physics/train.jsonl

{"question_id": "ADV-P01-Q26", "question": "A proton accelerated by a potential difference $V$ gets\ninto the uniform electric field of a parallel-plate capacitor whose\nplates extend over a length $l$ in the motion direction. The field\nstrength varies with time as $E = at$, where $a$ is a constant. Assuming\nthe proton to be non-relativistic, find the angle between the motion\ndirections of the proton before and after its flight through the\ncapacitor; the proton gets in the field at the moment $t = 0$. The edge\neffects are to be neglected.", "question_images": [], "answer": "$\\frac{al^{2}}{4}\\sqrt{\\frac{m}{2eV^{3}}}$", "solution": "The electric field inside the capacitor varies with time\nas,\n$$E = at$$\nHence, electric force on the proton,\n$$F = eat$$\nand subsequently, acceleration of the proton,\n$$w = \\frac{eat}{m}$$\nNow, if $t$ is the time elapsed during the motion of the proton\nbetween the plates, then $t = \\frac{l}{v_{||}}$, as no acceleration is\neffective in this direction. (Here $v_{||}$ is velocity along the\nlength of the plate.)\nFrom kinematics, $\\frac{dv_{\\bot}}{dt} = w$\nSo, $\\int_{0}^{t}{dv_{\\bot}} = \\int_{0}^{t}{wdt,}$\n(as initially, the component of velocity in the direction, $\\bot$ to\nplates, was zero.)\nOr\n$v_{\\bot} = \\int_{0}^{t}\\frac{ea}{m}\\frac{t^{2}}{2m} = \\frac{ea}{2m}\\frac{l^{2}}{v_{||}^{2}}$\nNow,\n$\\tan\\alpha = \\frac{v_{\\bot}}{v_{||}} = \\frac{eal^{2}}{2m\\ v_{||}^{3}}$\n$= \\frac{eal^{2}}{2m\\left( \\frac{2eV}{m} \\right)^{\\frac{3}{2}}},$ as\n$v_{||} = \\left( \\frac{2eV}{m} \\right)^{\\frac{1}{2}}$, from energy\nconservation.\n$$= \\frac{al^{2}}{4}\\sqrt{\\frac{m}{2eV^{3}}}$$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q24", "question": "A long solenoid of cross-sectional radius $a$ has a\nthin insulated wire ring tightly put on its winding; one half of the\nring has the resistance $\\eta$ times that of the other half. The\nmagnetic induction produced by the solenoid varies with time as\n$B = bt$, where $b$ is a constant. Find the magnitude of the electric\nfield strength in the ring.", "question_images": [], "answer": "$\\frac{1}{2}ab\\frac{\\eta - 1}{\\eta + 1}$", "solution": "The changing magnetic field will induce an e.m.f. in the\nring, which is obviously equal, in the two parts by symmetry (the e.m.f.\ninduced by electromagnetic induction does not depend on resistance). The\ncurrent, that will flow due to this, will be different in the two parts.\nThis will cause an acceleration of charge, leading to the setting up of\nan electric field $E$ which has opposite sign in the two parts. Thus,\n$\\frac{\\xi}{2} - \\pi aE = rl$ and, $\\frac{\\xi}{2} + \\pi aE = \\eta rI$\nwhere $\\xi$ is the total induced e.m.f. From this,\n$\\xi = (\\eta + 1)rI$,\nand\n$E = \\frac{1}{2\\pi a}(\\eta - 1)rI = \\frac{1}{2\\pi a}\\frac{\\eta - 1}{\\eta + 1}\\xi$\nBut by Faraday's law, $\\xi = \\pi a^{2}b$\nSo, $E = \\frac{1}{2}ab\\frac{\\eta - 1}{\\eta + 1}$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q20", "question": "A long dielectric cylinder of radius $R$ is statically\npolarized so that at all its points the polarization is equal to\n$P = \\alpha r$, where $\\alpha$ is a positive constant, and $r$ is the\ndistance from the axis. The cylinder is set into rotation about its axis\nwith an angular velocity $\\omega$. Find the magnetic induction $B$ at\nthe centre of the cylinder.", "question_images": [], "answer": "$B = B_{s} + B_{v} = 0$", "solution": "Because of polarization a space charge is present within\nthe cylinder. It's density is\n$$\\rho_{p} = - div\\overrightarrow{P} = - 2\\alpha$$\nSince the cylinder as a whole is neutral a surface charge density\n$\\sigma_{P}$ must be present on the surface of the cylinder also. This\nhas the magnitude (algebraically)\n$\\sigma_{p} \\times 2\\pi R = 2\\alpha\\pi R^{2}$ or,\n$\\sigma_{p} = \\alpha R$\nWhen the cylinder rotates, currents are set up which give rise to\nmagnetic fields. The contribution of $\\rho_{p}$ and $\\sigma_{p}$ can\nbe calculated separately and then added.\nFor the surface charge the current is (for a particular element)\n$$\\alpha R \\times 2\\pi R\\ dx \\times \\frac{\\omega}{2\\pi} = \\alpha R^{2}\\omega dx$$\nIts contribution to the magnetic field at the centre is\n$$\\frac{\\mu_{0}R^{2}\\left( \\alpha R^{2}\\omega dx \\right)}{2\\left( x^{2} + R^{2} \\right)^{3/2}}$$\nand the total magnetic field is\n$$B_{S} = \\int_{\\infty}^{- \\infty}\\frac{\\mu_{0}R^{2}\\left( \\alpha R^{2}\\omega dx \\right)}{2\\left( x^{2} + R^{2} \\right)^{3/2}} = \\frac{\\mu_{0}\\alpha R^{4}\\omega}{2}\\int_{- \\infty}^{\\infty}\\frac{dx}{\\left( x^{2} + R^{2} \\right)^{3/2}} = \\frac{\\mu_{0}\\alpha R^{4}\\omega}{2} \\times \\frac{2}{R^{2}} = \\mu_{0}\\alpha R^{2}\\omega$$\nAs for the volume charge density consider a circle of radius r, radial\nthickness $dr$ and length $dx$.\nThe current is\n$- 2\\alpha \\times 2\\pi rdrdx \\times \\frac{\\omega}{2\\pi} = - 2\\alpha r\\ dr\\ \\omega dx$\nThe total magnetic field due to the volume charge distribution is\n$$B_{v} = - \\int_{0}^{R}{dr}\\int_{- \\infty}^{\\infty}{dx}2\\pi r\\omega\\frac{\\mu_{0}r^{2}}{2\\left( x^{2} + r^{2} \\right)^{\\frac{3}{2}}} = - \\int_{0}^{R}{\\alpha\\mu_{0}\\omega r^{3}\\ dr}\\int_{- \\infty}^{\\infty}{dx}\\left( x^{2} + r^{2} \\right)^{3/2}$$\n$$= - \\int_{0}^{R}{\\alpha\\mu_{0}\\omega r}dr \\times 2 = - \\mu_{0}\\alpha\\omega R^{2}\\ \\ so,B = B_{s} + B_{v} = 0$$", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q12", "question": "Inside an infinitely long circular cylinder charged\nuniformly with volume density $\\rho$ there is a circular cylindrical\ncavity. The distance between the axes of the cylindrical cavity. The\ndistance between the axes of the cylinder and the cavity is equal to a.\nFind the electric field strength E inside the cavity. The permittivity\nis assumed to be equal to unity.", "question_images": [], "answer": "$\\overrightarrow{E} = \\overrightarrow{E} + \\overrightarrow{E} = \\frac{\\rho}{2\\varepsilon_{0}}\\left( \\overrightarrow{r_{+}} - \\overrightarrow{r_{-}} \\right) = \\frac{\\rho}{2\\varepsilon_{0}}\\overrightarrow{a}$", "solution": "Let us consider a cylindrical Gaussian surface of radius\n$r$ and height $h$ inside an infinitely long charged cylinder with\ncharge density $\\rho$. Now from Gauss theorem:\n$$E_{r}2\\pi rh = \\frac{q_{inclosed}}{\\varepsilon_{0}}$$\n(where $E_{r}$ is the field inside the cylinder at a distance $r$ from\nit axis.)\nOr, $E_{r}2\\pi rh = \\frac{\\rho\\pi r^{2}h}{\\varepsilon_{0}}$ or\n$E_{r} = \\frac{\\rho r}{2\\varepsilon_{0}}$\nNow, using the method of 3.28 field at a point $P$, inside the cavity,\nis\n$$\\overrightarrow{E} = \\overrightarrow{E} + \\overrightarrow{E} = \\frac{\\rho}{2\\varepsilon_{0}}\\left( \\overrightarrow{r_{+}} - \\overrightarrow{r_{-}} \\right) = \\frac{\\rho}{2\\varepsilon_{0}}\\overrightarrow{a}$$\n[IMAGE]", "solution_images": ["images/image9.png"], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q5", "question": "Find the pressure of air in a vessel being evacuated as\na function of evacuation time $t$. The vessel volume is $V$, the initial\npressure is $p_{0}$. The process is assumed to be isothermal, and the\nevacuation rate equal to $C$ and independent of pressure.\nNote: The evacuations rate is the gas volume being evacuated per\nunit time, with that volume being measured under the gas pressure\nattained by that moment.", "question_images": [], "answer": "$p = p_{0}e^{- Ct/V}$", "solution": "From the ideal gas equation $p = \\frac{m}{M}\\frac{RT}{V}$\n$\\frac{dp}{dt} = \\frac{RT}{MV}\\frac{dm}{dt}$ (1)\nIn each stroke, volume $v$ of the gas is ejected, where $v$ is given\nby\n$$v = \\frac{V}{m_{N}}\\left\\lbrack m_{N - 1} - m_{N} \\right\\rbrack$$\nIn case of continuous ejection, if $\\left( m_{N - 1} \\right)$\ncorresponds to mass of gas in the vessel at time $t$, time $m_{N}$ is\nthe mas at time $t + \\mathrm{\\Delta}t$, where $\\mathrm{\\Delta}t$, is\nthe time in which volume $v$ of the gas has come out. The rate of\nevacuation is therefore $\\frac{v}{\\mathrm{\\Delta}t}$ i.e.\n$$C = \\frac{v}{\\mathrm{\\Delta}t} = - \\frac{V}{m(t + \\mathrm{\\Delta}t)}.\\frac{m(t + \\mathrm{\\Delta}t) - m(t)}{\\mathrm{\\Delta}t}$$\nIn the limit $\\mathrm{\\Delta}t - 0$, we get\n$C = \\frac{V}{m}\\frac{dm}{dt}$ (2)\nFrom (1) and (2)\n$$\\frac{dp}{dt} = - \\frac{C}{V}\\frac{mRT}{MV} = - \\frac{C}{V}p\\ \\ \\ \\ or\\ \\ \\ \\frac{dp}{p} = - \\frac{C}{V}dt$$\nIntegrating\n$\\int_{p}^{p_{0}}\\frac{dp}{p} = - \\frac{C}{V}\\int_{t}^{0}{dt}$ or\n$\\ln\\frac{p}{p_{0}} = - \\frac{C}{V}t$\nThus $p = p_{0}e^{- Ct/V}$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q46", "question": "A cylinder rolls without slipping over a horizontal\nplane. The radius of the cylinder is equal to $r$. Find the curvature\nradii of trajectories traced out by the points $A$ and $B$ (see Fig.).\n[IMAGE]", "question_images": ["images/image43.png"], "answer": "$2\\sqrt{2}r$", "solution": "Let us draw the kinematical diagram of the rolling\ncylinder on the basis of the solution of problem 1.53.\n[IMAGE] Figure 1.54: Insert this diagram from the shared reference image ---\nrolling cylinder kinematics and acceleration directions.\nAs, an arbitrary point of the cylinder follows a curve, its normal\nacceleration and radius of curvature are related by the well known\nequation\n$$w_{n} = \\frac{v^{2}}{R}$$\nso, for point $A$,\n$$w_{A(n)} = \\frac{v_{A}^{2}}{R_{A}}$$\nor,\n$$R_{A} = \\frac{4v_{c}^{2}}{\\omega^{2}r} = 4r$$\nbecause $v_{c} = \\omega r$, for pure rolling.\nSimilarly for point $B$,\n$$w_{B(n)} = \\frac{v_{B}^{2}}{R_{B}}$$\n$$\\omega^{2}r\\cos 45^{\\circ} = \\frac{\\left( \\sqrt{2}v_{c} \\right)^{2}}{R_{B}},$$\nor,\n$$R_{B} = 2\\sqrt{2}\\frac{v_{c}^{2}}{\\omega^{2}r} = 2\\sqrt{2}r$$", "solution_images": ["images/image44.png"], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q27", "question": "A system consists of a long cylindrical anode of radius\n$a$ and a coaxial cylindrical cathode of radius $b(b < a)$. A filament\nlocated along the axis of the system carries a heating current $I$\nproducing a magnetic field in the surrounding space. Find the least\npotential difference between the cathode and anode at which the thermal\nelectrons leaving the cathode without initial velocity start reaching\nthe anode.", "question_images": [], "answer": "$V \\geq \\frac{1}{2}\\frac{|e|}{m}\\left( \\frac{\\mu_{0}I}{2\\pi} \\right)^{2}\\ln\\frac{a}{b}$", "solution": "When a current $I$ flows along the axis, a magnetic field\n$B_{\\varphi} = \\frac{\\mu_{0}I}{2\\pi\\rho}$ is set up where\n$\\rho^{2} = x^{2} + y^{2}$. In terms of components,\n$B_{x} = - \\frac{\\mu_{0}Iy}{2\\pi\\rho^{2}},B_{y} = \\frac{\\mu_{0}Ix}{2\\pi\\rho^{2}}$\nand $B_{z} = 0$\nSuppose a p.d. $V$ is set up between the inner cathode and the outer\nanode. This means a potential function of the form\n$$\\varphi = V\\frac{\\ln{\\rho/b}}{\\ln{a/b}},a > \\rho > b,$$\nas one can check by solving Laplace equation.\n[IMAGE] The electric field corresponding to this is,\n$$E_{x} = - \\frac{Vx}{\\rho^{2}\\ln{a/b}},E_{y} = - \\frac{Vy}{\\rho^{2}\\ln{a/b}},E_{z} = 0$$\nThe equations of motion are,\n$$\\frac{d}{dt}mv_{x} = + \\frac{|e|Vz}{\\rho^{2}\\ln{a/b}} + \\frac{|e|\\mu_{0}I}{2\\pi\\rho^{2}}x\\acute{z}$$\n$$\\frac{d}{dt}mv_{y} = + \\frac{|e|Vy}{\\rho^{2}\\ln{a/b}} + \\frac{|e|\\mu_{0}I}{2\\pi\\rho^{2}}y\\acute{z}$$\nand\n$\\frac{d}{dt}mv_{z} = - |e|\\frac{\\mu_{0}I}{2\\pi\\rho^{2}}\\left( x\\acute{x} + y\\acute{y} \\right) = - |e|\\frac{\\mu_{0}I}{2\\pi}\\frac{d}{dt}\\ln\\rho$\n$\\left( - |e| \\right)$ is the charge on the electron.\nIntegrating the last equation,\n$$mv_{z} = - |e|\\frac{\\mu_{0}I}{2\\pi}\\ln{\\rho/a} = m\\acute{z}$$\nsince $v_{z} = 0$ where $\\rho = a$. We now substitute this $\\acute{z}$\nin the other two equations to get\n$$\\frac{d}{dt}\\left( \\frac{1}{2}mv_{x}^{2} + \\frac{1}{2}mv_{y}^{2} \\right)$$\n$$= \\left\\lbrack \\frac{|e|V}{\\ln\\frac{a}{b}} - \\frac{|e|^{2}}{m}\\left( \\frac{\\mu_{0}I}{2\\pi} \\right)^{2}\\ln\\frac{\\rho}{b} \\right\\rbrack.\\frac{x\\acute{x} + y\\acute{y}}{\\rho^{2}}$$\n$$= \\left\\lbrack \\frac{|e|V}{\\ln\\frac{a}{b}} - \\frac{|e|^{2}}{m}\\left( \\frac{\\mu_{0}I}{2\\pi} \\right)^{2}\\ln\\frac{\\rho}{b} \\right\\rbrack \\bullet \\frac{1}{2\\rho^{2}}\\frac{d}{dt}\\rho^{2}$$\n$$= \\left\\lbrack \\frac{|e|V}{\\ln\\frac{a}{b}} - \\frac{|e|^{2}}{m}\\left( \\frac{\\mu_{0}I}{2\\pi} \\right)^{2}\\ln\\frac{\\rho}{b} \\right\\rbrack\\frac{d}{dt}\\ln\\frac{\\rho}{b}$$\nIntegrating and using $v^{2} = 0,$ at $\\rho = b,$ we get,\n$$\\frac{1}{2}mv^{2} = \\left\\lbrack \\frac{|e|V}{\\ln\\frac{a}{b}}\\ln\\frac{\\rho}{b} - \\frac{1}{2m}|e|^{2}\\left( \\frac{\\mu_{0}I}{2\\pi} \\right)^{2}\\left( \\ln\\frac{\\rho}{b} \\right) \\right\\rbrack$$\nThe RHS must be positive, for all $a > \\rho > b$. The condition for\nthis is,\n$$V \\geq \\frac{1}{2}\\frac{|e|}{m}\\left( \\frac{\\mu_{0}I}{2\\pi} \\right)^{2}\\ln\\frac{a}{b}$$", "solution_images": ["images/image22.png"], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q10", "question": "A standing wave $y = a\\sin{kx}.\\cos{\\omega t}$ is\nmaintained in a homogeneous rod with cross-sectional area $S$ and\ndensity $\\rho$. Find the total mechanical energy confined between the\nsections corresponding to the adjacent displacement nodes.", "question_images": [], "answer": "$E = T + U = \\frac{\\pi\\rho\\omega^{2}a^{2}S}{4k}$", "solution": "We have a standing wave given by the equation\n$$y = a\\sin{kx}\\cos{\\omega t}$$\nSo, $\\frac{\\partial\\xi}{\\partial t} = - a\\omega\\sin kx\\sin\\omega t$\n(1)\nand $\\frac{\\partial y}{\\partial t} = ak\\cos kx\\cos\\omega t$ (2)\nThe kinetic energy confined in an element of length $dx$ of the rod\n$$dT = \\frac{1}{2}(\\rho Sdx)\\left( \\frac{\\partial y}{\\partial t} \\right)^{2} = \\frac{1}{2}\\rho Sa^{2}\\omega^{2}\\sin^{2}kx\\ dx$$\nSo total kinetic energy confined into rod\n$$T = \\int_{}^{}{dT} = \\frac{1}{2}\\rho Sa^{2}\\omega^{2}\\sin^{2}{\\omega t}\\int_{0}^{\\lambda/2}{\\sin^{2}\\frac{2\\pi}{\\lambda}x}dx$$\nOr, $T = \\frac{\\pi Sa^{2}\\omega^{2}\\rho\\sin^{2}{\\omega t}}{4k}$ (3)\nThe potential energy in the above rod element\n$dU = \\int_{}^{}{\\partial U} = - \\int_{0}^{y}{F_{y}dy}$, where\n$F_{y} = (\\rho Sdx)\\frac{\\partial^{2}y}{\\partial t^{2}}$\nOr, $F_{y} = - (\\rho Sdx)\\omega^{2}y$\nSo, $dU = \\omega^{2}\\rho Sdx\\int_{0}^{y}{ydy}$\nOr,\n$dU = \\frac{\\rho\\omega^{2}Sy^{2}}{2}dx = \\frac{\\rho\\omega^{2}Sa^{2}\\cos^{2}\\ \\omega t\\sin^{2}{kx}dx}{2}$\nThus the total potential energy stored in the rod $U = \\int_{}^{}{dU}$\nOr,\n$U = \\rho\\omega^{2}Sa^{2}\\cos^{2}{\\omega t}\\int_{0}^{\\lambda/2}{\\sin^{2}{\\frac{2\\pi}{\\lambda}x}}dx$\nSo, $U = \\frac{\\pi\\rho Sa^{2}\\omega^{2}\\cos^{2}{\\omega t}}{4k}$\nTo find the potential energy stored in the rod element we may adopt an\neasier way. We know that the potential energy density confined in a\nrod under elastic force equals:\n$$U_{D} = \\frac{1}{2}(stress \\times strain) = \\frac{1}{2}\\sigma\\varepsilon = \\frac{1}{2}Y\\varepsilon^{2}$$\n$$= \\frac{1}{2}\\rho v^{2}\\varepsilon^{2} = \\frac{1}{2}\\frac{\\rho\\omega^{2}}{k^{2}}\\varepsilon^{2}$$\n$$= \\frac{1}{2}\\frac{\\rho\\omega^{2}}{k^{2}}\\left( \\frac{\\partial y}{\\partial x} \\right)^{2} = \\frac{1}{2}\\rho a^{2}\\omega^{2}\\cos^{2}{\\omega t}\\cos^{2}{kx}$$\nHence the total potential energy stored in the rod\n$$U = \\int_{}^{}U_{D}dV = \\int_{0}^{\\lambda/2}\\frac{1}{2}\\rho a^{2}\\omega^{2}\\cos^{2}{\\omega t}\\cos^{2}{kx}Sdx$$\n$$= \\frac{\\pi\\rho Sa^{2}\\omega^{2}\\cos^{2}{\\omega t}}{4k}$$\nHence the sought mechanical energy confined in the rod between the two\nadjacent nodes\n$$E = T + U = \\frac{\\pi\\rho\\omega^{2}a^{2}S}{4k}$$", "solution_images": [], "subject": "Physics", "topic": "Elasticity", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q30", "question": "A parallel-plate capacity whose electrodes are shaped\nas round discs is charged slowly. Demonstrate that the flux of the\nPoynting vector across the capacitor's lateral surface is equal to the\nincrement of the capacitor's energy per unit time. The dissipation of\nfield at the edge is to be neglected in calculations.", "question_images": [], "answer": "$2\\pi Rd \\times S = \\frac{Q\\dot{Q}d}{\\pi R^{2}\\varepsilon_{0}}$", "solution": "If the charge on the capacitor is $Q$, the rate of\nincrease of the capacitor's energy\n$$= \\frac{d}{dt}\\left( \\frac{1}{2}\\frac{Q^{2}}{C} \\right) = \\frac{Q\\dot{Q}}{C} = \\frac{d}{\\varepsilon_{0}\\pi R^{2}}Q\\dot{Q}$$\nNow electric field between the plates (inside it) is,\n$E = \\frac{Q}{\\pi R^{2}\\varepsilon_{0}}$.\nSo displacement current\n$= \\frac{\\partial D}{\\partial t} = \\frac{\\dot{Q}}{\\pi R^{2}}$\nThis will lead to a magnetic field, (circuital) inside the plates. At\na radial distance $r$\n$2\\pi rH_{\\theta}(r) = \\pi r^{2}\\frac{\\dot{Q}}{\\pi R^{2}}$ or\n$H_{\\theta} = \\frac{\\dot{Q}r}{2\\pi R^{2}}$\nHence $H_{\\theta}(R) = \\frac{Q}{2\\pi R}$ at the edge.\nThus inward Poynting vector\n$= S = \\frac{Q}{2\\pi R} \\times \\frac{Q}{\\pi R^{2}\\varepsilon_{0}}$\nTotal flow\n$= 2\\pi Rd \\times S = \\frac{Q\\dot{Q}d}{\\pi R^{2}\\varepsilon_{0}}$\nProved", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q17", "question": "An inhomogeneous poorly conducting medium fills up the\nspace between plates 1 and 2 of a parallel-plate capacitor. Its\npermittivity and resistivity vary from values $\\varepsilon_{1},\\rho_{1}$\nat plate 1 to values $\\varepsilon_{2},\\rho_{2}$ at plate 2. A dc voltage\nis applied to the capacitor through which a steady current I flows from\nplate 1 to plate 2. Find the total extraneous charge in the given\nmedium.", "question_images": [], "answer": "$\\varepsilon_{0}I\\left( \\varepsilon_{2}\\rho_{2} - \\varepsilon_{1}\\rho_{1} \\right)$", "solution": "By current conservation\n$$\\frac{E(x)}{\\rho(x)} = \\frac{E(x) + dE(x)}{\\rho(x) + d\\rho(x)} = \\frac{dE(x)}{d\\rho(x)}$$\nThis has the solution\n$$E(x) = C\\rho(x) = \\frac{I\\rho(x)}{A}$$\n[IMAGE] Hence charge induced in the slice per unit area\n$$d\\sigma = \\varepsilon_{0}\\frac{I}{A}\\left\\lbrack \\left\\{ \\varepsilon(x) + d\\varepsilon(x) \\right\\}\\left\\{ \\rho(x) + d\\rho(x) \\right\\} - \\varepsilon(x)\\rho(x) \\right\\rbrack = \\varepsilon_{0}\\frac{I}{A}d\\left\\lbrack \\varepsilon(x)\\rho(x) \\right\\rbrack$$\nThus,\n$dQ = \\varepsilon_{0}Id\\left\\lbrack \\varepsilon(x)\\rho(x) \\right\\rbrack$\nHence total charge induced, is by integration,\n$$Q = \\varepsilon_{0}I\\left( \\varepsilon_{2}\\rho_{2} - \\varepsilon_{1}\\rho_{1} \\right)$$", "solution_images": ["images/image14.png"], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q32", "question": "A man standing on the edge of a swimming pool looks at\na stone lying on the bottom. The depth of the swimming pool is equal to\n$h$. At what distance from the surface of water is the image of the\nstone formed if the line of vision makes an angle $\\theta$ with the\nnormal to the surface?", "question_images": [], "answer": "$\\frac{n^{2}h\\cos^{3}\\theta}{\\left( n^{2} - \\sin^{2}\\theta \\right)^{3/2}}$", "solution": "From the Fig.\n[IMAGE] $$\\sin{d\\alpha} = \\frac{MP}{OM} = \\frac{MN\\cos\\alpha}{h\\sec(\\alpha + d\\alpha)}$$\nAs $d\\alpha$ is very small, so\n$d\\alpha = \\frac{MN\\cos\\alpha}{h\\sec\\alpha} = \\frac{MN\\cos^{2}\\alpha}{h}$\n(1)\nSimilarly\n$d\\theta = \\frac{MN\\cos^{2}\\theta}{h'}$ (2)\nFrom Eqns (1) and (2)\n$\\frac{d\\alpha}{d\\theta} = \\frac{h^{'}\\cos^{2}\\alpha}{h\\cos^{2}\\theta}$\nor,\n$h^{'} = \\frac{h\\cos^{2}\\theta}{\\cos^{2}\\alpha}\\frac{d\\alpha}{d\\theta}$\n(3)\nFrom the law of refraction\n$$n\\sin\\alpha = \\sin\\theta$$\n$\\sin\\alpha = \\frac{\\sin\\theta}{n},$ so,\n$\\cos\\alpha = \\sqrt{\\frac{n^{2} - \\sin^{2}\\theta}{n^{2}}}$\nDifferentiating Eqn. (A)\n$n\\cos\\alpha d\\alpha = \\cos\\theta d\\theta$ or,\n$\\frac{d\\alpha}{d\\theta} = \\frac{\\cos\\theta}{n\\cos\\alpha}$ (4)\nUsing (4) and (3), we get\n$h^{'} = \\frac{h\\cos^{3}\\theta}{n\\cos^{3}\\alpha}$ (5)\nHence\n$h^{'} = \\frac{h\\cos^{3}\\theta}{n\\left( \\frac{n^{2} - \\sin^{2}\\theta}{n^{2}} \\right)^{3/2}} = \\frac{n^{2}h\\cos^{3}\\theta}{\\left( n^{2} - \\sin^{2}\\theta \\right)^{3/2}}$\n\\[Using Eqn. (B)\\]", "solution_images": ["images/image27.png"], "subject": "Physics", "topic": "Optics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q22", "question": "A metal rod of mass $m$ can rotate abut a horizontal\naxis $O$, sliding along a circular conductor of radius. The arrangement\nis located in a uniform magnetic field of induction $B$ directed\nperpendicular to the ring plane. The axis and the ring are connected to\nan emf source to form a circuit of resistance $R$. Neglecting the\nfriction, circuit inductance, and ring resistance, find the law\naccording to which the source emf must vary to make the rod rotate with\na constant angular velocity $\\omega$.\n[IMAGE]", "question_images": ["images/image19.png"], "answer": "$\\frac{1}{2aB}\\left( a^{3}B^{2}\\omega + 2mgR\\sin\\omega t \\right)$", "solution": "As the rod rotates, an emf.\n$$\\frac{d}{dt}\\frac{1}{2}a^{2}\\theta \\bullet B = \\frac{1}{2}a^{2}B\\omega$$\nis induced in it. The net current in the conductor is then\n$\\frac{\\varepsilon(t) - \\frac{1}{2}a^{2}B\\omega}{R}$\nA magnetic force will then act on the conductor of magnitude BI per\nunit length. Its direction will be normal to B and rod and its torque\nwill be\n$$\\int_{0}^{a}{\\left( \\frac{\\varepsilon(t) - \\frac{1}{2}a^{2}B\\omega}{R} \\right)dx}Bx$$\nObviously both magnetic and mechanical torque acting on the C.M. of\nthe rod must be equal but opposite in sense. Then\nfor equilibrium at constant $\\omega$\n$$\\frac{\\varepsilon(t) - \\frac{1}{2}a^{2}B\\omega}{R} \\bullet \\frac{Ba^{2}}{2} = \\frac{1}{2}mga\\sin{\\omega t}$$\nOr,\n$\\varepsilon(t) = \\frac{1}{2}a^{2}B\\omega + \\frac{mgR}{aB}\\sin{\\omega t} = \\frac{1}{2aB}\\left( a^{3}B^{2}\\omega + 2mgR\\sin\\omega t \\right)$\n(The answer given in the book is incorrect dimensionally.)", "solution_images": [], "subject": "Physics", "topic": "Waves and Oscillations", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q13", "question": "A parallel-plate capacitor was lowered into water in a\nhorizontal position, with water filling up the gap between the plates\n$d = 1.0\\ mm$ wide. Then a constant voltage $V = 500\\ V$ was applied to\nthe capacitor. Find the water pressure increment in the gap.", "question_images": [], "answer": "$\\mathrm{\\Delta}p = 7.17\\ kPa = 0.07\\ atm$", "solution": "When the capacitor which is immersed in water is connected\nto a constant voltage source, it gets charged. Suppose $\\sigma_{0}$ is\nthe free charge density on the condenser plates. Because water is a\ndielectric, bound charges also appear in it. Let $\\sigma'$ be the\nsurface density of bound charges. (Because of homogeneity of the medium\nand uniformity of the field when we ignore edge effects no volume\ndensity of bound charges exists.) The electric field due to free charges\nonly $\\frac{\\sigma_{0}}{\\varepsilon_{0}}$; that due to bound charges is\n$\\frac{\\sigma'}{\\varepsilon_{0}}$ and the total electric field a\n$\\frac{\\sigma_{0}}{\\varepsilon\\varepsilon_{0}}$. Recalling that the sign\nof bound charges is opposite of the free charges, we have\n$\\frac{\\sigma_{0}}{\\varepsilon\\varepsilon_{0}} = \\frac{\\sigma_{0}}{\\varepsilon_{0}} - \\frac{\\sigma^{'}}{\\varepsilon_{0}}$\nor,\n$\\sigma^{'} = \\left( \\frac{\\varepsilon - 1}{\\varepsilon} \\right)\\sigma_{0}$\nBecause of the field that exists due to the free charges (not the\ntotal field; the field due to the bound charges must be excluded for\nthis purpose as they only give rise to self energy effects), there is\na force attracting the bound charges to the near by plates. This force\nis\n$\\frac{1}{2}\\sigma^{'}\\frac{\\sigma_{0}}{\\varepsilon_{0}} = \\frac{(\\varepsilon - 1)\\sigma_{0}^{2}}{2\\varepsilon\\varepsilon_{0}}$\nper unit area.\nThe factor $\\frac{1}{2}$ needs an explanation. Normally the force on a\ntest charge is qE in an electric field $E$. But if the charge itself\nis produced by the electric filed then the force must be constructed\nbit by bit and is\n$$F = \\int_{0}^{E}{q\\left( E^{'} \\right)}dE'$$\nif $q(E') \\propto E'$ then we get\n$$F = \\frac{1}{2}q(E)E$$\nThis factor of $\\frac{1}{2}$ is well known. For example the energy of\na dipole of moment $\\overrightarrow{p}$ in an electric field\n${\\overrightarrow{E}}_{0}$ is\n$- \\overrightarrow{p} \\bullet {\\overrightarrow{E}}_{0}$ while the\nenergy per unit volume of a linear dielectric in an electric field is\n$- \\frac{1}{2}\\overrightarrow{P} \\bullet {\\overrightarrow{E}}_{0}$\nwhere $\\overrightarrow{P}$ is the polarization vector (i.e. dipole\nmoment per unit volume). Now the force per unit area manifests itself\nas excess pressure of the liquid.\nNoting that\n$\\frac{V}{d} = \\frac{\\sigma_{0}}{\\varepsilon\\varepsilon_{0}}$\nWe get\n$\\mathrm{\\Delta}p = \\frac{\\varepsilon_{0}\\varepsilon(\\varepsilon - 1)V^{2}}{2d^{2}}$\nSubstitution, using $\\varepsilon = 81$ for water, gives\n$\\mathrm{\\Delta}p = 7.17\\ kPa = 0.07\\ atm$.", "solution_images": [], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q4", "question": "A fluid with viscosity $\\eta$ fills the space between\ntwo long co-axial cylinders of radii $R_{1}$ and $R_{2}$, with\n$R_{1} < R_{2}$. The inner cylinder is stationary while the outer one is\nrotated with a constant angular velocity $\\omega_{2}$. The fluid flow is\nlaminar. Taking into account that the friction force acting on a unit\narea of a cylindrical surface of radius $r$ is defined by the formula\n$\\sigma = \\eta r(\\partial\\omega/\\partial r)$, find :\n\\(a\\) the angular velocity of the rotating fluid as a function of radius\n$r$;\n\\(b\\) the moment of the friction forces acting on a unit length of the\nouter cylinder.", "question_images": [], "answer": "(a)\n$\\omega = \\omega_{2}\\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}\\left\\lbrack \\frac{1}{R_{1}^{2}} - \\frac{1}{r^{2}} \\right\\rbrack$\n\\(b\\)\n$N_{1} = \\frac{N}{l} = 4\\pi\\eta\\omega_{2}\\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}$", "solution": "(a) Let us consider an elemental cylinder of radius $r$\nand thickness $dr$ then from Newton's formula\n$$F = 2\\pi rl\\eta r\\frac{d\\omega}{dr} = 2\\pi l\\eta r^{2}\\frac{d\\omega}{dr}$$\nand moment of this force acting on the element,\n$$N = 2\\pi r^{2}l\\eta\\frac{d\\omega}{dr}r = 2\\pi r^{3}l\\eta\\frac{d\\omega}{dr}$$\nOr, $2\\pi l\\eta d\\omega = N\\frac{dr}{r^{3}}$ (2)\n[IMAGE] As in the previous problem $N$ is constant when conditions are steady\nIntegrating,\n$2\\pi l\\eta\\int_{0}^{\\omega}{d\\omega} = N\\int_{R_{1}}^{r}\\frac{dr}{r^{3}}$\nOr,\n$2\\pi l\\eta\\omega = \\frac{N}{2}\\left\\lbrack \\frac{1}{R_{1}^{2}} - \\frac{1}{r^{2}} \\right\\rbrack$\n(3)\nPutting $r = R_{2},\\omega = \\omega_{2}$, we get\n$2\\pi l\\eta\\omega_{2} = \\frac{N}{2}\\left\\lbrack \\frac{1}{R_{1}^{2}} - \\frac{1}{R_{2}^{2}} \\right\\rbrack$\n(4)\nFrom (3) and (4),\n$$\\omega = \\omega_{2}\\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}\\left\\lbrack \\frac{1}{R_{1}^{2}} - \\frac{1}{r^{2}} \\right\\rbrack$$\n\\(b\\) From Eq. (4),\n$N_{1} = \\frac{N}{l} = 4\\pi\\eta\\omega_{2}\\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}$", "solution_images": ["images/image6.png"], "subject": "Physics", "topic": "Fluid Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q40", "question": "At what minimum kinetic energy must a hydrogen atom\nmove for its inelastic head on collision with another, stationary,\nhydrogen atom to make one of them capable of emitting a photon ? Both\natoms are supposed to be in the ground state prior to the collision.", "question_images": [], "answer": "$20.4\\ eV$", "solution": "Photon can be emitted in $H - H$ collision only if one of\nthe $H$ is excited to an $n = 2$ state which then dexcites to $n = 1$\nstate by emitting a photon. Let $v_{1}$ and $v_{2}$ be the velocities of\nthe two Hydrogen atoms after the collision and $M$ their masses. Then,\nenergy momentum conservation\n$$Mv_{1} + Mv_{2} = \\sqrt{2MT}$$\n(in the frame of the stationary $H$ atom)\n$$\\frac{1}{2}Mv_{1}^{2} + \\frac{1}{2}Mv_{2}^{2} + \\frac{3}{4}hR = T$$\n$\\frac{3}{4}hR = hR\\left( 1 - \\frac{1}{4} \\right)$ is the excitation\nenergy of the $n = 2$ state from the ground state.\nEliminating $v_{2}$\n$\\frac{1}{2}M\\left\\{ v_{1}^{2} + \\left( \\sqrt{\\frac{2T}{M}} - v_{1} \\right)^{2} \\right\\} + \\frac{3}{4}hR = T$\nOr\n$\\frac{1}{2}M\\left\\{ 2v_{1}^{2} - 2\\sqrt{\\frac{2T}{M}}v_{1} + \\frac{2T}{M} \\right\\} + \\frac{3}{4}hR = T$\n$M\\left\\{ \\left( v_{1} - \\frac{1}{2}\\sqrt{\\frac{2T}{M}} \\right)^{2} \\right\\} + \\frac{1}{2}T + \\frac{3}{4}hR = T$\nOr\n$M\\left\\{ \\left( v_{1} - \\frac{1}{2}\\sqrt{\\frac{2T}{M}} \\right) \\right\\} + \\frac{3}{4}hR = \\frac{1}{2}T$\nFor minimum $T$, the square on the left should vanish. Thus\n$T = \\frac{3}{2}hR = 20.4\\ eV$", "solution_images": [], "subject": "Physics", "topic": "Modern Physics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q39", "question": "An alpha particle with kinetic energy $T = 0.50\\ MeV$\nis deflected through an angle of $\\theta = 90{^\\circ}$ by the Coulomb\nfield of a stationary Hg nucleus. Find:\nAll the formulas in this Part are given the Gaussian system of units.\n\\(a\\) the least curvature radius of its trajectory;\n\\(b\\) the minimum approach distance between the particle and the\nnucleus.", "question_images": [], "answer": "(a) $0.231\\ pm$\n\\(b\\) $0.557\\ pm$", "solution": "We shall ignore the recoil of Hg nucleus.\n\\(a\\) Let $A$ be the point of closest approach to the centre\n$C,AC = r_{\\min}$. At $A$ the motion is instantaneously circular\nbecause the radial velocity vanishes. Then if $v_{0}$ is the speed of\nthe particle at $A,$ the following equations hold\n$\\Gamma = \\frac{1}{2}mv_{0}^{2} + \\frac{Z_{1}Z_{2}e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)r_{\\min}}$\n(1)\n$mv_{0}r_{\\min} = \\sqrt{2mT}b$ (2)\n$\\frac{mv_{0}^{2}}{\\rho_{\\min}} = \\frac{Z_{1}Z_{2}e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)r_{\\min}^{2}}$\n(3)\n[IMAGE] (This is Newton's law. Here $\\rho = \\rho_{\\min}$ is the radius of\ncurvature of the path at A and $\\rho$ is minimum at $A$ by symmetry.)\nFinally we have Eqn. (6.1 a) in the form\n$b = \\frac{Z_{1}\\ Z_{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)2T}\\cot\\frac{\\theta}{2}$\n(4)\nFrom (2) and (3)\n$\\frac{2Tb^{2}}{\\rho_{\\min}} = \\frac{Z_{1}\\ Z_{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)}$\nOr\n$\\rho_{\\min} = \\frac{Z_{1}\\ Z^{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)2T}\\cot^{2}\\frac{\\theta}{2},$\nwith $z_{1} = 2,z_{2} = 80$ we get\n$\\rho_{\\min} = 0.231\\ pm$\n\\(b\\) From (2) and (4) we write\n$r_{\\min} = \\frac{z_{1}\\ z_{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)\\sqrt{2m\\ T}}\\frac{\\cot{\\theta/2}}{v_{0}},$\nSubstituting in (1)\n$T = \\frac{1}{2}mv_{0}^{2} + \\sqrt{2\\ m\\ T}v_{0}\\tan{\\theta/2}$\nSolving for $v_{0}$ we get\n$v_{0} = \\sqrt{\\frac{2T}{m}}\\left( \\sec\\frac{\\theta}{2} - \\tan\\frac{\\theta}{2} \\right)$\nThen\n$r_{\\min} = \\frac{z_{1}\\ z_{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)\\ 2T}\\frac{\\cot\\frac{\\theta}{2}}{\\sec\\frac{\\theta}{2} - \\tan\\frac{\\theta}{2}}$\n$= \\frac{z_{1}\\ z_{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)2T}\\cot\\frac{\\theta}{2}\\left( \\sec\\frac{\\theta}{2} + \\tan\\frac{\\theta}{2} \\right)$\n$= \\frac{z_{1}\\ z_{2}\\ e^{2}}{\\left( 4\\pi\\varepsilon_{0} \\right)2T}\\left( 1 + {cosec}\\frac{\\theta}{2} \\right) = 0.557\\ pm$", "solution_images": ["images/image36.png"], "subject": "Physics", "topic": "Modern Physics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q11", "question": "Suppose the surface charge density over a sphere of\nradius $R$ depends on a polar angle $\\theta$ as\n$\\sigma = \\sigma_{0} = \\cos\\theta$, where $\\sigma_{0}$ is a positive\nconstant. Show that such a charge distribution can be represented as a\nresult of a small relative shift of two uniformly charged balls of\nradius $R$ whose charges are equal in magnitude and opposite in sign.\nResorting to this representation, find the electric field strength\nvector inside the given sphere.\n$$= - \\frac{\\rho\\overrightarrow{a}}{3\\varepsilon_{0}} = \\frac{\\sigma_{0}}{3\\varepsilon_{0}}\\overrightarrow{k}$$", "question_images": [], "answer": "$\\overrightarrow{E} = - \\frac{\\rho\\overrightarrow{a}}{3\\varepsilon_{0}} = \\frac{\\sigma_{0}}{3\\varepsilon_{0}}\\overrightarrow{k}$", "solution": "We start from two charged spherical balls each of radius\n$R$ with equal and opposite charge densities $+ \\rho$ and $- \\rho$. The\ncentre of the balls are at $+ \\frac{\\overrightarrow{a}}{2}$ and\n$- \\frac{\\overrightarrow{a}}{2}$ respectively so the equation of their\nsurface are\n$\\left| \\overrightarrow{r} - \\frac{\\overrightarrow{a}}{2} \\right| = R$\nor $r - \\frac{a}{2}\\cos{\\theta \\simeq R}$ and\n$r + \\frac{a}{2}\\cos\\theta \\simeq R$, considering a to be small. The\ndistance between the two surfaces in the radial direction at angle\n$\\theta$ is $\\left| a\\cos\\theta \\right|$ and does not depend on the\nazimuthal angle. It is seen from the diagram that the surface of the\nsphere has in effect a surface density $\\sigma = \\sigma_{0}\\cos\\theta$\nwhen\n$$\\sigma_{0} = \\rho a$$\nInside any uniformly charged spherical ball, the field is radial and\nhas the magnitude given by Gauss's theorm\n$$4\\pi r^{2}E = \\frac{4\\pi}{3}r^{3}\\rho/\\varepsilon_{0}$$\nOr $E = \\frac{\\rho r}{3\\varepsilon_{0}}$\n[IMAGE] In vector notation, using the fact the $V$ must be measured from the\ncentre of the ball, we get, for the present case\n$$\\overrightarrow{E} = \\frac{\\rho}{3\\varepsilon_{0}}\\left( \\overrightarrow{r} - \\frac{\\overrightarrow{a}}{2} \\right) - \\frac{\\rho}{3\\varepsilon_{0}}\\left( \\overrightarrow{r} + \\frac{\\overrightarrow{a}}{2} \\right)$$\n$$\\overrightarrow{E} = - \\frac{\\rho\\overrightarrow{a}}{3\\varepsilon_{0}} = \\frac{\\sigma_{0}}{3\\varepsilon_{0}}\\overrightarrow{k}$$\nWhen $\\overrightarrow{k}$ is the unit vector along the polar axis from\nwhich $\\theta$ is measured.", "solution_images": ["images/image8.png"], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q25", "question": "A thin wire ring of radius $a$ and resistance $r$ is\nlocated inside a long solenoid so that their axes coincide. The length\nof the solenoid is equal to $l$, its cross-sectional radius, to $b$. At\na certain moment the solenoid was connected to a source of constant\nvoltage $V$. The total resistance of the circuit is equal to $R$.\nAssuming the inductance of the ring to be negligible, find the maximum\nvalue of the radial force acting per unit length of the ring.", "question_images": [], "answer": "$\\frac{\\mu_{0}a^{2}V^{2}}{4r\\ Rlb^{2}}$", "solution": "The solenoid has an inductance,\n$$L = \\mu_{0}n^{2}\\pi b^{2}I,$$\nwhere $n =$ number of turns of the solenoid per unit length. When the\nsolenoid is connected to the source an e.m.f. is set up, which,\nbecause of the inductance and resistance, rises slowly, according to\nthe equation,\n$$RI + LI = V$$\nThis has the well known solution,\n$$I = \\frac{V}{R}\\left( 1 - e^{- t\\ R/L} \\right)$$\nCorresponding to this current, an e.m.f. is induced in the ring. Its\nmagnetic field $B = \\mu_{0}nI$ in the solenoid, produces a force per\nunit length, $\\frac{dF}{dl} = Bi = \\mu_{0}^{2}n^{2}\\pi a^{2}\\ II/r$\n$$= \\frac{\\mu_{0}^{2}\\pi a^{2}V^{2}}{r}\\left( \\frac{n^{2}}{RL} \\right)e^{- t\\ R/L}\\left( 1 - e^{- tR/L} \\right),$$\nacting on each segment of the ring. This force is zero initially and\nzero for large $t$. Its maximum value is for some finite $t$. The\nmaximum value of\n$e^{- tR/L}\\left( 1 - e^{- tR/L} \\right) = \\frac{1}{4} - \\left( \\frac{1}{2} - e^{- tR/L} \\right)^{2}$\nis $\\frac{1}{4}$.\nSo\n$\\frac{dF_{\\max}}{dl} = \\frac{\\mu_{0}^{2}\\pi a^{2}V^{2}}{r}\\frac{n^{2}}{4RL} = \\frac{\\mu_{0}a^{2}V^{2}}{4r\\ Rlb^{2}}$", "solution_images": [], "subject": "Physics", "topic": "Current Electricity", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q36", "question": "A plane light wave with wavelength\n$\\lambda = 0.70\\ \\mu m$ fails normally on the base of biprism made of\nglass $(n = 1.520)$ with refracting angle $\\theta = 5.0{^\\circ}$. Behind\nthe biprism Fig. there is a plane-parallel plate, with the space between\nthem filled up with benzene $\\left( n^{'} = 1.500 \\right)$. Find the\nwidth of a fringe on the screen $Sc$ placed behind this system.\n[IMAGE]", "question_images": ["images/image32.png"], "answer": "$0.2\\ mm$", "solution": "It will be assumed that the space between the biprism and\nthe glass plate filled with benzene constitutes complementary prisms as\nshown.\n[IMAGE] Then the two prisms being oppositely placed, the net deviation\nproduced by them is\n$$\\delta = (n - 1)\\theta - \\left( n^{'} - 1 \\right)\\theta = \\left( n - n^{'} \\right)\\theta$$\nHence as in the previous problem\n$$d = 2a\\delta = 2a\\theta\\left( n - n^{'} \\right)$$\nSo,\n$\\mathrm{\\Delta}x = \\frac{(a + b)\\lambda}{2a\\theta\\left( n - n^{'} \\right)}$\nFor plane incident wave we let $a \\rightarrow \\infty$\nSo, $\\mathrm{\\Delta}x = \\frac{\\lambda}{2\\theta(n - n')} = 0.2\\ mm$", "solution_images": ["images/image33.png"], "subject": "Physics", "topic": "Optics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q1", "question": "Find the velocity of the excited hydrogen atoms if the\nfirst line of the Lyman series is displaced by\n$\\mathrm{\\Delta}\\lambda = 0.20\\ nm$ when their radiation is observed at\nan angle $\\theta = 45{^\\circ}$ to their motion direction.", "question_images": [], "answer": "$7.0 \\times 10^{5}\\ m/s$", "solution": "Since\n$\\mathrm{\\Delta}\\lambda( = 0.20\\ nm) \\ll \\lambda( = 121\\ nm)$ of the\nfirst Lyman line of H atom, we need not worry about $v^{2}/c^{2}$\neffects. Then\n$$\\omega^{'} = \\frac{\\omega}{1 - \\beta\\cos\\theta},\\beta = \\frac{v}{c}$$\nHence\n$1 - \\beta\\cos\\theta = \\frac{\\omega}{\\omega^{'}} = \\frac{\\lambda^{'}}{\\lambda}$\nOr\n$\\beta\\cos\\theta = 1 - \\frac{\\lambda^{'}}{\\lambda} = \\frac{\\mathrm{\\Delta}\\lambda}{\\lambda}$\nBut $\\omega = \\frac{3}{4}R$ so\n$\\lambda = \\frac{2\\pi c}{\\omega} = \\frac{8\\pi c}{3R}$\nHence $v = c\\beta = \\frac{3R\\mathrm{\\Delta}\\lambda}{8\\pi\\cos\\theta}$\nSubstitution gives $\\left( \\cos\\theta = \\frac{1}{\\sqrt{2}} \\right)$\n$$v = 7.0 \\times 10^{5}\\ m/s$$", "solution_images": [], "subject": "Physics", "topic": "Modern Physics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q44", "question": "A particle moves along an are of a circle of radius $R$\naccording to the law $l = a\\sin{\\omega t}$, where $l$ is the\ndisplacement from the initial position measured along the arc, and $a$\nand $\\omega$ are constants. Assuming $R = 1.00\\ m,a = 0.80\\ m,$ and\n$\\omega = 2.00\\ rad/s,$ find:\n\\(a\\) the magnitude of the total acceleration of the particle at the\npoints $l = 0$ and $l = \\pm a;$\n\\(b\\) the minimum value of the total acceleration $w_{mn}$ and the\ncorresponding displacement $l_{m}$.", "question_images": [], "answer": "$\\left| w_{t} \\right| = a\\omega^{2}$", "solution": "From the equation\n$$l = a\\sin\\omega t$$\n$$\\frac{dl}{dt} = v = a\\omega\\cos\\omega t$$\nSo,\n$$w_{t} = \\frac{dv}{dt} = - a\\omega^{2}\\sin\\omega t\\quad(1)$$\nand\n$$w_{n} = \\frac{v^{2}}{R} = \\frac{a^{2}\\omega^{2}\\cos^{2}\\omega t}{R}\\quad(2)$$\n\\(a\\) At the point $l = 0$, $\\sin\\omega t = 0$ and\n$\\cos\\omega t = \\pm 1$ so, $\\omega t = 0,\\pi$ etc.\nHence\n$$w = w_{n} = \\frac{a^{2}\\omega^{2}}{R}$$\nSimilarly at $l = \\pm a$, $\\sin\\omega t = \\pm 1$ and\n$\\cos\\omega t = 0$, so, $w_{n} = 0$\nHence\n$$w = \\left| w_{t} \\right| = a\\omega^{2}$$", "solution_images": [], "subject": "Physics", "topic": "Waves and Oscillations", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q19", "question": "Calculate the magnetic moment of a thin wire with a\ncurrent $I = 0.8\\ A,$ wound tightly on half a tore Fig. The diameter of\nthe cross-section of the tore is equal to $d = 5.0\\ cm$, the number of\nturns is $N = 500.$\n[IMAGE]", "question_images": ["images/image16.png"], "answer": "$\\frac{1}{2}Nd^{2}I$", "solution": "Take an element of length $rd\\ \\theta$ containing\n$\\frac{N}{\\pi r}.rd\\theta$ turns. Its magnetic moment is\n$$\\frac{N}{\\pi}d\\theta.\\frac{\\pi}{4}d^{2}I$$\nNormal to the plane of cross section. We resolve it along OA and OB.\nThe moment along $OA$ integrates to\n$$\\int_{0}^{\\pi}{\\frac{N}{4}d^{2}Id\\theta}\\cos\\theta = 0$$\nwhile that along $OB$ gives\n$$p_{m} = \\int_{0}^{\\pi}\\frac{Nd^{2}I}{4}\\sin\\theta d\\theta = \\frac{1}{2}Nd^{2}I$$\n[IMAGE]", "solution_images": ["images/image17.png"], "subject": "Physics", "topic": "Magnetism", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q34", "question": "A parallel beam of light falls from vacuum on a surface\nenclosing a medium with refractive index $n$ Fig. Find the shape of that\nsurface, $x(r)$, if the beam is brought into focus at the point $F$ at a\ndistance $f$ from the crest $O$. What is the maximum radius of a beam\nthat can still be focussed ?\n[IMAGE]", "question_images": ["images/image29.jpeg"], "answer": "$r_{\\max} = f\\sqrt{(n - 1)/(n + 1)}$", "solution": "All rays focusing at a point must have traversed the same\noptical path. Thus\n$x + n\\sqrt{r^{2} + (f - x)^{2}} = nf$ or\n$(n - f - x)^{2} = n^{2}r^{2} + n^{2}(f - x)^{2}$\nOr,\n$n^{2}r^{2} = (nf - x)^{2} - \\left\\lbrack n(f - x) \\right\\rbrack^{2} = (nf - x + nf - nx)(nf - x - nf + nx)$\n$= x(n - 1)\\left( 2nf - (n + 1)x \\right)$\n$= 2n(n - 1)fx - (n + 1)(n - 1)x^{2}$\nThus, $(n + 1)(n - 1)x^{2} - 2n(n - 1)fx + n^{2}r^{2} = 0$\nSo,\n$x = \\frac{n(n - 1)f \\pm \\sqrt{n^{2}(n - 1)^{2}f^{2} - n^{2}r^{2}(n + 1)(n - 1)}}{(n + 1)(n - 1)}$\n$= \\frac{nf}{n + 1}\\left\\lbrack 1 \\pm \\sqrt{1 - \\frac{n + 1}{n - 1}\\frac{r^{2}\\ }{f^{2}}} \\right\\rbrack$\nRay must move forward so $x < f,$ for + sign $x > f$ for small $r$, so\n--sign.\n(Also $x \\rightarrow 0$ as $r \\rightarrow 0$)\n($x > f$ means ray turning back in the direction of incidence.\n[IMAGE] Hence\n$x = \\frac{nf}{n + 1}\\left\\lbrack 1 - \\sqrt{1 - \\frac{n + 1}{n - 1}\\frac{r^{2}}{f^{2}}} \\right\\rbrack$\nFor the maximum value of $r$,\n$$\\sqrt{1 - \\frac{n + 1}{n - 1}\\frac{r^{2}}{f^{2}}} = 0$$\nbecause the expression under the radical sign must be non-negative,\nwhich gives the maximum value of $r$.\nHence form Eqn.(A), $r_{\\max} = f\\sqrt{(n - 1)/(n + 1)}$", "solution_images": ["images/image30.jpeg"], "subject": "Physics", "topic": "Optics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q49", "question": "Three identical discs $A,B,$ and $C$ Fig. rest on a\nsmooth horizontal plane. The disc $A$ is set in motion with velocity $v$\nafter which it experiences an elastic collision simultaneously with the\ndiscs B and C. The distance between the centres of the latter discs\nprior to the collision is $\\eta$ times greater than the diameter of each\ndisc. Find the velocity of the disc A after the collision. At what value\nof $\\eta$ will the disc $A$ recoil after the collision; stop; move on?\n[IMAGE]", "question_images": ["images/image49.png"], "answer": "$\\eta = \\sqrt{2}$", "solution": "From the symmetry of the problem, the velocity of the disc\n$A$ will be directed either in the initial direction or opposite to it\njust after the impact. Let the velocity of the disc $A$ after the\ncollision be $v^{'}$ and be directed towards right after the collision.\nIt is also clear from the symmetry of problem that the discs $B$ and $C$\nhave equal speed (say $v\"$) in the directions, shown. From the condition\nof the problem.\n$\\cos{\\theta = \\frac{\\eta\\frac{d}{2}}{d}} = \\frac{\\eta}{2}$ so,\n$\\sin\\theta = \\sqrt{4 - \\eta^{2}}/2$ (1)\nFor the three discs, system, from the conservation of linear momentum\nin the symmetry direction (towards right)\n$mv = 2mv\"\\ sin\\theta\\ + mv'$ or, $v = 2v\"\\ sin\\ \\theta + v'$ (2)\nFrom the definition of the coefficient of restitution, we have for the\ndiscs $A$ and $B$ (or $C$)\n$$e = \\frac{v\" - v'\\sin\\theta}{v\\sin{\\theta - 0}}$$\nBut $e = 1$, for perfectly elastic collision,\nSo, $v\\sin\\theta = v\" - v'\\ sin\\ \\theta$ (3)\n[IMAGE] From (2) and (3),\n$$v^{'} = \\frac{v\\left( 1 - 2\\sin^{2}\\theta \\right)}{\\left( 1 + 2\\sin^{2}\\theta \\right)}$$\n$= \\frac{v\\left( \\eta^{2} - 2 \\right)}{6 - \\eta^{2}}$ {using (1)}\nHence, we have,\n$$v^{'} = \\frac{v\\left( \\eta^{2} - 2 \\right)}{6 - \\eta^{2}}$$\nTherefore, the disc $A$ will recoil if $\\eta < \\sqrt{2}$ and stop if\n$\\eta = \\sqrt{2}$.\nNote: One can write the equations of momentum conservation along\nthe direction perpendicular to the initial direction of disc $A$ and\nthe conservation of kinetic energy instead of the equation of\nrestitution.", "solution_images": ["images/image50.png"], "subject": "Physics", "topic": "Elasticity", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q42", "question": "In the arrangement shown in Fig. the masses m of the\nbar and M of the wedge, as well as the wedge angle α, are known.\n[IMAGE] The masses of the pulley and the thread are negligible. The friction\nis absent. Find the acceleration of the wedge M.", "question_images": ["images/image39.png"], "answer": "$\\frac{mg\\sin\\alpha}{M + 2m\\left( 1 - \\cos\\alpha \\right)}$", "solution": "To analyse the kinematic relations between the bodies,\nsketch the force diagram of each body as shown in the figure.\nOn the basis of force diagram, it is obvious that the wedge M will\nmove towards right and the block will move down along the wedge. As\nthe length of the thread is constant, the distance travelled by the\nblock on the wedge must be equal to the distance travelled by the\nwedge on the floor. Hence $ds_{mM} = ds_{M}$. As\n${\\overrightarrow{v}}_{mM}$ and ${\\overrightarrow{v}}_{M}$ do not\nchange their directions and acceleration that's why\n${\\overrightarrow{w}}_{mM} \\uparrow \\uparrow {\\overrightarrow{v}}_{mM}$\nand\n${\\overrightarrow{w}}_{M} \\uparrow \\uparrow {\\overrightarrow{v}}_{M}$\nand $w_{mM} = w_{M} = w$ (say) and accordingly the diagram of\nkinematical dependence is shown in figure.\n[IMAGE] As\n${\\overrightarrow{w}}_{m} = {\\overrightarrow{w}}_{mM} + {\\overrightarrow{w}}_{M}$,\nso from triangle law of vector addition.\n$w_{m} = \\sqrt{w_{M}^{2} + w_{mM}^{2} - 2w_{mM}w_{M}\\cos\\alpha} = w\\sqrt{2\\left( 1 - \\cos\\alpha \\right)}$\n(1)\nFrom $F_{x} = mw_{x},$ (for the wedge),\n$T = T\\cos\\alpha + N\\sin\\alpha = Mw$ (2)\nFor the bar m let us fix $(x - y)$ coordinate system in the frame of\nground Newton's law in projection form along $x$ and $y$ axes Fig.\ngives\n$$mg\\sin\\alpha - T = mw_{m(x)} = m\\left\\lbrack w_{mM(x)} + w_{M(x)} \\right\\rbrack$$\n$$= m\\left\\lbrack w_{mM} + w_{M}\\cos(\\pi - \\alpha) \\right\\rbrack = mw(1 - \\cos\\alpha)$$\n$$mg\\cos{\\alpha - N} = mw_{m(y)} = m\\left\\lbrack w_{mM(y)} + w_{M(y)} \\right\\rbrack = m\\left\\lbrack 0 + w\\sin\\alpha \\right\\rbrack$$\nSolving the above Eqs. simultaneously, we get\n$$w = \\frac{mg\\sin\\alpha}{M + 2m\\left( 1 - \\cos\\alpha \\right)}$$\nNote: We can study the motion of the block $m$ in the frame of\nwedge also, alternately we may solve this problem using conservation\nof mechanical energy.", "solution_images": ["images/image40.png"], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q31", "question": "A vertical shaft of light from a projector forms a\nlight spot $S = 100\\ cm^{2}$ in area on the ceiling of a round room of\nradius $R = 2.0\\ m$. The illuminance of the spot is equal to\n$E = 1000\\ lx$. The reflection coefficient of the ceiling is equal to\n$\\rho = 0.80$. Find the maximum illuminance of the wall produced by the\nlight reflected from the ceiling. The reflection is assumed to obey\nLambert's law.", "question_images": [], "answer": "$0.21\\ lux$", "solution": "The illuminated area acts as a Lambert source of\nluminosity$M = \\pi L$ where\n$MS = \\rho ES =$ total reflected light\nThus, the luminance\n$$L = \\frac{\\rho E}{\\pi}$$\n[IMAGE] The equivalent luminous intensity in the direction making an angle\n$\\theta$ from the vertical is\n$$LS\\cos\\theta = \\frac{\\rho ES}{\\pi}\\cos\\theta$$\nand the illuminance at the point P is\n$$\\frac{\\rho ES}{\\pi}\\cos\\theta\\sin\\theta/R^{2}{cosec}^{2}\\theta = \\frac{\\rho ES}{\\pi R^{2}}\\cos\\theta\\sin^{3}\\theta$$\nThis is maximum when\n$$\\frac{d}{d\\theta}\\left( \\cos\\theta\\sin^{3}\\theta \\right) = - \\sin^{4}\\theta = 3\\sin^{2}\\theta\\cos^{2}\\theta = 0$$\nOr $\\tan^{2}\\theta = 3 \\Rightarrow \\tan\\theta = \\sqrt{3}$\nThen the maximum illuminance is\n$$\\frac{3\\sqrt{3}}{16\\pi}\\frac{\\rho ES}{R^{2}}$$\nThis illuminance is obtained at a distance $R\\cot\\theta = R/\\sqrt{3}$\nfrom the ceiling. Substitution gives the value\n$$0.21\\ lux$$", "solution_images": ["images/image26.png"], "subject": "Physics", "topic": "Optics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q29", "question": "A circuit consists of a capacitor with capacitance $C$\nand a coil of inductance $L$ connected in series, as well as a switch\nand a resistance equal to the critical value for this circuit. With the\nswitch disconnected, the capacitor was charged to a voltage $V_{0}$, and\nat the moment $t = 0$ the switch was closed. Find the current $I$ in the\ncircuit as a function of time $t$. What is $I_{\\max}$ equal to ?", "question_images": [], "answer": "$\\frac{V_{0}}{e}\\sqrt{\\frac{C}{L}}$", "solution": "$o = \\frac{q}{C} + L\\frac{dI}{dt} + RI,I = + \\frac{dq}{dt}$\n[IMAGE] For the critical case $R = 2\\sqrt{\\frac{L}{C}}$\nThus $LC\\ddot{q} + 2\\sqrt{LC}\\dot{q} + q = 0$\nLook for a solution with $q\\alpha e^{\\alpha t}$\n$$\\alpha = - \\frac{1}{\\sqrt{LC}}$$\nAn independent solution is $te^{\\alpha t}$. Thus\n$$q = (A + Bt)e^{- t/\\sqrt{LC}},$$\nAt $t = 0\\ q = CV_{0}$ thus $A = CV_{0}$\nAlso at $t = 0\\ \\dot{q} = I = 0$\n$$0 = B - A\\frac{1}{\\sqrt{l\\ C}} \\Rightarrow B = V_{0}\\sqrt{\\frac{C}{L}}$$\nThus finally\n$I = \\frac{dq}{dt} = V_{0}\\sqrt{\\frac{C}{L}}e^{- t/\\sqrt{LC}}$\n$$- \\frac{1}{\\sqrt{LC}}\\left( CV_{0} + V_{0}\\sqrt{\\frac{C}{L}}t \\right)e^{- t/\\sqrt{LC}}$$\n$$= - \\frac{V_{0}}{L}t\\ e^{- t/\\sqrt{LC}}$$\nThe current has been defined to increase the charge. Hence the minus\nsign.\nThe current is maximum when\n$$\\frac{dI}{dt} = - \\frac{V_{0}}{e}e^{- \\frac{t}{\\sqrt{LC}}}\\left( 1 - \\frac{t}{\\sqrt{LC}} \\right) = 0$$\nThis gives $t = \\sqrt{LC}$ and the magnitude of the maximum current is\n$$\\left| I_{\\max} \\right| = \\frac{V_{0}}{e}\\sqrt{\\frac{C}{L}}$$", "solution_images": ["images/image25.png"], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q21", "question": "A current $I$ flows in a long thin walled cylinder of\nradius $R$. What pressure do the walls of the cylinder experience?", "question_images": [], "answer": "$\\frac{\\mu_{0}I^{2}}{8\\pi^{2}R^{2}}$", "solution": "Let $t =$ thickness of the wall of the cylinder. Then,\n$J = I/2\\pi Rt$ along z axis. The magnetic field due to this at a\ndistance $r\\left( R - \\frac{t}{2} < r < R + \\frac{t}{2} \\right)$, is\ngiven by,\n[IMAGE] $$B_{\\varphi}(2\\pi r) = \\mu_{0}\\frac{I}{2\\pi Rt}\\pi\\left\\{ r^{2} - \\left( R - \\frac{t}{2} \\right)^{2} \\right\\}$$\nOr,\n$B_{\\varphi} = \\frac{\\mu_{0}I}{4\\pi Rrt}\\left\\{ r^{2} - (R - t/2)^{2} \\right\\}$\nNow,\n$\\overrightarrow{F} = \\int_{}^{}{\\overrightarrow{J} \\times \\overrightarrow{B}dV}$\nand\n$p = \\frac{F_{r}}{2\\pi RL} = \\frac{1}{2\\pi RL}\\int_{R - \\frac{t}{2}}^{R + \\frac{t}{2}}\\frac{\\mu_{0}I^{2}}{8\\pi^{2}R^{2}t^{2}r}\\left\\{ r^{2} - \\left( R - \\frac{t}{2} \\right)^{2} \\right\\} \\times 2\\pi rLdr$\n$$= \\frac{\\mu_{0}I^{2}}{8\\pi^{2}R^{3}t^{2}}\\int_{R - \\frac{t}{2}}^{R + \\frac{t}{2}}{\\left\\{ r^{2} - \\left( R - \\frac{t}{2} \\right)^{2} \\right\\} dr} = \\frac{\\mu_{0}I^{2}}{8\\pi^{2}R^{3}t^{2}}\\left\\lbrack \\frac{\\left( R + \\frac{t}{2} \\right)^{3} - \\left( R - \\frac{t}{2} \\right)^{3}}{3} - \\left( R - \\frac{t}{2} \\right)^{2}t \\right\\rbrack$$\n$$= \\frac{\\mu_{0}I^{2}}{8\\pi^{2}R^{3}t}\\left\\lbrack Rt + 0\\left( t^{2} \\right) \\right\\rbrack = \\frac{\\mu_{0}I^{2}}{8\\pi^{2}R^{2}}$$", "solution_images": ["images/image18.png"], "subject": "Physics", "topic": "Thermodynamics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q23", "question": "A square wire frame with side a and straight conductor\ncarrying a constant current $I$ are located in the same plane Fig.\n[IMAGE] The inductance and the resistance of the frame are equal to L and R\nrespectively. The frame was turned through $180{^\\circ}$ about the axis\n$OO'$ separated from the current-carrying conductor by a distance $b$.\nFind the electric charge having flown through the frame.", "question_images": ["images/image20.png"], "answer": "$\\frac{\\mu_{0}ai}{2\\pi R}\\ln\\frac{b + a}{b - a}$", "solution": "According to Ohm's law and Faraday's law of induction, the\ncurrent $i_{0}$ appearing in the frame, during its rotation, is\ndetermined by the formula,\n$$i_{0} = - \\frac{d\\Phi}{dt} = - \\frac{Ldi_{0}}{dt}$$\nHence, the required amount of electricity (charge) is,\n$$q = \\int_{}^{}{i_{0}dt} = - \\frac{1}{R}\\int_{}^{}\\left( d\\Phi + Ldi_{0} \\right) = - \\frac{1}{R}\\left( \\mathrm{\\Delta}\\Phi + L\\mathrm{\\Delta}i_{0} \\right)$$\n[IMAGE] Since the frame has been stopped after rotation, the current in it\nvanishes, and hence $\\mathrm{\\Delta}i_{0} = 0$. It remains for us to\nfind the increment of the flux $\\mathrm{\\Delta}\\Phi$ through the frame\n$\\left( \\mathrm{\\Delta}\\Phi = \\Phi_{2} - \\Phi_{1} \\right)$.\nLet us choose the normal $\\overrightarrow{n}$ to the plane of the\nframe, for instance, so that in the final position,\n$\\overrightarrow{n}$ is directed behind the plane of the figure (along\n$\\overrightarrow{B}$).\nThen it can be easily seen that in the final position, $\\Phi_{2} > 0$,\nwhile in the initial position, $\\Phi_{1} < 0$ (the normal is opposite\nto $\\overrightarrow{B}$), and $\\mathrm{\\Delta}\\Phi$ turns out to be\nsimply equal to the fulx through the surface bounded by the final and\ninitial positions of the frame :\n$$\\mathrm{\\Delta}\\Phi = \\Phi_{2} + \\left| \\Phi_{1} \\right| = \\int_{b - a}^{b + a}{Ba\\ dr},$$\nwhere $B$ is a function of r, whose form can be easily found with the\nhelp of the theorem of circulation. Finally omitting the minus sign,\nwe obtain,\n$$q = \\frac{\\mathrm{\\Delta}\\Phi}{R} = \\frac{\\mu_{0}ai}{2\\pi R}\\ln\\frac{b + a}{b - a}$$", "solution_images": ["images/image21.jpeg"], "subject": "Physics", "topic": "Current Electricity", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q43", "question": "A horizontal smooth rod $AB$ rotates with a constant\nangular velocity $\\omega = 2.00\\ rad/s$ about a vertical axis passing\nthrough its end $A$. A feely sliding sleeve of mass m = 0.50 kg moves\nalong the rod from the point A with the initial velocity\n$v_{0} = 1.00\\ m/s,$. Find the Coriolis force acting on the sleeve (in\nthe reference frame fixed to the rotating rod) at the moment when the\nsleeve is located at the distance $r = 50\\ cm$ from the rotation axis.", "question_images": [], "answer": "2.83 N", "solution": "The sleeve is free to slide along the rod $AB$. Thus only\nthe centrifugal force acts on it.\nThe equation is,\n$m\\dot{v} = m\\omega^{2}r$ where $v = \\frac{dr}{dt}$\nBut\n$\\dot{v} = v\\frac{dv}{dt} = \\frac{d}{dr}\\left( \\frac{1}{2}v^{2} \\right)$\nSo, $\\frac{1}{2}v^{2} = \\frac{1}{2}\\omega^{2}r^{2} +$ constant\nOr, $v^{2} = v_{0}^{2} + \\omega^{2}r^{2}$\n$v_{0}$ being the initial velocity when $r = 0$. The Coriolis force is\nthen,\n$$2m\\omega\\sqrt{v_{0}^{2} + \\omega^{2}r^{2}} = 2m\\omega^{2}r\\sqrt{1 + v_{0}^{2}/\\omega^{2}r^{2}}$$\n= 2.83 N on putting the values.", "solution_images": [], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q47", "question": "A horizontal plane supports stationary vertical\ncylinder of radius $R$ and a disc $A$ attached to the cylinder by a\nhorizontal thread $AB$ of length $l_{0}$ (Fig. top view). An initial\nvelocity $v_{0}$ is imparted to the disc as shown in the figure. How\nlong will it move along the plane until it strikes against the cylinder?\nThe friction is assumed to be absent.\n[IMAGE]", "question_images": ["images/image45.jpeg"], "answer": "$\\frac{l_{0}^{2}}{2Rv_{0}}$", "solution": "Since the tension is always perpendicular to the velocity\nvector, the work done by the tension force will be zero. Hence,\naccording to the work energy theorem, the kinetic energy or velocity of\nthe disc will remain constant during it's motion. Hence, the sought time\n$t = \\frac{s}{v_{0}}$, where $s$ is the total distance traversed by the\nsmall disc during it's motion.\nNow, at an arbitrary position Fig.\n[IMAGE] $$ds = \\left( l_{0} - R\\theta \\right)d\\theta,$$\nSo, $s = \\int_{0}^{l/R}\\left( l_{0} - R\\theta \\right)d\\theta,$\nOr,\n$s = \\frac{l_{0}^{2}}{R} - \\frac{Rl_{0}^{2}}{2R^{2}} = \\frac{l_{0}^{2}}{2R}$\nHence, the required time, $t = \\frac{l_{0}^{2}}{2Rv_{0}}$\nIt should be clearly understood that the only uncompensated force\nacting on the disc A in this case is the tension $T$, of the thread.\nIt is easy to see that there is no point here, relative to which the\nmoment of force $T$ is invariable in the process of motion. Hence\nconservation of angular momentum is not applicable here.", "solution_images": ["images/image46.jpeg"], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q37", "question": "There are two cavities Fig. with small holes of equal\ndiameters $d = 1.0\\ cm$ and perfectly reflecting outer surfaces. The\ndistance between the holes is $l = 10\\ cm$. A constant temperature\n$T_{1} = 1700\\ K$ is maintained in cavity 1. Calculate the steady-state\ntemperature inside cavity 2.\nInstruction: Take into account that a black body radiation obeys\nthe cosine emission law.\n[IMAGE]", "question_images": ["images/image34.png"], "answer": "$380\\ K$", "solution": "Taking account of cosine low of emissions we write for the\nenergy radiated per second by the hole in cavity \\# 1 as\n$$dI(\\Omega) = A\\cos\\theta d\\Omega$$\nwhere $A$ is an constant, $d\\Omega$ is an element of solid angle\naround some direction defined by the symbol $\\Omega$. Integrating over\nthe whole forward hemisphere we get\n$$I = A\\int_{0}^{\\pi/2}{\\cos\\theta 2\\pi\\sin\\theta d\\theta} = \\pi A$$\n[IMAGE] We find $A$ by equating this to the quantity\n$\\sigma T_{1}^{4} \\bullet \\frac{\\pi d^{2}}{4}\\sigma$ is\nStefan-Boltaman constant and $d$ is the diameter of the hole.\nThen $A = \\frac{1}{4}\\sigma d^{2}T_{1}^{4}$\nNow energy reaching 2 from 1 is\n$\\left( \\cos{\\theta \\approx 1} \\right)$\n$$\\frac{1}{4}\\sigma d^{2}T_{1}^{4} \\bullet \\mathrm{\\Delta}\\Omega$$\nwhere\n$\\mathrm{\\Delta}\\Omega = \\frac{\\left( \\pi d^{2}/4 \\right)}{l^{2}}$ is\nthe solid angle subtended by the hole of 2 at 1. {We are assuming\n$d \\ll l$ so $\\mathrm{\\Delta}\\Omega =$ area of hole / (distance)^2^}\nThis must equal $\\sigma T_{2}^{4}\\pi d^{2}/4$\nwhich is the energy emitted by 2. Thus equating\n$$\\frac{1}{4}\\sigma d^{2}T_{1}^{4}\\frac{\\pi d^{2}}{4l^{2}} = \\sigma T_{2}^{4}\\frac{\\pi d^{2}}{4}$$\nOr $T_{2} = T_{1}\\sqrt{\\frac{d}{2l}}$\nSubstituting we get $T_{2} = 0.380\\ kK = 380\\ K$", "solution_images": ["images/image35.png"], "subject": "Physics", "topic": "Heat and Thermal Radiation", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q33", "question": "A light ray composed of two monochromatic components\npasses through a trihedral prism with refracting angle\n$\\theta = 60{^\\circ}$. Find the angle $\\mathrm{\\Delta}\\alpha$ between\nthe components of the ray after its passage through the prism if their\nrespective indices of refraction are equal to 1.515 and 1.520. The prism\nis oriented to provide the least deflection angle.", "question_images": [], "answer": "$0.44$", "solution": "From the Caubhy's formula, and also experimentally the\nR.I. of a medium depends upon the wavelength of the mochromatic ray i.e.\n$n = f(\\lambda)$. In the case of least deviation of a monochromatic ray\nthe passage a prism, we have :\n$n\\sin{\\frac{\\theta}{2} = \\sin\\frac{\\alpha + \\theta}{2}}$ (1)\nThe above equation tales us that we have $n = n(\\alpha)$, so we may\nwrite\n$\\mathrm{\\Delta}n = \\frac{dn}{d\\alpha}\\mathrm{\\Delta}\\alpha$ (2)\nFrom Eqn. (1)\n$$dn\\sin\\frac{\\theta}{2} = \\frac{1}{2}\\cos\\frac{\\alpha + \\theta}{2}d\\alpha$$\nOr,\n$\\frac{dn}{d\\alpha} = \\frac{\\cos\\frac{\\alpha + \\theta}{2}}{2\\sin\\frac{\\theta}{2}}$\n(3)\n[IMAGE] From Eqns (2) and (3)\n$$\\mathrm{\\Delta}n = \\frac{\\cos\\frac{\\alpha + \\theta}{2}}{2\\sin\\frac{\\theta}{2}}\\mathrm{\\Delta}\\alpha$$\nOr,\n$\\mathrm{\\Delta}n = \\frac{\\sqrt{1 - \\sin^{2}\\left( \\frac{\\alpha + \\theta}{2} \\right)}}{2\\sin\\frac{\\theta}{2}}\\mathrm{\\Delta}\\alpha = \\frac{\\sqrt{1 - n^{2}\\sin^{2}\\frac{\\theta}{2}}}{2\\sin\\frac{\\theta}{2}}\\mathrm{\\Delta}\\alpha$\n(Using Eqn. 1.)\nThus\n$\\mathrm{\\Delta}\\alpha = \\frac{2\\sin\\frac{\\theta}{2}}{\\sqrt{1 - n^{2}}\\sin^{2}\\frac{\\theta}{2}}\\mathrm{\\Delta}n = 0.44$", "solution_images": ["images/image28.png"], "subject": "Physics", "topic": "Optics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q45", "question": "A rotating disc Fig. moves in the positive direction of\nthe $x$ axis. Find the equation $y(x)$ describing the position of the\ninstantaneous axis of rotation, if at the initial moment the axis $C$ of\nthe disc was located at the point O after which it moved\n\\(a\\) with a constant velocity $v$, while the disc started rotating\ncounterclockwise with a constant angular acceleration $\\beta$ (the\ninitial angular velocity is equal to zero);\n\\(b\\) with a constant acceleration $w$ (and the zero initial\nvelocity), while the disc rotates counterclockwise with a constant\nangular velocity $\\omega$.\n[IMAGE]", "question_images": ["images/image41.png"], "answer": "(a) $\\frac{v^{2}}{\\beta x}$\n\\(b\\) $\\frac{\\sqrt{2wx}}{\\omega}$", "solution": "Rotating disc moves along the $x$-axis, in plane motion in\n$x - y$ plane. Plane motion of a solid\ncan be imagined to be in pure rotation about a point (say $I$) at a\ncertain instant known as instantaneous centre of rotation. The\ninstantaneous axis whose positive sense is directed along\n$\\overrightarrow{\\omega}$ of the solid and which passes through the\npoint $I$, is known as instantaneous axis of rotation.\nTherefore the velocity vector of an arbitrary point $(P)$ of the solid\ncan be represented as:\n$${\\overrightarrow{v}}_{P} = \\overrightarrow{\\omega} \\times {\\overrightarrow{r}}_{PI}\\quad(1)$$\nOn the basis of Eq. (1) for the C. M. $(C)$ of the disc\n$${\\overrightarrow{v}}_{c} = \\overrightarrow{\\omega} \\times {\\overrightarrow{r}}_{cI}\\quad(2)$$\nAccording to the problem\n${\\overrightarrow{v}}_{c} \\uparrow \\uparrow \\overrightarrow{i}$ and\n$\\overrightarrow{\\omega} \\uparrow \\uparrow \\overrightarrow{k}$\ni.e. $\\overrightarrow{\\omega}\\bot x - y$ plane, so to satisfy the Eqn.\n(2)\n${\\overrightarrow{r}}_{cI}$ is directed along\n$\\left( - \\overrightarrow{j} \\right)$. Hence point $I$ is at a\ndistance $r_{CI} = y$, above the centre of the disc\nalong $y$-axis. Using all these facts in Eq. (2), we get\n$$v_{c} = \\omega y\\quad\\text{or}\\quad y = \\frac{v_{c}}{\\omega}\\quad(3)$$\n[IMAGE] Figure 1.51: Insert this diagram from the shared reference image ---\ninstantaneous centre of rotation for the rolling disc.\n(a) From the angular kinematical equation\n$$\\omega_{z} = \\omega_{0z} + \\beta_{z}t\\quad(4)$$\n$$\\omega = \\beta t.$$\nOn the other hand $x = vt$, (where $x$ is the $x$ coordinate of the\nC.M.)\nor,\n$$t = \\frac{x}{v}\\quad(5)$$\nFrom Eqs. (4) and (5),\n$$\\omega = \\frac{\\beta x}{v}$$\nUsing this value of $\\omega$ in Eq. (3) we get\n$$y = \\frac{v_{c}}{\\omega} = \\frac{v}{\\beta x/v} = \\frac{v^{2}}{\\beta x}$$\n(hyperbola)\n(b) As centre $C$ moves with constant acceleration $w$, with zero\ninitial velocity\nSo,\n$$x = \\frac{1}{2}wt^{2}\\quad\\text{and}\\quad v_{c} = wt$$\nTherefore,\n$$v_{c} = w\\sqrt{\\frac{2x}{w}} = \\sqrt{2xw}$$\nHence\n$$y = \\frac{v_{c}}{\\omega} = \\frac{\\sqrt{2wx}}{\\omega}$$\n(parabola)", "solution_images": ["images/image42.png"], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q14", "question": "A parallel-plate capacitor is located horizontally so\nthat one of its plates is submerged into liquid while the other is over\nits surface Fig. The permittivity of the liquid is equal to\n$\\varepsilon$, its density is equal to $\\rho$. To what height will the\nlevel of the liquid in the capacitor rise after its plates get a charge\nof surface density $\\sigma$?\n[IMAGE]", "question_images": ["images/image10.png"], "answer": "$h = \\frac{(\\varepsilon - 1)\\sigma^{2}}{2\\varepsilon_{0}\\varepsilon\\rho g}$", "solution": "One way of doing this problem will be exactly as in the\nprevious case so let us try an alternative method based on energy.\nSuppose the liquid rises by a distance $h$. Then let us calculate the\nextra energy of the liquid as a sum of polarization energy and the\nordinary gravitational energy. The latter is\n$$\\frac{1}{2}h \\bullet \\rho g \\bullet Sh = \\frac{1}{2}\\rho gSh^{2}$$\nIf $\\sigma$ is the free charge surface density on the plate, the bound\ncharge density is, from the previous problem,\n$$\\sigma^{'} = \\frac{\\varepsilon - 1}{\\varepsilon}\\sigma$$\nThis is also the volume density of induced dipole moment i.e.\nPolarization. Then the energy is, as before\n$$- \\frac{1}{2} \\bullet \\sigma^{'}E_{0} = \\frac{- 1}{2} \\bullet \\sigma^{'}\\frac{\\sigma}{\\varepsilon_{0}} = \\frac{- (\\varepsilon - 1)\\sigma^{2}}{2\\varepsilon_{0}\\varepsilon}$$\nand the total polarization energy is\n$$- S(a + h)\\frac{(\\varepsilon - 1)\\sigma^{2}}{2\\varepsilon_{0}\\varepsilon}$$\n[IMAGE] Then, total energy is\n$$U(h) = - S(a + h)\\frac{(\\varepsilon - 1)\\sigma^{2}}{2\\varepsilon_{0}\\varepsilon} + \\frac{1}{2}\\rho gSh^{2}$$\nThe actual height to which the liquid rises is determined from the\nformula\n$$\\frac{dU}{dh} = U^{'}(h) = 0$$\nThis gives\n$h = \\frac{(\\varepsilon - 1)\\sigma^{2}}{2\\varepsilon_{0}\\varepsilon\\rho g}$", "solution_images": ["images/image11.png"], "subject": "Physics", "topic": "Electrostatics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q50", "question": "A particle of mass $m$ is located outside a uniform\nsphere of mass $M$ at a distance $r$ from its centre. Find:\n\\(a\\) the potential energy of gravitational interaction of the particle\nand the sphere;\n\\(b\\) the gravitational force which the sphere exerts on the particle.", "question_images": [], "answer": "(a) $- \\frac{\\gamma Mm}{r}$\n\\(b\\) $- \\frac{\\gamma mM}{r^{3}}\\overrightarrow{r}$", "solution": "Suppose that the sphere has a radius equal to $a$. We may\nimagine that the sphere is made up of concentric thin spherical shells\n(layers) with radii ranging from 0 to a, and each spherical layer is\nmade up of elementary bands (rings). Let us first calculate potential\ndue to an elementary band of spherical layer at the point of location of\nthe point mass $m$ (say point $P$) Fig. As all the points of the band\nare located at the distance $l$ from the point $P$, so,\n[IMAGE] $\\partial\\varphi = - \\frac{\\gamma\\partial M}{l}$ (where mass of the\nband) (1)\n$$\\partial M = \\left( \\frac{dM}{4\\pi a^{2}} \\right)\\left( 2\\pi a\\sin\\theta \\right)(a\\ d\\theta)$$\n$= \\left( \\frac{dM}{2} \\right)\\sin\\theta d\\theta$ (2)\nAnd $l^{2} = a^{2} + r^{2} - 2ar\\cos\\theta$ (3)\nDifferentiating Eq. (3), we get\n$ldl = ar\\sin\\theta d\\theta$ (4)\nHence using above equations\n$\\partial\\varphi = - \\left( \\frac{\\gamma dM}{2ar} \\right)dl$ (5)\nNow integrating this Eq. over the whole spherical layer\n$$d\\varphi = \\int_{}^{}{\\partial\\varphi} = - \\frac{\\gamma dM}{2ar}\\int_{r - a}^{r + a}\\ $$\nSo $d\\varphi = - \\frac{\\gamma dM}{r}$ (6)\nEquation (6) demonstrates that the potential produced by a thin\nuniform spherical layer outside the layer is such as if the whole mass\nof the layer were concentrated at it's centre;\nHence the potential due to the sphere at point P;\n$\\varphi = \\int_{}^{}{d\\varphi} = - \\frac{\\gamma}{r}\\int_{}^{}{dM} = - \\frac{\\gamma M}{r}$\n(7)\nThis expression is similar to that of Eq. (6)\nHence the sought potential energy of gravitational interaction of the\nparticle $m$ and the sphere,\n$$U = m\\varphi = - \\frac{\\gamma Mm}{r}$$\n\\(b\\) Using the Eq., $G_{r} = - \\frac{\\partial\\varphi}{\\partial r}$\n$G_{r} = - \\frac{\\gamma M}{r^{2}}$ (using Eq. 7)\nSo, $\\overrightarrow{G} = \\frac{\\gamma M}{r^{3}}\\overrightarrow{r}$\nand\n$\\overrightarrow{F} = m\\overrightarrow{G} = - \\frac{\\gamma mM}{r^{3}}\\overrightarrow{r}$\n(8)", "solution_images": ["images/image51.jpeg"], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q48", "question": "Two bars connected by a weightless spring of stiffness\n$x$ and length (in the non-deformed state) $l_{0}$ rest on a horizontal\nplane. A constant horizontal force $F$ starts acting on one of the bars\nas shown in Fig. Find the maximum and minimum distances between the bars\nduring the subsequent motion of the system, if the masses of the bars\nare:\n\\(a\\) equal;'\n\\(b\\) equal to $m_{1}$ and $m_{2}$, and the force $F$ is applied to\nthe bar of mass $m_{2}$.\n[IMAGE]", "question_images": ["images/image47.png"], "answer": "$l_{0} + \\frac{2m_{1}F}{k\\left( m_{1} + m_{2} \\right)}$", "solution": "Let us consider both blocks and spring as the physical\nsystem. The centre of mass of the system moves with acceleration\n$a = \\frac{F}{m_{1} + m_{2}}$ towards right. Let us work in the frame of\ncentre of mass. As this frame is a non-inertial frame (accelerated with\nrespect to the ground) we have to apply a pseudo force $m_{1}a$ towards\nleft on the block $m_{1}$ and $m_{2}$ a towards left on the block\n$m_{2}$.\nAs the center of mass is at rest in this frame, the blocks move in\nopposite directions and come to instantaneous rest at some instant.\nThe elongation of the spring will be maximum or minimum at this\ninstant. Assume that the block $m_{1}$ is displaced by the distance\n$x_{1}$ and the block $m_{2}$ through a distance $x_{2}$ from the\ninitial positions.\n[IMAGE] From the energy equation in the frame of C.M.\n$$\\mathrm{\\Delta}\\widetilde{T} + U = A_{ext},$$\n(where $A_{ext}$ also includes the work done by the pseudo forces)\nHere,\n$\\mathrm{\\Delta}\\widetilde{T} = 0,U = \\frac{1}{2}k\\left( x_{1} + x_{2} \\right)^{2}$\nand\n$$W_{ext} = \\left( \\frac{F - m_{2}F}{m_{1} + m_{2}} \\right)x_{2} + \\frac{m_{1}F}{m_{1} + m_{2}}x_{1} = \\frac{m_{1}F\\left( x_{1} + x_{2} \\right)}{m_{1} + m_{2}},$$\nOr,\n$\\frac{1}{2}k\\left( x_{1} + x_{2} \\right)^{2} = \\frac{m_{1}\\left( x_{1} + x_{2} \\right)F}{m_{1} + m_{2}}$\nSo, $x_{1} + x_{2} = 0$ or,\n$x_{1} + x_{2} = \\frac{2m_{1}F}{k\\left( m_{1} + m_{2} \\right)}$\nHence the maximum separation between the blocks equals:\n$l_{0} + \\frac{2m_{1}F}{k\\left( m_{1} + m_{2} \\right)}$\nObviously the minimum sepation corresponds to zero elongation and is\nequal to $l_{0}$", "solution_images": ["images/image48.png"], "subject": "Physics", "topic": "Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q3", "question": "A horizontally oriented tube $AB$ of length $l$ rotates\nwith a constant angular velocity $\\omega$ about a stationary vertical\naxis $OO'$ passing through the end $A$ Fig. The tube is filled with an\nideal fluid. The end $A$ of the tube is open, the closed end $B$ has a\nvery small orifice. Find the velocity of the fluid relative to the tube\nas a function of the column \"height\" $h$.\n[IMAGE]", "question_images": ["images/image4.png"], "answer": "$v = \\omega h\\sqrt{\\frac{2l}{h} - 1}$", "solution": "In a rotating frame (with constant angular velocity) the\nEulerian equation is\n$$- \\overrightarrow{\\nabla}p + p\\overrightarrow{g} + 2\\rho\\left( \\overrightarrow{v} \\times \\overrightarrow{\\omega} \\right) + \\rho\\omega^{2}\\overrightarrow{r} = \\rho\\frac{d\\overrightarrow{v}}{dt}$$\nIn the frame of rotating tube the liquid in the \"column\" is\npractically static because the orifice is sufficiently small. Thus the\nEulerian Eq. in projection form along $\\overrightarrow{r}$ (which is\nthe position vector of an arbitrary liquid element of length $dr$\nrelative to the rotation axis) reduces to\n[IMAGE] $$\\frac{- dp}{dr} + \\rho\\omega^{2}r = 0$$\nOr, $dp = \\rho\\omega^{2}rdr$\nSo, $\\int_{p_{0}}^{p}{dp} = \\rho\\omega^{2}\\int_{(l - h)}^{r}{rdr}$\nThus\n$p(r) = p_{0} + \\frac{\\rho\\omega^{2}}{2}\\left\\lbrack r^{2} - (l - h)^{2} \\right\\rbrack$\n(1)\nHence the pressure at the end $B$ just before the orifice i.e.\n$p(l) = p_{0} + \\frac{\\rho\\omega^{2}}{2}\\left( 2lh - h^{2} \\right)$\n(2)\nThen applying Bernoull's theorem at the orifice for the points just\ninside and outside of the end $B$\n$p_{0} + \\frac{1}{2}\\rho\\omega^{2}\\left( 2lh - h^{2} \\right) = p_{0} + \\frac{1}{2}\\rho v^{2}$\n(where $v$ is the sought velocity)\nSo, $v = \\omega h\\sqrt{\\frac{2l}{h} - 1}$", "solution_images": ["images/image5.png"], "subject": "Physics", "topic": "Fluid Mechanics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q28", "question": "In an oscillating circuit shown in Fig. the coil\ninductance is equal to $L = 2.5\\ mH$ and the capacitor have capacitances\n$C_{1} = 2.0\\ \\mu F$ and $C_{2} = 3.0\\ \\mu F$. The capacitor were\ncharged to a voltage $V = 180\\ V$, and then the switch $Sw$ was closed.\nFind :\n\\(a\\) the natural oscillation frequency;\n\\(b\\) the peak value of the current flowing through the coil.\n[IMAGE]", "question_images": ["images/image23.png"], "answer": "$0.7\\ ms,\\ 8.05\\ A$", "solution": "The equation of the $L - C$ circuit are\n$$L\\frac{d}{dt}\\left( I_{1} + I \\right) = \\frac{C_{1}V - \\int_{}^{}{I_{1}dt}}{C_{1}} = \\frac{C_{2}V - \\int_{}^{}I_{2}dt}{C_{2}}$$\nDifferentiating again\n$L\\left( I_{1} + I_{2} \\right) = - \\frac{1}{C_{1}}I_{1} = - \\frac{1}{C_{2}}I_{2}$\nThen\n$I_{1} = \\frac{C_{1}}{C_{1} - C_{2}}I,I_{2} = \\frac{C_{2}}{C_{1} + C_{2}}I,$\n$$I = I_{1} + I_{2}$$\nso $L\\left( C_{1} + C_{2} \\right)I + I = 0$\nor $I = I_{0}\\sin\\left( \\omega_{0}t + \\alpha \\right)$\nwhere $\\omega_{0}^{2} = \\frac{1}{L\\left( C_{1} + C_{2} \\right)}$ (Part\na)\n(Hence $T = \\frac{2\\pi}{\\omega_{0}} = 0.7\\ ms$)\n[IMAGE] At $t = 0,I = 0$ so $\\alpha = 0$\n$$I = I_{0}\\sin{\\omega_{0}t}$$\nThe peak value of the current is $I_{0}$ and it is related to the\nvoltage $V$ by the first equation\n$$LI = V - \\int_{}^{}Idt/\\left( C_{1} + C_{2} \\right)$$\nOr\n$+ L\\omega_{0}I_{0}\\cos\\omega_{0}t = V - \\frac{1}{C_{1} + C_{2}}\\int_{\\omega_{0}}^{t}I_{0}\\sin\\omega_{0}dt$\n(The P.D. across the inductance is $V$ at $t = 0$)\n$$= V + \\frac{1}{C_{1} + C_{2}} \\bullet \\frac{I_{0}}{\\omega_{0}}\\left( \\cos\\omega_{0}t - 1 \\right)$$\nHence\n$I_{0} = \\left( C_{1} + C_{2} \\right)\\omega_{0}V = V\\sqrt{\\frac{C_{1} + C_{2}}{L}} = 8.05\\ A$", "solution_images": ["images/image24.jpeg"], "subject": "Physics", "topic": "Waves and Oscillations", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q38", "question": "A photon is scattered at an angle\n$\\theta = 120{^\\circ}$ by a stationary free electron. As a result, the\nelectron acquires a kinetic energy $T = 0.45\\ MeV$. Find the energy that\nthe photon had prior to scattering.", "question_images": [], "answer": "$0.677\\ Mev$", "solution": "From the Compton formula\n$$\\lambda = \\lambda_{0} + \\frac{2\\pi h}{m\\ c}(1 - cos\\theta)$$\nFrom conservation of energy\n$$\\frac{2\\pi hc}{\\lambda_{0}} = \\frac{2\\pi hc}{\\lambda} + T = \\frac{2\\pi hc}{\\lambda_{0} + \\frac{2\\pi h}{mc}\\left( 1 - \\cos\\theta \\right)} + T$$\nOr\n$\\frac{4\\pi h}{mc}\\sin^{2}\\frac{\\theta}{2} = \\frac{T}{2\\pi hc}\\lambda_{0}\\left( \\lambda_{0} + \\frac{4\\pi h}{m\\ c}\\sin^{2}\\frac{\\theta}{2} \\right)$\nor introducing $h\\omega_{0} = 2\\pi hc/\\lambda_{0}$\n$$\\frac{2\\sin^{2}{\\theta/2}}{mc^{2}} = \\frac{T}{h\\omega_{0}}\\left( \\frac{1}{h\\omega_{0}} + \\frac{2}{mc^{2}}\\sin^{2}\\frac{\\theta}{2} \\right)$$\nHence\n$\\left( \\frac{1}{h\\omega_{0}} \\right)^{2} + 2\\frac{1}{h\\omega_{0}}\\frac{\\sin^{2}\\frac{\\theta}{2}}{mc^{2}} - \\frac{2\\sin^{2}\\frac{\\theta}{2}}{mc^{2}T} = 0$\n$$\\left( \\frac{1}{h\\omega_{0}} + \\frac{\\sin^{2}\\frac{\\theta}{2}}{mc^{2}} \\right)^{2} = \\frac{2\\sin^{2}\\frac{\\theta}{2}}{mc^{2}T} + \\left( \\frac{\\sin^{2}\\frac{\\theta}{2}}{mc^{2}\\ } \\right)^{2}$$\n$$\\frac{1}{h\\omega_{0}} = \\frac{\\sin^{2}\\frac{\\theta}{2}}{mc^{2}}\\left\\lbrack \\sqrt{1 + \\frac{2mc^{2}}{T\\sin^{2}\\frac{\\theta}{2}}} - 1 \\right\\rbrack$$\nOr\n$h\\omega_{0} = \\frac{mc^{2}/\\sin^{2}{\\theta/2}}{\\sqrt{1 + \\frac{2mc^{2}}{T\\sin^{2}\\frac{\\theta}{2}}} - 1}$\n$$= \\frac{T}{2}\\left\\lbrack \\sqrt{1 + \\frac{2mc^{2}}{T\\sin^{2}{\\theta/2}}} + 1 \\right\\rbrack$$\nSubstituting we get $h\\omega_{0} = 0.677\\ Mev$", "solution_images": [], "subject": "Physics", "topic": "Modern Physics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q6", "question": "An ideal gas of molar mass $M$ is contained in a very\ntall vertical cylindrical vessel in the uniform gravitational field in\nwhich the free-fall acceleration equals $g$. Assuming the gas\ntemperature to be the same and equal to $T$, find the height at which\nthe centre of gravity of the gas is located.", "question_images": [], "answer": "$h = \\frac{RT}{Mg}$", "solution": "As the gravitational field is constant the centre of\ngravity and the centre of mass are same.\nThe location of C.M.\n$$h = \\frac{\\int_{0}^{\\infty}hdm}{\\int_{0}^{\\infty}{dm}} = \\frac{\\int_{0}^{\\infty}{h\\rho}dh}{\\int_{0}^{\\infty}\\rho dh}$$\nBut from Barometric formula and gas law $\\rho = \\rho_{0}e^{- Mgh/RT}$\nSo,\n$h = \\frac{\\int_{0}^{\\alpha}{h\\left( e^{- Mgh/RT} \\right)}dh}{\\int_{0}^{\\alpha}\\left( e^{- Mg\\ h/RT} \\right)dh} = \\frac{RT}{Mg}$", "solution_images": [], "subject": "Physics", "topic": "Thermodynamics", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}
{"question_id": "ADV-P01-Q35", "question": "An object is placed in front of convex surface of a\nglass plano-convex lens of thickness $d = 9.0\\ cm$. The image of that\nobject is formed on the plane surface of the lens serving as a screen.\nFind:\n(a) the transverse magnification if the curvature radius of the lens's\n convex surface is\n$$R = 2.5\\ cm;$$\n(b) the image illuminance if the luminance of the object is\n $L = 7700\\ cd/m^{2}$ and the entrance aperture diameter of the lens\n is $D = 5.0\\ mm;$ losses of light are negligible.", "question_images": [], "answer": "(a) $1 - \\frac{d(n - 1)}{nR}$\n(b) $42\\ 1x$", "solution": "(a) The formation of the image of a source $S$, placed at\na distance $u$ from the pole of the convex surface of plano-convex lens\nof thickness $d$ is shown in the figure.\nOn applying the formula for refraction through spherical surface, we\nget\n$\\frac{n}{s'} - \\frac{1}{s} = (n - 1)/R$, (here $n_{2} = n$ and\n$n_{1} = 1$)\nOr, $\\frac{n}{d} - \\frac{1}{s} = (n - 1)/R$ or\n$\\frac{1}{s} = \\frac{n}{d} - \\frac{(n - 1)}{R}$\nOr,\n$\\frac{s'}{s} = s^{'}\\left\\{ \\frac{n}{d} - \\frac{(n - 1)}{R} \\right\\}$\n[IMAGE] But in this case optical path of the light, corresponding to the\ndistance $v$ in the medium is $v/n$, so the magnification produced\nwill be,\n$$\\beta = \\frac{s'}{ns} = \\frac{s'}{n}\\left\\{ \\frac{n}{d} - \\frac{(m - 1)}{R} \\right\\} = \\frac{d}{n}\\left\\{ \\frac{n}{d} - \\frac{(n - 1)}{R} \\right\\} = 1 - \\frac{d(n - 1)}{nR}$$\nSubstituting the values, we get magnification $\\beta = - 0.20.$\n\\(b\\) If the transverse area of the object is $A$ (assumed small), the\narea of the image is $\\beta^{2}A$.\nWe shall assume that $\\frac{\\pi D^{2}}{4} > A$. Then light falling on\nthe lens is : $LA\\frac{\\pi D^{2}/4}{s^{2}}$\nfrom the definition of luminance (See Eqn. (5.1 c) of the book; here\n$\\cos\\theta \\approx 1$ if $D^{2} \\ll s^{2}$ and\n$d\\Omega = \\frac{\\pi D^{2}/4}{s^{2}}$. Then the illuminance of the\nimage is\n$LA\\frac{\\pi D^{2}/4}{s^{2}}/\\beta^{2}A = Ln^{2}\\pi D^{2}/4d^{2}$\nSubstitution gives 42 1x.", "solution_images": ["images/image31.png"], "subject": "Physics", "topic": "Optics", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Manish_sir__Phy_1-50_with_img.docx"}