physics/test.jsonl

{"question_id": "ADV-P01-Q7", "question": "A piston can freely move inside a horizontal cylinder\nclosed from both ends. Initially, the piston separates the inside space\nof the cylinder into two equal parts each of volume $V_{0}$, in which an\nideal gas is contained under the same pressure $p_{0}$ and at the same\ntemperature. What work has to be performed in order to increase\nisothermally the volume of one part of gas $\\eta$ times compared to that\nof the other by slowly moving the piston?", "question_images": [], "answer": "$p_{0}V_{0}\\ln\\frac{(\\eta + 1)^{2}}{4\\eta}$", "solution": "Let the agent move the piston to the right by $x$. In\nequilibrium position,\n$p_{1}S + F_{agent} = p_{2}S,$ or,\n$F_{agent} = \\left( p_{2} - p_{1} \\right)S$\nWork done by the agent in an infinitesimal change $dx$ is\n$$F_{agent} \\bullet dx = \\left( p_{2} - p_{1} \\right)Sdx = \\left( p_{2} - p_{1} \\right)dV$$\nBy applying $pV =$ constant, for the two parts,\n$p_{1}\\left( V_{0} + Sx \\right) = p_{0}V_{0}$ and\n$p_{2}\\left( V_{0} - Sx \\right) = p_{0}V_{0}$\nSo,\n$p_{2} - p_{1} = \\frac{p_{0}V_{0}2Sx}{V_{0}^{2} - S^{2}x^{2}} = \\frac{2p_{0}V_{0}V}{V_{1}^{2} - V^{2}}$\n(where $Sx = V$)\nWhen the volume of the left end is $\\eta$ times the volume of the\nright end\n$\\left( V_{0} + V \\right) = \\eta\\left( V_{0} - V \\right),$ or,\n$V = \\frac{\\eta - 1}{\\eta + 1}V_{0}$.\n$$A = \\int_{0}^{V}\\left( p_{2} - p_{1} \\right)dv = \\int_{0}^{V}\\frac{2p_{0}V_{0}V}{V_{0}^{2} - V^{2}}dV = - p_{0}V_{0}\\left\\lbrack \\ln\\left( V_{0}^{2} - V^{2} \\right) \\right\\rbrack_{0}^{V}$$\n$$= - p_{0}V_{0}\\left\\lbrack \\ln{\\left( V_{0}^{2} - V^{2} \\right) - \\ln V_{0}^{2}} \\right\\rbrack$$\n$$= - p_{0}V_{0}\\left\\lbrack \\ln{\\left\\{ V_{0}^{2} - \\left( \\frac{\\eta - 1}{\\eta + 1^{2}} \\right)V_{0}^{2} \\right\\} - \\ln V_{0}^{2}} \\right\\rbrack$$\n$$= - p_{0}V_{0}\\left( \\ln\\frac{4\\eta}{(\\eta + 1)^{2}} \\right) = p_{0}V_{0}\\ln\\frac{(\\eta + 1)^{2}}{4\\eta}$$", "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-Q9", "question": "A pendulum clock is mounted in an elevator car which\nstarts going up with a constant acceleration $w$, with $w < g$. At a\nheight $h$ the acceleration of the car reverses, its magnitude remaining\nconstant. How soon after the start of the motion will the clock show the\nright time again?", "question_images": [], "answer": "$\\sqrt{\\frac{2h}{w}}\\frac{\\sqrt{1 + \\beta} - \\sqrt{1 - \\beta}}{1 - \\sqrt{1 - \\beta}}$", "solution": "Let the downward acceleration of the elevator car has\ncontinued for time $t'$, then the sought time\n$t = \\sqrt{\\frac{2h}{w}} + t'$, where obviously $\\sqrt{\\frac{2h}{w}}$\nis the time of upward acceleration of the elevator. One should note\nthat if the point of suspension of a mathematical pendulum moves with\nan acceleration $\\overrightarrow{w}$, then the time period of the\npendulum becomes\n$$2\\pi\\sqrt{\\frac{l}{\\left| \\overrightarrow{g} - \\overrightarrow{w} \\right|}}$$\nIn this problem the time period of the pendulum while it is moving\nupward with acceleration $w$ becomes $2\\pi\\sqrt{\\frac{l}{g + w}}$ and\nits time period while the elevator moves downward with the same\nmagnitude of acceleration becomes\n$$2\\pi\\sqrt{\\frac{l}{g - w}}$$\nAs the time of upward acceleration equal $\\sqrt{\\frac{2h}{w}}$, the\ntotal number of oscillations during this time equals\n$$\\frac{\\sqrt{2h/w}}{2\\pi\\sqrt{l/(g + w)}}$$\nThus, the indicated time\n$= \\frac{\\sqrt{2h/w}}{2\\pi\\sqrt{l/(g + w)}} \\bullet 2\\pi\\sqrt{l/g} = \\sqrt{2h/w}\\sqrt{(g + w)/g}$\nSimilarly, the indicated time for the time interval $t'$\n$$= \\frac{t'}{2\\pi\\sqrt{l/(g - w)}}2\\pi\\sqrt{l/g} = t^{'}\\sqrt{(g - w)/g}$$\nwe demand that\n$$\\sqrt{2h/w}\\ \\sqrt{(g + w)/g} + t^{'}\\sqrt{(g - w)/g} = \\sqrt{2h/w + t'}$$\nOr,\n$t^{'} = \\sqrt{2h/w}\\frac{\\sqrt{g + w} - \\sqrt{g}}{\\sqrt{g} - \\sqrt{g - w}}$\nHence the sought time\n$$t = \\sqrt{\\frac{2h}{w}} + t^{'} = \\sqrt{\\frac{2h}{w}}\\frac{\\sqrt{g + w} - \\sqrt{g - w}}{\\sqrt{g} - \\sqrt{g - w}}$$\n$= \\sqrt{\\frac{2h}{w}}\\frac{\\sqrt{1 + \\beta} - \\sqrt{1 - \\beta}}{1 - \\sqrt{1 - \\beta}}$,\nwhere $\\beta = w/g$", "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-Q15", "question": "There is an infinite wire grid with square cells Fig.\nThe resistance of each wire between neighbouring joint connections is\nequal to $R_{0}$. Find the resistance $R$ of the whole grid between\npoints $A$ and $B$.\nInstruction: Make use of principles of symmetry ad superposition.\n[IMAGE]", "question_images": ["images/image12.png"], "answer": "$R = \\frac{R_{0}}{2}$", "solution": "Suppose that the voltage $V$ is applied between the points\n$A$ and $B$ then\n$$V = IR = I_{0}R_{0}$$\nwhere $R$ is resistance of whole the grid, $I$, the current through\nthe grid and $I_{0}$, the current through the segment $AB$. Now from\nsymmetry, $I/4$ is the part of the current, flowing through all the\nfour wire segments, meeting at the point $A$ and similarly the amount\nof current flowing through the wires, meeting at $B$ is also $I/4$.\nThus, a current $I/2$ flows through the conductor $AB$, i.e.\n$$I_{0} = \\frac{I}{2}$$\nHence, $R = \\frac{R_{0}}{2}$", "solution_images": [], "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-Q16", "question": "A metal ball of radius $a$ is located at a distance $l$\nfrom an infinite ideally conducting plane. The space around the ball is\nfilled with a homogeneous poorly conducting medium with resistivity ρ.\nIn the case of $a \\ll l$ find:\n\\(a\\) the current density at the conducting plane as a function of\ndistance $r$ from the ball if the\npotential difference between the ball and the plane is equal to V;\n\\(b\\) the electric resistance of the medium between the ball and the\nplane.", "question_images": [], "answer": "(a) $j = \\frac{1}{\\rho}E = \\frac{2alV}{\\rho r^{3}}$\n\\(b\\) $R = \\frac{V}{I} = \\frac{\\rho}{2\\pi a}$", "solution": "(a) The potential in the unshaded region beyond the\nconductor as the potential of the given charge and its image and has the\nform\n$$\\varphi = A\\left( \\frac{1}{r_{1}} - \\frac{1}{r_{2}} \\right)$$\nwhere $r_{1},r_{2}$ are the distances of the point from the charge and\nits image. The potential has been\ntaken to be zero on the conducting plane and on the ball\n$$\\varphi \\approx A\\left( \\frac{1}{a} - \\frac{1}{2l} \\right) = V$$\n[IMAGE] So $A \\approx Va$. In this calculation the conditions $a \\ll l$ is\nused to ignore the variation of $\\varphi$ over\nthe ball.\nThe electric field at $P$ can be calculated similarly. The charge on\nthe ball is\n$$Q = 4\\pi\\varepsilon_{0}Va$$\nand $E_{P} = \\frac{Va}{r^{2}}2\\cos\\theta = \\frac{2alV}{r^{3}}$\nThen $j = \\frac{1}{\\rho}E = \\frac{2alV}{\\rho r^{3}}$ normal to the\nplane.\n\\(b\\) The total current flowing into the conducting plane is\n$I = \\int_{0}^{\\infty}{2\\pi x}dxj = \\int_{0}^{\\infty}{2\\pi x}dx\\frac{2alV}{\\rho\\left( \\pi^{2} + l^{2} \\right)^{3/2}}$\n(On putting $y = x^{2} + l^{2}$)\n$I = \\frac{2\\pi alV}{\\rho}\\int_{l^{2}}^{\\infty}\\frac{dy}{y^{3/2}} = \\frac{4\\pi aV}{\\rho}$\nHence $R = \\frac{V}{I} = \\frac{\\rho}{2\\pi a}$", "solution_images": ["images/image13.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-Q18", "question": "A cylindrical capacitor connected to a dc voltage\nsource V touches the surface of water with its end Fig. The separation d\nbetween the capacitor electrodes is substantially less than their mean\nradius. Find a height $h$ to which the water level in the will rise. The\ncapillary effects are to be neglected.\n[IMAGE]", "question_images": ["images/image15.jpeg"], "answer": "$\\frac{\\varepsilon_{0}(\\varepsilon - 1)V^{2}}{2\\rho gd^{2}}$", "solution": "If $C_{0}$ is the initial capacitance of the condenser\nbefore water rises in it then\n$U_{i} = \\frac{1}{2}C_{0}V^{2}$, where\n$C_{0} = \\frac{\\varepsilon_{0}2l\\pi R}{d}$\n($R$ is the mean radius and $l$ is the length of the capacitor\nplates.)\nSuppose the liquid rises to a height $h$ in it. Then the capacitor of\nthe condenser is\n$$C = \\frac{\\varepsilon\\varepsilon_{0}h2\\pi R}{d} + \\frac{\\varepsilon(l - h)2\\pi R}{d} = \\ \\frac{\\varepsilon_{0}2\\pi R}{d}\\left( l + (\\varepsilon - 1)h \\right)$$\nand energy of the capacitor and the liquid (including both\ngravitational and electrostatic contribution) is\n$$\\frac{1}{2}\\frac{\\varepsilon_{0}2\\pi R}{d}\\left( l + (\\varepsilon - 1)h \\right)V^{2} + \\rho g(2\\pi R\\ hd)\\frac{h}{2}$$\nIf the capacitor were not connected to a battery this energy would\nhave to be minimized. But the capacitor is connected to the battery\nand, in effect, the potential energy of the whole system has to be\nminimized. Suppose we increase $h$ by $\\delta h$. Then the energy of\nthe capacitor and the liquid increases by\n$$\\delta h\\left( \\frac{\\varepsilon_{0}2\\pi R}{2d}(\\varepsilon - 1)V^{2} + \\rho g(2\\pi Rd)h \\right)$$\nand that of the cell diminishes by the quantity $A_{cell}$ which is\nthe product of charge flown and $V$\n$$\\delta h\\frac{\\varepsilon_{0}(2\\pi R)}{d}(\\varepsilon - 1)V^{2}$$\nIn equilibrium, the two must balance; so\n$$\\rho gdh = \\frac{\\varepsilon_{0}(\\varepsilon - 1)V^{2}}{2d}$$\nHence $h = \\frac{\\varepsilon_{0}(\\varepsilon - 1)V^{2}}{2\\rho gd^{2}}$", "solution_images": [], "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-Q2", "question": "One end of a steel rectangular girder is embedded into a\nwall Fig. Due to gravity it sags slightly. Find the radius of curvature\nof the neutral layer (see the dotted line in the figure) in the vicinity\nof the point O if the length of the protruding section of the girder is\nequal to $l = 6.0\\ m$ and the thickness of the girder equals\n$h = 10\\ cm$.\n[IMAGE]", "question_images": ["images/image1.png"], "answer": "$(0.121\\ km)^{- 1}$", "solution": "A beam clamped at one end and supporting an applied load\nat the free end is called a cantilever. The theory of cantilevers is\ndiscussed in advanced text book on mechanics. The key result is that\nelastic forces in the beam generate a couple, whose moment, called the\nmoment of resistances, balances the external bending moment due to\nweight of the beam, load etc. The moment of resistance, also called\ninternal bending moment (I.B.M) is given by\n$$I.B.M. = EI/R$$\nHere $R$ is the radius of curvature of the beam at the representative\npoint $(x,y)$. I is called the geometrical moment of inertia\n$$I = \\int_{}^{}z^{2}dS$$\nof the cross section relative to the axis passing through the netural\nlayer which remains unstretched. Fig. The section of the beam beyond\n$P$ exerts the bending moment $N(x)$ and we have,\n$$\\frac{EI}{R} = N(x)$$\n[IMAGE] If there is no load other than that due to the weight of the beam,\nthen\n$$N(x) = \\frac{1}{2}\\rho g(l - x)^{2}bh$$\nwhere $\\rho =$ density of steel.\nHence, at $x = 0$\n$$\\left( \\frac{I}{R} \\right)_{0} = \\frac{\\rho gl^{2}bh}{2EI}$$\nHere $b =$ width of the beam perpendicular to paper.\nAlso, $I = \\int_{- h/2}^{h/2}z^{2}bdz = \\frac{bh^{3}}{12}$.\n[IMAGE]", "solution_images": ["images/image2.png", "images/image3.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-Q8", "question": "A heat-conducting piston can freely move inside a closed\nthermally insulated cylinder with an ideal gas. In equilibrium the\npiston divides the cylinder into two equal parts, the gas temperature\nbeing equal to $T_{0}$. The piston is slowly displaced. Find the gas\ntemperature as a function of the ratio $\\eta$ of the volume of the\ngreater and smaller sections. The adiabatic exponent of the gas is equal\nto $\\gamma$.", "question_images": [], "answer": "$T = T_{0}\\left( \\frac{(\\eta + 1)^{2}}{4\\eta} \\right)^{\\frac{\\gamma - 1}{2}}$", "solution": "Since here the piston is conducting and it is moved slowly\nthe temperature on the two sides increases and maintained at the same\nvalue.\nElementary work done by the agent = Work done in compression -- Work\ndone in expansion i.e.\n$dA = p_{2}dV - p_{1}dV = \\left( p_{2} - p_{1} \\right)dV$\nwhere $p_{1}$ and $p_{2}$ are pressures at any instant of the gas on\nexpansion and compression side respectively.\nFrom the gas law $p_{1}\\left( V_{0} + Sx \\right) = \\nu RT$ and\n$p_{2}\\left( V_{0} - Sx \\right) = \\nu RT$, for each section\n($x$ is the displacement of the piston towards section 2)\nSo,\n$p_{2} - p_{1} = \\nu RT\\frac{2Sx}{V_{0}^{2} - S^{2}x^{2}} = \\nu RT \\bullet \\frac{2V}{v_{0}^{2} - V^{2}}$\n(as $Sx = V$)\nSo $dA = \\nu RT\\frac{2V}{V_{0}^{2} - V^{2}}dV$\nAlso, from the first law of thermodynamics\n$dA = - dU = - 2\\nu\\frac{R}{\\gamma - 1}dT$ (as $dQ = 0$)\nSo, work done on the gas\n$= - dA = 2\\nu \\bullet \\frac{R}{\\gamma - 1}dT$\nThus\n$2\\nu\\frac{R}{\\gamma - 1}dT = \\nu RT\\frac{2V \\bullet dV}{V_{0}^{2} - V^{2}}$,\nOr, $\\frac{dT}{T} = \\gamma - 1\\frac{V\\ dV}{V_{0}^{2} - V^{2}}$\n[IMAGE] When the left end is $\\eta$ times the volume of the right end.\n$\\left( V_{0} + V \\right) = \\eta\\left( V_{0} - V \\right)$ or\n$V = \\frac{\\eta - 1}{\\eta + 1}V_{0}$\nOn integrating\n$\\int_{T_{0}}^{T}\\frac{dT}{T} = (\\gamma - 1)\\int_{0}^{V}\\frac{VdV}{V_{0}^{2} - V^{2}}$\nOr\n$\\ln\\frac{T}{T_{0}} = (\\gamma - 1)\\left\\lbrack - \\frac{1}{2}\\ln\\left( V_{0}^{2} - V^{2} \\right) \\right\\rbrack_{0}^{V}$\n$$= - \\frac{\\gamma - 1}{2}\\left\\lbrack \\ln{\\left( V_{0}^{2} - V^{2} \\right) - \\ln V_{0}^{2} - V^{2} - \\ln V_{0}^{2}} \\right\\rbrack$$\n$$= \\frac{\\gamma - 1}{2}\\left\\lbrack \\ln V_{0}^{2} - \\ln V_{0}^{2}\\left\\{ 1 - \\left( \\frac{\\eta - 1}{\\eta + 1} \\right)^{2} \\right\\} \\right\\rbrack = \\frac{\\gamma - 1}{2}\\ln\\frac{(\\eta + 1)^{2}}{4\\eta}$$\nHence\n$T = T_{0}\\left( \\frac{(\\eta + 1)^{2}}{4\\eta} \\right)^{\\frac{\\gamma - 1}{2}}$", "solution_images": ["images/image7.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-Q41", "question": "From point A located on a highway (Fig.) one has to get\nby car as soon as possible to point B located in the field at a distance\nl from the highway. It is known that the car moves in the field η time\nslower than on the highway. At what distance from point Done must turn\noff the highway?\n[IMAGE]", "question_images": ["images/image37.png"], "answer": "$\\frac{l}{\\sqrt{\\eta^{2} - 1}}$", "solution": "Let the car turn off the highway at a distance $x$ from\nthe point $D$.\nSo, $CD = x,$ and if the speed of the car in the field is $v$, then\nthe time taken by the car to cover the distance $AC = AD - x$ on the\nhighway\n[IMAGE] $t_{1} = \\frac{AD - x}{\\eta v}$ (1)\nand the time taken to travel the distance $CB$ in the field\n$t_{2} = \\frac{\\sqrt{l^{2} + x^{2}}}{v}$ (2)\nSo, the total time elapsed to move the car from point A to B\n$$t = t_{1} + t - 2 = \\frac{AD - x}{\\eta v} + \\frac{\\sqrt{l^{2} - x^{2}}}{v}$$\nFor $t$ to be minimum\n$$\\frac{dt}{dx} = 0\\ \\ \\ \\ or\\ \\ \\ \\frac{1}{v}\\left\\lbrack - \\frac{1}{\\eta} + \\frac{x}{\\sqrt{l^{2} + x^{2}}} \\right\\rbrack = 0$$\nOr $\\eta^{2}x^{2} = l^{2} + x^{2}$ or\n$x = \\frac{l}{\\sqrt{\\eta^{2} - 1}}$", "solution_images": ["images/image38.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"}