n=0.5V", "solution_images": [], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Potentiometer", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "Ph-25-Q2", "question": "The ratio of r.m.s. speed to the r.ms. angular speed of a\ndiatomic gas at certain temperature is:(assume m= mass of one molecule\nmass I = moment of inertia of the molecules)\newy ekfu, m =,d v.kq \nvk.kfod nzO;eku I = v.kqvksa", "question_images": [], "option_1": "$\\sqrt{\\frac{3}{2}}$", "option_2": "$\\sqrt{\\frac{3I}{2M}}$", "option_3": "$\\sqrt{\\frac{3I}{2m}}$", "option_4": "1", "correct_option": 3, "numerical_answer": null, "solution": "$\\frac{1}{2}{mV}^{2} = \\frac{3}{2}kT$\n$\\frac{1}{2}I\\omega^{2} = \\frac{2}{2}kT$\n$\\frac{V}{\\omega} = \\sqrt{\\frac{3I}{2m}}$", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "RMS speed", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-19-Q21", "question": "In a fluorescent lamp choke (a small transformer) 100 V of\nreverse voltage is produced when the choke current changes uniformly\nfrom 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the\nchoke(in mH) is estimated to be 2.5n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "$100 = \\frac{L(0.25)}{0.025} \\times 10^{3}$\n$\\therefore L = 100 \\times 10^{- 4}H$\n$= 10mH$", "solution_images": [], "subject": "Physics", "topic": "Electromagnetic induction", "subtopic": "Self inductance", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-22-Q2", "question": "In the circuit, if the forward voltage drop for the diode is\n[IMAGE] the current in the circuit will be (approximately)", "question_images": ["images/image6.png", "images/image7.png"], "option_1": "3.4 mA", "option_2": "2 mA", "option_3": "2.5 mA", "option_4": "3 mA", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image8.png"], "subject": "Physics", "topic": "electronic device", "subtopic": "Diodes", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-02-Q32", "question": "The frequency of incident light falling on photo sensitive\nsubstance is doubled. If E_1 is the kinetic energy of emitted electron\nin first case and E_2 is the kinetic energy in second case, then", "question_images": [], "option_1": "E_1 = E_2", "option_2": "E_2 = 2E_1", "option_3": "E_2 < 2E_1", "option_4": "E_2 > E_1", "correct_option": 4, "numerical_answer": null, "solution": ". [IMAGE] Dividing both equations,\n[IMAGE] Hence, E_2 > 2E_1", "solution_images": ["images/image11.png", "images/image12.png", "images/image13.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-10-Q16", "question": "A train of mass M is moving on a circular track of radius ' R\n' with constant speed V. The length of the train is half of the\nperimeter of the track. The linear momentum of the train will be.", "question_images": [], "option_1": "Zero", "option_2": "$\\frac{2MV}{\\pi}$", "option_3": "MVR", "option_4": "MV", "correct_option": 2, "numerical_answer": null, "solution": "If we treat the train as a ring of mass then its COM will\nbe at a distance$\\frac{2R}{\\pi}$from the centre of thecircle. Velocity\nof centre of mass is:\n$V_{CM} = R_{CM} \\cdot \\omega$\n$= \\frac{2R}{\\pi} \\cdot \\omega$\n$= \\frac{2R}{\\pi} \\cdot \\left( \\frac{V}{R} \\right)$\n$\\left( \\because\\omega = \\frac{V}{R} \\right)$\n$\\Rightarrow V_{CM} = \\frac{2V}{\\pi} \\Rightarrow MV_{CM} = \\frac{2MV}{\\pi}$\nAs the linear momentum of any system = MV_CM\n[IMAGE] The linear momentum of the train\n$= \\frac{2MV}{\\pi}$.", "solution_images": ["images/image59.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-21-Q10", "question": "In the circuit, if the forward voltage drop for the diode is\n0.5V, the current in the circuit will be(approximately) -\nifjiFk fy, vxzfoHko iru 0.5V gS] rc ifjiFk", "question_images": ["images/image28.png"], "option_1": "3.4 mA", "option_2": "2 mA", "option_3": "2.5 mA", "option_4": "3 mA", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image29.png"], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Diodes", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-11-Q2", "question": "A rod of length L has non-uniform linear mass density given by\n[IMAGE] where a and b are constants\nand[IMAGE]. The value of x for the centre of mass of the\nrod is at", "question_images": ["images/image14.png", "images/image15.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$x_{cm} = \\frac{1}{M}\\int_{0}^{l}x \\cdot dM$", "solution_images": ["images/image20.png", "images/image21.png", "images/image22.png", "images/image23.png", "images/image24.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Position of COM", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "Ph-27-Q35", "question": "A non-viscous ideal fluid is flowing vertically downwards in a\npipe. The area of cross-section at section is double that at\nsection Vertical distance between section A and section B is h\nand the height of water column in tube (2) is more than that in tube", "question_images": [], "option_1": "by [IMAGE] distance. If the velocity(m/s) of the\nfluid at section A is x, then find the value of x. (if h = 2.7 m),d nzo uyh esa uhps dh rjQ çokfgr gSAHkkx A ij\nvuqçLFk dkV\nchp nwjh h", "option_2": "is more than that in tube", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "Applying Bernoulli's equation between section A and B\nArFkk B", "solution_images": ["images/image228.png", "images/image229.png", "images/image230.png", "images/image231.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-19-Q1", "question": "Radiation, with wavelength falls on a metal surface to\nproduce photoelectrons. The electronsare made to enter a uniform\nmagnetic field of $3 \\times 10^{- 4}$T. If the radius of the largest\ncircular pathfollowed by the electrons is 10 mm, the work function of\nthe metal is close to", "question_images": [], "option_1": "1.8 eV", "option_2": "0.8 eV", "option_3": "1.6 eV", "option_4": "1.1 eV", "correct_option": 2, "numerical_answer": null, "solution": "${KE}_{\\max} = E - \\phi$\n$= \\frac{12400}{\\lambda(inÅ)} - \\phi$ (in eV)\n$r = \\frac{\\sqrt{2mKE}}{eB}$\n$KE_{\\max} = \\frac{r^{2}e^{2}B^{2}}{2m}$ (in J)\n$= \\frac{r^{2}eB^{2}}{2m}$ (in eV)", "solution_images": ["images/image1.png"], "subject": "Physics", "topic": "Modern physics", "subtopic": "Photoelectric effect", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-16-Q8", "question": "Three particles each of mass m can slide on fixed frictionless\nhorizontal circular tracks in the same horizontal plane as shown in the\nfigure. The coefficient of restitution being e=0.5. Assuming that\nm_2and m_3 are at rest initially and lie along a radial line before\nimpact and the spring is initially unstretched, then maximum extension\nin spring in subsequent motion.", "question_images": ["images/image16.png"], "option_1": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{m}{k}}$", "option_2": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{m}{5k}}$", "option_3": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{2m}{5}k}$", "option_4": "$\\frac{3}{5}V_{0}\\sqrt{\\frac{m}{k}}$", "correct_option": 2, "numerical_answer": null, "solution": "Velocity just after collision\n$V_{0} = V_{1} + V_{2}$.....(1)\n$\\frac{1}{2} = \\frac{- V_{1} + V_{2}}{V_{0}}$\n$- V_{1} + V_{2} = \\frac{V_{0}}{2}$.....(2)\nFrom (1) & (2)\n$\\Rightarrow V_{1} = \\frac{V_{0}}{4}\\& V_{2} = \\frac{3V_{0}}{4}$\nWhen maximum extension occurs then angular speed with rest to centre\nfor mass m_2& m_3 are same. Using angular momentum conservation\nabout centre of circle.\n$m\\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\\omega + m(2R)^{2}\\omega$\n$\\Rightarrow \\omega = \\frac{3}{10}\\frac{V_{0}}{R}$\nSo velocity of$m_{2} = \\frac{3}{5}V_{0}$\n& velocity of$m_{3} = \\frac{3}{10}V_{0}$\nWhen extension is maximum using energy conservation.\n$\\frac{1}{2}m{(\\frac{3}{4}V_{0})}^{2} = \\frac{1}{2}m{(\\frac{3}{5}V_{0})}^{2} + \\frac{1}{2}m{(\\frac{3}{10}V_{0})}^{2} + \\frac{1}{2}kx_{max.}^{2} \\Rightarrow x_{\\max} = \\frac{3}{4}V_{0}\\sqrt{\\frac{m}{5k}}$", "solution_images": [], "subject": "Physics", "topic": "Center of mass", "subtopic": "Collision", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-24-Q13", "question": "A container is filled with a liquid of density ρ. Container is\naccelerating on a horizontal surface with acceleration $\\frac{3g}{4}$\ntowards right hand side and liquid is at rest with respect to container\nas shown in figure. If there is a point A in liquid then, which of the\nfollowing is correct: (assume atmospheric pressure is zero and AB is\nvertical and AC is horizontal line)", "question_images": ["images/image43.png"], "option_1": "Pressure at A is $\\rho g\\mathcal{l}_{1}$", "option_2": "Pressure at A is $\\rho\\frac{3g}{4}\\mathcal{l}_{2}$", "option_3": "$\\theta = \\tan^{- 1}\\left( \\frac{3}{4} \\right)$", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "for point A from top surface\n$P_{A} = \\rho g \\cdot \\mathcal{l}_{1}$and from right wall\n$P_{A} = \\frac{\\rho^{3}g}{4} \\cdot \\mathcal{l}_{2}$\n$\\mathcal{l}_{1} = \\frac{3}{4}\\mathcal{l}_{2}$ and", "solution_images": ["images/image44.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Fluid in motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-26-Q27", "question": "[IMAGE] gas follows the process,\n[IMAGE] constant. If then the ratio of work done by gas\nto change in itsinternal energy is [IMAGE] then n is:\nvuqikr [IMAGE] gks rks n", "question_images": ["images/image118.png", "images/image119.png", "images/image120.png", "images/image118.png", "images/image119.png", "images/image120.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "", "solution_images": ["images/image121.png", "images/image122.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Internal energy", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-27-Q2", "question": "A slightly tapering wire of length[IMAGE] and end\nradii is subjected to tension F. Y is the\nYoung'sModulus. The extension produced in the wire is.,d FkksM+s Vsifjx rkj ftl rFkk fljks dh\nf= ij F ruko \nrkj", "question_images": ["images/image1.png", "images/image1.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image6.png", "images/image7.png"], "subject": "Physics", "topic": "Elasticity", "subtopic": "Extension in a wire of non-uniform cross-section", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-17-Q12", "question": "A small circular loop of conducting wire has radius a and\ncarries current I. It is placed in a uniform magnetic field B\nperpendicular to its plane such that when rotated slightly about its\ndiameter and released, it starts performing simple harmonic motion of\ntime period T. If the mass of the loop is $\\\\mathrm{m}\\$ then", "question_images": [], "option_1": "$T = \\sqrt{\\frac{\\pi m}{IB}}$", "option_2": "$T = \\sqrt{\\frac{2m}{IB}}$", "option_3": "$T = \\sqrt{\\frac{\\pi m}{2IB}}$", "option_4": "$T = \\sqrt{\\frac{2\\pi m}{IB}}$", "correct_option": 4, "numerical_answer": null, "solution": "$\\tau = MBsin\\theta = I\\alpha$\n$\\pi R^{2}IB\\theta = \\frac{{mR}^{2}}{2}\\alpha$.\n$\\omega = \\sqrt{\\frac{2\\pi IB}{m}} = \\frac{2\\pi}{T};T = \\sqrt{\\frac{2\\pi m}{IB}}$", "solution_images": [], "subject": "Physics", "topic": "Moving charges and", "subtopic": "Current carrying conductor in magnetic field", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-10-Q12", "question": "In the fig. shown a cart moves on a smooth horizontal surface\ndue to an external constant force of magnitude F. The initial mass of\nthe cart is M_0and velocity is zero. Sand falls on to the cart with\nnegligible velocity at constant rate and\nsticks to the cart. The velocity of the cart at time t is", "question_images": ["images/image44.png", "images/image45.png"], "option_1": "$\\frac{Ft}{M_{0} + \\mu t}$", "option_2": "$\\frac{F}{\\mu}\\mathcal{l}_{n}\\frac{m_{0} + \\mu t}{m_{0}}$", "option_3": "$\\frac{Ft}{M_{0}}$", "option_4": "$\\frac{Ft}{M_{0} + \\mu t}e^{\\mu t}$", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] Formula $F = m\\frac{dv}{dt} + (V - u)\\frac{dm}{dt}$\nHere, u = velocity of s and = O\nm = M_o[IMAGE] t = mass at time t\nand [IMAGE] $\\therefore$ $F = \\left( M_{0} + \\mu t \\right)\\frac{dv}{dt} + v\\mu$\n$(F - \\mu v)dt = \\left( M_{0} + \\mu t \\right)dv$\n$\\int_{0}^{t}\\mspace{2mu}\\frac{dt}{M_{0} + \\mu t} = \\int_{0}^{v}\\mspace{2mu}\\frac{dv}{F - \\mu v}$\n$\\frac{1}{\\mu}\\left\\lbrack \\mathcal{l}n\\left( M_{0} + \\mu t \\right) \\right\\rbrack_{0}^{t} = - \\frac{1}{\\mu}\\mathcal{\\lbrack l}n(F - \\mu v)\\rbrack_{0}^{v}$\n$\\mathcal{l}n\\frac{\\left( M_{0} + \\mu t \\right)}{M_{0}}\\mathcal{= l}n\\left\\{ \\frac{F}{F - \\mu v} \\right)$\n$F - \\mu v = \\frac{M_{0}F}{M_{0} + \\mu t} \\Rightarrow v = \\frac{Ft}{M_{0} + \\mu t}$", "solution_images": ["images/image46.png", "images/image47.png", "images/image48.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Variable mass system", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-23-Q20", "question": "Three particles each of mass m can slide on fixed frictionless\nhorizontal circular tracks in the same horizontal plane as shown in the\nfigure. The coefficient of restitution being e=0.5. Assuming that\nm_2and m_3 are at rest initially and lie along a radial line before\nimpact and the spring is initially unstretched, then maximum extension\nin spring in subsequent motion.", "question_images": ["images/image51.png"], "option_1": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{m}{k}}$", "option_2": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{m}{5k}}$", "option_3": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{2m}{5}k}$", "option_4": "$\\frac{3}{5}V_{0}\\sqrt{\\frac{m}{k}}$", "correct_option": 2, "numerical_answer": null, "solution": "Velocity just after collision\n$V_{0} = V_{1} + V_{2}$.....(1)\n$\\frac{1}{2} = \\frac{- V_{1} + V_{2}}{V_{0}}$\n$- V_{1} + V_{2} = \\frac{V_{0}}{2}$.....(2)\nFrom (1) & (2)\n$\\Rightarrow V_{1} = \\frac{V_{0}}{4}\\& V_{2} = \\frac{3V_{0}}{4}$\nWhen maximum extension occurs then angular speed with rest to centre\nfor mass m_2& m_3 are same. Using angular momentum conservation\nabout centre of circle.\n$m\\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\\omega + m(2R)^{2}\\omega$\n$\\omega = \\frac{3}{10}\\frac{V_{0}}{R}$\nSo velocity of$m_{2} = \\frac{3}{5}V_{0}$ & velocity\nof$m_{3} = \\frac{3}{10}V_{0}$\nWhen extension is maximum using energy conservation.\n$\\frac{1}{2}m{(\\frac{3}{4}V_{0})}^{2} = \\frac{1}{2}m{(\\frac{3}{5}V_{0})}^{2} + \\frac{1}{2}m{(\\frac{3}{10}V_{0})}^{2} + \\frac{1}{2}kx_{max.}^{2} \\Rightarrow x_{\\max} = \\frac{3}{4}V_{0}\\sqrt{\\frac{m}{5k}}$", "solution_images": [], "subject": "Physics", "topic": "Center of mass", "subtopic": "Collision", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-16-Q23", "question": "If the projectile hits the inclined plane perpendicularly when\nthrown horizontally with $v_{0}$, from a tower of height H as shown then\nthe value of $\\frac{4gH}{{(v_{0})}^{2}}$ is 2k. Find k.", "question_images": ["images/image41.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "$a_{y} = gcos30^{\\circ}$\n$u_{x} = V_{0}cos30^{\\circ}$\n$u_{y} = V_{0}sin30^{\\circ}$\n$V_{x} = u_{x} + a_{x}t$\n$0 = \\frac{V_{0}\\sqrt{3}}{2} - \\frac{g}{2}t$\n$t = \\frac{v_{0}\\sqrt{3}}{g}$....(i)\n$S_{y} = u_{y}t + a_{y}t^{2}$\n$(Hcos30^{\\circ} - 0) = V_{0}sin30^{\\circ}t + \\frac{1}{2}gcos30^{\\circ}t^{2}$....(ii)\nFrom (i) (ii)\n$\\Rightarrow \\frac{4gH}{{(v_{0})}^{2}} = 10$", "solution_images": ["images/image42.png"], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Horizonatal projectile", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "Ph-25-Q23", "question": "An ideal gas obey the law [IMAGE] constant. x is\nvalue for which it has zero molar specific heat at normal temperature.\nThen find 5 x. ([IMAGE] for the gas is 2.5 R)\ngS ftlds fy, ij bl j fof'k\"B A rc 5x\ndk eku Kkr djksa ¼xSl", "question_images": ["images/image165.png", "images/image166.png", "images/image165.png", "images/image166.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "7", "solution": "For a gas at normal temperature, ij xSl ds fy,[IMAGE] Specific heat for a polytropic process\ni‚ayhVªksfid çØe ds fy, fof'k\"B \n[IMAGE] According to question iz\"ukuqlkj C=0", "solution_images": ["images/image167.png", "images/image167.png", "images/image168.png", "images/image169.png", "images/image170.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic processes", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-18-Q20", "question": "A uniform disc of radius R lies in the x-y plane, with its\ncentre at origin. its moment of inertia about zaxis is equal to its\nmoment of inertia about line $y = x + c$. The value of c will be", "question_images": [], "option_1": "$- \\frac{R}{2}$", "option_2": "$\\pm \\frac{R}{\\sqrt{2}}$", "option_3": "$\\frac{+ R}{4}$", "option_4": "$- R$", "correct_option": 2, "numerical_answer": null, "solution": "$I_{PQR} = \\frac{1}{4}{MR}^{2} + M{(\\frac{C}{\\sqrt{2}})}^{2}$\nBut $I_{PQR} = \\frac{1}{2}{MR}^{2}$\n$\\therefore C = \\pm \\frac{R}{\\sqrt{2}}$\nHence (2) is correct.", "solution_images": ["images/image27.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-08-Q6", "question": "When the temperature of copper coin is raised by 80ºC, its\ndiameter increases by 0.2.", "question_images": [], "option_1": "Percentage rise in the area of a face is 0.4", "option_2": "Percentage rise in the volume is 0.6", "option_3": "Coefficient of linear expansion of copper is\n$0.25 \\times 10^{- 4}C^{- 1}$", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$100 \\times \\frac{\\Delta D}{D} = \\alpha\\Delta T \\times 100$\n$100 \\times \\left( \\frac{\\Delta A}{A} \\right) = \\beta\\Delta T \\times 100$\n$= 2\\alpha\\Delta T \\times 100$ $= 2(0.2)$ $= 0.4\\%$\n$100 \\times \\frac{\\Delta t}{t} = \\alpha\\Delta T \\times 100$\n$\\alpha = \\frac{\\frac{\\Delta d}{d}}{\\Delta T} = \\frac{0.2 \\times 10^{- 2}}{80}\\left( \\frac{\\Delta d}{d} \\times 100 = 0.2 \\right)$\n$\\alpha = 0.25 \\times 10^{- 4}{}^{\\circ}C^{- 1}$\n$100 \\times \\frac{\\Delta V}{V} = \\gamma\\Delta T \\times 100$\n$= 3(0.2) = 0.6\\%$", "solution_images": [], "subject": "Physics", "topic": "Heat", "subtopic": "Thermal expansion", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-22-Q3", "question": "The maximum peak to peak voltage of an AM wave is 24 mV and the\nminimum peak to peak voltage is 8 mV. The modulation factor is", "question_images": [], "option_1": "10", "option_2": "20", "option_3": "25", "option_4": "50", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image9.png", "images/image10.png"], "subject": "Physics", "topic": "Communication system", "subtopic": "Modulation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-23-Q11", "question": "[IMAGE] As shown in the figure, a battery of emf$\\varepsilon$ is connected to\nan inductor L and resistance R in series. The switch is closed at t=0.\nThe total charge that flows from the battery, between t=0 and\n$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is", "question_images": ["images/image28.png"], "option_1": "$\\frac{\\varepsilon L}{R^{2}}(1 - \\frac{1}{e})$", "option_2": "$\\frac{\\varepsilon R}{eL^{2}}$", "option_3": "$\\frac{\\varepsilon L}{R^{2}}$", "option_4": "$\\frac{\\varepsilon L}{eR^{2}}$", "correct_option": 4, "numerical_answer": null, "solution": "$q = \\int_{0}^{T_{C}}idt$\n$= \\frac{\\varepsilon}{R}{\\lbrack t - \\frac{e^{- t/T_{C}}}{\\frac{- 1}{T_{C}}}\\rbrack}_{0}^{T_{C}}$;\n$= \\frac{\\varepsilon}{R}\\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\\rbrack$\n$= \\frac{\\varepsilon}{R} \\times \\frac{1}{e} \\times \\frac{L}{R}$\n$= \\frac{\\varepsilon L}{R^{2}e}$", "solution_images": [], "subject": "Physics", "topic": "Electromagnetic induction", "subtopic": "LR circuit", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-01-Q5", "question": "The primary purpose of using a parabolic mirror in a reflecting\ntelescope is", "question_images": [], "option_1": "To collect more light from a distant galaxy", "option_2": "To increase the magnification", "option_3": "To overcome diffraction effects", "option_4": "To correct spherical aberration", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-22-Q6", "question": "An ideal gas at a pressure of 1 atmosphere andtemperature of 27°C\nis compressed adiabaticallyuntil its pressure becomes 8 times the\ninitialpressure. Then the final temperature is [IMAGE]", "question_images": ["images/image13.png"], "option_1": "627°C", "option_2": "527°C", "option_3": "427°C", "option_4": "327°C", "correct_option": 4, "numerical_answer": null, "solution": "Here, [IMAGE] As changes are adiabatic,", "solution_images": ["images/image14.png", "images/image15.png", "images/image16.png", "images/image17.png", "images/image18.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adaibaticproces", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-23-Q24", "question": "A steel wire is rigidly fixed at both ends. Its length, mass and\ncross sectional area are 1m, 0.01 kg and\n[IMAGE] respectively. Then the temperature of the\nwire is lowered by[IMAGE]. If the transverse waves are setup by\nplucking the wire at [IMAGE] from one end and assuming that wire\nvibrates with minimum number of loops possible for such a case. The\nfrequency of vibration is 44n Hz. Find n. [Coefficient of linear\nexpansion of steel [IMAGE] and Young's", "question_images": ["images/image64.png", "images/image65.png", "images/image66.png", "images/image67.png", "images/image68.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "[IMAGE] The mechanical strain [IMAGE] The tension in wire[IMAGE] speed of wave in wire\n[IMAGE] Since the wire is plucked at\n[IMAGE] from one end\nThe wire shall oscillate in\n[IMAGE] overtone (for minimum number of loops)\nTOPIC:Waves in a string\nSUB TOPIC: Wire fixed at both ends\nLEVEL:Moderate", "solution_images": ["images/image69.png", "images/image70.png", "images/image71.png", "images/image72.png", "images/image73.png", "images/image74.png", "images/image75.png", "images/image76.png", "images/image77.png", "images/image78.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-18-Q9", "question": "A system is shown in the figure. Block A is moving with 1 m /s\ntowards left. Wedge is moving with 1 m /s towards right. Then speed of\nthe block B will be", "question_images": ["images/image17.png"], "option_1": "1 m /s", "option_2": "2 m /s", "option_3": "", "option_4": "none of these", "correct_option": 3, "numerical_answer": null, "solution": "Velocity of block A w.r.t. wedge is 2 m So we have\n${\\overset{\\rightarrow}{V}}_{BW} = {\\overset{\\rightarrow}{V}}_{B} - {\\overset{\\rightarrow}{V}}_{W} \\Rightarrow {\\overset{\\rightarrow}{V}}_{B} = {\\overset{\\rightarrow}{V}}_{BW} + {\\overset{\\rightarrow}{V}}_{W}$\nSo\n$V_{B} = \\sqrt{{(V_{W})}^{2} + {(V_{BW})}^{2} + 2V_{W} \\times V_{BW}(cos120^{\\circ})}$\n$= \\sqrt{1^{2} + 2^{2} + 2 \\times 1 \\times 2 \\times ( - 1/2)} = \\sqrt{3}m/s$", "solution_images": ["images/image18.png", "images/image19.png"], "subject": "Physics", "topic": "Kinematics", "subtopic": "Relative motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-21-Q5", "question": "[IMAGE] A parallel plate capacitor has plates of area A separated by distance\n between them. It is filled with adielectric which has a\ndielectric constant that varies as k(x) = K(1 + $\\alpha$x) where \nis the distancemeasured from one of the plates. If ($\\alpha$d) << 1,\nthe total capacitance of the system is best given by theexpression:\nlekukUrj IysVksa ls cus,d la\nmuds chp dh nwjh A bu IysVksa ds chp,d ijkoS|qr inkFkZ Hkjk\ngqvk gS ftldk ijk oS|qrkad k(x) = K(1+ αx)\nmi;qDr eku gksxk", "question_images": ["images/image9.png"], "option_1": "$\\frac{A \\in_{0}K}{d}\\left( 1 + \\frac{\\alpha^{2}d^{2}}{2} \\right)b$", "option_2": "$\\frac{AK \\in_{0}}{d}\\left( 1 + \\frac{\\alpha d}{2} \\right)$", "option_3": "$\\frac{A \\in_{0}K}{d}\\left( 1 + \\left( \\frac{\\alpha d}{2} \\right)^{2} \\right)$", "option_4": "$\\frac{AK\\epsilon_{0}}{d}(1 + \\alpha d)$", "correct_option": 2, "numerical_answer": null, "solution": "Capacitance of element ?kVd Capacitance of element ?kVd \n$\\sum\\frac{1}{C^{'}} = \\int_{0}^{d}\\mspace{2mu}\\frac{dx}{K\\varepsilon_{0}A(1 + \\alpha x)}$\n$\\frac{1}{C} = \\frac{1}{K\\varepsilon_{0}A\\alpha}\\mathcal{l}n(1 + \\alpha d)$\nGiven αd << 1\n$\\frac{1}{C} = \\frac{1}{K\\varepsilon_{0}A\\alpha}\\left( \\alpha d - \\frac{\\alpha^{2}d^{2}}{2} \\right)$\n$\\frac{1}{C} = \\frac{d}{K\\varepsilon_{0}A}\\left( 1 - \\frac{\\alpha d}{2} \\right)$\n$C = \\frac{K\\varepsilon_{0}A}{d}\\left( 1 + \\frac{\\alpha d}{2} \\right)$", "solution_images": ["images/image10.png"], "subject": "Physics", "topic": "Capacitance", "subtopic": "Equivalent capacitance", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-16-Q1", "question": "A ball is thrown vertically upwards from the ground. It crosses a\npoint at the height of 25 m. twice at an interval of 4 secs. The ball\nwas thrown with the velocity of", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Let initial velocity of the ball = u\n$\\therefore 25 = ut - \\frac{1}{2}gt^{2} = ut - 5t^{2}$\n$\\Rightarrow 5t^{2} - ut + 25 = 0$\nIf t_1and t_2 are the two solutions,\n[IMAGE] Then, $t_{2} - t_{1} = \\sqrt{{(t_{2} + t_{1})}^{2} - 4t_{1}t_{2}}$\n$\\Rightarrow 4^{2} = \\frac{u^{2}}{25} - 4 \\times \\frac{25}{5}$\n$\\Rightarrow 16 = \\frac{u^{2}}{25} - 20 \\Rightarrow u^{2} = 25 \\times 36$\n$\\Rightarrow u = 30m/s$", "solution_images": ["images/image1.png"], "subject": "Physics", "topic": "Kinematics", "subtopic": "Motion under gravity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "Ph-26-Q10", "question": "A diatomic ideal gas is heated at constant volume until the\npressure is doubled and again heated at constant pressure until volume\nis doubled. Find out the average molar heat capacity for whole process.\nin K)\nnkc nqxuk ugha gks tk;s\nrd ml nqxuk ugha gks", "question_images": [], "option_1": "28.5", "option_2": "23.5", "option_3": "25.3", "option_4": "23.5", "correct_option": 3, "numerical_answer": null, "solution": "Let initial pressure, volume, temperature be P_0, V_0, T_0\nindicated by state A in P-V diagram. The gas is then isochorically taken\nto state B (2P_0, V_0, 2T_0) and then taken from state B to state C\n(2P_0, V_0, 4T_0) isobarically\nvxj bl:vkrh s P_0, V_0, T_0ekus tSlk\nfd uhps P-V fp=k bls fQj levk;rfud izfØ B (2P_0, V_0,\n2T_0) vkSj fQj Bls C (2P_0, V_0, 4T_0) (2P_0, V_0, 4T_0) rd\n[IMAGE] Total heat absorbed by 1 mole of gas (by definition of molar heat\ncapacity)\ndqy tks 1 eksy xSl }kjk xzg.k dh xbZ &.\n$\\Delta Q = C_{v}\\left( 2T_{0} - T_{0} \\right) + C_{p}\\left( 4T_{0} - 2T_{0} \\right)$\n$= \\frac{5}{2}RT_{0} + \\frac{7}{2}R \\times 2T_{0} = \\frac{19}{2}{RT}_{0}$\nTotal change in temperature from state A to C is\n dqy ifjorZu AvoLFkk ls Crd gS &\n$\\Delta T = 3T_{0}$\nMolar heat capacityeksyj m\"ek", "solution_images": ["images/image30.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Molar heat capacity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-20-Q18", "question": "The figure gives experimentally measured B vs. H variation in a\nferromagnetic material. The retentivity, coercivity and saturation,\nrespectively, of the material are", "question_images": ["images/image60.png"], "option_1": "1.0 T, 50 A/m and 1.5 T", "option_2": "150 A/m, 1.0 T and 1.5 T", "option_3": "1.5 T, 50 A/m and 1.0 T", "option_4": "1.5T, 50 A/m and 1.0 T", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] x = retentivity\ny = coercivity\nz = saturation magnetization", "solution_images": ["images/image61.png"], "subject": "Physics", "topic": "Magnetism", "subtopic": "Hysteresis loop", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "Ph-27-Q1", "question": "When the temperature of copper coin is raised by 80oC, its\ndiameter increases by 0.2.", "question_images": [], "option_1": "percentage rise in the area of a face is 0.4", "option_2": "percentage rise in the volume is 0.6", "option_3": "coefficient of linear expansion of copper is\n$0.25 \\times 10^{- 4}C^{- 1}$", "option_4": "All of the above\ntc rkacs ds flD eku 80ºC c<", "correct_option": 4, "numerical_answer": null, "solution": "$100 \\times \\frac{\\Delta D}{D} = \\alpha\\Delta T \\times 100$\n$100 \\times \\left( \\frac{\\Delta A}{A} \\right) = \\beta\\Delta T \\times 100$\n$= 2\\alpha\\Delta T \\times 100$\n$= 2(0.2)$\n$= 0.4\\%$\n$100 \\times \\frac{\\Delta t}{t} = \\alpha\\Delta T \\times 100$\n$\\alpha = \\frac{\\frac{\\Delta d}{d}}{\\Delta T} = \\frac{0.2 \\times 10^{- 2}}{80}\\left( \\frac{\\Delta d}{d} \\times 100 = 0.2 \\right)$\n$\\alpha = 0.25 \\times 10^{- 4}\\ ^{\\circ}C^{- 1}$\n$100 \\times \\frac{\\Delta V}{V} = \\gamma\\Delta T \\times 100$\n$= 3(0.2) = 0.6\\%$", "solution_images": [], "subject": "Physics", "topic": "Heat", "subtopic": "Thermal expansion", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "Ph-27-Q27", "question": "A simple seconds pendulum is constructed out of a very thin\nstring of thermal coefficient of linear expansion\n[IMAGE] and a heavy particle attached to one end. The free end of the string is\nsuspended from the ceiling of an elevator at rest. The pendulum keeps\ncorrect time at 0° C. When the temperature rises to 50° C, the elevator\noperator of mass 60 kg being a student of Physics accelerates the\nelevator vertically, to have the pendulum correct time. The apparent\nweight (in N) of the operator when the pendulum keeps correct time at 50\n°C is 330x. Find x. [IMAGE],d lSd.M ljy yksyd cgqr iryh NM+ ls cuk gS] ftldk çlkj\nxq.kkad α= 20 x \n0ºC ij lgh Zr tc rki 50 ºC c<+rk gS rks 60kg\n fy¶V pkyd tks HkkSfrd foKku dk Nk=k gS] yksyd lgh\n fy, fy¶V dks Rofjr tc yksyd 50 ºC ij lgh\nle; ç gks rks fy¶V pkyd dk vkHkklh Hkkj330x ( )\ngksrk", "question_images": ["images/image172.png", "images/image173.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "[IMAGE] when the temperature is raised length changes to l(1 + α ∆T)\ntc rki c<+ rc yEckbZ esa ifjorZu l(1 + α ΔT) gks A\nwhen the lift acceleration upwards\ntc fy¶V dh vksj Rofjr gSA\ng_eff = g + a\nnew period of pendulum,\nyksyd", "solution_images": ["images/image174.png", "images/image175.png", "images/image176.png", "images/image177.png", "images/image178.png", "images/image179.png"], "subject": "Physics", "topic": "Thermal properties of matter", "subtopic": "Thermal expansion", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-01-Q16", "question": "A uniform solid cylinder of mass M = 2 kg and radius R = 10 cm\nis connected about an axis through the centre of the cylinder to a\nhorizontal spring with spring constant 4 N/m. The cylinder is pulled\nback, stretching the spring 1 m from equilibrium. When released, the\ncylinder rolls without slipping. What is the speed of the center of the\ncylinder when it returns to equilibrium ?", "question_images": ["images/image76.jpeg"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "wsp = ∆k", "solution_images": ["images/image77.png", "images/image78.png", "images/image79.png", "images/image80.png", "images/image81.png", "images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "Ph-27-Q9", "question": "When a capillary tube is dipped in a liquid, the liquid rises to\na height h in the tube. The free liquidsurface inside the tube is\nhemispherical in shape. The tube is now pushed down so that the height\nofthe tube outside the liquid is less than h.", "question_images": [], "option_1": "The liquid will come out of the tube like in a small fountain.", "option_2": "The liquid will come out of the tube slowly.", "option_3": "The liquid will fill the tube but not come out of its upper end.", "option_4": "The free liquid surface inside the tube will not be hemispherical.\ntc,d s nzo uyh esa nzo h rd\nuhps rkfd nzo ds ckgj uyh", "correct_option": 3, "numerical_answer": null, "solution": "The angle of contact at the free liquid surface inside the\ncapillary tube will change such that the verticalcomponent of the\nsurface tension forces just balance the weight of the liquid column.\n vUnj nzo dh eqDr lrg ij lEidZ dks.k bl çdkj cny tk,xk fd\ni`\"B ruko cy dk ?kVd nzo LrEHk", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Capillary action", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-02-Q43", "question": "The variation of internal energy vs density of a diatomic gas\nis as shown in figure. The molar heat capacity of the gas is", "question_images": ["images/image57.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "R", "correct_option": 1, "numerical_answer": null, "solution": "According to graph\nTV = constant\nor PV^2 = constant", "solution_images": ["images/image61.png", "images/image62.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-08-Q13", "question": "For an adiabatic process graph between PV&V for a sample of\nideal gas will be", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$PV\\ \\alpha\\ T$\n for adiabatic process, \n${TV}^{\\gamma - 1} = \\ $constant", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-20-Q4", "question": "A thin uniform rod of mass is hinged at one\nend. This rod is maintained in horizontal position by colliding very\ntiny balls each of mass m/10 completely elastically 10 times per sec\nstriking at the opposite end as shown in figure. Find the speed of the\nball - (g = 10 m/sec^2)", "question_images": ["images/image12.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] Taking torque about hinge:\n$MgL/2 = FL$\n$mg\\frac{L}{2} = 2(\\frac{mV}{10})10L$\n$V = 2.5m/sec$", "solution_images": ["images/image13.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Torque", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-14-Q20", "question": "A parallel beam of light of wavelength [IMAGE] is\nincident on slit S. Another two slit\ndistance 2 m as shown in the figure. The minimum distance between\nintensity will occur at O is.", "question_images": ["images/image127.png", "images/image133.png", "images/image134.png", "images/image133.png", "images/image134.png", "images/image135.png"], "option_1": "", "option_2": "1 mm", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE], when n = 1", "solution_images": ["images/image139.png", "images/image140.png", "images/image141.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-11-Q12", "question": "Mass per unit area of a circular disc of radius a depends on the\ndistance r from its centre as [IMAGE] The moment of\ninertia of the disc about the axis, perpendicular to the plane and\npassing through its centre is", "question_images": ["images/image117.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image122.jpeg", "images/image123.png", "images/image124.png", "images/image125.png", "images/image126.png", "images/image127.png", "images/image128.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-14-Q6", "question": "In the shown figure a ray of light enters the face XY of a right\nangered prism at grazing incident. It emerges from the adjacent face. YZ\nat an angle[IMAGE]. If x is the critical angle of the\nprism held in water of refractive index [IMAGE], then the\nangle [IMAGE] with normal is", "question_images": ["images/image26.png", "images/image27.png", "images/image26.png", "images/image28.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] As at XY\n[IMAGE] At YZ\n[IMAGE] On solving", "solution_images": ["images/image33.png", "images/image34.png", "images/image35.png", "images/image36.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Prism", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-06-Q3", "question": "System shown in figure is in equilibrium. The magnitude of change\nin tension in the string just before and just after, when one of the\nspring is cut. Mass of both the blocks is same and equal to m and\nspringconstant of both springs is k. (Neglect any effect of rotation)", "question_images": ["images/image5.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "Before cutting the spring\nAfter cutting the spring\n[IMAGE] 2mg - mg = 2 ma\na = g/2\n$T_{3} = mg/2$\n$T_{2} - T_{3} = mg - mg/2 = mg/2$", "solution_images": ["images/image6.png", "images/image7.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Newton's second law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-14-Q24", "question": "A point object in air is in front of the curved surface of a\nplano-convex lens. The radius of curvature of the curved surface is 30\ncm and the refractive index of the lens material is 1.5, then the focal\nlength of the lens (in cm) is10n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image155.png", "images/image156.png", "images/image157.png", "images/image158.png", "images/image159.png", "images/image160.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Lens maker's formula", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-15-Q19", "question": "In the circuit, if the forward voltage drop for the diode is\n[IMAGE] the current in the circuit will be\n(approximately)", "question_images": ["images/image148.png", "images/image149.png"], "option_1": "3.4 mA", "option_2": "2 mA", "option_3": "2.5 mA", "option_4": "3 mA", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image150.png"], "subject": "Physics", "topic": "electronic device", "subtopic": "Diodes", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-22-Q22", "question": "In the given A.C. circuit, if battery voltage\nis[IMAGE], power delivered by battery will have a power\nfactor of[IMAGE]. Find n.", "question_images": ["images/image68.png", "images/image69.png", "images/image70.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "[IMAGE] Phasor diagram\n[IMAGE] TOPIC: Alternating current\nSUB TOPIC: LCR circuit\nLEVEL: Moderate", "solution_images": ["images/image71.png", "images/image72.png", "images/image73.png", "images/image74.png", "images/image75.png", "images/image76.png", "images/image77.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-22-Q14", "question": "the frequency of the light used in an experiment on\nphotoelectric effect. If the work function is $\\frac{x}{42}$ (in eV)\nthen value of x is. $\\left( h = 6.4 \\times 10^{- 34}Js \\right)$", "question_images": ["images/image39.png"], "option_1": "33.3", "option_2": "33.5", "option_3": "33.6", "option_4": "33.7", "correct_option": 3, "numerical_answer": null, "solution": "$V = \\frac{1.6}{4 \\times 10^{14}}\\left( v - 2 \\times 10^{14} \\right) = \\frac{1.6}{4 \\times 10^{14}}v - 0.8$\nCompare it with $V = \\frac{h}{e}v - \\frac{\\phi}{e}$\n$\\frac{\\phi}{e} = 0.8$\n$\\phi = 0.8eV$", "solution_images": [], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "Einstein's photoelectric equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-20-Q6", "question": "Take two infinite planes with same surface charge density. Place\nthem as shown it the figure. The electric field in the region between\nthem is given by", "question_images": ["images/image15.jpeg"], "option_1": "$\\frac{\\sigma}{2\\epsilon_{0}}\\lbrack(1 + \\sqrt{3})\\overset{\\hat{}}{y} - \\frac{\\overset{\\hat{}}{x}}{2}\\rbrack$", "option_2": "$\\frac{\\sigma}{2\\epsilon_{0}}\\lbrack(1 - \\frac{\\sqrt{3}}{2})\\overset{\\hat{}}{y} - \\frac{\\overset{\\hat{}}{x}}{2}\\rbrack$", "option_3": "$\\frac{\\sigma}{2\\epsilon_{0}}\\lbrack(1 + \\sqrt{3})\\overset{\\hat{}}{y} + \\frac{\\overset{\\hat{}}{x}}{2}\\rbrack$", "option_4": "$\\frac{\\sigma}{\\epsilon_{0}}\\lbrack(1 + \\frac{\\sqrt{3}}{2})\\overset{\\hat{}}{y} + \\frac{\\overset{\\hat{}}{x}}{2}\\rbrack$", "correct_option": 2, "numerical_answer": null, "solution": "$\\overset{\\rightarrow}{E} = \\frac{\\sigma}{2\\varepsilon_{0}}\\lbrack(1 - \\frac{\\sqrt{3}}{2})\\overset{\\hat{}}{y} - \\frac{1}{2}\\overset{\\hat{}}{x}\\rbrack$", "solution_images": ["images/image16.jpeg"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field due to infinite sheet of charge", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-18-Q17", "question": "In the figure $m_{A} = m_{B} = m_{C} = 60kg$. The co-efficient\nof friction between C and ground is 0.5, B and ground is 0.3, A& B is\n0.4. C is pulling the string with the maximum possible force without\nmoving. Then tension in the string connected to A will be", "question_images": ["images/image24.png"], "option_1": "120 N", "option_2": "60 N", "option_3": "100 N", "option_4": "zero", "correct_option": 4, "numerical_answer": null, "solution": "Maximum frictional force between C and ground =300 Nt\nMax. frictional force between B and ground =360 Nt\nSo man is unable to pull B Hence T=", "solution_images": [], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-25-Q19", "question": "One mole of an ideal gas is kept enclosed under a light piston\n(area=[IMAGE] ) connected by acompressed spring (spring\nconstant 100 N / m). The volume of gas is [IMAGE] and\nits temperature is 100K. The gas is heated so that it compresses the\nspring further by 0.1 m. Then the work done by the gas in the process\nis: (in Joule) (Take R =8.3 J / K -mole and suppose there is no\natmosphere).,d 100 N / m½\nls tqM+s gYds fiLVu ¼kjk\n ¼R =8.3 J ekusa fd ok;qe.My ugha", "question_images": ["images/image136.png", "images/image137.png", "images/image136.png", "images/image137.png", "images/image138.png"], "option_1": "1.0", "option_2": "1.5", "option_3": "2.0", "option_4": "1.8", "correct_option": 2, "numerical_answer": null, "solution": "Before heating let the pressure of gas be\n[IMAGE] from the equilibrium piston.\n -\n[IMAGE] Since during heating process, pwafd xeZ izfØ nkSjku,\nThe spring is compressed further by 0.1 m fLizax 0.1 m", "solution_images": ["images/image139.png", "images/image139.png", "images/image140.png", "images/image141.png", "images/image142.png", "images/image143.png", "images/image144.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Work done", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-16-Q15", "question": "A river is flowing with a speed of $1\\ kmh^{- 1}$. A swimmer\nwants to go to point c starting from A. He swims with a speed of\n$5\\ kmh^{- 1}$ at an angle $\\theta$w.r.t. the river flow, at what angle\nwith the river bank should the swimmer swim?", "question_images": ["images/image27.png"], "option_1": "30º", "option_2": "37º", "option_3": "53º", "option_4": "60º", "correct_option": 3, "numerical_answer": null, "solution": "Path of swimmer relative to ground is a straight line at 45º\nwith bank, therefore, x and y components of swimmer's resultant\nvelocity must be equal.\nLet velocity of swimmer$|{\\overset{\\rightarrow}{v}}_{m}| = v$\n$V_{x} = V_{r} + V_{m,r}cos\\theta$\n$V_{y} = V_{m,r}sin\\theta$\n[IMAGE] Condition for reaching the point C,\n$tan45^{\\circ} = \\frac{v_{y}}{v_{x}}$\n$V_{y} = V_{x}$.\n$(V_{r} + V_{m,r}cos\\theta) = V_{m,r}sin\\theta$\n$1 + 5cos\\theta = 5sin\\theta$\nOn squaring$1 + 25\\cos^{2}\\theta + 10cos\\theta = 25 - 25\\cos^{2}\\theta$\n$50\\cos^{2}\\theta + 10cos\\theta - 24 = 0$\nOn solving, we get$\\theta = 53^{\\circ}$", "solution_images": ["images/image28.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "River problem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-18-Q18", "question": "You have a ring balanced at the center of the table by three\nforces ${\\overset{\\rightarrow}{F}}_{1},{\\overset{\\rightarrow}{F}}_{2}$\nand ${\\overset{\\rightarrow}{F}}_{3}$. The forces\n${\\overset{\\rightarrow}{F}}_{1}$ and ${\\overset{\\rightarrow}{F}}_{2}$\nhave\ncomponents.$F_{1x} = 10N,F_{1y} = - 50N,F_{2x} = - 40N,F_{2y} = 100N$\nThe force ${\\overset{\\rightarrow}{F}}_{3}$ is required to make the\nringstationary at the center of the table. Which one ofthe 4 vectors\nin the diagram could represent${\\overset{\\rightarrow}{\\ F}}_{3}$?", "question_images": ["images/image25.png"], "option_1": "$\\overrightarrow{A}$", "option_2": "$\\overrightarrow{B}$", "option_3": "$\\overrightarrow{C}$", "option_4": "$\\overrightarrow{D}$", "correct_option": 4, "numerical_answer": null, "solution": "${\\overset{\\rightarrow}{F}}_{1} + {\\overset{\\rightarrow}{F}}_{2} + {\\overset{\\rightarrow}{F}}_{3} = 0 \\Rightarrow 10\\overset{\\hat{}}{i} - 50\\overset{\\hat{}}{j} - 40\\overset{\\hat{}}{j} + {\\overset{\\rightarrow}{F}}_{3} = 0$\n$\\Rightarrow {\\overset{\\rightarrow}{F}}_{3} = 30\\overset{\\hat{}}{i} - 50\\overset{\\hat{}}{j} \\Rightarrow IVth\\ quadrant.$", "solution_images": [], "subject": "Physics", "topic": "Vector", "subtopic": "Equilibrium of forces", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-05-Q14", "question": "If the velocity of light denoted by acceleration due to\ngravity are taken as fundamental\nunits, then the dimensions of length will be", "question_images": [], "option_1": "$c^{2}/g$", "option_2": "$c/p$", "option_3": "$g/c$", "option_4": "$c/g$", "correct_option": 1, "numerical_answer": null, "solution": "Let $L\\alpha c^{a}g^{b}p^{c}$\n$L = kc^{a}g^{b}p^{c}$\nPutting dimensions on both sides:\n$\\left\\lbrack M^{0}LT^{0} \\right\\rbrack = \\left\\lbrack LT^{- 1} \\right\\rbrack^{a}\\left\\lbrack LT^{- 2} \\right\\rbrack^{b}\\left\\lbrack ML^{- 1}T^{- 2} \\right\\rbrack^{c}$\n$\\left\\lbrack M^{0}LT^{0} \\right\\rbrack = \\left\\lbrack M^{c}L^{a + b - c}T^{- a - 2b - 2c} \\right\\rbrack$\nComparing dimensions of both sides\n$c = 0,\\ \\ \\ \\ \\ a + b - c = 1$\n$- a - 2b - 2c = 0$\nsolving we get $a = 2,b = - 1$\nhence$\\lbrack L\\rbrack = \\left\\lbrack c^{2}/g \\right\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "Unit and dimension", "subtopic": "Dimensional analysis", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-12-Q9", "question": "In the given circuit diagram, a wire is joining points B and D.\nThe current in this wire is", "question_images": ["images/image11.png"], "option_1": "Zero", "option_2": "2 A", "option_3": "0.4 A", "option_4": "4A", "correct_option": 2, "numerical_answer": null, "solution": "$i = \\frac{20}{2} = 10A$\n$1 = \\frac{41}{5} - \\frac{3i}{5}$\n$= + \\frac{i}{5} = 2A$", "solution_images": ["images/image12.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-23-Q28", "question": "Figure is the plot of the stopping potential versus the\nfrequency of the light used in an experiment on photoelectric effect. If\nthe work function is $\\frac{x}{5}($ in\n[IMAGE] then value of x is.", "question_images": ["images/image98.png", "images/image99.png", "images/image100.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "", "solution_images": ["images/image101.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "Einstein's photoelectric equation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-13-Q25", "question": "A uniform magnetic field B in positive z direction exists in a\ncircular region of radius R = 5 m. A loop of radius R = 5m lying in x -\ny plane encloses the magnetic field at t = 0 and then pulled at uniform. The emf induced (in volts) is the\nloop at t = 2 sec is 6V. Then magnitude of 4B is", "question_images": ["images/image25.png", "images/image26.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "$= VB2\\sqrt{R^{2} - \\left( \\frac{vt}{2} \\right)^{2}} = VB\\sqrt{4R^{2} - V^{2}t^{2}}$\n$6V = 4 \\times B\\sqrt{4 \\times 25 - 16 \\times 4}$\n$B = 0.25T$", "solution_images": ["images/image27.png", "images/image28.png"], "subject": "Physics", "topic": "EMI", "subtopic": "Induced emf", "difficulty": "Tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-15-Q5", "question": "An electron of mass m and magnitude of charge |e| initially at\nrest gets accelerated by a constant electric field E. The rate of change\nof de-Broglie wavelength of this electron at time t ignoring\nrelativistic effects is", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image42.png", "images/image43.png", "images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "De Broglie wavelength", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-05-Q5", "question": "A particle is projected up an inclined plane of inclination\n$\\beta$ at an elevation $\\alpha$ to the horizontal. If the particle\nstrikes the inclined plane at right angles then $\\tan\\alpha =$", "question_images": [], "option_1": "$(cot\\beta - 2tan\\beta)$", "option_2": "$(tan\\beta + 2cot\\beta)$", "option_3": "$(cot\\beta + 2tan\\beta)$", "option_4": "$(tan\\beta - 2cot\\beta)$", "correct_option": 3, "numerical_answer": null, "solution": "$T = \\frac{2usin(\\alpha - \\beta)}{gcos\\beta}$......(1)\n$O = ucos(\\alpha - \\beta) - (gsin\\beta)T$\n$\\Rightarrow T = \\frac{ucos(\\alpha - \\beta)}{gsin\\beta}$......(2)\nfrom (1) &(2)\n$2tan(\\alpha - \\beta) = cot\\beta$\n$\\Rightarrow tan\\alpha = (cot\\beta + 2tan\\beta)$", "solution_images": [], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Projectile up the inclined plane", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-09-Q11", "question": "A string of length L of variable density is clamped between two\nrigid supports. Whose density variation is shown in figure, ρ = density\n[IMAGE] Find the time after which the disturbance created at one end will reach\nother end.\n(A = area of cross section of string in [IMAGE], T =\ntension in the string is constant in N)", "question_images": ["images/image59.png", "images/image60.png", "images/image61.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image66.png", "images/image67.png"], "subject": "Physics", "topic": "Wave on a string", "subtopic": "speed of wave", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-02-Q44", "question": "For a particle moving in a straight line, the acceleration\nversus displacement graph is as shown. The relation between which of the\nfollowing quantities cannot be a straight line?", "question_images": ["images/image63.png"], "option_1": "Velocity vs displacement", "option_2": "Force vs velocity", "option_3": "Force vs displacement", "option_4": "Kinetic energy vs displacement", "correct_option": null, "numerical_answer": null, "solution": "D\nα = kx\nor Force = ma = kmx\n∴ (C) is a straight line.\nAgain [IMAGE] ∴ (A) is a straight line.\n∴ (B) is a straight line.\nor [IMAGE] (D) is a parabola.\n∴", "solution_images": ["images/image64.png", "images/image65.png", "images/image66.png", "images/image67.png", "images/image68.png", "images/image69.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-15-Q24", "question": "In an α -decay the Kinetic energy of α particle is\n[IMAGE] and Q -value of the reaction\nis[IMAGE]. If the mass number of the mother nucleus is\n20 N. Find [IMAGE]: (Assume that daughter nucleus is in\nground state)", "question_images": ["images/image178.png", "images/image179.png", "images/image180.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "", "solution_images": ["images/image181.png"], "subject": "Physics", "topic": "Nuclear physics", "subtopic": "Alpha decay", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-21-Q19", "question": "In the show circuit diagram all capacitors are initially\nuncharged and cells are ideal. If C_1 = 1 $\\mu F$,C_2 = 2 $\\mu F$ and\npotential of earth is taken zero. Choose the correct options.\n izkjEHk rFkk\n A;fn C_1 = 1 $\\mu F$,C_2 = 2 $\\mu F$ rFkk i`Foh", "question_images": ["images/image73.png"], "option_1": "Potential of point A is (4/3) volts\nfcUnq A", "option_2": "Potential of point B is (-13/3) volts\nfcUnq B", "option_3": "Potential of point C is (10/3) volts\nfcUnq C", "option_4": "Potential of point A is (-4/3) volts\nfcUnq Adk foHko (-4/3) oksYV gSA", "correct_option": 3, "numerical_answer": null, "solution": "$V_{C} = 5 - \\frac{5}{3} = \\frac{10}{3}$V\n$\\frac{10}{3} - V_{B} = 8$\n$V_{B} = \\frac{- 14}{3}$V\n$\\frac{- 14}{3} - V_{A} = \\frac{10}{3}$\n$V_{A} = \\frac{- 24}{3} = - 8$V", "solution_images": ["images/image74.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electrical potential in an electric circuit", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-17-Q6", "question": "If an object is placed at 20 cm in front of a half thin convex\nlens of focal length 10 cm as shown in figure then taking object\nposition as shown the.", "question_images": ["images/image7.png"], "option_1": "x- coordinate of image is 20 cm.", "option_2": "x-coordinate of image is 40 cm.", "option_3": "y- coordinate of image is 0.2 cm.", "option_4": "y-coordinate of image is 0.4 cm.", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{1}{v} + \\frac{1}{- 20} = \\frac{1}{10} \\Rightarrow v = 20$\n$m = - 1$", "solution_images": [], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Convex lens", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-05-Q19", "question": "A force F is given by: F = at + bt^2, where t represents time.\nWhat are dimensions of a and b?", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$\\lbrack F\\rbrack = \\lbrack at\\rbrack;\\ \\lbrack a\\rbrack = \\left\\lbrack \\frac{F}{t} \\right\\rbrack = \\left\\lbrack \\frac{{MLT}^{- 2}}{T} \\right\\rbrack = \\left\\lbrack {MLT}^{- 3} \\right\\rbrack$\n$$\\lbrack F\\rbrack = \\left\\lbrack \\lbrack b\\rbrack = \\left\\lbrack \\frac{F}{t^{2}} \\right\\rbrack = \\left\\lbrack \\frac{{MLT}^{- 2}}{T^{2}} \\right\\rbrack = \\left\\lbrack {MLT}^{- 4} \\right\\rbrack$$", "solution_images": [], "subject": "Physics", "topic": "Unit and dimension", "subtopic": "Dimensional analysis", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-13-Q23", "question": "Two identical wires A and B, each of length $\\mathcal{l}$, carry\nthe same current I. Wire A is bent into acircle of radius R and wire B\nis bent form a squareof side a. If B_Aand B_B are the values of\nmagnetic field at the centres of the circle and square respectively,\nthen the ratio B_A/ B_B is$\\pi^{2}:n\\sqrt{2}.\\ $What is the value of\nn.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "B at centre of a circle $= \\frac{\\mu_{0}I}{2R}$\nB at centre of a\nsquare $= 4 \\times \\frac{\\mu I}{4\\pi \\cdot \\frac{\\mathcal{l}}{2}}\\left( sin45^{\\circ} + sin45^{\\circ} \\right)$\n$= 4\\sqrt{2}\\frac{\\mu_{0}I}{2\\pi\\mathcal{l}}$\n Now, $R = \\frac{L}{2\\pi}\\ $and $\\mathcal{l =}\\frac{L}{4}(\\ as\\ L = 2\\pi R = 4\\mathcal{l)}$\n Where, L= length of wire \n$\\therefore{\\ \\ \\ B}_{A} = \\frac{\\mu_{0}L}{2 \\cdot \\frac{L}{2\\pi}} = \\frac{\\pi\\mu_{0}I}{L} = \\pi\\left( \\frac{\\mu_{0}I}{L} \\right)$\n$B_{B} = 4\\sqrt{2}\\frac{\\mu_{0}I}{2\\pi\\left( \\frac{L}{4} \\right)} = \\frac{8\\sqrt{2}\\mu_{0}I}{\\pi L} = \\frac{8\\sqrt{2}}{\\pi}\\left( \\frac{\\mu_{0}I}{L} \\right)$\n$\\therefore{\\ \\ \\ \\ B}_{A}:B_{B} = \\pi^{2}:8\\sqrt{2}$", "solution_images": [], "subject": "Physics", "topic": "Magnetic effect of current", "subtopic": "Magnetic field due to current carrying conductor", "difficulty": "Easy / Moderate/ Tough", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-13-Q6", "question": "A very long wire ABDMNDC is shown in figure carrying current I.\nAB and BC parts are straight, longand at right angle. At D wire forms a\ncircular turn DMND of radius R. AB, BC parts are tangential tocircular\nturn at N and D. Magnetic field at the centre of circle is", "question_images": ["images/image8.png"], "option_1": "$\\frac{\\mu_{0}I}{2\\pi R}\\left( \\pi + \\frac{1}{\\sqrt{2}} \\right)$", "option_2": "$\\frac{\\mu_{0}I}{2R}$", "option_3": "$\\frac{\\mu_{0}I}{2\\pi R}\\left( \\pi - \\frac{1}{\\sqrt{2}} \\right)$", "option_4": "$\\frac{\\mu_{0}I}{2\\pi R}(\\pi + 1)$", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{\\mu_{0}i}{4\\pi R}\\left\\lbrack \\sin 90^{\\circ} - \\sin 45^{\\circ} \\right\\rbrack \\otimes + \\frac{\\mu_{0}i}{2R} \\odot + \\frac{\\mu_{0}i}{4\\pi R}\\left( \\sin 45^{\\circ} + \\sin 90^{\\circ} \\right) \\odot$\n$= \\frac{- \\mu_{0}i}{4\\pi R}\\left\\lbrack 1 - \\frac{1}{\\sqrt{2}} \\right\\rbrack + \\frac{\\mu_{0}j}{2R} + \\frac{\\mu_{0}i}{4\\pi R}\\left\\lbrack \\frac{1}{\\sqrt{2}} + 1 \\right\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{4\\pi R}\\left\\lbrack - 1 + \\frac{1}{\\sqrt{2}} + 2\\pi + \\frac{1}{\\sqrt{2}} + 1 \\right\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{4\\pi R}\\lbrack\\sqrt{2} + 2\\pi\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{2\\pi R}\\left\\lbrack \\frac{1}{\\sqrt{2}} + \\pi \\right\\rbrack \\odot$", "solution_images": ["images/image9.png"], "subject": "Physics", "topic": "Magnetic effects of current", "subtopic": "Magnetic field due to current carrying conductor", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-13-Q10", "question": "If the magnetic field in a plane electromagnetic wave is given\nby\n$\\overset{\\rightarrow}{B} = 3 \\times 10^{- 8}sin\\left( 1.6 \\times 10^{3}x + 48 \\times 10^{10}t \\right)\\overset{\\hat{}}{j}\\ T$,\nthen what will be expression for electric field?", "question_images": [], "option_1": "$\\overset{\\rightarrow}{E} = \\left( 3 \\times 10^{- 8}sin\\left( 1.6 \\times 10^{3}x + 48 \\times 10^{10}t \\right)\\overset{\\hat{}}{j}V/m \\right)$", "option_2": "$\\overset{\\rightarrow}{E} = \\left( 3 \\times 10^{- 8}sin\\left( 1.6 \\times 10^{3}x + 48 \\times 10^{10}t \\right)\\overset{\\hat{}}{i}V/m \\right)$", "option_3": "$\\overset{\\rightarrow}{E} = \\left( 60sin\\left( 1.6 \\times 10^{3}x + 48 \\times 10^{10}t \\right)\\overset{\\hat{}}{k}V/m \\right)$", "option_4": "$\\overset{\\rightarrow}{E} = \\left( 9sin\\left( 1.6 \\times 10^{3}x + 48 \\times 10^{10}t \\right)\\overset{\\hat{}}{k}V/m \\right)$", "correct_option": 4, "numerical_answer": null, "solution": "$\\frac{E_{0}}{B_{0}} = C$(speed of light in vacuum)\n$E_{0} = B_{0}C = 3 \\times 10^{- 8} \\times 3 \\times 10^{8}$\n= 9 N/C\nSo, $E = 9\\sin\\left( 1.6 \\times 10^{3}x + 48 \\times 10^{10}t \\right)$", "solution_images": [], "subject": "Physics", "topic": "EM wave", "subtopic": "Relation $\\overset{\\rightarrow}{E}$ =", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-06-Q17", "question": "The potential energy of a particle varies with x according to\nthe relation $U(x) = x^{2} - 4x$. The point x=2 is a point of", "question_images": [], "option_1": "stable equilibrium", "option_2": "unstable equilibrium", "option_3": "neutral equilibrium", "option_4": "none of above", "correct_option": 1, "numerical_answer": null, "solution": "$U(x) = x^{2} - 4x$\n$F = 0$\n$\\frac{dU(x)}{dx} = 0$\n$2x - 4 = 0\\ gives\\ \\ \\ x = 2$\n$\\frac{d^{2}U}{{dx}^{2}} = 2 > 0$ i.e. U is minimum hence x = 2 is a\npoint of stable equilibrium.", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Potential energy curve", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-21-Q15", "question": "Figure shows two identical capacitors each of capacity C in a\ncircuit with two ideal diodes. A 100 voltbattery in connected to the\ninput terminals. The output voltage across MN is.", "question_images": ["images/image41.png"], "option_1": "50 V", "option_2": "25 V", "option_3": "100 V", "option_4": "75 V", "correct_option": 1, "numerical_answer": null, "solution": "When terminal x is positive the diode D2 gets reverse biased\nand D1 forward biased therefore circuit is\nTc fljk x rks rFkk D_1vxzck;l gks Hence vr V_MN = 50 V", "solution_images": ["images/image42.png"], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Diodes", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-12-Q7", "question": "A fiber sheet with relative permittivity 5 is inserted between\ntwo parallel metal plates 0.25 cm apart. Apotential difference of 2500 v\nis applied between the plates. If dielectric strength of air is 3 ×\n10^6 v/m then.", "question_images": ["images/image8.png"], "option_1": "Air will not breakdown", "option_2": "Air will breakdown", "option_3": "Potential gradient strength in fiber sheet exceeds dielectric\nstrength of air", "option_4": "All the above are correct", "correct_option": 2, "numerical_answer": null, "solution": "As V = V_1 + V_2\n2500 = 0.0002 E_1 + 0.0023 E_2\n2E_1 + 23 E_2 = 25 x 10^6\nAlso; E_1 = 5E_2\n \nE_1 = 3.788 x 10^6 V/m\nAs E_1> 3.6 x 10^6 Hence Air will breakdown", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Dielectric strength", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "Ph-25-Q26", "question": "Under an adiabatic process, the volume of an ideal gas gets\ndoubled. Consequently the mean collisiontime between the gas molecules\nchanges from. If\n[IMAGE] for this gas then a good estimate,d \nbl j.k mlds v.kqvksa esa gksus okyh VDdjksa", "question_images": ["images/image186.png", "images/image187.png", "images/image188.png", "images/image189.png", "images/image190.png", "images/image186.png", "images/image187.png", "images/image188.png", "images/image189.png", "images/image190.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "", "solution_images": ["images/image191.png", "images/image192.png", "images/image193.png", "images/image194.png", "images/image195.png", "images/image196.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-07-Q24", "question": "A constant torque acting on a uniform circular wheel changes its\nangular momentum from 10 J-sec to60 J-sec in 4 sec. The torque in Nm is\n2.5n. Find the value of n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "$\\tau = \\frac{dL}{dt} = \\frac{60 - 10}{4} = 12.5$ Nm", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Angular impulse", "difficulty": "Easy", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-18-Q23", "question": "At a height 0.4 m from ground, the velocity of projectile in\nvector form\nis$(\\overset{\\rightarrow}{v} = 6\\overset{\\hat{}}{i} + 2\\overset{\\hat{}}{j})$.\nThe angle of projection is π/n. Find n.", "question_images": [], "option_1": "6", "option_2": "3", "option_3": "2", "option_4": "4", "correct_option": 1, "numerical_answer": null, "solution": "${(v_{y})}^{2} = u_{y}^{2} + 2a_{y}S$\n$(2)^{2} = {(u_{y})}^{2} + 2 \\times ( - 10) \\times 0.4$\n$4 + 8 = u_{y}^{2}$\n$u_{y} = \\sqrt{12}$\n$tan\\theta = \\frac{u_{y}}{u_{x}}$\n$= \\frac{\\sqrt{12}}{6}$\n$= \\frac{1}{\\sqrt{3}}$\n$\\theta = 30^{\\circ}$", "solution_images": ["images/image30.png"], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Angle of projection", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-20-Q21", "question": "A long, straight wire of radius a carries a current distributed\nuniformly over its cross-section. The ratioof the magnetic fields due\nto the wire at distance[IMAGE] and 2a, respectively from\nthe axis of the wire is n/3. Find n.", "question_images": ["images/image77.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "Using Ampere's circuital law:\n$B_{B} = \\frac{\\mu_{0}i}{2\\pi(2a)}$\n$\\frac{B_{A}}{B_{B}} = \\frac{4}{6} = \\frac{2}{3}$", "solution_images": ["images/image78.png"], "subject": "Physics", "topic": "Magnetism", "subtopic": "Ampere's circuital law", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-19-Q16", "question": "A galvanometer having a coil resistance $100\\Omega$ gives a full\nscale deflection when a current of 1 mA is passed through it. What is\nthe value of the resistance which can convert this galvanometer into a\nvoltmeter giving full scale deflection for a potential difference of 10\nV?", "question_images": [], "option_1": "$8.9k\\Omega$", "option_2": "$10k\\Omega$", "option_3": "$9.9k\\Omega$", "option_4": "$7.9k\\Omega$", "correct_option": 3, "numerical_answer": null, "solution": "$V_{g} = i_{g}R_{g} = 0.1V$\n$V = 10V$\n$R = R_{g}(\\frac{V}{V_{g}} - 1)$ $= 100 \\times 99 = 9.9K\\Omega$", "solution_images": [], "subject": "Physics", "topic": "Electric current", "subtopic": "Converting galvanometer into voltmeter", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-24-Q25", "question": "A vessel contains two immiscible liquids of densities\n$\\rho_{1} = 1000\\ {kgm}^{- 3}$ and $\\rho_{2} = 1500\\ {kgm}^{- 3}$. A\nsolid block of volume $V = 10^{- 3}m^{3}$ and density\n$d = 800{kgm}^{- 3}$ is tied to one end of a string and other is tied to\nthe bottom of the vessel as shown in figure. The block is immersed with\n$\\frac{2^{th}}{5}$ of its volume in the liquid of higher density and\n$\\frac{3^{th}}{5}$ in the liquid of lower density. The entire system is\nkept in an elevator which is moving upwards with an acceleration of a =\ng/2. Find the tension in the string in Newton. (Take $g = 10ms^{- 2}$)\n(in N)", "question_images": ["images/image60.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "[IMAGE] We analysed this problem from the reference frame of Elevator. Total\nbuoyant force On the block,\n$F_{B} = \\left( \\frac{2}{5}V\\rho_{2} + \\frac{3}{5}V\\rho_{1} \\right)(g + a)$\nFor equilibrium,\n$F_{B} = T + Vd(g + a)$\nor $T = F_{B} - Vd(g + a)$\n$= (g + a)V\\left\\lbrack \\frac{2}{5}\\rho_{2} + \\frac{3}{5}\\rho_{1} - d \\right\\rbrack$\n$T = \\left( 10 + \\frac{10}{2} \\right) \\times 10^{- 3}\\left\\lbrack \\frac{2}{5} \\times 1500 + \\frac{3}{5} \\times 1000 - 800\\rbrack = 6\\text{N} \\right.\\ $", "solution_images": ["images/image61.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Archimedes principle", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-12-Q17", "question": "The current I_1 (in A) flowing through 1$\\Omega$ resistor in\nthe following circuit is", "question_images": ["images/image20.png"], "option_1": "0.2", "option_2": "0.4", "option_3": "0.5", "option_4": "0.25", "correct_option": 1, "numerical_answer": null, "solution": "$i = \\frac{I}{2} = 0.2A$", "solution_images": ["images/image21.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-12-Q23", "question": "The balancing length for a cell is 560 cm in a potentiometer\nexperiment. When an external resistance of10[IMAGE] is\nconnected in parallel to the cell, the balancing length changes by 60\ncm. If the internalresistance of the cell is [IMAGE],\nwhere N is an integer then value of N/2 is ______.", "question_images": ["images/image35.png", "images/image36.png"], "option_1": "After connecting the resistor\n$\\frac{\\varepsilon \\times 10}{10 + r} = 500x$......", "option_2": "from", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "Let the emf of cell is $\\varepsilon$ internal resistance is\n and potential gradient is x\nOnly cell connected:\n$\\epsilon = 560\\ x$......(1)\nAfter connecting the resistor\n$\\frac{\\varepsilon \\times 10}{10 + r} = 500x$......(2)\nfrom (1) and (2)\n$\\frac{560 \\times 10}{10 + r} = 500s$\n$56 = 50 + 5r$\n$r = \\frac{6}{5} = 1.2\\Omega$\nn = 12", "solution_images": ["images/image37.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Potentiometer", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-05-Q8", "question": "A man who can swim at the rate of $2km/hr$ (in still river)\ncrosses a river to a point exactly opposite on the other bank by\nswimming at an angle of 60º perpendicular direction of river flow of the\nwater in the river. The velocity of the water current in $km/hr$ is", "question_images": [], "option_1": "1", "option_2": "2", "option_3": "$1/2$", "option_4": "$\\sqrt{3}$", "correct_option": 4, "numerical_answer": null, "solution": "Let speed of current be u. For net velocity ofman to be normal\nto river flow.\n$2sin60^{\\circ} = u$\nor$u = \\sqrt{3}km/hr$", "solution_images": ["images/image12.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "River problem", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-05-Q22", "question": "A particle at a height from the ground is projected with\nan angle 30º fromthe horizontal, it strikes the ground making angle 45º\nwith horizontal. It is again projected from the same point with the same\nspeed but with an angle of 60º with horizontal. The tangent of the angle\nit makes with the horizontal is $\\sqrt{n}$when it strikes the ground", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "Using $v = \\sqrt{u^{2} + 2gh}$\n$v = \\sqrt{u^{2}\\sin^{2}\\theta + 2gh}$(Vertical comp. when striking)\n$ucos\\theta = \\sqrt{u^{2}\\sin^{2}\\theta + 2gh}$\n$u^{2}\\cos^{2}\\theta = u^{2}\\sin^{2}\\theta + 2gh$\n$u^{2}\\left( \\frac{3}{4} - \\frac{1}{4} \\right) = 2gh$\n$u^{2} = 4gh$\n$u = 2\\sqrt{gh}$", "solution_images": ["images/image18.png", "images/image19.png"], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Projectile projected obliquely from a height", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-20-Q15", "question": "In a double-slit experiment, at a certain point on the screen\nthe path difference between the two interfering waves is\n[IMAGE] of a wavelength. The ratio of the intensity of\nlight at that point to that at the centre of a bright fringe is", "question_images": ["images/image48.png"], "option_1": "0.568", "option_2": "0.760", "option_3": "0.853", "option_4": "0.672", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image49.png", "images/image50.png", "images/image51.png", "images/image52.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Interference", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-02-Q49", "question": "Consider two black bodies A and B. The surface temperature\nof black body A is twice that of B. The wavelength corresponding to\nmaximum spectral radiancy for bodies A and B are λ_A and λ_B\nrespectively. When radiations of wavelength λ_A is incident on a\nmetallic surface the kinetic energy of fastest photoelectron is K_1\nand when radiations of wavelength λ_B is incident on the same metallic\nsurface the kinetic energy of fastest photoelectron is K_2. The work\nfunction of the metal is", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": null, "solution": "D", "solution_images": ["images/image83.png", "images/image84.png", "images/image85.png", "images/image86.png", "images/image87.png", "images/image88.png", "images/image89.png", "images/image90.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-03-Q4", "question": "Find the time when the displacement of the particle becomes zero.\nAssume velocity and position at t = 0 to be 0 and -10 m respectively.", "question_images": ["images/image12.png"], "option_1": "1s", "option_2": "", "option_3": "5s", "option_4": "2s", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-24-Q21", "question": "The pressure gauge reading in meter of water column shown in the\ngiven figure will be", "question_images": ["images/image56.png"], "option_1": "3.20 m", "option_2": "2.72 m", "option_3": "2.52 m", "option_4": "1.52 m", "correct_option": 4, "numerical_answer": null, "solution": "$P_{0} + \\rho_{w}g1 + \\rho_{wg}0.2 - \\rho_{Hg}g0.2 = P_{0}$\n$P - P_{0} = \\rho_{Hg}g0.2 - \\rho_{w}g(1.2) = 14896$\n$14896 = \\rho_{wgh}$\nh = 1.52", "solution_images": [], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Hydrostatic pressure", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-28-Q18", "question": "Consider a sphere of radius R which carries a uniform charge\ndensity $\\rho$. If a sphere of radius$\\frac{R}{2}$iscarved out of it, as\nshown, the\nratio$\\left| \\frac{{\\overset{\\rightarrow}{E}}_{A}}{{\\overset{\\rightarrow}{E}}_{B}} \\right|$\nof magnitude of electric field of magnitude of electric field\n${\\overset{\\rightarrow}{E}}_{A}$and ${\\overset{\\rightarrow}{E}}_{B}$,\nrespectively, atpoints A and B due to the remaining portion is", "question_images": ["images/image170.png"], "option_1": "$\\frac{9}{17}$", "option_2": "$\\frac{21}{34}$", "option_3": "$\\frac{17}{54}$", "option_4": "$\\frac{18}{54}$", "correct_option": 1, "numerical_answer": null, "solution": "For a solid sphere\n$E = \\frac{\\rho r}{3\\varepsilon_{0}}$\n$E_{A} = \\frac{- \\rho R}{2\\left( 3\\varepsilon_{0} \\right)}$\n$\\left| E_{A} \\right| = \\frac{\\rho R}{6\\varepsilon_{0}}$\nElectric field at point B = E_B = E_1A + E_2A\nE_1A = Electric Field Due to solid sphere of radius R at point\n$B = \\frac{\\rho R}{3\\varepsilon_{0}}$\nE_2A = Electric Field Due to solid sphere of radius R/2 (which having\ncharge density -$\\rho$)", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field by uniformly charged sphere", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-19-Q2", "question": "An electron (mass m) with initial velocity\n$\\overset{\\rightarrow}{v} = v_{0}\\overset{\\hat{}}{i} + v_{0}\\overset{\\hat{}}{j}$\nis in an electric field\n$\\overset{\\rightarrow}{E} = - E_{0}\\overset{\\hat{}}{k}$. If\n[IMAGE] is initial de-Broglie wavelength of electron, its\nde-Broglie wave length at time t is given by", "question_images": ["images/image2.png"], "option_1": "$\\frac{\\lambda_{0}}{\\sqrt{1 + \\frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$", "option_2": "$\\frac{\\lambda_{0}}{\\sqrt{1 + \\frac{e^{2}E_{0}^{2}t^{2}}{2m^{2}v_{0}^{2}}}}$", "option_3": "$\\frac{\\lambda_{0}}{\\sqrt{2 + \\frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$", "option_4": "$\\frac{\\lambda_{0}\\sqrt{2}}{\\sqrt{1 + \\frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$", "correct_option": 2, "numerical_answer": null, "solution": "Initially m$(\\sqrt{2}v_{0}) = \\frac{h}{\\lambda_{0}}$\nVelocity as a function of\ntime$= v_{0}\\overset{\\hat{}}{i} + v_{0}\\overset{\\hat{}}{j} + \\frac{{eE}_{0}}{m}t\\overset{\\hat{}}{k}$\nso wavelength\n$\\lambda = \\frac{h}{m\\sqrt{2v_{0}^{2} + \\frac{e^{2}E_{0}^{2}}{m^{2}}t^{2}}}$\n$\\lambda = \\frac{\\lambda_{0}}{\\sqrt{1 + \\frac{e^{2}E_{0}^{2}}{2m^{2}v_{0}^{2}}t^{2}}}$", "solution_images": [], "subject": "Physics", "topic": "Modern physics", "subtopic": "De Broglie wavelength", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-17-Q10", "question": "There is a small source of light at some depth below the surface\nof water (refractive index = 4/3) in a tank of large cross sectional\nsurface area. Neglecting any reflection from the bottom and absorption\nby water, percentage of light that emerges out of surface is (nearly):\n[Use the fact that surface area of a spherical cap of height h and\nradius of curvature r is $2\\pi rh$]", "question_images": [], "option_1": "50", "option_2": "34", "option_3": "17", "option_4": "21", "correct_option": 3, "numerical_answer": null, "solution": "Solid angle $d\\Omega = 2\\pi R^{2}(1 - cos\\beta)$\nPercentage of light\n$= \\frac{2\\pi R^{2}(1 - cos\\beta)}{4\\pi R^{2}} \\times 100$\n$= \\frac{1 - cos\\beta}{2} \\times 100 = (\\frac{4 - \\sqrt{7}}{8}) \\times 100 \\approx 17\\%$", "solution_images": ["images/image10.jpeg"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "TIR", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "Ph-26-Q8", "question": "For two thermodynamic process temperature and volume diagram are\ngiven. In first process, it is a straight line having initial and final\ncoordinates as (V_0, T_0) and (2V_0,2T_0), where as in second\nprocess it is a rectangular hyperbola having initial and final\ncoordinates (V_0, T_0) and (2V_0, T_0/2). Then ratio of work done in\nthe two processes must be\nnks izfØ fy, rkieku rFkk (V_0, T_0) rFkk (2V_0, 2T_0), gS] tgk¡ nwljs", "question_images": ["images/image26.png"], "option_1": "1: 2", "option_2": "2: 1", "option_3": "1: 1", "option_4": "None of thesebueas", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] First process is constant pressure izFke\nHence vr ] $W_{1} = nR\\left( 2T_{0} - T_{0} \\right) = nRT_{0}$\nEquation of second process is$T = \\frac{c}{V}$", "solution_images": ["images/image27.png", "images/image28.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Work done in thermodynamic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-25-Q7", "question": "Figure below shows two paths that may be taken by a gas to go\nfrom a state A to a state C.\n[IMAGE] In process AB, 400J of heat is added to the system and in process BC,\n100 J of heat is added to the system. The heat absorbed by the system\nin the process AC will be:;gk¡ nks iFk ftu \nA ls voLFkk B rd ys tk;k tk ldrk gSA\n s AB] çØe esa 400 J rFkk çØe BC", "question_images": ["images/image25.png", "images/image26.png"], "option_1": "500 J", "option_2": "460 J", "option_3": "300 J", "option_4": "380 J", "correct_option": 2, "numerical_answer": null, "solution": "For a complete cycle\niw.kZ pfØ; çØe ds fy,\nQ_cycle = W_cycle\n$+ 400 + 100 + Q_{C \\rightarrow A} = \\frac{1}{2}\\left( 2 \\times 10^{- 3} \\right)\\left( 4 \\times 10^{4} \\right)$\n⇒ Q~C→ A~ = - 460 J\n⇒ Q_A→C = + 460 J", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "Ph-26-Q18", "question": "In a cyclic waste power plant for thermal decomposition of waste\nhot gases are supplied from a \"Kentuki Chamber\" where gases are heated\nunder kentuki process [IMAGE] where P, V and T are\npressure, volume and temperature of gases in chamber and k is positive\nconstant. Find work done by 5 mole of hot gases when temperature is\nraised from 500 K to 550 K:,d pØ IykUV fo?kVu\n\"dsUVqdh psEcj\" ls rIr xSals çnku dh tkrh\nçfØ xeZ tgk¡ P, V o T psEcj\neksyksa 500 K ls 550 K rd o`f)", "question_images": ["images/image41.png", "images/image41.png"], "option_1": "625 R", "option_2": "- 625 R", "option_3": "250 R", "option_4": "- 250 R", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image42.png", "images/image43.png", "images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Ideal gas equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-24-Q6", "question": "Water flows in a horizontal tube (see figure). The pressure of\nwater changes by 700 Nm-2 between A and B where the area of cross\nsection are [IMAGE] and [IMAGE], respectively. Find the rate of flow of water\nthrough the tube. (density of water [IMAGE] )", "question_images": ["images/image30.png", "images/image31.png", "images/image32.png", "images/image33.png"], "option_1": "$\\text{27}\\text{3}\\text{2}\\text{ cm}^{\\text{3}}/\\text{s}$", "option_2": "$2420{cm}^{3}/s$", "option_3": "$3020{cm}^{3}/s$", "option_4": "$1810{cm}^{3}/s$", "correct_option": 1, "numerical_answer": null, "solution": "using equation of continuity\n$40V_{A} = 20V_{B}$\n$\\Rightarrow 2V_{A} = V_{B}$\nUsing Bernoulli's equation\n$P_{A} + \\frac{1}{2}\\rho V_{A}^{2} = P_{B} + \\frac{1}{2}\\rho V_{B}^{2}$\n$\\Rightarrow P_{A} - P_{B} = \\frac{1}{2}\\rho\\left( V_{B}^{2} - V_{A}^{2} \\right)$\n$\\Rightarrow \\Delta P = \\frac{1}{2}1000\\left( V_{B}^{2} - \\frac{V_{B}^{2}}{4} \\right)$\n$\\Rightarrow \\Delta P = 500 \\times \\frac{3V_{B}^{2}}{4}$\n$\\Rightarrow V_{B} = \\sqrt{\\frac{(\\Delta P) \\times 4}{1500}} = \\sqrt{\\frac{(700) \\times 4}{1500}m/s}$\nVolume flow rate $= 20 \\times 100 \\times V_{B} = 2732{cm}^{3}/s$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-18-Q16", "question": "A uniform rod of mass 200 gram and length $L = \\frac{1}{3}m$ is\ninitially at rest in vertical position. The rod is hinged at centre such\nthat it can rotate freely without friction about a fixed horizontal axis\npassing through its centre. Two particles of mass m=100 gram each having\nhorizontal velocity of equal magnitude u=17 m /s strikes the rod at top\nand bottom simultaneously as shown and stick to the rod. Find the\nangular speed (in rad/s) of rod when it becomes horizontal.", "question_images": [], "option_1": "78.50", "option_2": "76.50", "option_3": "77.50", "option_4": "67.50", "correct_option": 2, "numerical_answer": null, "solution": "From conservation of angular momentum.\n$mu\\frac{L}{2} + mu\\frac{L}{2} = \\lbrack 2m\\frac{L^{2}}{12} + m\\{{(\\frac{L}{2})}^{2} + {(\\frac{L}{2})}^{2}\\}\\rbrack\\omega$\n$muL = \\lbrack\\frac{{mL}^{2}}{6} + \\frac{{mL}^{2}}{4} + \\frac{{mL}^{2}}{4}\\rbrack\\omega = \\frac{2{mL}^{2}}{3}\\omega$\nor $\\omega = \\frac{3u}{2L} = 76.50$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-28-Q13", "question": "In the given A.C. circuit, if battery voltage\nis[IMAGE], power delivered by battery will have a power\nfactor of[IMAGE]. Find n.", "question_images": ["images/image124.png", "images/image125.png", "images/image126.png"], "option_1": "7", "option_2": "3", "option_3": "5", "option_4": "6", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] Phasor diagram\n[IMAGE] TOPIC: Alternating current\nSUB TOPIC: LCR circuit\nLEVEL: Moderate", "solution_images": ["images/image127.png", "images/image128.png", "images/image129.png", "images/image130.png", "images/image131.png", "images/image132.png", "images/image133.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-17-Q2", "question": "In the given A.C. circuit, if battery voltage is\n$V = 200\\sqrt{2}sin(100\\pi t)$, power delivered by battery will have a\npower factor of", "question_images": ["images/image2.png"], "option_1": "$\\frac{1}{2}$", "option_2": "$\\sqrt{\\frac{3}{10}}$", "option_3": "$\\frac{3}{\\sqrt{10}}$", "option_4": "$\\frac{1}{\\sqrt{10}}$", "correct_option": 3, "numerical_answer": null, "solution": "$i_{1} = \\frac{200\\sqrt{2}}{\\sqrt{30^{2} + 40^{2}}}sin(\\omega t - \\tan^{- 1}(\\frac{40}{30}))$\n$= 4\\sqrt{2}sin(100\\pi t - 53^{\\circ})$\nPhasor diagram\n$\\Rightarrow tan\\theta = \\frac{R_{y}}{R_{x}} = \\frac{1}{3} \\Rightarrow cos\\theta = \\frac{3}{\\sqrt{10}}$", "solution_images": ["images/image3.png"], "subject": "Physics", "topic": "AC", "subtopic": "Power factor", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "Ph-25-Q13", "question": "For an adiabatic process graph between PV & V for a sample of\nideal gas will be:\nvkn'kZ xSl ds,d uewus dk:)ks\"k çØe", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] for adiabatic process,:)ks\"k çØe", "solution_images": ["images/image91.png", "images/image92.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-02-Q53", "question": "The positions of\n are shown on the\nbinding energy curve as shown in figure. The energy released in the\nfusion reaction will be closest to [IMAGE] will be\nclosest to( in MeV )", "question_images": ["images/image109.png", "images/image110.png", "images/image111.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "16", "solution": "Q = -2 × 1 - 7 × 5.5 + 8 × 7 = 56 - 2 - 38.5 = 15.5 ]", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-06-Q12", "question": "Initially the block is at rest acceleration of the block is", "question_images": ["images/image21.png"], "option_1": "$2\\ m/s^{2}$", "option_2": "$0\\ m/s^{2}$", "option_3": "$1\\ m/s^{2}$", "option_4": "$0.5\\ m/s^{2}$", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] $N + 24 - 100 = 0$ For vertical direction\n$\\therefore N = 76N$\nNow, $0 \\leq f_{s} \\leq \\mu_{s}N\\ $\n$0 \\leq f_{s} \\leq 76 \\times 0.5$\n$0 \\leq f_{s} \\leq 38N$\n$\\therefore 32 < 38\\text{ Hence }f = 32$\n$\\therefore\\ \\ $Acceleration of block is zero.", "solution_images": ["images/image22.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Static friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-17-Q16", "question": "In the shown figure three slits, separated by\n$d = \\sqrt{\\frac{2\\lambda D}{3}}$ and illuminated by monochromatic\nparallel beam of light of wavelength $\\lambda$. P is a point on the line\nperpendicular to the plane of slits through S_3. If I_0 is the\nintensity of the wave from each slit, the intensity at p is", "question_images": ["images/image14.png"], "option_1": "$I_{0}$", "option_2": "${2I}_{0}$", "option_3": "${4I}_{0}$", "option_4": "${3I}_{0}$", "correct_option": 4, "numerical_answer": null, "solution": "Phase difference between S_2 &\nS_3$\\ = \\frac{2\\pi}{\\lambda}(\\sqrt{D^{2} + d^{2}} - D)$\n$= \\frac{2\\pi}{\\lambda}(\\frac{D \\times d^{2}}{2D^{2}}) = \\frac{2\\pi}{3}$\nPhase difference between S_1 &\nS_3$\\ = \\frac{2\\pi}{\\lambda} = (\\sqrt{D^{2} + 4d^{2}} - D)$\n$= \\frac{2\\pi}{\\lambda} \\times D(\\frac{4d^{2}}{2D}) = \\frac{8\\pi}{3}$\n[IMAGE] Now using phase diagram\n$A_{R} = A\\sqrt{3}andIR = 3Io$", "solution_images": ["images/image15.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-04-Q22", "question": "A stone is dropped from a certain height and can reach the ground\nin 5s. But in its fall, the stone is stopped after, 3s of the fall for a\nmoment and is dropped again at once. Now the stone reaches the ground in\ntotal time of seconds. Find the value of t?\n 5s \n 3s \n \n t", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "07", "solution": "h = [IMAGE] × 10 × 52 = 125 m\nh1 = [IMAGE] × 10 × 32 = 45\n80 = [IMAGE] × 10 × t2\nt = 4 sec. Total time = 3 + 4 = 7 sec\nTOPIC:KINEMATIC\nSUB TOPIC: MOTION UNDER GRAVITY", "solution_images": ["images/image119.png", "images/image120.png", "images/image121.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-12-Q25", "question": "In a meter bridge experiment S is a standard resistance. R is a\nresistance wire. It is found thatbalancing length is $\\mathcal{l}$ = 25\ncm. If R is replaced by a wire of half length and half diameter that of\nR ofsame material, then the balancing distance is$\\mathcal{l}'$ (in cm).\nWhat is the value of $\\mathcal{l}^{'}/10$.", "question_images": ["images/image39.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "$\\frac{x}{R} = \\frac{75}{25} = 3$\n$R = \\frac{\\rho\\mathcal{l}}{A} = \\frac{4\\rho\\mathcal{l}}{\\pi d^{2}}$\n$R^{'} = \\frac{4\\rho\\left( \\frac{\\mathcal{l}}{2} \\right)}{\\pi\\left( \\frac{d}{2} \\right)^{2}} = 2R$\n then $\\frac{X}{R^{'}} = \\left( \\frac{100 - \\mathcal{l}}{\\mathcal{l}} \\right)$\n$\\frac{100 - \\mathcal{l}}{\\mathcal{l}} = \\frac{X}{2R} = \\frac{3}{2}$\n$\\mathcal{l =}40.00cm$", "solution_images": [], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Potentiometer", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-14-Q16", "question": "In a Young's double slit experiment, the separation between the\nslits is 0.15 mm. In the experiment, asource of light of wavelength 589\nmm is used and the interference pattern is observed on a screen kept1.5\nm away. The separation between the successive bright fringes on the\nscreen is", "question_images": [], "option_1": "4.9 mm", "option_2": "3.9 mm", "option_3": "5.9 mm", "option_4": "6.9 mm", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image114.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "Ph-26-Q33", "question": "A ring shaped tube contains two ideal gases with equal masses\nand atomic mass numbers M_1=32 and M_2=28. The gases are separated by\none fixed partition X and another movable conducting partition Y which\ncan move freely without friction inside the ring. The angle is α, in\nequilibrium as shown in thefigure (in degrees). Find the value of\n[IMAGE],d vkdkj dh Vîqc esa nks vkn'kZ xSl ftudk æO;eku leku rFkk\nijek.kq æ M_1=32 rFkk M_2=28\nfoHkktd X ls vyx&vyx dh tkrh gS rFkk nqljk foHkktd,d xfreku pkydfoHkktd\nX tks fcuk ?k\"kZ.k xfr dj ldrk gSA fp=kuqlkj\ndks.k α ¼fMxzh esa½ rks [IMAGE]", "question_images": ["images/image139.png", "images/image140.png", "images/image141.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "At equilibrium, pressure and temperature are same. rFkk rki leku gSA\n Also, let cross-sectional area of tube be A and radius of ring be r.\nekuk nkc", "solution_images": ["images/image142.png", "images/image143.png", "images/image144.png", "images/image145.png", "images/image146.png", "images/image147.png", "images/image148.png"], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Gases in equilibrium", "difficulty": "Tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-13-Q1", "question": "The magnetic field is increasing at a constant rate$\\alpha$\nthrough a circular loop of radius aandthe axis of magnetic field\ncoincides with axis of loop as shown in figure. Theresistance per unit\nlength of loop wire is$\\rho$. Choose correct option(s).", "question_images": ["images/image1.png"], "option_1": "Current in loop PQRS is $\\frac{a\\alpha}{2\\rho}$ anticlockwise.", "option_2": "Current in the loop PQRS is $\\frac{a\\alpha}{2\\rho}$ clock wise.", "option_3": "Current in wire PR is $\\frac{\\pi a\\alpha}{\\rho}$.", "option_4": "Current in wire PR is$\\frac{\\pi a\\alpha}{2\\rho}$.", "correct_option": 1, "numerical_answer": null, "solution": "Magnitude of Induced emf = $A\\frac{dB}{dt}$\n$\\varepsilon = \\pi a^{2}\\alpha$\n$I_{PQRS\\ } = \\frac{a\\alpha}{2\\rho}$\nI_PR=0", "solution_images": [], "subject": "Physics", "topic": "EMI", "subtopic": "Induced current in a loop", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-10-Q1", "question": "A ball of mass m is suspended from a fixed point by an\ninextensible string. A small particle of same mass m, moving downward at\nan angle 37º with the vertical, collides head on with v_0 velocity.\nCoefficient of restitution is 4/5.", "question_images": ["images/image1.png"], "option_1": "Velocity of ball after impact is [IMAGE]", "option_2": "Velocity of ball after impact is\n$\\left( \\frac{25}{34} \\right)\\text{v}_{\\text{0}}$", "option_3": "Impulse produced by tension is m v_0[IMAGE]", "option_4": "Impulse produced by tension is [IMAGE]", "correct_option": 3, "numerical_answer": null, "solution": "Say V_1 is velocity attained by ball after collision\nand $\\frac{4}{5} = \\frac{\\frac{3V_{1}}{5} + V}{V_{0}}$\n$V_{1} = \\left( \\frac{27}{34} \\right)v_{0}:V = \\frac{11}{34}v_{0}$\nImpulse produced by tension $= mv_{0}cos37^{\\circ} + mvcos37^{\\circ}$", "solution_images": ["images/image5.png", "images/image6.png"], "subject": "Physics", "topic": "Collision", "subtopic": "Inelastic collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-21-Q23", "question": "In the figure, potential difference (in V) between A and B is\n2n. Find n:\n fp=k", "question_images": ["images/image79.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "Diode is in forward bias, so it will behave as simple wire so,\n voLFkk vr;g lk\n[IMAGE] So vc $V_{ab} = \\frac{30}{5 + 10} \\times 5 = 10V$", "solution_images": ["images/image80.png"], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Diodes", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "Ph-25-Q6", "question": "Two moles of hydrogen are mixed with n moles of helium. The root\nmean square speed of gas molecules in the mixture is $\\sqrt{2}$ times\nthe speed of sound in the mixture. Then n is.\ngkbMªkstu ds nks eksyksa n eksyksa ds lkFk", "question_images": ["images/image24.png"], "option_1": "3", "option_2": "2", "option_3": "1.5", "option_4": "2.5", "correct_option": 2, "numerical_answer": null, "solution": "$V_{rms} = \\sqrt{\\frac{3RT}{M_{mix}}}$\n$V_{sound\\ } = \\sqrt{\\frac{\\gamma RT}{M_{mix}}}$\n$V_{rms} = \\sqrt{2}V_{sound\\ }$\n$\\sqrt{\\frac{3RT}{M_{mix}}} = \\sqrt{2}\\sqrt{\\frac{\\gamma RT}{M_{mix}}}$\n$\\gamma_{mix} = \\frac{3}{2}$\n$\\gamma_{mix} = \\frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$\n$\\frac{3}{2} = \\frac{2 \\times \\frac{7R}{2} + n \\times \\frac{5R}{2}}{2 \\times \\frac{5R}{2} + n \\times \\frac{3R}{2}} \\Rightarrow \\frac{3}{2} = \\frac{14 + 5n}{10 + 3n} \\Rightarrow 30 + 9n = 28 + 10n \\Rightarrow n = 2$", "solution_images": [], "subject": "Physics", "topic": "Kinetic theory of gases", "subtopic": "Root mean squared speed", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "Ph-28-Q15", "question": "A reflecting surface is represented by the equation\n[IMAGE] A ray travelling horizontally becomes vertical after reflection. The\nco-ordinates of points where this ray incident is", "question_images": ["images/image147.png", "images/image148.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "The slope of curve at such points is [IMAGE] or [IMAGE] TOPIC: Ray Optics\nSUB TOPIC:Reflection by mirror\nLEVEL:Moderate", "solution_images": ["images/image153.png", "images/image154.png", "images/image155.png", "images/image156.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-17-Q8", "question": "In a meter bridge experiment S is a standard resistance. R is a\nresistance wire. It is found that balancing length is\n$\\mathcal{l =}25cm$. If R is replaced by a wire of half length and half\ndiameter that of R of same material, then the balancing distance\n$\\mathcal{l}'$ (in cm) will now be_______.", "question_images": ["images/image8.jpeg"], "option_1": "40.00", "option_2": "41.00", "option_3": "42.50", "option_4": "40.50", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{X}{R} = \\frac{75}{25} = 3$\n$R = \\frac{\\rho\\mathcal{l}}{A} = \\frac{4\\rho\\mathcal{l}}{\\pi d^{2}}$\n$R^{'} = \\frac{4\\rho(\\frac{\\mathcal{l}}{2})}{\\pi{(\\frac{d}{2})}^{2}} = 2R$\nThen $\\frac{X}{R^{'}} = (\\frac{100 - \\mathcal{l}}{\\mathcal{l}})$\n$\\frac{100 - \\mathcal{l}}{\\mathcal{l}} = \\frac{X}{2R} = \\frac{3}{2}$\n$\\mathcal{l =}40.00cm$", "solution_images": [], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Meter bridge", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-02-Q54", "question": "In the circuit shown, equivalent resistance between A and B is\nnR,then value of n is", "question_images": ["images/image113.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "02", "solution": "", "solution_images": ["images/image114.png", "images/image115.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-11-Q3", "question": "A particle of mass m is projected with a speed u from the ground\nat an angle [IMAGE] horizontal (x-axis). When it has\nreached its maximum height, it collides completely inelastically with\nanother particle of the same mass and velocity[IMAGE].\nThe horizontal distance covered by the combined mass before reaching the\nground is", "question_images": ["images/image25.png", "images/image26.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] so horizontal range after collision[IMAGE]", "solution_images": ["images/image31.png", "images/image32.png", "images/image33.png", "images/image34.png", "images/image35.png", "images/image36.png", "images/image37.png"], "subject": "Physics", "topic": "Collision", "subtopic": "Projectile", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-14-Q21", "question": "The distance between two slits in a Young's double slit\nexperiment is 3 mm. The distance of the screen from the slits is 1 m.\nMicrowaves of wavelength 1 mm are incident on the plane of the slits\nnormally. The distance (in cm) of the first maxima on the screen from\nthe central maxima is $\\frac{100}{\\sqrt{n}}$. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "At first maxima [IMAGE]", "solution_images": ["images/image142.png", "images/image143.png", "images/image144.png", "images/image145.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-24-Q11", "question": "When a capillary tube is dipped in a liquid, the liquid rises to\na height h in the tube. The free liquid surface inside the tube is\nhemispherical in shape. The tube is now pushed down so that the height\nof the tube outside the liquid is less than h.", "question_images": [], "option_1": "The liquid will come out of the tube like in a small fountain.", "option_2": "The liquid will come out of the tube slowly.", "option_3": "The liquid will fill the tube but not come out of its upper end.", "option_4": "The free liquid surface inside the tube will be hemispherical.", "correct_option": 3, "numerical_answer": null, "solution": "The angle of contact at the free liquid surface inside the\ncapillary tube will change such that the vertical component of the\nsurface tension forces just balance the weight of the liquid column.", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Capillary action", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-10-Q4", "question": "A bullet of mass m hits the hanging block of mass M at its centre\ncompletely inelastically with a velocity v and gets embedded in it as\nshown, then choose the correct option", "question_images": ["images/image22.png"], "option_1": "The tension in the string just before and after the collision changes\nby$\\frac{m^{2}v^{2}}{(m + M)^{\\mathcal{l}}}$", "option_2": "The tension in the string just before and after the collision changes\nby\n$$\\left\\lbrack g + \\right\\rbrack m$$", "option_3": "Tension in the string just after the collision\nbecomes$m\\left\\lbrack g + \\frac{mv^{2}}{(m + M)^{\\mathcal{l}}} \\right\\rbrack$", "option_4": "Tension in the string will not change after the collision", "correct_option": 2, "numerical_answer": null, "solution": "As T_1 = Mg\n$T_{2} = (M + m)g + \\frac{m^{2}v^{2}}{(M + m)^{\\mathcal{l}}}$\n$\\therefore T_{2} - T_{1} = m\\left\\lbrack g + \\frac{mv^{2}}{(m + M)^{\\mathcal{l}}} \\right\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "Collision", "subtopic": "Perfectly inelastic collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-22-Q24", "question": "A steel wire is rigidly fixed at both ends. Its length, mass and\ncross sectional area are 1m, 0.01 kg and\n[IMAGE] respectively. Then the temperature of the wire is\nlowered by[IMAGE]. If the transverse waves are setup by\nplucking the wire at [IMAGE] from one end and assuming\nthat wire vibrates with minimum number of loops possible for such a\ncase. The frequency of vibration is 44n Hz. Find n. [Coefficient of\nlinear expansion of steel [IMAGE] and Young's", "question_images": ["images/image88.png", "images/image89.png", "images/image90.png", "images/image91.png", "images/image92.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "[IMAGE] The mechanical strain [IMAGE] The tension in wire[IMAGE] speed of wave in wire\n[IMAGE] Since the wire is plucked at [IMAGE] from one end\nThe wire shall oscillate in [IMAGE] overtone (for minimum\nnumber of loops)\nTOPIC:Waves in a string\nSUB TOPIC: Wire fixed at both ends\nLEVEL:Moderate", "solution_images": ["images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png", "images/image97.png", "images/image98.png", "images/image99.png", "images/image100.png", "images/image101.png", "images/image102.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-08-Q7", "question": "Select the correct alternative for an ideal gas", "question_images": [], "option_1": "The change in internal energy in a constant pressure process from\ntemperature $T_{1}\\ to\\ T_{2}\\ $is\nequal to$\\ nC_{v}\\left( T_{2} - T_{1} \\right)$ where $C_{v}\\ $is the\nmolar specific heat at constant volume and n the number of moles of\nthe gas.", "option_2": "The change in internal energy of the gas and the work done by the\ngas are equal in magnitude in an adiabatic process.", "option_3": "The internal energy does not change in an isothermal process and No\nheat is added or removed in an adiabatic process.", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$\\Delta U = n\\frac{f}{2}R\\left( T_{2} - T_{1} \\right) = nC_{v}\\left( T_{2} - T_{1} \\right)$\n For adiabatic process $Q_{12} = 0$\n$Q_{12} = \\Delta U + W_{12}$\n$W_{12} = - \\Delta U$\nFor Isothermal process Temperature is constant.\nSo, change internal energy is zero.", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic processes", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-11-Q5", "question": "A body A of mass m is moving in a circular orbit of radius R\nabout a planet. Another body B of mass [IMAGE] collides\nwith A with a velocity which is half [IMAGE] the\ninstantaneous velocity [IMAGE] of A. The collision\niscompletely inelastic. Then, the combined body", "question_images": ["images/image45.png", "images/image46.png", "images/image47.png"], "option_1": "Escapes from the planet's Gravitational field", "option_2": "Continues to move in a circular orbit", "option_3": "Falls vertically downwards towards the planet", "option_4": "Starts moving in a elliptical orbit around the planet", "correct_option": 4, "numerical_answer": null, "solution": "Conserving momentum\n[IMAGE] thus the combined mass will go on to an\nelliptical path", "solution_images": ["images/image48.png", "images/image49.png", "images/image50.png"], "subject": "Physics", "topic": "Gravitation", "subtopic": "Conservation of momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-01-Q23", "question": "Find the resultant focal length (in cm ) for following system\nwhere the common radius of curvature is 15 cm. Glass has refractive\nindex of 1.5 and water has refractive index.", "question_images": ["images/image97.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "40", "solution": "", "solution_images": ["images/image98.png", "images/image99.png", "images/image100.png", "images/image101.png", "images/image102.png", "images/image103.png", "images/image104.png", "images/image105.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-21-Q2", "question": "A fiber sheet with relative permittivity 5 is inserted between\ntwo parallel metal plates 0.25 cm apart. Apotential difference of 2500 v\nis applied between the plates. If dielectric strength of air is 3 ×\n10^6 v/mthan.,d Qkbcj dh ifV~Vdk ftl fo|qr'khyrk 5 gS 0-25 lseh-\nfo|qr foHko 2500 v o ok;q", "question_images": ["images/image2.png"], "option_1": "Air will not breakdown\nok;q", "option_2": "Air will breakdown\nok;q", "option_3": "Potential gradient strength in fiber sheet exceeds dielectric\nstrength of air\nQkbcj ifV~Vdk", "option_4": "All the above are correct\nlHkh fodYi lgh gSA", "correct_option": 2, "numerical_answer": null, "solution": "As D;ksafd V = V_1 + V_2\n2500 = 0.0002 E_1 + 0.0023 E_2\n2E_1 + 23 E_2 = 25 x 10^6\nAlso rFkk; E_1 = 5E_2\nSolving gy E_2 = 0 757 x 10^6 V/m\nE_1 = 3.788 x 10^6 V/m\nAs E_1> 3.6 x 10^6 Hence Air will breakdown", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Dielectric strength", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-19-Q17", "question": "Consider a sphere of radius R which carries a uniform charge\ndensity$\\rho$. If a sphere of radius $\\frac{R}{2}$is carved out of it,\nas shown, the ratio\n$|\\frac{{\\overset{\\rightarrow}{E}}_{A}}{{\\overset{\\rightarrow}{E}}_{B}}|$\nof magnitude of electric field\n${\\overset{\\rightarrow}{E}}_{A}$and${\\overset{\\rightarrow}{E}}_{B}$,\nrespectively, at points A and B due to the remaining portion is", "question_images": ["images/image26.jpeg"], "option_1": "$\\frac{18}{34}$", "option_2": "$\\frac{21}{34}$", "option_3": "$\\frac{17}{54}$", "option_4": "$\\frac{18}{54}$", "correct_option": 1, "numerical_answer": null, "solution": "For a solid sphere\n$E = \\frac{\\rho r}{3\\varepsilon_{0}}$\n$E_{A} = \\frac{- \\rho R}{2(3\\varepsilon_{0})}$\n$|E_{A}| = \\frac{\\rho R}{6\\varepsilon_{0}}$\nElectric field at point$B = E_{B} = E_{1A} + E_{2A}$\n$E_{1A}$Electric Field Due to solid sphere of radius R at point\nB$= \\frac{\\rho R}{3\\varepsilon_{0}}$\n$E_{2A}$Electric Field Due to solid sphere of radius $R/2$ (which having\ncharge density $- \\rho$)\n$= - \\frac{\\rho R}{54\\varepsilon_{0}}$\n$E_{B} = E_{1A} + E_{2A} = \\frac{\\rho R}{3\\varepsilon_{0}} - \\frac{\\rho R}{54\\varepsilon_{0}} = \\frac{17\\rho R}{54\\varepsilon_{0}}$\n$\\frac{|E_{A}|}{|E_{B}|} = \\frac{9}{17}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field due to uniformly charged sphere", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-07-Q2", "question": "A bead can slide on a smooth straight wire and a particle of mass\nm attached to the bead by a lightstring of length L. The particle is\nheld in contact with the wire and with the string taut and is then let\nfall.\nIf the bead has mass 2 m then when the string makes an angle $\\theta$\nwith the wire, the bead will haveslipped a distance", "question_images": ["images/image2.png", "images/image3.png"], "option_1": "$L(1 - cos\\theta)$", "option_2": "$\\ (L/2)\\ (1 - cos\\theta)$", "option_3": "$(L/3)(1 - cos\\theta)$", "option_4": "$(L/6)(1 - cos\\theta)$", "correct_option": 3, "numerical_answer": null, "solution": "Since the centre of mass will not move\n2mx=my$\\Rightarrow$y=2x\nx + y = $L - L\\cos\\theta = L\\left( 1 - \\cos\\theta \\right)$\n$3x = L(1 - cos\\theta)$\n⇒ $x = \\frac{L(1 - cos\\theta)}{3}$", "solution_images": ["images/image3.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Center of mass at rest in the horizontal direction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-19-Q7", "question": "A charged particle of mass moving under\nthe influence of uniform electric field $E\\overset{\\rightarrow}{i}$and a\nuniform magnetic field $B\\overset{\\rightarrow}{k}$follows a trajectory\nfrom point P to Q as shown in figure. The velocitiesat P and Q are\nrespectively,\n$v\\overset{\\rightarrow}{i}$and$- 2v\\overset{\\rightarrow}{j}$. Then which\nof the following statements (A, B, C, D) are the correct? (Trajectory\nshown is schematic and not to scale)", "question_images": ["images/image10.jpeg"], "option_1": "$E = \\frac{3}{4}(\\frac{mv^{2}}{qa})$", "option_2": "Rate of work done by the electric field at P\nis$\\frac{3}{4}\\left( \\frac{{mv}^{3}}{a} \\right)$and at Q is zero.", "option_3": "The difference between the magnitude of angular momentum of the\nparticle at P and Q is 2mav", "option_4": ". All of the above", "correct_option": 4, "numerical_answer": null, "solution": "(A) by work energy theorem\n$0 + {qE}_{0}2a = \\frac{3}{2}{mv}^{2}$\n$$E = \n(B) Rate of work done at A= power of electric force\n$= {qE}_{0}V$\n$= \\frac{3}{4}\\frac{{mv}^{3}}{a}$\n(C) at$Q,\\frac{dW}{dt} = 0$for both forces\n(D)\n$\\Delta\\overset{\\rightarrow}{L} = ( - m2v2a\\overset{\\hat{}}{k}) - ( - mva\\overset{\\hat{}}{k})$\n$|\\Delta\\overset{\\rightarrow}{L}| = 3mva$", "solution_images": [], "subject": "Physics", "topic": "Magnetics", "subtopic": "Cross field", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-23-Q25", "question": "Consider a particle initially moving with a velocity 7 m/s\nstarts decelerating at a constant rate of$\\ 2m/s^{2}$.\nThe distance travelled in the fourth second is$\\frac{x}{10}m$. Find\nthe value of x.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "Here particle is decelerating so velocity of particle becomes\n after t=3.5 sec. the particle has a turning point at t=3.5 sec.\nV=U+at\n$0 = 7 - 2t$\n$t = 3.5\\sec$\n$y_{2} = 7 \\times 3.5 - \\frac{1}{2} \\times 2 \\times (3.5)^{2} = \\frac{49}{4}m$\n$y_{2} - x_{1} = \\frac{49}{4} - 12 = \\frac{1}{4}m$.\nDue to symmetry displacement of the particle at t=3 and t=4sec. is same.\nso total distance traveled in the fourth sec\nis$= \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}m$", "solution_images": ["images/image79.png"], "subject": "Physics", "topic": "Kinematics", "subtopic": "Uniform accelerated motion", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-23-Q17", "question": "The rod is uniform & weigh 500 N. Find magnitude of tension so\nthat tension in both strings (S_1, S_2) aresame (strings ideal).", "question_images": ["images/image40.png"], "option_1": "1000 N", "option_2": "1500 N", "option_3": "2000 N", "option_4": "500 N", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] Taking torque about W.\n- T_1 (0.4)+T_2 (0.3)+500(0.2)=0\nT = 1000", "solution_images": ["images/image41.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Torque", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "Ph-28-Q11", "question": "A Young's double slit experiment is conducted in water\n[IMAGE] as shown in the figure, and a glass plate of\nthickness t and refractive index [IMAGE] is placed in the\npath of[IMAGE]. The magnitude of the phase difference at\nO is[IMAGE]. Find n. (Assume that\n is the wavelength of light in air). O is\nsymmetrical w.r.t. [IMAGE] and [IMAGE].", "question_images": ["images/image100.png", "images/image101.png", "images/image102.png", "images/image103.png", "images/image104.png", "images/image105.png", "images/image102.png", "images/image106.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Here path difference will be:\n[IMAGE] TOPIC:Waves Optics\nSUB TOPIC:Optical path difference\nLEVEL:Moderate", "solution_images": ["images/image110.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-24-Q20", "question": "For a fluid which is flowing steadily, the level in the vertical\ntubes is best represented by", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "From continuity equation, velocity at cross-section (1) is more\nthan that at cross-section (2).\nHence $P_{1} < P_{2}$\nHence option (1) is correct.", "solution_images": [], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Buoyancy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-02-Q47", "question": "The P-V diagram of a cyclic thermodynamic process for an\nideal gas is as shown in figure. Its P-T curve will be", "question_images": ["images/image72.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": null, "solution": "A\nProcess A → B is isochoric process, therefore, options (C) and (D)\ncannot be correct. Again B → C is an adiabatic process, therefore, B → C\nshould be described by equati­on [IMAGE] constant.\nAs γ is always greater than", "solution_images": ["images/image77.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-08-Q3", "question": "$\\ $Figure below shows two paths that may be taken by a gas to go\nfrom a state A to a state C.\n[IMAGE] In process AB, 400J of heat is added to the system and in process BC,\n100J of heat is added to the system.The heat absorbed by the system in\nthe process AC will be", "question_images": ["images/image1.png"], "option_1": "500 J", "option_2": "460 J", "option_3": "300 J", "option_4": "380 J", "correct_option": 2, "numerical_answer": null, "solution": "For a complete cycle\n$Q_{cycle\\ } = W_{cycle\\ }$\n$+ 400 + 100 + Q_{C \\rightarrow A} = \\frac{1}{2}\\left( 2 \\times 10^{- 3} \\right)\\left( 4 \\times 10^{4} \\right)$\n$\\Rightarrow \\ \\ \\ \\ Q_{C \\rightarrow A} = - 460J$\n$\\Rightarrow \\ \\ \\ \\ Q_{A \\rightarrow C} = + 460J$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic cycle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-20-Q5", "question": "The length of a potentiometer wire is 120 cm and it carries a\ncurrent of 60 mA. For a cell of emf 5V and internal resistance of\n$20\\Omega$, the null point on it is found to be at 1000 cm. The\nresistance of whole wire is", "question_images": [], "option_1": "$60\\Omega$", "option_2": "$120\\Omega$", "option_3": "$100\\Omega$", "option_4": "$80\\Omega$", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Potential gradient\n$= \\frac{5}{1000} = \\frac{V_{p}}{1200}$\n$V_{P} = 6V$\nAnd $R_{P} = \\frac{V_{P}}{I} = \\frac{6}{60 \\times 10^{- 3}} = 100\\Omega$", "solution_images": ["images/image14.jpeg"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Potentiometer", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "Ph-27-Q22", "question": "In the figure shown cross-section area of holes are same and\nvery small in comparison to area of cross-section of water tank:\n(tanks are the open from the top)\ngq, gS)", "question_images": ["images/image112.png", "images/image112.png", "images/image112.png"], "option_1": "", "option_2": "", "option_3": "$x_{1}:x_{2}:x_{3} = \\sqrt{3}:\\sqrt{6}:\\sqrt{5}$", "option_4": "$x_{1}:x_{2}:x_{3} = \\sqrt{2}:\\sqrt{5}:\\sqrt{3}$", "correct_option": 1, "numerical_answer": null, "solution": "For case-1: fLFkfr-1 ds fy, [IMAGE] For case-2: fLFkfr-2 ds fy, [IMAGE] For case-3: fLFkfr-3", "solution_images": ["images/image115.png", "images/image116.png", "images/image117.png", "images/image118.png", "images/image119.png", "images/image120.png", "images/image121.png", "images/image122.png", "images/image123.png", "images/image124.png", "images/image125.png", "images/image126.png", "images/image127.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-24-Q18", "question": "The speed of water flowing out of orifice after the cylinder is\nkept inside it", "question_images": [], "option_1": "$\\ \\sqrt{3gH}$", "option_2": "$\\ \\sqrt{2gH}$", "option_3": "$\\ \\sqrt{\\frac{3gH}{2}}$", "option_4": "$\\ \\sqrt{\\frac{gH}{2}}$", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Let h′ be height of water column just after putting cylinder,\n$\\ \\rho h'\\left( A - \\frac{A}{3} \\right) = \\left( \\frac{H}{2} \\right)A\\rho$\n$\\ \\Rightarrow h^{'} = \\frac{3}{4}H$\n$\\ V' = \\sqrt{2{gh}^{'}} = \\sqrt{\\frac{3}{2}gH}$", "solution_images": ["images/image50.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Velocity of efflux", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-17-Q19", "question": "In the figure shown $C_{1} = 11\\ \\mu F$ and $C_{2} = 5\\ \\mu F$\nthen at steady state", "question_images": ["images/image20.png"], "option_1": "the potential difference across C_1 is 5V.", "option_2": "the potential difference across C_2 is 2V.", "option_3": "the potential difference$\\ V_{a} - V_{b} = - 4V$.", "option_4": "the potential difference between terminals of 15V battery is 9V.", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] at steady state\n$I(3) + 1(2) = 15$\n$I = 3$\nkvL\n$C \\rightarrow D \\rightarrow E \\rightarrow a \\rightarrow b \\rightarrow C$\n$V_{c} - I(3) + \\frac{q}{11} - 7 + \\frac{q}{5} = V_{c}$\n$\\frac{q}{11} + \\frac{q}{5} = 7 + 3 \\times 3 = 16$\n$q = 55\\mu c$\nNow kvL for $a \\rightarrow b$\n$v_{a} - 7 + \\frac{q}{5} = v_{b}$\n$\\Rightarrow V_{a} - V_{b} = 7 - \\frac{q}{5}$\n$= 7 - \\frac{55}{5}$\n$= - \\frac{20}{5} = - 4V$", "solution_images": ["images/image21.png"], "subject": "Physics", "topic": "Capacitance", "subtopic": "KVL", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "Ph-28-Q36", "question": "A solid sphere of mass 10 kg is placed over two smooth inclined\nplanes as shown in figure. The normal reactions at 2 is 10x Newton. Find", "question_images": ["images/image323.png", "images/image324.jpeg"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "[IMAGE] Solving above equation\n[IMAGE] =10 x\nx = 5\nTOPIC:Newton's first Law\nSUB TOPIC:Free body diagram", "solution_images": ["images/image325.jpeg", "images/image326.png", "images/image327.png", "images/image328.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-27-Q7", "question": "During an experiment an ideal gas is found to obey condition\n[IMAGE] constant. [IMAGE] density of gas).\nThe gas is initially at a temperature [IMAGE], pressure\n[IMAGE] and density [IMAGE]. The gas\nexpands such that densitychanges to [IMAGE] then which of\nfollowing is/are correct?,d iz;ksx", "question_images": ["images/image27.png", "images/image28.png", "images/image29.png", "images/image30.png", "images/image31.png", "images/image32.png", "images/image27.png", "images/image28.png", "images/image29.png", "images/image30.png", "images/image31.png", "images/image32.png"], "option_1": "The pressure of gas changes to [IMAGE]", "option_2": "The temperature of gas changes to [IMAGE].", "option_3": "The graph of above process on the P - T diagram is parabola.", "option_4": "The graph of above process on the P - T diagram is not rectangular\nhyperbola.", "correct_option": 2, "numerical_answer": null, "solution": "PT = constant", "solution_images": ["images/image35.png", "images/image36.png", "images/image37.png", "images/image38.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Ideal Gas", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-07-Q12", "question": "When a person throws a meter stick it is found that the centre\nof the stick is moving with speed 10 m/s and left end of stick with\nspeed 20m/s. Both points move vertically upwards at that moment. Then\nangular speed of the stick is", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these", "correct_option": 1, "numerical_answer": null, "solution": "Angular", "solution_images": ["images/image24.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Angular velocity of rigid body", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-10-Q6", "question": "A uniform disc of radius R lies in the x-y plane, with its centre\nat origin. Its moment of inertia about z-axis is equal to its moment of\ninertia about line y = x + c. The value of c will be", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "- R", "correct_option": 2, "numerical_answer": null, "solution": "$I_{PQR} = \\frac{1}{4}{MR}^{2} + M\\left( \\frac{C}{\\sqrt{2}} \\right)^{2}$.\nBut $I_{PQR} = \\frac{1}{2}{MR}^{2}$\n$C = \\pm \\frac{R}{\\sqrt{2}}$.\nHence is (B) is correct.", "solution_images": ["images/image30.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-13-Q5", "question": "The electric fields of two plane electromagnetic plane waves in\nvacuum are given by:\n${\\overset{\\rightarrow}{E}}_{1} = E_{0}\\overset{\\hat{}}{j}cos(\\omega t - kx)\\ and\\ {\\overset{\\rightarrow}{E}}_{2} = E_{0}\\overset{\\hat{}}{k}cos(\\omega t - ky)$\nAt t = 0, a particle of charge q is at origin with a velocity\nj$\\overset{\\rightarrow}{v} = 0.8c\\overset{\\hat{}}{j}$(c is the speed\nof the light in vacuum). The instantaneous force experienced by the\nparticle is", "question_images": [], "option_1": "$E_{0}q(0.8\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j} + 0.2\\overset{\\hat{}}{k})$", "option_2": "$E_{0}q( - 0.8\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j} + \\overset{\\hat{}}{k})$", "option_3": "$E_{0}q(0.4\\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} + 0.8\\overset{\\hat{}}{k})$", "option_4": "$E_{0}q(0.8\\overset{\\hat{}}{i} - \\overset{\\hat{}}{j} + 0.4\\overset{\\hat{}}{k})\\ $", "correct_option": 1, "numerical_answer": null, "solution": "Magnetic field vectors associated with this electromagnetic\nwave are given by\n${\\overset{\\rightarrow}{B}}_{1} = \\frac{E_{0}}{c}\\overset{\\hat{}}{k}cos(kx - \\omega t)\\&{\\overset{\\rightarrow}{B}}_{2} = \\frac{E_{0}}{c}\\overset{\\hat{}}{i}cos(ky - \\omega t)$\n$\\overset{\\rightarrow}{F} = q\\overset{\\rightarrow}{E} + q\\left( \\overset{\\rightarrow}{V} \\times \\overset{\\rightarrow}{B} \\right)$\n$= q\\left( {\\overset{\\rightarrow}{E}}_{1} + {\\overset{\\rightarrow}{E}}_{2} \\right) + q\\left( \\overset{\\rightarrow}{V} \\times \\left( {\\overset{\\rightarrow}{B}}_{1} + {\\overset{\\rightarrow}{B}}_{2} \\right) \\right)$\nBy putting the value of\n${\\overset{\\rightarrow}{E}}_{1},{\\overset{\\rightarrow}{E}}_{2},{\\overset{\\rightarrow}{B}}_{1}\\&{\\overset{\\rightarrow}{B}}_{2}$\nThe net Lorentz force on the charged particle is\n$\\overset{\\rightarrow}{F} = {qE}_{0}\\lbrack 0.8cos(kx - \\omega t)\\overset{\\hat{}}{i} + cos(kx - \\omega t)\\overset{\\hat{}}{j} + 0.2cos(ky - \\omega t)\\overset{\\hat{}}{k}\\rbrack$\nAt t = 0 and at x = y = 0\n$\\overset{\\rightarrow}{F} = qE_{0}\\lbrack 0.8\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j} + 0.2\\overset{\\hat{}}{k}\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "EM wave", "subtopic": "Lorentz force on a charged particle due to EM wave.", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-05-Q9", "question": "Two particle A and B are located in x-y plane at points (0, 0)\nand (0, 4 m). They simultaneously start moving with velocities.\n${\\overset{\\rightarrow}{v}}_{A} = 3\\overset{\\hat{}}{j}m/s$and\n${\\overset{\\rightarrow}{v}}_{B} = 4\\overset{\\hat{}}{i}m/s$. The shortest\ndistance between them will be", "question_images": ["images/image13.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "zero", "correct_option": 1, "numerical_answer": null, "solution": "Assuming B to be at rest, A will move with velocity\n${\\overset{\\rightarrow}{v}}_{AB}$ in the direction shown in figure. The\ndistance between them will first decrease from A to Cand then increase\nbeyond C.\n[IMAGE] Minimum distance between them is BC which is equal to\n$4\\ sin53^{\\circ} = 16/\\mathbf{5}\\mathbf{m}$", "solution_images": ["images/image14.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "Shortest distance", "difficulty": "Hard", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-15-Q2", "question": "Suppose the potential energy between electron and proton at a\ndistance r is given by [IMAGE] where k is constant, e is\ncharge of proton and r is radius of orbit. Application of Bohr's theory\nto hydrogen atom in this case shows that", "question_images": ["images/image10.png"], "option_1": "energy of electron in [IMAGE] orbit is proportional", "option_2": "energy of electron in [IMAGE] orbit is proportional\nto [IMAGE] mass of electron).", "option_3": "energy of electron in [IMAGE] orbit is proportional", "option_4": "energy of electron in [IMAGE] orbit is proportional", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] From (1) and (2)", "solution_images": ["images/image19.png", "images/image20.png", "images/image21.png", "images/image22.png", "images/image23.png", "images/image24.png", "images/image25.png", "images/image26.png"], "subject": "Physics", "topic": "Atomic Physics", "subtopic": "Bohr's model of hydrogen atom", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-11-Q4", "question": "The dependence of acceleration due to gravity g on the distance r\nfrom the centre of the earthassumed to be a sphere of radius R of\nuniformdensity is as shown in figure below:-\n[IMAGE] The correct figure is", "question_images": ["images/image38.png"], "option_1": "(i)", "option_2": "(ii)", "option_3": "(iii)", "option_4": "(iv)", "correct_option": 4, "numerical_answer": null, "solution": "The acceleration due to gravity at a depth d below the surface\nof earth is\nWhere R-d=r= distance of location from thecentre of the earth. When\n[IMAGE] From [IMAGE] till R=r, for which\n[IMAGE] For [IMAGE] or [IMAGE] Here, R+h=r Therefore, the variation of g with distance r from centre\nof earth will be as shown in figure(4). Thus, option (4) is correct", "solution_images": ["images/image39.png", "images/image40.png", "images/image41.png", "images/image42.png", "images/image43.png", "images/image44.png"], "subject": "Physics", "topic": "Collision", "subtopic": "Inelastic collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-06-Q20", "question": "A block of mass 1 kg initially at rest is dropped from a height\nh on the a spring of force constant k=50N/m. The maximum compression in\nthe spring is x=1 m then height hof the block is", "question_images": ["images/image28.png"], "option_1": "3 m", "option_2": "$\\frac{3}{2}$ m", "option_3": "$\\frac{5}{2}$ m", "option_4": "5 m", "correct_option": 2, "numerical_answer": null, "solution": "Change in gravitational potential energy\n= Elastic potential energy stored in-compressed spring\n$\\Rightarrow mg(h + x) = \\frac{1}{2}kx^{2}$\n$\\Rightarrow 1 \\times 10(h + 1) = \\frac{1}{2} \\times 50 \\times 1^{2}$\n$\\Rightarrow h = \\frac{3}{2}$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Conservation of mechanical energy", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-06-Q13", "question": "A block of mass 10 kg is to be pulled on a horizontal rough\nsurface (µ =0.75) the minimum magnitude of applied force is", "question_images": [], "option_1": "75 N", "option_2": "80 N", "option_3": "60 N", "option_4": "None of these", "correct_option": 3, "numerical_answer": null, "solution": "$Fcos\\theta - \\mu N = 0$\n$N = mg - {Fsin}\\theta\\ldots\\ldots(2)$\nFrom (1) and (2)\n${Fcos}\\theta = \\mu\\left( mg - {Fsin}\\theta \\right)$\n$K = \\frac{\\mu mg}{cos\\theta + \\mu sin\\theta}$\n$F_{\\min} = \\frac{\\mu mg}{\\sqrt{1 + \\mu^{2}}}\\left( \\because y = Asin\\theta + Bcos\\theta;ymax = \\sqrt{A^{2} + B^{2}} \\right)$\n$F_{\\min} = \\frac{0.75 \\times 10 \\times 10}{\\sqrt{1 + (0.75)^{2}}}$", "solution_images": ["images/image23.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Static friction", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-08-Q16", "question": "The internal energy of a system in adiabatic expansion", "question_images": [], "option_1": "Decreases", "option_2": "Increases", "option_3": "Remains unchanged", "option_4": "Nothing can be said", "correct_option": 2, "numerical_answer": null, "solution": "Adiabatic expansion is adiabatic cooling.\n$\\because\\ $temp. Decreases $\\Rightarrow$ internal energy decreases", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-05-Q1", "question": "Position of a particle moving on x-axis as a function of time t\nis given by\n$x = \\left( t^{3} - 6t^{2} + 3t + 4 \\right)$meter.\nWhat is the velocity of the particle when its acceleration is zero?", "question_images": [], "option_1": "$- 12m/s$", "option_2": "$- 6m/s$", "option_3": "$- 3m/s$", "option_4": "$- 9m/s$", "correct_option": 4, "numerical_answer": null, "solution": "$v = 3t^{2} - 12t + 3$\n$a = 6t - 12 = 0$\n$\\Rightarrow t = 2\\sec$\n$v(t = 2sec) = - 9m/s$", "solution_images": [], "subject": "Physics", "topic": "Motion in 1D", "subtopic": "Non-uniform acceleration", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-16-Q9", "question": "A body of mass 6 kg is acted upon by a force which causes a\ndisplacement in it given by $x = \\frac{t^{2}}{4}$ metre where t is the\ntime in second. The work done by the force is 2 seconds is", "question_images": [], "option_1": "12 J", "option_2": "9 J", "option_3": "6 J", "option_4": "3 J", "correct_option": 4, "numerical_answer": null, "solution": "The velocity of the body a time t is given by\n$= \\frac{dx}{dt} = \\frac{d}{dt}(\\frac{t^{2}}{4}) = \\frac{t}{2}$\nAt $t = 0, = u = 0\\ and\\ t = 2s, = 1{ms}^{- 1}$, Now, work done =\nincrease in KE\n$= \\frac{1}{2}m - \\frac{1}{2}mu^{2} = \\frac{1}{2}m - 0$\n$= \\frac{1}{2}m = \\frac{1}{2} \\times 6 \\times (1)^{2} = 3J$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work -energy theorem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "Ph-26-Q35", "question": "A Carnot engine operates between two reservoirs of temperatures\n900 K and 300 K. The engineperforms 1200 J of work per cycle. The heat\nenergy (in J) delivered by the engine to the lowtemperature reservoir,\nin a cycle, is 100n. Find n,d dkuksZ bUtu dks 900 K vkSj 300 K ds nks HkaMkjks a ds chp pyk;k\nfuEu rki okys HkaMkj esa izfr pØ 100n (J esa) NksM+rk gSFind\nn", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image151.png", "images/image152.png", "images/image153.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Carnot engine", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-28-Q20", "question": "A particle moving with initial velocity 3 m/s along x-axis\nfrom origin. Its acceleration is varying with x as\n[IMAGE] as shown in figure. At\n[IMAGE] tangent to the graph makes an angle 60° with\npositive x-axis as shown in diagram. Then at [IMAGE]", "question_images": ["images/image181.png", "images/image182.png", "images/image183.png", "images/image184.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "TOPIC: Rectilinear motion\nSUB TOPIC: Acceleration-displacement graph\nLEVEL: Moderate", "solution_images": ["images/image189.png", "images/image190.png", "images/image191.png", "images/image192.png", "images/image193.png", "images/image194.png", "images/image195.png", "images/image196.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-16-Q20", "question": "You are shown a photo of a car driven on the inside wall of a\nhuge vertical cylinder of radius 50 m. The coefficient of static\nfriction between the car tyres and the cylinder is $\\mu_{s} = 0.8$. The\nminimum speed in (m/s), at which the car can be driven like that is x.\nFind x.", "question_images": ["images/image37.png"], "option_1": "23", "option_2": "26", "option_3": "25", "option_4": "22", "correct_option": 3, "numerical_answer": null, "solution": "F. B. D. Of car:-\n[IMAGE] $\\mu N = mg$....(1)\n$N = \\frac{mv^{2}}{R}$....(2)\nFrom (1) and (2)\n$V = 25m/s$", "solution_images": ["images/image38.png"], "subject": "Physics", "topic": "Center of mass", "subtopic": "Centripetal force", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-09-Q3", "question": "Sinusoidal waves 5.00 cm in amplitude are to be transmitted along\na string having a linear mass density [IMAGE]. If the\nsource can deliver an average power of 90 W and the string is under a\ntension of 100 N, then the frequency at which the source can operate is\n(take [IMAGE] )", "question_images": ["images/image12.png", "images/image13.png"], "option_1": "45.3 Hz", "option_2": "50 Hz", "option_3": "30 Hz", "option_4": "62.3 Hz", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] using data f = 30 Hz", "solution_images": ["images/image14.png", "images/image15.png", "images/image16.png"], "subject": "Physics", "topic": "Wave on a string", "subtopic": "Power delivered to the string", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-06-Q5", "question": "A frictionless track ABCDE ends in a circular loop of radius R. A\nbody slides down the track from point A which is at height $h = 82cm$.\nThe maximum value of R for a body to complete the loop successfully is", "question_images": ["images/image8.png"], "option_1": "22.8 cm", "option_2": "32.8 cm", "option_3": "12.8 cm", "option_4": "42.8 cm", "correct_option": 2, "numerical_answer": null, "solution": "Using energy conservation:\n$\\frac{1}{2}mv_{B}^{2} = mgh$\n$V_{B} = \\sqrt{\\frac{2mgh}{m}}$\n$v_{B} = \\sqrt{2hg}\\ \\ldots(1)$\nAlso to complete vertical circle $v_{B} = \\sqrt{5gR}$...(2)\n$\\therefore R = \\frac{2}{5}h = 32.8cm$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-08-Q22", "question": "A ring shaped tube contains two ideal gases with equal masses\nand atomic mass numbers M_1­=32 and M_2=28. The gases are separated by\none fixed partition X and another movable conducting partition Y which\ncan move freely without friction inside the ring. The angle is $\\alpha$,\nin equilibrium as shown in thefigure (in degrees). Find the value\nof$\\frac{\\alpha}{48}$.", "question_images": ["images/image13.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "At equilibrium, pressure and temperature are same.\n$\\therefore P_{1} = P_{2}$ and $T_{1} = T_{2}$\nAlso, let cross-sectional area of tube be A and radius of ring be r.\n$\\therefore\\frac{V_{1}}{n_{1}} = \\frac{V_{2}}{n_{2}}$\n$\\Rightarrow \\frac{360 - \\alpha}{n_{1}} = \\frac{\\alpha}{n_{2}}$\n$\\left( \\therefore V_{1} = (360 - \\alpha)rA \\right.\\ $ and $V_{2} = \\alpha rA$\n$\\Rightarrow M_{1}(360 - \\alpha) = M_{2}\\alpha\\lbrack\\therefore n_{1} = \\frac{m}{M_{1}}andn_{2} = \\frac{m}{M_{2}}\\rbrack$\n$\\alpha = \\frac{360M_{1}}{M_{1} + M_{2}} = 192^{\\circ}$", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Gases in equilibrium", "difficulty": "Tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-16-Q18", "question": "A particle is moved under a force\n$\\overset{\\rightarrow}{F} = x\\overset{\\hat{}}{j}N$(x in meter) from O to\nA by two paths: path 1 is OBA andpath 2 is straight line OA. If w_1 and\nw_2 are work done by the force $\\overset{\\rightarrow}{F}$ along the two\npaths respectivelythe value of $\\frac{w_{1}}{w_{2}}$is", "question_images": ["images/image34.png"], "option_1": "2", "option_2": "5", "option_3": "3", "option_4": "1", "correct_option": 1, "numerical_answer": null, "solution": "work done in path 2\n$W_{2} = \\int\\overset{\\rightarrow}{F} \\cdot (dx\\overset{\\hat{}}{i} + dy\\overset{\\hat{}}{j}) = \\int x\\overset{\\hat{}}{j} \\cdot (dx\\overset{\\hat{}}{i} + dy\\overset{\\hat{}}{j})$as\nthe line: y = x\n$= \\int_{0}^{1}{xdx} = \\frac{1}{2}J$\nwork done in path 1\n$w_{OB} = 0$ (force is zero)\n$W_{1} = \\int\\overset{\\rightarrow}{F} \\cdot dx\\overset{\\hat{}}{i} + \\int\\overset{\\rightarrow}{F} \\cdot dy\\overset{\\hat{}}{j}$\n$= 0 + \\int_{y = 0}^{1}{(1\\overset{\\hat{}}{j}) \\cdot dy\\overset{\\hat{}}{j} = 1J}$\n$\\frac{w_{1}}{w_{2}} = 2$", "solution_images": ["images/image35.png"], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work done by variable force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "Ph-28-Q28", "question": "In an [IMAGE] the Kinetic energy of\n[IMAGE] is 48 MeV and Q-value of the reaction is 50 MeV.\nIf the mass number of the mother nucleus is 100N. Find N: (Assume that\ndaughter nucleus is in ground state)", "question_images": ["images/image251.png", "images/image252.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "We have\n[IMAGE] TOPIC:Modern physics\nSUB TOPIC:Alpha decay\nLEVEL: Moderate", "solution_images": ["images/image253.png", "images/image254.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-25-Q30", "question": "[IMAGE] gas follows the process,\n[IMAGE] constant. If then the ratio of work done by gas\nto change in itsinternal energy is [IMAGE] then n is:\nvuqikr [IMAGE] gks rks n", "question_images": ["images/image220.png", "images/image221.png", "images/image222.png", "images/image220.png", "images/image221.png", "images/image222.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "", "solution_images": ["images/image223.png", "images/image224.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Internal energy", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-09-Q5", "question": "The particle displacement (in cm) in a stationary wave is given\nby [IMAGE] Find the distance (in cm) between a node and\nthe next antinode.", "question_images": ["images/image21.png"], "option_1": "18.75 cm", "option_2": "36.5 cm", "option_3": "9.375 cm", "option_4": "73 cm", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] compare with [IMAGE] λ = 75 cm\n$\\frac{\\lambda}{4} = 18.75cm$", "solution_images": ["images/image22.png", "images/image23.png", "images/image24.png"], "subject": "Physics", "topic": "Wave on a string", "subtopic": "Speed wave", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-12-Q22", "question": "Four resistances of [IMAGE] and\n[IMAGE] respectively in cyclic order to form Wheat\nstone's network. The resistance (in ohms) that is to be connected in\nparallel with the resistance of [IMAGE] to balance the\nnetwork is2n. Find n.", "question_images": ["images/image30.png", "images/image31.png", "images/image32.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "on solving", "solution_images": ["images/image33.png", "images/image34.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Wheatstone bridge", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "Ph-27-Q37", "question": "A non-isotropic solid metal cube has coefficients of linear\nexpansion as:$5 \\times \\ 10^{–5}\\ /{^\\circ}C$ along the x-axis and\n$5 \\times 10^{–6}\\ /{^\\circ}C$ along the y and the z-axis. If the\ncoefficient of volume expansion of the solid is\n$C \\times 10^{–5}\\ /{^\\circ}C$ then the value of C is.........\nizdkj $5 \\times \\ 10^{–5}\\ /{^\\circ}C$ x- rFkk\nizlkj xq.kkad $C \\times 10^{–5}\\ /{^\\circ}C$ gks] rks C", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "$V = 2\\alpha_{2} + \\alpha_{1}$\n$= 10 \\times 10^{- 6} + 5 \\times 10^{- 5}$\n$= 60 \\times 10^{- 6}/\\ ^{\\circ}C$", "solution_images": [], "subject": "Physics", "topic": "Thermal properties of matter", "subtopic": "Thermal expansion", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-19-Q22", "question": "Both the diodes used in the circuit shown are assumed to be\nideal and have negligible resistance whenthese are forward biased. built\nin potential in each diode is 0.7 V. For the input voltages shown in\nthefigure, the voltage (in volts) at point A is 6n. Find", "question_images": ["images/image29.jpeg"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "Right diode is reversed biased and left diode is forward biased\n$\\therefore V_{E} = 12.7 - 0.7 = 12Volt$", "solution_images": ["images/image30.jpeg"], "subject": "Physics", "topic": "Semiconductors", "subtopic": "Diodes in circuit", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-21-Q8", "question": "Which of the following gives a reversible operation ?", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "A logic gate is reversible if we can recover input data from\nthe outputeg. NOT gate,d rdZ gksxk ladsr\nizkIr fd;k tk ldsA\neg. NOT }kj", "solution_images": [], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Gates", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "Ph-27-Q13", "question": "The speed of water flowing out of orifice after the cylinder is\nkept inside it\ncsyu dks VSd ckn fNnz", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Let h' be height of water column just after putting cylinder,\nekuk csyu", "solution_images": ["images/image65.png", "images/image66.png", "images/image67.png", "images/image68.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Velocity of efflux", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-08-Q24", "question": "An engine has an efficiency of 0.25 when temperature of sink is\nreduced by58ºC, if ts efficiency is doubled, then the temperature of the\nsource in °C is 29n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "Here,$\\eta_{1} = 1 - \\frac{T_{2}}{T_{1}}$\nor $\\ 0.25 = 1 - \\frac{T_{2}}{T_{1}} \\Rightarrow \\frac{1}{4} = 1 - \\frac{T_{2}}{T_{1}}$\n$\\Rightarrow \\frac{T_{2}}{T_{1}} = 1 - \\frac{1}{4} = \\frac{3}{4}$.....(i)\nAccording to question,\n$\\eta_{2} = 2\\eta_{1},$ and $T_{2} = T_{2} - 58^{\\circ}C$\n$\\therefore 2 \\times \\frac{1}{4} = 1 - \\frac{\\left( T_{2} - 58^{\\circ}C \\right)}{T_{1}}$\n$\\Rightarrow 1 - \\frac{1}{2} = \\frac{T_{2} - 58^{\\circ}C}{T_{1}}$\n$\\frac{1}{2} = \\frac{T_{2}}{T_{1}} - \\frac{58^{\\circ}}{T_{1}} \\Rightarrow \\frac{3}{4} - \\frac{1}{2} = \\frac{58}{T_{1}}$\n$\\Rightarrow T_{1} = 232^{\\circ}C$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Carnot engine", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-06-Q23", "question": "A uniform sphere of weight w and radius 3 m is being held by a\nstring of length 2 m. attached to a frictionless wall as shown in the\nfigure. The tension in the string is $\\frac{5w}{n}$ then find the value\nof n.", "question_images": ["images/image32.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "[IMAGE] From geometry:\n$\\cos\\theta = \\frac{3}{5}$\n$\\sin\\theta = \\frac{4}{5}$\nAs sphere is at equilibrium\n$T\\sin\\theta = w$\n$T\\left( \\frac{4}{5} \\right) = w$\n$T = \\frac{5w}{4}$", "solution_images": ["images/image33.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Equilibrium of forces", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-17-Q5", "question": "A conducting circular loop is pulled with a constant velocity v\ntowards a region of uniform magnetic field of induction B as shown in\nfigure then current induced in the loop is (d > 2r).", "question_images": ["images/image6.png"], "option_1": "clockwise while entering.", "option_2": "anti clockwise while entering.", "option_3": "zero when completely inside.", "option_4": "clockwise while leaving.", "correct_option": 2, "numerical_answer": null, "solution": "By lenz's law induced current will oppose change in flux.", "solution_images": [], "subject": "Physics", "topic": "EMI", "subtopic": "Lenz's law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-16-Q11", "question": "The particle of mass 1 kg is acted upon by a\nforce$\\overset{\\rightarrow}{F} = F_{x}\\overset{\\hat{}}{i} + F_{y}\\overset{\\hat{}}{j}$\nfrom t = 0 to t = 4 sec. Find the velocityof the particle at t=4 sec.,\nif its initial velocity is\n$\\overset{\\rightarrow}{u} = - \\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}$and\nthe variation of applied force with time is as shown in the graphs", "question_images": ["images/image21.png"], "option_1": "$\\overset{\\rightarrow}{v} = - \\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}$", "option_2": "$\\overset{\\rightarrow}{v} = - \\overset{\\hat{}}{i} + 11\\overset{\\hat{}}{j}$", "option_3": "$\\overset{\\rightarrow}{v} = 10\\overset{\\hat{}}{i}$", "option_4": "$\\overset{\\rightarrow}{v} = 2\\overset{\\hat{}}{i} + 10\\overset{\\hat{}}{j}$", "correct_option": 2, "numerical_answer": null, "solution": "$\\int_{0}^{4}{F_{x}dt = \\Delta P_{x} = m(v_{x} - u_{x})}$\n$\\Rightarrow 0 = m(v_{x} - u_{x})$.\n$\\Rightarrow V_{x} = u_{x} = - 1$\n$\\int_{0}^{4}{F_{y}dt = \\Delta P_{y} = m(v_{y} - u_{y})}$\n$\\Rightarrow \\frac{1}{2} \\times 4 \\times 5 = 1(v_{y} - 1) \\Rightarrow v_{y} = 11$\n$\\overset{\\rightarrow}{v} = - \\overset{\\hat{}}{i} + 11\\overset{\\hat{}}{j}$", "solution_images": [], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Impulse", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-03-Q8", "question": "A projectile is projected at an angle of 37° to the horizontal. When\nit is projected at 53°, its time of flight on level ground would have\nbeen 1 sec. more. At what angle should it be projected so that its range\nis 10 m less than either of the situations ?", "question_images": [], "option_1": "26.5°", "option_2": "30°", "option_3": "18°", "option_4": "15°\n 37° 53° \n 1 \n 10 m ?", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] T1 =[IMAGE] (sin 53° - sin 37°) = 1\nR =[IMAGE] = 60 m\nR' = 50 =[IMAGE] = sin 2θ\n2θ = 53°\nTOPIC:KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png", "images/image48.png", "images/image49.png", "images/image50.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-14-Q23", "question": "Light of wavelength 600 mm is incident upon a single slit with\nwidth[IMAGE]. The figure shows the pattern observed on a\nscreen positioned 2 m from the slits. Determine the distance s (in cm).", "question_images": ["images/image152.png", "images/image153.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image154.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Diffraction", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-12-Q12", "question": "Effective capacitance of parallel combination of two capacitors\nC1 and C2 is 10$\\mu F$. When these capacitorsare individually connected\nto a voltage source of 1V, the energy stored in the capacitor C_2 is 4\ntimes thatof C_1. If these capacitors are connected in series, their\neffective capacitance will be", "question_images": [], "option_1": "$4.2\\mu F$", "option_2": "$8.4\\mu F$", "option_3": "$3.2\\mu F$", "option_4": "$1.6\\mu F$", "correct_option": 4, "numerical_answer": null, "solution": "Given C_1 + C_2$= 10\\mu F$....(i)\n$4\\left( \\frac{1}{2}C_{1}V^{2} \\right) = \\frac{1}{2}C_{2}V^{2}$\n$\\Rightarrow$ 4C_1=C_2....(ii)\nfrom equation (i) and (ii)\n$C_{1} = 2\\mu F$\n$C_{2} = 8\\mu F$\nIf they are in series\n$C_{eq} = \\frac{C_{1}C_{2}}{C_{1} + C_{2}} = 1.6\\mu F$", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Equivalent capacitance", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-12-Q18", "question": "[IMAGE] A parallel plate capacitor has plates of area A separated by distance\n between them. It is filled with adielectric which has a\ndielectric constant that varies as k(x) = K(1 + $\\alpha$x) where \nis the distancemeasured from one of the plates. If ($\\alpha$d) << 1,\nthe total capacitance of the system is best given by theexpression", "question_images": ["images/image22.png"], "option_1": "$\\frac{A \\in_{0}K}{d}\\left( 1 + \\frac{\\alpha^{2}d^{2}}{2} \\right)b$", "option_2": "$\\frac{AK \\in_{0}}{d}\\left( 1 + \\frac{\\alpha d}{2} \\right)$", "option_3": "$\\frac{A \\in_{0}K}{d}\\left( 1 + \\left( \\frac{\\alpha d}{2} \\right)^{2} \\right)$", "option_4": "$\\frac{AK\\epsilon_{0}}{d}(1 + \\alpha d)$", "correct_option": 2, "numerical_answer": null, "solution": "Capacitance of element$= \\frac{k\\varepsilon_{0}A}{dx}$\n[IMAGE] Capacitance of\nelement$C^{'} = \\frac{K(1 + \\alpha x)\\varepsilon_{0}A}{dx}$\n$\\sum\\frac{1}{C^{'}} = \\int_{0}^{d}\\mspace{2mu}\\frac{dx}{K\\varepsilon_{0}A(1 + \\alpha x)}$\n$\\frac{1}{C} = \\frac{1}{K\\varepsilon_{0}A\\alpha}\\mathcal{l}n(1 + \\alpha d)$\n Given $- \\ \\alpha d < < 1$\n$\\frac{1}{C} = \\frac{1}{K\\varepsilon_{0}A\\alpha}\\left( \\alpha d - \\frac{\\alpha^{2}d^{2}}{2} \\right)$\n$\\frac{1}{C} = \\frac{d}{K\\varepsilon_{0}A}\\left( 1 - \\frac{\\alpha d}{2} \\right)$\n$C = \\frac{K\\varepsilon_{0}A}{d}\\left( 1 + \\frac{\\alpha d}{2} \\right)$", "solution_images": ["images/image23.png"], "subject": "Physics", "topic": "Capacitance", "subtopic": "Equivalent capacitance", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-18-Q22", "question": "There are two vectors\n$\\overset{\\rightarrow}{OA}\\&\\overset{\\rightarrow}{OB}$. Vector\n$\\overset{\\rightarrow}{OA}$ is inxy plane and of magnitude 5 units and\nvector $\\overset{\\rightarrow}{OB}$ isx-z plane and of magnitude\n$\\sqrt{2}$units as shown. Theangle between the vectors\n$\\overset{\\rightarrow}{OA}\\&\\overset{\\rightarrow}{OB}$is$\\theta = \\cos^{- 1}(\\frac{3}{n\\sqrt{2}})$.\nFind n.", "question_images": ["images/image29.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": null, "solution": "$\\overset{\\rightarrow}{OA} = 5cos53^{\\circ}\\overset{\\hat{}}{i} + 5sin53^{\\circ}\\overset{\\hat{}}{j} = 3\\overset{\\hat{}}{i} + 4\\overset{\\hat{}}{j}$\n$\\overset{\\rightarrow}{OB} = \\sqrt{2}cos45^{\\circ}\\overset{\\hat{}}{i} + \\sqrt{2}sin45^{\\circ}\\overset{\\hat{}}{k} = 1\\overset{\\hat{}}{i} + 1\\overset{\\hat{}}{k}$\n$cos\\theta = \\frac{\\overset{\\rightarrow}{OA} \\cdot \\overset{\\rightarrow}{OB}}{|\\overset{\\rightarrow}{OA}||\\overset{\\rightarrow}{OB}|} = \\frac{3 + 0}{5 \\cdot \\sqrt{2}} = \\frac{3}{5\\sqrt{2}}$\n$\\Rightarrow \\theta = \\cos^{- 1}(\\frac{3}{5\\sqrt{2}})$", "solution_images": [], "subject": "Physics", "topic": "Vector", "subtopic": "Angle between vectors", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-16-Q13", "question": "An object A is dropped from rest from the top of a 30 m high\nbuilding and at the same moment another object B is projected vertically\nupwards with an initial speed of 15 m/s from the base of the building.\nMass of the object A is 2 kg while mass of the object B is 4kg. The\nmaximum height above the ground level attained by the centre of mass of\nthe A and B system is (take$g = 10m/s^{2}$)", "question_images": [], "option_1": "15 m", "option_2": "25 m", "option_3": "30 m", "option_4": "35 m", "correct_option": 1, "numerical_answer": null, "solution": "Using\n$\\frac{m_{1}{\\overset{\\rightarrow}{x}}_{1} + m_{2}{\\overset{\\rightarrow}{x}}_{2}}{m_{1} + m_{2}} = X_{cm}$\nBy putting values $X_{cm} = 10\\ m$from base.\n[IMAGE] Initial velocity of\nCOM$\\Rightarrow \\frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}} = 10m/supward$\nAlso, acceleration of COM is g thus using\n$v^{2} = u^{2} + 2$as for COM\n$0 = 10^{2} + 2 \\times ( - 10) \\times h_{\\max}$\n$h_{\\max} = 5m$from the initial position of COM\nHence total height from base$= 10 + 5m = 15m$", "solution_images": ["images/image23.png"], "subject": "Physics", "topic": "Center of mass", "subtopic": "Motion of center of mass", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-21-Q7", "question": "In the given circuit, value of Y is:\n ifjiFk", "question_images": ["images/image12.png"], "option_1": "Toggles between 0 and 1 0", "option_2": "0", "option_3": "will not execute ifjiFk dk;kZ fuor ugha gksxk", "option_4": "1", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image13.png", "images/image14.png", "images/image15.png", "images/image16.png", "images/image17.png"], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Gates", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-22-Q9", "question": "Consider a sphere of radius R which carries a uniform charge\ndensity$\\rho$. If a sphere of radius$\\frac{R}{2}$iscarved out of it, as\nshown, the\nratio$\\left| \\frac{{\\overset{\\rightarrow}{E}}_{A}}{{\\overset{\\rightarrow}{E}}_{B}} \\right|$\nof magnitude of electric field of magnitude of electric field\n${\\overset{\\rightarrow}{E}}_{A}\\ $and ${\\overset{\\rightarrow}{E}}_{B}$,\nrespectively, atpoints A and B due to the remaining portion is", "question_images": ["images/image28.png"], "option_1": "$\\frac{9}{17}$", "option_2": "$\\frac{21}{34}$", "option_3": "$\\frac{17}{54}$", "option_4": "$\\frac{18}{54}$", "correct_option": 1, "numerical_answer": null, "solution": "For a solid sphere\n$E = \\frac{\\rho r}{3\\varepsilon_{0}}$\n$E_{A} = \\frac{- \\rho R}{2\\left( 3\\varepsilon_{0} \\right)}$\n$\\left| E_{A} \\right| = \\frac{\\rho R}{6\\varepsilon_{0}}$\nElectric field at point B = E_B = E_1A + E_2A\nE_1A = Electric Field Due to solid sphere of radius R at point\n$B = \\frac{\\rho R}{3\\varepsilon_{0}}$\nE_2A = Electric Field Due to solid sphere of radius R/2 (which having\ncharge density -$\\rho$)$= - \\frac{\\rho R}{54\\varepsilon_{0}}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field by uniformly charged sphere", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-11-Q6", "question": "A uniform sphere of mass 500 g rolls without slipping on a plane\nhorizontal surface with its centre moving at a speed\nof[IMAGE]. Its kinetic energy is", "question_images": ["images/image51.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "K.E. of the sphere = Translational K.E + Rotational K.E.\n[IMAGE] K = Radius of gyration", "solution_images": ["images/image57.png", "images/image58.png", "images/image59.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Kinetic energy in combined motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-24-Q12", "question": "A small ball of mass m is released from rest from a height h\nabove the liquid surface. The density of ball is $\\frac{\\rho}{3}$:\n(assume no splash)", "question_images": ["images/image41.png"], "option_1": "If $h \\geq H$, then the ball will be able to enter the liquid of\nhigher density", "option_2": "If $h \\geq \\frac{7H}{2}$, then the ball will be able to strike the\nbottom of the vessel", "option_3": "If h = 2H, then the ball will cross the interface of two liquids\nwith kinetic energy = mgH and If h = 0 then ball will float on the\nsurface.", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$mg\\left( h + \\frac{H}{2} \\right) = \\lbrack v\\rho g\\rbrack\\frac{H}{2}$\nh= H\n$mg\\left( h + \\frac{H}{2} + \\frac{H}{2} \\right) = \\lbrack v\\rho g\\rbrack\\frac{H}{2} + \\lbrack v2\\rho g\\rbrack\\frac{H}{2}$\n$mg\\left( \\frac{5H}{2} \\right) - \\frac{3}{2}mgH = K.E$\nK.E. = mgH.", "solution_images": ["images/image42.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Buoyant force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-16-Q16", "question": "For the situation shown in figure below, there is no friction\nbetween pulley and string. For which of the following values of mass of\nblock A, the system will stay at rest $(g = 10m/s^{2})$", "question_images": ["images/image29.png"], "option_1": "4 Kg", "option_2": "8 Kg", "option_3": "16 Kg", "option_4": "30 Kg", "correct_option": 4, "numerical_answer": null, "solution": "Tension in string =100 N in equilibrium F.B.D. of A\n$F_{s} = 80 \\leq \\mu N$\n$80 \\leq \\frac{1}{2}(mg - 60)$\n$mg \\geq 220$\n$m \\geq 22$", "solution_images": ["images/image30.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-15-Q22", "question": "When a metallic surface is illuminated with monochromatic light\nof wavelength [IMAGE] the stopping potential is\n[IMAGE] When the same surface is illuminated with light\nof wavelength [IMAGE] the stopping potential is\n[IMAGE] Then the work function of the metallic surface\nis $\\frac{hc}{x\\lambda},$ where value of x", "question_images": ["images/image165.png", "images/image166.png", "images/image167.png", "images/image168.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image169.png", "images/image170.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "Einstein's photoelectric equation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-05-Q20", "question": "The pressure on a square plate is measured by measuring the\nforce on the plate and the lengthof the sides of plate. If the maximum\nerror in themeasurement of force and length are respectively 4 and 2 then what is the maximum percentage error in themeasurement of pressure?", "question_images": [], "option_1": "1", "option_2": "2", "option_3": "6", "option_4": "8", "correct_option": 4, "numerical_answer": null, "solution": "$P = \\frac{F}{A};\\ \\ \\lbrack P\\rbrack = \\left\\lbrack \\frac{F}{A} \\right\\rbrack = \\left\\lbrack \\frac{{MLT}^{- 2}}{L^{2}} \\right\\rbrack = \\left\\lbrack {ML}^{- 1}T^{- 2} \\right\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "Error analysis", "subtopic": "", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-19-Q23", "question": "The first member of the Balmer series of hydrogen atom has a\nwavelength of The wavelength of the second member of the Balmer\nseries (in nm) is 81n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "$\\frac{1}{\\lambda} = RZ^{2}(\\frac{1}{n_{1}^{2}} - \\frac{1}{n_{2}^{2}})$\n$\\frac{1}{\\lambda_{1}} = R(1)^{2}(\\frac{1}{2^{2}} - \\frac{1}{3^{2}}) = \\frac{5R}{36}$\n$\\frac{1}{\\lambda_{2}} = R(1)^{2}(\\frac{1}{2^{2}} - \\frac{1}{4^{2}}) = \\frac{3R}{16}$\n$\\frac{\\lambda_{2}}{\\lambda_{1}} = \\frac{20}{27}$\n$\\lambda_{2} = \\frac{20}{27} \\times 6561A = 4860Å$\n$= 486nm$", "solution_images": [], "subject": "Physics", "topic": "Modern physics", "subtopic": "Balmer spectral series", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-08-Q12", "question": "V - T diagram for a process of a given mass of ideal gas is as\nshown in the figure. During the process pressure of gas.", "question_images": ["images/image5.png"], "option_1": "First increases then decreases", "option_2": "Continuously decreases", "option_3": "Continuously increases", "option_4": "First decreases then increases.", "correct_option": 2, "numerical_answer": null, "solution": "$V = KT + C$\n$P = \\frac{nRT}{V}$\n$\\Rightarrow P = \\frac{nRT}{KT + C}$\n$\\frac{dP}{dT} = \\frac{nRC}{(KT + C)^{2}}$\nAs C < 0 by diagram\n$\\Rightarrow \\frac{dP}{dT} < 0$ for all T\n$\\Rightarrow$ P continuously decreases", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "V-T diagram", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-12-Q10", "question": "Consider a sphere of radius R which carries a uniform charge\ndensity$\\rho$. If a sphere of radius$\\frac{R}{2}$iscarved out of it, as\nshown, the\nratio$\\left| \\frac{{\\overset{\\rightarrow}{E}}_{A}}{{\\overset{\\rightarrow}{E}}_{B}} \\right|$\nof magnitude of electric field of magnitude of electric field\n${\\overset{\\rightarrow}{E}}_{A}$ and ${\\overset{\\rightarrow}{E}}_{B}$,\nrespectively, atpoints A and B due to the remaining portion is", "question_images": ["images/image13.png"], "option_1": "$\\frac{18}{34}$", "option_2": "$\\frac{21}{34}$", "option_3": "$\\frac{17}{54}$", "option_4": "$\\frac{18}{54}$", "correct_option": 1, "numerical_answer": null, "solution": "For a solid sphere\n$E = \\frac{\\rho r}{3\\varepsilon_{0}}$\n$E_{A} = \\frac{- \\rho R}{2\\left( 3\\varepsilon_{0} \\right)}$\n$\\left| E_{A} \\right| = \\frac{\\rho R}{6\\varepsilon_{0}}$\nElectric field at point B = E_B = E_1A + E_2A\nE_1A = Electric Field Due to solid sphere of radius R at point\n$B = \\frac{\\rho R}{3\\varepsilon_{0}}$\nE_2A = Electric Field Due to solid sphere of radius R/2 (which having\ncharge density -$\\rho$)\n$= - \\frac{{KQ}^{'} \\times 4}{{OR}^{2}} = - \\frac{\\rho R}{54\\varepsilon_{0}}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field by uniformly charged sphere", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-11-Q1", "question": "A uniformly thick wheel with moment of inertial I and radius R is\nfree to rotate about its centre of mass (see figure). A massless string\nis wrapped over its rim and two blocks of masses [IMAGE] and [IMAGE] are attached to the ends of the string. The\nsystem is released from rest. The angular speed of the wheel\nwhen[IMAGE] descents by a distance his", "question_images": ["images/image1.png", "images/image2.png", "images/image3.png", "images/image4.jpeg"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image9.png", "images/image10.png", "images/image11.png", "images/image12.png", "images/image13.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "Ph-28-Q22", "question": "The following figure depicts a wave travelling in a medium.\nWhich pair of particles are in phase.", "question_images": ["images/image212.png"], "option_1": "A and G", "option_2": "B and F", "option_3": "C and E", "option_4": "B and F", "correct_option": 1, "numerical_answer": null, "solution": "Point is moving down though\nthey are at the same distance from equilibrium position at the instant\nshown. Therefore B and F are not in phase.\nBy same argument, A and G are in phase.\nTOPIC:Waves in a string\nSUB TOPIC: Particles in phase", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-17-Q25", "question": "The radius of a coil of wire with 21 turns is 0.22 m, and i_2\ncurrent flows clockwise in the coil as shown. A long straight wire\ncarrying a current is toward the left is located 0.05 m from the edge of\nthe coil. The magnetic field at the centre of the coil is zero tesla.\nThe ratio $\\frac{i_{1}}{4i_{2}}$ is (Use$\\pi = 22/7$)", "question_images": ["images/image29.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "20.25", "solution": "$\\frac{\\mu_{0} \\cdot i_{1}}{2\\pi(d + r)} = \\frac{N \\times \\mu_{0}i_{2}}{2r}$\n$\\Rightarrow N = \\frac{i_{1}}{i_{2}} \\cdot \\frac{r}{\\pi(d + r)}$\n$\\frac{i_{1}}{i_{2}} = 81$", "solution_images": [], "subject": "Physics", "topic": "Moving charges and", "subtopic": "Magnetic field due to current carrying conducting", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-07-Q16", "question": "A railway flat car has an artillery gun installed on it and\ncontains shells for fring The combined system excluding shell has a mass\nM and moves with a velocity $v_{0}$. The barrel of the gun makes an\nangle $\\alpha$ with the horizontal. A shell of mass m leaves the barrel\nat a speed relative to initial state of barrel in the forward\ndirection. The speed of the flat car so that it may stop after the\nfiring is: (neglect frinstion)", "question_images": [], "option_1": "$\\ \\frac{mu}{M + m}$", "option_2": "$(\\frac{Mu}{M + m})cos\\alpha$", "option_3": "$(\\frac{mu}{M})cos\\alpha$", "option_4": "$\\ (M + m)ucos\\alpha$", "correct_option": 3, "numerical_answer": null, "solution": "u - velocity of shell with respect to gun.\nIn horizontal direction ${\\overset{\\rightarrow}{F}}_{ext} = 0$, so\ncentre of mass move with constant velocity in horizontal plane.\n$(M + m)v_{0} = m(v_{0} + ucos\\alpha)$\n$V_{0} = \\frac{mucos\\alpha}{M}$", "solution_images": [], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Conservation of momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-09-Q12", "question": "Sound of frequency 1000 Hz from a stationary source is reflected\nfrom an object approaching the source at 30 m/s, back to a stationary\nobserver located at the source. The speed of sound in air is 330 m/s.\nThe frequency of the sound heard by the observer is", "question_images": [], "option_1": "1200Hz", "option_2": "1200Hz", "option_3": "1090Hz", "option_4": "1100Hz", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image68.png", "images/image69.png", "images/image70.png", "images/image71.png", "images/image72.png"], "subject": "Physics", "topic": "Sound wave", "subtopic": "Doppler effect", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-23-Q15", "question": "A solid sphere of mass m and radius r is gently placed on a\nconveyer belt moving with constant velocityV. If the coefficient of\nfriction between the belt and sphere is$\\frac{2}{7}$, the distance\ntravelled by the centre of the sphere before it starts pure rolling is.", "question_images": ["images/image36.png"], "option_1": "$\\frac{v^{2}}{7g}$", "option_2": "$\\frac{2V^{2}}{49g}$", "option_3": "$\\frac{2V^{2}}{5g}$", "option_4": "$\\frac{2V^{2}}{7g}$", "correct_option": 1, "numerical_answer": null, "solution": "Initial situation w.r.t. the conveyer belt:\n$\\mu mgR = \\frac{2}{5}{mR}^{2} \\cdot \\alpha$ $\\Rightarrow$\n$\\alpha = \\frac{5}{7} \\cdot \\frac{g}{R}$\nAt pure rolling:\n$v^{'} = v - (\\mu g)t$and $\\omega^{'} = 0 + \\left( \\frac{5}{7} \\cdot \\frac{g}{R} \\right)t$\n(Since at pure rolling)\n$\\Rightarrow$ $v - \\mu gt = \\frac{5}{7} \\cdot g \\cdot t$\n$\\Rightarrow$ $t = \\frac{v}{g}\\left( \\because\\mu = \\frac{2}{7} \\right)$\nFor centre of the sphere:$S = v\\left( \\frac{v}{g} \\right) - \\frac{1}{2} \\cdot \\left( \\frac{2}{7} \\cdot g \\right)\\left( \\frac{v^{2}}{g^{2}} \\right)$\n$S = \\frac{v^{2}}{g} - \\frac{v^{2}}{7g}$ (w.r.t. belt)\nIn this time, the belt with speed will move a distance\nof$s_{b} = vt = v\\left( \\frac{v}{g} \\right) = \\frac{v^{2}}{g}$\n[IMAGE] Distance travelled w.r.t. ground will be:$S_{g} = \\left| S - S_{b} \\right| = \\left| \\left( \\frac{v^{2}}{g} - \\frac{v^{2}}{7g} \\right) - \\left( \\frac{v^{2}}{g} \\right) \\right| = \\frac{v^{2}}{7g}$", "solution_images": ["images/image37.png"], "subject": "Physics", "topic": "Rotational motion", "subtopic": "Rolling with sliding", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "Ph-26-Q22", "question": "A leak proof cylinder of length 1m, made of a metal which has\nvery low coefficient of expansion is floating vertically in water at 0ºC\nsuch that its height above the water surface is 20 cm. When the\ntemperature of water is increased to 4ºC, the height of the cylinder\nabove the water surface becomes 21 cm. The density of water at T = 4ºC,\nrelative to the density at T = 0ºC is close to:,d yhd iwQ 1m yEck csyukdkj crZu,d,slh cuk gqvk gS ftldk\n4ºC rd c<+k rks blds ckgj jgus okys Hkkx", "question_images": [], "option_1": "1.03", "option_2": "1.04", "option_3": "1.26", "option_4": "1.01", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image76.png", "images/image77.png", "images/image78.png", "images/image79.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Buoyancy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-19-Q18", "question": "In a double-slit experiment, at a certain point on the screen\nthe path difference between the two interfering waves is\n$\\frac{1}{8}$thof a wavelength. The ratio of the intensity of light at\nthat point to that at the centre of a bright fringe is", "question_images": [], "option_1": "0.568", "option_2": "0.760", "option_3": "0.853", "option_4": "0.672", "correct_option": 3, "numerical_answer": null, "solution": "$I = I_{0}\\cos^{2}(\\frac{\\Delta\\phi}{2})$\n$\\frac{I}{I_{0}} = \\cos^{2}\\lbrack\\frac{\\frac{2\\pi}{\\lambda} \\times \\Delta x}{2}\\rbrack = \\cos^{2}(\\frac{\\pi}{8})$;\n$\\frac{I}{I_{0}} = 0.853$", "solution_images": [], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-13-Q15", "question": "The coercivity of a small magnet where the ferromagnet gets\ndemagnetized is 3 × 10^3 Am^-1. The current required to be passed in\na solenoid of length 10 cm and number of turns 100, so that the magnet\ngets demagnetized when inside the solenoid, is", "question_images": [], "option_1": "30 mA", "option_2": "60 mA", "option_3": "3 A", "option_4": "6 A", "correct_option": 3, "numerical_answer": null, "solution": "For solenoid\n$\\frac{B}{\\mu_{0}} = H$ or $B = \\mu_{0}nI$\n$H = nI$\n$\\Rightarrow 3 \\times 10^{3} = \\frac{100}{0.1} \\times I$\nΙ = 3A", "solution_images": [], "subject": "Physics", "topic": "Magnetism", "subtopic": "Coercivity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "Ph-25-Q29", "question": "Figure shows the variation of internal energy \"U\" with the\ndensity \"$\\rho$ \" of one mole of ideal diatomic gas. Process BA is a\npart of rectangular hyperbola. If the work done by gas in the process BA\nis 7n joules. Find n ?\n &ijek.kqd xSl ds,d fLFkfrt \n ?kuRo [IMAGE]", "question_images": ["images/image209.png", "images/image210.png", "images/image211.png", "images/image211.png", "images/image212.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "XY = constant constant (isobaric process) fu;rkad¼lenkch", "solution_images": ["images/image213.png", "images/image214.png", "images/image215.png", "images/image216.png", "images/image217.png", "images/image218.png", "images/image219.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Isobaric process", "difficulty": "tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "Ph-25-Q12", "question": "V-T diagram for a process of a given mass of ideal gas is as\nshown in the figure. During the process pressure of gas.\nçfØ;k", "question_images": ["images/image81.png"], "option_1": "first increases then decreases", "option_2": "continuously decreases", "option_3": "continuously increases", "option_4": "first decreases then increases.", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] As C<0 by diagram\ntSlk fd fp= }kjkC<0\n[IMAGE] P continuously decreasesyxkrkj ?kVrk gSA", "solution_images": ["images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png", "images/image86.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "V-T diagram", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-01-Q14", "question": "Which one of the following choices is closest to the order of\nthe angular momentum of the Earth (expressed in base MKS units)\nassociated with its rotation about its axis ?", "question_images": [], "option_1": "1038", "option_2": "1034", "option_3": "1030", "option_4": "1022", "correct_option": 2, "numerical_answer": null, "solution": "m = 6 × 1024 kg\nR = 6.4 × 106 m\nI = [IMAGE] mR2\nL = Iω = [IMAGE] × 6 × 1024 × (6.4 × 106)2\n×[IMAGE] = [IMAGE] × 0.39 × 1034", "solution_images": ["images/image67.png", "images/image68.png", "images/image69.png", "images/image70.png", "images/image71.png", "images/image72.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-24-Q2", "question": "A non-viscous fluid of density [IMAGE] is flowing\nin a tube as shown in figure. Area of section", "question_images": ["images/image12.png"], "option_1": "is double that of\nsection", "option_2": ". Centre of mass of section", "option_3": "Work done by gravitational force per unit volume from\nsection", "option_4": "Work done by elastic forces (pressure) per unit volume from\nsection", "correct_option": 2, "numerical_answer": null, "solution": "Applying Bernoulli is equation from section-(1) and (2)\nSolving we get,\n[IMAGE] (3) Work done by gravitation force per unit volume = negative of the\nincrease in gravitational potential energy per unit volume =\n$- (\\rho gh)$\n(4) Work done by elastic force per unit volume = decrease in\npressure = [IMAGE] = $2\\rho gh$\nTOPIC:Fluid dynamics\nSUB TOPIC:Bernoulli's theorem", "solution_images": ["images/image18.png", "images/image19.png", "images/image20.png", "images/image21.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-26-Q16", "question": "The stress versus strain graphs for wires of two materials A and\nB are as shown in the figure. If $Y_{A}$ and $Y_{B}$ are the Young's\nmodulii of the materials, then\nizfrcy rFkk fod`fr xzkQksa dks nks inkFkksZ A rFkk B", "question_images": ["images/image38.png"], "option_1": "$Y_{B} = 2Y_{A}$", "option_2": "$Y_{A} = Y_{B}$", "option_3": "$Y_{B} = 3Y_{A}$", "option_4": "$Y_{A} = 3Y_{B}$", "correct_option": 4, "numerical_answer": null, "solution": "$\\frac{\\ Stress\\ }{\\ Strain\\ } = Y = \\ slope\\ of\\ graph\\ $", "solution_images": ["images/image39.png"], "subject": "Physics", "topic": "Mechanical properties of solid", "subtopic": "Young's modulus", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-12-Q21", "question": "The current (in A) flowing through battery is2n. Find the value", "question_images": ["images/image27.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "The diode D_1is reverse-biased and acts as an open switch as\nshown in Fig. So, there is no currentthrough D_1and the\nHowever, D_2is forward biased and acts like a short-current or closed\nswitch. The current drawn isI = 12/(2 + 4) = 2A.", "solution_images": ["images/image28.png", "images/image29.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-19-Q4", "question": "A particle of mass m and charge q is released from rest in a\nuniform electric field. If there is no other force on the particle, the\ndependence of its speed v on the distance x travelled by it is correctly\ngiven by (graphs are schematic and not drawn to scale)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$V^{2} = \\frac{2qE}{m}x$\n$$", "solution_images": [], "subject": "Physics", "topic": "Magnetics", "subtopic": "Motion of charged particle in uniform electric field", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-13-Q2", "question": "What would be the magnetic moment of given current distribution", "question_images": ["images/image2.png"], "option_1": "$\\frac{3\\pi{IR}^{2}}{4} \\odot$", "option_2": "$\\frac{3\\pi{IR}^{2}}{4} \\otimes$", "option_3": "$\\frac{3\\pi{IR}^{2}}{2} \\odot$", "option_4": "$\\frac{3\\pi{IR}^{2}}{2} \\otimes$", "correct_option": 2, "numerical_answer": null, "solution": "$M = \\frac{I}{4}\\left\\{ \\pi(2R)^{2} - \\pi(R)^{2} \\right\\}$ =\n$\\frac{3\\pi{IR}^{2}}{4}$", "solution_images": [], "subject": "Physics", "topic": "Magnetism", "subtopic": "Magnetic moment", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "Ph-26-Q24", "question": "Water flows in a horizontal tube (see figure). The pressure of\nwater changes by [IMAGE] between Aand B where the area of\ncross section are [IMAGE] respectively. Find the rate of\nflow ofwater through the tube.\n(density of water [IMAGE],d {kSfrt uyh esa ikuh cg jgk gS ¼fp= ns[ksa½A bl uyh esa A ls B ds\nchp ikuh ds ncko vUrj\ndh vuqçLFk dkV A uyh esa ikuh", "question_images": ["images/image81.png", "images/image82.png", "images/image83.png", "images/image81.png", "images/image84.png", "images/image85.png", "images/image83.png", "images/image86.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "Using equation of continuity lehdj.k Using Bernoullies equation cjuksyh lehdj.k", "solution_images": ["images/image91.png", "images/image92.png", "images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png", "images/image97.png", "images/image98.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-28-Q14", "question": "A particle is revolving in a circle of radius r and centre at\n'O' with uniform angular velocity [IMAGE]. Let angular\nvelocity at A about O, B, and C is[IMAGE],\nthe correct option", "question_images": ["images/image134.png", "images/image135.png", "images/image136.png", "images/image137.png", "images/image138.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these.", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] TOPIC:Rigid body dynamics\nSUB TOPIC: Angular velocity", "solution_images": ["images/image144.png", "images/image145.png", "images/image146.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-26-Q5", "question": "The ratio of r.m.s. speed to the r.ms. angular speed of a\ndiatomic gas at certain temperature is:(assume m= mass of one molecule\nmass I = moment of inertia of the molecules)\newy ekfu, m =,d v.kq \nvk.kfod nzO;eku I = v.kqvksa", "question_images": [], "option_1": "$\\sqrt{\\frac{3}{2}}$", "option_2": "$\\sqrt{\\frac{3I}{2M}}$", "option_3": "$\\sqrt{\\frac{3I}{2m}}$", "option_4": "1", "correct_option": 3, "numerical_answer": null, "solution": "$\\frac{1}{2}{mV}^{2} = \\frac{3}{2}kT$\n$\\frac{1}{2}I\\omega^{2} = \\frac{2}{2}kT$\n$\\frac{V}{\\omega} = \\sqrt{\\frac{3I}{2m}}$", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "RMS speed", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-07-Q8", "question": "Two men of equal masses stand at opposite ends of the diameter of\na turntable disc of a certain mass, moving with some angular velocity.\nThe two men start moving towards the center of the turntable atequal\nrates. Then choose the correct option", "question_images": [], "option_1": "Kinetic energy of the system will increase.", "option_2": "Kinetic energy of system will decrease.", "option_3": "Angular momentum of the system will increase.", "option_4": "Angular momentum of the system will decrease.", "correct_option": 1, "numerical_answer": null, "solution": "$I_{1}\\omega_{1} = I_{2}\\omega_{2}$\nSince, men move towards middle of turn table Ι decreases hence\n$\\omega_{2}$increases.\n$\\therefore$ $KE = \\frac{L^{2}}{2I};L =$constant and I decreases so KE\nincreases", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-08-Q19", "question": "The expansion of unit mass of an ideal gas atconstant pressure\nis show in the diagram. Here", "question_images": ["images/image10.png"], "option_1": "a= volume,b=K (Kelvine) temperature c", "option_2": "a= volume,b= ºC temperature", "option_3": "a= ºC temperature, b= volume", "option_4": "a=K temperature, b= volume", "correct_option": 3, "numerical_answer": null, "solution": "$PV = nRT(K)$\n$V = \\frac{nR}{P}T(K)$\n$V = \\frac{nR}{P}\\lbrack T_{c}^{o} + 273\\rbrack$\n$V = \\frac{nR}{P}T_{c}^{o} + 273\\frac{nR}{P}$\n$\\Rightarrow T_{c}^{o} = \\frac{P}{nR}V - 273$\n$y = mx - c$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Ideal gas equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-04-Q14", "question": "A observer standing in a cart which has acceleration g on\nhorizontal road is observing vertical circular motion of a pendulum. One\nend of the pendulum string is at rest at O with respect to the cart.\nChoose the possible position of minimum speed for the pendulum with\nrespect to cart during complete vertical circular motion from the figure\ngiven below", "question_images": ["images/image87.png"], "option_1": "1", "option_2": "2", "option_3": "3", "option_4": "4\n g \n O", "correct_option": 3, "numerical_answer": null, "solution": "Opposite to geffect.\nTOPIC: CIRCULAR MOTION\nSUB TOPIC: DYNAMICS", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "Ph-26-Q40", "question": "If N Joules of heat is to be transferred to nitrogen in the\nisobaric heating process for that gas to perform the work A =2.0 J then\nthe value of N is\n[IMAGE] xSl dks nh x;h dk eku N twy gksrk gS] rks\nN", "question_images": ["images/image185.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "7", "solution": "W = 2J", "solution_images": ["images/image186.png", "images/image187.png"], "subject": "Physics", "topic": "Isobaric process", "subtopic": "Thermodynamics", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-24-Q8", "question": "Two liquids of densities $\\rho_{1}\\ $and\n$\\rho_{2}\\left( \\rho_{2} = 2\\rho_{1} \\right)$ are filled up behind a\nsquare wall of side 10 m as shown in figure. Each liquid has a height of\n5 m. The ratio of the forces due to these liquids exerted on upper part\nMN to that at the lower part NO is (Assume that the liquids are not\nmixing)", "question_images": ["images/image37.png"], "option_1": "$\\frac{1}{2}$", "option_2": "$\\frac{1}{3}$", "option_3": "$\\frac{1}{4}$", "option_4": "$\\frac{2}{3}$", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image38.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-05-Q6", "question": "A balloon rises from rest on the ground with constant\nacceleration $\\left( \\frac{g}{8} \\right)$. A stone is dropped when the\nballoon has risen to a height H metre. Then, the time taken by the stone\nto reach the ground is: (t=0 is the moment when stone is dropped)", "question_images": [], "option_1": "$\\frac{3}{2}\\sqrt{\\frac{H}{g}}$", "option_2": "$\\frac{1}{2}\\sqrt{\\frac{H}{g}}$", "option_3": "$\\sqrt{\\frac{H}{g}}$", "option_4": "$2\\sqrt{\\frac{H}{g}}$", "correct_option": 4, "numerical_answer": null, "solution": "$v^{2} = 0 + 2 \\times \\frac{g}{8} \\times H \\Rightarrow v = \\frac{\\sqrt{gH}}{2}$\n$- H = \\frac{\\sqrt{gH}}{2}t - \\frac{1}{2}{gt}^{2} \\Rightarrow t = \\frac{\\sqrt{gH} \\pm 3\\sqrt{gH}}{2g}$\n$\\Rightarrow t = 2\\sqrt{\\frac{H}{g}}$ (taking positive value)", "solution_images": [], "subject": "Physics", "topic": "Motion in 1D", "subtopic": "Motion under gravity", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-24-Q17", "question": "An open rectangular tank of height 3m and length and width\nrespectively 5m and 3m contains water up to height 2m. It is accelerated\nhorizontally along the longer side. Find the acceleration of the tank so\nthat 10 of water spills", "question_images": ["images/image48.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$\\ \\Rightarrow y = \\frac{12}{5}$\n$\\ tan\\theta = \\frac{y}{5} = \\frac{a}{g}$\n$\\ \\Rightarrow a = \\frac{24}{5}$", "solution_images": ["images/image49.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Fluid in motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-07-Q19", "question": "Two blocks of mass 10 kg and 4 kg are connected by a spring of\nnegligible mass and are placed on a frictionless horizontal surface. An\nimpulse gives a speed of $14{ms}^{- 1}$ to the heavier block in the\ndirection of the lighter block. Then, the velocity of the centre of mass\nis", "question_images": [], "option_1": "$30\\ {ms}^{- 1}$", "option_2": "$20\\ {ms}^{- 1}$", "option_3": "$10\\ {ms}^{- 1}$", "option_4": "$5\\ {ms}^{- 1}$", "correct_option": 3, "numerical_answer": null, "solution": "$= \\frac{10 \\times 14 + 4 \\times 0}{10 + 4} = 10m/s$.", "solution_images": ["images/image33.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Velocity of center of mass", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-23-Q3", "question": "An asteroid is moving directly towards the centre of the earth.\nwhen at a distance of 10R (R is theradius of the earth) from the earths\ncentre, it has a speed of 12 km/s. Neglecting the effect of\nearth'satmosphere, what will be the speed of the asteroid when it hits\nthe surface of the earth (escape velocityfrom the earth is 11.2", "question_images": [], "option_1": "14", "option_2": "12", "option_3": "16", "option_4": "18", "correct_option": 3, "numerical_answer": null, "solution": "KE_i + PE_i = KE_f + PE_i\n$\\frac{1}{2}mu_{0}^{2} + \\left( - \\frac{GMm}{10R} \\right) = \\frac{1}{2}mv^{2} + \\left( - \\frac{GMm}{R} \\right)$\n$v^{2} = u_{0}^{2} + \\frac{2GM}{R}\\left\\lbrack 1 - \\frac{1}{10} \\right\\rbrack$\n$v = \\sqrt{u_{0}^{2} + \\frac{9}{5}\\frac{GM}{R}}$\n$= \\sqrt{12^{2} + \\left( \\frac{9}{5} \\right)\\frac{(11.2)^{2}}{2}}$\n$= \\sqrt{144 + 0.9(11.2)^{2}} = \\sqrt{256.896}$\n= 16.028 km/s $\\simeq 16$", "solution_images": [], "subject": "Physics", "topic": "Gravitation", "subtopic": "Conservation of mechanical energy", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-04-Q18", "question": "A body is acted upon by a force which is inversely proportional to\nthe distance covered. The work done will be proportional to", "question_images": [], "option_1": "s", "option_2": "s2", "option_3": "", "option_4": "None", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] TOPIC: WPE\nSUB TOPIC: WORK DONE BY VARIABLE FORCE", "solution_images": ["images/image97.png", "images/image98.png", "images/image99.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-05-Q3", "question": "Two seconds after projection, a projectile is travelling in a\ndirection inclined at 30º to the horizontal after one more second, it is\ntravelling horizontally. Then the speed of projection(u) and angle of\nprojection ($\\theta$) are given by", "question_images": [], "option_1": "$u = 10\\sqrt{3}m/s$and$\\theta = 45^{\\circ}$", "option_2": "$u = 10\\sqrt{3}m/s$and$\\theta = 60^{\\circ}$", "option_3": "$u = 20\\sqrt{3}m/s$and$\\theta = 60^{\\circ}$", "option_4": "$u = 20\\sqrt{3}m/s$and$\\theta = 45^{\\circ}$", "correct_option": 3, "numerical_answer": null, "solution": "$T = 6sec = \\frac{2usin\\theta}{g} \\Rightarrow usin\\theta = 30$\n$\\tan 30^{\\circ} = \\frac{{usin}\\theta - gt}{{ucos}\\theta} = \\frac{30 - 20}{{ucos}\\theta}$which\ngives $ucos\\theta = 10\\sqrt{3}$\n$$tan\\theta =", "solution_images": [], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Velocity of projectile at an instant", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-10-Q5", "question": "A body of mass m and radius R rolling horizontally without\nslipping at a speed v climbs a ramp to a height[IMAGE].\nThe rolling body can be.", "question_images": ["images/image23.png"], "option_1": "A sphere", "option_2": "A circular ring", "option_3": "A spherical shell", "option_4": "A circular disc", "correct_option": 4, "numerical_answer": null, "solution": "From energy\n$K = \\frac{R}{\\sqrt{2}}$\n[IMAGE] Body is a disc", "solution_images": ["images/image24.png", "images/image25.png", "images/image26.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Pure rolling", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "Ph-25-Q28", "question": "A litre of hydrogen at [IMAGE] and\n[IMAGE] pressure expands isothermally until its volume\nbecomes double and then expands adiabatically until it becomes four\ntimes. [IMAGE]. Final temperature (in\n27°C o 106 izlkj tc rd bl nqxquk gks rFkk fQj:)ks\"e izØe }kjk\npkj xqus vk;ru rd izlkfjr A vfUre", "question_images": ["images/image200.png", "images/image201.png", "images/image202.png", "images/image203.png", "images/image204.png", "images/image204.png", "images/image203.png", "images/image205.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "", "solution_images": ["images/image206.png", "images/image207.png", "images/image208.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-08-Q10", "question": "An ideal gas whose internal energy U is at absolute zero\ntemperature is equal to zero undergoes areversible adiabatic\ncompression. If U, P, V, T represents the internal energy, pressure,\nvolume and absolute temperature respectively for ideal gas\nthen $\\left( \\gamma = \\frac{C_{P}}{C_{v}} \\right)$.", "question_images": [], "option_1": "$UV^{\\gamma} =$constant", "option_2": "${UP}^{\\gamma} =$constant", "option_3": "${TU}^{\\gamma - 1} =$constant", "option_4": "${VU}^{\\frac{1}{\\gamma - 1}} =$constant", "correct_option": 4, "numerical_answer": null, "solution": "$U\\ \\alpha\\ T$\nFor adiabatic process ${TV}^{\\gamma - 1} =$ constant\n$\\therefore{UV}^{\\gamma - 1} =$constant\n$\\Rightarrow {VU}^{\\frac{1}{\\gamma - 1}} =$constant", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-21-Q25", "question": "The current (in A) flowing through battery in the given circuit\n ifjiFk", "question_images": ["images/image81.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "The diode D_1is reverse-biased and acts as an open switch as\nshown in Fig. So, there is no currentthrough D_1and the\n/kkjk izokfgr ugha gksxhA\nHowever, D_2is forward biased and acts like a short-current or closed\nswitch. The current drawn isI = 12/(2 + 4) = 2A.;)fi, D_2 vxz", "solution_images": ["images/image82.png", "images/image83.png", "images/image84.png"], "subject": "Physics", "topic": "Electronic devices", "subtopic": "Diodes", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "Ph-27-Q10", "question": "A small ball of mass m is released from rest from a height h\nabove the liquid surface. The density of ballis $\\frac{\\rho}{3}$:\n(assume no splash)\n xsan dks fLFkjkoLFkk", "question_images": ["images/image45.png"], "option_1": ") If h ≥ H, then the ball will be able to enter the liquid of\nhigher density\n ?kuRo", "option_2": "If h ≥ $\\frac{7H}{2}$, then the ball will be able to strike the\nbottom of the vessel;fn h ≥ $\\frac{7H}{2}$, gks rks xsan ik=k", "option_3": "If h = 2H, then the ball will cross the interface of two liquids\nwith kinetic energy = mgH and if h = 0 then ball will float on the\nsurface;fn h = 2H, gks rks xsan nksuks nzoks lrg ij xfrt = mgH h = 0 gks rks xasn lrg ij rSjxhA", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] K.E. = mgH.", "solution_images": ["images/image46.png", "images/image47.png", "images/image48.png", "images/image49.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Buoyant force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-06-Q1", "question": "In the system shown in figure mA = 4m, mB = 3m and mC = 8m.\nFriction is absent everywhere. String is light and inextensible. If the\nsystem is released from rest find the acceleration of block B", "question_images": ["images/image1.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$T = m_{B}b$[Newtons II law for B in horizontal direction]\n$T = \\left\\lbrack m_{A} + m_{e} \\right\\rbrack c$ I law for A\nand C in horizontal direction]\n$m_{A}g - T = m_{A}a$ [Newton's II law for A in vertical direction].\n$\\frac{m_{A}g - T}{m_{A}} = \\frac{T}{m_{B}} + \\frac{T}{m_{A} + m_{C}}$\n$\\frac{4mg - T}{4m} = \\frac{T}{3m} + \\frac{T}{12m} \\Rightarrow T = \\frac{3mg}{2} \\Rightarrow b = \\frac{g}{2}$", "solution_images": ["images/image2.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Newton's second law", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-14-Q3", "question": "If an object is placed at 20 cm in front of a half thin convex\nlens of focal length 10 cm as shown in figure then taking object\nposition as shown the", "question_images": ["images/image15.png"], "option_1": "x - coordinate of image is 20 cm", "option_2": "x - coordinate of image is 40 cm", "option_3": "y - coordinate of image is 0.2 cm", "option_4": "y - coordinate of image is 0.4 cm", "correct_option": 3, "numerical_answer": null, "solution": "$\\frac{1}{\\text{v}} - \\frac{1}{- 20} = \\frac{1}{10} \\Rightarrow \\text{v} = 20$", "solution_images": ["images/image16.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Refraction (Lens)", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-08-Q25", "question": "Figure shows the variation of internal energy \"U\" with the\ndensity \"$\\rho$ \" of one mole of ideal diatomic gas. Process BA is a\npart of rectangular hyperbola. If the work done by gas in the process BA\nis 7n joules. Find n ?", "question_images": ["images/image14.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "For rectangular hyperbola\nXY = constant\n$U\\rho =$ constant\n$nC_{V}T\\rho =$ constant\n$n\\frac{5}{2}RT\\frac{M}{V} =$constant\n$\\frac{5}{2}P\\frac{VM}{V} =$constant $\\Rightarrow$ P = constant\n(isobaric process)\n$\\Delta U_{B \\rightarrow A} = 37 - 2 = 35J = \\frac{5}{2}nR\\Delta T$\n$14J = nR\\Delta T$\n$\\therefore$ $W = 14J$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Isobaric process", "difficulty": "tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-15-Q6", "question": "A particle moving with kinetic energy E has de Broglie\nwavelength[IMAGE]. If energy [IMAGE] is\nadded to its energy, the wavelength becomes[IMAGE]. Value\nof[IMAGE], is", "question_images": ["images/image48.png", "images/image49.png", "images/image50.png", "images/image51.png"], "option_1": "4E", "option_2": "3E", "option_3": "2E", "option_4": "E", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image52.png", "images/image53.png", "images/image54.png", "images/image55.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "De Broglie wavelength", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-04-Q7", "question": "A particle is released from a certain height H = 400 m. Due to the\nwind the particle gathers the horizontal velocity component vx = ay\nwhere [IMAGE] and y is the vertical displacement of the\nparticle from point of release, then the horizontal drift of the\nparticle when it strikes the ground is", "question_images": ["images/image38.png"], "option_1": "2.67 km", "option_2": "8.67 km", "option_3": "1.67 km", "option_4": "5.1 km\n H=400m \n vx = ay [IMAGE], \n y,,", "correct_option": 1, "numerical_answer": null, "solution": "Vy = [IMAGE]... (1)\nVx =[IMAGE]... (2)\n[IMAGE] Dividing (1) by (2)\n[IMAGE] or [IMAGE] dy = 2dx\nIntegrating\n[IMAGE] ∴ x = 2.67 km\nTOPIC:KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image39.png", "images/image40.png", "images/image41.png", "images/image42.png", "images/image43.png", "images/image44.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-01-Q12", "question": "A horizontal conductor is oriented north south and carries some\ncurrent. A positively charged particle located vertically above it and\nhaving a velocity directed northward experiences an upward force. What\nis the direction of the force if this charged particle were located to\nthe east of the conductor and had a velocity directed towards the\nconductor.", "question_images": [], "option_1": "North", "option_2": "South", "option_3": "Up", "option_4": "Down", "correct_option": 1, "numerical_answer": null, "solution": "12 in first case according to [IMAGE] Magnetic field\nwill be towards west at position of charge particle for this current in\nconductor should be towards south in second case magnetic field at\nposition of particular is upward so according to\n[IMAGE] force will be towards north.", "solution_images": ["images/image57.png", "images/image57.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-09-Q1", "question": "A disc of mass is attached to a spring of\nstiffness \"k\". During its motion, the disc rolls on the ground. When\nreleased from some stretched position, the center of the disc will\nexecute simple harmonic motion with a time period of", "question_images": ["images/image1.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image6.png", "images/image7.png", "images/image8.png", "images/image9.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Time period of oscillation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-04-Q6", "question": "A particle is projected at an angle tan-1[IMAGE].\nFind the height at which velocity of the particle is\n[IMAGE] (m/sec). (g = 10 ms-2)", "question_images": ["images/image30.png", "images/image31.png"], "option_1": "1.8 m", "option_2": "0.9 m", "option_3": "0.7 m", "option_4": "0.11 m\n tan-1 (m/sec) (g = 10\nms-2)", "correct_option": 2, "numerical_answer": null, "solution": "tan θ\n[IMAGE] Uy [IMAGE] 9 = 27 + 2(-10)Sy\nSy = 0.9", "solution_images": ["images/image32.png", "images/image33.png", "images/image34.png", "images/image35.png", "images/image36.png", "images/image37.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-04-Q1", "question": "[IMAGE] = [IMAGE] X", "question_images": ["images/image1.png", "images/image2.png"], "option_1": "", "option_2": "1", "option_3": "2", "option_4": "- 1", "correct_option": 1, "numerical_answer": null, "solution": "L.H.S. is dimensionless. So \nTOPIC: UNIT & DIMENSION\nSUB TOPIC: DIMENSIONS", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-05-Q11", "question": "The velocity of an object moving rectilinearly is given as a\nfunction of time by $v = 4t - 3t^{2}$, where v is in m/s and t is in\nseconds. The average velocity of particle between t=0 to t=2 seconds is", "question_images": [], "option_1": "0", "option_2": "$- 2m/s$", "option_3": "$- 4m/s$", "option_4": "none of these", "correct_option": 1, "numerical_answer": null, "solution": "$< v > = \\frac{\\int_{0}^{2}\\mspace{2mu} vdt}{\\int_{0}^{2}\\mspace{2mu} dt} = \\frac{\\int_{0}^{2}\\mspace{2mu}\\left( 4t - 3t^{2} \\right)dt}{2} = 0$", "solution_images": [], "subject": "Physics", "topic": "Motion in 1D", "subtopic": "Non uniform acceleration", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-16-Q2", "question": "A small block of mass m is released from rest from point A inside\na smooth hemisphere bowl of radius R, which is fixed on ground such that\nOA is horizontal. The ratio (x) of magnitude of centripetal force and\nnormal reaction on the block at any point B varies with $\\theta$ as", "question_images": ["images/image2.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{mv^{2}}{R} = N - mg\\ sin\\theta$\nBy energy conservation,\n$mgR\\ sin\\theta = \\frac{1}{2}{mv}^{2}$\n$\\frac{{mv}^{2}}{R} = 2mg\\ sin\\theta$\n$N = 3mg\\ sin\\theta$\n$Ratio = \\frac{{mv}^{2}}{RN} = \\frac{2}{3}$ (Constant)\n$x = \\frac{2}{3}$", "solution_images": ["images/image7.png"], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-18-Q12", "question": "Projectiles are simultaneously projected from the same point\nwith speeds u and 2 u, at angles $\\theta$ and $2\\theta$ respectively\nwith horizontal. The locus of second projectile with respect to the\nfirst is", "question_images": [], "option_1": "$y = (\\frac{2cos2\\theta - cos\\theta}{2sin2\\theta - sin\\theta})x$", "option_2": "$y = {(\\frac{2sin2\\theta - sin\\theta}{2cos2\\theta - cos\\theta})}^{x}$", "option_3": "$y = (\\frac{2cos2\\theta + cos\\theta}{2sin2\\theta + sin\\theta})x$", "option_4": "$y = (\\frac{2sin2\\theta + sin\\theta}{2cos2\\theta + cos\\theta})x$", "correct_option": 2, "numerical_answer": null, "solution": "For first projectile$y_{1} = usin\\theta t - \\frac{1}{2}gt^{2}$\n$y_{2} = 2usin2\\theta t - \\frac{1}{2}{gt}^{2}$\n$x_{1} = ucos\\theta t$ $x_{2} = 2ucos2\\theta t$\n$y = y_{2} - y_{1} = t\\lbrack 2usin2\\theta - usin\\theta\\rbrack$\n$x = x_{2} - x_{1} = t\\lbrack 2ucos2\\theta - ucos\\theta\\rbrack$\n$\\frac{y}{x} = \\frac{2usin2\\theta - usin\\theta}{2ucos2\\theta - ucos\\theta}$\n$y = x\\lbrack\\frac{2sin2\\theta - sin\\theta}{2cos2\\theta - cos\\theta}\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Equation of trajectory", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-21-Q1", "question": "In the figure shown a parallel plate capacitor has a dielectric\nof width d/2 and dielectric constant K = 2.The other dimensions of the\ndielectric are same as that of the plates. The plates P_1 and P_2 of\nthecapacitor have area each. The energy of the capacitor is:\nfp=k lekUrj iê la/kkfj=k esa ijkoS|qr dh pkSM+kbZ d/2\nrFkk ijkoS|qrkad K = 2 gSAijkoS|qr", "question_images": ["images/image1.png"], "option_1": "$\\frac{\\epsilon_{0}AV^{2}}{3d}$", "option_2": "$\\frac{2\\epsilon_{0}AV^{2}}{d}$", "option_3": "$\\frac{3}{2}\\frac{\\epsilon_{0}AV^{2}}{d}$", "option_4": "$\\frac{2\\epsilon_{0}AV^{2}}{3d}$", "correct_option": 4, "numerical_answer": null, "solution": "$U = \\frac{1}{2}C_{eq}v^{2}$\n$C_{1} = \\frac{k\\varepsilon_{0}A}{d/2} = \\frac{2\\varepsilon_{0}A}{(d/2)}$\nand $C_{2} = \\frac{\\varepsilon_{0}A}{d/2}$\n$C_{eq} = \\frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \\frac{\\left( 2\\frac{\\varepsilon_{0}A}{d/2} \\right) \\times \\frac{\\varepsilon_{0}}{d/2}}{3\\frac{\\varepsilon_{0}}{d/2}}$\n$= \\frac{4}{3}\\frac{\\varepsilon_{0}A}{d}\\ or\\ C_{eq} = \\frac{\\varepsilon_{0}A}{d - \\frac{d}{2} + \\frac{d}{4}} = \\frac{4\\varepsilon_{0}A}{3d}$\n$U = \\frac{1}{2}\\left( \\frac{4}{3}\\frac{\\varepsilon_{0}A}{d} \\right)V^{2} = \\frac{2}{3}\\left( \\frac{\\varepsilon_{0}}{d} \\right){AV}^{2}$", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Capacitor with dielectric", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-12-Q1", "question": "In the figure shown a parallel plate capacitor has a dielectric\nof width d/2 and dielectric constant K = 2.The other dimensions of the\ndielectric are same as that of the plates. The plates P_1 and P_2 of\nthecapacitor have area each. The energy of the capacitor is", "question_images": ["images/image1.png"], "option_1": "$\\frac{\\epsilon_{0}AV^{2}}{3d}$", "option_2": "$\\frac{2\\epsilon_{0}AV^{2}}{d}$", "option_3": "$\\frac{3}{2}\\frac{\\epsilon_{0}AV^{2}}{d}$", "option_4": "$\\frac{2\\epsilon_{0}AV^{2}}{3d}$", "correct_option": 4, "numerical_answer": null, "solution": "$U = \\frac{1}{2}C_{eq}v^{2}$\n$C_{1} = \\frac{k\\varepsilon_{0}A}{d/2} = \\frac{2\\varepsilon_{0}A}{(d/2)}$\nand $C_{2} = \\frac{\\varepsilon_{0}A}{d/2}$\n$C_{eq} = \\frac{C_{1}C_{2}}{C_{1} + C_{2}};$\n$C_{eq} = \\frac{\\left( 2\\frac{\\varepsilon_{0}A}{d/2} \\right) \\times \\frac{\\varepsilon_{0}}{d/2}}{3\\frac{\\varepsilon_{0}}{d/2}}$\n$= \\frac{4}{3}\\frac{\\varepsilon_{0}A}{d}\\ or\\ C_{eq}$\n$= \\frac{\\varepsilon_{0}A}{d - \\frac{d}{2} + \\frac{d}{4}} = \\frac{4\\varepsilon_{0}A}{3d}$\n$U = \\frac{1}{2}\\left( \\frac{4}{3}\\frac{\\varepsilon_{0}A}{d} \\right)V^{2}$\n$= \\frac{2}{3}\\left( \\frac{\\varepsilon_{0}}{d} \\right){AV}^{2}$", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Capacitor with dielectric", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-07-Q21", "question": "A sphere of mass m is given some angular velocity about a\nhorizontal axis through its centre, andgently placed on a plank of mass\nm. The coefficient of friction between the two is $\\mu$. The plank rests\non a smooth horizontal surface. The initial acceleration of the sphere\nrelative to the plank will be $x(\\mu g)$. Find the value of x.", "question_images": ["images/image34.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "$a_{sphere\\ \\ } = \\frac{\\mu mg}{m} = \\mu g( + \\widehat{i})$\n$a_{plank\\ \\ } = \\frac{\\mu mg}{m} = \\mu g\\left( - \\widehat{i} \\right)$\n${\\ a}_{relative}\\ \\ = 2\\mu g$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Rolling over a movable platform", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-03-Q20", "question": "has four choices", "question_images": [], "option_1": ",", "option_2": ",", "option_3": ",", "option_4": "out of which ONLY ONE is correct]", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "Ph-27-Q25", "question": "A container is filled with a liquid of density ρ. Container is\naccelerating on a horizontal surface with acceleration\n[IMAGE] towards right hand side and liquid is at\nrest with respect to container as shown in figure. If there is a point A\nin liquid then, which of the following is correct: (assume atmospheric\npressure is zero and AB is vertical and AC is horizontal line),d ik=$\\rho$?kuRo ds nzo ls Hkjk gSA ik=k {kSfrt lrg\nij$\\frac{3g}{4}$Roj.k vksj Rofjr gS rFkk nzo fp=kkuqlkj ik=k ds\nfodYi gS P_atm = 0 o AB", "question_images": ["images/image154.png", "images/image155.png"], "option_1": "Pressure at A is[IMAGE] Aij nkc[IMAGE] gksxkA", "option_2": "Pressure at A is [IMAGE] Aij nkc[IMAGE] gksxkA", "option_3": "", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "For point A from top surface PA=[IMAGE] and from\nright wall P_A=[IMAGE] lrg ls fcUnq A ds fy, PA = [IMAGE] rFkk nka;h nhokj\nlsP_A=[IMAGE] and rFkk", "solution_images": ["images/image159.png", "images/image160.png", "images/image159.png", "images/image160.png", "images/image161.png", "images/image162.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Fluid in motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-20-Q13", "question": "A uniform cylinder floats with its axis vertical with\n[IMAGE] parts of its length in the upper liquid and\n[IMAGE] part of its length in lower liquid. If the\ndensities of upper and lower liquids are [IMAGE] then the\ndensity of the material of the cylinder is", "question_images": ["images/image28.png", "images/image29.png", "images/image30.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "Let [IMAGE] be the be the length of the cylinder\nthen\n is the density of the cylinder.\nTOPIC:Fluid\nSUB TOPIC:Bouyancy\nLEVEL: Moderate", "solution_images": ["images/image35.png", "images/image36.png", "images/image37.png", "images/image38.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-07-Q23", "question": "A disc has mass 30 kg and radius 1m. The moment of inertia in\nkgm^2about a tangent to its rim in itsplane is 7.5n. Find the value on\nn.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "Moment of inertia of a disc about its diameter is\n$I_{d} = \\frac{1}{4}{MR}^{2}$\nNow, according to perpendicular axis theorem moment of inertia of disc\nabout a tangent passingthrough rim and in the plane of disc is\n$$= = 37.5$$", "solution_images": ["images/image36.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Easy", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-18-Q13", "question": "The potential energy of a particle of mass 1 kg in a\nconservative field is given as $U = (3x^{2}y^{2} + 6x)J$, where x and y\nare measured in meter. Initially particle is at (1,1) & at rest then", "question_images": [], "option_1": "Initial acceleration of particle is $6\\sqrt{5}{ms}^{2}.$", "option_2": "If particle is left free it does not move in a straight line.", "option_3": "Work done to slowly bring the particle to origin is - 9 J.", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$\\overset{\\rightarrow}{F} = - \\frac{\\partial u}{\\partial x}\\overset{\\hat{}}{i} - \\frac{\\partial u}{\\partial y}\\overset{\\hat{}}{j}$\n$\\overset{\\rightarrow}{F} = - (6xy^{2} + 6)\\overset{\\hat{}}{i} - (6x^{2}y)\\overset{\\hat{}}{j}$\nAt (1,1)\n$\\overset{\\rightarrow}{F} = - 12\\overset{\\hat{}}{i} - 6\\overset{\\hat{}}{j}$\n$|\\overset{\\rightarrow}{a}| = \\frac{|\\overset{\\rightarrow}{F}|}{m}$ =\n$\\sqrt{12^{2} + 6^{2}} = 6\\sqrt{5}ms^{- 2}$\nAlso,\n$w_{ext} = \\Delta u = u_{f} - u_{i} = 0 - \\left( 3.1^{2}.1^{2} + 6.(1) \\right) = - 9J$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Potential energy", "difficulty": "Hard", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-08-Q1", "question": "A monoatomic ideal gas follows the process:\nTV^3/2 = constant\nThe molar specific heat for this process is: (R = gas constant).", "question_images": [], "option_1": "$\\frac{5R}{2}$", "option_2": "$\\frac{5R}{6}$", "option_3": "$\\frac{3R}{2}$", "option_4": "R", "correct_option": 2, "numerical_answer": null, "solution": "TV^3/2 = constant\n$\\frac{(P \\vee )V^{3/2}}{nR} =$constant\n${PV}^{\\frac{5}{2}} =$constant \n${PV}^{x} =$constant\n$x = \\frac{5}{2}$\n$C = C_{v} + \\frac{R}{1 - x}$\nFor polytropic process,\n$C = \\frac{3}{2}R + \\frac{R}{1 - \\frac{5}{2}} = \\frac{5R}{6}$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-06-Q22", "question": "A thin rod of length 1 m is kept vertical and is hinged from the\nlower end Its linear mass density varies with the distance (x)\nfrom the end $H^{'}$ as $\\lambda = (12x)kg/m$, where x is distance along\nrod from point H. The potential energy (in Joules) of the rod is 5 N.\nFind N (at H the potential energy is zero)\n$\\left( g = 10m/s^{2} \\right)$.", "question_images": ["images/image30.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "Lets take an element of dx width at a distance x from the lowest\nend.\n$dU = (dm)gx,U = 12g\\int_{x = 0}^{x - 1}\\mspace{2mu} x^{2}dx = 40J$", "solution_images": ["images/image31.png"], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Potential energy", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-23-Q23", "question": "A ball is given velocity\n[IMAGE] as shown. If the ratio of centripetal acceleration totangential\nacceleration at the point where the ball leaves the circular path is\n[IMAGE] then n is: [Neglect size of the ball].", "question_images": ["images/image54.png", "images/image55.png", "images/image56.png"], "option_1": "and [IMAGE].....", "option_2": "[IMAGE] Now TOPIC: Circular motion\nSUB TOPIC: Vertical circle\nLEVEL: Moderate", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "and [IMAGE].....(2)\n[IMAGE] Now TOPIC: Circular motion\nSUB TOPIC: Vertical circle\nLEVEL: Moderate", "solution_images": ["images/image57.png", "images/image58.png", "images/image59.png", "images/image60.png", "images/image61.png", "images/image62.png", "images/image63.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-23-Q6", "question": "An ideal gas at a pressure of 1 atmosphere andtemperature of 27°C\nis compressed adiabaticallyuntil its pressure becomes 8 times the\ninitialpressure. Then the final temperature is", "question_images": ["images/image9.png"], "option_1": "627°C", "option_2": "527°C", "option_3": "427°C", "option_4": "327°C", "correct_option": 4, "numerical_answer": null, "solution": "Here, [IMAGE] As changes are adiabatic,", "solution_images": ["images/image10.png", "images/image11.png", "images/image12.png", "images/image13.png", "images/image14.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adaibaticproces", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "Ph-27-Q38", "question": "A vertical capillary tube with inner radius 0.5 mm, is submerged\ninto water so that the length of its part above the water surface is h =\n25 mm. Radius of curvature of meniscuses formed is\n[IMAGE]. Here α is an integer. Find α. (surface\ntension of water = 0.075 N/m. Density of water\n[IMAGE], angle of contact = 0°)\n0.5 mm vkUrfjd f=,d ikuh esa bl çdkj Mwch\ngqbZ gS fd ikuh dh lrg ds blds Hkkx dh yEckbZ h = 25 mm gSA;gk¡α,d iw.kkZad gSAα dk eku Kkr dhft,A (ikuh dk i`\"B ruko = 0.075\nN/m, ty", "question_images": ["images/image238.png", "images/image239.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image240.png", "images/image241.png", "images/image242.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Capillary tube", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-03-Q20", "question": "If the block in the shown arrangement is acted upon by a constant\nforce F on a frictionless surface for t ≥ 0, its maximum speed will be", "question_images": ["images/image120.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE], t ≥ 0 \nF,", "correct_option": 1, "numerical_answer": null, "solution": "For vmax, [IMAGE] ⇒ Fnet = 0\n⇒ Kx = F\nx =[IMAGE] W = ∆KE\n[IMAGE] + Fx = [IMAGE] mv2 - 0\nvmax =[IMAGE] TOPIC: WPE\nSUB TOPIC: WORK ENERGY THEOREM", "solution_images": ["images/image125.png", "images/image126.png", "images/image127.png", "images/image128.png", "images/image129.png", "images/image130.png", "images/image131.png", "images/image132.png", "images/image133.png", "images/image134.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-14-Q25", "question": "In the arrangement shown in figure, wavelength of light used\nis[IMAGE]. The distance between slits\n[IMAGE] is [IMAGE] The distance between\n[IMAGE] If the ratio of maximum to minimum intensity\nreceived on screen P is K. Then K is", "question_images": ["images/image161.png", "images/image162.png", "images/image163.png", "images/image164.png", "images/image165.png", "images/image166.png", "images/image167.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "9", "solution": "If intensity of light incident on slits [IMAGE] is [IMAGE] The intensity through [IMAGE] will be", "solution_images": ["images/image162.png", "images/image168.png", "images/image165.png", "images/image169.png", "images/image170.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "Ph-25-Q22", "question": "Two moles of an ideal gas with [IMAGE] are mixed\nwith 3 moles of another ideal gas with [IMAGE] Thevalue\nof [IMAGE] for the mixture is\neksy=", "question_images": ["images/image160.png", "images/image161.png", "images/image162.png", "images/image160.png", "images/image161.png", "images/image162.png"], "option_1": "1.42", "option_2": "1.47", "option_3": "1.50", "option_4": "1.45", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] On rearranging we getiqu ij", "solution_images": ["images/image163.png", "images/image164.png"], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Heat capacity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-23-Q14", "question": "In a double-slit experiment, at a certain point on the screen\nthe path difference between the two interfering waves is\na wavelength. The ratio of the intensity of light at that point to that\nat the centre of a bright fringe is", "question_images": ["images/image33.png"], "option_1": "0.568", "option_2": "0.760", "option_3": "0.853", "option_4": "0.672", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image34.png", "images/image35.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Interference", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-01-Q18", "question": "A positively charged particle is moved in a straight line to\nthe right from position x = -4 m to position x = -2 m by an external\nagent. The electric potential at these positions in space are V(x =\n-4m) = -4 volts and V(x = -2m) = -2 volts. Which statement is true\nabout the work done by the external agent moving the charge between\nthese positions and the electric field component parallel to the\ndirection of motion that the moving charged particle experiences? Assume\nthe particle is moved at constant speed.", "question_images": [], "option_1": "The external agent does negative work and the electric field is\ndirected to the right.", "option_2": "The external agent does positive work and the electric field is\ndirected to the right.", "option_3": "The external agent does negative work and the electric field is\ndirected to the left.", "option_4": "The external agent does positive work and the electric field is\ndirected to the left.", "correct_option": 4, "numerical_answer": null, "solution": "wext = q (vs - vi) = q (-4 - (-2)) = -2q = -ve\nE goes from higher potential to lower potential.", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-11-Q19", "question": "A rocket is fired vertically from the surface of the earthwith a\nspeed v. How far from the earth does the rocketgo before returning to\nthe earth (where [IMAGE] is the radius of the earth and\ng is acceleration due to gravity)", "question_images": ["images/image183.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Let the rocket reached a height h from thesurface of earth.\nTotal energy at the surface of the earth is [IMAGE] Where m and [IMAGE] are the masses of rocket andearth\nrespectively.\nAt highest point, the velocity of the rocketbecomes zero.\nTotal energy at the highest point is\n[IMAGE] According to law of conservation of energy,", "solution_images": ["images/image188.png", "images/image189.png", "images/image190.png", "images/image191.png", "images/image192.png", "images/image193.png", "images/image194.png", "images/image195.png", "images/image196.png", "images/image197.png", "images/image198.png"], "subject": "Physics", "topic": "Gravitation", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-03-Q21", "question": "One unit of power of a motor is 150 Watts in some system of units.\nIf the unit of force is doubled and unit of velocity tripled, then one\nnew unit of power of motor is ( in W )\n 1 150,", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "25", "solution": "P = f v\nnu = constant\nn = no. attached, u = size of unit\nP' =[IMAGE] so, size of power will become 6 times\nTOPIC: UNIT & DIMENSION\nSUB TOPIC: MAGNITUDE", "solution_images": ["images/image135.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Easy", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-14-Q19", "question": "The first diffraction minima due to a single slit diffraction is\nat [IMAGE] for a light of\nwavelength[IMAGE]. The width of the slit is", "question_images": ["images/image126.png", "images/image127.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image132.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Diffraction", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-02-Q42", "question": "Two tuning forks are producing 6 beats/s. The frequency of\nfirst tuning fork is 256 Hz. Now, first tuning fork is loaded with wax.\nAs a result the beat frequency becomes 4 beats/s. Now some portion of\nthe wax is removed from first tuning fork. As a result the beat\nfrequency becomes 0 beats/s. Then the frequency of first tuning fork\nafter waxing (the condition when beat frequency is 4 beats/s) will be", "question_images": [], "option_1": "250 Hz", "option_2": "262 Hz", "option_3": "254 Hz", "option_4": "246 Hz", "correct_option": 4, "numerical_answer": null, "solution": "Since beat frequency is 6 Hz, therefore, frequency of second\nfork will be either 262 Hz or 250 Hz. Since on loading with wax beat\nfrequency decreases, frequency of second fork should be 250 Hz. So after\nwaxing, the frequency of first tuning fork should be either 252 or 246\nHz. Since after removing a portion of wax from first fork beat frequency\nbecomes zero. Therefore frequency of first fork after waxing should be\n246 Hz.", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "Ph-26-Q21", "question": "In a cylinder-piston arrangement, air is under a pressure P_1.\nA soap bubble of radius r lies inside the cylinder, soap bubble has\nsurface tension T. The radius of soap bubble is to be reduced to half,\nThe new pressure P_2 to which air should be compressed isothermally.\n(Assume r is very small as compared to height of cylinder),d csyu&fiLVu mifLFkr gS rFkk;g P_1 nkc ij gSA blesa,d lkcqu dk cqycqyk ftldh mifLFkr gSA;fn\nlkcqu ds cqycqys dh f= djuh gks rks] csyu \neku r csyu", "question_images": ["images/image68.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Isothermal process. ¼lerkih; çØe", "solution_images": ["images/image73.png", "images/image74.png", "images/image75.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-16-Q14", "question": "Slider block A moves to the right with constant velocity 6 m/s\nin the arrangement as shown in the figure. C and D are points on the\nstring. If, and are the velocities of B, C and D respectively then,\nchoose", "question_images": ["images/image24.png"], "option_1": "$|\\overset{\\rightarrow}{V_{C}} - \\overset{\\rightarrow}{V_{D}}| = 4|\\overset{\\rightarrow}{V_{B}}|$", "option_2": "$|\\overset{\\rightarrow}{V_{B}} - \\overset{\\rightarrow}{V_{C}}| = |\\overset{\\rightarrow}{V_{B}} - \\overset{\\rightarrow}{V_{D}}| = 2|\\overset{\\rightarrow}{V_{B}}|$", "option_3": "$|\\overset{\\rightarrow}{V_{C}}| - |\\overset{\\rightarrow}{V_{D}}| = 2|\\overset{\\rightarrow}{V_{B}}|$", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "Clearly $v_{C} = 6m/s \\downarrow$\nBy string constraint,\n$–6 + V_{B} + V_{B} + V_{B} = 0 \\Rightarrow V_{B} = 2m/s \\downarrow$\nAlso, $\\frac{V_{C} + V_{D}}{2} = V_{B}$\n$\\Rightarrow V_{D} = - 2m/s$\n$\\therefore V_{D} = 2m/s \\uparrow$\n$|\\overset{\\rightarrow}{V_{C}} - \\overset{\\rightarrow}{V_{D}}| = 6 + 2 = 8m/s$\n$|\\overset{\\rightarrow}{V_{B}} - \\overset{\\rightarrow}{V_{C}}| = 4m/s$\n$|\\overset{\\rightarrow}{V_{B}} - \\overset{\\rightarrow}{V_{D}}| = 4m/s$\n$|\\overset{\\rightarrow}{V_{c}}| - |\\overset{\\rightarrow}{V_{D}}| = 6 - 2 = 4m/s$", "solution_images": ["images/image26.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Constraint motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-12-Q2", "question": "In the figure shown $C_{1} = 11\\mu F$ and $C_{2} = 5\\mu F$ then\nat steady state", "question_images": ["images/image2.png"], "option_1": "The potential difference across C_1 is 5V", "option_2": "The potential difference between terminals of 15V battery is 9V", "option_3": "The potential difference V_a - V_b = - 4V", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] at steady state\n$I(3) + 1(2) = 15$\nI = 3\n$kvL$\n$C \\rightarrow D \\rightarrow E \\rightarrow a \\rightarrow b \\rightarrow C$\n$V_{c} - I(3) + \\frac{q}{11} - 7 + \\frac{q}{5} = V_{c}$\n$\\frac{q}{11} + \\frac{q}{5} = 7 + 3 \\times 3 = 16$\n$q = 55\\mu C$\nNow KVL for a$\\rightarrow$b\n$V_{a} - 7 + \\frac{q}{5} = V_{b}$\n$\\Rightarrow V_{a} - V_{b} = 7 - \\frac{q}{5}$\n$= 7 - \\frac{55}{5}$\n$= - \\frac{20}{5} = - 4V$", "solution_images": ["images/image3.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-09-Q8", "question": "A car moves towards a hill with speed c v. It blows a horn of\nfrequency f which is heard by an observer following the car with speed\nofv. The speed of sound in air is v. then,", "question_images": [], "option_1": "the wavelength of sound reaching the hill is [IMAGE]", "option_2": "the wavelength of sound reaching the hill is [IMAGE]", "option_3": "the beat frequency observed by the observe [IMAGE]", "option_4": "the beat frequency observed by the observer is [IMAGE]", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] Beat frequency head by observer = f′′−f′′′", "solution_images": ["images/image42.png", "images/image43.png", "images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png", "images/image48.png"], "subject": "Physics", "topic": "Sound wave", "subtopic": "Doppler effect", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-11-Q22", "question": "The time period of a satellite of earth is 5 hours.If the\nseparation between the centre of earth and the satellite is increased to\n4 times the previous value, the new time period in hours will become\n10n. Find the value of n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "For satellite", "solution_images": ["images/image216.png", "images/image217.png"], "subject": "Physics", "topic": "Gravitation", "subtopic": "Kepler's laws", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-09-Q16", "question": "A particle is performing small oscillation about X & Y axis as\n[IMAGE] then path of particle will look like", "question_images": ["images/image93.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these", "correct_option": 1, "numerical_answer": null, "solution": "$x = \\sin wt;y = \\sin 3wt;w = \\frac{\\pi}{6}$", "solution_images": ["images/image97.png", "images/image98.png"], "subject": "Physics", "topic": "Oscillation", "subtopic": "Position of particle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-04-Q1", "question": "If equation[IMAGE] = [IMAGE] is correct.\nFind the value of X.", "question_images": ["images/image1.png", "images/image2.png"], "option_1": "Zero", "option_2": "1", "option_3": "2", "option_4": "- 1", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-05-Q24", "question": "A plane is flying with an air speed 10 m/s toward north but\nsuddenly encounters a wind of 10 m/ s at 30º north of east. If angle\nmade by new direction of velocity of plane with respect to ground from\nnorth direction is $\\frac{\\pi}{n}$ then find the value of n", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "The angle θ = $\\pi/6$", "solution_images": ["images/image20.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "Plane problem", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-13-Q16", "question": "Two concentric rings of radius r and R are in same plane such\nthat R >> r. Current Ι is maintained in thering of radius r. Flux\nassociated to the ring of radius R due to this current is (use proper\napproximation).", "question_images": [], "option_1": "$\\frac{\\pi\\mu_{0}{Ir}^{2}}{2R}$", "option_2": "$\\frac{\\pi\\mu_{0}Ir}{2}$", "option_3": "$\\frac{\\pi\\mu_{0}{IR}^{2}}{2r}$", "option_4": "$\\frac{\\pi\\mu_{0}IR}{2}$", "correct_option": 1, "numerical_answer": null, "solution": "If current I is maintained in ring of radius R the flux\nassociated with ring of radius r will be:$\\frac{\\pi\\mu_{0}{Ir}^{2}}{2R}$\n[IMAGE] Since M_12 = M_21 so same flux would be associated to the ring of\nradius R if I is maintained in ring ofradius r.", "solution_images": ["images/image18.png"], "subject": "Physics", "topic": "EMI", "subtopic": "Mutual induction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-03-Q11", "question": "Find the acceleration of each block.", "question_images": ["images/image59.png"], "option_1": "", "option_2": "", "option_3": "zero", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Tendency of 2 kg block is of coming down and of 1 kg is of going\nup.\n[IMAGE] Where F1 is friction between 1 kg and 2 kg block and F2 is friction\nbetween 2 kg and inclined surface.\nFor 2 kg block 2 × 10 ×[IMAGE] - T - F1 - F2 = 2a\nFor 1 kg block T - 1 × 10 ×[IMAGE] × F1 = a\n⇒ Add both 10 - 5 - 2F1 - F2 = 3a\n5 - 2F1 - F2 = 3a\nThe max. value F1 = 0.1 × 1 × 10[IMAGE] F2max = 3 × 10 × 0.2 ×[IMAGE] Max (2F1 + F2) =[IMAGE] ⇒ Static friction will act. ⇒ a = 0\nTOPIC: NLM\nSUB TOPIC: FRICTION", "solution_images": ["images/image63.png", "images/image64.png", "images/image65.png", "images/image66.png", "images/image67.png", "images/image68.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-07-Q10", "question": "A particle of mass 1 kg is thrown vertically upwards with speed\n100 m/s. After 5s it explodes into twoparts. One part of mass 400g comes\nback with speed 25 m/s, what is the speed of other part just\nafterexplosion?", "question_images": [], "option_1": "100 m/s upwards", "option_2": "600 m/s upwards", "option_3": "100 m/s downwards", "option_4": "300 m/s upwards", "correct_option": 1, "numerical_answer": null, "solution": "Velocity of particle after 5 s\nv = u - gt\nv = 100 - 10 × 5\n= 100 - 50 = 50 m/s (upwards)\nConservation of linear momentum gives\nMv = m1v1 + m2v2......(i)\nTaking upward direction positive, the velocity v_1 will be negative.\n$\\therefore v_{1} = - 25m/s,v = 50m/s$\n$\\ $ Also M=1kg,m_1=400g=0.4kg\nand m_2=(M - m_1)=1 - 0.4=0.6kg\nThus,Eq.(i) becomes,\n$1*50 = 0.4 \\times ( - 25) + 0.6v_{2}$\nor$\\ 5$``0= - 10+0.6v_2\nor 0.6v_2=60\nor$v_{2} = \\frac{60}{0.6} = 100m/s$\nAs v_2 is positive, therefore the other part will move upwards with a\nvelocity 100 m/s.", "solution_images": [], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Conservation of linear momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-20-Q16", "question": "An object is gradually moving away from the focal point of a\nconcave mirror along the axis of the mirror. The graphical\nrepresentation of the magnitude of linear magnification (m) versus\ndistance of the object from the mirror (x) is correctly given by (Graphs\nare drawn schematically and are not to scale)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "At focus, magnification is[IMAGE].", "solution_images": ["images/image57.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Linear magnification", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-01-Q22", "question": "Gravitational force between two identical uniform solid gold\nspheres of radius r each in contact is proportional to n power of r.\nThen value of n is.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "04", "solution": "[IMAGE] F = [IMAGE] m = ρ × [IMAGE] πr3\nF = Gρ2 × [IMAGE] × [IMAGE] ∝ r4", "solution_images": ["images/image92.png", "images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "Ph-25-Q24", "question": "A Carnot engine operates between two reservoirs of temperatures\n900 K and 300 K. The engineperforms 1200 J of work per cycle. The heat\nenergy (in J) delivered by the engine to the lowtemperature reservoir,\nin a cycle, is 100n. Find n,d dkuksZ bUtu dks 900 K vkSj 300 K ds nks HkaMkjks a ds chp pyk;k\nfuEu rki okys HkaMkj", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image171.png", "images/image172.png", "images/image173.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Carnot engine", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-20-Q8", "question": "[IMAGE] A and d are the plate area and distance between plates of a parallel\nplate capacitor repectively. The dielectric constant of dielectric\nfilled inside a parallel plate capacitor is variable and given as\n$k(x) = K(1 + \\gamma x$) where is the distancemeasured from the\nleft plate as shown in the figure. If$(\\gamma d) < < 1$, then find the\ncapacitance of capacitor with dielectric.", "question_images": ["images/image20.png"], "option_1": "$\\frac{A\\epsilon_{0}K}{d}\\left( 1 + \\frac{\\gamma^{2}d^{2}}{2} \\right)$", "option_2": "$\\frac{AK \\in_{0}}{d}\\left( 1 + \\frac{\\gamma d}{2} \\right)$", "option_3": "$\\frac{A \\in_{0}K}{d}\\left( 1 + \\left( \\frac{\\gamma d}{2} \\right)^{2} \\right)$", "option_4": "$\\frac{AK\\epsilon_{0}}{d}(1 + \\gamma d)$", "correct_option": 2, "numerical_answer": null, "solution": "Capacitance of element$= \\frac{k\\varepsilon_{0}A}{dx}$\n$\\sum\\frac{1}{C^{'}} = \\int_{0}^{d}\\mspace{2mu}\\frac{dx}{K\\varepsilon_{0}A(1 + \\gamma x)}$\n$\\frac{1}{C} = \\frac{1}{K\\varepsilon_{0}A\\alpha}ln(1 + \\gamma d)$\n$${\\ Given\\ \\ \\gamma d < < 1\n$$C = \\right)$$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Equivalent capacitance", "difficulty": "Difficult", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-15-Q18", "question": "[IMAGE] is maximum is", "question_images": ["images/image140.png", "images/image141.png", "images/image142.png", "images/image143.png", "images/image144.png"], "option_1": "at no time this is possible", "option_2": "2", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "$\\frac{dN_{2}}{dt} = 0\\text{\\quad} \\Rightarrow \\text{\\quad}\\lambda N_{1} = 2\\lambda N_{2}\\text{\\quad}or\\frac{N_{1}}{N_{2}} = 2$", "solution_images": ["images/image146.png", "images/image147.png"], "subject": "Physics", "topic": "Nuclear physics", "subtopic": "Radioactivity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-02-Q51", "question": "Photons are emitted at the rate of 1.6 × 10^16 per second by a\nlaser source operating at the wavelength of 632.8 nm. The power of\nsource is ( in mW )", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "05", "solution": "Energy of each photon [IMAGE] = 3.14 ´\n10^-19 J\nTotal energy emitted in one second = 3.14 × 10^-19 × 1.6 × 10^16 = 5\nmW\nPower = Energy emitted per second = 5 mW", "solution_images": ["images/image100.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-02-Q20", "question": "has four choices", "question_images": [], "option_1": ",", "option_2": ",", "option_3": ",", "option_4": "out of which ONLY ONE is correct]", "correct_option": null, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-23-Q26", "question": "A uniform magnetic field B in positive z direction exists in a\ncircular region of radius R = 5 m. A loop of radius R = 5m lying in x -\ny plane encloses the magnetic field at t = 0 and then pulled at uniform\nvelocity[IMAGE]. The emf induced (in volts) is the loop\nat t = 2 sec is 6V. Then magnitude of 4B is", "question_images": ["images/image80.png", "images/image81.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "[IMAGE] TOPIC:Electromagnetic induction\nSUB TOPIC:Faraday's law\nLEVEL:Moderate", "solution_images": ["images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png", "images/image86.png", "images/image87.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "Ph-26-Q30", "question": "Two liquids of densities [IMAGE] are filled up\nbehind a square wall of side 10m as shown in figure. Each liquid has a\nheight of 5 m. The ratio of the forces due to these liquids exerted on\nupper part MN to that at the lower part NO is 1/n. Find n. (Assume that\nthe liquids are not mixing):\nfHkUu ?kuRoksa [IMAGE] rFkk nks\næo 10m yEckbZ dh,d oxkZdkj nhokj ds ihNs Hkjs gq, gSa ¼fp= ns[ksa½A\nçR;sd æo dh 5 m gSA rc bu æoksa }kjk nhokj ds Hkkx MN rFkk\nfupys Hkkx NO ij yxkus okys cyksa dk vuqikr 1/n gksxkrks n dk eku Kkr\ndjksaA æo fefJr ugha gksrs gSa½", "question_images": ["images/image123.png", "images/image124.png", "images/image125.png", "images/image126.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "", "solution_images": ["images/image127.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-10-Q23", "question": "The moment of inertia of a thin uniform square sheet of mass M\nand side a about the axis AB which is in the plane of sheet is\n$\\frac{nMa^{2}}{12}$. Find n", "question_images": ["images/image71.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "7", "solution": "$I = \\frac{{ma}^{2}}{12} + m\\left( \\frac{a}{\\sqrt{2}} \\right)^{2} = \\frac{7ma^{2}}{12}$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-16-Q7", "question": "Initially spring is in natural length and system is at rest.\nSystem is released when a particle of mass m collides vertically to\nblock B with velocity v and sticks to it. Find maximum extension in the\nspring. (Pulley and string are ideal). (Given that\n$m = 1kg,v = 10\\sqrt{2}m/sec$ and spring constant$k = 10N/m$)\n($g = 10m/s^{2})$", "question_images": ["images/image15.png"], "option_1": "$(2 + \\sqrt{19})m$", "option_2": "5m", "option_3": "1m", "option_4": "None of these", "correct_option": 2, "numerical_answer": null, "solution": "Just after collision velocity of (block B+\nparticle)$= \\frac{v}{2}$\nAt the time of maximum extension both A and B block will have same\nvelocity (say v')\nConserving angular momentum about centre of pulley of system of (A+B+\nparticle)\n$v^{'} = \\frac{v}{4}$\nBy energy conservation,\n$\\frac{1}{2} \\times 2m \\times {(\\frac{v}{2})}^{2} = 2 \\times \\frac{1}{2} \\times 2m \\times {(\\frac{v}{4})}^{2} + \\frac{1}{2} \\times 10 \\times A^{2} - 2mgA$\nOn solving,\nA=5m", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-13-Q14", "question": "In an electromagnetic wave, the electric field oscillates\nsinusoidal with amplitude 48 Vm^-1, the RMSvalue of oscillating\nmagnetic field will be nearly equal to", "question_images": [], "option_1": "1.6 × 10^-8T", "option_2": "16 × 10^-9 T", "option_3": "144 × 10^8T", "option_4": "11.3 × 10^-8T", "correct_option": 4, "numerical_answer": null, "solution": "$B_{0} = \\frac{E_{0}}{c} = \\frac{48}{3 \\times 10^{8}} = 16 \\times 10^{- 8}T$\n$B_{rms} = \\frac{B_{0}}{\\sqrt{2}} = \\frac{16 \\times 10^{- 8}}{\\sqrt{2}} = 8\\sqrt{2} \\times 10^{- 8}T$", "solution_images": [], "subject": "Physics", "topic": "EM wave", "subtopic": "RMS value of oscillating magnetic field", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-14-Q17", "question": "If we need a magnification of 375 from a compound microscope of\ntube length 150 mm and an objective of focal length 5 mm, the focal\nlength of the eye-piece, should be close to", "question_images": [], "option_1": "2 mm", "option_2": "33 mm", "option_3": "12 mm", "option_4": "22 mm", "correct_option": 4, "numerical_answer": null, "solution": "Case I\nIf final image is at least distance of clear vision\nIf final image is at infinity", "solution_images": ["images/image115.png", "images/image116.png", "images/image117.png", "images/image118.png", "images/image119.png", "images/image120.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Optical instruments", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-18-Q3", "question": "A plank having mass M is placed on smooth horizontal surface.\nBlock of mass m is placed on it coefficient of friction between block\nand plank is $\\mu_{0} + kx$,where k is constant and x is relative\ndisplacement of block w.r.t. plank. A force F is applied on block where\nF=at, where a=10; t is in second. Find$t_{0}$ to when relative motion\nwill occur between block and plank (use $g = 10m/s^{2}$).", "question_images": ["images/image2.png"], "option_1": "$\\mu_{0}M + \\frac{\\mu_{0}M^{2}}{m}$", "option_2": "$\\mu_{0}m + \\frac{\\mu_{0}M^{2}}{m}$", "option_3": "$\\mu_{0}m + \\frac{\\mu_{0}m^{2}}{M}$", "option_4": "$\\mu_{0}M + \\frac{\\mu_{0}m^{2}}{M}$", "correct_option": 3, "numerical_answer": null, "solution": "Let at time t_o relative motion will occur.\n[IMAGE] $\\mu_{0}mg = Ma$..... (1)\n[IMAGE] $10t_{0} - \\mu_{0}mg = ma$..... (2)\nFrom (1) and (2);\n$t_{0} = \\mu_{0}m + \\frac{\\mu_{0}m^{2}}{M}$", "solution_images": ["images/image3.png", "images/image4.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-26-Q32", "question": "The surface tension of a liquid is 5 Newton per metre. If a film\nof this liquid is held on a ring of area 0.02 m^2, if its surface\nenergy is about ([IMAGE] ) Joule then n is:\nesa bl æo dh fQYe mifLFkr gSrks bldh i`\"Bh; \n([IMAGE] ) gks rks n", "question_images": ["images/image132.png", "images/image132.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "We know that surface energy\n[IMAGE] Here, as 2 films are formed because of ring.\n[IMAGE] ge tkurs gS fd i`\"B;gk¡ oy", "solution_images": ["images/image133.png", "images/image134.png", "images/image135.png", "images/image136.png", "images/image137.png", "images/image138.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-20-Q2", "question": "A thin walled hollow cylinder is closed from both the ends. The\nmass of the curved part is m and the mass of each circular part is also\nm. The radius of the cylinder is R and its length is 2R. What will be\nthe moment of inertia of this hollow cylinder about the axis,\nperpendicular to its length and passing through its centre of mass?", "question_images": ["images/image3.png"], "option_1": "$\\frac{10}{3}mR^{2}$", "option_2": "$\\frac{5}{3}mR^{2}$", "option_3": "$\\frac{5}{6}mR^{2}$", "option_4": "$\\frac{11}{3}mR^{2}$", "correct_option": 1, "numerical_answer": null, "solution": "First find I of the hollow tube\n$I_{1} = \\int\\frac{{dmR}^{2}}{2} + \\int_{x = - \\frac{\\mathcal{l}}{2}}^{x = + \\frac{\\mathcal{l}}{2}}{(\\frac{m}{\\mathcal{l}}dx)}x^{2}$\n$I_{1} = \\frac{{mR}^{2}}{2} + \\frac{m^{\\mathcal{l}^{2}}}{12}$ where\n$\\mathcal{l} = 2R$\nSo $I_{1} = \\frac{5}{6}{mR}^{2}$\nNow moment of inertias of each circular plate about that axis.\n$I = \\frac{mR^{2}}{4} + mR^{2} = \\frac{5}{4}mR^{2}$\n$I_{total} = \\frac{5}{6}{mR}^{2} + 2 \\times \\left( \\frac{5}{4}{mR}^{2} \\right)$\n$I_{total} = \\frac{10}{3}{mR}^{2}$", "solution_images": ["images/image4.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-04-Q23", "question": "At t = 2 sec, a projectile is moving upwards. Its speed is 20 m/s\nand it is making at angle 30° with horizontal. What is angle of\nprojection of the projectile from ground ( in degree ) ?\nt = 2 sec 20 m/sec \n 30° ( )", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "60", "solution": "[IMAGE] u cos θ = 20 cos 30°\nvy = uy + ayt\n20 sin 30° = u sin θ - g × 2\nu sin θ = 30 m/s\ntan θ = [IMAGE] TOPIC:KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image122.png", "images/image123.png", "images/image124.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-12-Q24", "question": "A 60 pF capacitor is fully charged by a 20V supply. It is then\ndisconnected from the supply and isconnected to another uncharged 60 pF\ncapacitor in parallel. The electrostatic energy that is lost in\nthisprocess by the time the charge is redistributed between them is (in\nnJ)________.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "[IMAGE] V_0 = 20 V\nHeat loss = Ui - Uf\n$= \\frac{1}{2}CV_{0}^{2} - 2\\left\\lbrack \\frac{1}{2}C\\left( \\frac{V_{0}}{2} \\right)^{2} \\right\\rbrack$\n$= \\frac{CV_{0}^{2}}{4}$\n$= \\frac{\\left( 60 \\times 10^{- 12} \\right)(20)^{2}}{4}J$\n= 6 × 10^-9 J = 6 nJ", "solution_images": ["images/image38.png"], "subject": "Physics", "topic": "Capacitance", "subtopic": "Energy stored in capacitor", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-06-Q9", "question": "In which of the following cases the magnitude of acceleration of\nthe block A will be maximum (Neglect friction and mass of pulley and\nstring)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "(i) $a = \\frac{2mg - mg}{3m} = \\frac{g}{3}$ (ii)\n$a = \\frac{2mg - mg}{m} = g$\n(iii) $a = \\frac{2mg}{m} = 2g$ (iv) $a = \\frac{2g}{3}$", "solution_images": [], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Newton's second law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-06-Q4", "question": "A force\n$\\overrightarrow{F} = \\left( 3t\\widehat{i} + 5\\widehat{j} \\right)N$ acts\non a body due to which its position varies as\n$\\overrightarrow{s} = \\left( 2t^{2}\\widehat{i} - 5\\widehat{j} \\right)$.\nWork done by this force in first two seconds is", "question_images": [], "option_1": "23 J", "option_2": "32 J", "option_3": "zero", "option_4": "can't be obtained", "correct_option": 2, "numerical_answer": null, "solution": "$W = \\int\\overline{F} \\cdot d\\overline{s} = \\int(3t\\ \\widehat{i} + 5\\widehat{j}).(4t\\ dt\\ \\widehat{i})$\n$= \\int_{0}^{2}\\mspace{2mu} 12t^{2}dt = \\frac{12\\left\\lbrack t^{3} \\right\\rbrack_{0}^{2}}{3} = 32\\ J$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work done by variable force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-19-Q24", "question": "A point object in air is in front of the curved surface of a\nplano-convex lens. The radius of curvature of the curved surface is 30\ncm and the refractive index of the lens material is 1.5, then the focal\nlength of the lens (in cm) is10n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "$\\frac{1}{f} = (\\mu - 1)(\\frac{1}{R_{1}} - \\frac{1}{R_{2}})$\n$R_{1} = \\infty$ $R_{2} = - 30cm$\n$\\frac{1}{f} = (1.5 - 1)(\\frac{1}{\\infty} - \\frac{1}{- 30})$\n$\\frac{1}{f} = \\frac{0.5}{30}$ $f = 60cm$", "solution_images": [], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Lens Maker's formula", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-16-Q12", "question": "Two inclined planes OA and OB of inclinations to the horizontal\n$\\alpha\\ and\\ \\beta$, each equal to 30º are placed as shown in the\nfigure. A particle is projected at an angle of 90º with plane OA from\npoint A and its strikes the plane OB at point B normally. If the speed\nof projection in m/s is x. Fill the value of 8 x. (given that OA=OB=20\ncm and $g = 10m/s^{2})$", "question_images": ["images/image22.png"], "option_1": "16", "option_2": "18", "option_3": "12", "option_4": "15", "correct_option": 1, "numerical_answer": null, "solution": "Since $\\alpha = \\beta$\n$AO = BO = 20cm = d$\n$cos120^{\\circ} = \\frac{d^{2} + d^{2} - (AB)^{2}}{2d^{2}}$\n$AB = \\sqrt{3}d$\nNow ABis the range of projectile from A to B\n$\\sqrt{3}d = \\frac{u^{2}sin2\\theta}{g}$ $u = 2m/s$", "solution_images": [], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Projectile on an inclined plane", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-15-Q21", "question": "The maximum peak to peak voltage of an AM wave is\n[IMAGE] and the minimum peak to peak voltage\nis[IMAGE]. The modulation factor is 10x. Find x.", "question_images": ["images/image161.png", "images/image162.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "", "solution_images": ["images/image163.png", "images/image164.png"], "subject": "Physics", "topic": "Communication system", "subtopic": "Modulation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-03-Q2", "question": "If in a system force (F), pressure (P) and time (T) is\nfundamental unit, then dimensional formulae for length will be", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these\n (F), (P) (T)", "correct_option": 3, "numerical_answer": null, "solution": "x + y = 0 x - y = 1 - 2x - 2y + z = 0\nx = [IMAGE] y = - [IMAGE] z = 0\nTOPIC: UNIT & DIMENSION\nSUB TOPIC: DIMENSIONS", "solution_images": ["images/image6.png", "images/image7.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-24-Q4", "question": "A string is holding a solid block below the surface of the liquid\nas shown in figure. If the system is given an upward acceleration a,\nthen as compared to previous state, choose the correct option(s)", "question_images": ["images/image28.png"], "option_1": "Tension in string will be$\\left( 1 + \\frac{a}{g} \\right)$ times", "option_2": "Tension in string will be $\\left( 1 - \\frac{a}{g} \\right)$ times", "option_3": "Upthrust force on block become 2$\\left( 1 + \\frac{a}{g} \\right)$\ntimes", "option_4": "Upthrust force on block becomes $\\left( 1 - \\frac{a}{g} \\right)$\ntimes", "correct_option": 1, "numerical_answer": null, "solution": "Let σ is density of liquid &ρ is density at object\n$B_{i} = \\sigma vg\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ B_{f} = \\sigma v(g + a) = B_{i}\\left( \\frac{g + a}{g} \\right)$\n$T_{i} = (\\sigma - \\rho)vg\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ T_{f} = (\\sigma - \\rho)v(g + a) = T_{i}\\left( \\frac{g + a}{g} \\right)$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Buoyant force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-17-Q4", "question": "A plane electromagnetic wave of wavelength $\\\\lambda\\$ has an\nintensity I. It is propagating along the positive Y-direction. The\nallowed expressions for the electric and magnetic fields are given by", "question_images": [], "option_1": "$\\overset{\\rightarrow}{E} = \\sqrt{\\frac{2I}{\\varepsilon_{0}c}}cos\\lbrack\\frac{2\\pi}{\\lambda}(y - ct)\\rbrack\\;\\ \\overset{\\rightarrow}{B} = + \\frac{1}{c}E\\overset{\\hat{}}{i}$", "option_2": "$\\overset{\\rightarrow}{E} = \\sqrt{\\frac{I}{\\varepsilon_{0}c}}cos\\lbrack\\frac{2\\pi}{\\lambda}(y - ct)\\rbrack\\overset{\\hat{}}{k}\\;\\ \\overset{\\rightarrow}{B} = + \\frac{1}{c}E\\overset{\\hat{}}{i}$", "option_3": "$\\overset{\\rightarrow}{E} = \\sqrt{\\frac{2I}{\\varepsilon_{0}c}}cos\\lbrack\\frac{2\\pi}{\\lambda}(y - ct)\\rbrack\\overset{\\hat{}}{k}\\;\\ \\overset{\\rightarrow}{B} = - \\frac{1}{c}E\\overset{\\hat{}}{i}$", "option_4": "$\\overset{\\rightarrow}{E} = \\sqrt{\\frac{I}{\\varepsilon_{0}c}}cos\\lbrack\\frac{2\\pi}{\\lambda}(y - ct)\\rbrack\\;\\ \\overset{\\rightarrow}{B} = \\frac{1}{c}E$", "correct_option": 1, "numerical_answer": null, "solution": "If E_0 is magnitude of electric field then\n$\\frac{1}{2}\\varepsilon_{0}E^{2} \\times C = I$\n$E_{0} = \\sqrt{\\frac{2I}{C\\varepsilon_{0}}}$\n$B_{0} = \\frac{E_{0}}{C}$\nDirection of $\\overset{\\rightarrow}{E} \\times \\overset{\\rightarrow}{B}$\nwill be along$+ \\overset{\\hat{}}{j}$.", "solution_images": [], "subject": "Physics", "topic": "E-M waves", "subtopic": "Electric and magnetic field", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-03-Q13", "question": "A block of mass m is at rest on an incline having an angle of\ninclination θ with the horizontal. The incline is now being rotated with\nconstant angular velocity ω about an axis passing through the\nperpendicular side. If the surface of the incline is rough having\ncoefficient of friction as µ, then find the distance along the\nincline so that block just remains stationary with respect to incline\nand is on the verge of slipping.\n from the top of the incline as shown in the figure.]", "question_images": ["images/image70.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE] m θ \n ω µ \n \n )", "correct_option": 3, "numerical_answer": null, "solution": "N + mrω2 sin θ = mg cos θ\nmg sin θ + mrω2 cos θ = Fdown\nµ (mg cos θ - mrω2 sin θ) = Fup\nFdown = Fup\nr =[IMAGE] x cos θ = r\nx =[IMAGE] =[IMAGE] =[IMAGE] TOPIC: CIRCULAR MOTION\nSUB TOPIC: DYNAMICS", "solution_images": ["images/image76.png", "images/image77.png", "images/image78.png", "images/image79.png", "images/image80.png", "images/image81.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-03-Q1", "question": "In new system of unit 1 unit mass is 5 kg 1 unit length is 20\nmeter and 1 unit time is 2 second then 1 unit force in new system is\nhow much newton ?", "question_images": [], "option_1": "100 N", "option_2": "50 N", "option_3": "30 N", "option_4": "25 N\n 1 5 kg 1 20 1\n 2 1", "correct_option": 4, "numerical_answer": null, "solution": "1 unit of mass = 5kg\n1 unit of length = 20 m\n1 unit of time = 2 sec\n1 unit of F = [IMAGE] = [IMAGE] = 25 N\nTOPIC: UNIT & DIMENSION\nSUB TOPIC: DIMENSIONS", "solution_images": ["images/image1.png", "images/image2.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-14-Q5", "question": "A reflecting surface is represented by the equation\n[IMAGE] A ray travelling horizontally becomes vertical after reflection. The\nco-ordinates of points where this ray incident is", "question_images": ["images/image20.png", "images/image21.png"], "option_1": "$\\left( \\frac{L}{4},\\frac{\\sqrt{2}L}{\\pi} \\right)$", "option_2": "$\\left( \\frac{L}{3},\\frac{\\sqrt{3}L}{2\\pi} \\right)$", "option_3": "$\\left( \\frac{3L}{4},\\frac{\\sqrt{2}L}{\\pi} \\right)$", "option_4": "$\\left( \\frac{2L}{3},\\frac{\\sqrt{3}L}{\\pi} \\right)$", "correct_option": 4, "numerical_answer": null, "solution": "The slope of curve at such points is", "solution_images": ["images/image22.png", "images/image23.png", "images/image24.png", "images/image25.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Reflection (mirror)", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-17-Q18", "question": "In the potentiometer circuit shown, AB is a $10\\Omega$ uniform\npotentiometer wire of length 50 cm, E_1 is 2V accumulator of negligible\ninternal resistance. R_1 and R_2 are $15\\Omega$ and 5$\\Omega$\nrespectively. When K_1 and K_2 are both open, the galvanometer shows\nno deflection when AJ is 31.25 cm. When K_1 and K_2 both are closed\nthe balance length AJ =5 cm. The emf of cell E_2is ____", "question_images": ["images/image18.png"], "option_1": "0.5 V", "option_2": "1 V", "option_3": "1.5 V", "option_4": "2 V", "correct_option": 1, "numerical_answer": null, "solution": "$V_{AB} = i10 = \\frac{20}{25}$\n$E_{2} = \\frac{V_{AB}}{50} \\times 31.25$\n$= \\frac{20}{25} \\times \\frac{31.25}{50}$\n$= 0.5V$", "solution_images": ["images/image19.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Potentiometer", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-11-Q23", "question": "One end of a straight uniform [IMAGE] long bar\nis pivoted on horizontal table. It is released from rest when itmakes an\nangle [IMAGE] from the horizontal (see figure). Its\nangular speed when it hits the table is given as$\\sqrt{3n}s^{- 1},$where\nn is an integer. The value of n is.......", "question_images": ["images/image218.png", "images/image219.png", "images/image220.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "", "solution_images": ["images/image221.png", "images/image222.png", "images/image223.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-14-Q7", "question": "In the shown figure three slits, separated by\n[IMAGE] and illuminated by monochromatic parallel beam of\nlight of wavelength[IMAGE]. P is a point on the line\nperpendicular to the plane of slits through[IMAGE]. If\n[IMAGE] is the intensity of the wave from each slit, the\nintensity at p is", "question_images": ["images/image37.png", "images/image38.png", "images/image39.png", "images/image40.png", "images/image41.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "Phase difference between\n[IMAGE] &[IMAGE] Phase difference between [IMAGE] &[IMAGE] Now using phase diagram", "solution_images": ["images/image46.png", "images/image47.png", "images/image48.png", "images/image49.png", "images/image50.png", "images/image51.png", "images/image52.png", "images/image53.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-17-Q13", "question": "An electron gun is placed inside a long solenoid of radius R on\nits axis. The solenoid has n turns/length and carries a current I. The\nelectron gun shoots an electron along the radius of the solenoid with\nspeed\nv. If the electron does not hit the surface of the solenoid, maximum\npossible value of v is (all symbols have their standard meaning)", "question_images": [], "option_1": "$\\frac{e\\mu_{0}nIR}{2m}$", "option_2": "$\\frac{e\\mu_{0}nIR}{m}$", "option_3": "$\\frac{2e\\mu_{0}nIR}{m}$", "option_4": "$\\frac{e\\mu_{0}nIR}{4m}$", "correct_option": 1, "numerical_answer": null, "solution": "$R_{\\max} = \\frac{R}{2} = \\frac{mv_{\\max}}{e\\mu_{0}in}$\n$V_{\\max} = \\frac{Re\\mu_{0}in}{2m}$", "solution_images": [], "subject": "Physics", "topic": "Moving charges and", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "Ph-27-Q29", "question": "A Carnot engine having an efficiency of$\\frac{1}{10}$is being\nused as a refrigerator. If the work done on therefrigerator is 10J, the\namount of heat absorbed from the reservoir at lower temperature is 10x.\nFind x.,d dkuksZa bUtu $\\frac{1}{10}$ bl s,d jsfÝtjsVj ds:i\n tk jgk\n10J gks rks fuEurki okys rkidq.M tkus okyh", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "9", "solution": "For Carnot engine using as refrigerator\ndkuksZV btau dks jsfÝtjsVj dh Hkkfr fy,\n$W = Q_{2}\\left( \\frac{T_{1}}{T_{2}} - 1 \\right)$\n$\\eta = 1 - \\frac{T_{2}}{T_{1}}$ $\\Rightarrow$\n$\\frac{T_{2}}{T_{1}} = \\frac{9}{10}$\nSo, vr Q_2 = 90 J (as W = 10 J)", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Carnot engine", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-13-Q24", "question": "The loss of energy in KWH in a specimen of Iron is\n$\\frac{n}{25}$. The hysteresis loop of which is equivalent in area equal\nto 250 Jm^-3 cycle^-1and frequency 50 min^-1 and specific gravity\nof iron is 7.5 and mass of iron is 3.6kg. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "9", "solution": "Hysteresis loss = 250 cycle\nNo. of cycle per hour = 50 ×60 = 3000\nHence total loss per hour = 250 × (volume) × (cycles)\n$= 250 \\times \\frac{3.6}{7.5 \\times 1000} \\times 3000$\n= 360 WH = 0.36 kWH", "solution_images": [], "subject": "Physics", "topic": "Magnetism", "subtopic": "Hysteresis loss", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-13-Q3", "question": "A plane electromagnetic wave is propagating along the\ndirection$\\frac{\\overset{\\hat{}}{\\mathbf{i}} + \\overset{\\hat{}}{\\mathbf{j}}}{\\sqrt{2}}$,\nwith its polarization along the direction$\\widehat{k}$. The correct form\nof the magnetic field of the wave would be (here B0 is an\nappropriateconstant)", "question_images": [], "option_1": "$B_{0}\\frac{\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}}{\\sqrt{2}}cos\\left( \\omega t - k\\frac{\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}}{\\sqrt{2}} \\right)$", "option_2": "$B_{0}\\frac{\\overset{\\hat{}}{i} - \\overset{\\hat{}}{j}}{\\sqrt{2}}cos\\left( \\omega t + k\\frac{\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}}{\\sqrt{2}} \\right)$", "option_3": "$B_{0}\\frac{\\overset{\\hat{}}{i} - \\overset{\\hat{}}{j}}{\\sqrt{2}}cos\\left( \\omega t - k\\frac{\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}}{\\sqrt{2}} \\right)$", "option_4": "$B_{0}\\overset{\\hat{}}{k}cos\\left( \\omega t - k\\frac{\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}}{\\sqrt{2}} \\right)$", "correct_option": 3, "numerical_answer": null, "solution": "EM wave is in\ndirection$\\ \\rightarrow \\frac{\\overset{\\hat{}}{i} + \\overset{\\hat{}}{j}}{\\sqrt{2}}$.\nElectric field is in direction$\\ \\rightarrow \\overset{\\hat{}}{k}$.\n$\\overset{\\rightarrow}{E} \\times \\overset{\\rightarrow}{B} \\rightarrow$\nDirection of propagation of EM wave\nTherefore, B is along\n$\\frac{\\overset{\\hat{}}{i} - \\overset{\\hat{}}{j}}{\\sqrt{2}}$.", "solution_images": [], "subject": "Physics", "topic": "EM wave", "subtopic": "Magnetic field form of the EM wave.", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-06-Q19", "question": "A uniform thick rope of length 5m is kept on frictionless\nsurface and a force of 5N is applied to one of its end. Find tension in\nthe rope at 1 m from this end", "question_images": [], "option_1": "1 N", "option_2": "3 N", "option_3": "4 N", "option_4": "5 N", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Equation of motion\n$$F - T = \\times a\\ldots.(i)$$\n$$T = \\times a\\ldots(ii)$$\nSolving (i) and (ii)\nT = 4 N\nTOPIC: Newton's second law\nSUB TOPIC: Tension in a rope", "solution_images": ["images/image27.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-15-Q1", "question": "Nuclei of radioactive element A are produced at rate \nat any time t. The element A has decay constant. Let N be the number of\nnuclei of element A at any time t. At time [IMAGE] is\nminimum. Then the number of nuclei of element A at time", "question_images": ["images/image1.png", "images/image2.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{dN}{dt} = t^{2} - \\lambda N$", "solution_images": ["images/image6.png", "images/image7.png", "images/image8.png", "images/image9.png"], "subject": "Physics", "topic": "Nuclear Physics", "subtopic": "Radioactivity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-11-Q18", "question": "Two semicircular rings of linear mass densities\neach are joined to form a complete ring. The distance of the center of\nthe mass of complete ring from its geometrical centre is", "question_images": ["images/image174.png", "images/image175.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "none of these", "correct_option": 2, "numerical_answer": null, "solution": "Let the two half rings be placed in left and right of y-axis\nwith centre as shown in figure.\n[IMAGE] Then the coordinate of the centre of mass of left and right half rings\nX-coordinates of the centre of mass of complete ring is", "solution_images": ["images/image179.png", "images/image180.png", "images/image181.png", "images/image182.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Position of COM", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "Ph-27-Q20", "question": "The figure shows the P-V plot of an ideal gas taken through a\ncycle DCBAD then\ndks pØ DCBAD", "question_images": ["images/image99.png"], "option_1": "maximum temperature will occur during the process C → B → A.", "option_2": "negative work is done by the gas in the cycle DCBAD.", "option_3": "heat flows out of the gas during C → B → A.", "option_4": "All of the above.", "correct_option": 4, "numerical_answer": null, "solution": "Since temperature at C and A are same so maximum temperature\noccurs at a state between them during process CBA.\nSince cycle in PV graph is anticlockwise so work done by gas in\nnegative.\nDuring CBA work done by gas is negative and [IMAGE] so\n[IMAGE]. So heat is released by gas.\nD;ksafdC rFkk A ij rkieku leku gS rks eku bu izØe CBA\nds nkSjku gksxkA D;ksfd PV vkjs[k okekorZ", "solution_images": ["images/image100.png", "images/image101.png", "images/image100.png", "images/image101.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Cyclic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-19-Q6", "question": "A capacitor is made of two square plates each of side \nmaking a very small angle $\\alpha$ between them, as shown in figure. The\ncapacitance will be close to", "question_images": ["images/image8.jpeg"], "option_1": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{2d})$", "option_2": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 + \\frac{\\alpha a}{2d})$", "option_3": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{4d})$", "option_4": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{3\\alpha a}{2d})$", "correct_option": 1, "numerical_answer": null, "solution": "$\\Rightarrow c = \\frac{\\varepsilon_{0}a}{\\alpha}\\lbrack ln(d + \\alpha x)\\rbrack_{0}^{a}$\n$= \\frac{\\varepsilon_{0}a}{\\alpha}In(1 + \\frac{\\alpha a}{d}) \\approx \\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{2d})$", "solution_images": ["images/image9.jpeg"], "subject": "Physics", "topic": "Capacitor", "subtopic": "Capacitance", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-06-Q10", "question": "Starting from rest, a body slides down a $45^{\\circ}$ inclined\nplane in twice the time it takes to slide down the same distance in the\nabsence of friction. The coefficient of friction between the body and\nthe inclined plane is", "question_images": [], "option_1": "0.33", "option_2": "0.25", "option_3": "0.75", "option_4": "0.80", "correct_option": 3, "numerical_answer": null, "solution": "$\\mu = tan\\theta\\left( 1 - \\frac{1}{n^{2}} \\right)$\n$\\theta = 45^{\\circ}$ and $n = 2$ (Given)\n$\\therefore\\mu = tan45^{\\circ}\\left( 1 - \\frac{1}{2^{2}} \\right) = 1 - \\frac{1}{4} = \\frac{3}{4} = 0.75$", "solution_images": [], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Inclined plane with friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-08-Q4", "question": "A system undergoes three quasi-static processes sequentially as\nindicated in the given figure. $1 \\rightarrow 2$is an isobaric process,\n$2 \\rightarrow 3$ is a polytrophic process with\n$\\gamma = \\frac{4}{3}$and $3 \\rightarrow 1$ is a process in which PV=\nconstant.\n$P_{2} = P_{1} = 4 \\times 10^{5}Nm^{- 2},P_{3} = 1 \\times 10^{5}Nm^{- 2}$and$\\ V_{1} = 1m^{3}$.\nThe heat transfer of the cycle is $\\Delta Q,$ change in internal\nenergy is $\\Delta U$ and the work done is $\\Delta W$. for polytropic\nprocess $PV^{\\gamma} =$ constant.\n(Take $\\mathcal{l}$n2 = 0.693). Then", "question_images": ["images/image2.png"], "option_1": "$\\Delta W = 0$", "option_2": "$\\Delta Q = 1.08 \\times 10^{5}J$", "option_3": "$\\Delta U = 0$", "option_4": "$\\Delta Q > \\Delta W$", "correct_option": 2, "numerical_answer": null, "solution": "For the process $3 \\rightarrow 1$\n$P_{1}V_{1} = P_{3}V_{3} \\Rightarrow V_{3} = \\frac{P_{1}V_{1}}{P_{3}} = \\frac{4 \\times 10^{5} \\times 1}{1 \\times 10^{5}} = 4m^{3}$\nFor the process $2 \\rightarrow 3$\n$P_{2}V_{2}^{\\gamma} = P_{3}V_{3}^{\\gamma} \\Rightarrow V_{2} = \\left( \\frac{P_{3}}{P_{2}} \\right)^{\\frac{1}{\\gamma}}V_{3} = 4\\left( \\frac{1}{4} \\right)^{\\frac{3}{4}} = 4^{\\frac{1}{4}} = \\sqrt{2}m^{3}$\nFor the process $1 \\rightarrow 2$\n$W_{12} = P_{1}\\left( V_{2} - V_{1} \\right) = \\left( \\sqrt{2} - 1 \\right)4 \\times 10^{5}J$\nFor the process $2 \\rightarrow 3$\n$W_{23} = \\frac{\\left( P_{2}V_{2} - P_{3}V_{3} \\right)}{(\\gamma - 1)} = 12(\\sqrt{2} - 1) \\times 10^{5}J$\nFor the process $3 \\rightarrow 1$,\n$w_{31} = - P_{1}V_{1}ln\\frac{V_{3}}{V_{1}} - 4 \\times 10^{5} \\times 1 \\times ln(4)$\nFor cyclic process, $\\Delta\\ U = 0$\n$\\therefore\\Delta Q = \\Delta U + \\Delta W = \\Delta W = 4 \\times 10^{5}(\\sqrt{2} - 1)$\n$1 \\rightarrow 2 \\rightarrow 3 \\rightarrow 1$\n$\\Delta Q = + 12 \\times 10^{5}(\\sqrt{2} - 1) - 4 \\times 10^{5} \\times 2 \\times 0.693$\n$= 1.08 \\times 10^{5}J$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic process", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-05-Q15", "question": "Rain is falling with constant and uniform velocity v at angle\n37º vertical. An observer starts from rest and moves with uniform\nacceleration $2m/s^{2}$ in horizontal direction (with the rain). After 6\nsec. velocity of rain drop again appear at angle 37º from vertical.\nVelocity of rain fall with respect to ground\nis:$\\left( tan37^{\\circ} \\right.\\ = 3/4)$", "question_images": ["images/image16.png"], "option_1": "$10m/s$", "option_2": "$20m/s$", "option_3": "$15m/s$", "option_4": "$30m/s$", "correct_option": 1, "numerical_answer": null, "solution": "w.r.t observer\nlet final velocity of rain be v'\ninitial velocity of\nrain$= \\frac{3v}{5}\\overset{\\hat{}}{i} - \\frac{4v}{5}\\overset{\\hat{}}{j}$\n[IMAGE] w.r.t observer.\nacceleration of rain$= - 2\\overset{\\hat{}}{j}m/s^{2}$\nfinal velocity of\nrain$= - \\frac{3v^{'}}{5}\\overset{\\hat{}}{i} - \\frac{4v^{'}}{5}\\overset{\\hat{}}{j}$\nin y- direction\n$V_{y\\ final\\ } = V_{y\\ initial\\ }$\n$v^{'} = v$", "solution_images": ["images/image17.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "Rain problem", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-21-Q16", "question": "An electric field\n$\\overset{\\rightarrow}{E} = 4x\\overset{\\hat{}}{i} - (y^{2} + 1)jN/C$\npasses through the box shown in figure. The flux of the electric field\nthrough surface ABCDand BCGF and marked as $\\phi_{1}$ and $\\phi_{2}$\nrespectively. The difference between\n$(\\phi_{2} - \\phi_{1})is\\ (in{Nm}^{2}/C)$\n[IMAGE] fudyrk gSA;fn cD ABCD rFkk BCGF leryksa", "question_images": ["images/image43.png", "images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png"], "option_1": "47", "option_2": "46", "option_3": "42", "option_4": "48", "correct_option": 4, "numerical_answer": null, "solution": "Flux via ABCD ls ¶yDl\n$\\phi_{1} = \\int\\overset{\\rightarrow}{E} \\cdot d\\overset{\\rightarrow}{A} = 0$\n[IMAGE] Flux via BCEF ls ¶yDl\n$= 16x,x = 3$\n$\\phi_{2} = 48\\frac{N - m^{2}}{C};\\phi_{2} - \\phi_{1} = 48\\frac{N - m^{2}}{C}$", "solution_images": ["images/image48.png", "images/image49.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Gauss's law", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-07-Q18", "question": "A bar of mass M & length L is in pure translatory motion with\nits centre of mass velocity V. It collides with and sticks to a second\nidentical bar which is initially at rest. (Assume that it becomes one\ncomposite bar of length 2 L). The angular velocity of the composite bar\nwill be", "question_images": ["images/image31.png", "images/image32.png"], "option_1": "", "option_2": "", "option_3": "3V/4L counter clockwise", "option_4": "V/L counter clockwise", "correct_option": 3, "numerical_answer": null, "solution": "Cons. of ang. momentum about P gives\n$MV\\frac{L}{2} = \\frac{(2M)(2L)^{2}}{12}w$\n$\\Rightarrow \\frac{V}{2} = \\frac{2Lw}{3}$\n$w = \\frac{3V}{4L}$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-17-Q21", "question": "In the arrangement shown in figure, wavelength of light used is\n$\\lambda$. The distance between slits S_1 and S_2 is $d(d < < D)$. The\ndistance between S_3 and S_4 is $u = \\frac{\\lambda D}{3d}$. If the\nratio of maximum to minimum intensity received on screen P is K. Then 10\nK is", "question_images": ["images/image23.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "90", "solution": "If intensity of light incident on slits S_1 and S_2 is\n$I_{0}$\nThe intensity through S_4 will be $I_{1} = 4I_{0}$\nThe intensity through S_3 will be\n$I_{2} = 4I_{0}\\cos^{2}(\\frac{\\phi}{2}) = I_{0}$\n$K = \\frac{I_{\\max}}{I_{\\min}} = \\frac{{(\\sqrt{4I_{0}} + \\sqrt{I_{0}})}^{2}}{{(\\sqrt{4I_{0}} - \\sqrt{I_{0}})}^{2}} = {(\\frac{3}{1})}^{2} = \\frac{9}{1}$", "solution_images": [], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-21-Q14", "question": "Three metallic plates are given and middle plate is given charge\nQ, as shown in the figure. The outer plates can be earthed with switches\nS1 and S2. The area of plates is same.The charge that will flow to earth\nwhen only switch S1 is closed and right most plate is uncharged is", "question_images": ["images/image35.png"], "option_1": "$- \\left( \\frac{Q}{2} \\right)$", "option_2": "$\\left( \\frac{Q}{2} \\right)$", "option_3": "Q", "option_4": "- Q", "correct_option": 3, "numerical_answer": null, "solution": "At point D,[IMAGE] must be zero because potential\nof left most plate is zero.\n[IMAGE] So, vr x = 0\n[IMAGE] (conductor pkyd)\nx = y + Q\ny = -Q", "solution_images": ["images/image38.png", "images/image38.png", "images/image39.png", "images/image40.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Capacitance", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-10-Q24", "question": "A uniform semi circular ring of radius R = 0.4m and mass m is\nfree to rotate in vertical plane about fixed horizontal axis passing\nthrough A as shown in the figure. The angular acceleration (in\nrad/sec^2) of ring immediately after it is released from rest from the\nposition", "question_images": ["images/image72.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "${mgR}\\left( 1 - \\frac{2}{\\pi} \\right) = I\\alpha$\n$I = {mR}^{2} - m\\left( \\frac{2R}{\\pi} \\right)^{2} + m\\left( R - \\frac{2R}{\\pi} \\right)^{2}$\n$I = 2{mR}^{2}\\left( 1 - \\frac{2}{\\pi} \\right)$\n${mgR}\\left( 1 - \\frac{2}{\\pi} \\right) = 2\\ m\\ R^{2}\\left( 1 - \\frac{2}{\\pi} \\right)\\alpha$\n$\\Rightarrow \\alpha = \\frac{g}{2R} = \\frac{10}{2 \\times 0.4} = 12.50rad/\\sec^{2}$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Torque", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-20-Q23", "question": "The magnifying power of a telescope with tube length 60 cm is5.\nThe focal length in cm of its eye piece is 2f. Find f?", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "", "solution_images": ["images/image81.png", "images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png", "images/image86.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Optical instruments", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-11-Q15", "question": "A particle of mass m is attached to an end of a uniform rod of\nmass [IMAGE] and length [IMAGE] which is\nsuspended through its mid point by an inextensible string as shown.\nInitially the rod is in horizontal position and at rest. The system is\nreleased from this position. Just after the release", "question_images": ["images/image149.png", "images/image150.png", "images/image151.png"], "option_1": "The angular acceleration of the system is [IMAGE]", "option_2": "The angular acceleration of the system is [IMAGE]", "option_3": "The Tension in the string is [IMAGE]", "option_4": "The Tension in the string is [IMAGE]", "correct_option": 3, "numerical_answer": null, "solution": "$\\frac{mgl}{2} = \\left( \\frac{2ml^{2}}{12} + \\frac{{ml}^{2}}{4} \\right)\\alpha$\n$\\alpha = \\frac{6\\text{g}}{5l}$\n$\\text{a}_{\\text{cm}} = \\frac{\\alpha l}{6} = \\frac{\\text{g}}{5}$", "solution_images": ["images/image156.png", "images/image157.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Torque", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-17-Q14", "question": "In a single slit diffraction of different light of wavelength\n$\\lambda$ by a slit of with t_1, the size of the central maximum on a\nscreen at a distance t_2 is", "question_images": ["images/image12.png"], "option_1": "$2t_{2}\\lambda + t_{1}$", "option_2": "$\\frac{2t_{2}\\lambda}{t_{1}}$", "option_3": "$\\frac{2t_{2}\\lambda}{t_{1}} + t_{1}$", "option_4": "$\\frac{2t_{2}\\lambda}{t_{1}} - t_{1}$", "correct_option": 3, "numerical_answer": null, "solution": "$t_{1}sin\\theta = \\lambda$\n$\\theta = \\lambda/t_{1}$\nWidth of central Max. $y = 2t_{2}\\theta + t_{1}$\n$y = \\frac{2t_{2}\\lambda}{t_{1}} + t_{1}$", "solution_images": [], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Diffraction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-21-Q6", "question": "A 60 pF capacitor is fully charged by a 20V supply. It is then\ndisconnected from the supply and isconnected to another uncharged 60 pF\ncapacitor in parallel. The electrostatic energy that is lost in\nthisprocess by the time the charge is redistributed between them is (in\nnJ)________.\n A rRi'pkr~ bls lzksr ls gVkdj 60 pF ds,d nwljs =k (parallel connection) esa tksM+k tc\n esa forfjr gks tk;s rks bl\nizfØ fLFkj oS|qr", "question_images": [], "option_1": "3", "option_2": "4", "option_3": "5", "option_4": "6", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] V_0 = 20 V\nHeat loss gkfu = U_i - U_f\n$= \\frac{1}{2}CV_{0}^{2} - 2\\left\\lbrack \\frac{1}{2}C\\left( \\frac{V_{0}}{2} \\right)^{2} \\right\\rbrack$\n$= \\frac{CV_{0}^{2}}{4}$\n$= \\frac{\\left( 60 \\times 10^{- 12} \\right)(20)^{2}}{4}J$\n= 6 × 10^-9 J = 6 nJ", "solution_images": ["images/image11.png"], "subject": "Physics", "topic": "Capacitance", "subtopic": "Energy stored in capacitor", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-12-Q3", "question": "The charge appearing on outer surface of external plates when\nS_1 and S_2 are open and external plates are uncharged.", "question_images": [], "option_1": "$- \\left( \\frac{Q}{2} \\right)$", "option_2": "$\\left( \\frac{Q}{2} \\right)$", "option_3": "Q", "option_4": "- Q", "correct_option": 2, "numerical_answer": null, "solution": "${\\overrightarrow{E}}_{A} = 0$ (Conductor)\n-x = x + Q + y - y\n2x = - Q\n$x = \\frac{- Q}{2}$\n$- x = \\frac{Q}{2}$", "solution_images": ["images/image4.png"], "subject": "Physics", "topic": "Capacitance", "subtopic": "Distribution of charge on plates", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-13-Q8", "question": "Consider a circular coil of wire carrying constant current I,\nforming a magnetic dipole. The magnetic flux through an infinite plane\nthat contains the circular coil and excluding the circular coil area is\ngiven by $\\phi_{i}$.The magnetic flux through the area of the circular\ncoil area is given by $\\phi_{0}$. Which of the following optionis\ncorrect?", "question_images": [], "option_1": "$\\phi > \\phi_{0}$", "option_2": "$\\phi_{1} < \\phi_{0}$", "option_3": "$\\phi_{i} = - \\phi_{0}$", "option_4": "$\\phi_{1} = \\phi_{0}$", "correct_option": 3, "numerical_answer": null, "solution": "As magnetic field lines always form a closed loop, hence every\nmagnetic field line creating magnetic flux in the inner region must be\npassing through the outer region. Since flux in two regions are in\nopposite direction,\n$\\therefore$ $\\phi_{i} = - \\phi_{0}$", "solution_images": [], "subject": "Physics", "topic": "Magnetic effect of current", "subtopic": "Magnetic flux", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-17-Q17", "question": "In the shown figure a ray of light enters the face XY of a right\nangered prism at grazing incident. It emerges from the adjacent face. YZ\nat an angle$\\ \\theta$. If x is the critical angle of the prism held in\nwater of refractive index $\\frac{4}{3}$, then the angle$\\ \\theta$ with\nnormal is", "question_images": ["images/image16.png"], "option_1": "$\\theta = \\sin^{- 1}\\lbrack tanx\\rbrack$", "option_2": "$\\theta = \\sin^{- 1}\\lbrack\\frac{4}{3}tanx\\rbrack$", "option_3": "$\\theta = \\sin^{- 1}\\lbrack\\frac{3}{4}cotx\\rbrack$", "option_4": "$\\theta = \\sin^{- 1}\\lbrack cotx\\rbrack$", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] As at XY\n$\\frac{4}{3}sin90 = \\mu sinx$\nAt YZ\n$\\mu cosx = \\frac{4}{3}sin\\theta$\nOn solving\n$\\theta = \\sin^{- 1}(cotx)$", "solution_images": ["images/image17.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Prism", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-01-Q4", "question": "A mixture of two moles of hydrogen and one mole of argon gas is\ntaken in a closed container at room temperature. Consider the following\ntwo statements\n(i) The average kinetic energy of each molecule of H2 and Ar are the\nsame.\n(ii) The partial pressure due to argon gas is more than that due to\nhydrogen gas.", "question_images": [], "option_1": "Both statement (i) and (ii) are correct", "option_2": "Statement (i) is correct while statement (ii) is incorrect", "option_3": "Both statement (i) and (ii) are incorrect", "option_4": "Statement (i) is incorrect while statement (ii) is correct", "correct_option": 3, "numerical_answer": null, "solution": "Kav = [IMAGE] kT \nKav = [IMAGE] kT \np1 = [IMAGE] × p\nn1 > n2 ⇒ p1 > p2", "solution_images": ["images/image19.png", "images/image20.png", "images/image21.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-02-Q41", "question": "A charge q is projected from origin with a velocity along\npositive x-axis in a region having uniform magnetic field directed\ntowards negative y-axis. If T is the time period of circular motion then\nthe velocity vector of charge q at some instant t where\n[IMAGE] may be", "question_images": ["images/image49.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] The charge executes circular motion in x-z plane with its center lying\non the negative z-axis. At time t where[IMAGE] the charge\nhas completed [IMAGE] th of the circle and so its velocity\nvector is pointing towards first quadrant in x-z plane.\n⇒ V_x is positive and V_z is positive and V_y is zero.", "solution_images": ["images/image54.png", "images/image55.png", "images/image56.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-15-Q25", "question": "Both the diodes used in the circuit shown are assumed to be\nideal and have negligible resistance when these are forward biased.\nBuilt in potential in each diode is[IMAGE]. For the\ninput voltages shown in the figure, the voltage (in volts) at point A is\n6x. Find x.", "question_images": ["images/image182.png", "images/image183.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "Let [IMAGE] Right diode is reversed biased and left diode is forward biased", "solution_images": ["images/image184.png", "images/image185.png"], "subject": "Physics", "topic": "Electronic device", "subtopic": "Diodes", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-18-Q5", "question": "In the arrangement shown in the figure there is friction only\nbetween the light string and the sharp spike at the point P. When the\nblocks are in motion, the tensions in the vertical and horizontal\nsections of the string are 4 N & 3 N respectively. The co-efficient of\nfriction between the string and the spike is equal to", "question_images": ["images/image6.png"], "option_1": "0.1", "option_2": "0.2", "option_3": "0.5", "option_4": "1", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] N = 5\non string\n$4 - 3 - \\mu N = 0$\n$\\mu N = 1$\n$\\mu = 0.2$", "solution_images": ["images/image7.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-22-Q23", "question": "A ball is given velocity [IMAGE] as shown. If the\nratio of centripetal acceleration totangential acceleration at the point\nwhere the ball leaves the circular path is [IMAGE] then n\nis", "question_images": ["images/image78.png", "images/image79.png", "images/image80.png"], "option_1": "and [IMAGE].....", "option_2": "[IMAGE] Now TOPIC: Circular motion\nSUB TOPIC: Vertical circle\nLEVEL: Moderate", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "and [IMAGE].....(2)\n[IMAGE] Now TOPIC: Circular motion\nSUB TOPIC: Vertical circle\nLEVEL: Moderate", "solution_images": ["images/image81.png", "images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png", "images/image86.png", "images/image87.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-06-Q18", "question": "A block is moved very slowly from A to D with the help of a\nvariable force on block which is always tangential to the surface of the\nfixed curve path. Coefficient of friction between block and surface is\nµ. (The block always remain in contact with plane). Work done by\nfriction in this process is", "question_images": ["images/image25.png"], "option_1": "$- \\mu mg\\mathcal{l}$", "option_2": "$- 2\\mu mg\\mathcal{l}$", "option_3": "$- 3\\mu mg\\mathcal{l}$", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Work done by\nfriction$= - \\int\\mu mgcos\\theta d\\mathcal{l = - \\int}\\mu mgdx = - 3\\mu mg\\mathcal{l}$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work done by a variable force", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-13-Q20", "question": "A conducting circular loop of copper is placed as shown in\nfigure. Cross section area of part abc is A and part adc is A/3 then\nmagnetic field at point is", "question_images": ["images/image23.png"], "option_1": "$\\frac{\\mu_{0}I}{8R} \\otimes$", "option_2": "$\\frac{3\\mu_{0}I}{11R} \\otimes$", "option_3": "$\\frac{\\mu_{0}I}{16R} \\odot$", "option_4": "Zero", "correct_option": 1, "numerical_answer": null, "solution": "$R = \\frac{\\rho l}{A} \\Rightarrow \\frac{R_{abc}}{R_{adc}} = \\left( \\frac{\\mathcal{l}}{A} \\right)_{abc} \\times \\left( \\frac{A}{\\mathcal{l}} \\right)_{adc}$\n$= \\frac{3/4\\mathcal{l}}{A} \\times \\frac{A/3}{\\mathcal{l/}4} = \\frac{1}{1}$\n$\\frac{R_{abc}}{R_{adc}} = \\frac{1}{1} \\Rightarrow I_{abc} = I_{adc} = \\frac{I}{2}$\n$B_{0} = \\frac{\\mu_{0}I_{abc}\\alpha_{1}}{4\\pi R} - \\frac{\\mu_{0}I_{adc}\\alpha_{2}}{4\\pi R}$\n$= \\frac{\\mu_{0}\\left( \\frac{I}{2} \\right)}{4\\pi R}\\left( \\alpha_{1} - \\alpha_{2} \\right) = \\frac{\\mu_{0}\\left( \\frac{I}{2} \\right)}{4\\pi R}\\left( \\frac{3\\pi}{2} - \\frac{\\pi}{2} \\right)$\n$= \\frac{\\mu_{0}I}{8R} \\otimes$", "solution_images": [], "subject": "Physics", "topic": "Magnetic effect of current", "subtopic": "Magnetic field due to current carrying conductor", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-14-Q9", "question": "A vessel of depth 2h is half filled with a liquid of refractive\nindex[IMAGE] and the upper half with another liquid of\nrefractive index[IMAGE]. The liquids are immiscible. The\napparent depth of the inner surface of the bottom of vessel will be", "question_images": ["images/image65.png", "images/image66.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image71.png", "images/image72.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Apparent shift", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "Ph-26-Q36", "question": "One mole of an ideal gas is kept enclosed under a light piston\n(area=[IMAGE] ) connected by acompressed spring (spring\nconstant 100 N / m). The volume of gas is [IMAGE] and\nits temperature is 100K. The gas is heated so that it compresses the\nspring further by 0.1 m. Then the work done (in Joule)by the gas in the\nprocess is 1.5n. Find n: (Take R =8.3 J / K -mole and suppose there is\nno atmosphere).,d 100 N / m½", "question_images": ["images/image154.png", "images/image155.png", "images/image154.png", "images/image155.png", "images/image156.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "Before heating let the pressure of gas be\n[IMAGE] from the equilibrium piston.\n -\n[IMAGE] Since during heating process, pwafd xeZ izfØ nkSjku,\nThe spring is compressed further by 0.1 m fLizax 0.1 m", "solution_images": ["images/image157.png", "images/image157.png", "images/image158.png", "images/image159.png", "images/image160.png", "images/image161.png", "images/image162.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Work done", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-08-Q17", "question": "Power radiated by a black body is P_0 and the wavelength\ncorresponding to the maximum energyis around$\\lambda_{0}$. On changing\nthe temperature of the black body, it is observed that the power\nradiatedis increased to$\\frac{256}{81}P_{0}$. The shift in the\nwavelength\ncorresponding to the maximum energy will be", "question_images": [], "option_1": "$+ \\frac{\\lambda_{0}}{4}$", "option_2": "$\\frac{\\lambda_{0}}{3}$", "option_3": "$–\\frac{\\lambda_{0}}{4}$", "option_4": "$- \\frac{\\lambda_{0}}{3}$", "correct_option": 3, "numerical_answer": null, "solution": "$P = \\sigma{AT}^{4}$\n$\\lambda_{m}T = b$\n$T = \\frac{b}{\\lambda_{m}} \\Rightarrow P = \\frac{\\sigma{Ab}^{4}}{\\lambda_{m}^{4}}$\n$\\Rightarrow P \\propto \\frac{1}{\\lambda_{m}^{4}} \\Rightarrow \\frac{P_{i}}{P_{2}} = {(\\frac{\\lambda_{m_{2}}}{\\lambda_{m_{4}}})}^{4}$\n$\\frac{P_{0}}{\\frac{256}{81}P_{0}} = {(\\frac{\\lambda_{m_{2}}}{\\lambda_{0}})}^{4} \\Rightarrow \\lambda_{m_{2}} = \\frac{3}{4}\\lambda_{0}$\n$\\Delta\\lambda = \\frac{3}{4}\\lambda_{0} - \\lambda_{0} = - \\frac{\\lambda_{0}}{4}$", "solution_images": [], "subject": "Physics", "topic": "Heat", "subtopic": "Stefan's law of radiation", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-22-Q15", "question": "A solid sphere of mass m and radius r is gently placed on a\nconveyer belt moving with constant velocityV. If the coefficient of\nfriction between the belt and sphere is$\\frac{2}{7}$, the distance\ntravelled by the centre of the sphere before it starts pure rolling is.", "question_images": ["images/image40.png"], "option_1": "$\\frac{v^{2}}{7g}$", "option_2": "$\\frac{2V^{2}}{49g}$", "option_3": "$\\frac{2V^{2}}{5g}$", "option_4": "$\\frac{2V^{2}}{7g}$", "correct_option": 1, "numerical_answer": null, "solution": "Initial situation w.r.t. the conveyer belt:\n$\\mu mgR = \\frac{2}{5}{mR}^{2} \\cdot \\alpha$ $\\Rightarrow$\n$\\alpha = \\frac{5}{7} \\cdot \\frac{g}{R}$\nAt pure rolling:\n$v^{'} = v - (\\mu g)t$and $\\omega^{'} = 0 + \\left( \\frac{5}{7} \\cdot \\frac{g}{R} \\right)t$\n(Since at pure rolling)\n$\\Rightarrow$ $v - \\mu gt = \\frac{5}{7} \\cdot g \\cdot t$\n$\\Rightarrow$ $t = \\frac{v}{g}\\left( \\because\\mu = \\frac{2}{7} \\right)$\nFor centre of the sphere:$S = v\\left( \\frac{v}{g} \\right) - \\frac{1}{2} \\cdot \\left( \\frac{2}{7} \\cdot g \\right)\\left( \\frac{v^{2}}{g^{2}} \\right)$\n$S = \\frac{v^{2}}{g} - \\frac{v^{2}}{7g}$ (w.r.t. belt)\nIn this time, the belt with speed will move a distance\nof$s_{b} = vt = v\\left( \\frac{v}{g} \\right) = \\frac{v^{2}}{g}$\n[IMAGE] Distance travelled w.r.t. ground will be:$S_{g} = \\left| S - S_{b} \\right| = \\left| \\left( \\frac{v^{2}}{g} - \\frac{v^{2}}{7g} \\right) - \\left( \\frac{v^{2}}{g} \\right) \\right| = \\frac{v^{2}}{7g}$", "solution_images": ["images/image41.png"], "subject": "Physics", "topic": "Rotational motion", "subtopic": "Rolling with sliding", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-05-Q12", "question": "Initially two particles A and B are present at (0,0) and (d, 0)\nrespectively. They start moving with speed\n${\\overset{\\rightarrow}{v}}_{A} = v\\overset{\\hat{}}{i} + v\\overset{\\hat{}}{j}$and$V_{B} = - v\\overset{\\hat{}}{j}$.\nIf R is the separation between them and $T_{0}$ be the time when\nseparationbetween them is minimum, then", "question_images": [], "option_1": "$T_{0} = \\frac{d}{5v}$", "option_2": "$R_{\\min} = \\frac{d}{\\sqrt{5}}$", "option_3": "Graph of R vs time is straight line", "option_4": "Graph of R vs time is Circle", "correct_option": 1, "numerical_answer": null, "solution": "$V_{BA} = - v\\overset{\\hat{}}{i} - 2v\\overset{\\hat{}}{j}$\n$tan\\theta = 2$\n[IMAGE] at any time t\n$x = d - vt$\n$y = - 2vt$\n$\\therefore R^{2} = (d - vt)^{2} + 4v^{2}t^{2}$\n$R = \\sqrt{5v^{2}t^{2} - 2vdt + d^{2}}$\nIt is neither st. line nor circle\n$R_{\\min} = dsin\\theta = \\frac{2d}{\\sqrt{5}}$\nDistance $BC = dcos\\theta$\n$= \\frac{d}{\\sqrt{5}}$\ntime, $T_{0} = \\frac{d}{\\sqrt{5}v_{BA}}$\n$T_{0} = \\frac{d}{\\sqrt{5} \\times \\sqrt{5v}};\\ T_{0} = \\frac{d}{5v}$", "solution_images": ["images/image15.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "Minimum separation", "difficulty": "Hard", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-24-Q19", "question": "In the figure shown cross-section area of holes are same and\nvery small in comparison to area of cross-section of water tank:\n(tanks are the open from the top)", "question_images": ["images/image51.png"], "option_1": "$x_{1}:x_{2}:x_{3} = \\sqrt{5}:\\sqrt{6}:\\sqrt{3}$", "option_2": "$\\ x_{1}:x_{2}:x_{3} = \\sqrt{6}:\\sqrt{5}:\\sqrt{3}$", "option_3": "$x_{1}:x_{2}:x_{3} = \\sqrt{3}:\\sqrt{6}:\\sqrt{5}$", "option_4": "$x_{1}:x_{2}:x_{3} = \\sqrt{2}:\\sqrt{5}:\\sqrt{3}$", "correct_option": 1, "numerical_answer": null, "solution": "For case -1\n$\\rho gh = 1/2\\rho v_{1}^{2}$\n$V_{1} = \\sqrt{2gh}$\n$t_{1} = \\sqrt{(10h/g)}$\nFor case -2:\n$\\rho g2h + 2\\rho gh = \\frac{1}{2}2\\rho v_{2}^{2}$\n$V_{2} = \\sqrt{4gh}$\n$t_{2} = \\sqrt{\\frac{6h}{g}}$\nFor case-3:\n$\\rho g2h + 2\\rho gh + 3\\rho gh = \\frac{1}{2}3\\rho v_{3}^{2}$\n$V_{3} = \\sqrt{6gh}$\n$t_{3} = \\sqrt{\\frac{2h}{g}}$\n$x_{1} = v_{1}t_{1} = \\sqrt{20}$ $x_{2} = \\sqrt{24}$ $x_{3} = \\sqrt{12}$\n$x_{1}:x_{2}:x_{3} = \\sqrt{20}:\\sqrt{24}:\\sqrt{12}$\n$x_{1}:x_{2}:x_{3} = \\sqrt{10}:\\sqrt{12}:\\sqrt{6}$\n$x_{1}:x_{2}:x_{3} = \\sqrt{5}:\\sqrt{6}:\\sqrt{3}$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-15-Q7", "question": "Radiation, with wavelength 656.1 nm falls on a metal surface to\nproduce photoelectrons. The electrons are made to enter a uniform\nmagnetic field of[IMAGE]. If the radius of the largest\ncircular path followed by the electrons is[IMAGE], the\nwork function of the metal is close to", "question_images": ["images/image56.png", "images/image57.png"], "option_1": "1.8 eV", "option_2": "0.8 eV", "option_3": "1.6 eV", "option_4": "1.1 eV", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image58.png", "images/image59.png", "images/image60.png", "images/image61.png", "images/image62.png", "images/image63.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "Einstein's photoelectric equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-14-Q8", "question": "In a single slit diffraction of different light of wavelength\n[IMAGE] by a slit of with t, the size of the central\nmaximum on a screen at a distance[IMAGE] is", "question_images": ["images/image54.png", "images/image55.png", "images/image56.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Width of central Max. [IMAGE]", "solution_images": ["images/image61.png", "images/image62.png", "images/image63.png", "images/image64.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Diffraction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "Ph-27-Q36", "question": "Two immiscible liquids are poured in a U-tube having densities\n[IMAGE]. If the ratio of heights (of the liquids\nabove their interface)\n[IMAGE] is x, then find the value of x.\nU-ufy nks vfeJ.kh; nzo ftuds ?kuRoρ_1= 1.0 × 10^3\nkg/m^3rFkkρ_2= 3.0 × 10^3 kg/m^3gSA;fn (nksuks nzoks ds vUr i`\"B ls) nzoksa dh vuqikr $\\frac{h_{1}}{h_{2}}$dk eku\nx", "question_images": ["images/image232.png", "images/image233.png", "images/image234.png", "images/image235.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "", "solution_images": ["images/image236.png", "images/image237.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Hydrostatic pressure", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-05-Q23", "question": "A wire has a mass $0.3\\ \\pm \\ 0.003$ g, radius\n$0.5\\ 0 \\pm 0.005$cm and length $6 \\pm 0.06$cm. What is the maximum\npercentage error in the measurement of density?", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "$\\frac{d\\rho}{A} = \\frac{dm}{m} + 2\\frac{dr}{r} + \\frac{dl}{l}$\n$\\%\\frac{d\\rho}{A} = \\left( \\frac{0.003}{0.3} + 2*\\frac{0.005}{0.5} + \\frac{0.06}{6} \\right)*100 = 4\\%$", "solution_images": [], "subject": "Physics", "topic": "Error analysis", "subtopic": "Maximum percentage error", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "Ph-28-Q19", "question": "A non-viscous fluid of density [IMAGE] is\nflowing in a tube as shown in figure. Area of section", "question_images": ["images/image171.png"], "option_1": "is double\nthat of section", "option_2": ". Centre of mass of section", "option_3": "Work done by gravitational force per unit volume from\nsection", "option_4": "Work done by elastic forces (pressure) per unit volume from\nsection", "correct_option": 2, "numerical_answer": null, "solution": "Applying Bernoulli is equation from section-(1) and (2)\n[IMAGE] and $P_{1} - P_{2} = \\rho g(2h)$\nSolving we get, [IMAGE] (3) Work done by gravitation force per unit volume = negative of the\nincrease in gravitational potential energy per unit volume =\n$- (\\rho gh)$\n(4) Work done by elastic force per unit volume = decrease in\npressure = [IMAGE] = $2\\rho gh$\nTOPIC: Fluid dynamics\nSUB TOPIC: Bernoulli's theorem", "solution_images": ["images/image177.png", "images/image178.png", "images/image179.png", "images/image180.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-11-Q7", "question": "Two particles of equal mass m have respective initial\nvelocities[IMAGE] and [IMAGE]. They collide\ncompletely inelastically. The energy lost in the process is", "question_images": ["images/image60.png", "images/image61.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "Conserving momentum\n[IMAGE] On solving", "solution_images": ["images/image66.png", "images/image67.png", "images/image68.png", "images/image69.png", "images/image70.png"], "subject": "Physics", "topic": "Collision", "subtopic": "Perfectly inelastic collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-12-Q4", "question": "The charge that will flow to earth when only switch S1 is closed\nand right most plate is uncharged is", "question_images": [], "option_1": "$- \\left( \\frac{Q}{2} \\right)$", "option_2": "$\\left( \\frac{Q}{2} \\right)$", "option_3": "Q", "option_4": "- Q", "correct_option": 3, "numerical_answer": null, "solution": "At point D,$\\overrightarrow{E}$must be zero because potential\nof left most plate is zero.\nSo, x = 0\n${\\overrightarrow{E}}_{A} = 0$ (Conductor)\nx = y + Q\ny = - Q", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Distribution of charge on plates", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-12-Q13", "question": "If finding the electric field using gauss law the\nformula$|\\overset{\\rightarrow}{E}| = \\frac{q_{enc}}{\\varepsilon_{0}|A|}$is\napplicable. In the formula $\\varepsilon_{0}$ is permittivity of free\nspace, A is the area of Gaussian surface and q_enc is charge enclosed\nby the Gaussiansurface. This equation can be used in which of the\nfollowing situation?", "question_images": [], "option_1": "Only when the Gaussian surface is an equipotential surface.", "option_2": "Only when $|\\overset{\\rightarrow}{E}| =$ constant on the\nsurface.", "option_3": "Only when the Gaussian surface is an equipotential\nsurface and $\\left| \\overset{\\rightarrow}{E} \\right|$is constant of\nthe surface.", "option_4": "For any choice of Gaussian surface.", "correct_option": 3, "numerical_answer": null, "solution": "Magnitude of electric field is constant & the surface is\nequipotential", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Gauss law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-14-Q2", "question": "In YDSE arrangement, white light is used to illuminate the slits.\nAt point on the screen directly in front of slits, which of these\nwavelengths is missing?\n([IMAGE] = wavelength of light used)\n(d = distance between slits)\n(D = distance between slits and screen)\n(Given d< tan(\\frac{\\pi}{4} - \\frac{\\theta}{2})$", "option_2": "$\\mu < tan(\\frac{\\pi}{4} - \\frac{\\theta}{2})$", "option_3": "$\\mu < tan\\theta$", "option_4": "$\\mu > tan\\frac{\\pi}{4}$", "correct_option": 2, "numerical_answer": null, "solution": "Force applied to drag$F_{1} = Wsin\\theta + \\mu Wcos\\theta$\nForce applied to lift $F_{2} = W$\nAs, $F_{1} < F_{2} \\Rightarrow Wsin\\theta + \\mu Wcos\\theta < W$\n$\\Rightarrow \\mu < \\frac{1 - sin\\theta}{cos\\theta} = \\frac{\\sin^{2}\\frac{\\theta}{2} + \\cos^{2}\\frac{\\theta}{2} - 2sin\\frac{\\theta}{2}cos\\frac{\\theta}{2}}{\\cos^{2}\\frac{\\theta}{2} - \\sin^{2}\\frac{\\theta}{2}}$\n$= \\frac{{(cos\\frac{\\theta}{2} - sin\\frac{\\theta}{2})}^{2}}{(\\cos^{2}\\frac{\\theta}{2} - \\sin^{2}\\frac{\\theta}{2})} = \\frac{cos\\frac{\\theta}{2} - sin\\frac{\\theta}{2}}{cos\\frac{\\theta}{2} + sin\\frac{\\theta}{2}} = \\frac{1 - tan\\frac{\\theta}{2}}{1 + tan\\frac{\\theta}{2}}$\n$\\Rightarrow \\mu < tan(\\frac{\\pi}{4} - \\frac{\\theta}{2})$", "solution_images": [], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-02-Q45", "question": "A convex lens is placed between a fixed object and a fixed\nscreen. The distance between the object and the screen is 150 cm. The\nreal images of the object are formed on the screen for two successive\npositions of the lens separated by a distance of 50 cm. The focal length\nof the lens is", "question_images": [], "option_1": "25 cm", "option_2": "33.3 cm", "option_3": "45 cm", "option_4": "66.6 cm", "correct_option": null, "numerical_answer": null, "solution": "B\nAccording to displacement method\n[IMAGE] = 33.3 cm\n∴ (B)", "solution_images": ["images/image70.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-10-Q25", "question": "A uniform sphere of mass 20 kg and radius 10 cm is placed on a\nrough horizontal surface. The coefficient of friction between the sphere\nand the surface is 0.5. A horizontal force of magnitude 14 N is applied\non the sphere as shown. If the friction force (in N) acting on the\nsphere is F the value of F is (g = 10 ms^-2)", "question_images": ["images/image73.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "$f_{e} = \\mu mg = 100N$\n$F - f = ma$\n$fR = I\\alpha \\Rightarrow f = \\frac{2}{5}ma$\n$\\therefore f = \\frac{2}{5} \\times 20 \\times 0.5 = 4N$", "solution_images": ["images/image74.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Rolling", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-19-Q13", "question": "A very long wire ABDMNDC is shown in figure carrying current I.\nAB and BC parts are straight, long and at right angle. At D wire forms a\ncircular turn DMND of radius R. AB, BC parts are tangential to circular\nturn at N and D. Magnetic field at the centre of circle is", "question_images": ["images/image18.jpeg"], "option_1": "$\\frac{\\mu_{0}I}{2\\pi R}(\\pi + \\frac{1}{\\sqrt{2}})$", "option_2": "$\\frac{\\mu_{0}I}{2R}$", "option_3": "$\\frac{\\mu_{0}I}{2\\pi R}(\\pi - \\frac{1}{\\sqrt{2}})$", "option_4": "$\\frac{\\mu_{0}I}{2\\pi R}(\\pi + 1)$", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{\\mu_{0}i}{4\\pi R}\\lbrack sin90^{\\circ} - sin45^{\\circ}\\rbrack \\otimes + \\frac{\\mu_{0}I}{2R} \\odot + \\frac{\\mu_{0}i}{4\\pi R}(sin45^{\\circ} + sin90^{\\circ}) \\odot$\n$= \\frac{- \\mu_{0}i}{4\\pi R}\\lbrack 1 - \\frac{1}{\\sqrt{2}}\\rbrack + \\frac{\\mu_{0}i}{2R} + \\frac{\\mu_{0}i}{4\\pi R}\\lbrack\\frac{1}{\\sqrt{2}} + 1\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{4\\pi R}\\lbrack - 1 + \\frac{1}{\\sqrt{2}} + 2\\pi + \\frac{1}{\\sqrt{2}} + 1\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{4\\pi R}\\lbrack\\sqrt{2} + 2\\pi\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{2\\pi R}\\lbrack\\frac{1}{\\sqrt{2}} + \\pi\\rbrack \\odot$", "solution_images": ["images/image19.jpeg"], "subject": "Physics", "topic": "Magnetics", "subtopic": "Magnetic field due to current carrying conductors", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-24-Q1", "question": "A uniform cylinder floats with its axis vertical with\n[IMAGE] parts of its length in the upper liquid and\n[IMAGE] part of its length in lower liquid. If the\ndensities of upper and lower liquids are [IMAGE] then the\ndensity of the material of the cylinder is", "question_images": ["images/image1.png", "images/image2.png", "images/image3.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "Let [IMAGE] be the be the length of the cylinder then\n is the density of the cylinder.\nTOPIC: Hydrostatics\nSUB TOPIC:Bouyancy\nLEVEL: Moderate", "solution_images": ["images/image8.png", "images/image9.png", "images/image10.png", "images/image11.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-23-Q21", "question": "A 20 gm particle is subjected to two simple harmonic motions\nx_1 = 2 sin 10 t,\n$x_{2} = 4sin\\left( 10t + \\frac{\\pi}{3} \\right)$where x_1& x_2 are in\nmetre& t is in sec.", "question_images": [], "option_1": "The displacement of the particle at t = 0 will be $2\\sqrt{3}m$", "option_2": "Maximum speed of the particle will be m/s. $20\\sqrt{7}m$", "option_3": "Magnitude of maximum acceleration of the particle will be", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "At t = 0\nDisplacement $x = x_{1} + x_{2} = 4sin\\frac{\\pi}{2} = 2\\sqrt{3}m$\nResulting Amplitude\n$A = \\sqrt{2^{2} + 4^{2} + 2(2)(4)cos\\pi/3} = \\sqrt{4 + 16 + 8} = \\sqrt{28} = 2\\sqrt{7}m$\nMaximum speed $= A\\omega = 20\\sqrt{7}m/s$\nMaximum acceleration $= A\\omega^{2} = 200\\sqrt{7}m/s^{2}$", "solution_images": [], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "kinematics of SHM", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-18-Q2", "question": "Kinetic energy of a particle of mass 2 kg moving along a circle\nvaries with time t as $K = 3t^{2}$, where t in second and K is in\nJoules. Then rate at which work is being done on particle at t=1 s in\nwatt is", "question_images": [], "option_1": "6", "option_2": "4", "option_3": "5", "option_4": "2", "correct_option": 1, "numerical_answer": null, "solution": "$K = \\frac{1}{2}{mv}^{2} = 3t^{2}$\n$v = \\sqrt{3}t$\n$a_{t} = dv/dt = \\sqrt{3}$\n$= ma_{t}v$\n$= 2.\\sqrt{3} \\cdot \\sqrt{3} \\times 1 = 6watt$", "solution_images": ["images/image1.png"], "subject": "Physics", "topic": "Circular motion", "subtopic": "Tangential acceleration", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-25-Q8", "question": "A system undergoes three quasi-static processes sequentially as\nindicated in the given figure. $1 \\rightarrow 2$is an isobaric process,\n$2 \\rightarrow 3$ is a polytrophic process with\n$\\gamma = \\frac{4}{3}$and $3 \\rightarrow 1$ is a process in which PV=\nconstant.\n$P_{2} = P_{1} = 4 \\times 10^{5}Nm^{- 2},P_{3} = 1 \\times 10^{5}Nm^{- 2}$and$\\ V_{1} = 1m^{3}$.\nThe heat transfer of the cycle is $\\Delta Q,$ change in internal\nenergy is $\\Delta U$ and the work done is $\\Delta W$. for polytropic\nprocess $PV^{\\gamma} =$ constant.\n(Take $\\mathcal{l}$n2 = 0.693). Then,d s fp=kuqlkj rhu vnZ~/k&fLFkj çØeks ls xqtkjk A\n vkarfjd", "question_images": ["images/image27.png", "images/image28.png", "images/image29.png", "images/image30.png", "images/image31.png", "images/image32.png", "images/image33.png", "images/image34.png", "images/image35.png", "images/image36.png", "images/image37.png", "images/image38.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "For the process izØe ds fy, 3 → 1\n[IMAGE] For the processizØe ds fy, 2→ 3\n[IMAGE] For the processizØe ds fy, 1 → 2\n[IMAGE] For the processizØe ds fy, 2 → 3\n[IMAGE] For the process 3 → 1\n[IMAGE] For cyclic process pØh\n$= 1.08 \\times 10^{5}J$", "solution_images": ["images/image43.png", "images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png", "images/image48.png", "images/image49.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic process", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-10-Q9", "question": "A rod AB is free to rotate on a smooth horizontal plane about one\nof its end A which is hinged. Initially the rod is at rest and D and E\nare fixed points located in horizontal plane in such a way that DA is\nperpendicular to rod and AE is along the length of the rod. A particle P\nhits the rod perpendicularly at point C some distance from point A and\ngets imbedded in it, then", "question_images": ["images/image38.png"], "option_1": "Angular momentum of rod plus particle system about point C must\nincrease after collision", "option_2": "Angular momentum of rod plus particle\nsystem about point D is conserved during collision.", "option_3": "Angular momentum of rod plus particle system about point D is not\nconserved during collision.", "option_4": "Angular momentum of rod plus particle system about point E is\nconserved", "correct_option": 2, "numerical_answer": null, "solution": "The force applied by hinge on rod may be towards point D or\naway from D or zero depending on distance of point C from A. Reaction at\nhinge has torque about E therefore, angular momentum is not conserved.", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-12-Q14", "question": "Three charged particles A, B and C with charges -4q, 2q and\n-2q are present on the circumference of acircle of radius d. The\ncharged particles A, C and centre O of the circle formed an equilateral\ntriangle asshown in figure. Electric field at O along x-direction is", "question_images": ["images/image15.png"], "option_1": "$\\frac{\\sqrt{3}q}{\\pi\\varepsilon_{0}d^{2}}$", "option_2": "$\\frac{\\sqrt{3}q}{4\\pi\\varepsilon_{0}d^{2}}$", "option_3": "$\\frac{3\\sqrt{3}q}{4\\pi\\varepsilon_{0}d^{2}}$", "option_4": "$\\frac{2\\sqrt{3}q}{\\pi\\varepsilon_{0}d^{2}}$", "correct_option": 1, "numerical_answer": null, "solution": "$= \\frac{q\\sqrt{3}}{\\pi\\varepsilon_{0}d^{2}}$", "solution_images": ["images/image16.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electic field", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "Ph-28-Q38", "question": "In a certain series RLC circuit it is found that at resonance,\nthe inductive reactance is equal to the resistance. It is also found\nthat rms current in this circuit at resonance was\n[IMAGE]. If now angular frequency of source is doubled\nwithout changing any other parameter, what will be new rms current in\nampere in circuit?", "question_images": ["images/image342.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "At resonance:\n[IMAGE] Atangular frequency of source is double of resonance frequency:\n[IMAGE] TOPIC:Alternating current\nSUB TOPIC:RLC circuit", "solution_images": ["images/image343.png", "images/image344.png", "images/image345.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-25-Q10", "question": "An ideal gas whose internal energy U is at absolute zero\ntemperature is equal to zero undergoes areversible adiabatic\ncompression. If U, P, V, T represents the internal energy, pressure,\nvolume andabsolute temperature respectively for ideal gas then,d ije ij mRØe.kh;\nfy, U, P, V, T Øe'k vkUrfjd rFkk ije rki", "question_images": ["images/image61.png", "images/image61.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] For adiabatic process [IMAGE] constant", "solution_images": ["images/image66.png", "images/image67.png", "images/image67.png", "images/image68.png", "images/image69.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Adiabatic process", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-06-Q25", "question": "Two forces of 6 N and 3 N are acting on the two blocks of 2 kg\nand 1 kg kept on frictionless floor. The force exerted on 2 kg block by\n1 kg block is 2 n then find the value of n.", "question_images": ["images/image36.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "[IMAGE] Both blocks are constrained to move with same acceleration.\n$6 - N = 2a\\ \\ \\ \\ \\ \\ \\ $[Newtons II law for 2kg block]\n$N - 3 = 1a$ [Newton's II law for 1 kg block]\nThis gives a= 1m/s^2\nTherefore, N =4 N", "solution_images": ["images/image37.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Newton's second law", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-23-Q13", "question": "A force F= 4t acts on a particle of mass 1kg. If the particle\nstarts from rest the work done by the force during the first 2 second\nwill be 8n J, find n.", "question_images": [], "option_1": "4", "option_2": "6", "option_3": "5", "option_4": "3", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{dv}{dt} = 4t$ $\\int_{0}^{v}{dv} = 4\\int_{0}^{t}{t\\ dt}$\n$V = 2t^{2}$\n$ds = 2t^{2}dt$\n$W = \\int_{}^{}{Fds\\ cos0^{\\circ}}$\n$= \\int_{0}^{2}\\mspace{2mu} 4t \\cdot 2t^{2}dt$\n$= 8\\left\\lbrack \\frac{t^{4}}{4} \\right\\rbrack_{0}^{2}$ $= 32J$", "solution_images": ["images/image31.png", "images/image32.png"], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work done by variable force", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-02-Q35", "question": "An object is moving along x-axis. It is definitely moving\ntowards origin if", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "When x > 0, the particle will be moving towards origin if\n[IMAGE] ⇒ [IMAGE] When x < 0, the particle will be moving towards origin if\n[IMAGE] ⇒ [IMAGE] Hence in both cases [IMAGE]", "solution_images": ["images/image21.png", "images/image22.png", "images/image23.png", "images/image24.png", "images/image25.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-11-Q17", "question": "Four point mass, each of mass m are connected at a corner of a\nsquare of side by massless rods as shown in the figure. x and y\naxis are in the plane of the system and z axis is perpendicular to the\nplane and passing through the centre of the square.", "question_images": ["images/image163.png"], "option_1": "Moment of inertia of the system about x axis is", "option_2": "Moment of inertia of the system about y axis", "option_3": "Moment of inertia of the system about the diagonal axis", "option_4": "Moment of inertia of the system about z axis is", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image169.png", "images/image170.png", "images/image171.png", "images/image172.png", "images/image173.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Moment of inertia", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-15-Q17", "question": "The activity of a radioactive sample falls from\n[IMAGE] to [IMAGE] in 30 minutes. Its\nhalf life is close to", "question_images": ["images/image135.png", "images/image136.png"], "option_1": "62 min", "option_2": "66 min", "option_3": "72 min", "option_4": "52 min", "correct_option": 1, "numerical_answer": null, "solution": "$\\Rightarrow \\text{\\quad}\\frac{ln2}{ln(7/5)} = \\frac{t_{\\frac{1}{2}}}{(30\\min)} \\Rightarrow \\text{\\quad}(2.06004)30 = t_{\\frac{1}{2}} = 61.8\\text{min}$", "solution_images": ["images/image137.png", "images/image138.png", "images/image139.png"], "subject": "Physics", "topic": "Nuclear Physics", "subtopic": "Radioactivity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "Ph-27-Q11", "question": "Consider a mixture of n moles of helium gas and 2n moles of\noxygen gas (molecules taken to be rigid)as an ideal gas. Its CP/CV value\nwill be:\nds 2n eksYl dh feJ.k xSl ekus rks bl feJ.k", "question_images": [], "option_1": "$\\frac{23}{15}$", "option_2": "$\\frac{40}{27}$", "option_3": "$\\frac{67}{45}$", "option_4": "$\\frac{19}{13}$", "correct_option": 4, "numerical_answer": null, "solution": "$\\gamma_{mix} = \\frac{n_{1}C_{p_{1}} + n_{2}C_{p_{2}}}{n_{1}C_{v_{1}} + n_{2}c_{v_{2}}}$\n$= \\frac{n\\left( \\frac{5}{2}R \\right) + 2n\\left( \\frac{7}{2}R \\right)}{n\\left( \\frac{3}{2}R \\right) + 2n\\left( \\frac{5}{2}R \\right)}$\n$= \\frac{5 + 14}{3 + 10} = \\frac{19}{13}$", "solution_images": [], "subject": "Physics", "topic": "Kinetic theory of gasses", "subtopic": "Heat Capacity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-02-Q33", "question": "The temperature needed to excite the hydrogen atoms to first\nexcited level is:\n(Boltzman constant = 1.381 × 10^-23 J/K)", "question_images": [], "option_1": "7.88 × 10^4 K", "option_2": "3.2 × 10^4 K", "option_3": "0.85 × 10^4 K", "option_4": "13.6 ×\n10^4 K", "correct_option": 1, "numerical_answer": null, "solution": "According to kinetic interpretation of temperature, the average\nkinetic energy per atom\n[IMAGE] Energy required to excite the atoms to first excited state is\n[IMAGE] 10.2 × 1.6 × 10^-19\nT = 7.88 ´ 10^4 K", "solution_images": ["images/image14.png", "images/image15.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-21-Q20", "question": "Four identical plates (equally spaced) and a battery are\nconnected as shown. If the capacitancebetween two consecutive plates is\nC then choose the correct statement\npkj,d leku IysVs( leku nwjh ij),oa,d cSVjh ls fp=kuqlkj tqM+h gSA;fn nks Øekxr IysVksa C", "question_images": ["images/image75.png"], "option_1": "Energy supplied by the battery isCV^2.\ncSVjh }kjk lIykbZ CV^2", "option_2": "Energy linked in the space between plates 1 and 2 is$\\frac{1}{6}$\nCV^2", "option_3": "Potential difference between plates 2 and 4 is V/2\nIysV 2 rFkk 4", "option_4": "The surface charge density on plate 3 on its right side is equal to\nthat on its left side.\nIysV 3 ij i`\" ?kuRo ml rFkk cka;h lrg ij leku gSA", "correct_option": 1, "numerical_answer": null, "solution": "Equivalent Circuit rqY; ifjiFk\n[IMAGE] Combination of Capacitors (1) and (2) are short circuited.\nenergy supplied by battery cSVjh }kjk lIykbZ \n= QV\n= CV.V\n= CV^2\nPotential difference between 1 and 3 = 0\n1 rFkk 3 foHkokUrj= 0\nHence, potential difference between 2 and 4 = V\nvr IysV 2 rFkk 4 foHkokUrj= V\nCharge is only on right side of plate", "solution_images": ["images/image76.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Energy stored in capacitor", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-06-Q16", "question": "Under the action of a force, a 2 kg body moves such that its\nposition x as a function of time is given by $x = t^{3}/3$,where xis in\nmeters and t is in seconds. The work done by the force in the first two\nseconds starting form rest is", "question_images": [], "option_1": "1600 J", "option_2": "160 J", "option_3": "16 J", "option_4": "1.6 J", "correct_option": 3, "numerical_answer": null, "solution": "$x = \\frac{t^{3}}{3}$\n$\\frac{dx}{dt} = \\frac{3t^{2}}{3}$\n$V = t^{2}$\n$V(2) = 4$\n$V(0) = 0$\n$\\Rightarrow \\Delta K = \\frac{1}{2}m\\left( 4^{2} \\right) - \\frac{1}{2}m\\left( 0^{2} \\right)$\n$\\Delta K = 16\\ J$\nFrom work energy theorem\nWork done by force = $\\Delta K = 16\\ J$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Work energy theorem", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-04-Q13", "question": "If a particle of mass m start from A along curved circular path\nshown in figure with constant tangential acceleration a, then net force\nat B on the particle has the magnitude", "question_images": ["images/image78.png"], "option_1": "zero", "option_2": "", "option_3": "", "option_4": "[IMAGE] m a \n A, B", "correct_option": 3, "numerical_answer": null, "solution": "Fnet = matotal = [IMAGE] at = a\nac = [IMAGE] v2 = 0 + 2a[IMAGE] ⇒ ac =[IMAGE] TOPIC: CIRCULAR MOTION\nSUB TOPIC: KIMATIC", "solution_images": ["images/image82.png", "images/image83.png", "images/image84.png", "images/image85.png", "images/image86.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-03-Q24", "question": "In a uniform gravitational field g, which is acting vertically\ndownward, a ball is thrown at an angle with horizontal such that the\ninitial (immediately after projection) radius of curvature is 8 times\nthe minimum radius of curvature. If angle of projection is θ°, then find\nthe value of θ.\n g, \n ( ) \n 8 θ°, θ", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "60", "solution": "TOPIC: CIRCULAR MOTION\nSUB TOPIC: RADIUS OF CURVATURE", "solution_images": ["images/image144.png", "images/image145.png", "images/image146.png", "images/image147.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-10-Q14", "question": "A sphere of mass is given some angular velocity about a\nhorizontal axis through its centre and gently placed on a plank of mass\n The co-efficient of friction between sphere and plank\nis[IMAGE]. The plank rests on a smooth horizontal\nsurface. The initial acceleration of the centre of sphere relative to\nthe plank will be", "question_images": ["images/image51.png", "images/image52.png"], "option_1": "Zero", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "FBD for sphere & block\n$a_{2} = \\frac{f_{r}}{m} = \\frac{\\mu mg}{m}$\n${\\overrightarrow{a}}_{1} = \\mu g\\overset{\\hat{}}{i}$\n${\\overrightarrow{a}}_{2} = - \\mu g\\overset{\\hat{}}{i}$\n${\\overset{\\rightarrow}{a}}_{rel} = {\\overset{\\rightarrow}{a}}_{1} - {\\overset{\\rightarrow}{a}}_{2} = 2\\mu g\\overset{\\hat{}}{i}$\n$a_{rel} = 2\\mu g$.", "solution_images": ["images/image56.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Pure rolling on a plank", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "Ph-27-Q32", "question": "A container filled with viscous liquid is moving vertically\ndownwards with constant speed\n[IMAGE]. At the instant shown, a sphere of radius\nr is moving vertically downwards (in liquid) has speed\n[IMAGE]. The coefficient of viscosity is η. There is no relative motion between\nthe liquid and the container. Then at the shown instant, the magnitude\nof viscous force acting on sphere is[IMAGE]. Find x.,d ik=k tks fd,d pky 3v_0ls\nf= xksyk uhps dh rjQ (nzo esa) tk jgk gS mlds xksys dh\nugh gSA rks bl.k ij xksys ij cy", "question_images": ["images/image206.png", "images/image207.png", "images/image208.png", "images/image208.png", "images/image209.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "Relative to liquid, the velocity of sphere is\n[IMAGE] viscous force on sphere\n$= 12\\pi\\eta\\text{r}\\text{v}_{0}$downward\nnzo xksys dk osx dh rjQ gSA\n$\\therefore$xksys ij nzo = 6 πη r2v_0uhps", "solution_images": ["images/image210.png", "images/image211.png", "images/image212.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Terminal velocity", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-22-Q20", "question": "Three particles each of mass m can slide on fixed frictionless\nhorizontal circular tracks in the same horizontal plane as shown in the\nfigure. The coefficient of restitution being e=0.5. Assuming that\nm_2and m_3 are at rest initially and lie along a radial line before\nimpact and the spring is initially unstretched, then maximum extension\nin spring in subsequent motion.", "question_images": ["images/image54.png"], "option_1": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{m}{k}}$", "option_2": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{m}{5k}}$", "option_3": "$\\frac{3}{4}V_{0}\\sqrt{\\frac{2m}{5}k}$", "option_4": "$\\frac{3}{5}V_{0}\\sqrt{\\frac{m}{k}}$", "correct_option": 2, "numerical_answer": null, "solution": "Velocity just after collision $V_{0} = V_{1} + V_{2}$.....(1)\n$\\Rightarrow \\frac{1}{2} = \\frac{- V_{1} + V_{2}}{V_{0}}$\n$- V_{1} + V_{2} = \\frac{V_{0}}{2}$.....(2)\nFrom (1) & (2)\n$\\Rightarrow V_{1} = \\frac{V_{0}}{4}\\& V_{2} = \\frac{3V_{0}}{4}$\nWhen maximum extension occurs then angular speed with rest to centre\nfor mass m_2& m_3 are same. Using angular momentum conservation\nabout centre of circle.\n$m\\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\\omega + m(2R)^{2}\\omega$\n$\\Rightarrow \\omega = \\frac{3}{10}\\frac{V_{0}}{R}$\nSo velocity of$m_{2} = \\frac{3}{5}V_{0}$\n& velocity of$m_{3} = \\frac{3}{10}V_{0}$\nWhen extension is maximum using energy conservation.\n$\\frac{1}{2}m{(\\frac{3}{4}V_{0})}^{2} = \\frac{1}{2}m{(\\frac{3}{5}V_{0})}^{2} + \\frac{1}{2}m{(\\frac{3}{10}V_{0})}^{2} + \\frac{1}{2}kx_{max.}^{2} \\Rightarrow x_{\\max} = \\frac{3}{4}V_{0}\\sqrt{\\frac{m}{5k}}$", "solution_images": [], "subject": "Physics", "topic": "Center of mass", "subtopic": "Collision", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "Ph-27-Q6", "question": "32 g of [IMAGE] is contained in a cubical\ncontainer of side 1 m and maintained at a temperature of\n[IMAGE]. The isothermal bulk modulus of elasticity of the\ngas is [IMAGE]. Then the value of K is:[IMAGE] 32 s 1m Hkqtk ds ?kukdkj crZu esa 127°C A rks K", "question_images": ["images/image21.png", "images/image22.png", "images/image23.png", "images/image24.png", "images/image21.png", "images/image23.png", "images/image24.png"], "option_1": "32.30", "option_2": "33.40", "option_3": "33.20", "option_4": "33.00", "correct_option": 3, "numerical_answer": null, "solution": "Bulk modulus of elasticity at constant temperature ij", "solution_images": ["images/image25.png", "images/image26.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Bulk modulus", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-20-Q11", "question": "The activity of a radioactive sample falls from\n[IMAGE] to [IMAGE] in 30 minutes. Its half\nlife is close to", "question_images": ["images/image21.png", "images/image22.png"], "option_1": "62 min", "option_2": "66 min", "option_3": "72 min", "option_4": "52 min", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] From (i)(ii)\n$\\Rightarrow \\text{\\quad}\\frac{ln2}{ln(7/5)} = \\frac{t_{\\frac{1}{2}}}{(30\\min)} \\Rightarrow \\text{\\quad}(2.06004)30 = t_{\\frac{1}{2}} = 61.8\\text{min}$", "solution_images": ["images/image23.png", "images/image24.png", "images/image25.png"], "subject": "Physics", "topic": "Modern physics", "subtopic": "Radioactivity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-15-Q15", "question": "The time period of revolution of electron in its ground state\norbit in a hydrogen atom is[IMAGE]. The frequency of\nrevolution of the electron in its first excited state (in\n[IMAGE] ) is", "question_images": ["images/image112.png", "images/image113.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image118.png", "images/image119.png", "images/image120.png", "images/image121.png", "images/image122.png"], "subject": "Physics", "topic": "Atomic physics", "subtopic": "Bohr's mode of hydrogen atom", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-14-Q22", "question": "A convex lens forms inverted image of a real object on a fixed\nscreen. The size of image is 12 cm. When lens is displaced 20 cm along\nprinciple axis it again forms a real image of size 3 cm on the screen.\nFocal length (in cm) of the lens is 40/n. Find n. (Assume image\nformation only by paraxial rays)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "since [IMAGE] For first position of the lens\n[IMAGE] From displacement method\nUsing\n$f = \\frac{D^{2} - \\Delta d^{2}}{4D} = \\frac{60^{2} - 20^{2}}{4 \\times 60} = \\frac{40}{3}\\text{cm}$", "solution_images": ["images/image146.png", "images/image147.png", "images/image148.png", "images/image149.png", "images/image150.png", "images/image151.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Refraction (Lens)", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-23-Q18", "question": "The dependence of potential with distance r from the centre of a\nnegatively charged non-conducting solid sphere is given by the following\ncurve", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Where, r → distance from circle", "solution_images": ["images/image46.png", "images/image47.png", "images/image48.png", "images/image49.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric potential due to non-conducting solid sphere", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "Ph-26-Q25", "question": "A small spherical droplet of density d is floating exactly half\nimmerse in a liquid of density [IMAGE] and surface\ntension T. The radius of the droplet is (take note that the surface\ntension applies an upward force on the droplet):\n?kuRo d dh,d NksVh xksykdkj cwan ?kuRo [IMAGE] rFkk\ni`\"B ruko T ds æo esa Bhd vk\ndk eku gS nsa fd i`\"B ruko cwan ij", "question_images": ["images/image99.png", "images/image99.png", "images/image100.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image105.png", "images/image106.png", "images/image107.png", "images/image108.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-09-Q22", "question": "A street car moves rectilinearly from station A (here car stops)\nto the next station B (here also car stops) with an acceleration varying\naccording to the law f = a - bx, where a and b are positive constants\nand x is the distance from station A. The distance between the two\nstations & the maximum velocity are\n$x = \\frac{na}{b};v_{\\max} = \\frac{na}{2\\sqrt{b}}$. Find n", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "f = a - bx means particle will do SHM.\nAt mean position; f = 0\n[IMAGE] In the figure shown, 'C' is the mean position and A & B are extreme\npositions", "solution_images": ["images/image137.png", "images/image138.png", "images/image139.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Position and velocity of a particle", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-20-Q12", "question": "A uniform elastic rod of cross-section area A, natural length L\nand Young's modulus Y is placed on asmooth horizontal surface. Now two\nhorizontal forces (of magnitude F and 3F) directed along the lengthof\nrod and in opposite direction act at two of its ends as shown. After the\nrod has acquired steady state,the extension of the rod will be.", "question_images": ["images/image26.png"], "option_1": "$\\frac{2F}{YA}L$", "option_2": "$\\frac{4F}{YA}L$", "option_3": "$\\frac{F}{YA}L$", "option_4": "$\\frac{3F}{2YA}L$", "correct_option": 1, "numerical_answer": null, "solution": "Tension in rod at a distance x from right edge is\n$T = F\\left( 3 - 2\\frac{x}{L} \\right)$\n[IMAGE] $\\therefore$net extension in\nrod $= \\int_{0}^{L}\\mspace{2mu}\\frac{T}{YA}dx = \\frac{2F}{YA}L$", "solution_images": ["images/image27.png"], "subject": "Physics", "topic": "", "subtopic": "Elongation in rod due to unbalanced force", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-16-Q17", "question": "All surfaces are smooth. Find the acceleration (in$m/s^{2}$) of\nsphere in the given figure. $(g = 10m/s^{2})$", "question_images": ["images/image31.png"], "option_1": "2", "option_2": "4", "option_3": "5", "option_4": "3", "correct_option": 3, "numerical_answer": null, "solution": "Suppose of acceleration of wedge a_1, towards right and\nacceleration of sphere a_2 downward\n[IMAGE] $mg - Ncos\\theta = {ma}_{2}$.....(1)\n$Nsin\\theta = {ma}_{1}$.....(2)\n[IMAGE] From wedge constraint\n$a_{2}cos\\theta = a_{1}sin\\theta$\nFrom (i),(ii) & (iii)\n$a_{2} = g\\sin^{2}\\theta$\n$= 5m/s^{2}$", "solution_images": ["images/image32.png", "images/image33.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Newton's second law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-19-Q15", "question": "In the given circuit diagram, a wire is joining points B and D.\nThe current in this wire is", "question_images": ["images/image24.jpeg"], "option_1": "zero", "option_2": "2 A", "option_3": "0.4 A", "option_4": "4A", "correct_option": 2, "numerical_answer": null, "solution": "$i = \\frac{20}{2} = 10A$\n$I = \\frac{4i}{5} - \\frac{3i}{5} = + \\frac{i}{5} = 2A$", "solution_images": ["images/image25.jpeg"], "subject": "Physics", "topic": "Electric current", "subtopic": "Ohm's law", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "Ph-26-Q31", "question": "A U tube filled with a liquid is accelerating horizontally with\nan acceleration a. The acceleration (in m/s^2) of the tube is:\næo ls Hkjh gqbZ,d U uyha Roj.k Rofjr gSA ufydk", "question_images": ["images/image128.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "For free surface eqDr lrg", "solution_images": ["images/image129.png", "images/image130.png", "images/image131.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Fluid in motion", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-08-Q14", "question": "When a gas in a vessel expands, its internal energy decreases.\nThe process involved is", "question_images": [], "option_1": "Isothermal", "option_2": "Adiabatic", "option_3": "Isobaric", "option_4": "Isochoric", "correct_option": 2, "numerical_answer": null, "solution": "Internal energy decreases $\\Rightarrow$ temperature drops.\nExpansion $\\Rightarrow$ not isochoric\n Cooling $\\Rightarrow$ not isothermal\nIn isobaric process, $V \\propto T \\Rightarrow \\ $expansion raisestemp.", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic processes", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-01-Q1", "question": "A metal sample carrying a current along x-axis with density Jx\nis subjected to a magnetic field Bz (along z-axis). The electric field\nEy (Hall field) developed along y-axis is directly proportional to Jx as\nwell as Bz. The constant of proportionality (Hall coefficient) has SI\nunit", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "Jx = Jx [IMAGE] I = JxA\nevdBz = eEy\nEy = vdBz\nneAvd = JxA\nEy = [IMAGE] Constant of proportionality = [IMAGE] =", "solution_images": ["images/image2.png", "images/image3.png", "images/image4.png", "images/image5.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-21-Q17", "question": "An electric dipole of moment\n$\\overset{\\rightarrow}{p} = ( - \\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} + 2\\overset{\\hat{}}{k}) \\times 10^{- 29}$\ncm. is at the origin (0,0,0). The electric fielddue to this dipole at\n$\\overset{\\rightarrow}{r} = ( + \\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} + 5\\overset{\\hat{}}{k})$\n(note that$\\overset{\\rightarrow}{r}.\\overset{\\rightarrow}{p} = 0$) is\nparallel to,d fo|qr f}/kzqo ftldk vk?kw.kZ ¼moment) [IMAGE] C.m.\n[IMAGE] ij cuus okys fo|qr", "question_images": ["images/image50.png", "images/image51.png", "images/image52.png"], "option_1": "$( + \\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "option_2": "$( - \\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "option_3": "$( + \\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "option_4": "$( - \\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} + 2\\overset{\\hat{}}{k})$", "correct_option": 1, "numerical_answer": null, "solution": "Since [IMAGE] must be antiparallel to [IMAGE], izfrlekUrj gksxk\nSo,[IMAGE] Where [IMAGE] is a arbitrary positive constant\nLet\n$\\overset{\\rightarrow}{A} = a\\overset{\\hat{}}{i} + b\\overset{\\hat{}}{j} + c\\overset{\\hat{}}{k}$\nis parallel to $\\overset{\\rightarrow}{E}$\nSo, blfy, [IMAGE]", "solution_images": ["images/image53.png", "images/image54.png", "images/image55.png", "images/image54.png", "images/image55.png", "images/image56.png", "images/image57.png", "images/image58.png", "images/image59.png", "images/image60.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field due to electric dipole", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-04-Q2", "question": "An unknown quantity is expressed as\nα =[IMAGE], where m = mass,\na = acceleration, = length. The unit of α should be", "question_images": ["images/image3.png"], "option_1": "meter", "option_2": "", "option_3": "", "option_4": "s-1", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "Ph-28-Q24", "question": "A light ray parallel to x-axis is incident on a parabolic\nconcave mirror in XY plane. The focus of this parabola is at (1,0). The\nunit vector along the reflected ray will be", "question_images": ["images/image222.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "After reflection reflected ray passes through focus\nSo vector along reflected ray[IMAGE] TOPIC:Ray optics\nSUB TOPIC:Vectors", "solution_images": ["images/image227.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-23-Q10", "question": "A capacitor is made of two square plates each of side \nmaking a very small angle $\\alpha$ between them, as shown in figure. The\ncapacitance will be close to", "question_images": ["images/image26.jpeg"], "option_1": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{2d})$", "option_2": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 + \\frac{\\alpha a}{2d})$", "option_3": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{4d})$", "option_4": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{3\\alpha a}{2d})$", "correct_option": 1, "numerical_answer": null, "solution": "$\\Rightarrow c = \\frac{\\varepsilon_{0}a}{\\alpha}\\lbrack ln(d + \\alpha x)\\rbrack_{0}^{a}$\n$= \\frac{\\varepsilon_{0}a}{\\alpha}In(1 + \\frac{\\alpha a}{d}) \\approx \\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{2d})$", "solution_images": ["images/image27.jpeg"], "subject": "Physics", "topic": "Capacitor", "subtopic": "Capacitance", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "Ph-26-Q3", "question": "A uniform elastic rod of cross-section area A, natural length L\nand Young's modulus Y is placed on a smooth horizontal surface. Now two\nhorizontal forces (of magnitude F and 3F) directed along the length of\nrod and in opposite direction act at two of its ends as shown. After the\nrod has acquired steady state, the extension of the rod will be,d,dleku izR;kLFk NM+ ftl V {ks=k AgS rFkk izkd`frd yEckbZ LrFkk\n xq.kkad YgS bldks,d fpdus {kSfrt lrg ij j\nvc nks {kSfrt cy ¼ftuds ifj.kke Fo 3FgS½ dks NM+ ds yEckbZ \nfp=kkuqlkj,d nwljs ds foijhr fljksa ij fp=kkuqlkj A tc NM+ LFkk;h voLFkk izkIr dj ysrh", "question_images": ["images/image1.png"], "option_1": "$\\frac{2F}{YA}L$", "option_2": "$\\frac{4F}{YA}L$", "option_3": "$\\frac{F}{YA}L$", "option_4": "$\\frac{3F}{2YA}L$", "correct_option": 1, "numerical_answer": null, "solution": "Tension in rod at a distance x from right edge is\nnka;s fljs ls NM+ ij xnwjh ij ruko\n$T = F\\left( 3 - 2\\frac{x}{L} \\right)$\n[IMAGE] net extension in\nrod$= \\int_{0}^{L}\\mspace{2mu}\\frac{T}{YA}dx = \\frac{2F}{YA}L$\nNM+", "solution_images": ["images/image2.png"], "subject": "Physics", "topic": "Mechanical properties of matter", "subtopic": "Elongation in rod due to unbalanced force", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-18-Q8", "question": "A particle is dropped from a certain height onto a fixed surface\nat which it collides inelastically. Which of the following represents\ntrue phase-space (momentum-position) graph? Taking upward direction as\npositive and x=0 at the surface", "question_images": ["images/image12.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Initial velocity is zero and x=+H\nJust before collision velocity hence momentum is negative and position\nis zero. Just after collision, particle will change its direction. So,\noption (C) is correct.", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Potential energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-04-Q25", "question": "A point object moves on a circular path such that distance covered\nby it is given by function [IMAGE] meter (t in second).\nThe ratio of the magnitude of tangential and radial acceleration at t =\n2 sec. is 1: 2 then radius of the circle is ( in meter )\n \n[IMAGE] (t ) t = 2 sec. \n 1:2", "question_images": ["images/image132.png", "images/image132.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "08", "solution": "S = [IMAGE] + 2t\nv = t + 2\n[IMAGE] R = 8 m\nTOPIC: CIRCULAR MOTION\nSUB TOPIC: KINEMATIC", "solution_images": ["images/image133.png", "images/image134.png", "images/image135.png", "images/image136.png", "images/image137.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "Ph-27-Q17", "question": "Figure shows a capillary tube C dipped in a liquid that wets it.\nThe liquid rises to a point A. If we blow airthrough the horizontal tube\nH, what will happen to the liquid column in the capillary tube?\niznf'kZr fp=k esa C æo ds vUnj Mwch gqbZ,d dSiyjh V~ A æo\nfcUnq A rd mBrk gSA;fn ge kjk gok izokfgr djsa\nrks dSiyjh V~;wc", "question_images": ["images/image87.png"], "option_1": "Level will rise above A Lrj A", "option_2": "Level will fall below A Lrj A", "option_3": "Level will remain at A Lrj A ij gh jgsxk", "option_4": "can not say dqN ugha dgk tk ldrk", "correct_option": 1, "numerical_answer": null, "solution": "Level rise because to compensate for the reduction in pressure\nabove A, due to high velocity (fromBernoulli's principle pressure\ndecreases if velocity increases.)\nmPp osx j.k nkc A ds ?kVsxk \n¼cjukWyh", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Application of Bernoulli's principle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-22-Q11", "question": "[IMAGE] As shown in the figure, a battery of emf $\\varepsilon$ is connected to\nan inductor L and resistance R in series. The switch is closed at t=0.\nThe total charge that flows from the battery, between t=0 and\n$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is", "question_images": ["images/image31.png"], "option_1": "$\\frac{\\varepsilon L}{R^{2}}(1 - \\frac{1}{e})$", "option_2": "$\\frac{\\varepsilon R}{eL^{2}}$", "option_3": "$\\frac{\\varepsilon L}{R^{2}}$", "option_4": "$\\frac{\\varepsilon L}{eR^{2}}$", "correct_option": 4, "numerical_answer": null, "solution": "$q = \\int_{0}^{T_{C}}idt$\n$= \\frac{\\varepsilon}{R}{\\lbrack t - \\frac{e^{- t/T_{C}}}{\\frac{- 1}{T_{C}}}\\rbrack}_{0}^{T_{C}}$;\n$= \\frac{\\varepsilon}{R}\\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\\rbrack$\n$= \\frac{\\varepsilon}{R} \\times \\frac{1}{e} \\times \\frac{L}{R}$\n$= \\frac{\\varepsilon L}{R^{2}e}$", "solution_images": [], "subject": "Physics", "topic": "Electromagnetic induction", "subtopic": "LR circuit", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-08-Q23", "question": "An ideal gas obeys the law $PV^{x} =$ constant. x is value for\nwhich it has zero molar specific heat at normal temperature. Then find 5\nx. (Cv for the gas is 2.5 R)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "7", "solution": "For a gas at normal temperature,$\\ c_{v} = \\frac{5}{2}R$\nSpecific heat for a polytropicprocess\n$C = C_{v} + \\frac{R}{1 - x} = \\frac{5}{2}R + \\frac{R}{1 - x}$\nAccording to question C = 0\n$\\frac{5(1 - x) + 2}{2(1 - x)} = 0\\ or\\ \\frac{7 - 5x}{1 - x} = 0 \\Rightarrow x = 1.4$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic processes", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-04-Q2", "question": "[IMAGE] m =\n a = = ", "question_images": ["images/image3.png"], "option_1": "meter", "option_2": "", "option_3": "", "option_4": "s-1", "correct_option": 1, "numerical_answer": null, "solution": "α] =[IMAGE] TOPIC: UNIT & DIMENSION\nSUB TOPIC: DIMENSIONS", "solution_images": ["images/image4.png", "images/image5.png", "images/image6.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-10-Q7", "question": "Mass of plank shown is 102.5 kg. If plank remains horizontal and\nslipping does not take place at any contact, Radius of uniform\ncylindrical rollers are 10 cm and 5 cm and masses 40 kg and 20 kg\nrespectively, then choose the correct options. Given that during the\nfirst second starting from rest plank gets displaced by 10 cm.", "question_images": ["images/image31.png"], "option_1": "Ratio of magnitude of acceleration of centre of mass of bigger and\nsmaller rollers is 1", "option_2": "Magnitude of friction between bigger roller and ground is more than\nthat between smaller roller and ground.", "option_3": "Magnitude of friction between plank and bigger roller is more than\nthat between plank and smaller roller.", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$S = ut + \\frac{1}{2}at^{2}$\na = 0.2 m/s^2\n$\\alpha_{1} = \\frac{a_{1}}{2R},\\ $ $\\alpha_{2} = \\frac{a_{2}}{2r}$\n$\\alpha_{1}:\\alpha_{2} = 1:2$\n$a_{1} = a_{2} = 0.1$\n[IMAGE] Solving f_1 = 3N\nf_3 = 1N\nf_2 = 1.5 N\nf_4 = 0.5 N", "solution_images": ["images/image32.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Rolling", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "Ph-28-Q23", "question": "An object of melting point [IMAGE] is at\ntemperature of [IMAGE] initially. When\n[IMAGE] amount of heat is supplied to it then find its\ntemperature. If it is known that heat is\nrequired to completely melt it and the ratio of specific heat to latent\nheat is 1: 240", "question_images": ["images/image213.png", "images/image214.png", "images/image215.png", "images/image216.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Let specific heat =S\nLatent heat = 240 s\n$$2\\theta = m(240s) + ms(120) = 360ms$$\n[IMAGE] Only '120ms' of heat is required to increase\ntemp of object to its melting point. Hence final temp is\nTOPIC:Thermometry\nSUB TOPIC:Heat transfer", "solution_images": ["images/image220.png", "images/image221.png", "images/image213.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-01-Q15", "question": "An electromagnetic wave has an electric field given by the\nexpression (in Cartesian co-ordinates):\n[IMAGE] What is the direction of the magnetic field at time t = 0 and position x\n= 0 ?", "question_images": ["images/image73.png"], "option_1": "-x", "option_2": "+x", "option_3": "-y", "option_4": "+y", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image74.png", "images/image75.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-02-Q55", "question": "In the circuit shown V_d - V_a is ( in V )", "question_images": ["images/image116.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "02", "solution": "[IMAGE] Along agbfcd\nV_a + 1 - 2 - 4 + 8 - 1 = V_d\n⇒ V_a + 2 = V_d ⇒ V_d - V_a = 2 V", "solution_images": ["images/image117.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-21-Q24", "question": "For an amplitude modulated wave maximum and minimum amplitude\nare 10 V and 2 V respectively.Modulation index is n/3. Find n:\n 10 V rFkk 2 V", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "A_max = A_m + A_c = 10\nA_min = A_c - A_m = 2\nA_c = 6 V\nA_m = 4 V\n$m = \\frac{A_{m}}{A_{c}} = \\frac{4}{6} = \\frac{2}{3}$", "solution_images": [], "subject": "Physics", "topic": "Communication system", "subtopic": "Modulation", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-18-Q19", "question": "A bullet of mass m hits the hanging block of mass M at its\ncentre completely inelastically with a velocity v and gets embedded in\nit as shown, then choose the correct option", "question_images": ["images/image26.png"], "option_1": "The tension in the string just before and after the collision changes\nby$\\frac{m^{2}v^{2}}{(m + M)\\mathcal{l}}$", "option_2": "The tension in the string just before and after the collision changes\nby $m\\lbrack g + \\frac{mv^{2}}{(m + M)\\mathcal{l}}\\rbrack$", "option_3": "Tension in the string just after the collision\nbecomes$m\\lbrack g + \\frac{mv^{2}}{(m + M)\\mathcal{l}}\\rbrack$", "option_4": "Tension in the string will not change after the collision", "correct_option": 2, "numerical_answer": null, "solution": "As $T_{1} = Mg$\nAfter hitting\n$T_{2} = (M + m)g + \\frac{m^{2}v^{2}}{(M + m)\\mathcal{l}}$\n$\\therefore T_{2} - T_{1} = m\\lbrack g + \\frac{mv^{2}}{(m + M)\\mathcal{l}}\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "System of particle", "subtopic": "Collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-03-Q22", "question": "A particle is projected from point P with velocity\n[IMAGE] ms-1 perpendicular to the surface of a hollow\nright angled cone as shown in figure. It collides at Q normally. The\nvelocity with which it collides at Q will be ( in m/sec )\n[IMAGE] ms-1 \nP Q Q", "question_images": ["images/image136.png", "images/image137.png", "images/image137.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "15", "solution": "[IMAGE] u cos 30° = V cos 60°\n[IMAGE] TOPIC:KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image138.png", "images/image139.png", "images/image140.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-10-Q20", "question": "A body stretches a spring by x_0 at the equator of the earth of\naverage radius R. The height at which it will stretch the spring by the\nsame amount at North Pole is (assume [IMAGE] to be\nangular velocity of the earth and g is value of gravity on the surface\nof the earth)", "question_images": ["images/image63.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these", "correct_option": 3, "numerical_answer": null, "solution": "$\\frac{GM}{R^{2}} - m\\omega^{2}R = kx_{0}$.............(i)\nNone at height h $\\frac{GmM}{(R + h)^{2}} = kx_{0}$............(ii)\nEquating 1 & 2[IMAGE]", "solution_images": ["images/image67.png"], "subject": "Physics", "topic": "Gravitation", "subtopic": "Acceleration due to gravity at poles and equator", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "Ph-26-Q28", "question": "A non-isotropic solid metal cube has coefficients of linear\nexpansion as:$5 \\times \\ 10^{–5}\\ /{^\\circ}C$ along the x-axis and\n$5 \\times 10^{–6}\\ /{^\\circ}C$ along the y and the z-axis. If the\ncoefficient of volume expansion of the solid is\n$C \\times 10^{–5}\\ /{^\\circ}C$ then the value of C is.........\nizdkj $5 \\times \\ 10^{–5}\\ /{^\\circ}C$ x- rFkk\nizlkj xq.kkad $C \\times 10^{–5}\\ /{^\\circ}C$ gks] rks C", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "$V = 2\\alpha_{2} + \\alpha_{1}$\n$= 10 \\times 10^{- 6} + 5 \\times 10^{- 5}$\n$= 60 \\times 10^{- 6}/\\ ^{\\circ}C$", "solution_images": [], "subject": "Physics", "topic": "Thermal properties of matter", "subtopic": "Thermal expansion", "difficulty": "Moderate", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-25-Q4", "question": "A litre of dry air at STP expands adiabatically to a volume of 3\nlitres. If [IMAGE] the work done by air is:\n air to be an ideal gas]\n1 yhVj gok tks fd ekud rki o nkcSTP ij gS]:)ks\"e\nçfØ çlkfjr gksdj 3 yhVjvk;ru", "question_images": ["images/image6.png", "images/image7.png", "images/image6.png", "images/image7.png"], "option_1": "88.75 J", "option_2": "48 J", "option_3": "60.7 J", "option_4": "100.8 J", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image8.png", "images/image9.png", "images/image10.png", "images/image11.png", "images/image12.png", "images/image13.png"], "subject": "Physics", "topic": "Thermodynamic", "subtopic": "Adiabatic process", "difficulty": "Hard", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-08-Q2", "question": "Two moles of hydrogen are mixed with n moles of helium. The root\nmean square speed of gas molecules in the mixture is $\\sqrt{2}$ times\nthe speed of sound in the mixture. Then n is.", "question_images": [], "option_1": "3", "option_2": "2", "option_3": "1.5", "option_4": "2.5", "correct_option": 2, "numerical_answer": null, "solution": "$V_{rms} = \\sqrt{\\frac{3RT}{M_{mix}}}$\n$V_{sound\\ } = \\sqrt{\\frac{\\gamma RT}{M_{mix}}}$\n$V_{rms} = \\sqrt{2}V_{sound\\ }$\n$\\sqrt{\\frac{3RT}{M_{mix}}} = \\sqrt{2}\\sqrt{\\frac{\\gamma RT}{M_{mix}}}$\n$\\gamma_{mix} = \\frac{3}{2}$\n$\\gamma_{mix} = \\frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$\n$\\frac{3}{2} = \\frac{2 \\times \\frac{7R}{2} + n \\times \\frac{5R}{2}}{2 \\times \\frac{5R}{2} + n \\times \\frac{3R}{2}}$\n$\\Rightarrow \\frac{3}{2} = \\frac{14 + 5n}{10 + 3n}$\n$\\Rightarrow 30 + 9n = 28 + 10n$\n$\\Rightarrow n = 2$", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Root mean square speed", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "Ph-25-Q16", "question": "During an experiment an ideal gas is found to obey condition\ngas). The gas is initially at a temperature [IMAGE],\npressure [IMAGE] and density [IMAGE].\nThe gas expands such that density changes to [IMAGE] then which of following is/are correct?,d iz;ksx", "question_images": ["images/image114.png", "images/image115.png", "images/image116.png", "images/image117.png", "images/image118.png", "images/image119.png", "images/image114.png", "images/image115.png", "images/image116.png", "images/image117.png", "images/image118.png", "images/image119.png"], "option_1": "The pressure of gas changes to [IMAGE]", "option_2": "The temperature of gas changes to [IMAGE].", "option_3": "The graph of above process on the P - T diagram is parabola.", "option_4": "The graph of above process on the P - T diagram is not rectangular\nhyperbola.", "correct_option": 2, "numerical_answer": null, "solution": "PT = constant", "solution_images": ["images/image122.png", "images/image123.png", "images/image124.png", "images/image125.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Ideal Gas", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-06-Q8", "question": "Two blocks of masses $M_{1}$ and $M_{2}$are connected to each\nother through a light spring as shown in figure. If we push mass $M_{1}$\nwith force F and cause acceleration $a_{1}$ in mass $M_{1}$, what will\nbe the magnitude of acceleration in $M_{2}$?", "question_images": ["images/image13.png"], "option_1": "$F/M_{2}$", "option_2": "$F/\\left( M_{1} + M_{2} \\right)$", "option_3": "$a_{1}$", "option_4": "$\\left( F - M_{1}a_{1} \\right)/M_{2}$", "correct_option": 4, "numerical_answer": null, "solution": "$kx = M_{2}a_{2}$ [Newton's II law for $M_{2}$]\nBy adding both equations.\n$F = M_{1}a_{1} + M_{2}a_{2} \\Rightarrow a_{2} = \\frac{F - M_{1}a_{1}}{M_{2}}$", "solution_images": ["images/image14.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Newton's second law for a system", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "Ph-28-Q9", "question": "A uniform magnetic field B in positive z direction exists in a\ncircular region of radius R = 5 m. A loop of radius R = 5m lying in x -\ny plane encloses the magnetic field at t = 0 and then pulled at uniform\nvelocity[IMAGE]. The emf induced (in volts) is the loop\nat t = 2 sec is 6V. Then magnitude of 4B is", "question_images": ["images/image82.png", "images/image83.png"], "option_1": "2", "option_2": "3", "option_3": "1", "option_4": "5", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] TOPIC:Electromagnetic induction\nSUB TOPIC:Faraday's law\nLEVEL:Moderate", "solution_images": ["images/image84.png", "images/image85.png", "images/image86.png", "images/image87.png", "images/image88.png", "images/image89.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-25-Q9", "question": "70 calories of heat is required to raise the temperature of 2\nmoles of an ideal gas at constant pressure from 30°C to 35°C (R= 2 cal\n/mol-°C) The gas may be:\nls] bl gks tkrk\ngs rks xSl gks ldrh", "question_images": ["images/image50.png", "images/image51.png", "images/image52.png"], "option_1": "", "option_2": "He", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] So gas is", "solution_images": ["images/image56.png", "images/image57.png", "images/image58.png", "images/image59.png", "images/image60.png", "images/image60.png"], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Ideal Gas", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "Ph-27-Q16", "question": "The speed of water flowing out of the orifice before the\ncylinder kept inside the tank\ncsyu dks VSd iwoZ fNnz", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image86.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Velocity of efflux", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-10-Q19", "question": "A satellite is orbiting around the earth in a circular orbit\nvery close to surface of earth then comparing with other satellites of\nsame mass moving in circular orbits around the earth.", "question_images": [], "option_1": "The total energy of earth plus given satellite system is minimum", "option_2": "The time period of revolution of given satellite is minimum", "option_3": "The orbital speed of given satellite is maximum", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$v_{0} = \\sqrt{\\frac{GM}{(R + h)}}$\n$T = \\frac{2\\pi(R + h)}{v_{0}} = \\frac{2\\pi(R + h)^{3/2}}{\\sqrt{GM}}$\n$E = \\frac{- GMm}{2(R + h)}$", "solution_images": [], "subject": "Physics", "topic": "Gravitation", "subtopic": "Satellite orbiting around earth", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-05-Q2", "question": "Two seconds after projection, a projectile is travelling in a\ndirection inclined at 30º to the horizontal after one more second, it is\ntravelling horizontally. Then the speed of projection(u) and angle of\nprojection ($\\theta$) are given by", "question_images": [], "option_1": "$u = 10\\sqrt{3}m/s$and$\\theta = 45^{\\circ}$", "option_2": "$u = 10\\sqrt{3}m/s$and$\\theta = 60^{\\circ}$", "option_3": "$u = 20\\sqrt{3}m/s$and$\\theta = 60^{\\circ}$", "option_4": "$u = 20\\sqrt{3}m/s$and$\\theta = 45^{\\circ}$", "correct_option": 3, "numerical_answer": null, "solution": "$T = 6sec = \\frac{2usin\\theta}{g} \\Rightarrow usin\\theta = 30$\n$\\tan 30^{\\circ} = \\frac{{usin}\\theta - gt}{{ucos}\\theta} = \\frac{30 - 20}{{ucos}\\theta}$which\ngives $ucos\\theta = 10\\sqrt{3}$\n$$tan\\theta =", "solution_images": [], "subject": "Physics", "topic": "Projectile Motion", "subtopic": "Velocity of projectile at an instant", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-14-Q13", "question": "The magnifying power of a telescope with tube length 60 cm is5.\nWhat is the focal length of its eye piece?", "question_images": [], "option_1": "30 cm", "option_2": "10 cm", "option_3": "40 cm", "option_4": "50 cm", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image91.png", "images/image92.png", "images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Optical instruments", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "P-17-Q3", "question": "In the figure shown a parallel plate capacitor has a dielectric\nof width d/2 and dielectric constant K=2. The other dimensions of the\ndielectric are same as that of the plates. The plates P_1 and P_2 of\nthe capacitor have area each. The energy of the capacitor is", "question_images": [], "option_1": "$\\frac{\\epsilon_{0}AV^{2}}{3d}$", "option_2": "$\\frac{2\\epsilon_{0}AV^{2}}{d}$", "option_3": "$\\frac{3}{2}\\frac{\\epsilon_{0}AV^{2}}{d}$", "option_4": "$\\frac{2\\epsilon_{0}AV^{2}}{3d}$", "correct_option": 4, "numerical_answer": null, "solution": "$U = \\frac{1}{2}C_{eq}V^{2}$\n$C_{1} = \\frac{k\\varepsilon_{0}A}{d/2} = \\frac{2\\varepsilon_{0}A}{(d/2)}$\nand $C_{2} = \\frac{\\varepsilon_{0}\\ A}{d/2}$\n$C_{eq} = \\frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \\frac{(2\\frac{\\varepsilon_{0}}{d/2}) \\times \\frac{\\varepsilon_{0}}{d/2}}{3\\frac{\\varepsilon_{0}\\ A}{d/2}}$\n$= \\frac{4}{3}\\frac{\\varepsilon_{0}\\ A}{d}$ or\n$C_{eq} = \\frac{\\varepsilon_{0}A}{d - \\frac{d}{2} + \\frac{d}{4}} = \\frac{4\\varepsilon_{0}A}{3d}$\n$U = \\frac{1}{2}(\\frac{4}{3}\\frac{\\varepsilon_{0}A}{d}) \\vee^{2} = \\frac{2}{3}(\\frac{\\varepsilon_{0}A}{d})v^{2}$", "solution_images": [], "subject": "Physics", "topic": "Capacitance", "subtopic": "Dielectric", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-03-Q7", "question": "The position of a projectile launched from the origin at t = 0 is\ngiven by [IMAGE] at\nt = 2s. If the projectile was launched at an angle θ from the\nhorizontal, then θ is (take g=10 ms-2)", "question_images": ["images/image36.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE] t = 0 t = 2s \n  θ  θ  (g = 10 ms-2 )", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] 40 = Sx = V cos θ × t\n50 = Sy = V sin θt - [IMAGE] gt2\nt = 2 given\nθ = tan-1 [IMAGE] TOPIC: KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image41.png", "images/image42.png", "images/image43.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-04-Q16", "question": "A massless thread passes through a ring (smooth). The ring is in a\ncircular path with both part of the thread in tension. Which of the\nfollowing values of θ and α are possible?", "question_images": ["images/image92.png"], "option_1": "θ > α", "option_2": "α > θ", "option_3": "θ = α", "option_4": "All of these\n () \n θ α ?", "correct_option": 2, "numerical_answer": null, "solution": "Tension will be same in both parts.\n[IMAGE] Tcosθ = mg + Tcosα\n⇒ α > θ", "solution_images": ["images/image93.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-03-Q3", "question": "A particle is projected upwards with velocity v = 30 m/s. There are\ntwo points A and B at heights hA = 25 m and hB = 40m. If the particle\ncrosses point B twice in a time interval ∆t. Then, it will also cross A\ntwice in a time interval equal to", "question_images": ["images/image8.png"], "option_1": "2∆t", "option_2": "1.5 ∆t", "option_3": "3∆t", "option_4": "2.5 ∆t\n v = 30 m/s. hA = 25 hB = 40\n A B B ∆t \n A", "correct_option": 1, "numerical_answer": null, "solution": "hmax = [IMAGE] = 45m\nTime to go from C to B,\n5 = [IMAGE] gt2\nt = 1s, So ∆t = 2s\nTime to go from C to A,\n20 =[IMAGE] gt2\nt = 2s\n∆tA = 2 × 2 = 4s = 2∆t\nTOPIC: KINEMATIC\nSUB TOPIC: MOTION UNDER GRAVITY", "solution_images": ["images/image9.png", "images/image10.png", "images/image11.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-19-Q19", "question": "[IMAGE] As shown in the figure, a battery of emf $\\varepsilon$ is connected to\nan inductor L and resistance R in series. The switch is closed at t=0.\nThe total charge that flows from the battery, between t=0 and\n$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is", "question_images": ["images/image27.png"], "option_1": "$\\frac{\\varepsilon L}{R^{2}}(1 - \\frac{1}{e})$", "option_2": "$\\frac{\\varepsilon R}{eL^{2}}$", "option_3": "$\\frac{\\varepsilon L}{R^{2}}$", "option_4": "$\\frac{\\varepsilon L}{eR^{2}}$", "correct_option": 4, "numerical_answer": null, "solution": "$q = \\int_{0}^{T_{C}}idt$\n$= \\frac{\\varepsilon}{R}{\\lbrack t - \\frac{e^{- t/T_{C}}}{\\frac{- 1}{T_{C}}}\\rbrack}_{0}^{T_{C}}$;\n$= \\frac{\\varepsilon}{R}\\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\\rbrack$\n$= \\frac{\\varepsilon}{R} \\times \\frac{1}{e} \\times \\frac{L}{R}$\n$= \\frac{\\varepsilon L}{R^{2}e}$", "solution_images": [], "subject": "Physics", "topic": "Electromagnetic induction", "subtopic": "LR circuit", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-17-Q11", "question": "In LC circuit the inductance L= 40 mH and\ncapacitance$C = 100\\mu F$. If a voltage V(t)=10 sin (314 t) is applied\nto the circuit, the current in the circuit is given as", "question_images": [], "option_1": "$0.52\\ cos314t$", "option_2": "$10\\ cos314t$", "option_3": "$5.2\\ cos314t$", "option_4": "$0.52\\ sin314t$", "correct_option": 1, "numerical_answer": null, "solution": "$R = 0$\n$z = X_{C} - X_{L}$\n$= \\frac{1}{\\omega C} - \\omega L$\n$= \\frac{1}{314 \\times 100 \\times 10^{- 6}} - 314 \\times 40 \\times 10^{- 3}$\n$= 31.84 - 12.56$\n$= 19.28\\Omega$\n$X_{C} > X_{L}$", "solution_images": ["images/image11.jpeg"], "subject": "Physics", "topic": "AC", "subtopic": "LCR", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-23-Q19", "question": "In the given diagram aproton is given a speed v at Pin the\nhorizontal direction then the time after which it comesout of the\nmagnetic field B. m is the mass of proton.", "question_images": ["images/image50.png"], "option_1": "$\\frac{m\\pi}{eB}$", "option_2": "$\\frac{2m\\pi}{eB}$", "option_3": "$\\frac{3m\\pi}{4eB}$", "option_4": "It will never come out of the magnetic field", "correct_option": 1, "numerical_answer": null, "solution": "Radius of charge particle\n$r = \\frac{mV}{eB}$\nhere $r < \\frac{d}{2}$, proton will complete a semicircle and comes out\nof the magnetic field.", "solution_images": [], "subject": "Physics", "topic": "Mangetism", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-15-Q3", "question": "When an electron of hydrogen like atom jumps from a higher energy\nlevel to lower energy level, then", "question_images": [], "option_1": "angular momentum of the electron about nucleus remains constant", "option_2": "kinetic energy increases", "option_3": "de-Broglie wavelength, associated with motion of electron\nincreases", "option_4": "none of the above", "correct_option": 2, "numerical_answer": null, "solution": "Angular momentum [IMAGE] Kinetic energy [IMAGE] De-Broglie wavelength [IMAGE] So as v increases [IMAGE] decreases.", "solution_images": ["images/image27.png", "images/image28.png", "images/image29.png", "images/image30.png"], "subject": "Physics", "topic": "Atomic Physics", "subtopic": "Bohr's model of hydrogen atom", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "P-18-Q7", "question": "In the figure shown a block A of mass m is kept on block B of\nmass m. A is given a velocity towards right as shown. The coefficient of\nfriction between A and B is $\\frac{3\\mu}{2}$ and that between B and\nground is$\\ \\mu$.", "question_images": ["images/image10.png"], "option_1": "B will accelerate towards right with acceleration$\\frac{\\mu g}{2}$.", "option_2": "B will accelerate towards left with acceleration$\\frac{\\mu g}{2}$", "option_3": "B will not accelerate at all.", "option_4": "B will move towards right with acceleration$\\ \\mu g$.", "correct_option": 3, "numerical_answer": null, "solution": "as$f_{k^{'}\\max} > \\frac{3}{2}\\mu g$\nSo it will not move", "solution_images": ["images/image11.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-27-Q33", "question": "A Carnot engine operates between two reservoirs of temperatures\n900 K and 300 K. The engineperforms 1200 J of work per cycle. The heat\nenergy (in J) delivered by the engine to the lowtemperature reservoir,\nin a cycle, is 100n. Find n,d dkuksZ bUtu dks 900 K vkSj 300 K ds nks HkaMkjks a ds chp pyk;k\nfuEu rki okys HkaMkj esa izfr pØ 100n (J esa) NksM+rk gSFind\nn", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "6", "solution": "", "solution_images": ["images/image213.png", "images/image214.png", "images/image215.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Carnot engine", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-04-Q12", "question": "If the mass of block is 1 kg and a force of [IMAGE] N\nis applied horizontally on the block as shown in the figure. The\nfrictional force acting on the block is", "question_images": ["images/image72.png", "images/image73.png"], "option_1": "zero", "option_2": "", "option_3": "", "option_4": "5 N\n 1 kg [IMAGE]", "correct_option": 1, "numerical_answer": null, "solution": "Net force along the plane\n= mg sin 30 - FH cos 30\n[IMAGE] So friction force is zero.\nTOPIC: NLM\nSUB TOPIC: FRICTION", "solution_images": ["images/image76.png", "images/image77.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-08-Q9", "question": "A container of fixed volume has a mixture of one mole of hydrogen\nand one mole of helium in equilibrium at temperature T. Assuming the\ngases are ideal, the correct statement(s) is (are)", "question_images": [], "option_1": "The average energy per mole of the gas mixture is 2RT.", "option_2": "The ratio of speed of sound in the gas mixture to that in helium\ngas is $\\sqrt{6/5}$", "option_3": "The ratio of the rms speed of helium atoms to that of hydrogen\nmolecules is $1/\\sqrt{2}$", "option_4": "$\\ $All of the above", "correct_option": 4, "numerical_answer": null, "solution": "$\\ $For hydrogen,$\\ n_{1} = 1,C_{v,} = \\frac{5}{2}R$\n For mixture of gases,\n$C_{v} = \\frac{n_{1}C_{V,} + n_{2}C_{V_{2}}}{n_{1} + n_{2}}$\n$= \\frac{1 \\times \\frac{5}{2}R + 1 \\times \\frac{3}{2}R}{1 + 1} = 2R$\n$C_{P} = C_{v} + R = 3R,\\gamma_{mix} = \\frac{C_{P}}{C_{v}} = \\frac{3}{2}$\nAlso, $\\gamma_{mx} = 1 + \\frac{2}{f} \\Rightarrow \\frac{3}{2} = 1 + \\frac{2}{f}\\therefore f = 4$\n$\\therefore$ Average energy per mole $= \\frac{1}{2}fRT = 2RT$\n$M_{\\max} = \\frac{n_{1}M_{1} + n_{2}M_{2}}{n_{1} + n_{2}} = \\frac{1 \\times 2 + 1 \\times 4}{1 + 1} = 3gm{ol}^{- 1}$\n Speed of sound in a gas, $v = \\sqrt{\\frac{\\gamma RT}{M}}$\nFor a given value of T,v$\\propto \\sqrt{\\frac{\\gamma}{M}}$\n Rms speed of a gas molecule at temperature T is given by \n$T,v \\propto \\sqrt{\\frac{\\gamma}{M}}$\n For a given value of T,$\\ v_{rms} \\propto \\frac{1}{\\sqrt{M}}$\n$\\frac{\\left( v_{rms\\ } \\right)He}{\\left( v_{rms\\ } \\right)H_{2}} = \\sqrt{\\frac{M_{H_{2}}}{M_{He}}} = \\sqrt{\\frac{2}{4}} = \\frac{1}{\\sqrt{2}}$\n So, options (a), (b) and (c) are correct", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "RMS speed", "difficulty": "Tough", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-17-Q7", "question": "In a Young's double slit experiment 15 fringes are observed on a\nsmall portion of the screen when light of wavelength 500 nm is used. Ten\nfringes are observed on the same section of the screen when another\nlight source of wavelength $\\lambda$ is used Then the value of\n$\\\\lambda\\$ is (in nm)______.", "question_images": [], "option_1": "850 nm", "option_2": "650 nm", "option_3": "750 nm", "option_4": "550 nm", "correct_option": 3, "numerical_answer": null, "solution": "$15 \\times 500 \\times \\frac{D}{d} = 10 \\times \\lambda_{2} \\times \\frac{D}{d}$\n$\\lambda_{2} = 15 \\times 50nm$\n$\\lambda_{2} = 750nm$", "solution_images": [], "subject": "Physics", "topic": "Wave Optics", "subtopic": "YDSE", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-23-Q27", "question": "In the given A.C. circuit, if battery voltage\nis[IMAGE], power delivered by battery will have a power\nfactor of[IMAGE]. Find n.", "question_images": ["images/image88.png", "images/image89.png", "images/image90.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "[IMAGE] Phasor diagram\n[IMAGE] TOPIC: Alternating current\nSUB TOPIC: LCR circuit\nLEVEL: Moderate", "solution_images": ["images/image91.png", "images/image92.png", "images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png", "images/image97.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-10-Q13", "question": "A particle of mass m is made to move with uniform speed v along\nthe perimeter of a regular polygon of n sides inscribed in a circle of\nradius a. The magnitude of impulse applied at each corner of the polygon\nis", "question_images": [], "option_1": "$2\\ mv\\ sin\\frac{\\pi}{n}$", "option_2": "$mv\\ sin\\frac{\\pi}{n}$", "option_3": "$mv\\ sin\\frac{2\\pi}{n}$", "option_4": "$2mv\\ sin\\frac{n}{\\pi}$", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] Magnitude of the impulse at each corner is simply the change in momentum\nof the particle.\nMagnitude of final momentum = mV\nMagnitude of initial momentum = mV\nAngle between two momentums =[IMAGE] Hence,\n$\\left| {\\overset{\\rightarrow}{P}}_{final\\ } - {\\overset{\\rightarrow}{P}}_{initial\\ } \\right| = |mpulse = \\left| \\sqrt{P^{2} + P^{2} - 2P^{2}\\cos\\left( \\frac{2\\pi}{n} \\right)} \\right|$\n$\\Rightarrow$ Impulse $\\left. \\ = mV \\mid \\sqrt{2\\left( 1 - cos\\left( \\frac{2\\pi}{n} \\right) \\right.\\ } \\right)| = mv|\\sqrt{4\\sin^{2}\\left( \\frac{\\pi}{n} \\right)} \\mid = 2mvsin\\frac{\\pi}{n}$", "solution_images": ["images/image49.png", "images/image50.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Impulse momentum equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-20-Q1", "question": "A uniform rod of length L and mass M is placed on a smooth\nhorizontal surface. The rod is hinged at the end A and is free to rotate\nin horizontal plane about a vertical axis passing through A. As shown in\nfigure there is a nail N at a perpendicular distance $\\frac{L}{4}$ from\nend A of rod. At t = 0 and impulse P_0 is applied at a distance\n$\\frac{3L}{4}$ from end A. The impulse is in the horizontal plane and is\nperpendicular to the rod. At $t = \\frac{8\\pi ML}{27P_{0}}$, the rod\nreturns to its initial position. Find the impulse of the force exerted\nby the nail on the rod.", "question_images": ["images/image1.png"], "option_1": "$11P_{0}$", "option_2": "$12P_{0}$", "option_3": "$10P_{0}$", "option_4": "$13P_{0}$", "correct_option": 2, "numerical_answer": null, "solution": "$P_{0}\\frac{3L}{4} = \\frac{M\\mathcal{l}^{2}}{3}\\omega_{0}$\n$t_{1} = \\frac{\\pi}{2\\omega_{0}}$\n$t_{2} = \\frac{8\\pi ML}{27P_{0}} - t_{1}$\n$\\omega = \\frac{\\pi}{2t_{2}}$\n$P\\frac{L}{4} = \\frac{M\\mathcal{l}^{2}}{3}(\\omega_{0} + \\omega)$\n$P = 12P_{0}$", "solution_images": [], "subject": "Physics", "topic": "Mechanics", "subtopic": "Angular impulse", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-05-Q25", "question": "The velocity time relation of an electron starting from rest is\ngiven by v=K t, where K=2 m/s^2. The distance traversed in 3 sec is 3n\nthen the find the value of n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": null, "solution": "$a = \\frac{dv}{dt} = k = 2m/s^{2}$\n$S = 0 + \\frac{1}{2}(2)(3)^{2} = 9m$", "solution_images": [], "subject": "Physics", "topic": "Motion in 1D", "subtopic": "Uniform acceleration", "difficulty": "Easy", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-09-Q9", "question": "A horizontal spring-block system of mass 2kg executes S.H.M.\nWhen the block is passing through its equilibrium position, an object of\nmass 1kg is put gently on it and the two move together. The new\namplitude of vibration is (A being its initial amplitude)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "Conserving momentum: 2V = 3V′ [IMAGE]", "solution_images": ["images/image53.png", "images/image54.png", "images/image55.png", "images/image56.png", "images/image57.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Amplitude of SHM", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-23-Q29", "question": "The maximum peak to peak voltage of an AM wave is\n[IMAGE] and the\nminimum peak to peak voltage is[IMAGE]. The modulation factor is 10x. Find x.", "question_images": ["images/image102.png", "images/image103.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "", "solution_images": ["images/image104.png", "images/image105.png"], "subject": "Physics", "topic": "Communication system", "subtopic": "Modulation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-11-Q25", "question": "A uniform rod of length L and mass M is placed on a smooth\nhorizontal surface. The rod is hinged at the end A and is free to rotate\nin horizontal plane about a vertical axis passing through A. As shown\ninfigure there is a nail N at a perpendicular distance\n[IMAGE] from end A of rod. At t = 0 and impulse\n[IMAGE] isapplied at a distance [IMAGE] from end A. The impulse is in the horizontal plane and is perpendicular\nto therod. At [IMAGE] the rod returns to its initial\nposition. If the impulse of the force exerted by the nail on\nthe rod is $\\frac{36\\text{P}_{0}}{\\text{x}},$then value of x is", "question_images": ["images/image236.png", "images/image237.png", "images/image238.png", "images/image239.png", "images/image240.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "", "solution_images": ["images/image241.png", "images/image242.png", "images/image243.png", "images/image244.png", "images/image245.png", "images/image246.png", "images/image247.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Angular impulse", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-10-Q8", "question": "A disc of radius R rolls on a horizontal surface with linear\nvelocity V and angular velocity[IMAGE]. There is a point\nP on circumference of disc at angle [IMAGE] with upward\nvertical diameter measured anticlockwise (see figure), which has a\nvertical velocity. Here [IMAGE] is equal to.", "question_images": ["images/image33.png", "images/image34.png", "images/image35.png", "images/image36.png"], "option_1": "$\\pi \\pm \\sin^{- 1}\\left( \\frac{V}{\\omega R} \\right)$", "option_2": "$\\pi \\pm \\cos^{- 1}\\left( \\frac{V}{\\omega R} \\right)$", "option_3": "$2\\pi \\pm \\cos^{- 1}\\left( \\frac{V}{\\omega R} \\right)$", "option_4": "$\\frac{3\\pi}{2} \\pm \\cos^{- 1}\\left( \\frac{V}{\\omega R} \\right)$", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] For vertical velocity of point P\n$\\omega{Rcos}\\theta = - V$\n$\\Rightarrow$\n$\\omega{Rcos}(\\pi - \\theta) = V\\ or\\ \\omega{Rcos}(\\pi + \\theta) = V$\n$\\therefore$\n$\\theta = \\pi + \\cos^{- 1}\\left( \\frac{V}{\\omega R} \\right)$\nor $\\theta = \\pi - \\cos^{- 1}\\left( \\frac{V}{\\omega R} \\right)$", "solution_images": ["images/image37.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Pure rolling", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-21-Q18", "question": "For a system of two\n as shown in the\nfigure (both are at origin and perpendicular toeach other along x and y\naxes respectively) (P_1 and P_2denotes magnitudes\n and r is quite large in\ncomparison to dimensions of dipoles,\n[IMAGE] is resultant electric field and QPR is a quarter\nof circle whose centre is at O)\nfp=kuqlkj nks\nfy, ¼nksuksa ewy fcUnq ij gS rFkk,d nwljs ds yEcor~ gS rFkk Øe'k x\no y chp dh nwjh dh rqyuk esa r \npkSFkkbZ Hkkx", "question_images": ["images/image61.png", "images/image62.png", "images/image63.png", "images/image64.png", "images/image65.png", "images/image61.png", "images/image66.png", "images/image67.png", "images/image68.png", "images/image69.png", "images/image70.png"], "option_1": "Work done in taking electron from P to R on QPR = 0\nP ls R rd QPR ij ys tkus", "option_2": "$tan\\alpha = \\frac{\\left( P_{1} + P_{2} \\right)}{2\\left( P_{1} - P_{2} \\right)}$", "option_3": "$tan\\alpha = \\frac{\\left( P_{1} - P_{2} \\right)}{2\\left( P_{1} + P_{2} \\right)}$", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "$E_{1} = \\frac{2k\\frac{\\left( P_{1} + P_{2} \\right)}{\\sqrt{2}}}{r3}$\n$E_{2} = \\frac{k\\frac{\\left( P_{1} - P_{2} \\right)}{\\sqrt{2}}}{r3}$\n$tan\\alpha = \\frac{E_{2}}{E_{1}}$\n$= \\frac{P_{1} - P_{2}}{2\\left( P_{1} + P_{2} \\right)}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field due to electric dipole", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "Ph-28-Q12", "question": "A uniform solid sphere of radius R and mass M purely rolls down\nan inclined plane. The coefficient of friction between the sphere and\nthe inclined plane is[IMAGE]. If the maximum value of\nangle of inclination for this to be possible is\n[IMAGE] then n is: (Take[IMAGE] )", "question_images": ["images/image111.png", "images/image112.png", "images/image113.png", "images/image114.png"], "option_1": "7", "option_2": "3", "option_3": "5", "option_4": "6", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] put in (1)\n[IMAGE] Now [IMAGE] TOPIC:Rotaional dynamics\nSUB TOPIC: Rolling down the incline\nLEVEL: Moderate", "solution_images": ["images/image115.png", "images/image116.png", "images/image117.png", "images/image118.png", "images/image119.png", "images/image120.png", "images/image121.png", "images/image122.png", "images/image123.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-14-Q4", "question": "In a double-slit experiment, at a certain point on the screen the\npath difference between the two interfering waves is\n[IMAGE] of a wavelength. The ratio of the intensity of\nlight at that point to that at the centre of a bright fringe is", "question_images": ["images/image17.png"], "option_1": "0.568", "option_2": "0.760", "option_3": "0.853", "option_4": "0.672", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": ["images/image18.png", "images/image19.png"], "subject": "Physics", "topic": "Wave Optics", "subtopic": "Interference", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "Ph-26-Q19", "question": "A container of fixed volume has a mixture of one mole of\nhydrogen and one mole of helium in equilibrium at temperature T.\nAssuming the gases are ideal, the correct statement(s) is (are)", "question_images": [], "option_1": "The average energy per mole of the gas mixture is 2RT.", "option_2": "The ratio of speed of sound in the gas mixture to that in helium\ngas is $\\sqrt{6/5}$", "option_3": "The ratio of the rms speed of helium atoms to that of hydrogen\nmolecules is $1/\\sqrt{2}$", "option_4": "$\\ $All of the above,d ik= esa 1 eksy gkbMªkstu rFkk 1", "correct_option": 4, "numerical_answer": null, "solution": "$\\ $For hydrogen,$\\ n_{1} = 1,C_{v,} = \\frac{5}{2}R$\ngkbMªkstu ds fy, [IMAGE] For mixture of gases,\n$C_{v} = \\frac{n_{1}C_{V,} + n_{2}C_{V_{2}}}{n_{1} + n_{2}}$\n$= \\frac{1 \\times \\frac{5}{2}R + 1 \\times \\frac{3}{2}R}{1 + 1} = 2R$\n[IMAGE] T ds fn, x, eku ds fy,[IMAGE] T ds fn, x, eku ds fy,[IMAGE] Rms speed of a gas molecule at temperature Tis given by\nT rkieku ij xSl dhRms pky\n[IMAGE] T", "solution_images": ["images/image50.png", "images/image51.png", "images/image52.png", "images/image53.png", "images/image54.png", "images/image55.png", "images/image56.png", "images/image57.png", "images/image58.png", "images/image59.png", "images/image60.png", "images/image61.png", "images/image62.png", "images/image63.png"], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "RMS speed", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-03-Q14", "question": "A rotor is rotating with a constant angular velocity ω about its\nown axis. At the centre of the rotor a spring is fixed of natural length\n(l0) whose other end is connected to a block of mass m. Find the\nminimum value of coefficient of friction between block and rotor wall\nfor which spring remains horizontal. The radius of rotor is R (R >\nl0).", "question_images": ["images/image82.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE] ω (l0)\n m \n \nR (R > l0)", "correct_option": 2, "numerical_answer": null, "solution": "N = mω2R - K(R - l0)... (1)\nFor the block to remains in the vertical equilibrium,\n[IMAGE] f = mg\n⇒ N = mg\n⇒ µ(mω2R - K (R - l0) = mg \n∴ µ =[IMAGE] TOPIC: CIRCULAR MOTION\nSUB TOPIC: DYNAMICS", "solution_images": ["images/image87.png", "images/image88.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "Ph-26-Q20", "question": "M grams of steam at 100°C is mixed with 200g of ice at its\nmelting point in a thermally insulated container. If it produces liquid\nwater at 40°C and heat of\nfusion of ice is 80 the value of M is______.\n100°C rkieku dh M xzke ok\"i dks 200g xzke cQZ esa,d crZu\n A ok\"i igys cQZ eku vius xyukad ds\ncjkcj FkkA izfØ vUr esa 40°C", "question_images": [], "option_1": "40", "option_2": "80", "option_3": "120", "option_4": "160", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image64.png", "images/image65.png", "images/image66.png", "images/image67.png"], "subject": "Physics", "topic": "Thermal properties of matter", "subtopic": "Latent heat and specific heat", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-03-Q6", "question": "A projectile is thrown, from the origin along X-Y plane to have\nmaximum possible horizontal range of 100 m. Then the position vector of\nthe projectile when its speed is minimum is (take Y on the vertical\naxis)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-03-Q17", "question": "A person projected a small ball with speed U along the floor from\npoint A. If x = 3R, determine the required speed U so that the ball\nreturn to A after moving the circular surface in the vertical plane from\nB to C and becoming projectile at C.", "question_images": ["images/image105.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE] A U \nx = 3R, U B C \n C A", "correct_option": 4, "numerical_answer": null, "solution": "V2 = V2 - 4gR\n3R = V [IMAGE] V2 = U2 - 2g2R\nU2 = 9gR / 4 + 4gR\nU2 = 25 gR / 4\nU =[IMAGE] TOPIC: WPE\nSUB TOPIC: VERTICAL CIRCULAR MOTION", "solution_images": ["images/image110.png", "images/image111.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-01-Q7", "question": "Statement-1: In a dynamo, it is necessary that coil should\nbe moving and magnet should be stationary and not vice versa.\nStatement-2: Magnetic force is perpendicular to the direction of\nmotion of charges.\nWhich of the following options is correct?", "question_images": [], "option_1": "Statement is incorrect.", "option_2": "Statement is correct.", "option_3": "Both the statements are correct and is the correct reason of", "option_4": "Both the statements are correct and is not the reason of", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "Ph-28-Q10", "question": "In a meter bridge, the wire of length 1m has a non-uniform\ncross-section such that, the variation [IMAGE] of its\nresistance R with length.\nTwo equal resistances are connected as shown in the figure. The\ngalvanometer has zero deflection when the jockey is at point P. What is\nthe eight times the lengthAP ?", "question_images": ["images/image90.png", "images/image91.png", "images/image92.png", "images/image93.png"], "option_1": "0.2 m", "option_2": "0.3 m", "option_3": "0.25 m", "option_4": "0.35 m", "correct_option": null, "numerical_answer": "0.25", "solution": "[IMAGE] Acording to Question\n[IMAGE] TOPIC:Current and electricity\nSUB TOPIC: Potentiometer\nLEVEL:Moderate", "solution_images": ["images/image94.png", "images/image95.png", "images/image96.png", "images/image97.png", "images/image98.png", "images/image99.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-09-Q25", "question": "A uniform solid hemisphere of mass m and radius R is attached to\nthe roof with a chord of torsional constant C and performing torsional\nSHM. If time period (in seconds) of SHM is x, then find the value of 5x.", "question_images": ["images/image152.png", "images/image153.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "20", "solution": "", "solution_images": ["images/image154.png", "images/image155.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Time period", "difficulty": "Easy", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-01-Q21", "question": "Four metal rods all of identical dimensions and made of same\nmaterial are welded together at a single point. The configuration is\nsuch that any two rods are oriented at 120°. The far ends of three of\nthe rods are maintained at 90°C temperature and that of the fourth rod\nis maintained at 30°C temperature. The temperature of the junction point\nis ( in degree Celsius)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "75", "solution": "For equilibrium condition, heat flow in and out of junction\nshould be same. Let k represent some constant representing heat content,\nwhich may depend on dimensions and material of the rods.\nk(T - 30) = 3k(90 - T)\n4T = 300\nT = 75°C", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-14-Q12", "question": "An object is gradually moving away from the focal point of a\nconcave mirror along the axis of the mirror. The graphical\nrepresentation of the magnitude of linear magnification (m) versus\ndistance of the object from the mirror (x) is correctly given by (Graphs\nare drawn schematically and are not to scale)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "At focus, magnification is[IMAGE].", "solution_images": ["images/image90.png"], "subject": "Physics", "topic": "Ray Optics", "subtopic": "Linear magnification", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-14 Physics paper 7 Nov.docx"} {"question_id": "Ph-25-Q14", "question": "One mole of an ideal gas at pressure P_0, volume V_0 and\ntemperature T_0 is expanded isothermally to twice its volume and then\ncompressed at constant pressure to (V_0/2) and the gas is brought back\nto original state by a process in which P ∝ V (Pressure is directly\nproportional to volume). The correct representation of process is\n[IMAGE] rFkk rkieku nqxqus:i ls çlkfjr dh tkrh gSrFkk fQj lenkoh çØe rd laihfMr dh tkrh gS rFkk xSl dks okil çkjfEHkd\nvoLFkk esa,sls çØe lsykrs gS ftl ¼ lekuqikrh gksrk gS½ gS rks mijksä çØe", "question_images": ["images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png", "images/image97.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Process AB is isothermal expansion BC is isobaric compression\nand in process CA\nizfØ;k AB lerkih izlkj gS rFkk BC lenkch laihMu", "solution_images": ["images/image102.png", "images/image103.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "P-V diagram", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-09-Q20", "question": "A simple pendulum A and a homogeneous rod B hinged at its end\nare released from horizontal position as shown in figure", "question_images": [], "option_1": "The times period of swing of simple pendulum is greater than that\nof rod if L = [IMAGE]", "option_2": "The time period of swing of simple pendulum is equal to that of\nrod if L = [IMAGE]", "option_3": "The time period of swing of simple pendulum is greater than that\nof rod if [IMAGE]", "option_4": "The time period of swing of simple pendulum is equal to that of\nrod if L[IMAGE]", "correct_option": 1, "numerical_answer": null, "solution": "from conservation of energy\nfor simple pendulum\n[IMAGE] for homogeneous rod\n[IMAGE] When [IMAGE] at all time\nso time periods of swing are equal.", "solution_images": ["images/image128.png", "images/image129.png", "images/image130.png", "images/image131.png", "images/image132.png", "images/image133.png", "images/image134.png"], "subject": "Physics", "topic": "Oscillation", "subtopic": "Time period", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-09 Physics paper 26 October.docx"} {"question_id": "P-10-Q18", "question": "India's Mangalyan was sent to the Mars by launching it into a\ntransfer orbit EOM around the sun. It leaves the earth at E and meets\nMars at M. If the radius of Earth's orbit is a_e = 1.5 × 10^11 m,\nthat of laws give the\nestimate of time for Mangalyan to reach Mars from Earth to be close to\n(The time period of revolution of earth around sun is 1 year)", "question_images": ["images/image61.png"], "option_1": "$\\frac{1}{2}\\left( \\frac{4}{3} \\right)^{3/2}$ year", "option_2": "$\\frac{1}{2}\\left( \\frac{5}{3} \\right)^{3/2}$ year", "option_3": "$\\frac{1}{2}\\left( \\frac{3}{2} \\right)^{3/2}$ year", "option_4": "$\\frac{3}{2}$ year", "correct_option": 1, "numerical_answer": null, "solution": "$\\left( \\frac{T_{1}}{T_{2}} \\right)^{2} = \\left( \\frac{a_{1}}{a_{2}} \\right)^{3}$\n${\\left( \\frac{1}{T} \\right)}^{2} = \\left\\lbrack \\frac{1.5 \\times 10^{11}}{2 \\times 10^{11}} \\right\\rbrack^{3}$\n(semi major axis for the path from EOM$= \\frac{a_{E} + a_{M}}{2}$)\n$T = \\left( \\frac{4}{3} \\right)^{3/2}$ years", "solution_images": [], "subject": "Physics", "topic": "Gravitation", "subtopic": "Kepler's law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-05-Q13", "question": "The velocity v (in cm/sec) of a particle is given in terms of\ntime t (in sec) by the equation:$v = at + \\frac{b}{t\\ + \\ c}$\nThe dimensions of a, b and c are:\n ------------------------------------\n a b c", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "$\\lbrack v\\rbrack = \\lbrack at\\rbrack;\\ \\lbrack a\\rbrack = \\left\\lbrack \\frac{v}{t} \\right\\rbrack = \\left\\lbrack \\frac{{LT}^{- 1}}{T} \\right\\rbrack = \\left\\lbrack {LT}^{- 2} \\right\\rbrack$\n$\\lbrack t\\rbrack = \\lbrack c\\rbrack = \\lbrack T\\rbrack$\n$\\left\\lbrack \\frac{b}{t + c} \\right\\rbrack = \\lbrack v\\rbrack\\ \\Rightarrow \\ \\frac{\\lbrack b\\rbrack}{\\lbrack t\\rbrack} = \\lbrack v\\rbrack \\Rightarrow \\lbrack b\\rbrack = \\lbrack vt\\rbrack = \\left\\lbrack {LT}^{- 1}T \\right\\rbrack = \\lbrack L\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "Error analysis", "subtopic": "Dimensional analysis", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-19-Q3", "question": "An electric dipole of moment\n$\\overset{\\rightarrow}{p} = ( - \\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} + 2\\overset{\\hat{}}{k}) \\times 10^{- 29}$\ncm. is at the origin (0,0,0). The electric fielddue to this dipole at\n$\\overset{\\rightarrow}{r} = ( + \\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} + 5\\overset{\\hat{}}{k})$\n(note that$\\overset{\\rightarrow}{r}.\\overset{\\rightarrow}{p} = 0$) is\nparallel to", "question_images": [], "option_1": "$( + \\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "option_2": "$( - \\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "option_3": "$( + \\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "option_4": "$( - \\overset{\\hat{}}{i} - 3\\overset{\\hat{}}{j} + 2\\overset{\\hat{}}{k})$", "correct_option": 1, "numerical_answer": null, "solution": "$\\overset{\\rightarrow}{p} \\cdot \\overset{\\rightarrow}{r} = 0$\n$\\overset{\\rightarrow}{E}$must be antiparallel\nto$\\overset{\\rightarrow}{p}$\nSo, $\\overset{\\rightarrow}{E} = - \\lambda(\\overset{\\rightarrow}{p})$\nWhere $\\lambda$ is a arbitrary positive constant\nLet\n$\\overset{\\rightarrow}{A} = a\\overset{\\hat{}}{i} + b\\overset{\\hat{}}{j} + c\\overset{\\hat{}}{k}$\nis parallel to $\\overset{\\rightarrow}{E}$\n$\\frac{a}{\\lambda} = \\frac{b}{3\\lambda} = \\frac{c}{- 2\\lambda} = k$\nSo\n$\\overset{\\rightarrow}{A} = \\lambda k(\\overset{\\hat{}}{i} + 3\\overset{\\hat{}}{j} - 2\\overset{\\hat{}}{k})$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric dipole in electric field", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "P-21-Q21", "question": "A $1\\ µF$ capacitor is connected in the circuit shown below. The\ne.m.f. of the cell is 3 volts and internalresistance is 0.5 ohms. The\nresistors R_1and R_2have values 4 ohms and 1 ohm respectively.\nThecharge (in$µC)$ on the capacitor in steady state must be", "question_images": ["images/image77.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "In steady state no current flows through capacitor. The\npotential difference across capacitor andresistor of resistance R_2is\nsame.\n$\\mathbf{\\therefore}$ charge on capacitor\n$= CV = C \\times 3 = 1\\mu F \\times 3 = 3\\mu C$.\nLFkk;h voLFkk esa] la", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Charge on capacitor", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-18-Q14", "question": "A solid cylinder and a solid sphere, both having the same mass\nand radius, are released from an inclined plane having angle of\ninclination $\\theta$ one by one. They roll on the incline without\nslipping. The statement that holds good in this motion is", "question_images": [], "option_1": "the force of friction that acts on the two is the same", "option_2": "the force of friction is greater in case of a sphere than for a\ncylinder.", "option_3": "the force of friction is greater in case of cylinder than for\nsphere.", "option_4": "the force of friction will depend on the nature of the surface\nof the body that is moving and that of the inclined surface, and is\nindependent of the shape and size of the moving body.", "correct_option": 3, "numerical_answer": null, "solution": "Equation of motion\n$mg\\ sin\\theta - f = ma$....(i)\n$fR = I\\alpha = I\\frac{a}{R}$\n$\\Rightarrow f = I\\frac{a}{R^{2}}$....(ii)\nfrom (i) and (ii)\n$mgsin\\theta - f = m.\\frac{fR^{2}}{I}$\n$mgsin\\theta = f(1 + m.\\frac{R^{2}}{I}$)\n$f = \\frac{mgsin\\theta}{1 + \\frac{mR^{2}}{I}}$\nMoment of inertia of solid cylinder is greater, so friction force f will\nbe greater.", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Pure rolling down the incline", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-28-Q37", "question": "A small sphere tied to a string of length 0.8 m is describing a\nvertical circle so that the maximum and minimum tensions in the string\nare in the ratio 3: 1. The fixed end of the string is at a height of\n5.8 m above ground. Find the velocity of the sphere at the lowest\nposition. (Take g = 10 m/s^2)", "question_images": [], "option_1": "Using conservation of energy\nKE from bottom to the top =Gain in GPE", "option_2": "Using equations", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "let u and v be the speeds of the sphere at the bottom and the\ntop positions and m be themass\n[IMAGE] Radius of circle = length of string =r=0.8 m\n[IMAGE] =tension at the bottom or the maximum tension\n[IMAGE] =tension at the top or the minimum tension\nUsing conservation of energy\nKE from bottom to the top =Gain in GPE\nUsing equations (1) and (2) we get[IMAGE] TOPIC:Circular motion\nSUB TOPIC:Vertical circle", "solution_images": ["images/image329.png", "images/image330.png", "images/image331.png", "images/image332.png", "images/image333.png", "images/image334.png", "images/image335.png", "images/image336.png", "images/image337.png", "images/image338.png", "images/image339.png", "images/image340.png", "images/image341.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-25-Q27", "question": "If N Joules of heat is to be transferred to nitrogen in the\nisobaric heating process for that gas to perform the work A =2.0 J then\nthe value of N is\n[IMAGE] xSl dks nh x;h dk eku N twy gksrk gS] rks\nN", "question_images": ["images/image197.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "7", "solution": "W = 2J", "solution_images": ["images/image198.png", "images/image199.png"], "subject": "Physics", "topic": "Isobaric process", "subtopic": "Thermodynamics", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-13-Q9", "question": "A uniform magnetic field parallel to the plane of paper, existed\nin space initially directed from left to right. When a bar of soft iron\nis placed in the field parallel to it, the lines of force passing\nthrough it will be represented by figure", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "When a bar of soft iron is placed in the uniform magnetic field\nwhich is parallel to it because of large permeability of soft iron,\nmagnetic lines of force prefer to pass through it. Concentration of\nlines in soft iron bar increases as shown in figure.", "solution_images": ["images/image14.png"], "subject": "Physics", "topic": "Magnetism", "subtopic": "Properties of ferromagnetic material", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-08-Q20", "question": "When 300 J of heat is added to 25 gm of sampleof a material its\ntemperature rises from 25ºC to45ºC. The thermal capacity of the sample\nandspecific heat of the material are respectively given by", "question_images": [], "option_1": "$15J/ºC,600J/kg - ºC$", "option_2": "$600J/ºC,15J/ºC - kg$", "option_3": "$150J/ºC,60J/kg - ºC$", "option_4": "None of these", "correct_option": 1, "numerical_answer": null, "solution": "$\\Delta\\phi = ms\\Delta T$\n$\\therefore$ Thermal\ncapacity $= ms = \\frac{\\Delta Q}{\\Delta T} = \\frac{300J}{45 - 25} = 15J/ºC$\n and specific\nheat, $s = \\frac{\\Delta Q}{m\\Delta T} = \\frac{300}{25 \\times 10^{- 3} \\times (45 - 25)} = \\frac{600J}{kg} - C$", "solution_images": [], "subject": "Physics", "topic": "Heat", "subtopic": "Specific heat", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-08 Physics paper 23 October.docx"} {"question_id": "P-24-Q14", "question": "Figure shows a capillary tube C dipped in a liquid that wets it.\nThe liquid rises to a point A. If we blow air through the horizontal\ntube H, what will happen to the liquid column in the capillary tube?", "question_images": ["images/image45.png"], "option_1": "Level will rise above A", "option_2": "Level will fall below A", "option_3": "Level will remain at A", "option_4": "Cannot say", "correct_option": 1, "numerical_answer": null, "solution": "Level rise because to compensate for the reduction in pressure\nabove A, due to high velocity (from Bernoulli's principle pressure\ndecreases if velocity increases.)", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Application of Bernoulli's principle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-26-Q14", "question": "A long capillary tube of mass$'\\pi^{'}\\ gm$, radius 2 mm and\nnegligible thickness, is partially immersed in a liquid of surface\ntension 0.1 N/m. Take angle of contact zero and neglect buoyant force of\nliquid. The force required to hold the tube vertically, will be-\n$\\left( g = 10m/s^{2} \\right)$\n eksVkbZ dh,d yEch\n i`\"B ruko okys æo:i ls Mqcksrs\ngSaA s.k ysrs gq, rFkk æo ds mRIykou cy dks ux.;\nekurs cy", "question_images": [], "option_1": "$10.4\\ \\pi mN$", "option_2": "$10.8\\ \\pi mN$", "option_3": "$0.8\\ \\pi mN$", "option_4": "$4.8\\ \\pi mN$", "correct_option": 2, "numerical_answer": null, "solution": "The free body diagram of the capillary tube is as shown in the\nfigure. Net force F required to hold tube is\nF = force due to surface tension at cross-section\n$\\left( S_{1} + S_{2} \\right) +$weight of tube.\ndsf'kdk uyh dk cy js\nfy, dqy cy F gS\nF = vuqizLFk dkV ij i`\"B ruko j.k cy\n$\\left( S_{1} + S_{2} \\right) +$V~ Hkkj\n$= 4\\pi \\times 2 \\times 10^{- 3} \\times 0.1 + \\pi \\times 10^{- 3} \\times 10 = 10.8\\pi mN$", "solution_images": ["images/image37.png"], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-17-Q24", "question": "In the given circuit, resistance of potentiometer wire AB is\n$10\\Omega$ and its length is 1m. Rest of quantities is shown in figure.\nThe null point will be at a separation x m from point A, then find value\nof 50 x.", "question_images": ["images/image28.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "18.75", "solution": "$k = (\\frac{E}{R + R_{p}})\\frac{R_{p}}{\\mathcal{l}} = \\frac{2 \\times 10}{(10 + 15) \\times 1} = 0.8 \\vee /m = \\frac{8}{10}V/m$\n$\\frac{8}{10} \\times AJ = 0.3$\n$AJ = x = \\frac{3}{8}m$", "solution_images": [], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Potentiometer", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-17 Physics paper 26 Nov Second.docx"} {"question_id": "P-22-Q7", "question": "Two uniform solid spheres of equal radii R, butmass M and 4M have\na centre to centre separation 6R, as shown in figure. A projectile of\nmass m in projected from the surface of thesphere of mass M directly\ntowards the centre of the second sphere. The minimum speed ofthe\nprojectile so that it reaches the surface of the second sphere\nis[IMAGE]. What is n?", "question_images": ["images/image19.png", "images/image20.png"], "option_1": "4", "option_2": "5", "option_3": "3", "option_4": "1", "correct_option": null, "numerical_answer": "3", "solution": "[IMAGE] The projectile need to reach a point on the line joining the centers\nof sphere where net electric force on it is zero.\nLet the point N has no net electric field due to spheres.\nTherefore, $\\frac{GM}{x^{2}} = \\frac{G.4M}{{(6R - x)}^{2}}$ x = 2R\nApply conservation of mechanical energy at projectile initial position\nand particle at N\nE_i = E_f\nSUB TOPIC: Conservation of mechanical energy", "solution_images": ["images/image21.png", "images/image22.png", "images/image23.png", "images/image24.png", "images/image25.png", "images/image26.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "Ph-28-Q40", "question": "A person with a normal near point (25 cm) is using a compound\nmicroscope in normal adjustment with objective focal length\n[IMAGE] and an eye-piece of focal length\n[IMAGE] can bring an object placed\n[IMAGE] from the objective in sharp focus. The\nmagnifying power of the microscope is 11n. Find n", "question_images": ["images/image354.png", "images/image355.png", "images/image356.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "8", "solution": "Angular magnification of the eye-piece for image at\n[IMAGE] TOPIC: Optical instruments\nSUB TOPIC: Microscope\nLEVEL: Moderate", "solution_images": ["images/image357.png", "images/image358.png", "images/image359.png", "images/image360.png", "images/image361.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-07-Q9", "question": "A uniform metre stick is held vertically with one end on the\nfloor and is allowed to fall. The speed of theother end when it hits the\nfloor assuming that the end at the floor does not slip", "question_images": [], "option_1": "$\\sqrt{4g}$", "option_2": "$\\sqrt{3g}$", "option_3": "$\\sqrt{5g}$", "option_4": "$\\sqrt{g}$", "correct_option": 2, "numerical_answer": null, "solution": "using energy conservation\n$mg\\frac{\\mathcal{l}}{2} = \\frac{1}{2}I\\omega^{2}$\n$mg\\frac{\\mathcal{l}}{2} = \\frac{1}{2} \\cdot \\frac{m\\mathcal{l}^{2}}{3}\\omega^{2}$\n$= 1m$ $\\omega = \\sqrt{\\frac{3g}{}}$\n$V_{A} = \\omega = \\sqrt{3g} = (\\sqrt{3g})$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of mechanical energy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-23-Q4", "question": "Two moles of hydrogen are mixed with n moles of helium. The root\nmean square speed of gas molecules in the mixture is $\\sqrt{2}$ times\nthe speed of sound in the mixture. Then n is.", "question_images": [], "option_1": "3", "option_2": "2", "option_3": "1.5", "option_4": "2.5", "correct_option": 2, "numerical_answer": null, "solution": "$V_{rms} = \\sqrt{\\frac{3RT}{M_{mix}}}$\n$V_{sound\\ } = \\sqrt{\\frac{\\gamma RT}{M_{mix}}}$\n$V_{rms} = \\sqrt{2}V_{sound\\ }$\n$\\sqrt{\\frac{3RT}{M_{mix}}} = \\sqrt{2}\\sqrt{\\frac{\\gamma RT}{M_{mix}}}$\n$\\gamma_{mix} = \\frac{3}{2}$\n$\\gamma_{mix} = \\frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$\n$\\frac{3}{2} = \\frac{2 \\times \\frac{7R}{2} + n \\times \\frac{5R}{2}}{2 \\times \\frac{5R}{2} + n \\times \\frac{3R}{2}} \\Rightarrow \\frac{3}{2} = \\frac{14 + 5n}{10 + 3n} \\Rightarrow 30 + 9n = 28 + 10n \\Rightarrow n = 2$", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "Root mean square speed", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-20-Q3", "question": "Planet P has mass M_P and radius R_P. Planet Q has mass M_P/2.\nIf the escape velocities from the Planets P and Q are equal, then what\nis the radius of planet Q?", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image9.png", "images/image10.png", "images/image11.png"], "subject": "Physics", "topic": "Mechanics", "subtopic": "Escape velocity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-05-Q21", "question": "Error in the measurement of radius of a sphere is1.What is the\npercentage error in the calculated value of its volume?", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "3", "solution": "$V = \\frac{4}{3}\\pi r^{3}$\nTake natural log on both side\n$ln(V) = ln\\left( \\frac{4}{3}\\pi \\right) + 3ln(r)$\nDifferentiate both side\n$\\frac{\\Delta V}{V} = 3\\frac{\\Delta r}{r}$\n$\\%\\frac{\\Delta V}{V} = 3 \\times 1 = 3\\%$", "solution_images": [], "subject": "Physics", "topic": "Error analysis", "subtopic": "Percentage error", "difficulty": "Easy", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-05-Q18", "question": "The van der Waals' equation of state for somegases can be\nexpressed as:\n$$\\left( P + \\right)(V - b) = RT$$\nWhere P is the pressure, V the molar volume andT is the absolute\ntemperature of the given sampleof gas, a, b, and R constants. The\ndimensions is", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "$\\lbrack P\\rbrack = \\left\\lbrack \\frac{a}{V^{2}} \\right\\rbrack\\lbrack a\\rbrack = \\lbrack P\\rbrack\\left\\lbrack V^{2} \\right\\rbrack = \\left\\lbrack M^{1}L^{- 1}T^{- 2} \\right\\rbrack\\left\\lbrack L^{6} \\right\\rbrack = \\left\\lbrack M^{1}L^{5}T^{- 2} \\right\\rbrack$", "solution_images": [], "subject": "Physics", "topic": "Unit and dimension", "subtopic": "Dimensional analysis", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-22-Q12", "question": "In the fig. shown a cart moves on a smooth horizontal surface\ndue to an external constant force ofmagnitude F. The initial mass of the\ncart is M_0and velocity is zero. Sand falls on to the cart\nwithnegligible velocity at constant rate and\nsticks to the cart. The velocity of the cart at time t is", "question_images": ["images/image32.png", "images/image33.png"], "option_1": "$\\frac{Ft}{M_{0} + \\mu t}$", "option_2": "$\\frac{F}{\\mu}\\mathcal{l}_{n}\\frac{m_{0} + \\mu t}{m_{0}}$", "option_3": "$\\frac{Ft}{M_{0}}$", "option_4": "$\\frac{Ft}{M_{0} + \\mu t}e^{\\mu t}$", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] Formula $F = m\\frac{dv}{dt} + (V - u)\\frac{dm}{dt}$\nHere, u = velocity of sand = O\nm = M_o[IMAGE] t = mass at time t\nand [IMAGE] $\\therefore$ $F = \\left( M_{0} + \\mu t \\right)\\frac{dv}{dt} + v\\mu$\n$(F - \\mu v)dt = \\left( M_{0} + \\mu t \\right)dv$\n$\\int_{0}^{t}\\mspace{2mu}\\frac{dt}{M_{0} + \\mu t} = \\int_{0}^{v}\\mspace{2mu}\\frac{dv}{F - \\mu v}$\n$\\frac{1}{\\mu}\\left\\lbrack \\mathcal{l}n\\left( M_{0} + \\mu t \\right) \\right\\rbrack_{0}^{t} = - \\frac{1}{\\mu}\\mathcal{\\lbrack l}n(F - \\mu v)\\rbrack_{0}^{v}$\n$\\mathcal{l}n\\frac{\\left( M_{0} + \\mu t \\right)}{M_{0}}\\mathcal{= l}n\\left\\{ \\frac{F}{F - \\mu v} \\right)$\n$F - \\mu v = \\frac{M_{0}F}{M_{0} + \\mu t} \\Rightarrow v = \\frac{Ft}{M_{0} + \\mu t}$", "solution_images": ["images/image34.png", "images/image35.png", "images/image36.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Variable mass system", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-03-Q15", "question": "A bob of mass m is connected to a spring of spring constant k and a\nstring which is fixed to the ceiling as shown in figure. Original length\nof the spring is L and the bob rotates in a horizontal circle centred at\nthe fixed rod. Assuming the spring remains horizontal, then elongation\nin the spring is", "question_images": ["images/image89.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE] m k \n L,", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] T cos θ = mg... (1)\nT sin θ + kx = mw2 (L + x)... (2)\nfrom (1) and (2)\nx =[IMAGE] TOPIC: CIRCULAR MOTION\nSUB TOPIC: DYNAMICS", "solution_images": ["images/image94.png", "images/image95.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-20-Q14", "question": "A disc of mass is attached to a spring of\nstiffness During its motion, the disc rolls on the ground. When\nreleased from some stretched position, the centre of the disc will\nexecute simple harmonic motion with a time period of", "question_images": ["images/image39.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 3, "numerical_answer": null, "solution": "Torque = Iα about the point of contact", "solution_images": ["images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Time period of SHM", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-04-Q4", "question": "The vertical height between two points A and B is 30 m. A particle\nis projected downward from A and at the same time another particle is\nprojected upward from B with the same speed. Both particles reach the\nground simultaneously. The velocity of the projection of the particle is\n(B is on ground)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "A B 30m A \n B: (B )", "correct_option": 2, "numerical_answer": null, "solution": "S = 0 = ut -[IMAGE] gt2\n[IMAGE] 30 = ut + [IMAGE] gt2 =[IMAGE] by adding\n30 = 2ut =[IMAGE] u2 =[IMAGE] ⇒[IMAGE] = u\nTOPIC:KINEMATIC\nSUB TOPIC: MOTION UNDER GRAVITY", "solution_images": ["images/image17.png", "images/image18.png", "images/image19.png", "images/image20.png", "images/image21.png", "images/image22.png", "images/image23.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "Ph-27-Q23", "question": "On heating water, bubbles beings formed at the bottom of the\nvessel detach and rise. Take the bubbles to be spheres of radius R and\nmaking a circular contact of radius r with the bottom of the vessel. If\nr << R, and the surface tension of water is T, value of r just before\nbubbles detach is: (density of water is\nikuh dks xeZ\ngksdj dh vksj mBrs gSaA cqycqyksa dks f= xksyk eku ysa\nikuh dk i`\"B ruko T gSa, rc cqycyksa ds cl foyXu gksus ls tjk igys r", "question_images": ["images/image128.png", "images/image129.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] The bubble will detach if -\nBuoyant force ≥ Surface tension force\n[IMAGE] Solving [IMAGE] cqycqyk vyx gksxk;fn -\nmRIykou cy≥i`\"B ruko", "solution_images": ["images/image134.png", "images/image135.png", "images/image136.png", "images/image137.png", "images/image138.png", "images/image135.png", "images/image138.png"], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "Ph-26-Q13", "question": "The pressure gauge reading in metre of water column shown in the\ngiven figure will be:\n fp= esa ty LrEHk", "question_images": ["images/image36.png"], "option_1": "3.20 m", "option_2": "2.72 m", "option_3": "2.52 m", "option_4": "1.52 m", "correct_option": 4, "numerical_answer": null, "solution": "$P_{0} + \\rho_{w}g1 + \\rho_{w}g0.2 - \\rho_{Hg}g0.2 = P_{0}$\n$P - P_{0} = \\rho_{Hg}g0.2 - \\rho_{w}g(1.2) = 14896$\n$14896 = \\rho_{w}gh$\nh = 1.52", "solution_images": [], "subject": "Physics", "topic": "Mechanical properties of liquid", "subtopic": "Hydrostatic pressure", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-10-Q11", "question": "A uniform disc of mass M and radius R is lifted using a string\nas shown in the figure. Then,", "question_images": ["images/image42.png"], "option_1": "Its linear acceleration is g upward", "option_2": "Its rate of change of angular momentum is MgR", "option_3": "Its angular acceleration is[IMAGE]", "option_4": "All of the above", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{3Mg}{2} + \\frac{Mg}{2} - Mg = Ma$\n$a = g \\uparrow$\n$\\tau = I\\alpha$\n$\\frac{3Mg}{2}R - \\frac{Mg}{2}R = \\frac{{MR}^{2}}{2} \\cdot \\alpha$\n$\\alpha = \\frac{2g}{R}$\n$\\frac{dL}{dt} = \\tau = MgR$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Pure rolling", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-03-Q5", "question": "40 m \n80 m", "question_images": [], "option_1": "45°", "option_2": "90°", "option_3": "37°", "option_4": "53°", "correct_option": 1, "numerical_answer": null, "solution": "With standard kinematics, we get\n[IMAGE] y = [IMAGE] gt2, t = [IMAGE] = 2.86 s\nx = vxt, vx =[IMAGE] = 28.0 m/s\nvy2 - v02 = 2a(y - y0)\nvy = -[IMAGE] = -[IMAGE] = 28.0 m/s\nθ = A tan[IMAGE] = 315°\nTOPIC:KINEMATIC\nSUB TOPIC: 2 D", "solution_images": ["images/image20.png", "images/image21.png", "images/image22.png", "images/image23.png", "images/image24.png", "images/image25.png", "images/image26.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-10-Q10", "question": "A solid cylinder and a solid sphere, both having the same mass\nand radius, are released from an inclined plane having angle of\ninclination[IMAGE] one by one. They roll on the incline\nwithout slipping. The statement that holds good in this motion is", "question_images": ["images/image39.png"], "option_1": "The force of friction that acts on the two is the same", "option_2": "The force of friction is greater in case of a sphere than for a\ncylinder.", "option_3": "The force of friction is greater in case of cylinder than for sphere.", "option_4": "The force of friction will depend on the nature of the surface of\nthe body that is moving and that of the inclined surface, and is\nindependent of the shape and size of the moving body.", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Equation of motion\nmg sin [IMAGE]....(i)\n$fR = I\\alpha = I\\frac{a}{R}$\n$\\Rightarrow f = I\\frac{a}{R^{2}}$....(ii)\nFrom (i) and (ii)\n$mgsin\\theta - f = m\\frac{{fR}^{2}}{I} \\Rightarrow mgsin\\theta = f\\left\\{ 1 + \\frac{mR^{2}}{I} \\right)$\n$\\Rightarrow$ $\\ \\ f = \\frac{mgsin\\theta}{1 + \\frac{mR^{2}}{I}}$\nMoment of inertia of solid cylinder is greater, so friction force f will\nbe greater.", "solution_images": ["images/image40.png", "images/image41.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Pure rolling", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-13-Q4", "question": "A long, straight wire of radius a carries a current distributed\nuniformly over its cross-section. The ratioof the magnetic fields due\nto the wire at distance[IMAGE] and 2a, respectively from\nthe axis of the wire is", "question_images": ["images/image3.png"], "option_1": "", "option_2": "2", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "Using Ampere's circuital law:\n$B_{B} = \\frac{\\mu_{0}i}{2\\pi(2a)}$\n$\\frac{B_{A}}{B_{B}} = \\frac{4}{6} = \\frac{2}{3}$", "solution_images": ["images/image7.png"], "subject": "Physics", "topic": "Magnetic effect of current", "subtopic": "Ampere's circuital law", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-15-Q12", "question": "The graph which depicts the results of Rutherford gold foil\nexperiment with α -particles is:\nα -particles is:\nY = Number of scattered alpha -particles detected\n(Plots are schematic and not to scale)", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "", "solution_images": ["images/image92.png"], "subject": "Physics", "topic": "Atomic physics", "subtopic": "Alpha particle scattering", "difficulty": "Easy", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "Ph-25-Q1", "question": "The following graph shows two isotherms for a fixed mass of an\nideal gas. The ratio of r.m.s speed of the molecules at temperatures\nT_1 and T_2 is", "question_images": ["images/image1.png"], "option_1": "", "option_2": "", "option_3": "2", "option_4": "4", "correct_option": 3, "numerical_answer": null, "solution": "$PV = \\frac{1}{3}M\\left( v_{rms} \\right)_{1}^{2} = RT$\nOn lower curve, we have V=1m^3, corresponding to P=1×10^5Pa = 1\natmosphere. At T_1, we have\nde oØ tks P=1×10^5Pa = 1 okrkoj.k ds\nvuq:i gSA T_1 ij] gekjs ikl \n$1 \\times 1 = \\frac{1}{3}M\\left( v_{rms} \\right)_{1}^{2} = {RT}_{1}$\nor$\\left( v_{rms} \\right)_{1}^{2} = \\frac{3}{M}$\nSimilarly, for upper curve we have\nblh rjg] oØ ds fy, gekjs ikl gS\nV=1m^3, P = 2 atmosphere\n$\\therefore 2 \\times 2 = \\frac{1}{3}M\\left( v_{ans} \\right)_{2}^{2} = {RT}_{2}$\n$\\therefore\\left( v_{mas} \\right)_{2}^{2} = \\frac{4 \\times 3}{M} = \\frac{12}{M}$\n$\\therefore\\ Required\\ ratio\\ = \\sqrt{\\frac{12/M}{3/M}} = 2$", "solution_images": [], "subject": "Physics", "topic": "Kinetic Theory of Gases", "subtopic": "R.M.S. speed", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-01-Q19", "question": "A hollow conducting sphere of inner radius R and outer radius\n2R is given a charge Q as shown in the figure, then the", "question_images": ["images/image87.png"], "option_1": "potential at A and B is different", "option_2": "potential at O and B is\ndifferent", "option_3": "potential at O and C is different", "option_4": "potential at A, B, C and O\nis different", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] charge is on the outer surface hence [IMAGE] remains\nconstant.", "solution_images": ["images/image88.png", "images/image89.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-20-Q17", "question": "A very long wire ABDMNDC is shown in figure carrying current I.\nAB and BC parts are straight, longand at right angle. At D wire forms a\ncircular turn DMND of radius R. AB, BC parts are tangential tocircular\nturn at N and D. Magnetic field at the centre of circle is", "question_images": ["images/image58.png"], "option_1": "$\\frac{\\mu_{0}I}{2\\pi R}\\left( \\pi + \\frac{1}{\\sqrt{2}} \\right)$", "option_2": "$\\frac{\\mu_{0}I}{2R}$", "option_3": "$\\frac{\\mu_{0}I}{2\\pi R}\\left( \\pi - \\frac{1}{\\sqrt{2}} \\right)$", "option_4": "$\\frac{\\mu_{0}I}{2\\pi R}(\\pi + 1)$", "correct_option": 1, "numerical_answer": null, "solution": "$\\frac{\\mu_{0}i}{4\\pi R}\\left\\lbrack \\sin 90^{\\circ} - \\sin 45^{\\circ} \\right\\rbrack \\otimes + \\frac{\\mu_{0}i}{2R} \\odot + \\frac{\\mu_{0}i}{4\\pi R}\\left( \\sin 45^{\\circ} + \\sin 90^{\\circ} \\right) \\odot$\n$= \\frac{- \\mu_{0}i}{4\\pi R}\\left\\lbrack 1 - \\frac{1}{\\sqrt{2}} \\right\\rbrack + \\frac{\\mu_{0}j}{2R} + \\frac{\\mu_{0}i}{4\\pi R}\\left\\lbrack \\frac{1}{\\sqrt{2}} + 1 \\right\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{4\\pi R}\\left\\lbrack - 1 + \\frac{1}{\\sqrt{2}} + 2\\pi + \\frac{1}{\\sqrt{2}} + 1 \\right\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{4\\pi R}\\lbrack\\sqrt{2} + 2\\pi\\rbrack \\odot$\n$= \\frac{\\mu_{0}i}{2\\pi R}\\left\\lbrack \\frac{1}{\\sqrt{2}} + \\pi \\right\\rbrack \\odot$", "solution_images": ["images/image59.png"], "subject": "Physics", "topic": "Magnetism", "subtopic": "Magnetic field due to current carrying conductor", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-13-Q21", "question": "A 100 turns closely wound current carrying coilof radius 10 cm\nis placed in a vertical plane. Thecoil carries a current of 3.2 A and\nfree to rotateabout a horizontal axis which coincides with itsdiameter.\nA uniform horizontal magnetic field of2 Tesla exists in the region. Due\nto magnetic torquecoil starts rotating. The angular velocity in\nrad/secof the coilwhen it has rotated by $\\frac{\\pi}{2}$ is 10n. Find n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "2", "solution": "$\\tau = MBsin\\theta$\n$I\\alpha = MBsin\\theta$\n$\\alpha = \\frac{MB}{I}\\sin\\theta$\n$\\frac{d\\omega}{d\\theta}*\\frac{d\\theta}{dt} = \\frac{MB}{I}sin\\theta$\n$\\int_{0}^{\\omega}\\mspace{2mu}\\omega d\\omega = \\int_{0}^{\\frac{\\pi}{2}}\\mspace{2mu}\\frac{MB}{I}\\sin{\\theta d\\theta}$\n$\\left( \\frac{\\omega^{2}}{2} \\right)_{0}^{\\omega} = \\frac{MB}{I}\\lbrack - cos\\theta\\rbrack_{0}^{\\pi/2}$\n$\\omega = \\sqrt{\\frac{2MB}{I}} = \\sqrt{\\frac{2NIAB}{I_{rot}}}\\ $\n$= \\sqrt{\\frac{2 \\times 100 \\times 3.2 \\times \\pi \\times 100 \\times 10^{- 4} \\times 2}{0.1}}$\n= 20 rad/sec.", "solution_images": [], "subject": "Physics", "topic": "Magnetism", "subtopic": "Torque on current carrying loop in magnetic", "difficulty": "Tough", "question_type": "numerical", "has_image": false, "exam": "JEE Main", "source_paper": "P-13 Physics paper 4 Nov.docx"} {"question_id": "P-01-Q20", "question": "has four choices", "question_images": [], "option_1": ",", "option_2": ",", "option_3": ",", "option_4": "out of which ONLY ONE is correct]", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-22-Q10", "question": "A capacitor is made of two square plates each of side \nmaking a very small angle $\\alpha$ between them, as shown in figure. The\ncapacitance will be close to", "question_images": ["images/image29.jpeg"], "option_1": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{2d})$", "option_2": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 + \\frac{\\alpha a}{2d})$", "option_3": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{4d})$", "option_4": "$\\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{3\\alpha a}{2d})$", "correct_option": 1, "numerical_answer": null, "solution": "$\\Rightarrow c = \\frac{\\varepsilon_{0}a}{\\alpha}\\lbrack ln(d + \\alpha x)\\rbrack_{0}^{a}$\n$= \\frac{\\varepsilon_{0}a}{\\alpha}In(1 + \\frac{\\alpha a}{d}) \\approx \\frac{\\varepsilon_{0}a^{2}}{d}(1 - \\frac{\\alpha a}{2d})$", "solution_images": ["images/image30.jpeg"], "subject": "Physics", "topic": "Capacitor", "subtopic": "Capacitance", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-22-Q19", "question": "Consider a particle initially moving with a velocity 7 m/s\nstarts decelerating at a constant rate of$2m/s^{2}$.\nThe distance travelled in the fourth second is$\\frac{x}{10}m$. Find\nthe value of x.", "question_images": [], "option_1": "5", "option_2": "6", "option_3": "7", "option_4": "8", "correct_option": 1, "numerical_answer": null, "solution": "Here particle is decelerating so velocity of particle becomes\n after t=3.5 sec. the particle has a turning point at t=3.5 sec.\nV=U+at\n$0 = 7 - 2t$\n$t = 3.5\\sec$\n$y_{2} = 7 \\times 3.5 - \\frac{1}{2} \\times 2 \\times (3.5)^{2} = \\frac{49}{4}m$\n$y_{2} - x_{1} = \\frac{49}{4} - 12 = \\frac{1}{4}m$.\nDue to symmetry displacement of the particle at t=3 and t=4sec. is same.\nso total distance traveled in the fourth sec\nis$= \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}m$", "solution_images": ["images/image53.png"], "subject": "Physics", "topic": "Kinematics", "subtopic": "Uniform accelerated motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-07-Q5", "question": "A ball of mass M collides elastically with another\nidentical ball at rest as shown in figure. Initially velocity of\nball After collision", "question_images": ["images/image11.png"], "option_1": "Speed of ball B is u $\\cos\\theta$", "option_2": "Speed of ball A is u $\\sin\\theta$", "option_3": "Speed of ball B is u $\\sin\\theta$", "option_4": "None of these", "correct_option": 3, "numerical_answer": null, "solution": "[IMAGE] Velocity along line of impact will get exchanged because both bodies are\nidentical. $v_{B} = ucos\\theta$\n$V_{A} = usin\\theta$", "solution_images": ["images/image12.png"], "subject": "Physics", "topic": "Centre of Mass", "subtopic": "Collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-15-Q23", "question": "Figure is the plot of the stopping potential versus the\nfrequency of the light used in an experiment on photoelectric effect. If\nthe work function is $\\frac{x}{5}($ in [IMAGE] then value\nof x is. [IMAGE]", "question_images": ["images/image171.png", "images/image172.png", "images/image173.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "", "solution_images": ["images/image174.png", "images/image175.png", "images/image176.png", "images/image177.png"], "subject": "Physics", "topic": "Dual Nature of Matter", "subtopic": "Einstein's photoelectric equation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-15 Physics paper 11 Nov.docx"} {"question_id": "Ph-25-Q18", "question": "In a cyclic waste power plant for thermal decomposition of waste\nhot gases are supplied from a \"Kentuki Chamber\" where gases are heated\nunder kentuki process [IMAGE] where P, V and T are\npressure, volume and temperature of gases in chamber and k is positive\nconstant. Find work done by 5 mole of hot gases when temperature is\nraised from 500 K to 550 K:,d pØ IykUV fo?kVu\n\"dsUVqdh psEcj\" ls rIr xSals çnku dh tkrh\nçfØ xeZ tgk¡ P, V o T psEcj\neksyksa 500 K ls 550 K rd o`f)", "question_images": ["images/image129.png", "images/image129.png"], "option_1": "625 R", "option_2": "- 625 R", "option_3": "250 R", "option_4": "- 250 R", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image130.png", "images/image131.png", "images/image132.png", "images/image133.png", "images/image134.png", "images/image135.png"], "subject": "Physics", "topic": "Behaviour of perfect gases", "subtopic": "Ideal gas equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "Ph-28-Q16", "question": "A uniform cylinder floats with its axis vertical with\n[IMAGE] parts of its length in the upper liquid and\n[IMAGE] part of its length in lower liquid. If the\ndensities of upper and lower liquids are [IMAGE] then\nthe density of the material of the cylinder is", "question_images": ["images/image157.png", "images/image158.png", "images/image159.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "Let [IMAGE] be the be the length of the cylinder\nthen\n is the density of the cylinder.\nTOPIC: Hydrostatics\nSUB TOPIC:Bouyancy\nLEVEL: Moderate", "solution_images": ["images/image164.png", "images/image165.png", "images/image166.png", "images/image167.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-25-Q11", "question": "P-V diagram of ideal monoatomic gas is as shown. Find net heat\ngiven to the gas in the process ABC.", "question_images": ["images/image70.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 4, "numerical_answer": null, "solution": "", "solution_images": ["images/image75.png", "images/image76.png", "images/image77.png", "images/image78.png", "images/image79.png", "images/image80.png"], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "P-V diagram and first law of thermodynamics", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-25 Physics paper 7 Jan.docx"} {"question_id": "P-12-Q20", "question": "n a meter bridge, the wire of length 1m has a non-uniform\ncross-section such that, the variation$\\frac{dR}{d^{\\mathcal{l}}}$ of\nits resistance R with\nlength$\\mathcal{l}$ is $\\frac{dR}{d\\mathcal{l}} \\propto \\frac{1}{\\sqrt{\\mathcal{l}}}$.\nTwo equal resistances are connected as shown in thefigure. The\ngalvanometer has zero deflection when the jockey is at point P. What is\nthe length AP ?", "question_images": ["images/image26.png"], "option_1": "0.2 m", "option_2": "0.3 m", "option_3": "0.25 m", "option_4": "0.35 m", "correct_option": 3, "numerical_answer": null, "solution": "According to Question\n$\\int_{0}^{\\mathcal{l}}\\mspace{2mu} c\\frac{d\\mathcal{l}}{\\sqrt{\\mathcal{l}}} = \\int_{\\mathcal{l}}^{1}\\mspace{2mu} c\\frac{d\\mathcal{l}}{\\sqrt{\\mathcal{l}}}$\n$(2\\sqrt{\\mathcal{l}})_{0}^{\\mathcal{l}} = (2\\sqrt{\\mathcal{l}})_{t}^{1}$\n$2\\sqrt{\\mathcal{l}} = 2 - 2\\sqrt{\\mathcal{l}}$\n$2\\sqrt{\\mathcal{l}} = 2 - 2\\sqrt{\\mathcal{l}}$\n$4\\sqrt{\\mathcal{l}} = 2$\n$\\mathcal{l =}\\frac{1}{4} = 0.25m$.", "solution_images": [], "subject": "Physics", "topic": "Current Electricity", "subtopic": "Meter bridge", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-12 Physics paper 2 Nov.docx"} {"question_id": "P-04-Q24", "question": "A block of weight 10 N produces an extension of 9 cm when it is\nhung by an elastic spring of length 60 cm and is in equilibrium. The\nspring is cut into two parts, one of length 40 cm and the other of\nlength 20 cm. Another weight of 20 N hangs in equilibrium supported by\nboth parts as shown in the figure. Find the total extension of the two\nparts of the spring (in cm) now.\n10N 60 cm 9cm, 40\ncm 20 cm 20N \n (cm )", "question_images": ["images/image125.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "04", "solution": "10 = k ×[IMAGE] k =[IMAGE] k1 × 4 = k × 60 =[IMAGE] × 60\nk1 =[IMAGE] k2 × 20 = k × 60 ⇒ k2 =[IMAGE] kq = k1 + k2 = 500\n10 = 500 x ⇒ x =[IMAGE] m = 2 cm\nTOPIC: NLM\nSUB TOPIC: SPRING", "solution_images": ["images/image126.png", "images/image127.png", "images/image128.png", "images/image129.png", "images/image130.png", "images/image131.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Tough", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-21-Q11", "question": "The maximum peak to peak voltage of an AM wave is 24 mV and the\nminimum peak to peak voltage is8 mV. The modulation factor is.", "question_images": [], "option_1": "10", "option_2": "20", "option_3": "25", "option_4": "50", "correct_option": 4, "numerical_answer": null, "solution": "Here ¼;gk¡½, $V_{\\max} = \\frac{24}{2}$=12mV and¼rFkk½\n$V_{\\min} = \\frac{8}{2} = 4mV$", "solution_images": ["images/image30.png"], "subject": "Physics", "topic": "Communication system", "subtopic": "Modulation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-07-Q22", "question": "A particle moving on a circular path travels first one third\npart of circumference in 2 sec & nextone third part in 1 sec. Average\nangular velocity of the particle is $\\frac{n\\pi}{9}$ (in rad/sec). Then\nvalueof n.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "$\\omega_{avg} = < \\omega > = \\frac{\\ total\\ \\theta}{\\ total\\ \\ time\\ }$", "solution_images": ["images/image35.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Kinematics in rotation", "difficulty": "Easy", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-21-Q4", "question": "Two infinite planes each with uniform surface charge density\n$+ \\sigma$ are kept in such a way that the anglebetween them is 30°. The\nelectric field in the region shown between them is given by\nvuur yEckbZ vkSj pkSM+kbZ okys nks leryksa", "question_images": ["images/image7.png"], "option_1": "$\\frac{\\sigma}{2\\epsilon_{0}}\\left\\lbrack (1 + \\sqrt{3})\\overset{\\hat{}}{y} - \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "option_2": "$\\frac{\\sigma}{2\\epsilon_{0}}\\left\\lbrack \\left( 1 - \\frac{\\sqrt{3}}{2} \\right)\\overset{\\hat{}}{y} - \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "option_3": "$\\frac{\\sigma}{2\\epsilon_{0}}\\left\\lbrack (1 + \\sqrt{3})\\overset{\\hat{}}{y} + \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "option_4": "$\\frac{\\sigma}{\\epsilon_{0}}\\left\\lbrack \\left( 1 + \\frac{\\sqrt{3}}{2} \\right)\\overset{\\hat{}}{y} + \\frac{\\overset{\\hat{}}{x}}{2} \\right\\rbrack$", "correct_option": 2, "numerical_answer": null, "solution": "$\\overset{\\rightarrow}{E} = \\frac{\\sigma}{2\\varepsilon_{0}}\\left\\lbrack \\left( 1 - \\frac{\\sqrt{3}}{2} \\right)\\overset{\\hat{}}{y} - \\frac{1}{2}\\overset{\\hat{}}{x} \\right\\rbrack$.", "solution_images": ["images/image8.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Electric field due to infinite sheet of charge", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "Ph-26-Q1", "question": "A coal based thermal power plant producing electricity operates\nbetween the temperatures 27ºC and227ºC. The plant works at 80 of its\nmaximum theoretical efficiency. Complete burring of 1 kg of coalyields\n3600 KJ of heat. A house needs 10 units of electricity each day. Coal\nused for supplying theamount of energy for the house in one year is,d 1 kg iw.kZ tyus ij 3600 KJ mRiUu gksrh,d o\"kZ rd nsus", "question_images": [], "option_1": "1141 kg", "option_2": "580 kg", "option_3": "605 kg", "option_4": "765 kg", "correct_option": 1, "numerical_answer": null, "solution": "Maximum efficiency of engine\n$= \\left( 1 - \\frac{T_{2}}{T_{1}} \\right) \\times 100 = 40\\%$\n$= \\left( 1 - \\frac{T_{2}}{T_{1}} \\right) \\times 100 = 40\\%$\n$\\therefore$overall efficiency = 0.4 x 0.8 = 0.32\n$\\therefore$dqy n{krk = 0.4 x 0.8 = 0.32\nRequirement of house in one year,d lky esa ?kj \n= 10 × 365 units = 3650 kwh = 3650 × 3600 kJ\n$\\therefore$ in put =$\\frac{3650 \\times 3600}{0.32}$kJ\n$\\therefore$ coalrequired =$\\frac{\\ In\\ put\\ }{36000}$=1141kg", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Heat engine", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "P-01-Q3", "question": "In a Young's double slit experiment sources of equal\nintensities are used. Distance between slits is d and wavelength of\nlight used is λ(λ << d). Angular separation of the nearest points on\neither side of central maximum where intensities becomes half of the\nmaximum value is", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] I + I + 2 [IMAGE] cos φ\n2I = 2I + 2I cos φ \ncos φ = 0\nφ = [IMAGE] ∆x = [IMAGE] x = [IMAGE] θ =[IMAGE]", "solution_images": ["images/image11.png", "images/image12.png", "images/image13.png", "images/image14.png", "images/image15.png", "images/image16.png", "images/image17.png", "images/image18.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "Ph-27-Q5", "question": "A thin metal rod of length L_0 is shaped into a ring with a\nsmall gap x as shown. On heating the system", "question_images": ["images/image20.png"], "option_1": "x decreases, r and d increase", "option_2": "x and r increase, d decreases", "option_3": "x, r and d all increase", "option_4": "Data insufficient to arrive at a conclusion\nL_0 yackbZ dh,d iryh NM+ dks vYi varjky x", "correct_option": 3, "numerical_answer": null, "solution": "On uniform heating of an object which is free to expand,\ndistance between any two points increases.\noLrq tks çlkfjr gksus ds fy, LorU= gS]", "solution_images": [], "subject": "Physics", "topic": "Mechanical properties of solid", "subtopic": "Thermal expansion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-07-Q1", "question": "A uniform disc of mass m and radius R is resting on a smooth\nhorizontal surface and it is free to rotateabout its own axis which is\nvertical. A small particle of mass m collides its periphery with a\nvelocity of Vin tangential direction. If the collision is perfectly in\nelastic and the particle sticks to the periphery, theangular velocity of\nthe system just after the collision will be", "question_images": ["images/image1.png"], "option_1": "$\\frac{V}{3R}$", "option_2": "$\\frac{2V}{3R}$", "option_3": "$\\frac{3V}{2R}$", "option_4": "$\\frac{V}{R}$", "correct_option": 2, "numerical_answer": null, "solution": "Applying angular momentum conservation about the rotating axis\nL_i = L_f\nO + mv R= $\\left( \\frac{mR^{2}}{2} + mR^{2} \\right)\\omega_{f}$\n$\\omega = \\frac{2V}{3R}$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-07 Physics paper 17 October.docx"} {"question_id": "P-18-Q11", "question": "Two particle A and B are located in x-y plane at points (0,0)\nand (0,4 m). They simultaneously start moving with velocities.\n${\\overset{\\rightarrow}{v}}_{A} = 2\\overset{\\hat{}}{j}m/s$and${\\overset{\\rightarrow}{v}}_{B} = 2\\overset{\\hat{}}{i}m/s$.\nSelect the correct alternative(s).", "question_images": [], "option_1": "Time after which they are at minimum distance is 1 s", "option_2": "The distance between them first decreases and then increases", "option_3": "The shortest distance between them is $2\\sqrt{2}m$", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "${\\overset{\\rightarrow}{v}}_{AB} = {\\overset{\\rightarrow}{v}}_{A} - {\\overset{\\rightarrow}{v}}_{B} = 2(j - \\overset{\\hat{}}{i})m/s\\ \\ \\ or\\ |{\\overset{\\rightarrow}{v}}_{AB}| = 2\\sqrt{2}m/s$\nAssuming B to be at rest, A will move with velocity\n${\\overset{\\rightarrow}{v}}_{AB}$ in the direction shown in figure.\nThe distance between them will first decrease from A to Cand then\nincrease beyond C.\n[IMAGE] Minimum distance between them is BC which is equal to\n$\\frac{4}{\\sqrt{2}}$ or $2\\sqrt{2}$ and the time after which they are\nat closest distance is:\n$t = \\frac{AC}{|{\\overset{\\rightarrow}{v}}_{AB}|} = \\frac{2\\sqrt{2}}{2\\sqrt{2}} = 1s$", "solution_images": ["images/image22.png"], "subject": "Physics", "topic": "Kinematics", "subtopic": "Relative motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "Ph-27-Q21", "question": "A certain quantity of oxygen [IMAGE] is\ncompressed isothermally until its pressure is doubled\n[IMAGE]. The gas is then allowed to expand adiabatically\nuntil its original volume is restored. Then the final pressure(P) in\nterms of initial pressure [IMAGE] is:\nvkWDlhtu,d izkjfEHkd nkc", "question_images": ["images/image102.png", "images/image103.png", "images/image104.png", "images/image102.png", "images/image103.png", "images/image104.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "None of these", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] (isothermal process) ¼lerkih", "solution_images": ["images/image108.png", "images/image109.png", "images/image110.png", "images/image111.png"], "subject": "Physics", "topic": "Thermodynamic", "subtopic": "Thermodynamic processes", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "Ph-28-Q39", "question": "Radiation, with wavelength [IMAGE] falls on a\nmetal surface to produce photoelectrons. The electronsare made to enter\na uniform magnetic field of[IMAGE]. If the radius of the\nlargest circular pathfollowed by the electrons is 10 mm, the work\nfunction of the metal is closest to what integral value in eV", "question_images": ["images/image346.png", "images/image347.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "1", "solution": "[IMAGE] (in eV)\n[IMAGE] (in eV)\n[IMAGE] TOPIC:Dual nature of particle\nSUB TOPIC:Photoelectric effect", "solution_images": ["images/image348.png", "images/image349.png", "images/image350.png", "images/image351.png", "images/image352.png", "images/image353.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-28-Q1", "question": "A sound wave of frequency n travels horizontally to right. It is\nreflected from a large vertical wall moving to the left with a speed V.\nThe speed of sound in medium is C. Then", "question_images": [], "option_1": "the frequency of reflected wave for stationary listener is", "option_2": "the frequency of reflected wave for stationary listener is", "option_3": "the frequency of reflected wave for stationary listener is", "option_4": "the number of beats heard by stationary listener to the left of\nreflecting wall is [IMAGE]", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] (number of waves striking the wall per sec)\n[IMAGE] (frequency observed by stationary listener)\nTOPIC:Sound waves\nSUB TOPIC:Doppler effect\nLEVEL:Moderate\n2. In YDSE arrangement, white light is used to illuminate the slits. At\npoint on the screen directly in front of slits, which of these\nwavelength is missing?\n([IMAGE] = wavelength of light used)\n(d = distance between slits)\n(D = distance between slits and screen)\n(given d << D)", "solution_images": ["images/image5.png", "images/image6.png", "images/image7.png", "images/image8.png", "images/image9.png", "images/image10.png", "images/image11.png", "images/image12.png", "images/image13.png", "images/image14.png", "images/image15.png", "images/image16.png", "images/image17.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-26-Q37", "question": "A rubber cord has a cross-section area 1 cm^2 and natural\nlength of [IMAGE] It is stretched by\n[IMAGE] to fire a small object of\nmass[IMAGE]. If the Young's modulus (Y) is\n[IMAGE] Speed acquired by the object in (m/s) is 2n. Find\nn. Assume Hook's law is valid and neglect gravity.,d jcj dh Mksjh dk vuqçLFk dkV kjk çkIr\nosx (m/s) esa½2n.nKkr djsaA ekfu,sa fd gqd", "question_images": ["images/image163.png", "images/image164.png", "images/image165.png", "images/image166.png", "images/image167.png", "images/image165.png", "images/image164.png", "images/image168.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "v = 10^3 × 10^-2\nv = 10.00 m/s", "solution_images": ["images/image169.png", "images/image170.png", "images/image171.png"], "subject": "Physics", "topic": "Elasticity", "subtopic": "Potential energy stored in elongated rubber cord", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-28-Q4", "question": "A small ball of mass 1 kg strikes a wedge of mass 4kg\nhorizontally with a velocity of 10 m/s. Friction is absent everywhere\nand collision is elastic. Select the correct answers", "question_images": ["images/image32.png"], "option_1": "The linear momentum of the ball along common normal and common\ntangent will not change during the collision.", "option_2": "The linear momentum of the ball along common normal and common\ntangent will change during the collision.", "option_3": "The linear momentum of the ball along common normal will not\nchange but remain same along common tangent during the collision.", "option_4": "The linear momentum of the ball along common normal will change\nbut remain same along common tangent during the collision.", "correct_option": 4, "numerical_answer": null, "solution": "Since impulse acts along common normal the linear momentum of the\nball along common normal will change. There is no impulse along common\ntangent therefore linear momentum of the ballwill not change during the\ncollision.\nTOPIC: Collision\nSUB TOPIC:Elastic collision\nLEVEL:Moderate", "solution_images": [], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-05-Q7", "question": "A particle is projected vertically upwards. It will be at\n$\\left( \\frac{3}{4} \\right)$^th of its greatest height at time\n$t = t_{1}$and $t = t_{2}$. $\\left( t_{1} < t_{2} \\right)$. Then the\nratio $\\left( t_{1}:t_{2} \\right)$ is", "question_images": [], "option_1": "$\\frac{1}{3}$", "option_2": "$\\frac{1}{2}$", "option_3": "$\\frac{1}{4}$", "option_4": "$\\frac{1}{9}$", "correct_option": 1, "numerical_answer": null, "solution": "$H = \\frac{u^{2}}{2g}$;$\\frac{3}{4}\\left( \\frac{u^{2}}{2g} \\right) = ut - \\frac{1}{2}gt^{2} \\Rightarrow t = \\frac{u}{g} \\pm \\frac{u}{2g}$\n$\\Rightarrow t_{1} = \\frac{u}{2g}$and$t_{2} = \\frac{3u}{2g} \\Rightarrow t_{1}:t_{2} = 1:3$", "solution_images": [], "subject": "Physics", "topic": "Motion in 1D", "subtopic": "Motion under gravity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-01-Q11", "question": "For D1 forward bias resistance is 20 Ohm and reverse bias\nresistance is 600 Ohm. For D2 forward bias resistance is 30 Ohm and\nreverse bias resistance is 200 Ohm. Find the equivalent resistance\nbetween A and B if (i) A is at a higher potential (ii) B is at a higher\npotential", "question_images": ["images/image47.png"], "option_1": "", "option_2": "", "option_3": "50 Ω and 800 Ω", "option_4": "220 Ω and 630 Ω", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image52.png", "images/image53.png", "images/image54.png", "images/image55.png", "images/image56.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-01 Physics Paper-01_Update.docx"} {"question_id": "P-23-Q8", "question": "A charged particle of mass moving under\nthe influence of uniform electric field $E\\overset{\\rightarrow}{i}$and a\nuniform magnetic field $B\\overset{\\rightarrow}{k}$follows a trajectory\nfrom point P to Q as shown in figure. The velocitiesat P and Q are\nrespectively,\n$v\\overset{\\rightarrow}{i}$and$- 2v\\overset{\\rightarrow}{j}$. Then which\nof the following statements (A, B, C, D) are thecorrect ? (Trajectory\nshown is schematic and not to scale)", "question_images": ["images/image23.jpeg"], "option_1": "$E = \\frac{3}{4}(\\frac{mv^{2}}{qa})$", "option_2": "Rate of work done by the electric field at P\nis$\\frac{3}{4}\\left( \\frac{{mv}^{3}}{a} \\right)$and at Q is zero.", "option_3": "The difference between the magnitude of angular momentum of the\nparticle at P and Q is 2mav", "option_4": ". All of the above", "correct_option": 4, "numerical_answer": null, "solution": "(A) by work energy theorem\n$0 + {qE}_{0}2a = \\frac{3}{2}{mv}^{2}$\n$$E = \n(B) Rate of work done at A= power of electric force\n$= {qE}_{0}V$\n$= \\frac{3}{4}\\frac{{mv}^{3}}{a}$\n(C) at$Q,\\frac{dW}{dt} = 0$for both forces\n(D)\n$\\Delta\\overset{\\rightarrow}{L} = ( - m2v2a\\overset{\\hat{}}{k}) - ( - mva\\overset{\\hat{}}{k})$\n$|\\Delta\\overset{\\rightarrow}{L}| = 3mva$", "solution_images": [], "subject": "Physics", "topic": "Magnetics", "subtopic": "Cross field", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-23 Physcis Paper 31 Dec FST.docx"} {"question_id": "P-10-Q22", "question": "A uniform rod of mass 200 gram and length\n[IMAGE] is initially at rest in vertical position. The rod\nis hinged at centre such that it can rotate freely without friction\nabout a fixed horizontal axis passing through its centre. Two particles\nof mass m = 100 gram each having horizontal velocity of equal magnitude\nu = 17 m/s strikes the rod at top and bottom simultaneously as shown and\nstick to the rod.\nThe angular speed (in rad/s) of rod when it becomes horizontal is 8.5n.\nFind n", "question_images": ["images/image69.png", "images/image70.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "9", "solution": "From conservation of angular momentum.\n$mu\\frac{L}{2} + mu\\frac{L}{2} = \\left\\lbrack 2m\\frac{L^{2}}{12} + m\\left\\{ \\left( \\frac{L}{2} \\right)^{2} + \\left( \\frac{L}{2} \\right)^{2} \\right\\} \\right\\rbrack\\omega$\n$muL = \\left\\lbrack \\frac{{mL}^{2}}{6} + \\frac{{mL}^{2}}{4} + \\frac{{mL}^{2}}{4} \\right\\rbrack\\omega$\n$= \\frac{2{mL}^{2}}{3}\\omega$\nor $\\omega = \\frac{3u}{2L}$\n$= 76.50$", "solution_images": [], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-10 Physics paper 1 30 October Jee Main.docx"} {"question_id": "P-22-Q8", "question": "A charged particle of mass moving under\nthe influence of uniform electric field $E\\overset{\\rightarrow}{i}$and a\nuniform magnetic field $B\\overset{\\rightarrow}{k}$follows a trajectory\nfrom point P to Q as shown in figure. The velocitiesat P and Q are\nrespectively,\n$v\\overset{\\rightarrow}{i}$and$- 2v\\overset{\\rightarrow}{j}$. Then which\nof the following statements (A, B, C, D) are the correct ? (Trajectory\nshown is schematic and not to scale)", "question_images": ["images/image27.jpeg"], "option_1": "$E = \\frac{3}{4}(\\frac{mv^{2}}{qa})$", "option_2": "Rate of work done by the electric field at P\nis$\\frac{3}{4}\\left( \\frac{{mv}^{3}}{a} \\right)$and at Q is zero.", "option_3": "The difference between the magnitude of angular momentum of the\nparticle at P and Q is 2mav", "option_4": ". All of the above", "correct_option": 4, "numerical_answer": null, "solution": "(A) by work energy theorem\n$0 + {qE}_{0}2a = \\frac{3}{2}{mv}^{2}$\n$$E = \n(B) Rate of work done at A= power of electric force\n$= {qE}_{0}V$\n$= \\frac{3}{4}\\frac{{mv}^{3}}{a}$\n(C) at$Q,\\frac{dW}{dt} = 0$for both forces\n(D)\n$\\Delta\\overset{\\rightarrow}{L} = ( - m2v2a\\overset{\\hat{}}{k}) - ( - mva\\overset{\\hat{}}{k})$\n$|\\Delta\\overset{\\rightarrow}{L}| = 3mva$", "solution_images": [], "subject": "Physics", "topic": "Magnetics", "subtopic": "Cross field", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "P-19-Q20", "question": "A LCR circuit behaves like a damped harmonic oscillator.\nComparing it with a physical spring-mass damped oscillator having\ndamping constant the correct equivalence would be", "question_images": [], "option_1": "$L \\leftrightarrow \\frac{1}{b},C \\leftrightarrow \\frac{1}{m},R \\leftrightarrow \\frac{1}{k}$", "option_2": "$L \\leftrightarrow k,C \\leftrightarrow b,R \\leftrightarrow m$", "option_3": "$L \\leftrightarrow m,C \\leftrightarrow \\frac{1}{k},R \\leftrightarrow b$", "option_4": "$L \\leftrightarrow m,C \\leftrightarrow k,R \\leftrightarrow b$", "correct_option": 3, "numerical_answer": null, "solution": "In damped oscillation\n$ma + bv + kx = 0$\n[IMAGE] $m\\frac{d^{2}x}{{dt}^{2}} + b\\frac{dx}{dt} + kx = 0$....(i)\nIn the circuit\n$- iR - L\\frac{di}{dt} - \\frac{q}{c} = 0$\n$L\\frac{d^{2}q}{{dt}^{2}} + R\\frac{dq}{dt} + \\frac{1}{c} \\cdot q = 0$....(ii)\nComparing equation (i) and (ii)\n$m = L,b = R,k = \\frac{1}{c}$", "solution_images": ["images/image28.jpeg"], "subject": "Physics", "topic": "AC circuit", "subtopic": "LCR circuit", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-19 Physics paper 28 Nov Fourth.docx"} {"question_id": "Ph-27-Q19", "question": "A non-viscous fluid of density ρ is flowing in a tube as shown\nin figure. Area of section", "question_images": [], "option_1": "is double that of section", "option_2": ". Centre\nof mass of section", "option_3": "Work done by gravitational force per unit volume from section", "option_4": "Work done by elastic forces (pressure) per unit volume from\nsection", "correct_option": 2, "numerical_answer": null, "solution": "Applying Bernoulli is equation from section-(1) and (2)\n[k.M-(1).M-(2) esa cjukWyh lehdj.k yxkus ij\n[IMAGE] andrFkkP_1- P_2= $\\rho$g(2h)\nSolving we get, gy djus ij çkIr gksxk\n$V = \\sqrt{\\frac{2gh}{3}}$\n(C) Work done by gravitation force per unit value\n[IMAGE] decrease in gravitational\nPT_value = $\\rho{gh}_{1} - \\rho{gh}_{2}$\n \nPT_value = $\\rho{gh}_{1}\\ - \\rho{gh}_{2}$\n$W_{gr} = 0 - \\rho gh$\n(D) Work done by elastic force volume, çR;kLFk cyksa }kjk çfr bdkbZ\nW_e=decrease in elastic P.E._vol = decrease in pressure\nW_e=çR;kLFk fLFkfrt P.E._vol= nkc", "solution_images": ["images/image96.png", "images/image97.png", "images/image98.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Bernoulli's equation", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-16-Q6", "question": "A small smooth disc of mass m and radius r moving with an initial\nvelocity along the positive x-axis collided with a big disc of mass\n2 m and radius 2 r which was initially at rest with its centre at origin\nas show in figure.\n[IMAGE] If the coefficient of restitution is 0 then velocity of larger disc\nafter collision is", "question_images": ["images/image13.png"], "option_1": "$\\frac{8v}{27}\\overset{\\hat{}}{i} - \\frac{2\\sqrt{2}}{27}v\\overset{\\hat{}}{j}$", "option_2": "$\\frac{8v}{27}\\overset{\\hat{}}{i} + \\frac{2\\sqrt{2}}{27}v\\overset{\\hat{}}{j}$", "option_3": "$\\frac{v}{3}\\overset{\\hat{}}{i}$", "option_4": "$\\frac{2\\sqrt{2}}{27}v\\overset{\\hat{}}{i} - \\frac{8v}{27}\\overset{\\hat{}}{j}$", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] The larger sphere will move along line of impact.\ne=0, magnitude of velocity of larger sphere\n$v^{'} = \\frac{mvcos\\theta}{m + 2m} = \\frac{vcos\\theta}{3}$\nCos$\\theta$= $2\\sqrt{2}/3$\nSin $\\theta$= 1/3\nVelocity of larger sphere\n$= v^{'}cos\\theta\\overset{\\hat{}}{i} - v^{'}sin\\theta\\overset{\\hat{}}{j}$\n$= \\frac{v}{3}\\cos^{2}\\theta\\overset{\\hat{}}{i} - \\frac{v}{3}sin\\theta cos\\theta\\overset{\\hat{}}{j} = \\frac{8}{27}v\\overset{\\hat{}}{i} - \\frac{2\\sqrt{2}}{27}v\\overset{\\hat{}}{j}$", "solution_images": ["images/image14.png"], "subject": "Physics", "topic": "Center of mass", "subtopic": "Collision", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-16-Q10", "question": "Initially spring is in its natural length and blocks A and B are\nat rest. Find maximum value of constant force F that can be applied on B\nsuch that block A remains at rest. (g = 10 ms-2).", "question_images": ["images/image20.png"], "option_1": "2", "option_2": "3", "option_3": "4", "option_4": "5", "correct_option": 2, "numerical_answer": null, "solution": "Value of $\\mu mg$ for A and B are 2N and 2N.\nWork energy theorem on B gives\nFx - 2x - (k/2) x^2 = 0\nF = 2 + kx/2....................(i)\nFor equilibrium of A, kx = $\\mu mg = 2$\nSo, F = 2 + 1 = 3N", "solution_images": [], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Friction", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-03-Q10", "question": "In the system shown in the figure the friction and mass of rope is\nnegligible, then acceleration of 2 m is", "question_images": ["images/image54.png"], "option_1": "", "option_2": "", "option_3": "0", "option_4": "[IMAGE], 2 m", "correct_option": 3, "numerical_answer": null, "solution": "2mg - 2T = 2ma\n[IMAGE] 2 [T - mg = m (2a)]\n0 = 6 ma ∴ a = 0\nTOPIC: NLM\nSUB TOPIC: EQUILIBRIUM", "solution_images": ["images/image58.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx"} {"question_id": "P-24-Q22", "question": "In a cylinder-piston arrangement, air is under a pressure P_1.\nA soap bubble of radius r lies inside the cylinder, soap bubble has\nsurface tension T. The radius of soap bubble is to be reduced to half,\nThe new pressure P_2 to which air should be compressed isothermally.\n(Assume r is very small as compared to height of cylinder)", "question_images": ["images/image57.png"], "option_1": "$\\ P_{1} + \\frac{4T}{r}$", "option_2": "$\\ 4P_{1} + \\frac{12T}{r}$", "option_3": "$\\ 8P_{1} + \\frac{24T}{r}$", "option_4": "$\\ P_{1} + \\frac{2T}{r}$", "correct_option": 3, "numerical_answer": null, "solution": "Isothermal process.\n$\\left( P_{1} + \\frac{4T}{r} \\right)\\left( \\frac{4}{3}\\pi r^{3} \\right) = \\left( P_{2} + \\frac{4T}{r/2} \\right)\\left( \\frac{4}{3}\\pi(r/2)^{3} \\right)$\n$P_{2} = 8P_{1} + \\frac{24T}{r}$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Surface tension", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "Ph-28-Q3", "question": "A man sitting inside a train moving with uniform acceleration\n[IMAGE]. Man throws a ball from point O (on the floor of\ntrain) with relative velocity [IMAGE] (where\n[IMAGE] = velocity of ball with respect to train\n[IMAGE] Ball falls on the train at point A (where OA = R) after time T.\nChoose correct option", "question_images": ["images/image18.png", "images/image19.png", "images/image20.png", "images/image21.png", "images/image22.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "[IMAGE] Time of flight = T = $\\frac{2\\ (20\\sqrt{}2)sin45{^\\circ}}{10} = 4s$\nSolving we get[IMAGE] TOPIC:Projectile motion\nSUB TOPIC:Relative motion\nLEVEL:Moderate", "solution_images": ["images/image27.png", "images/image28.png", "images/image29.png", "images/image30.png", "images/image31.png", "images/image23.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "Ph-28-Q5", "question": "Suppose the potential energy between electron and proton at a\ndistance r is given by [IMAGE] where k is constant, e is\ncharge of proton and r is radius of orbit. Application of Bohr's theory\nto hydrogen atom in this case shows that", "question_images": ["images/image33.png"], "option_1": "energy of electron in [IMAGE] orbit is\nproportional to [IMAGE] and [IMAGE] (m =\nmass of electron).", "option_2": "energy of electron in [IMAGE] orbit is\nproportional to [IMAGE] and [IMAGE] (m =\nmass of electron).", "option_3": "energy of electron in [IMAGE] orbit is\nproportional to [IMAGE] and [IMAGE] (m =\nmass of electron).", "option_4": "energy of electron in [IMAGE] orbit is\nproportional to [IMAGE] and [IMAGE] (m =\nmass of electron).", "correct_option": 1, "numerical_answer": null, "solution": "From (1) and (2)\n[IMAGE] TOPIC:Structure of Atom\nSUB TOPIC:Bohr's model of hydrogen\nLEVEL:Moderate", "solution_images": ["images/image41.png", "images/image42.png", "images/image43.png", "images/image44.png", "images/image45.png", "images/image46.png", "images/image47.png", "images/image48.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"} {"question_id": "P-06-Q21", "question": "A uniform rope of linear mass density $\\lambda$ and length 1 is\ncoiled on a smooth horizontal surface. One end is pulled up with\nconstant velocity v. Then the average power applied by the external\nagent in pulling the entire rope just off the ground is", "question_images": ["images/image29.png"], "option_1": "$\\frac{1}{2}\\lambda\\mathcal{l}v^{2} + \\frac{\\lambda\\mathcal{l}^{2}g}{2}$", "option_2": "$\\lambda\\mathcal{l}gv$", "option_3": "$\\frac{1}{2}\\lambda v^{3} + \\frac{\\lambda\\mathcal{l}vg}{2}$", "option_4": "$\\lambda\\mathcal{l}gv + \\frac{1}{2}\\lambda v^{3}$", "correct_option": 3, "numerical_answer": null, "solution": "Initial K.E. = 0\nInitial P.E. = 0\nWhen the rope is just pulled off the table.\nFinal K.E. = $\\frac{1}{2}(\\lambda l)v^{2}$\nFinal P.E$= \\frac{\\left( \\lambda\\mathcal{l} \\right)g\\mathcal{l}}{2}$\nTime taken $t = 1/v$\nAverage power = $\\frac{\\text{ net change in energy }}{\\text{ time }}$\n$= \\frac{\\frac{1}{2}\\lambda\\mathcal{l}v^{2} + \\lambda\\mathcal{l}g\\mathcal{l/}2}{\\mathcal{l/}v} = \\frac{1}{2}\\lambda v^{3} + \\frac{\\lambda\\mathcal{l}vg}{2}$", "solution_images": [], "subject": "Physics", "topic": "Work Power & Energy", "subtopic": "Average power", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "P-05-Q3", "question": "A particle moves along x-axis with initial position x=0. Its\nvelocity varies with x-coordinate as shown in graph. The acceleration\n'a\" of this particle varies with x as", "question_images": ["images/image1.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "The linear relationship between V and x is $V = - mx + C$\nwhere m and C are positive constants.\nAcceleration\n$a = v\\frac{dV}{dx} = - m( - mx + C)$\nHence the graph relating a tox is.", "solution_images": ["images/image6.png"], "subject": "Physics", "topic": "Motion in 1D", "subtopic": "Uniform accelerated motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-05 Physics paper 10 October.docx"} {"question_id": "P-02-Q52", "question": "A particle is projected vertically upward from the surface of\nthe earth with a speed of [IMAGE], R being the radius\nof the earth and g is the acceleration due to gravity on the surface of\nthe earth. Then the maximum height ascended is nR,then value of n\nis(neglect cosmic dust resistance)", "question_images": ["images/image101.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "03", "solution": "[IMAGE] and g = [IMAGE] Therefore [IMAGE] ×[IMAGE] × gR - Rg =\n- [IMAGE] We get h = 3R", "solution_images": ["images/image102.png", "images/image103.png", "images/image104.png", "images/image105.png", "images/image106.png", "images/image107.png", "images/image108.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-02 Physics Paper-02_Update.docx"} {"question_id": "P-04-Q9", "question": "On a smooth pulley of radius R hangs a homogeneous flexible rope of\nmass / length λ and length  (Fig.). Find the maximum tension in the\nrope.", "question_images": ["images/image52.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "[IMAGE]  λ / R \n ()", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] Tmax - T0 = λgR [1 - cos θ]\nTmax = λgR + λg[IMAGE] = λ \nTOPIC: NLM\nSUB TOPIC: EQUILIBRIUM", "solution_images": ["images/image57.png", "images/image58.png", "images/image59.png", "images/image60.png", "images/image61.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx"} {"question_id": "P-20-Q22", "question": "The minimum time in which the potential energy of a particle\nexecuting SHM changes from maximum to minimum is 5 s. The time period of\nSHM in seconds is 4T. Find T.", "question_images": [], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "5", "solution": "P.E. is maximum at extreme position and minimum at mean\nposition.\nTime to go from extreme position to mean position is,\n where T is time period of SHM.", "solution_images": ["images/image79.png", "images/image80.png"], "subject": "Physics", "topic": "Simple Harmonic Motion", "subtopic": "Time period of oscillation", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-20 Physcis Paper 12 Dec.docx"} {"question_id": "P-21-Q13", "question": "Three metallic plates out of which middle is given a charge Q,\nas shown in the figure. The outer plates can be earthed with switches S1\nand S2. The area of plates is same.The charge appearing on outer surface\nof external plates when S_1 and S_2 are open and external platesare\nuncharged.\nCkgjh ifV~Vdkvksa dh ckgjh lrgks ij S_1 rFkk S_2", "question_images": ["images/image35.png"], "option_1": "$- \\left( \\frac{Q}{2} \\right)$", "option_2": "$\\left( \\frac{Q}{2} \\right)$", "option_3": "Q", "option_4": "- Q", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] (conductor pkyd)\n[IMAGE] So vr -x = x + Q + y - y\n2x = - Q\n$x = \\frac{- Q}{2}$\n$- x = \\frac{Q}{2}$", "solution_images": ["images/image36.png", "images/image37.png"], "subject": "Physics", "topic": "Electrostatics", "subtopic": "Capacitance", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-21 Physcis Paper 19 Dec Hindi Eng..docx"} {"question_id": "P-06-Q6", "question": "System is shown in the figure. Assume that cylinder remains in\ncontact with the two wedges. The velocity of cylinder is", "question_images": ["images/image9.png"], "option_1": "$\\sqrt{19 - 4\\sqrt{3}}\\frac{u}{2}\\ m/s$", "option_2": "$\\frac{\\sqrt{13}u}{2}\\ m/s$", "option_3": "$\\sqrt{3}u\\ m/s$", "option_4": "$\\sqrt{7}u\\ m/s$", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] As cylinder will remains in contact with wedge A\nVx = 2u\nAs it also remain in contact with wedge B\nu sin 30° = Vy cos30° - Vxsin30°\nVy = Vx$\\frac{Sin30{^\\circ}}{Cos30{^\\circ}}$ +\n$u\\frac{Sin30{^\\circ}}{Cos30{^\\circ}}$\nVy = Vx tan30° + u tan 30°\nVy = 3utan30° = $\\sqrt{3}$u\nV = $\\sqrt{7}$u", "solution_images": ["images/image10.png"], "subject": "Physics", "topic": "Newton's Laws of Motion", "subtopic": "Constraint motion", "difficulty": "Tough", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-06 Physics paper 14 October.docx"} {"question_id": "Ph-26-Q23", "question": "A thin metal rod of length L_0 is shaped into a ring with a\nsmall gap x as shown. On heating the system", "question_images": ["images/image80.png"], "option_1": "x decreases, r and d increase", "option_2": "x and r increase, d decreases", "option_3": "x, r and d all increase", "option_4": "Data insufficient to arrive at a conclusion\nL_0 yackbZ dh,d iryh NM+ dks vYi varjky x", "correct_option": 3, "numerical_answer": null, "solution": "On uniform heating of an object which is free to expand,\ndistance between any two points increases.\noLrq tks çlkfjr gksus ds fy, LorU= gS]", "solution_images": [], "subject": "Physics", "topic": "Mechanical properties of solid", "subtopic": "Thermal expansion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-26 Physics paper 1 15 Jan.docx"} {"question_id": "Ph-27-Q18", "question": "Which of the following is an equivalent cyclic process\ncorresponding to the thermodynamic cyclic givenin the figure ? where, 1\n→ 2 is adiabatic. (Graphs are schematic and are not to scale)", "question_images": ["images/image88.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 1, "numerical_answer": null, "solution": "For process A - B izØe ds fy,; Volume is constant \nPV = nRT T increases cM+rk gS\nFor process B - C izØe \n$\\Rightarrow {TV}^{\\gamma - 1}$ = Constant \nFor process C - A izØe ds fy,; pressure is constant", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "subtopic": "Thermodynamic cycle", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-27 Physics paper 2 15 Jan.docx"} {"question_id": "P-16-Q24", "question": "Two stones are in the same horizontal line and are$50\\sqrt{5}m$\naway. They are projected with a velocity of 10 \nrespectively and with an angle of 37º and 53º with horizontal\nrespectively. The minimum distance between them in meters is 5n. Find s.\nSuppose they don't strike the ground during the motion.\n(take$g = 10m/s^{2}$)", "question_images": ["images/image43.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": null, "numerical_answer": "4", "solution": "$d_{\\min} = 50\\sqrt{5}\\ \\ \\ \\ \\ \\ \\ \\ sin\\theta = 50\\sqrt{5} \\times \\frac{2}{5\\sqrt{5}} = 20m$", "solution_images": ["images/image44.png"], "subject": "Physics", "topic": "Relative motion", "subtopic": "closest distance of approach", "difficulty": "Moderate", "question_type": "numerical", "has_image": true, "exam": "JEE Main", "source_paper": "P-16 Physics paper 25 Nov First.docx"} {"question_id": "P-24-Q9", "question": "A leak proof cylinder of length 1 m, made of a metal which has\nvery low coefficient of expansion is floating vertically in water at 0°C\nsuch that its height above the water surface is 20 cm. When the\ntemperature of water is increased to 4°C, the height of the cylinder\nabove the water surface becomes 21 cm. The density of water at T = 4°C,\nrelative to the density at T = 0°C is close to", "question_images": [], "option_1": "1.03", "option_2": "1.04", "option_3": "1.26", "option_4": "1.01", "correct_option": 4, "numerical_answer": null, "solution": "$mg = A(79)\\rho_{4^{\\circ}C}g$\n$\\frac{\\rho_{4^{\\circ}C}}{\\rho_{0^{\\circ}C}} = \\frac{80}{79} = 1.01$", "solution_images": ["images/image39.png", "images/image40.png"], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Buoyancy", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-18-Q4", "question": "A particle of mass $m = \\frac{5}{42}kg$ is tied to an end of an\ninextensible string and is whirled in a vertical circle as shown below.\nThen, the difference in tension of string at A and B i.e.\n$T_{A} - T_{B}$ in Newton is: (particle completes the circle)", "question_images": ["images/image5.png"], "option_1": "2", "option_2": "4", "option_3": "3", "option_4": "5", "correct_option": 4, "numerical_answer": null, "solution": "$\\Delta T = \\frac{3mg\\Delta h}{I}$\n$h_{B} - h_{A} = I(\\frac{3}{5} + \\frac{4}{5}) = \\frac{7I}{5}$\n$T_{A} - B_{B} = \\frac{3mg}{I}(h_{B} - h_{A})$\n$= \\frac{3mg}{I} \\times \\frac{7I}{5}$\n$= \\frac{21mg}{5} = \\frac{21}{5} \\times \\frac{5}{42} \\times 10$\n$= 5N$", "solution_images": [], "subject": "Physics", "topic": "Circular motion", "subtopic": "Vertical circular motion", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-18 Physics paper 27 Nov Third.docx"} {"question_id": "P-24-Q10", "question": "An ideal fluid flows (laminar flow) through a pipe of\nnon-uniform diameter. The maximum and minimum diameters of the pipes are\n6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the\nmaximum velocities of fluid in this pipe is", "question_images": [], "option_1": "$\\frac{9}{16}$", "option_2": "$\\frac{\\sqrt{3}}{2}$", "option_3": "$\\frac{81}{256}$", "option_4": "$\\frac{8}{4}$", "correct_option": 1, "numerical_answer": null, "solution": "Using equation of continuity\n$\\frac{V_{1}}{V_{2}} = \\frac{A_{2}}{A_{2}} = \\left( \\frac{4.8}{6.4} \\right)^{2} = \\frac{9}{16}$", "solution_images": [], "subject": "Physics", "topic": "Mechanical Properties of Fluids", "subtopic": "Equation of continuity", "difficulty": "Moderate", "question_type": "single_correct", "has_image": false, "exam": "JEE Main", "source_paper": "P-24 Physcis Paper 31 Dec.docx"} {"question_id": "P-11-Q9", "question": "Consider a uniform rod of mass [IMAGE] and length\n[IMAGE] pivoted about its centre. A mass m moving\nwithvelocity v making angle [IMAGE] to the rod's long\naxis collides with one end of the rod and sticks to it. The angular\nspeed of the rod-mass system just after the collision is", "question_images": ["images/image86.png", "images/image87.png", "images/image88.png"], "option_1": "", "option_2": "", "option_3": "", "option_4": "", "correct_option": 2, "numerical_answer": null, "solution": "", "solution_images": ["images/image93.png", "images/image94.png", "images/image95.png", "images/image96.png"], "subject": "Physics", "topic": "Rigid Body Dynamics", "subtopic": "Conservation of angular momentum", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-11 Physics paper-2 1Nov Jee Main.docx"} {"question_id": "P-22-Q18", "question": "A uniform magnetic field B in positive z direction exists in a\ncircular region of radius R = 5 m. A loop of radius R = 5m lying in x -\ny plane encloses the magnetic field at t = 0 and then pulled at uniform\nvelocity[IMAGE]. The emf induced (in volts) is the loop\nat t = 2 sec is 6V. Then magnitude of 4B is", "question_images": ["images/image45.png", "images/image46.png"], "option_1": "3", "option_2": "1", "option_3": "7", "option_4": "8", "correct_option": 2, "numerical_answer": null, "solution": "[IMAGE] TOPIC:Electromagnetic induction\nSUB TOPIC:Faraday's law\nLEVEL: Moderate", "solution_images": ["images/image47.png", "images/image48.png", "images/image49.png", "images/image50.png", "images/image51.png", "images/image52.png"], "subject": "Physics", "topic": "", "subtopic": "", "difficulty": "", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "P-22 Physcis Paper 19 Dec Eng..docx"} {"question_id": "Ph-28-Q17", "question": "In the figure shown $C_{1} = 11\\mu F$ and $C_{2} = 5\\mu F$ then\nat steady state", "question_images": ["images/image168.png"], "option_1": "The potential difference across C_1 is 5V", "option_2": "The potential difference between terminals of 15V battery is 9V", "option_3": "The potential difference V_a - V_b = - 4V", "option_4": "All of the above", "correct_option": 4, "numerical_answer": null, "solution": "[IMAGE] at steady state\n$I(3) + 1(2) = 15$\nI = 3\n$kvL$\n$C \\rightarrow D \\rightarrow E \\rightarrow a \\rightarrow b \\rightarrow C$\n$V_{c} - I(3) + \\frac{q}{11} - 7 + \\frac{q}{5} = V_{c}$\n$\\frac{q}{11} + \\frac{q}{5} = 7 + 3 \\times 3 = 16$\n$q = 55\\mu C$\nNow KVL for a$\\rightarrow$b\n$V_{a} - 7 + \\frac{q}{5} = V_{b}$\n$\\Rightarrow V_{a} - V_{b} = 7 - \\frac{q}{5}$\n$= 7 - \\frac{55}{5}$ $= - \\frac{20}{5} = - 4V$", "solution_images": ["images/image169.png"], "subject": "Physics", "topic": "Current Electricity", "subtopic": "KVL", "difficulty": "Moderate", "question_type": "single_correct", "has_image": true, "exam": "JEE Main", "source_paper": "Ph-28 Physics paper FST 16 Jan.docx"}