[
    {
        "question": "Q1: A gas can be taken from A to B via two different processes ACB and ADB. When the path ACB is used $60 \\mathrm{~J}$ of heat flows into the system and $30 \\mathrm{~J}$ of work is done by the system. If path ADB is used, work done by the system is $10 \\mathrm{~J}$. The heat flow into the system in path ADB is",
        "input": "(a) $100 \\mathrm{~J}$ (b) $80 \\mathrm{~J}$ (c) $40 \\mathrm{~J}$ (d) $20 \\mathrm{~J}$",
        "answer": "(c) $40 \\mathrm{~J}$",
        "solution": "$\\Delta \\mathrm{Q}=\\Delta \\mathrm{U}+\\Delta \\mathrm{W}$ $\\Delta \\mathrm{U}=\\Delta \\mathrm{Q}-\\Delta \\mathrm{W}$ $(\\Delta \\mathrm{U}){\\mathrm{ACB}}=(\\Delta \\mathrm{U}){\\mathrm{ADB}}$ $60-30=\\Delta \\mathrm{Q}-10$ $\\Delta \\mathrm{Q}=40 \\mathrm{~J}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q2: In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $V T=K$, where $K$ is a constant. In this process, the temperature of the gas is increased by $T$. The amount of heat absorbed by gas is given by ( $\\mathrm{R}$ is gas constant)",
        "input": "(a) $(2 \\mathrm{K} / 3) \\Delta \\mathrm{T}$ (b) $1 / 2 R \\Delta T$ (c) $(3 / 2) \\mathrm{R} \\Delta \\mathrm{T}$ (d) $(1 / 2) \\mathrm{KR} \\Delta \\mathrm{T}$",
        "answer": "(b) $1 / 2 R \\Delta T$",
        "solution": "$\\mathrm{VT}=\\mathrm{K}$, $\\mathrm{V}(\\mathrm{PV} / \\mathrm{R})=\\mathrm{K}$ $\\mathrm{PV}^{2}=$ constant For a polytropic process $\\mathrm{C}=(\\mathrm{R} / 1-\\mathrm{x})+\\mathrm{C}_{\\mathrm{v}}=(\\mathrm{R} / 1-2)+3 \\mathrm{R} / 2=\\mathrm{R} / 2$ $\\Delta \\mathrm{Q}=\\mathrm{n} \\mathrm{C} T=(1 / 2) \\mathrm{R} \\Delta \\mathrm{T}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q3: A cylinder with a fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by $20^{\\circ} \\mathrm{C}$ is [Given that $R=8.31 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{K}^{-1}$ ]",
        "input": "(a) $350 \\mathrm{~J}$ (b) $700 \\mathrm{~J}$ (c) $748 \\mathrm{~J}$ (d) 374 J",
        "answer": "(c) 748 J",
        "solution": "Number of moles of gas, $\\mathrm{n}=(67.2 / 22.4)=3 \\mathrm{~mol}$ $\\Delta \\mathrm{Q}=\\mathrm{nC}_{\\mathrm{v}} \\Delta \\mathrm{T}=3 \times(3 / 2) \\mathrm{R} \times \\Delta \\mathrm{T}=3 \times(3 / 2) \times(8.31) \times 20=747.9=748 \\mathrm{~J}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q4: $200 \\mathrm{~g}$ water is heated from $40^{\\circ} \\mathrm{C}$ to $60^{\\circ} \\mathrm{C}$. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water $=4184 \\mathrm{~J} / \\mathrm{kg} / \\mathrm{K}$ )",
        "input": "(a) $167.4 \\mathrm{~kJ}$ (b) $8.4 \\mathrm{~kJ}$ (c) $4.2 \\mathrm{~kJ}$ (d) $16.7 \\mathrm{~kJ}$",
        "answer": "(d) 16.7 kJ",
        "solution": "For isochoric process, $\\Delta \\mathrm{U}=\\mathrm{Q}=\\mathrm{ms} \\Delta \\mathrm{T}$ Here, $\\mathrm{m}=200 \\mathrm{~g}=0.2 \\mathrm{~kg}, \\mathrm{~s}=4184 \\mathrm{~J} / \\mathrm{kg} / \\mathrm{K}$ $\\Delta \\mathrm{T}=60^{\\circ} \\mathrm{C}-40^{\\circ} \\mathrm{C}=20^{\\circ} \\mathrm{C}$ $\\Delta \\mathrm{U}=0.2 \times 4184 \times 20=16736 \\mathrm{~J}=16.7 \\mathrm{~kJ}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q5: A diatomic gas with rigid molecules does $10 \\mathrm{~J}$ of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?",
        "input": "(a) $25 \\mathrm{~J}$ (b) $30 \\mathrm{~J}$ (c) $35 \\mathrm{~J}$ (d) $40 \\mathrm{~J}$",
        "answer": "(c) 35 J",
        "solution": "Given that the process is isobaric. Therefore, heat energy absorbed by the gas is $\\Delta \\mathrm{Q}=\\mathrm{nC}_{\\mathrm{P}} \\mathrm{T} \\ldots(1)$ Also, work done by the gas is $\\mathrm{W}=\\mathrm{nRT}=10 \\mathrm{~J}$ (given). Since, $C_{P}=(7 / 2) R$ for a diatomic gas $\\Delta \\mathrm{Q}=\\mathrm{n}(7 / 2) \\mathrm{R} \\Delta \\mathrm{T}$ (Using 1) $\\Delta Q=(7 / 2) n R \\Delta T=(7 / 2) \\times 10$ (Using 2) $\\Delta \\mathrm{Q}=35 \\mathrm{~J}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q8: A heat source at $T=103 \\mathrm{~K}$ is connected to another heat reservoir at $T=102 \\mathrm{~K}$ by a copper slab that is $1 \\mathrm{~mm}$ thick. Given that the thermal conductivity of copper is $0.1 \\mathrm{WK}-1 \\mathrm{~m}-1$, the energy flux through it in the steady state is",
        "input": "(1) $90 \\mathrm{Wm}^{-2}$ (2) $120 \\mathrm{Wm}^{-2}$ (3) $65 \\mathrm{Wm}^{-2}$ (4) $200 \\mathrm{Wm}^{-2}$",
        "answer": "(1) $90 \\mathbf{Wm}^{-2}$",
        "solution": "\\\\section*\\{Temp. of \\\\ heat source $(\\mathrm{dQ} / \\mathrm{dt})=(\\mathrm{kA} \\Delta \\mathrm{T}) / \\mathrm{l}$ Energy flux, $\\frac{1}{A}\\left(\\frac{d Q}{d t}\\right)=\\frac{k \\Delta T}{l}$ $=(0.1)(900) / 1=90 \\mathrm{~W} / \\mathrm{m}^{2}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q10: A cylinder of radius $R$ is surrounded by a cylinder shell of inner radius $2 R$. The thermal conductivity of the material of the cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is",
        "input": "(1) $\\left(\\mathrm{K}{1}+\\mathrm{K}{2}\right) / 2$ (2) $\\mathrm{K}{1}+\\mathrm{K}{2}$ (3) $\\left(2 \\mathrm{~K}{1}+3 \\mathrm{~K}{2}\right) / 5$ (4) $\\left(\\mathrm{K}{1}+3 \\mathrm{~K}{2}\right) / 4$",
        "answer": "(4) $\\left(\\mathrm{K}{1}+3 \\mathrm{~K}{2}\right) / 4$",
        "solution": "$\\mathrm{K}{\\mathrm{eq}}=\\left(\\mathrm{K}{1} \\mathrm{A}{1}+\\mathrm{K}{2} \\mathrm{A}{2}\right) /\\left(\\mathrm{A}{1}+\\mathrm{A}{2}\right)$ $\\mathrm{K}{\\mathrm{eq}}=\\mathrm{K}{1} \\pi \\mathrm{R}^{2}+\\mathrm{K}{2}\\left[\\pi(2 \\mathrm{R})^{2}-\\pi \\mathrm{R}^{2}\right] / \\pi(2 \\mathrm{R})^{2}$ $=\\mathrm{K}{1} \\pi \\mathrm{R}^{2}+\\mathrm{K}{2}\\left[3 \\pi \\mathrm{R}^{2}\right] / \\pi 4 \\mathrm{R}^{2}$ $=\\left(\\mathrm{K}{1}+3 \\mathrm{~K}{2}\right) / 4$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q11: Heat given to a body which raises its temperature by $1^{\\circ} \\mathrm{C}$ is",
        "input": "(a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient",
        "answer": "(b) thermal capacity",
        "solution": "",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q12: The temperature of the two outer surfaces of a composite slab, consisting of two materials having\\ coefficients of thermal conductivity $K$ and $2 K$ and thickness $x$ and $4 x$, respectively are $T_{2}$ and $T_{1}\\left(T_{2}>T_{1}\right)$. The rate of heat transfer through the slab, in a steady-state, is $\\left[A\\left(T_{2}-T_{1}\right) K / x\right] f$, with $f$ equal to",
        "input": "(1) 1 (2) $1 / 2$ (3) $2 / 3$ (4) $1 / 3$",
        "answer": "(4) $1 / 3$",
        "solution": "For the first surface, $\\mathrm{Q}{1}=\\mathrm{KA}\\left(\\mathrm{T}{2}-\\mathrm{T}\right) \\mathrm{t} / \\mathrm{x}$ For second surface, $\\mathrm{Q}{2}=(2 \\mathrm{~K}) \\mathrm{A}\\left(\\mathrm{T}-\\mathrm{T}{1}\right) \\mathrm{t} /(4 \\mathrm{x})$ At steady state, $\\mathrm{Q}{2}=\\mathrm{Q}{1} \\Rightarrow \\mathrm{KA}\\left(\\mathrm{T}{2}-\\mathrm{T}\right) \\mathrm{t} / \\mathrm{x}=(2 \\mathrm{~K}) \\mathrm{A}\\left(\\mathrm{T}-\\mathrm{T}{1}\right) \\mathrm{t} /(4 \\mathrm{x})$ Or $2\\left(\\mathrm{T}{2}-\\mathrm{T}\right)=\\left(\\mathrm{T}-\\mathrm{T}{1}\right)$ $\\mathrm{T}=\\left(2 \\mathrm{T}{2}+\\mathrm{T}{1}\right) / 3$ $\\mathrm{Q}{1}=\\mathrm{KA}\\left(\\mathrm{T}{2}-\\left[\\left(2 \\mathrm{T}{2}+\\mathrm{T}{1}\right) / 3\right]\right) t / \\mathrm{x}$ $\\mathrm{H}=\\mathrm{Q}{1} / \\mathrm{t}=\\mathrm{KA}\\left(\\mathrm{T}{2}-\\left[\\left(2 \\mathrm{T}{2}+\\mathrm{T}{1}\right) / 3\right]\right) / \\mathrm{x}$\\ $=\\mathrm{KA}\\left(\\mathrm{T}{2-} \\mathrm{T}{1}\right) / 3 \\mathrm{x}$ $\\left[\frac{A\\left(T_{2}-T_{1}\right) K}{x}\right] f=\frac{K A}{x}\\left[\frac{T_{2}-T_{1}}{3}\right]$ $\\mathrm{f}=1 / 3$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q13: According to Newton's law of cooling, the rate of cooling of a body is proportional to $(\\Delta \\Theta)^{\text {n }}$, where is the difference of the temperature of the body and the surroundings, and $\\mathbf{n}$ is equal to",
        "input": "(a) two (b) three (c) four (d) one",
        "answer": "(d) one",
        "solution": "According to Newton's law of cooling, the rate of cooling is proportional to $\\Delta \\Theta(\\Delta \\Theta)^{n}=(\\Delta \\Theta)$ or $n=1$.",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q14: When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is",
        "input": "(a) $2 / 5$ (b) $3 / 5$ (c) $3 / 7$ (d) $5 / 7$",
        "answer": "(d) $5 / 7$",
        "solution": "$\\Delta \\mathrm{U}=\\mathrm{nC}{\\mathrm{v}} \\Delta \\mathrm{T}$ $\\Delta \\mathrm{Q}=\\mathrm{nC}{\\mathrm{p}} \\Delta \\mathrm{T}$ Therefore, $\\Delta \\mathrm{U} / \\Delta \\mathrm{Q}=\\mathrm{nC}{\\mathrm{v}} \\Delta \\mathrm{T} / \\mathrm{nC}{\\mathrm{p}} \\Delta \\mathrm{T}=\\mathrm{C}{\\mathrm{v}} / \\mathrm{C}{\\mathrm{p}}=1 / \\gamma=5 / 7$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q15: 70 calories of heat are required to raise the temperature of 2 moles of an ideal gas at constant pressure from $30^{\\circ} \\mathrm{C}$ to $35^{\\circ} \\mathrm{C}$. The amount of heat required (in calorie) to raise the temperature of the same gas through the same range $\\left(30^{\\circ} \\mathrm{C}\\ text { to } \\left.35^{\\circ} \\mathrm{C}\right)$ at constant volume is",
        "input": "(1) 30 (2) 50 (3) 70 (4) 90",
        "answer": "(2) 50",
        "solution": "$\\mathrm{Q}=\\mathrm{nC}{\\mathrm{p}} \\mathrm{dT}$ $\\mathrm{C}{\\mathrm{p}}=\\mathrm{Q} / \\mathrm{ndT}=70 / 2 \\mathrm{x}(35-30)$ $\\mathrm{C}{\\mathrm{p}}=70 /(2 \times 5)$ $\\mathrm{C}{\\mathrm{p}}=7 \\mathrm{cal} / \\mathrm{mol} \times \\mathrm{K}$ Now, $\\mathrm{C}{\\mathrm{v}}=\\mathrm{C}{\\mathrm{p}}-\\mathrm{R}$ $\\mathrm{C}{\\mathrm{v}}=7-2=5 \\mathrm{cal} / \\mathrm{mol} \times \\mathrm{K}$ $\\mathrm{Q}^{\\prime}=\\mathrm{nC}{\\mathrm{v}} \\mathrm{dT}=2 \times 5 \times 5=50 \\mathrm{cal}$",
        "topic": "Heat Transfer"
    },
    {
        "question": "Q1: A wave $y=a \\sin (\\omega t-k x)$ on a string meets with another wave producing a node at $x=0$. Then the equation of the unknown wave is",
        "input": "(a) $y=\\operatorname{asin}(\\omega t+k x)$\n\n(b) $y=-\\operatorname{asin}(\\omega t+k x)$\n\n(c) $\\mathrm{y}=\\operatorname{asin}(\\omega \\mathrm{t}-\\mathrm{kx})$",
        "answer": "(b) $y=-\\operatorname{asin}(\\omega t+k x)$",
        "solution": "Consider option (a)\n\nStationary wave: $\\mathrm{Y}=\\operatorname{asin}(\\omega \\mathrm{t}+\\mathrm{kx})+\\operatorname{asin}(\\omega \\mathrm{t}-\\mathrm{kx})$\n\nwhen, $\\mathrm{x}=0, \\mathrm{Y}$ is not zero.\n\nThe option is not acceptable.\n\nConsider option (b)\n\nStationary wave: $\\mathrm{Y}=\\operatorname{asin}(\\omega \\mathrm{t}-\\mathrm{kx})-\\operatorname{asin}(\\omega \\mathrm{t}+\\mathrm{kx})$ At $\\mathrm{x}=0, \\mathrm{Y}=\\mathrm{a} \\sin \\omega \\mathrm{t}-\\mathrm{a} \\sin \\omega \\mathrm{t}=\\mathrm{zero}$\n\nThis option holds good.\n\nOption (c) gives $\\mathrm{Y}=\\operatorname{asin}(\\omega \\mathrm{t}-\\mathrm{kx})+\\operatorname{asin}(\\omega \\mathrm{t}-\\mathrm{kx})$\n\n$\\mathrm{Y}=2 \\mathrm{asin} \\omega \\mathrm{t}($ At $\\mathrm{x}=0, \\mathrm{Y}$ is not zero)\n\nHence only option (b) holds good",
        "topic": "Waves on String"
    },
    {
        "question": "Q2: The displacement $y$ of a wave travelling in the $x$-direction is given by\n\n$y=10^{-4} \\sin (600 t-2 x+\\pi / 3)$ metre, where $x$ is expressed in metre and $t$ in second. The speed of the wave motion, in $\\mathrm{m} \\mathrm{s}^{-1}$ is",
        "input": "(a) 300\n\n(b) 600\n\n(c) 1200\n\n(d) 200",
        "answer": "(a) 300",
        "solution": "Given wave equation: $\\mathrm{y}=10^{-4} \\sin (600 \\mathrm{t}-2 \\mathrm{x}+\\pi / 3) \\mathrm{m}$\n\nStandard wave equation: $\\mathrm{y}=\\operatorname{asin}(\\omega \\mathrm{t}-\\mathrm{kx}+\\Phi)$\n\nCompare them\n\nAngular speed, $\\omega=600 \\sec ^{-1}$\n\nPropagation constant, $\\mathrm{k}=2 \\mathrm{~m}^{-1}$\n\n$\\omega / \\mathrm{k}=(2 \\pi \\mathrm{f}) /(2 \\pi / \\lambda)=\\mathrm{f} \\lambda=$ velocity\n\nSince velocity $=\\omega / \\mathrm{k}=600 / 2=300 \\mathrm{~m} / \\mathrm{sec}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q3: A string is stretched between fixed points separated by $75 \\mathrm{~cm}$. It is observed to have resonant frequencies of $420 \\mathrm{~Hz}$ and $315 \\mathrm{~Hz}$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is",
        "input": "(a) $10.5 \\mathrm{~Hz}$\n\n(b) $105 \\mathrm{~Hz}$\n\n(c) $1.05 \\mathrm{~Hz}$\n\n(d) $1050 \\mathrm{~Hz}$.",
        "answer": "(b) $105 \\mathrm{~Hz}$",
        "solution": "For the string fixed at both the ends, resonant frequency are given by $\\mathrm{f}=\\mathrm{nv} / 2 \\mathrm{~L}$, where symbols have their usual meanings. It is given that $315 \\mathrm{~Hz}$ and $420 \\mathrm{~Hz}$ are two consecutive resonant frequencies, let these be nth and ( $\\mathrm{n}+$ $1)$ the harmonics.\n\n$315=$ nv/2L--------(1)\n\n$420=(n+1) v / 2 L--(2)$\n\nDividing equa (1) by equa (2) we get\n\n$\\mathrm{n}=3$\n\nFrom equa (1), lowest resonant frequency\n\n$\\mathrm{f}_{0}=\\mathrm{v} / 2 \\mathrm{~L}=315 / 3=105 \\mathrm{~Hz}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q4: The transverse displacement $y(x, t)$ of a wave on a string is given by\n\n$$\ny(x, t)=e^{-\\left(a x^{2}+b t^{2}+2 \\sqrt{a b} x t\\right)}\n$$\n\n\\section*{This represents a}",
        "input": "(a) wave moving in $+\\mathrm{x}$-direction with speed $\\sqrt{\\frac{a}{b}}$\n\n(b) wave moving in $-\\mathrm{x}$-direction with speed $\\sqrt{\\frac{b}{a}}$\n\n(c) standing wave of frequency $\\sqrt{b}$\n\n(d) standing wave of frequency $1 / \\sqrt{ } b$",
        "answer": "(b) wave moving in $-\\mathrm{x}$-direction with speed $\\sqrt{\\frac{b}{a}}$",
        "solution": "Given\n\n$$\n\\begin{aligned}\n& y(x, t)=e^{-\\left(a x^{2}+b t^{2}+2 \\sqrt{a b} x t\\right)} \\end{aligned}\n$$\n\nComparing equation (1) with standard equation $y(x, t)=f(a x+b t)$\n\nAs there is a positive sign between $\\mathrm{x}$ and $\\mathrm{t}$ terms, hence wave travel in $-\\mathrm{x}$ direction.\n\nWave speed $=$ Coefficient of $\\mathrm{t} /$ Coefficient of $\\mathrm{x}=\\sqrt{\\frac{b}{a}}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q5: The equation of a wave on a string of linear mass density $0.04 \\mathrm{~kg} \\mathrm{~m}^{-1}$ is given by\n\n$$\ny=0.02(m) \\sin 2 \\pi\\left(\\frac{t}{0.04(s)}-\\frac{x}{0.50(m)}\\right)\n$$\n\nThe tension in the string is",
        "input": "(a) $6.25 \\mathrm{~N}$\n\n(b) $4.0 \\mathrm{~N}$\n\n(c) $12.5 \\mathrm{~N}$\n\n(d) $0.5 \\mathrm{~N}$",
        "answer": "(a) $6.25 \\mathrm{~N}$",
        "solution": "Here, linear mass density $\\mu=0.04 \\mathrm{~kg} \\mathrm{~m}^{-1}$\n\nThe given equation of a wave is\n\n$$\ny=0.02(m) \\sin \\left[2 \\pi\\left(\\frac{t}{0.04}-\\frac{x}{0.50}\\right)\\right]\n$$\n\nCompare it with the standard wave equation $\\mathrm{y}=\\mathrm{A} \\sin (\\omega \\mathrm{t}-\\mathrm{kx})$\n\nwe get, $\\omega=2 \\pi / 0.04 \\mathrm{rad} \\mathrm{s}^{-1}, \\mathrm{k}=2 \\pi / 0.5 \\mathrm{rad} \\mathrm{m}^{-1}$\n\nWave velocity, $\\mathrm{v}=\\omega / \\mathrm{k}=(2 \\pi / 0.04) /(2 \\pi / 0.5) \\mathrm{ms}^{-1}$\n\n$v=\\sqrt{\\frac{T}{\\mu}}$\n\nAlso,\n\n\nHere, $\\mathrm{T}$ is the tension in the string and $\\mu$ is the linear mass density\n\nEquating equations (1) and (2), we get\n\n$$\n\\frac{\\omega}{k}=\\sqrt{\\frac{T}{\\mu}}\n$$\n\n$\\mathrm{T}=\\mu \\omega^{2} / \\mathrm{k}^{2}$\n\n$\\mathrm{T}=\\left[0.04 \\times(2 \\pi / 0.04)^{2}\\right] /(2 \\pi / 0.5)^{2}=6.25 \\mathrm{~N}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q6: A wave travelling along the $x$-axis is described by the equation $y(x, t)=0.005 \\cos (\\alpha x-\\beta t)$. If the wavelength and the time period of the wave are $0.08 \\mathrm{~m}$ and $2.0 \\mathrm{~s}$, respectively, then $\\alpha$ and $\\beta$ in appropriate units are",
        "input": "(a) \\alpha=12.50 \\pi, \\beta=\\pi / 2.0\n(b) \\alpha=25.00 \\pi, \\beta=\\pi\n(c) \\alpha=0.08 / \\pi, \\beta=2.0 / \\pi\n(d) \\alpha=0.04 / \\pi, \\beta=1.0 / \\pi",
        "answer": "(b) \\alpha=25.00 \\pi, \\beta=\\pi",
        "solution": "The wave travelling along the $\\mathrm{x}$-axis is given by\n\ny(x, t)=0.005 \\cos (\\alpha x-\\beta t)\n\nTherefore, $\\alpha=\\mathrm{k}=2 \\pi / \\lambda$ As $\\lambda=0.08 \\mathrm{~m}$\n$\\alpha=2 \\pi / 0.08=\\pi / 0.04 \\Rightarrow \\alpha=(\\pi / 4) \\times 100=25.00 \\pi$\n\n$\\omega=\\beta \\Rightarrow 2 \\pi / 2=\\pi$\n\n$\\therefore \\alpha=25.00 \\pi$\n\n$\\beta=\\pi$",
        "topic": "Waves on String"
    },
    {
        "question": "Q7: A uniform string of length $\\mathbf{2 0} \\mathrm{m}$ is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (take $\\mathbf{g}=\\mathbf{1 0} \\mathbf{m ~ s}^{-2}$ )",
        "input": "(a) $5 \\pi \\sqrt{ } 5 \\mathrm{~s}$\n(b) $2 \\mathrm{~s}$\n(c) $2 \\sqrt{ } 2 \\mathrm{~s}$\n(d) $\\sqrt{ } 5$",
        "answer": "(c) $2 \\sqrt{2} s$",
        "solution": "A uniform string of length $20 \\mathrm{~m}$ is suspended from a rigid support.\n\nAs $\\mu$ is mass per unit length of the rope, then $\\mu=\\mathrm{m} / \\mathrm{L}$\n$v=\\sqrt{\\frac{T}{\\mu}}$\n\n$\\frac{d x}{d t}=\\sqrt{\\frac{m g x / L}{m / L}}=\\sqrt{g} \\sqrt{x}$\n\n$\\frac{d x}{\\sqrt{x}}=\\sqrt{g} d t$\n\nIntegrating, on both sides, we get\n\n$\\int_{0}^{L} x^{-1 / 2} d x=\\sqrt{g} \\int_{0}^{L} d t$\n\n$t=2 \\sqrt{L} / \\sqrt{g}=2 \\times \\sqrt{\\frac{20}{10}}=2 \\sqrt{2} s$",
        "topic": "Waves on String"
    },
    {
        "question": "Q8: A sonometer wire of length $1.5 \\mathrm{~m}$ is made of steel. The tension in it produces an elastic strain of $1 \\%$. What is the fundamental frequency of steel if the density and elasticity of steel are $7.7 \\times 10^{3} \\mathbf{~ k g} / \\mathrm{m}^{3}$ and $2.2 \\times 10^{11} \\mathrm{~N} / \\mathrm{m}^{2}$ respectively?",
        "input": "(a) $770 \\mathrm{~Hz}$\n(b) $188.5 \\mathrm{~Hz}$\n(c) $178.2 \\mathrm{~Hz}$\n(d) 200.5",
        "answer": "(c) $178.2 \\mathrm{~Hz}$",
        "solution": "Frequency, $\\mathrm{f}=\\mathrm{v} / 21$\n\n$=\\frac{1}{2 I} \\sqrt{\\frac{T}{\\mu}}=\\frac{1}{2 I} \\sqrt{\\frac{T}{A d}}$\n\nWhere, $\\mathrm{T}$ is tension in the wire and $\\mu$ is the mass per unit length of wire.\n\nAlso, Young's modulus, $\\mathrm{Y}=\\mathrm{Tl} / \\mathrm{A} \\Delta 1$\n\n$\\mathrm{T} / \\mathrm{A}=\\mathrm{Y} \\Delta \\mathrm{I} / 1$\n\nSubstituting this value in frequency expression, we get\n\n$f=\\frac{1}{2 l} \\sqrt{\\frac{y \\Delta l}{l d}}$\n\nGiven, $1=1.5 \\mathrm{~m}, \\Delta \\mathrm{l} / 1=0.01$\n\n$\\mathrm{d}=7.7 \\times 10^{3} \\mathrm{~kg} / \\mathrm{m}^{3}$\n\n$\\mathrm{Y}=2.2 \\times 10^{11} \\mathrm{~N} / \\mathrm{m}^{2}$\n\nSubstituting these values we have\n\n$f=\\frac{1}{2 l} \\sqrt{\\frac{2.2 \\times 10^{11} \\times 0.01}{7.7 \\times 10^{3}}}$\n\n$\\mathrm{f}=178.2 \\mathrm{~Hz}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q9: Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats of frequency $5 \\mathrm{~Hz}$. The tension of the string $B$ is slightly increased and the beat frequency is found to decrease by $3 \\mathrm{~Hz}$. If the frequency of $A$ is $425 \\mathrm{~Hz}$, the original frequency of $B$ is",
        "input": "(a) $428 \\mathrm{~Hz}$\n(b) $430 \\mathrm{~Hz}$\n(c) $420 \\mathrm{~Hz}$\n(d) $422 \\mathrm{~Hz}$",
        "answer": "(c) $420 \\mathrm{~Hz}$",
        "solution": "Frequency of sitar string B is either $420 \\mathrm{~Hz}$ or $430 \\mathrm{~Hz}$. As tension in string B is increased, its frequency will increase. If the frequency is $430 \\mathrm{~Hz}$, the beat frequency will increase. If the frequency is $420 \\mathrm{~Hz}$, the beat frequency will decrease; hence the correct answer is $420 \\mathrm{~Hz}$.",
        "topic": "Waves on String"
    },
    {
        "question": "Q10: A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by $y(x, t)=0.5 \\sin [(5 \\pi / 4) x] \\cos (200 \\pi t)$. What is the speed of the travelling wave moving in the positive $x$-direction? ( $x$ and $t$ are in meter and second, respectively)",
        "input": "(a) $180 \\mathrm{~m} / \\mathrm{s}$\n(b) $160 \\mathrm{~m} / \\mathrm{s}$\n(c) $120 \\mathrm{~m} / \\mathrm{s}$\n(d) $90 \\mathrm{~m} / \\mathrm{s}$",
        "answer": "(b) $160 \\mathrm{~m} / \\mathrm{s}$",
        "solution": "Given $\\mathrm{y}(\\mathrm{x}, \\mathrm{t})=0.5 \\sin [(5 \\pi / 4) \\mathrm{x}] \\cos (200 \\pi \\mathrm{t})$\n\nComparing this equation with the standard equation of a standing wave, $\\mathrm{y}(\\mathrm{x}, \\mathrm{t})=2 \\mathrm{a} \\sin \\mathrm{kx} \\cos \\omega \\mathrm{t}$, we get, $\\mathrm{k}=5 \\pi / 4$ $\\mathrm{rad} / \\mathrm{m}$, $\\omega=200 \\mathrm{rad} / \\mathrm{s}$\n\nSpeed of the travelling wave, $v=\\omega / \\mathrm{k}=200 \\pi /(5 \\pi / 4)=160 \\mathrm{~m} / \\mathrm{s}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q11: Length of a string tied to two rigid supports is $40 \\mathrm{~cm}$. Maximum length (wavelength in $\\mathrm{cm}$ ) of a stationary wave produced on it is",
        "input": "(a) 20\n(b) 80\n(c) 40\n(d) 120",
        "answer": "(b) 80",
        "solution": "$\\lambda_{\\max } / 2=40 \\Rightarrow \\lambda_{\\max }=80 \\mathrm{~cm}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q12: Statement-1: Two longitudinal waves given by equations $y_{1}(x, t)=2 a \\sin (\\omega t-k x)$ and $y_{2}(x, t)=2 \\operatorname{asin}(2 \\omega t-2 \\mathrm{kx}$ ) will have equal intensity.\n\nStatement-2: Intensity of waves of given frequency in the same medium is proportional to square of amplitude only.",
        "input": "(a) Statement-1 is false, statement-2 is true\n(b) Statement-1 is true, statement-2 is false\n(c) Statement-1 is true, statement-2 true; statement-2 is the correct explanation of statement-1\n(d) Statement-1 is true, statement-2 is true; statement-2 is not correct explanation of statement-1",
        "answer": "(b) Statement-1 is true, statement-2 is false",
        "solution": "$\\mathrm{y}_{1}(\\mathrm{x}, \\mathrm{t})=2 \\mathrm{asin}(\\omega \\mathrm{t}-\\mathrm{kx})$\n\n$\\mathrm{y}_{2}(\\mathrm{x}, \\mathrm{t})=2 \\mathrm{asin}(2 \\omega \\mathrm{t}-2 \\mathrm{kx})$\n\nBut Intensity, $I=1 / 2\\left(\\rho \\omega^{2} \\mathrm{~A}^{2} v\\right)$\n\nHere, $\\rho=$ density of the medium,\n\n$\\mathrm{A}=$ amplitude\n\n$\\mathrm{v}=$ velocity of the wave\n\nIntensity depends upon amplitude, frequency, and velocity of the wave.\n\nAlso, $\\mathrm{I}_{1}=\\mathrm{I}_{2}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q13: A pipe open at both ends has a fundamental frequency ' $f$ ' in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now",
        "input": "(a) $\\mathrm{f}$\n(b) $\\mathrm{f} / 2$\n(c) $3 \\mathrm{f} / 4$\n(d) $2 \\mathrm{f}$",
        "answer": "(a) f",
        "solution": "When the pipe is open at both ends\n\n$\\lambda / 2=1$\n\n$\\lambda=21$\n\n$\\mathrm{v}=\\mathrm{f} \\lambda$\n\n$\\mathrm{f}=\\mathrm{v} / \\lambda=\\mathrm{v} / 21$\n\nWhen the pipe is dipped vertically in the water so that half of the pipe is in water\n\n$\\lambda / 4=1 / 2$\n\n$\\lambda=21 \\Rightarrow \\mathrm{v}=\\mathrm{f}^{\\prime} \\lambda$\n\n$\\mathrm{f}^{\\prime}=\\mathrm{v} / \\lambda=\\mathrm{v} / 21=\\mathrm{f}--$\n\nThus, the fundamental frequency of the air column is now,\n\n$\\mathrm{f}=\\mathrm{f}^{\\prime}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q14: A string of length $1 \\mathrm{~m}$ and mass $5 \\mathrm{~g}$ is fixed at both ends. The tension in the string is $8.0 \\mathrm{~N}$. The string is set into vibration using an external vibrator of frequency $100 \\mathrm{~Hz}$. The separation between successive nodes on the string is close to",
        "input": "(a) $10 \\mathrm{~cm}$\n(b) $33.3 \\mathrm{~cm}$\n(c) $16.6 \\mathrm{~cm}$\n(d) $20.0 \\mathrm{~cm}$",
        "answer": "(d) $20.0 \\mathrm{~cm}$",
        "solution": "Velocity of the wave on the string\n\n$$\n\\begin{aligned}\n& V= \n& \\sqrt{\\frac{t}{\\mu}} \n& = \n& \\sqrt{\\frac{8}{5} \\times 1000}\n\\end{aligned}\n$$\n\n$\\mathrm{V}=40 \\mathrm{~m} / \\mathrm{s}$\n\nHere, $\\mathrm{T}=$ tension and $\\mu=$ mass/length\n\nWavelength of the wave $\\lambda=\\mathrm{v} / \\mathrm{n}=40 / 100$\n\nSeparation between successive nodes,\n\n$\\lambda / 2=40 /(2 \\mathrm{x} 100)=20 \\mathrm{~cm}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q15: Equation of a travelling wave on a stretched string of linear density $5 \\mathrm{~g} / \\mathrm{m}$ is $y=0.03 \\sin (450 t-9 x)$ where distance and time are measured in SI units. The tension in the string is",
        "input": "(a) $10 \\mathrm{~N}$\n(b) $7.5 \\mathrm{~N}$\n(c) $12.5 \\mathrm{~N}$\n(d) $5 \\mathrm{~N}$",
        "answer": "(c) $12.5 \\mathrm{~N}$",
        "solution": "We have given,\n\ny=0.03 \\sin (450 t-9 x)\n\nComparing it with standard equation of wave, we get\n\n\\omega=450 \\mathrm{k}=9\n\n\\mathrm{v}=\\omega / \\mathrm{k}=450 / 9=50 \\mathrm{~m} / \\mathrm{s}\n\nVelocity of the travelling wave on a string is given by\n\n$$\n\\begin{aligned}\n& \\mathrm{v}= \n& \\sqrt{\\frac{t}{\\mu}}\n\\end{aligned}\n$$\n\n$$\n50=\\sqrt{\\frac{T}{50 \\times 10^{-3}}}\n$$\n\n\\mu=$ linear mass density\n\n\\mathrm{T}=2500 \\times 5 \\times 10^{-3}\n\n\\mathrm{T}=12.5 \\mathrm{~N}$",
        "topic": "Waves on String"
    },
    {
        "question": "Q1: A series AC circuit containing an inductor ( $20 \\mathrm{mH}$ ), a capacitor ( $120 \\mu \\mathrm{F}$ ) and a resistor ( $60 \\Omega$ ) is driven by an AC source of $24 \\mathrm{~V} / 50 \\mathrm{~Hz}$. The energy dissipated in the circuit in $60 \\mathrm{~s}$ is",
        "input": "(a) $3.39 \\times 10^{3} \\mathrm{~J}$\n(b) $5.65 \\times 10^{2} \\mathrm{~J}$\n(c) $2.26 \\times 10^{3} \\mathrm{~J}$\n(d) $5.17 \\times 10^{2} \\mathrm{~J}$",
        "answer": "(d) $5.17 \\times 10^{2} \\mathrm{~J}$",
        "solution": "Impedance, $\\mathrm{Z}=\\left(\\mathrm{R}^{2}+\\left(\\mathrm{X}_{\\mathrm{C}}-\\mathrm{X}_{\\mathrm{L}}\\right)^{2}\\right)^{1 / 2}$\n$\\mathrm{X}_{\\mathrm{L}}=\\omega \\mathrm{L}=(2 \\pi \\mathrm{vL})$\n$\\mathrm{X}_{\\mathrm{L}}=6.28 \\times 50 \\times 20 \\times 10^{-3}=6.28 \\Omega$\n$\\mathrm{X}_{\\mathrm{C}}=(1 / \\omega \\mathrm{C})=1 /(2 \\pi \\mathrm{vC})=1 /\\left(6.28 \\times 120 \\times 10^{-6} \\times 50\\right)=26.54 \\Omega$\n$\\mathrm{X}_{\\mathrm{C}}-\\mathrm{X}_{\\mathrm{L}}=26.54-6.28=20.26$\n$\\mathrm{Z}=\\left((60)^{2}+(20.26)^{2}\\right)^{1 / 2}$\n$\\mathrm{Z}^{2}=4010 \\Omega^{2}$\nAverage power dissipated, $\\operatorname{Pav}=\\varepsilon_{\\mathrm{rms}} \\mathrm{I}_{\\mathrm{rms}} \\cos \\Phi$\n$\\mathrm{P}_{\\mathrm{av}}=\\varepsilon_{\\text {rms }} \\times\\left(\\varepsilon_{\\text {rms }} / \\mathrm{Z}\\right) \\times(\\mathrm{R} / \\mathrm{Z})$\n$\\mathrm{P}_{\\mathrm{av}}=\\left(\\varepsilon_{\\text {rms }}{ }^{2} / \\mathrm{Z}^{2}\\right) \\times \\mathrm{R}=\\left[(24)^{2} / 4010\\right] \\times 60 \\mathrm{~W}=8.62 \\mathrm{~W}$\nEnergy dissipated in $60 \\mathrm{~S}=8.62 \\times 60=5.17 \\times 10^{2} \\mathrm{~J}$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q2: In an AC generator, a coil with \\(\\mathbf{N}\\) turns, all of the same area \\(A\\) and total resistance \\(R\\), rotates with frequency \\(\\omega\\) in a magnetic field \\(B\\). The maximum value of emf generated in the coil is",
        "input": "(a) NAB\n(b) NABR\n(c) NAB \\omega\n(d) NABRc",
        "answer": "(c) In an a.c. generator, maximum emf $=\\mathbf{N A B} \\omega$",
        "solution": "",
        "topic": "Alternating Current"
    },
    {
        "question": "Q3: The phase difference between the alternating current and emf is \\(\\pi / 2\\). Which of the following cannot be the constituent of the circuit?",
        "input": "(a) LC\n(b) L alone\n(c) C alone\n(d) R, L",
        "answer": "(d) R, L",
        "solution": "$\\mathrm{R}$ and $\\mathrm{L}$ cause phase difference to lie between 0 and $\\pi / 2$ but never 0 and $\\pi / 2$ at extremities",
        "topic": "Alternating Current"
    },
    {
        "question": "Q4: Alternating current cannot be measured by D.C ammeter because",
        "input": "(a) A.C cannot pass through D.C ammeter\n(b) A.C changes direction\n(c) The average value of current for the complete cycle is zero\n(d) D.C. ammeter will get damaged",
        "answer": "(c) The average value of current for the complete cycle is zero",
        "solution": "The average value of A.C for the complete cycle is zero. Hence A.C cannot be measured by D.C ammeter",
        "topic": "Alternating Current"
    },
    {
        "question": "Q5: A power transmission line feeds input power at $2300 \\mathrm{~V}$ to a step-down transformer with its primary windings having 4000 turns. The output power is delivered at $230 \\mathrm{~V}$ by the transformer. If the current in the primary of the transformer is $5 \\mathrm{~A}$ and its efficiency is $90 \\%$, the output current would be",
        "input": "(a) $25 \\mathrm{~A}$\n(b) $50 \\mathrm{~A}$\n(c) $45 \\mathrm{~A}$\n(d) $35 \\mathrm{~A}$",
        "answer": "(c) 45 A",
        "solution": "Given $\\epsilon_{\\mathrm{p}}=2300 \\mathrm{~V}, \\mathrm{~N}_{\\mathrm{p}}=4000$\n$\\epsilon_{\\mathrm{s}}=230 \\mathrm{~V}$,\n$\\mathrm{I}_{\\mathrm{p}}=5 \\mathrm{~A}$,\n$\\eta=90 \\%=0.9$\n$\\eta=\\mathrm{P}_{\\mathrm{o}} / \\mathrm{P}_{\\mathrm{i}}=\\left(\\epsilon_{\\mathrm{s}} \\mathrm{I}_{\\mathrm{s}}\\right) /\\left(\\epsilon_{\\mathrm{p}} \\mathrm{I}_{\\mathrm{p}}\\right)$\n$\\mathrm{I}_{\\mathrm{s}}=\\eta \\epsilon_{\\mathrm{p}} \\mathrm{I}_{\\mathrm{p}} / \\epsilon_{\\mathrm{s}}=(0.9 \\times 2300 \\times 5) / 230=45 \\mathrm{~A}$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q6: A circuit connected to an ac source of emf $e=e_{0} \\sin (100 t)$ with $t$ in seconds, gives a phase difference of $\\pi / 4$ between the emf $\\mathrm{e}$ and current $\\mathrm{I}$. Which of the following circuits will exhibit this?",
        "input": "(a) $\\mathrm{RC}$ circuit with $\\mathrm{R}=1 \\mathrm{k} \\Omega$ and $\\mathrm{C}=10 \\mu \\mathrm{F}$\n(b) $\\mathrm{RL}$ circuit with $\\mathrm{R}=1 \\mathrm{k} \\Omega$ and $\\mathrm{L}=10 \\mathrm{mH}$\n(c) $\\mathrm{RC}$ circuit with $\\mathrm{R}=1 \\mathrm{k} \\Omega$ and $\\mathrm{C}=1 \\mu \\mathrm{F}$\n(d) $\\mathrm{RL}$ circuit with $\\mathrm{R}=1 \\mathrm{k} \\Omega$ and $\\mathrm{L}=1 \\mathrm{mH}$",
        "answer": "(a) $RC$ circuit with $R=1 \\mathrm{k} \\Omega$ and $C=10 \\mu \\mathrm{F}$",
        "solution": "$\\mathrm{X}_{\\mathrm{c}}=\\mathrm{R}$\n$1 / \\omega \\mathrm{C}=\\mathrm{R}$\n$\\omega=100 \\mathrm{rad} / \\mathrm{s}$\nTherefore, $1 / 100=\\mathrm{RC}$\nSubstituting values from option (a)\n$\\mathrm{R}=1 \\mathrm{k} \\Omega=10^{3} \\Omega$\n$\\mathrm{C}=10 \\mu \\mathrm{F}=10 \\times 10^{-6} \\mathrm{~F}$\n$\\mathrm{RC}=10^{3} \\times 10 \\times 10^{-6}=1 / 100$\nTherefore, option (a) is correct",
        "topic": "Alternating Current"
    },
    {
        "question": "Q7: In an a.c. circuit, the instantaneous e.m.f. and current is given by $e=100 \\sin 30 t, i=20 \\sin (30 t-\\pi / 4)$. In one cycle of A.C, the average power consumed by the circuit and the wattless current are, respectively",
        "input": "(a) 50,10\n(b) $1000 / \\sqrt{2}, 10$\n(c) $50 / \\sqrt{2},0$\n(d) 50,0",
        "answer": "(b) 1000/ $\\sqrt{2}, 10$",
        "solution": "$\\mathrm{P}_{\\mathrm{avg}}=\\mathrm{V}_{\\mathrm{rms}} \\mathrm{I}_{\\mathrm{rms}} \\cos \\theta$\n$\\mathrm{P}_{\\text {avg }}=\\left(\\mathrm{V}_{0} / \\sqrt{2}\\right)\\left(\\mathrm{I}_{0} / \\sqrt{2}\\right) \\cos \\theta$\n$=(100 / \\sqrt{2})(20 / \\sqrt{2}) \\cos 45^{\\circ}$\n$P_{\\text {avg }}=1000 / \\sqrt{2}$ watt\nWattless current $=\\mathrm{I}_{\\mathrm{rms}} \\sin \\theta$\nWattless current $=\\left(\\mathrm{I}_{0} / \\sqrt{2}\\right) \\sin \\theta$\n$=(20 / \\sqrt{2}) \\sin 45^{0}$\n$=10 \\mathrm{amp}$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q8: A coil having $\\mathbf{n}$ turns and resistance $R \\Omega$ is connected with a galvanometer of resistance 4R $\\Omega$. This combination is moved in time $t$ seconds from a magnetic field $W_{1}$ weber to $W_{2}$ weber. The induced current in the circuit is",
        "input": "(a) $-\\left(\\mathrm{W}_{2}-\\mathrm{W}_{1}\\right) / 5 \\mathrm{Rnt}$\n(b) $-\\mathrm{n}\\left(\\mathrm{W}_{2}-\\mathrm{W}_{1}\\right) / 5 \\mathrm{Rt}$\n(c) $-\\left(\\mathrm{W}_{2}-\\mathrm{W}_{1}\\right) / \\mathrm{Rnt}$\n(d) $-\\left(\\mathrm{W}_{2}-\\mathrm{W}_{1}\\right) / 5 \\mathrm{Rnt}$",
        "answer": "(b) $-\\mathbf{n}\\left(\\mathbf{W}_{2}-\\mathbf{W}_{1}\\right) / 5 R t$",
        "solution": "The emf induced in the coil is $\\mathrm{e}=-\\mathrm{n}(\\mathrm{d} \\Phi / \\mathrm{dt})$\nInduced current, $\\mathrm{I}=\\mathrm{e} / \\mathrm{R}^{\\prime}=-\\left(\\mathrm{n} / \\mathrm{R}^{\\prime}\\right)(\\mathrm{d} \\Phi / \\mathrm{dt})$\nGiven, $\\mathrm{R}^{\\prime}=\\mathrm{R}+4 \\mathrm{R}=5 \\mathrm{R}$\n$\\mathrm{d} \\Phi=\\mathrm{W}_{2}-\\mathrm{W}_{1}$\n$\\mathrm{dt}=\\mathrm{t}$\n(here, $\\mathrm{W}_{1}$ and $\\mathrm{W}_{2}$ are flux associated with one turn)\nSubstituting the given values in equa(1) we get\n$\\mathrm{I}=(-\\mathrm{n} / 5 \\mathrm{R})(\\mathrm{W}_{2}-\\mathrm{W}_{1} / \\mathrm{t})$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q9: An alternating voltage $V(t)=220 \\sin 100 \\pi t$ volts is applied to a purely resistive load of $50 \\Omega$. The time taken for the current to rise from half of the peak value is",
        "input": "(a) $5 \\mathrm{~ms}$\n(b) $2.2 \\mathrm{~ms}$\n(c) $7.2 \\mathrm{~ms}$\n(d) $3.3 \\mathrm{~ms}$",
        "answer": "(d) $3.3 \\mathrm{~ms}$",
        "solution": "As $v(t)=220 \\sin 100 \\pi t$\nSo $\\mathrm{I}(\\mathrm{t})=(220 / 50) \\sin 100 \\pi \\mathrm{t}$\nI.e., $\\mathrm{I}=\\mathrm{I}_{\\mathrm{m}}=\\sin 100 \\pi \\mathrm{t}$\nFor $\\mathrm{I}=\\mathrm{I}_{\\mathrm{m}}$\n$\\mathrm{t}_{1}=(\\pi / 2) \\times(1 / 100 \\pi)=(1 / 200) \\mathrm{sec}$,\nAnd for $\\mathrm{I}=\\mathrm{I}_{\\mathrm{m}} / 2$\n$\\Rightarrow(\\mathrm{I}_{\\mathrm{m}} / 2)=\\mathrm{I}_{\\mathrm{m}} \\sin \\left(100 \\pi \\mathrm{t}_{2}\\right)$\n$\\Rightarrow(\\pi / 6)=100 \\pi \\mathrm{t}_{2}$\n$\\Rightarrow \\mathrm{t}_{2}=(1 / 600) \\mathrm{s}$\n$\\therefore \\mathrm{t}_{\\text {req }}=(1 / 200)-(1 / 600)=(2 / 600)=(1 / 300) \\mathrm{s}=3.3 \\mathrm{~ms}$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q10: An arc lamp requires a direct current of $10 \\mathrm{~A}$ at $80 \\mathrm{~V}$ to function. If it is connected to a $220 \\mathrm{~V}$ (RMS), 50 $\\mathrm{Hz}$ AC supply, the series inductor needed for it to work is close to",
        "input": "(a) $0.065 \\mathrm{H}$\n(b) $80 \\mathrm{H}$\n(c) $0.08 \\mathrm{H}$\n(d) $0.044 \\mathrm{H}$",
        "answer": "(a) $0.065 \\mathrm{H}$",
        "solution": "$\\mathrm{I}=10 \\mathrm{~A}$\n$\\mathrm{V}=80 \\mathrm{~V}$\n$\\mathrm{R}=8 \\Omega$\n$10=220 /\\left(8^{2}+\\mathrm{X}_{\\mathrm{L}}{ }^{2}\\right)^{1 / 2}$\n$64+\\mathrm{X}_{\\mathrm{L}}{ }^{2}=484$\n$X_{L}=\\sqrt{ } 420$\n$2 \\pi \\times 50 \\mathrm{~L}=\\sqrt{ } 420$\n$\\mathrm{L}=\\sqrt{ } 420 / 100 \\pi$\n$\\mathrm{L}=0.065 \\mathrm{H}$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q11: In a series, LCR circuit $R=200 \\Omega$ and the voltage and the frequency of the main supply is $220 \\mathrm{~V}$ and $50 \\mathrm{~Hz}$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $30^{\\circ}$. The power dissipated in the LCR circuit is",
        "input": "(a) $242 \\mathrm{~W}$\n(b) $305 \\mathrm{~W}$\n(c) $210 \\mathrm{~W}$\n(d) Zero",
        "answer": "(a) $242 \\mathrm{~W}$",
        "solution": "Tan $\\Phi=\\left(\\mathrm{X}_{\\mathrm{L}}-\\mathrm{X}_{\\mathrm{c}}\\right) / \\mathrm{R}$\n$\\operatorname{Tan} 30^{\\circ}=\\mathrm{X}_{\\mathrm{c}} / \\mathrm{R}=\\mathrm{X}_{\\mathrm{c}}=\\mathrm{R} / \\sqrt{3}$\n$\\operatorname{Tan} 30^{\\circ}=\\mathrm{X}_{\\mathrm{L}} / \\mathrm{R}=\\mathrm{X}_{\\mathrm{L}}=\\mathrm{R} / \\sqrt{3}$\n$\\mathrm{X}_{\\mathrm{L}}=\\mathrm{X}_{\\mathrm{c}} \\Rightarrow$ Condition for resonance\nSo $\\Phi=0^{0}$\\n$\\mathrm{P}=\\mathrm{VI} \\cos 0^{0}$\n$\\mathrm{P}=\\mathrm{V}^{2} / \\mathrm{R}=(220)^{2} / 200=242 \\mathrm{~W}$",
        "topic": "Alternating Current"
    },
    {
        "question": "Q1: A parallel plate capacitor with plates of area $1 \\mathrm{~m}^{2}$ each are at a separation of $0.1 \\mathrm{~m}$. If the electric field between the plates is $100 \\mathrm{~N} \\mathrm{C}^{-1}$, the magnitude of charge on each plate is",
        "input": "(a) $8.85 \\times 10^{-10} \\mathrm{C}$\n(b) $7.85 \\times 10^{-10} \\mathrm{C}$\n(c) $9.85 \\times 10^{-10} \\mathrm{C}$\n(d) $6.85 \\times 10^{-10} \\mathrm{C}$",
        "answer": "(a) $8.85 \\times 10^{-10} \\mathrm{C}$",
        "solution": "The electric field between two plates is\n$\\mathrm{E}=\\sigma / \\varepsilon_{0}=\\mathrm{q} / \\mathrm{A} \\varepsilon_{0} \\Rightarrow \\mathrm{q}=\\mathrm{EA} \\varepsilon_{0}$\n$\\mathrm{q}=(100)(1)\\left(8.85 \\times 10^{-12}\\right)=8.85 \\times 10^{-10} \\mathrm{C}$",
        "topic": "Capacitor"
    },
    {
        "question": "Q2: The parallel combination of two air filled parallel plate capacitors of capacitance $\\mathbf{C}$ and $\\mathbf{n C}$ is connected to a battery of voltage, $\\mathrm{V}$. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is",
        "input": "(a) $\\mathrm{V}$\n(b) $(\\mathrm{n}+1) \\mathrm{V} /(\\mathrm{K}+\\mathrm{n})$\n(c) $(\\mathrm{nV}) /(\\mathrm{K}+\\mathrm{n})$\n(d) $\\mathrm{V} /(\\mathrm{K}+\\mathrm{n})$",
        "answer": "(b) $(\\mathrm{n}+1) \\mathrm{V} /(\\mathrm{K}+\\mathrm{n})$",
        "solution": "For parallel combination, $\\mathrm{C}_{\\mathrm{eq}}=\\mathrm{C}+\\mathrm{nC}=\\mathrm{C}(\\mathrm{n}+1)$\nCharge on capacitor, $\\mathrm{q}=\\mathrm{C}_{\\mathrm{eq}} \\mathrm{V}=\\mathrm{CV}(\\mathrm{n}+1)$\nNow, after removing the battery, dielectric material is placed.\nThen, $\\mathrm{C}_{\\mathrm{eq}}=\\mathrm{KC}+\\mathrm{nC}=\\mathrm{C}(\\mathrm{K}+\\mathrm{n})$\nNew potential difference, $\\mathrm{V}^{\\prime}=\\mathrm{q} / \\mathrm{C}_{\\mathrm{eq}}=\\mathrm{CV}(\\mathrm{n}+1) / \\mathrm{C}(\\mathrm{K}+\\mathrm{n})=(\\mathrm{n}+1) \\mathrm{V} /(\\mathrm{K}+\\mathrm{n})$",
        "topic": "Capacitor"
    },
    {
        "question": "Q3: A parallel plate capacitor is made by stacking $\\mathbf{n}$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $\\mathrm{C}$ then the resultant capacitance is",
        "input": "(a) $\\mathrm{C}$\n(b) $\\mathrm{nC}$\n(c) $(n-1) C$\n(d) $(\\mathrm{n}+1) \\mathrm{C}$",
        "answer": "(c) $(n-1) C$",
        "solution": "$\\mathrm{n}$ plates connected alternately give rise to $(\\mathrm{n}-1$ ) capacitors connected in parallel\nResultant capacitance $=(n-1) C$.",
        "topic": "Capacitor"
    },
    {
        "question": "Q4: Capacitance (in $F$ ) of a spherical conductor with radius $1 \\mathrm{~m}$ is",
        "input": "(a) $1.1 \\times 10^{-10}$\n(b) $10^{-6}$\n(c) $9 \\times 10^{-9}$\n(d) $10^{-3}$",
        "answer": "(a) $1.1 \\times 10^{-10}$",
        "solution": "$\\mathrm{C}=4 \\pi \\varepsilon_{0} \\mathrm{R}=1 / 9 \\times 10^{9}=1.1 \\times 10^{-10} \\mathrm{~F}$",
        "topic": "Capacitor"
    },
    {
        "question": "Q5: If there are $\\mathbf{n}$ capacitors in parallel connected to $\\mathbf{V}$ volt source, then the energy stored is equal to",
        "input": "(a) $\\mathrm{CV}$\n(b) $(1 / 2) \\mathrm{nCV}^{2}$\n(c) $\\mathrm{CV}^{2}$\n(d) $(1 / 2 \\mathrm{n}) \\mathrm{CV}^{2}$",
        "answer": "(b) $(1 / 2) n C V^{2}$",
        "solution": "Total capacity $=\\mathrm{nC}$\n$\\therefore$ Energy $=1 / 2 \\mathrm{nCV}{ }^{2}$",
        "topic": "Capacitor"
    },
    {
        "question": "Q6: A parallel plate capacitor with air between the plates has a capacitance of $9 \\mathrm{pF}$. The separation between its plates is $\\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_{1}=3$ and thickness $d / 3$ while the other one has dielectric constant $k_{2}=6$ and thickness $2 \\mathrm{~d} / 3$. The capacitance of the capacitor is now",
        "input": "(a) $20.25 \\mathrm{pF}$\n(b) $1.8 \\mathrm{pF}$\n(c) $45 \\mathrm{pF}$\n(d) $40.5 \\mathrm{pF}$",
        "answer": "(d) $40.5 \\mathrm{pF}$",
        "solution": "$$\\mathrm{K}_{1}=3 \\quad \\mathrm{~K}_{2}=3$$\n\n$$\\mathrm{C}=\\varepsilon_{0} \\mathrm{~A} / \\mathrm{d}=9 \\times 10^{-12} \\mathrm{~F}$$\n\nWith dielectric, $$\\mathrm{C}=\\varepsilon_{0} \\mathrm{Ak} / \\mathrm{d}$$\n\n$$\\mathrm{C}_{1}=\\varepsilon_{0} \\mathrm{~A} 3 /(\\mathrm{d} / 3)=9 \\mathrm{C}$$\n\n$$\\mathrm{C}_{2}=\\varepsilon_{0} \\mathrm{~A} 6 /(2 \\mathrm{~d} / 3)=9 \\mathrm{C}$$\n\n$$\\mathrm{C}_{\\text {total }}=\\mathrm{C}_{1} \\mathrm{C}_{2} /(\\mathrm{C}_{1}+\\mathrm{C}_{2})$$ as they are in series\n\n$$\\mathrm{C}_{\\text {total }}=(9 \\mathrm{C} \\times 9 \\mathrm{C}) / 18 \\mathrm{C}=(9 / 2) \\mathrm{C}=(9 / 2)(9 \\times 10^{-12} \\mathrm{~F})$$\n\n$$\\mathrm{C}_{\\text {total }}=40.5 \\mathrm{pF}$$",
        "topic": "Capacitor"
    },
    {
        "question": "Q7: A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is",
        "input": "(a) zero\n(b) $1 / 2(\\mathrm{~K}-1) \\mathrm{CV}^{2}$\n(c) $\\mathrm{CV}^{2}(\\mathrm{~K}-1) / \\mathrm{K}$\n(d) $(\\mathrm{K}-1) \\mathrm{CV}^{2}$",
        "answer": "(a) zero",
        "solution": "The potential energy of a charged capacitor\n\n$$\\mathrm{U}_{\\mathrm{i}}=\\mathrm{q}^{2} / 2 \\mathrm{C}$$\\where $$\\mathrm{U}_{\\mathrm{i}}$$ is the initial potential energy.\n\nIf a dielectric slab is slowly introduced, the energy $$=\\mathrm{q}^{2} / 2 \\mathrm{KC}$$\n\nOnce it is taken out, again the energy increases to the old value. Therefore after it is taken out, the potential energy comes back to the old value.\n\nTotal work done $$=$$ zero\n\nAnswer: (a) zero",
        "topic": "Capacitor"
    },
    {
        "question": "Q8: A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be",
        "input": "(a) $1 / 2$\n(b) 1\n(c) 2\n(d) $1 / 4$",
        "answer": "(a) $1 / 2$",
        "solution": "Let $$\\mathrm{E}$$ be emf of the battery\n\nWork done by the battery $$\\mathrm{W}=\\mathrm{CE}^{2}$$\n\nEnergy stored in the capacitor\n\n$$\\mathrm{U}=1 / 2 \\mathrm{CE}^{2}$$\n\n$$\\mathrm{U} / \\mathrm{W}=1 / 2 \\mathrm{CE}^{2} / \\mathrm{CE}^{2}=1 / 2$$\n\nAnswer: (a) $1 / 2$",
        "topic": "Capacitor"
    },
    {
        "question": "Q9: A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor",
        "input": "(a) decreases\n(b) remains unchanged\n(c) becomes infinite\n(d) increases",
        "answer": "(b) remains unchanged",
        "solution": "With air as dielectric, $$\\mathrm{C}=\\varepsilon_{0} \\mathrm{~A} / \\mathrm{d}$$\n\nWith space partially filled, $$\\mathrm{C}^{\\prime}=\\varepsilon_{0} \\mathrm{~A} /(\\mathrm{d}-\\mathrm{t})=\\varepsilon_{0} \\mathrm{~A} / \\mathrm{d}=\\mathrm{C}$$\n\nAluminium is a good conductor. Its sheet introduced between the plates of a capacitor is of negligible thickness. Therefore, capacity remains unchanged.\n\nAnswer: (b) remains unchanged",
        "topic": "Capacitor"
    },
    {
        "question": "Q10: If the electric flux entering and leaving an enclosed surface respectively is $\\Phi_{1}$ and $\\Phi_{2}$, the electric charge inside the surface will be",
        "input": "(a) $(\\Phi_{2}-\\Phi_{1}) \\varepsilon_{0}$\n(b) $(\\Phi_{1}+\\Phi_{2}) / \\varepsilon_{0}$\n(c) $(\\Phi_{2}-\\Phi_{1}) / \\varepsilon_{0}$\n(d) $(\\Phi_{1}+\\Phi_{2}) \\varepsilon_{0}$",
        "answer": "(a) $(\\Phi_{2}-\\Phi_{1}) \\varepsilon_{0}$",
        "solution": "According to Gauss theorem,\n\n$(\\Phi_{2}-\\Phi_{1})=\\mathrm{Q} / \\varepsilon_{0}$\n\n$\\Rightarrow \\mathrm{Q}=(\\Phi_{2}-\\Phi_{1}) \\varepsilon_{0}$\n\nThe flux enters the enclosure if one has a negative charge $(-\\mathrm{q}_{2})$ and flux goes out if one has a + ve charge $(+\\mathrm{q}_{1})$. As one does not know whether $\\Phi_{1}>\\Phi_{2}, \\Phi_{2}>\\Phi_{1}, \\mathrm{Q}=\\mathrm{q} 1 \\sim \\mathrm{q} 2$\n\nAnswer: (a) $(\\Phi_{2}-\\Phi_{1}) \\varepsilon_{0}$",
        "topic": "Capacitor"
    },
    {
        "question": "Q14: Voltage rating of a parallel plate capacitor is $500 \\mathrm{~V}$. Its dielectric can withstand a maximum electric field of $10^{6} \\mathrm{~V} \\mathrm{~m}^{-1}$. The plate area is $10^{-4} \\mathrm{~m}^{2}$. What is the dielectric constant if the capacitance is $15 \\mathrm{pF}$ ? (given $\\varepsilon_{0}=8.86 \\times 10^{-12} C^{2} \\mathrm{~N}^{-1} \\mathrm{~m}^{-2}$ )",
        "input": "(a) 3.8\n(b) 8.5\n(c) 6.2\n(d) 4.5",
        "answer": "(b) 8.5",
        "solution": "$\\mathrm{C}=\\mathrm{K} \\varepsilon_{0} \\mathrm{~A} / \\mathrm{d}$ or $\\mathrm{K}=\\mathrm{CV} / \\varepsilon_{0} \\mathrm{AE}_{\\max }$\n\n$\\mathrm{K}=(15 \\times 10^{-12} \\times 500) /(8.86 \\times 10^{-12} \\times 10^{-4} \\times 10^{6})=8.5$\n\nAnswer: (b) 8.5",
        "topic": "Capacitor"
    },
    {
        "question": "Q15:In free space, a particle $A$ of charge $1 \\mu \\mathrm{C}$ is held fixed at a point $\\mathrm{P}$. Another particle $B$ of the same charge and mass $4 \\mathrm{~g}$ is kept at a distance of $1 \\mathrm{~mm}$ from $P$. If $B$ is released, then its velocity at a distance of 9 $\\mathrm{mm}$ from $P$ is\n\n(take $1 / 4 \\pi \\varepsilon_{0}=9 \\times 10^{9} \\mathrm{Nm}^{2} \\mathrm{C}^{-2}$ )\n\n(a) $3.0 \\times 10^{4} \\mathrm{~m} / \\mathrm{s}$\n\n(b) $1.0 \\mathrm{~m} / \\mathrm{s}$\n\n(c) $1.5 \\times 10^{2} \\mathrm{~m} / \\mathrm{s}$\n\n(d) $2.0 \\times 10^{3} \\mathrm{~m} / \\mathrm{s}$\n\n\\section*{Solution}\nUsing work energy theorem\n\n$\\mathrm{W}_{\\mathrm{E}}=\\mathrm{U}_{\\mathrm{i}}-\\mathrm{U}_{\\mathrm{f}}=(1 / 2) \\mathrm{mv}^{2}$ or $\\mathrm{kq}_{1} \\mathrm{q}_{2}[\\left(1 / \\mathrm{r}_{1}\\right)-\\left(1 / \\mathrm{r}_{2}\\right)]=(1 / 2) \\mathrm{mv}^{2}$\n\n$$\n\\begin{aligned}\n& \\mathrm{W}_{\\mathrm{E}}=\\left(9 \\times 10^{9}\\right) \\times\\left(1 \\times 10^{-6}\\right)^{2}\\left\\{\\left(1 / 10^{-3}\\right)-\\left(1 /\\left(9 \\times 10^{-3}\\right)\\right\\}=(1 / 2) \\mathrm{mv}^{2}\\right. \\\\\n& \\Rightarrow\\left(9 \\times 10^{9} \\times 2 \\times 10^{-12}\\right) /\\left(4 \\times 10^{-6}\\right)=\\mathrm{v}^{2} \\Rightarrow \\mathrm{v}=2.0 \\times 10^{3} \\mathrm{~ms}^{-1}\n\\end{aligned}\n$$\n\nAnswer: (d) $2.0 \\times 10^{3} \\mathrm{~m} / \\mathrm{s}$",
        "input": "",
        "answer": "(d) $2.0 \\times 10^{3} \\mathrm{~m} / \\mathrm{s}$",
        "solution": "Using work energy theorem\n\n$\\mathrm{W}_{\\mathrm{E}}=\\mathrm{U}_{\\mathrm{i}}-\\mathrm{U}_{\\mathrm{f}}=(1 / 2) \\mathrm{mv}^{2}$ or $\\mathrm{kq}_{1} \\mathrm{q}_{2}[\\left(1 / \\mathrm{r}_{1}\\right)-\\left(1 / \\mathrm{r}_{2}\\right)]=(1 / 2) \\mathrm{mv}^{2}$\n\n$$\n\\begin{aligned}\n& \\mathrm{W}_{\\mathrm{E}}=\\left(9 \\times 10^{9}\\right) \\times\\left(1 \\times 10^{-6}\\right)^{2}\\left\\{\\left(1 / 10^{-3}\\right)-\\left(1 /\\left(9 \\times 10^{-3}\\right)\\right\\}=(1 / 2) \\mathrm{mv}^{2}\\right. \\\\\n& \\Rightarrow\\left(9 \\times 10^{9} \\times 2 \\times 10^{-12}\\right) /\\left(4 \\times 10^{-6}\\right)=\\mathrm{v}^{2} \\Rightarrow \\mathrm{v}=2.0 \\times 10^{3} \\mathrm{~ms}^{-1}\n\\end{aligned}\n$$\n\nAnswer: (d) $2.0 \\times 10^{3} \\mathrm{~m} / \\mathrm{s}$",
        "topic": "Capacitor"
    },
    {
        "question": "Q1: The wavelength of the carrier waves in a modern optical fibre communication network is close to",
        "input": "(a) 600 nm\n(b) 2400 nm\n(c) 1500 nm\n(d) 900 nm\n",
        "answer": "Answer: (c) 1500 nm",
        "solution": "Fibre optics communication is mainly conducted in a wavelength range from 1260 nm to 1625 nm",
        "topic": "Communication System"
    },
    {
        "question": "Q2: The physical sizes of the transmitter and receiver antenna in a communication system are",
        "input": "(a) inversely proportional to the modulation frequency\n(b) proportional to the carrier frequency\n(c) independent of both carrier and modulation frequency\n(d) inversely proportional to the carrier frequency\n",
        "answer": "Answer: (d) inversely proportional to the carrier frequency",
        "solution": "The physical size of the transmitter and receiver antenna is inversely proportional to the carrier frequency.",
        "topic": "Communication System"
    },
    {
        "question": "Q3: A telephonic communication service is working at a carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?",
        "input": "(a) 2 \u00d7 10^3\n(b) 2 \u00d7 10^4\n(c) 2 \u00d7 10^5\n(d) 2 \u00d7 10^6\n",
        "answer": "Answer: (c) 2 \u00d7 10^5",
        "solution": "Frequency of carrier wave = 10 \u00d7 10^9 Hz\nAvailable bandwidth 10% of 10 \u00d7 10^9 Hz = 10^9 Hz\nBandwidth for each telephonic channel 5 kHz = 5 \u00d7 10^3 Hz\nNumber of channels = 10^9 / (5 \u00d7 10^3) = 2 \u00d7 10^5",
        "topic": "Communication System"
    },
    {
        "question": "Q4: A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given: radius of earth = 6.4 \u00d7 10^6 m )",
        "input": "(a) 65 km\n(b) 48 km\n(c) 40 km\n(d) 80 km\n",
        "answer": "Answer (a) 65 km",
        "solution": "Maximum distance upto which signal can be broadcasted is\nd_max = \u221a(2 R h_T) + \u221a(2 R h_R)\nwhere h_T and h_R are heights of transmitter tower and height of receiver respectively.\nPutting all values, we get\nd_max = \u221a(2 \u00d7 6.4 \u00d7 10^6) [\u221a140 + \u221a40]\nd_max = 65 km",
        "topic": "Communication System"
    },
    {
        "question": "Q5: A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal are",
        "input": "(a) 2005 kHz, 2000 kHz, and 1995 kHz\n(b) 2000 kHz and 1995 kHz\n(c) 2 MHz only\n(d) 2005 kHz and 1995 kHz\n",
        "answer": "Answer: (a) 2005 kHz, 2000 kHz, and 1995 kHz",
        "solution": "Given, f_m = 5 kHz, f_c = 2 MHz = 2000 kHz\nThe frequencies of the resultant signal are f_c + f_m = (2000 + 5) kHz = 2005 kHz, f_c = 2000 kHz, and f_c - f_m = (2000 - 5) kHz = 1995 kHz",
        "topic": "Communication System"
    },
    {
        "question": "Q6: Consider telecommunication through optical fibres. Which of the following statements is not true?",
        "input": "(a) Optical fibres can be of graded refractive index\n(b) Optical fibres are subject to electromagnetic interference from outside\n(c) Optical fibres have extremely low transmission loss\n(d) Optical fibres may have a homogeneous core with suitable cladding\n",
        "answer": "Answer: (b) Optical fibres are not subject to electromagnetic interference from outside",
        "solution": "",
        "topic": "Communication System"
    },
    {
        "question": "Q7: The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for a licence, what broadcast frequency will you allot?",
        "input": "(a) 2750 kHz\n(b) 2900 kHz\n(c) 2250 kHz\n(d) 2000 kHz\n",
        "answer": "Answer: (d) 2000 kHz",
        "solution": "10% of f_c = 250 kHz\nHence, range of signal = (2500 \u00b1 250 kHz) = 2250 kHz to 2750 kHz\n10% of 2000 kHz = 200 kHz\nRange is 1800 kHz to 2200 kHz\nHence, allocated broadcast frequency will be 2000 kHz",
        "topic": "Communication System"
    },
    {
        "question": "Q8: In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is (Take velocity of light c=3\u00d710^8 m/s, h=6.6\u00d710^(-34) Js)",
        "input": "(a) 3.75 \u00d7 10^6\n(b) 4.87 \u00d7 10^5\n(c) 6.25 \u00d7 10^5\n(d) 3.86 \u00d7 10^6\n",
        "answer": "Answer: (c) 6.25 \u00d7 10^5",
        "solution": "f = C / \u03bb = (3 \u00d7 10^8) / (8 \u00d7 10^-7) = 3.75 \u00d7 10^14 Hz\n1% of f = 3.75 \u00d7 10^17 Hz = 3.75 \u00d7 10^6 MHz\nNumber of channels = (3.75 \u00d7 10^6) / 6\nNumber of channels = 6.25 \u00d7 10^5",
        "topic": "Communication System"
    },
    {
        "question": "Q9: A signal of frequency 20 kHz and peak voltage of 5 volts is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 volts. Choose the correct statement.",
        "input": "(a) Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz\n(b) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz\n(c) Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz\n(d) Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz\n",
        "answer": "Answer: (b) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz",
        "solution": "Modulation index, m = (V_m / V_c) = (5 / 25) = 0.2\nFrequency of carrier wave, f_c = 1.2 \u00d7 10^3 kHz = 1200 kHz\nFrequency of modulate wave = 20 kHz\nf1 = f_c - f_m = 1200 - 20 = 1180 kHz\nf2 = f_c + f_m = 1200 + 20 = 1220 kHz",
        "topic": "Communication System"
    },
    {
        "question": "Q10: An audio signal consists of two distinct sounds: one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high-frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech is",
        "input": "(a) 2\n(b) 5\n(c) 6\n(d) 3\n",
        "answer": "Answer: (c) 6",
        "solution": "Band width for both signals = 15200 Hz - 200 Hz = 15000 Hz\nBand width for human speed 2700 Hz - 200 Hz = 2500 Hz\nThe ratio = 15000 / 2500 = 6",
        "topic": "Communication System"
    },
    {
        "question": "Q11: In an amplitude modulator circuit, the carrier wave is given by, C(t)=4 sin(20000\u03c0t), while modulating signal is given by, m(t)=2 sin(2000\u03c0t). The values of modulation index and lower sideband frequency are",
        "input": "(a) 0.4 and 10 kHz\n(b) 0.5 and 9 kHz\n(c) 0.3 and 9 kHz\n(d) 0.5 and 10 kHz\n",
        "answer": "Answer: (b) 0.5 and 9 kHz",
        "solution": "Given C(t) = 4 sin(20000\u03c0t) \u21d2 A_c = 4\nm(t) = 2",
        "topic": "Communication System"
    },
    {
        "question": "Q1: A compressive force, $F$ is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $\\Delta T$. The net change in its length is zero. Let $L$ be the length of the rod, $A$ is its area of cross-section. $Y$ is Young's modulus, and $\u0007lpha$ is its coefficient of linear expansion. Then, $F$ is equal to",
        "input": "a. $\\mathrm{L}^{2} \\mathrm{Y} \u0007lpha \\Delta \\mathrm{T}$\nb. $\\mathrm{AY} / \u0007lpha \\Delta \\mathrm{T}$\nc. AY $\u0007lpha \\Delta \\mathrm{T}$\nd. LAY $\u0007lpha \\Delta \\mathrm{T}$",
        "answer": "c",
        "solution": "Thermal expansion, $\\Delta \\mathrm{L}=\\mathrm{L} \u0007lpha \\Delta \\mathrm{T}$\\ Let $\\Delta \\mathrm{L}'$ be the compression produced by applied force\\ $\\mathrm{Y}=\\mathrm{FL} / \\mathrm{A} \\Delta \\mathrm{L}' \\Rightarrow \\mathrm{F}=\\mathrm{YA} \\Delta \\mathrm{L}' / \\mathrm{L}$\\ Net change in length $=0 \\Rightarrow \\Delta \\mathrm{L}'=$\\ From (1),(2) and (3)\\ $\\mathrm{F}=\\mathrm{YAx}(\\mathrm{L} \u0007lpha \\Delta \\mathrm{T}) / \\mathrm{L}=\\mathrm{YA} \u0007lpha \\Delta \\mathrm{T}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \\mathrm{~N}$ to the lower end. The weight stretches the wire by $1 \\mathrm{~mm}$. Then the elastic energy stored in the wire is",
        "input": "a. $0.2 \\mathrm{~J}$\nb. $10 \\mathrm{~J}$\nc. $20 \\mathrm{~J}$\nd. $0.1 \\mathrm{~J}$",
        "answer": "d",
        "solution": "Elastic energy per unit volume $=1 / 2 \\mathrm{x}$ stress $\\mathrm{x}$ strain\\ Elastic Energy $=1 / 2 \\mathrm{x}$ stress $\\mathrm{x}$ strain $\\mathrm{x}$ volume\\ $=1 / 2 \times F / A \times(\\Delta L / L) \times(A L)$\\ $=1 / 2 \times$ F $\\Delta \\mathrm{L}$\\ $=1 / 2 \times 200 \times 10^{-3}$\\ Elastic Energy $=0.1 \\mathrm{~J}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q3: A rod of length $L$ at room temperature and uniform area of cross-section $A$, Is made of a metal having a coefficient of linear expansion $\u0007lpha$. It is observed that an external compressive force $F$ is applied to each of its ends, prevents any change in the length of the rod when its temperature rises by $\\Delta T \\mathrm{~K}$. Young's modulus, $Y$ for this metal is",
        "input": "a.F/A $\u0007lpha \\Delta \\mathrm{T}$\nb.F/A $\u0007lpha(\\Delta \\mathrm{T}-273)$\nc. F/2A $\u0007lpha \\Delta$\nd. $2 \\mathrm{~F} / \\mathrm{A} \u0007lpha \\Delta \\mathrm{T}$",
        "answer": "a",
        "solution": "Young's Modulus $\\mathrm{Y}=$ stress/strain $=(\\mathrm{F} / \\mathrm{A}) /(\\Delta l / l)$\\ Substituting the coefficient of linear expansion\\ $\u0007lpha=\\Delta l /(l \\Delta \\mathrm{T})$\\ $\\Delta l / l=\u0007lpha \\Delta \\mathrm{T}$\\ $\\mathrm{Y}=(\\mathrm{F} / \\mathrm{A} \u0007lpha \\Delta \\mathrm{T})$",
        "topic": "Elasticity"
    },
    {
        "question": "Q4: Young's moduli of two wires A and B are in the ratio 7:4. Wire A is $2 \\mathrm{~m}$ long and has radius $R$. Wire $B$ is $1.5 \\mathrm{~m}$ long and has a radius of $2 \\mathrm{~mm}$. If the two wires stretch by the same length for a given load, then the value of $R$ is close $t$",
        "input": "a. $1.5 \\mathrm{~mm}$\nb. $1.9 \\mathrm{~mm}$\nc. $1.7 \\mathrm{~mm}$\nd. $1.3 \\mathrm{~mm}$",
        "answer": "c",
        "solution": "$\\Delta_{1}=\\Delta_{2}$\\ $\\left(\\mathrm{Fl}{1} / \\pi \\mathrm{r}{1}^{2} \\mathrm{y}{1}\right)=\\left(\\mathrm{Fl}{2} / \\pi \\mathrm{r}{2}^{2} \\mathrm{y}{2}\right)$\\ $2 /(\\mathrm{R}^{2} \times 7)=1.5 /(2^{2} \times 4)$\\ $\\mathrm{R}=1.75 \\mathrm{~mm}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a $400 \\mathrm{~N}$ load without exceeding its elastic limit?",
        "input": "a. $1 \\mathrm{~mm}$\nb. $1.15 \\mathrm{~mm}$\nc. $0.90 \\mathrm{~mm}$\nd. $1.36 \\mathrm{~mm}$",
        "answer": "b",
        "solution": "Stress $=\\mathrm{F} / \\mathrm{A}$\\ Stress $=400 \times 4 / \\pi \\mathrm{d}^{2}$\\ $=379 \times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}$\\ $\\mathrm{d}^{2}=(400 \times 4) /(379 \times 10^{6} \\pi)$\\ $\\mathrm{d}=1.15 \\mathrm{~mm}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q6: A uniform cylindrical rod of length $L$ and radius $r$, is made from a material whose Young's modulus of Elasticity equals $Y$. When this rod is heated by temperature $T$ and simultaneously subjected to a net longitudinal compressional force $F$, its length remains unchanged. The coefficient of volume expansion, of the material of the rod is nearly equal to",
        "input": "a. $9 \\mathrm{~F} /(\\pi \\mathrm{r}^{2} \\mathrm{YT})$\nb. $6 \\mathrm{~F} /(\\pi \\mathrm{r}^{2} \\mathrm{YT})$\nc. $3 \\mathrm{~F} /(\\pi \\mathrm{r}^{2} \\mathrm{YT})$\nd. $\\mathrm{F} /(3 \\pi \\mathrm{r}^{2} \\mathrm{YT})$",
        "answer": "c",
        "solution": "$\\mathrm{Y}=\\left(\\mathrm{F} / \\pi \\mathrm{r}^{2}\right) \times \\mathrm{L} / \\Delta \\mathrm{L}$\\ $\\Delta \\mathrm{L}=\\mathrm{F} / / \\pi \\mathrm{r}^{2} \\mathrm{Y}-------(1)$\\ Change in length due to temperature change\\ $\\Delta \\mathrm{L}=\\mathrm{L} \u0007lpha \\Delta \\mathrm{T}$\\ From equa (1) and (2)\\ $\\mathrm{L} \u0007lpha \\Delta \\mathrm{T}=\\mathrm{FL} / \\mathrm{AY}$\\ $\u0007lpha=\\mathrm{F} / \\mathrm{AY} \\Delta \\mathrm{T}$\\ $\u0007lpha=\\mathrm{F} / \\pi \\mathrm{r}^{2} \\mathrm{YT}$\\ Coefficient of volume expansion\\ $3 \\propto=3 \\mathrm{~F} / \\pi \\mathrm{r}^{2} \\mathrm{YT}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q7: The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?",
        "input": "(a) length $=200 \\mathrm{~cm}$, diameter $=2 \\mathrm{~mm}$\n(b) length $=300 \\mathrm{~cm}$, diameter $=3 \\mathrm{~mm}$\n(c) length $=50 \\mathrm{~cm}$, diameter $=0.5 \\mathrm{~mm}$\n(d) length $=100 \\mathrm{~cm}$, diameter $=1 \\mathrm{~mm}$",
        "answer": "c",
        "solution": "Since all four wires are made from the same material Young's modulus will be the same.\\ $\\Delta \\mathrm{L} \\propto \\mathrm{L} / \\mathrm{D}^{2}$\\ In (a) $\\mathrm{L} / \\mathrm{D}^{2}=200 /(0.2)^{2}=5 \times 10^{3} \\mathrm{~cm}^{-1}$\\ In (b) $\\mathrm{L} / \\mathrm{D}^{2}=300 /(0.3)^{2}=3.3 \times 10^{3} \\mathrm{~cm}^{-1}$\\ In (c) $\\mathrm{L} / \\mathrm{D}^{2}=50 /(0.5)^{2}=20 \times 10^{3} \\mathrm{~cm}^{-1}$\\ In (d) $\\mathrm{L} / \\mathrm{D}^{2}=100 /(0.1)^{2}=10 \times 10^{3} \\mathrm{~cm}^{-1}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q8: A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of",
        "input": "(a) $1 / 9$\n(b) 81\n(c) $1 / 81$\n(d) 9",
        "answer": "d",
        "solution": "Stress $=$ Force $/$ Area\\ Stress $=$ Force $/ L^{2}$\\ Now, dimensions increases by a factor of 9\\ Now, $\\mathrm{S}=$ (volume $\\mathrm{x}$ density) $\\mathrm{xg} / \\mathrm{L}^{2}$\\ $\\mathrm{S}=\\mathrm{L}^{3} \times \rho \\mathrm{g} / \\mathrm{L}^{2}=\\mathrm{L} \rho \\mathrm{g}$\\ Stress S $\\propto \\mathrm{L}$\\ $\\mathrm{S}{2} / \\mathrm{S}{1}=\\mathrm{L}{2} \\mathrm{~L}{1}=9 \\mathrm{~L}{1} / \\mathrm{L}{1}=9$",
        "topic": "Elasticity"
    },
    {
        "question": "Q9. A solid sphere of radius $r$ made of a soft material of bulk modulus $K$ is surrounded by a liquid in a cylindrical container. A massless piston of the area a floats on the surface of the liquid, covering an entire cross-section of the cylindrical container. When a mass $m$ is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere is $\\mathrm{Mg} / \u0007lpha \\mathrm{AB}$. Find the value of $\u0007lpha$.",
        "input": "a. 4\nb. 5\nc. 3\nd. 2",
        "answer": "c",
        "solution": "Increase in pressure is $\\Delta \\mathrm{p}=\\mathrm{Mg} / \\mathrm{A}$\\ Bulk modulus is $\\mathrm{B}=\\Delta \\mathrm{p} /(\\Delta \\mathrm{V} / \\mathrm{V})$\\ $\\Delta \\mathrm{V} / \\mathrm{V}=\\Delta \\mathrm{p} / \\mathrm{B}=\\mathrm{Mg} / \\mathrm{AB}-----(1)$\\ The volume of the sphere is $\\mathrm{V}=(4 / 3) \\pi \\mathrm{R}^{3}$\\ $\\Delta \\mathrm{V} / \\mathrm{V}=3(\\Delta \\mathrm{R} / \\mathrm{R})$\\ From equation (1) we get\\ $\\mathrm{Mg} / \\mathrm{AB}=3(\\Delta \\mathrm{R} / \\mathrm{R})$\\ Therefore $\u0007lpha=3$",
        "topic": "Elasticity"
    },
    {
        "question": "Q10: A steel wire having a radius of $2.0 \\mathrm{~mm}$, carrying a load of $4 \\mathrm{~kg}$, is hanging from a ceiling. Given that $\\mathrm{g}$ $=3.1 \\pi \\mathrm{m} / \\mathrm{s}^{2}$, what will be the tensile stress that would be developed in the wire?",
        "input": "a. $4.8 \times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}$\nb. $3.1 \times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}$\nc. $6.2 \times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}$\nd. $5.2 \times 10^{6} \\mathrm{~N} / \\mathrm{m}^{2}$",
        "answer": "b",
        "solution": "Tensile stress $=$ Force/Area\\ Tensile stress $=(4)(3.1 \\pi) / \\pi\\left(2 \times 10^{-3}\right)^{2}$\\ Tensile stress $=3.1 \times 10^{6} \\mathrm{Nm}^{-2}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q11: A steel rail of length $5 \\mathrm{~m}$ and area of cross-section $40 \\mathrm{~cm}^{2}$ is prevented from expanding along its length while the temperature rises by $10^{\\circ} \\mathrm{C}$. If the coefficient of linear expansion and Young's modulus of steel is $1.2 \times 10^{-5} \\mathrm{~K}^{-1}$ and $2 \times 10^{11} \\mathrm{Nm}^{-2}$ respectively, the force developed in the rail is approximately",
        "input": "a. $2 \times 10^{9} \\mathrm{~N}$\nb. $3 \times 10^{-5} \\mathrm{~N}$\nc. $2 \times 10^{7} \\mathrm{~N}$\nd. $1 \times 10^{5} \\mathrm{~N}$",
        "answer": "d",
        "solution": "$\\mathrm{A}=40 \\mathrm{~cm}^{2}=4 \times 10^{-3} \\mathrm{~m}^{2}$\\ $\\Delta \\mathrm{T}=10^{\\circ} \\mathrm{C}$\\ $\\mathrm{Y}=2 \times 10^{11} \\mathrm{Nm}^{-2}$\\ $\u0007lpha=1.2 \times 10^{-5} \\mathrm{~K}^{-1}$\\ Force $=\\mathrm{YA} \u0007lpha \\Delta \\mathrm{T}$\\ Force $=\\left(2 \times 10^{11}\right)\\left(4 \times 10^{-3}\right)\\left(1.2 \times 10^{-5}\right)(10)=9.6 \times 10^{4} \\mathrm{~N}$\\ Force $\u0007pprox 1 \times 10^{5} \\mathrm{~N}$",
        "topic": "Elasticity"
    },
    {
        "question": "Q12: If $\\mathrm{S}$ is the stress and $\\mathrm{Y}$ is Young's Modulus of the material of the wire, the energy stored in the wire per unit volume is",
        "input": "a. $2 \\mathrm{Y} / \\mathrm{S}$\nb. $\\mathrm{S} / 2 \\mathrm{Y}$\nc. $2 S^{2} Y$\nd. $\\mathrm{S}^{2} / 2 \\mathrm{Y}$",
        "answer": "d",
        "solution": "Young's modulus, $\\mathrm{Y}=$ Stress $/$ Strain\\ $\\Rightarrow$ Strain $=$ Stress $/ \\mathrm{Y}=\\mathrm{S} / \\mathrm{Y}$\\ Energy stored per unit volume $=1 / 2 \\mathrm{x}$ stress $\\mathrm{x}$ strain\\ $=$ Stress $\\mathrm{x}$ Stress $/ 2 \\mathrm{Y}=\\mathrm{S}^{2} / 2 \\mathrm{Y}$ (since Young's modulus, $\\mathrm{Y}=$ Stress $/$ Strain)",
        "topic": "Elasticity"
    },
    {
        "question": "Q13Two wires are made of the same material and have the same volume. However, wire 1 has a cross-sectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of the wire 1 increases by $\\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount?",
        "input": "a. F\nb. $4 F$\nc. $6 F$\nd. $9 F$",
        "answer": "d",
        "solution": "For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire $\\mathrm{L}{1}$ is and that of $\\mathrm{L}{2}$ is $3 A$\\ $V=V_{1}=V_{2}$\\ $V=A \times L_{1}=3 A \times L_{2} \\Rightarrow L_{2}=L_{1} / 3$\\ $Y=(F / A) /(\\Delta L / L)$\\ $F_{1}=YA(\\Delta L_{1} / L_{1})$\\ $F_{2}=Y 3 A(\\Delta L_{2}, L_{2})$\\ Given $\\Delta L_{1}=\\Delta L_{2}=\\Delta x$ (for the same extension)\\ $F_{2}=Y 3 A(\\Delta x /(L_{1} / 3))=9 (YA \\Delta x / L_{1})=9 F_{1}=9 F$",
        "topic": "Elasticity"
    },
    {
        "question": "Q14 A wire elongates by $l \\mathrm{~mm}$ when a load $W$ is hanged from it. If the wire goes over a pulley and two weighs $W$ each is hung at the two ends, the elongation of the wire will be (in $\\mathrm{mm}$ )",
        "input": "a. $l / 2$\nb. l\nc. 2\nd. Zero",
        "answer": "b",
        "solution": "$Y=(\text{Force} \times L) /(A \times l)=WL / Al$\\ $l=WL / AY$\\ Due to the arrangement of the pulley, the length of wire is $L / 2$ on each side and so the elongation will be $l / 2$. For both sides, elongation $=l$\\ Answer: (b) $l$",
        "topic": "Elasticity"
    },
    {
        "question": "Q1: The mean intensity of radiation on the surface of the Sun is about $10^{8} \\mathrm{~W} / \\mathrm{m}^{2}$. The RMS value of the corresponding magnetic field is closest to",
        "input": "a. $10^{-2} \\mathrm{~T}$\nb. $1 \\mathrm{~T}$\nc. $10^{-4} \\mathrm{~T}$\nd. $10^{2} \\mathrm{~T}$",
        "answer": "c",
        "solution": "Mean intensity $\\mathrm{I}=(\\mathrm{B}_{\\text {rms }}^{2} / \\mu_{0}) \\mathrm{c}$\\ $\\mathrm{B}_{\\text {rms }}^{2}=(10^{8} \\times 4 \\pi \\times 10^{-7}) /(3 \\times 10^{8})$\\ $\\mathrm{B}_{\\mathrm{rms}} \\approx 10^{-4} \\mathrm{~T}$\\ Answer: (c) $10^{-4} \\mathrm{~T}$",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q2: A plane electromagnetic wave travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time is $E=6 \\mathrm{~V} \\mathrm{~m}^{-1}$ along the $y$-direction. Its corresponding magnetic field component, $B$ would be",
        "input": "a. $6 \\times 10^{-8} \\mathrm{~T}$ along the $\\mathrm{x}$-direction\nb. $2 \\times 10^{-8} \\mathrm{~T}$ along the $\\mathrm{y}$-direction\nc. $2 \\times 10^{-8} \\mathrm{~T}$ along the $\\mathrm{z}$-direction\nd. $6 \\times 10^{-8} \\mathrm{~T}$ along the $\\mathrm{z}$-direction",
        "answer": "c",
        "solution": "The direction of electromagnetic wave travelling is given by\\ $$\\bar{k}=\\bar{E} \\times \\bar{B}$$\\ As the wave is travelling along $\\mathrm{x}$-direction and $\\mathrm{E}$ is along the $\\mathrm{y}$-direction.\\ So B must point towards the z-direction.\\ The magnetic field, $\\mathrm{B}=\\mathrm{E} / \\mathrm{c}=6 /(3 \\times 10^{8})=2 \\times 10^{-8} \\mathrm{~T}$\\ Answer: (c) $2 \\times 10^{-8} \\mathrm{~T}$ along the z-direction",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q3: $50 \\mathrm{~W} / \\mathrm{m}^{2}$ energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy ( $\\mathbf{2 5 \\%}$ ) is reflected from the surface and the rest is absorbed. The force exerted on $1 \\mathrm{~m}^{2}$ surface area will be close to $\\left(\\mathrm{c}=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right.$ )",
        "input": "a. $20 \\times 10^{-8} \\mathrm{~N}$\nb. $10 \\times 10^{-8} \\mathrm{~N}$\nc. $35 \\times 10^{-8} \\mathrm{~N}$\nd. $15 \\times 10^{-8} \\mathrm{~N}$",
        "answer": "a",
        "solution": "Given energy density $=50 \\mathrm{~W} / \\mathrm{m}^{2}$\\ Here, change in momentum $\\Delta p=p_{f}-p_{i}$\\ $\\Delta \\mathrm{p}=\\left(-\\mathrm{p}_{\\mathrm{i}} / 4\\right)-\\mathrm{p}_{\\mathrm{i}}$\\ $\\Delta \\mathrm{p}=-5 \\mathrm{p}_{\\mathrm{i}} / 4 \\because \\mathrm{p}_{\\mathrm{i}}=\\mathrm{E} / \\mathrm{c}=50 \\mathrm{~W} / \\mathrm{s} /(3 \\times 10^{8})$\\ $\\Delta \\mathrm{p} / \\Delta \\mathrm{t}=\\mathrm{F}=\\left|-5 \\mathrm{p}_{\\mathrm{i}} / 4\\right|=(5 / 4) \\times\\left(50 /(3 \\times 10^{8}=20.8 \\times 10^{-8} \\mathrm{~N} \\approx 20 \\times 10^{-8} \\mathrm{~N}\\right.$\\ Answer: (a) $20 \\times 10^{-8} \\mathrm{~N}$",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q4: A red LED emits light at 0.1 watts uniformly around it. The amplitude of the electric field of the light at a distance of $1 \\mathrm{~m}$ from the diode is",
        "input": "a. $5.48 \\mathrm{~V} / \\mathrm{m}$\nb. $7.75 \\mathrm{~V} / \\mathrm{m}$\nc. $1.73 \\mathrm{~V} / \\mathrm{m}$\nd. $2.45 \\mathrm{~V} / \\mathrm{m}$",
        "answer": "d",
        "solution": "The intensity of light, $\\mathrm{I}=\\mathrm{u}_{\\mathrm{av}} \\mathrm{c}$\\ Also, $\\mathrm{I}=\\mathrm{P} / 4 \\pi \\mathrm{r}^{2}$ and $\\mathrm{u}_{\\mathrm{av}}=1 / 2 \\varepsilon_{0} \\mathrm{E}_{0}{ }^{2}$\\ $\\therefore \\mathrm{P} / 4 \\pi \\mathrm{r}^{2}=(1 / 2) \\varepsilon_{0} \\mathrm{E}_{0}{ }^{2} \\mathrm{c}$\\ Or $\\quad E_{0}=\\sqrt{\\frac{2 P}{4 \\pi \\epsilon_{0} r^{2} c}}$\\ Here, $\\mathrm{P}=0.1 \\mathrm{~W}, \\mathrm{r}=1 \\mathrm{~m}, \\mathrm{c}=3 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$\\ $1 / 4 \\pi \\varepsilon_{0}=9 \\times 10^{9} \\mathrm{~N} \\mathrm{C}^{-2} \\mathrm{~m}^{2}$\\ $$ E_{0}=\\sqrt{\\frac{2 \\times 0.1 \\times 9 \\times 10^{9}}{1^{2} \\times 3 \\times 10^{8}}} =\\sqrt{6}$$\\ $=2.45 \\mathrm{~V} \\mathrm{~m}^{-1}$\\ Answer: (d) $2.45 \\mathrm{~V} / \\mathrm{m}$",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q5: Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists",
        "input": "a. (i) (ii) (iii) (iv)\nb. (iv) (iii) (ii) (i)\nc. (i) (ii) (iv) (iii)\nd. (iii) (ii) (i) (iv)",
        "answer": "a",
        "solution": "Infrared waves are used to treat muscular strain. Radio waves are used for broadcasting. X-rays are used to detect the fracture of bones. Ultraviolet rays are absorbed by the ozone layer of the atmosphere.\\ Answer: (a)",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q6: During the propagation of electromagnetic waves in a medium",
        "input": "a. both electric and magnetic energy densities are zero\nb. electric energy density is double of the magnetic energy density\nc. electric energy density is half of the magnetic energy density\nd. electric energy density is equal to the magnetic energy density",
        "answer": "d",
        "solution": "Solution: In an em wave, energy is equally divided between the electric and the magnetic fields.\\ Answer: (d) The electric energy density is equal to the magnetic energy density",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q7: The magnetic field in a travelling electromagnetic wave has a peak value of $20 \\mathrm{nT}$. The peak value of electric field strength is",
        "input": "a. $12 \\mathrm{~V} / \\mathrm{m}$\nb. $3 \\mathrm{~V} / \\mathrm{m}$\nc. $6 \\mathrm{~V} / \\mathrm{m}$\nd. $9 \\mathrm{~V} / \\mathrm{m}$",
        "answer": "c",
        "solution": "In an electromagnetic wave, the peak value of the electric field $\\left(\\mathrm{E}_{0}\\right)$ and peak value of magnetic field $\\left(\\mathrm{B}_{0}\\right)$ are related by\\ $\\mathrm{E}_{0}=\\mathrm{B}_{0} \\mathrm{C}$\\ $\\mathrm{E}_{0}=\\left(20 \\times 10^{-9} \\mathrm{~T}\\right)\\left(3 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}\\right)=6 \\mathrm{~V} / \\mathrm{m}$\\ Answer: (c) 6 V/m",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q8: An electromagnetic wave of frequency $=3.0 \\mathrm{MHz}$ passes from vacuum into a dielectric medium with permittivity $=4.0$. Then",
        "input": "a. wavelength is doubled and the frequency remains unchanged\nb. wavelength is doubled and frequency becomes half\nc. wavelength is halved and frequency remains unchanged\nd. wavelength and frequency both remain unchanged",
        "answer": "c",
        "solution": "Solution: In an em wave, energy is equally divided between the electric and the magnetic fields.\\ Answer: (d) The electric energy density is equal to the magnetic energy density",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q9: Electromagnetic waves are transverse in nature is evident by",
        "input": "a. polarization\nb. interference\nc. reflection\nd. diffraction",
        "answer": "a",
        "solution": "Answer: (a) Polarization proves the transverse nature of electromagnetic waves.",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q10: The energy associated with electric field is $\\left(U_{E}\right)$ and with magnetic field is $\\left(U_{B}\right)$ for an electromagnetic wave in free space. Then",
        "input": "a. $U_{E}>U_{B}$\nb. $U_{E}=U_{B} / 2$\nc. $U_{E}=U_{B}$\nd. $U_{E}<U_{B}$",
        "answer": "c",
        "solution": "Answer: (c) $\\mathbf{U}_{\\mathrm{E}}=\\mathbf{U}_{\\mathrm{B}}$",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q11: A $27 \\mathrm{~mW}$ laser beam has a cross-sectional area of $10 \\mathrm{~mm}^{2}$. The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $\\varepsilon_{0}=9 \\times 10^{-12}$ SI units, speed of light $\\mathbf{c}=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$ ]",
        "input": "a. $2 \\mathrm{kV} / \\mathrm{m}$\nb. $0.7 \\mathrm{kV} / \\mathrm{m}$\nc. $1 \\mathrm{kV} / \\mathrm{m}$\nd. $1.4 \\mathrm{kV} / \\mathrm{m}$",
        "answer": "d",
        "solution": "The intensity of the electromagnetic wave is given by\\ $\\mathrm{I}=\\operatorname{Power}(\\mathrm{P}) / \\operatorname{Area}(\\mathrm{A})=1 / 2 \\varepsilon_{0} \\mathrm{E}^{2} \\mathrm{C}$\\ $$ \\begin{aligned} & E=\\sqrt{\\frac{2 P}{A E_{0} c}}=\\sqrt{\\frac{2 \\times 27 \\times 10^{-3}}{10 \\times 10^{-6} \\times 9 \\times 10^{-12} \\times 3 \\times 10^{8}}} \\ & E=\\sqrt{0.2 \\times 10^{7}}=\\sqrt{2 \\times 10^{6}} \\end{aligned} $$\\ $\\mathrm{E}=1.4 \\mathrm{kV} / \\mathrm{m}$\\ \\section*{Answer: (d) $1.4 \\mathrm{kV} / \\mathrm{m}$}",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q12: The RMS value of the electric field of the light coming from the sun is $720 \\mathrm{~N} / \\mathrm{C}$. The average total energy density of the electromagnetic wave is",
        "input": "a) $3.3 \\times 10^{-3} \\mathrm{~J} / \\mathrm{m}^{3}$\nb) $4.58 \\times 10^{-6} \\mathrm{~J} / \\mathrm{m}^{3}$\nc) $6.37 \\times 10^{-9} \\mathrm{~J} / \\mathrm{m}^{3}$\nd) $81.35 \\times 10^{-12} \\mathrm{~J} / \\mathrm{m}^{3}$",
        "answer": "b",
        "solution": "$\\mathrm{U}=\\left(8.85 \\times 10^{-12}\\right) \\times(720)^{2}=4.58 \\times 10^{-6} \\mathrm{Jm}^{-1}$\\ \\section*{Answer (b) $4.58 \\times 10^{-6} \\mathrm{~J} / \\mathrm{m}^{3}$}",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q13: Which of the following are not electromagnetic waves?",
        "input": "a) cosmic rays\nb) gamma rays\nc) $\\beta$-rays\nd) X-rays",
        "answer": "c",
        "solution": "Answer: (c) $\\beta$-rays are not electromagnetic waves",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q14: If a source of power $4 \\mathrm{~kW}$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called",
        "input": "a) microwaves\nb) y-rays\nc) X-rays\nd) ultraviolet rays",
        "answer": "c",
        "solution": "Solution\\ \\section*{Answer: (c) X-rays}",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q15: An EM wave from the air enters a medium. The electric fields are\\ $$\\begin{aligned} \\vec{E}_{1} & =E_{01} \\hat{x} \\cos \\left[2 \\pi f\\left(\\frac{z}{c}-t\\right)\\right] \\quad \\text { in air and } \\\\ \\vec{E}_{2} & =E_{02} \\hat{x} \\cos [k(2 z-c t)] \\end{aligned}$$\\ in medium, where the wavenumber $k$ and frequency $f$ refer to their values in air. The medium is non-magnetic. If $\\varepsilon_{\\mathrm{r} 1}$ and $\\varepsilon_{\\mathrm{r} 2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?",
        "input": "a) $\\varepsilon_{\\mathrm{r} 1} / \\varepsilon_{\\mathrm{r} 2}=4$\nb) $\\varepsilon_{\\mathrm{r} 1} / \\varepsilon_{\\mathrm{r} 2}=2$\nc) $\\varepsilon_{\\mathrm{r} 1} / \\varepsilon_{\\mathrm{r} 2}=1 / 4$\nd) $\\varepsilon_{\\mathrm{r} 1} / \\varepsilon_{\\mathrm{r} 2}=1 / 2$",
        "answer": "c",
        "solution": "During refraction, frequency remains unchanged, whereas the wavelength gets changed\\ $\\mathrm{k}^{\\prime}=2 \\mathrm{k}$ (From equations)\\ $2 \\pi / \\lambda^{\\prime}=2\\left(2 \\pi / \\lambda_{0}\\right)=\\lambda^{\\prime}=\\lambda_{0} / 2$\\ Since, $v=c / 2$\\ $$ \\frac{1}{\\sqrt{\\mu_{0} \\epsilon_{r 2}}}=\\frac{1}{2} \\times \\frac{1}{\\sqrt{\\mu_{0} \\epsilon_{r 1}}} $$\\ $\\varepsilon_{\\mathrm{r} 1} / \\varepsilon_{\\mathrm{r} 2}=1 / 4$\\ \\section*{Answer: (c) $\\varepsilon_{\\mathrm{r} 1} / \\varepsilon_{\\mathrm{r} 2}=1 / 4$}",
        "topic": "Electromagnetic Waves"
    },
    {
        "question": "Q1: A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction.",
        "input": "due north\n30\u00b0 east of north\n30\u00b0 west of north\n60\u00b0 east of north",
        "answer": "(1) due north",
        "solution": "Let the time taken to cross the river be t\nt=d/V_s cos\u03b8\nFor the time t to be minimum, cos\u03b8 should be maximum\n\u03b8 should be 0\nTherefore swimmer should swim due north",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q2: A particle P is sliding down a frictionless hemispherical bowl. It passes point A at t=0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t= 0 along the horizontal string AB, with speed v. Friction between the bead and the string may be neglected. Let t_p and t_q be the respective times taken by P and Q to reach point B.",
        "input": "t_p < t_Q\nt_p > t_Q\nt_p = t_Q\nt_p / t_Q = (length of arc ABC / length of chord AB )",
        "answer": "(1) t_P < t_Q",
        "solution": "Horizontal displacements of the particle (P) and bead (Q) are equal. Bead Q moves along AB with a constant velocity v. For particle P, motion between AC will be under acceleration while between CB will be under retardation. The horizontal component of its velocity may be greater than or equal to v. It will not be less than v. Obviously, t_P < t_Q.",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q3: A train is moving along a straight line with a constant acceleration 'a'. A girl standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60\u00b0 to the horizontal. The girl has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s\u00b2, is",
        "input": "",
        "answer": "a = 5 m/s\u00b2",
        "solution": "R = 1 / 2 at\u00b2 + 1.15\nu\u00b2 sin 2\u03b8 / g = 1 / 2 a (2 u sin \u03b8 / g)\u00b2 + 1.15\n100 \u00d7 \u221a3 / 2 / 10 = 1 / 2 a (4 \u00d7 100 \u00d7 3 / 4 / 100) + 1.15\n5 \u221a3 = 3 a / 2 + 1.15\n2 \u00d7 5 \u221a3 = 3 a + 2.30\n10 \u221a3 = 3 a + 2.30\n10 \u221a3 - 2.30 = 3 a\n10 \u00d7 1.73 - 2.3 = 3 a\n17.3 - 2.3 = 3 a\n15 / 3 = a\na = 5 m/s\u00b2",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q5: A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river in the shortest time. The velocity of river water in km/hr is",
        "input": "1\n3\n4\n\u221a41",
        "answer": "(2) 3 km/h",
        "solution": "Time t = 15 min = 15 / 60 = 1 / 4 hour\nVelocity = distance / time\nV_b cos \u03b8 = 1 / (1 / 4) = 4 km/h\n5 cos \u03b8 = 4 km/h\ncos \u03b8 = 4 / 5\nsin \u03b8 = 3 / 5\nConsider the triangle of velocity ABC,\nV_r / V_b = 3 / 5\nV_r / 5 = 3 / 5\nOr V_r = 3 km/h",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q8: A ball is projected from the ground at an angle of 45\u00b0 with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30\u00b0 with the horizontal surface. The maximum height it reaches after the bounce, in metres, is",
        "input": "",
        "answer": "H\u2082 = 30 m",
        "solution": "H\u2081 = u\u00b2 sin\u00b2 45 / 2g = 120\nu\u00b2 / 2g = 120\nWhen half of the kinetic energy is lost v = u / \u221a2\nH\u2082 = (u / \u221a2)\u00b2 sin\u00b2 30 / 2g = u\u00b2 / 16g\nFrom (1) and (2)\nH\u2082 = H\u2081 / 4 = 30 m",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q10: Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance). Is the statement true or false?",
        "input": "",
        "answer": "True",
        "solution": "$v^2=u^2-2gH$\nAs the body falls back to the point of projection, $H=0$\n$$v^2=u^2-0=u^2$$\n$$\\operatorname{Or} v=u$$\nHence the statement is true",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q11: A projectile fired from the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path. Is the statement true?",
        "input": "",
        "answer": "True",
        "solution": "The horizontal velocity of the projectile $=$ $u \\cos \\theta$\nIt remains constant throughout the motion\nThe vertical velocity becomes zero at the top of the path\nTherefore the velocity is minimum at the top\nHence the statement is true",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q12: Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. Is the statement true?",
        "input": "",
        "answer": "False",
        "solution": "One train is moving in the direction of the earth's rotation and the other moves in the opposite direction. Hence they will exert different pressures on the rails. The statement is false.",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q13: A particle moves in a circle of radius R. In half the period of revolution its displacement is and distance covered is",
        "input": "",
        "answer": "Displacement $AB=2r$\nDistance $ACB=2\\pi r / 2=\\pi r$",
        "solution": "Displacement $AB=2r$\nDistance $ACB=2\\pi r / 2=\\pi r$",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q14: Four persons $K, L, M, N$ are initially at the four corners of a square of side $d$. Each person now moves with a uniform speed $V$ in such a way that $K$ always moves directly towards $L$, $L$ directly towards $M$, $M$ directly towards $N$, and $N$ directly towards $K$. The four persons will meet at a time",
        "input": "",
        "answer": "Time Taken $=(d / \\sqrt{2}) / (v / \\sqrt{2})=d / v$",
        "solution": "It is inferred on the basis of symmetry that all four persons will meet at $O$, the centre of the square.\nAt any instant, the component velocity along\n$$KO=v \\cos 45^\\circ=v / \\sqrt{2}$$\nDistance $KO=d \\cos 45^\\circ=d / \\sqrt{2}$\nTime taken $=$ Distance $KO$ / velocity along $KO$\nTime Taken $=(d / \\sqrt{2}) / (v / \\sqrt{2})=d / v$",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q15: The trajectory of a projectile in a vertical plane is $y=ax-bx^{2}$, where $a, b$ are constants, and $x$ and $y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is. and the angle of projection from the horizontal is",
        "input": "",
        "answer": "Theta=\\tan ^{-1} a$$",
        "solution": "$y=ax-bx^{2}$\nComparing the given equation with\n$$y=(\\tan \\theta)x-\\left(\\frac{g}{2 u^{2} \\cos ^{2} \\theta}\\right) x^{2}$$\nThen $a=\\tan \\theta$ and\n$$b=\\frac{g}{2 u^{2} \\cos ^{2} \\theta}$$\nMaximum height $=a^{2} / 4b$ and angle of projection,\n$$\\tan \\Theta=a$$\n$$\\Theta=\\tan ^{-1} a$$",
        "topic": "Kinematics 2D"
    },
    {
        "question": "Q1: Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section $5 \\mathrm{~mm}^{2}$ is $\\mathrm{v}$. If the electron density of copper is $9 \\times 10^{28} / \\mathrm{m}^{3}$ the value of $v$ in $\\mathrm{mm} / \\mathrm{s$ is close to (Take charge of electron to be $=1.6 \\times 10^{-19} \\mathrm{C}$)",
        "input": "3\n0.2\n2\n0.02",
        "answer": "(d) 0.02",
        "solution": "As $I=$ neAvd $=$ neAv\n$v=1 / n e A=1.5 /\\left(9 \\times 10^{28} \\times 1.6 \\times 10^{-19} \\times 5 \\times 10^{-6}\\right)$\n$v=0.02 \\times 10^{-3} \\mathrm{~m} / \\mathrm{s}=0.02 \\mathrm{~mm} / \\mathrm{s$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q2: Two equal resistances when connected in series to a battery, consume electric power of $60 \\mathrm{~W}$. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be",
        "input": "$240 \\mathrm{~W}$\n$120 \\mathrm{~W}$\n$60 \\mathrm{~W}$\n$30 \\mathrm{~W}$",
        "answer": "(a) $240 \\mathrm{~W}$",
        "solution": "The power consumed when two resistance are in series combination is $\\mathrm{V}^{2} / 2 \\mathrm{R}=60 \\mathrm{~W} \\Rightarrow \\mathrm{V}^{2} / \\mathrm{R}=120 \\mathrm{~W}$\nWhen the two resistance are connected in parallel combination, power consumed is $2 \\mathrm{~V}^{2} / \\mathrm{R}=120(2)=240 \\mathrm{~W}$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q3: A current of $2 \\mathrm{~mA}$ was passed through an unknown resistor which dissipated a power of $4.4 \\mathrm{~W}$. Dissipated power when an ideal power supply of $11 \\mathrm{~V}$ is connected across it is",
        "input": "$11 \\times 10^{-4} \\mathrm{~W}$\n$11 \\times 10^{-5} \\mathrm{~W}$\n$11 \\times 10^{5} \\mathrm{~W}$\n$11 \\times 10^{-3} \\mathrm{~W}$",
        "answer": "(b) $11 \\times 10^{-5} \\mathrm{~W}$",
        "solution": "Case (1)\nAs $I^{2} R=P$\n$R=P / I^{2}$\n$R=(4.4) /\\left(2 \\times 10^{.3}\\right)^{2}=1.1 \\times 10^{6} \\Omega$\nCase (2)\n$P=V^{2} / R=(1.1) 2 /\\left(1.1 \\times 10^{6}\\right)=11 \\times 10^{-5} \\mathrm{~W}$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q4: An ideal battery of $4 \\mathrm{~V}$ and resistance $\\mathrm{R}$ are connected in series in the primary circuit of a potentiometer of length $1 \\mathrm{~m}$ and resistance 5 . The value of $R$, to give a potential difference of $5 \\mathrm{mV}$ across $10 \\mathrm{~cm}$ of potentiometer wire is",
        "input": "490\n495\n395\n480",
        "answer": "(c) 395",
        "solution": "Let I be the current in the circuit. $4=(5+R)$ \\qquad\nAccording to given condition,\n$5 \\times 10^{-3}=(10 / 100)(5)(I)$\n$\\mathrm{I}=10^{-2} \\mathrm{~A}$\nUsing (1) and (2), $5+R=400 R=395$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q5: A cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when",
        "input": "$R=0.001 r$\n$R=r$\n$R=2 r$\n$R=1000 r$",
        "answer": "(b) $\\mathrm{R}=\\mathrm{r}$",
        "solution": "The power delivered to resistance is $I^{2} R$\ni.e., $P=\\left[\\varepsilon^{2} /(R+r)^{2}\\right] R$\nFor the maximum power, $\\mathrm{dP} / \\mathrm{dR}=0$\n$\\Rightarrow-2 R+(R+r)=0$ or $R=r$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q6: A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^{\\circ}$ at the centre, the equivalent resistance between these two points will be",
        "input": "$(7 / 2) \\Omega$\n$(5 / 2) \\Omega$\n$(12 / 5) \\Omega$\n$(5 / 3) \\Omega$",
        "answer": "(d) (5/3) $\\Omega$",
        "solution": "$R=3 \\Omega=\\rho(1 / A)=\\rho\\left(I^{2} / V\\right)$\n$R^{\\prime}=\\rho\\left(l^{\\prime} / V\\right) \\Rightarrow R^{\\prime}=(21)^{2} / /^{2} \\times 3 \\Rightarrow R^{\\prime}=12 \\Omega$\nEquivalent resistance, $\\mathrm{R}_{\\text {eq }}=(10 \\times 2) /(10+2)=(5 / 3) \\Omega$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q7: On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10 \\mathrm{~cm}$. The resistance of the combination is $1 \\mathrm{k} \\Omega$. How much was the resistance on the left slot before the interchange?",
        "input": "990\n505\n550\n910",
        "answer": "(c) 550",
        "solution": "Let $\\mathrm{R}_{1}$ (left slot) and $\\mathrm{R}_{2}$ (right slot) be two resistances in two slots of a meter bridge. Initially I be the balancing length\nThen, $\\mathrm{R}_{1} / \\mathrm{R}_{2}=\\mathrm{I} /(100-\\mathrm{I})$\n$\\mathrm{R}_{1}+\\mathrm{R}_{2}=1000 \\Omega$\nOn interchanging the resistances, balancing length becomes (I -10 ), so\n$\\left(R_{2} / R_{1}\\right)=(I-10) /(110-I)$\nUsing (1)\n$(100-I) / I=(I-10) /(110-I)$\n$11000+\\left.\\right|^{2}-210|=|^{2}-10 \\mid$\n$200 \\mathrm{I}=11000, \\mathrm{I}=55 \\mathrm{~cm}$\nFrom equa (1) $R_{1} / R_{2}=55 / 45$\nUsing (2)\n$\\mathrm{R}_{1}=(55 / 45)\\left(1000-\\mathrm{R}_{1}\\right)$\n$\\mathrm{R}_{1}+(55 / 45) \\mathrm{R}_{1}=(1000) \\times(55 / 45)$\n$100 \\mathrm{R}_{1}=1000 \\times 55$\n$\\mathrm{R}_{1}=550 \\Omega$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q8: A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be",
        "input": "Increased 8 times\nUnchanged\nDoubled\nHalved",
        "answer": "(a) Increased 8 times",
        "solution": "Rate of heat developed, $\\mathrm{P}=\\mathrm{V}^{2} / \\mathrm{R}$\nFor given $V, P \\propto 1 / R=A / \\rho l=\\pi r^{2} / \\rho l$\nNow, $\\mathrm{P}_{1} / \\mathrm{P}_{2}=\\left(\\mathrm{r}_{1}^{2} / \\mathrm{r}_{2}^{2}\\right)\\left(\\mathrm{I}_{2} / I_{1}\\right)$\nAs per question, $l_{2}=I_{1} / 2$ and $r_{2}=2 r_{1}$\n$P_{1} / P_{2}=(1 / 4) \\times(1 / 2)=1 / 8$\n$\\mathrm{P}_{2}=8 \\mathrm{P}_{1}$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q9: A heating element has a resistance of $100 \\Omega$ at room temperature. When it is connected to a supply of $220 \\mathrm{~V}$, a steady current of $2 \\mathrm{~A}$ passes in it and the temperature is $500^{\\circ} \\mathrm{C}$ more than room temperature. What is the temperature coefficient of resistance of the heating element?",
        "input": "$1 \\times 10^{-4}{ }^{\\circ} \\mathrm{C}^{-1}$\n$2 \\times 10^{-4}{ }^{\\circ} \\mathrm{C}^{-1}$\n$0.5 \\times 10^{-4}{ }^{\\circ} \\mathrm{C}^{-1}$\n$5 \\times 10^{-4}{ }^{\\circ} \\mathrm{C}^{-1}$",
        "answer": "(b) $2 \\times 10^{-4}{ }^{\\circ} \\mathrm{C}^{-1}$",
        "solution": "Resistance after temperature increases by $500^{\\circ} \\mathrm{C}$,\n$\\mathrm{R}_{\\mathrm{T}}=$ Voltage applied/Current $=220 / 2=110$\nAlso, $\\mathrm{R}_{\\mathrm{T}}=\\mathrm{R}_{0}(1+\\alpha \\Delta \\mathrm{T})$\n$110=100(1+(\\alpha \\times 500))$\n$\\alpha=10 /(100 \\times 500)=2 \\times 10^{-4} \\mathrm{C}^{-1}$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q10: A uniform wire of length I and radius $r$ has a resistance of $100 \\Omega$. It is recast into a wire of radius $\\mathrm{r} / 2$.The resistance of new wire will be",
        "input": "$400 \\Omega$\n$100 \\Omega$\n$200 \\Omega$\n$1600 \\Omega$",
        "answer": "(d) $1600 \\Omega$",
        "solution": "Resistance of a wire of length I and radius $r$ is given by\n$R=\\rho I / A=(\\rho I / A) x(A / A)$\n$\\mathrm{R}=\\left(\\rho \\mathrm{I} / \\mathrm{A}^{2}\\right)=\\left(\\rho \\mathrm{V} / \\pi^{2} \\mathrm{r}^{4}\\right)(\\because \\mathrm{V}=\\mathrm{Al})$\ni.e., $R \\propto 1 / r^4$\n$R_{1} / R_{2}=\\left(r_{2} / r_{1}\\right)^{4}$\nHere, $R_{1}=100 \\Omega, r_{1}=r, r_{2}=r / 2$\n$R_{2}=R_{1}\\left(r_{1} / r_{2}\\right)^{4}=16 R_{1}=1600 \\Omega$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q11: A 2 W carbon resistor is colour coded with green, black, red and brown, respectively. The maximum current which can be passed through this resistor is",
        "input": "$20 \\mathrm{~mA}$\n$0.4 \\mathrm{~mA}$\n$100 \\mathrm{~mA}$\n$63 \\mathrm{~mA}$",
        "answer": "(a) $20 \\mathrm{~mA}$",
        "solution": "The resistance of the resistor is $50 \\times 10^{2} \\Omega$. So, the maximum current that can be passed through it is\n$$\n\\sqrt{\\frac{P}{R}}=\\sqrt{\\frac{2}{50 \\times 10^{2}}} A=20 \\mathrm{~mA}$$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q12: In a large building, there are 15 bulbs of $40 \\mathrm{~W}, 5$ bulbs of $100 \\mathrm{~W}, 5$ fans of $80 \\mathrm{~W}$ and 1 heater of 1 kW. The voltage of the electric mains is $220 \\mathrm{~V}$. The minimum capacity of the main fuse of the building will be",
        "input": "$14 \\mathrm{~A}$\n$8 \\mathrm{~A}$\n$10 \\mathrm{~A}$\n$12 \\mathrm{~A}$",
        "answer": "(d) $12 \\mathrm{~A}$",
        "solution": "Power of 15 bulbs of $40 \\mathrm{~W}=15 \\times 40=600 \\mathrm{~W}$\nPower of 5 bulbs of $100 \\mathrm{~W}=5 \\times 100=500 \\mathrm{~W}$\nPower of 5 fans of $80 \\mathrm{~W}=5 \\times 80=400 \\mathrm{~W}$\nPower of 1 heater of $1 \\mathrm{~kW}=1000$\nTotal power, $P=600+500+400+1000=2500 \\mathrm{~W}$\nWhen these combinations of bulbs, fans and heater are connected to $220 \\mathrm{~V}$ mains, current in the main fuse of the building is given by\n$\\mathrm{I}=\\mathrm{P} / \\mathrm{N}=2500 / 220=11.36 \\mathrm{~A} \\approx 12 \\mathrm{~A}$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q13: If a wire is stretched to make it $0.1 \\%$ longer, its resistance will",
        "input": "increase by $0.05 \\%$\nincrease by $0.2 \\%$\ndecrease by $0.2 \\%$\ndecrease by $0.05 \\%$",
        "answer": "(b) increase by $0.2 \\%$",
        "solution": "Resistance of wire $\\mathrm{R}=\\rho / / \\mathrm{A}$\nOn stretching, volume $(\\mathrm{V})$ remains constant.\nSo $\\mathrm{V}=\\mathrm{Al}$ or $\\mathrm{A}=\\mathrm{V} / \\mathrm{I}$\nTherefore, $R=\\rho l^{2} / V$ (Using (1))\nTaking logarithm on both sides and differentiating we get,\n$\\Delta R / R=2 \\Delta I / I$ (Since $V$ and $\\rho$ are constants)\n$(\\Delta R / R) \\%=(2 \\Delta I / l) \\%$\nHence, when wire is stretched by $0.1 \\%$ its resistance will increase by $0.2 \\%$",
        "topic": "Current Electricity"
    },
    {
        "question": "Q14: A thermocouple is made from two metals, antimony and bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will",
        "input": "flow from antimony to bismuth at the cold junction\nflow from antimony to bismuth at the hot junction\nflow from bismuth to antimony at the cold junction\nnot flow through the thermocouple.",
        "answer": "(a) flow from antimony to bismuth at the cold junction",
        "solution": "Antimony-bismuth couple is $A B C$ couple. It means that current flows from $A$ to $B$ at a cold junction",
        "topic": "Current Electricity"
    },
    {
        "question": "Q15: The resistance of a wire is 5 ohm at $50^{\\circ} \\mathrm{C}$ and 6 ohm at $100^{\\circ} \\mathrm{C}$. The resistance of the wire at $0^{\\circ} \\mathrm{C}$ will be",
        "input": "3 ohm\n2 ohm\n1 ohm\n4 ohm",
        "answer": "(d) 4 ohm",
        "solution": "$R_{t}=R_{0}(1+\\alpha t)$\n$R_{t}$ is the resistance of wire at $t^{0} \\mathrm{C}$\n$R_{0}$ is the resistance of wire at $0^{\\circ} \\mathrm{C}$\n$\\alpha$ is the temperature coefficient of resistance\n$R_{50}=R_{0}[1+\\alpha(50)]$\nAnd $R_{100}=R_{0}[1+\\alpha(100)]$\nOr $\\mathrm{R}_{50}-\\mathrm{R}_{0}=\\mathrm{R}_{0} \\alpha(50)$\n$R_{100}-R_{0}=R_{0} \\alpha(100)$\nDividing (1) by (2)\n$\\left(5-\\mathrm{R}_{0}\\right) /\\left(\\left(6-\\mathrm{R}_{0}\\right)=1 / 2\\right.$\n$\\mathrm{R}_{0}=4$ ohm",
        "topic": "Current Electricity"
    },
    {
        "question": "Q1: A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity 100 m / s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g=10 m / s)",
        "input": "(a) 10 m\n(b) 30 m\n(c) 20 m\n(d) 40 m",
        "answer": "(d) 40 m",
        "solution": "Suppose both collide at the point P after time t. Time taken for the particles to collide, t=d / v_rel = 100 / 100 = 1 s\nSpeed of wood just before collision = g t = 10 m / s\nSpeed of bullet just before collision v - g t = 100 - 10 = 90 m / s\nBefore 0.03 kg \u2193 10 m / s\n0.02 kg \u2191 90 m / s\nAfter \u4e2a v 0.05 kg\nNow, the conservation of linear momentum just before and after the collision - (0.03)(10) + (0.02)(90) = (0.05) v \u21d2 v = 30 m / s\nThe maximum height reached by the body a = v\u00b2 / 2 g = (30)\u00b2 / 2(10) = 45 m\n(100 - h) = 1 / 2 g t\u00b2 = 1 / 2 \u00d7 10 \u00d7 1 \u21d2 h = 95 m\nHeight above tower = 40 m",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q2: A passenger train of length 60 m travels at a speed of 80 km / hr. Another freight train of length 120 m travels at a speed of 30 km / hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction, is",
        "input": "(a) 25 / 11\n(b) 3 / 2\n(c) 5 / 2\n(d) 11 / 5",
        "answer": "(d) 11 / 5",
        "solution": "The total distance to be travelled by train is 60 + 120 = 180 m.\nWhen the trains are moving in the same direction, the relative velocity is v\u2081 - v\u2082 = 80 - 30 = 50 km hr\u207b\u00b9. So time taken to cross each other, t\u2081 = 180 / (50 \u00d7 10\u00b3 / 3600) = [(18 \u00d7 18) / 25] s\nWhen the trains are moving in opposite direction, relative velocity is |v\u2081 - (- v\u2082)| = 80 + 30 = 110 km hr\u207b\u00b9\nSo time taken to cross each other t\u2082 = 180 / (110 \u00d7 10\u00b3 / 3600) = [(18 \u00d7 36) / 110] s\nt\u2081 / t\u2082 = [(18 \u00d7 18) / 25] / [(18 \u00d7 36) / 110] = 11 / 5",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q3: A particle has an initial velocity 3\\hat{i}+4\\hat{j} and an acceleration of 0.4\\hat{i}+0.3\\hat{j}. Its speed after 10 s is",
        "input": "(a) 7 units\n(b) 8.5 units\n(c) 10 units\n(d) 7\\sqrt{2} units",
        "answer": "(d) 7\\sqrt{2} units",
        "solution": "v = u + a t\nu = 3 i + 4 j + (0.4 i + 0.3 j) \u00d7 10 = 3 i + 4 j + 4 i + 3 i\nu = 7 i + 7 j\n|u| = \\sqrt{7\u00b2 + 7\u00b2}",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q4: An automobile, travelling at 40 km / h, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km / h, the minimum stopping distance, in metres, is (assume no skidding)",
        "input": "(a) 100 m\n(b) 75 m\n(c) 160 m\n(d) 150 m",
        "answer": "(c) 160 m",
        "solution": "Using v\u00b2 = u\u00b2 - 2 a s\n0 = u\u00b2 - 2 a s\nS = u\u00b2 / 2 a\nS\u2081 / S\u2082 = u\u2081\u00b2 / u\u2082\u00b2\nS\u2082 = (u\u2081\u00b2 / u\u2082\u00b2) S\u2081 = (2)\u00b2 (40) = 160 m",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q7. A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m / s\u00b2. He reaches the ground with a speed of 3 m / s. At what height, did he bailout?",
        "input": "(a) 293 m\n(b) 111 m\n(c) 91 m\n(d) 182 m",
        "answer": "(a) 293 m",
        "solution": "Initially, the parachutist falls under gravity u\u00b2 = 2 a h = 2 \u00d7 9.8 \u00d7 50 = 980 m\u00b2 s\u207b\u00b2\nHe reaches the ground with speed = 3 m / s,\na = -2 m s\u207b\u00b2 \u21d2 (3)\u00b2 = u\u00b2 - 2 \u00d7 2 \u00d7 h\u2081\n9 = 980 - 4 h\u2081\nh\u2081 = 971 / 4\nh\u2081 = 242.75 m\nTotal height = 50 + 242.75 = 292.75 = 293 m.",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q8: A car, starting from rest, accelerates at the rate f through a distance s, then continues at a constant speed for time t and then decelerates at the rate f / 2 to come to rest. If the total distance traversed in 15 s, then",
        "input": "(a) s = 1 / 2 ft\u00b2\n(b) s = (1 / 4) ft\u00b2\n(c) s = ft\n(d) s = (1 / 72) ft\u00b2",
        "answer": "(d) s = (1 / 72) ft\u00b2",
        "solution": "For the first part of the journey, s = s\u2081,\ns\u2081 = 1 / 2 ft\u2081\u00b2\nv = f t\u2081\nFor second part of journey, s\u2082 = v t or s\u2082 = f t\u2081 t\nFor the third part of the journey, s\u2083 = 1 / 2(f / 2)(2 t\u2081)\u00b2 = 1 / 2 \u00d7 ft\u2081\u00b2\ns\u2083 = 2 s\u2081 = 2 s\ns\u2081 + s\u2082 + s\u2083 = 15 s\nOrs s + ft\u2081 t + 2 s = 15 s\nft\u2081 t = 12 s\nFrom (1) and (5) we get (s / 12 s) = ft\u2081\u00b2 / (2 \u00d7 ft\u2081 t)\nOr t\u2081 = t / 6\nOr s = 1 / 2 ft\u2081\u00b2\ns = 1 / 2 f(t / 6)\u00b2\ns = ft\u00b2 / 72",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q10: Which of the following statements is false for a particle moving in a circle with a constant angular speed?",
        "input": "(a) The velocity vector is tangent to the circle\n(b) The acceleration vector is tangent to the circle\n(c) The acceleration vector points to the centre of the circle\n(d) The velocity and acceleration vectors are perpendicular to each other",
        "answer": "(b) The acceleration vector acts along the radius of the circle. The given statement is false.",
        "solution":"",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q11: From a building, two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then",
        "input": "(a) $v_{B}>v_{A}$\n(b) $v_{A}=v_{B}$\n(c) $v_{A}>v_{B}$\n(d) their velocities depend on their masses",
        "answer": "(b) $v_{A}=v_{B}$",
        "solution": "Ball A projected upwards with velocity $u$, falls back with velocity $u$ downwards. It completes its journey to the ground under gravity.\n$v_{A}^{2}=u^{2}+2 g h \\ldots(1)$\nBall B starts with downwards velocity $u$ and reaches ground after travelling a vertical distance $h \\quad v_{B} 2=$ $u^{2}+2 g h$\nFrom (1) and (2)\n$v_{A}=v_{B}$",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q12: If a body loses half of its velocity on penetrating $3 \\mathrm{~cm}$ in a wooden block, then how much will it penetrate more before coming to rest?",
        "input": "(a) $1 \\mathrm{~cm}$\n(b) $2 \\mathrm{~cm}$\n(c) $3 \\mathrm{~cm}$\n(d) $4 \\mathrm{~cm}$",
        "answer": "(a) $1 \\mathrm{~cm}$",
        "solution": "For first part of penetration, by equation of motion, $(u / 2)^{2}=u^{2}-2 a(3)$ $3 u^{2}=24 a \\Rightarrow u^{2}=8 a$ For latter part of penetration, $0=(u / 2)^{2}-2 a x$ or $u^{2}=8 a x$.\nFrom (1) and (2)\n$8 \\mathrm{ax}=8 \\mathrm{a} \\quad \\mathrm{x}=1 \\mathrm{~cm}$",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q13: A car is standing $\\mathbf{2 0 0} \\mathrm{m}$ behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2 \\mathrm{~m} / \\mathrm{s}^{2}$ and the car has acceleration $4 \\mathrm{~m} / \\mathrm{s}^{2}$. The car will catch up with the bus after a time of",
        "input": "(a) $120 \\mathrm{~s}$\n(b) $15 \\mathrm{~s}$\n(c) [latex]2\\textbackslash sqrt\\{2\\}[/latex] s\n(d) $110 \\mathrm{~s}$",
        "answer": "(c) [latex]2\\textbackslash sqrt\\{2\\}[/latex] s",
        "solution": "Acceleration of car, $\\mathrm{ac}_{\\mathrm{c}}=4 \\mathrm{~m} \\mathrm{~s}^{-2}$\nAcceleration of bus, $a_{B}=2 \\mathrm{~m} \\mathrm{~s}^{-2}$\nInitial separation between the bus and car, $S_{C B}=200 \\mathrm{~m}$\nAcceleration of car with respect to bus, $a_{C B}=a_{c}-a_{B}=2 \\mathrm{~m} \\mathrm{~s}^{-2}$\nInitial velocity $u_{C B}=0, t=$ ?\nAs, $s_{C B}=u_{C B} \\times t+1 / 2 a_{C B} t^{2}$\n$200=0 \\times t+1 / 2 \\times 2 \\times t^{2}$\ni.e., $\\mathrm{t}^{2}=200$\n$\\mathrm{t}=[2\\textbackslash sqrt{2}]$",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q14: From a tower of height $H$, a particle is thrown vertically upwards with a speed $u$. The time taken by the particle, to hit the ground is $n$ times that taken by it to reach the highest point of its path. The relation between $\\mathrm{H}, \\mathrm{u}$ and $\\mathrm{n}$ is",
        "input": "(a) $g H=(n-2) u^{2}$\n(b) $2 g H=n 2 u^{2}$\n(c) $g H=(n-2) 2 u^{2}$\n(d) $2 g H=n u^{2}(n-2)$",
        "answer": "(d) $2 g H=n u^{2}(n-2)$",
        "solution": "",
        "topic": "Kinematics 1D"
    },
    {
        "question": "Q1: A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is",
        "input": "(a) 20 RT\n(b) 12 RT\n(c) 4 RT\n(d) 15 RT",
        "answer": "(d) 15 RT",
        "solution": "$U=\\left(f_{1} / 2\\right) n_{1} R T+\\left(f_{2} / 2\\right) n_{2} R T$\n$U=(5 / 2)(3 R T)+(3 / 2)(5 R T)$\n$U=15 R T$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q2: An ideal gas is enclosed in a cylinder at a pressure of 2 atm and temperature, $300 \\mathrm{~K}$. The mean time between two successive collisions is $6 \\times 10^{-8} \\mathrm{~s}$. If the pressure is doubled and the temperature is increased to $500 \\mathrm{~K}$, the mean time between two successive collisions will be close to",
        "input": "(a) $4 \\times 10^{-8} \\mathrm{~s}$\n(b) $3 \\times 10^{-6} \\mathrm{~s}$\n(c) $0.5 \\times 10^{-8} \\mathrm{~S}$\n(d) $2 \\times 10^{-7} \\mathrm{~s}$",
        "answer": "(a) $4 \\times 10^{-8} \\mathrm{~s}$",
        "solution": "Root mean square velocity\n$$\n\\operatorname{Vrms}=\\sqrt{\\frac{3 R T}{M}}=\\sqrt{\\frac{3 P}{\\rho}}\n$$\nHere,\n$R$ is the universal gas constant\n$M$ is the molar mass\n$\\mathrm{P}$ is the pressure due to gas\n$\\rho$ is the density\nVrms $\\propto \\sqrt{T}$\n$\\mathrm{V}_{\\mathrm{rms}} \\propto$ mean free path/ time between successive collisions\nAnd mean free path,\n$$\nY=\\frac{k T}{\\sqrt{2 \\pi \\sigma^{2} P}}\n$$\nVrms $\\propto \\mathrm{Y} / \\mathrm{b}$\n$$\n\\begin{aligned}\n& \\text { Vrms } \\propto T / \\sqrt{p} \\times t---(1) \\\\\n& \\text { But Vrms } \\propto \\sqrt{T}----(2)\n\\end{aligned}\n$$\n$\\sqrt{T} \\propto T / \\sqrt{p} \\times t$\n$\\mathrm{t} \\propto \\sqrt{T} / p \\quad \\frac{t_{2}}{t_{1}}=\\sqrt{\\frac{T_{2}}{T_{1}} \\times \\frac{P_{1}}{P_{2}}} \\frac{t_{2}}{t_{1}}$\n$=\\sqrt{\\frac{500}{300} \\times \\frac{P_{1}}{2 P_{1}}}=\\sqrt{\\frac{5}{6}} t_{2}=\\sqrt{\\frac{5}{6}} t_{1}$\n$\\mathrm{t}_{2} \\approx 4 \\times 10^{-8} \\mathrm{~s}$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q3: The specific heats, $C_{P}$ and $C_{V}$ of gas of diatomic molecules, $A$, is given (in units of $\\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ ) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then",
        "input": "(a) A has a vibrational mode but $\\mathrm{B}$ has none.\n(b) A has one vibrational mode and $B$ has two.\n(c) $A$ is rigid but $B$ has a vibrational mode.\n(d) Both $A$ and $B$ have a vibrational mode each.",
        "answer": "(a) A has a vibrational mode but $B$ has none.",
        "solution": "Here $\\mathrm{Cp}$ and $\\mathrm{Cv}$ of A are 29 and 22 and $\\mathrm{Cp}$ and $\\mathrm{Cv}$ of $\\mathrm{B}$ are 30 and 21 .\n$\\nu=\\mathrm{Cp} / \\mathrm{Cv}=1+2 / \\mathrm{f}$\nFor $\\mathrm{A}, \\mathrm{Cp} / \\mathrm{Cv}=1+2 / \\mathrm{f} \\Rightarrow \\mathrm{f}=6$\nMolecules $A$ has 3 translational, 2 rotational and 1 vibrational degree of freedom\nFor $B, C p / C v=1+2 / f \\Rightarrow f=5$\ni.e., $B$ has 3 translational and 2 rotational degrees of freedom.",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q4: Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? ( $R=8.3 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$ )",
        "input": "(a) $21.6 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$\n(b) $19.7 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$\n(c) $15.7 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$\n(d) $17.4 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$",
        "answer": "(d) $17.4 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$",
        "solution": "$\\mathrm{C}_{\\mathrm{v} 1}$ of helium $=(3 / 2) \\mathrm{R}$\n$\\mathrm{C}_{\\mathrm{v} 2}$ of hydrogen $=(5 / 2) \\mathrm{R}$\n$$\n\\mathrm{C}_{\\mathrm{v}} \\text { of mixture }=\\frac{2 \\times \\frac{3}{2} R+3 \\times \\frac{5}{2} R}{(2+3)}\n$$\n$C_{v}$ of mixture $=17.4 \\mathrm{~J} / \\mathrm{mol} \\mathrm{K}$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q5: The mass of a hydrogen molecule is $3.32 \\times 10^{-27} / \\mathrm{kg}$. If $10^{23}$ hydrogen molecules strike, per second, a fixed wall of area $2 \\mathrm{~cm}^{2}$ at an angle of $45^{\\circ}$ to the normal, and rebound elastically with a speed of $10^{3} \\mathrm{~m} \\mathrm{~s}-1$, then the pressure on the wall is nearly",
        "input": "(a) $2.35 \\times 10^{3} \\mathrm{~N} \\mathrm{~m}^{-2}$\n(b) $4.70 \\times 10^{3} \\mathrm{~N} \\mathrm{~m}^{-2}$\n(c) $2.35 \\times 10^{2} \\mathrm{~N} \\mathrm{~m}^{-2}$\n(d) $4.70 \\times 10^{2} \\mathrm{~N} \\mathrm{~m}^{-2}$",
        "answer": "(c) $2.35 \\times 10^{2} \\mathrm{~N} \\mathrm{~m}^{-2}$",
        "solution": "As $\\mathrm{p}_{\\mathrm{i}}=\\mathrm{p}_{\\mathrm{f}}$\nNet force on the wall,\n$\\mathrm{F}=\\mathrm{dp} / \\mathrm{dt}=2 \\mathrm{np} \\mathrm{f}_{\\mathrm{f}} \\cos 45^{\\circ}=2 \\mathrm{nmv} \\cos 45^{\\circ}$\nHere, $\\mathrm{n}$ is the number of hydrogen molecules striking per second.\nPressure $=F / A=\\left(2 n m v \\cos 45^{\\circ}\\right) /$ Area\n$$\n=\\frac{2 \\times 10^{23} \\times 3.32 \\times 10^{-27} \\times 10^{3} \\times(1 / \\sqrt{2})}{2 \\times 10^{-4}}=2.35 \\times 10^{3} \\mathrm{~N} \\mathrm{~m}^{-2}\n$$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q6: In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on $\\mathrm{T}$ as $\\mathrm{T}^{q}$. A good estimate for $q$ is",
        "input": "(a) 2\n(b) 1\n(c) $4 / 5$\n(d) $4 / 7$",
        "answer": "(b) 1",
        "solution": "Pressure, $\\mathrm{P}=1 / 3(\\mathrm{mN} / \\mathrm{V}) \\mathrm{V}_{\\text {rms }}^{2}$\n$\\mathrm{P}=(\\mathrm{mN}) \\mathrm{T} / \\mathrm{V}$\nIf the gas mass and temperature are constant then\n$\\mathrm{P} \\propto\\left(\\mathrm{V}_{\\mathrm{rms}}\\right)^{2} \\propto \\mathrm{T}$\nSo, force $\\propto\\left(\\mathrm{V}_{\\mathrm{rms}}\\right)^{2} \\propto \\mathrm{T}$ I.e., Value of $q=1$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q7: One $\\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \\times 10^{4} \\mathrm{~N} / \\mathrm{m}^{2}$. The density of the gas is $4 \\mathrm{~kg} / \\mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?",
        "input": "(a) $3 \\times 10^{4} \\mathrm{~J}$\n(b) $5 \\times 10^{4} \\mathrm{~J}$\n(c) $6 \\times 10^{4} \\mathrm{~J}$\n(d) $7 \\times 10^{4} \\mathrm{~J}$",
        "answer": "(b) $5 \\times 10^{4} \\mathrm{~J}$",
        "solution": "The thermal energy or internal energy is $52 \\mathrm{U} T$ for diatomic gases. (degree of freedom for diatomic gas $=5$ )\nBut PV $=$ RT\n$V=$ mass $/$ density $=1 \\mathrm{~kg} /\\left(4 \\mathrm{~kg} / \\mathrm{m}^{3}\\right)=(1 / 4) \\mathrm{m}^{3}$\n$P=8 \\times 10^{4} \\mathrm{~N} / \\mathrm{m}^{2}$\n$U=(5 / 2) \\times 8 \\times 10^{4} \\times 1 / 4=5 \\times 10^{4} \\mathrm{~J}$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q8: A gaseous mixture consists of $16 \\mathrm{~g}$ of helium and $16 \\mathrm{~g}$ of oxygen. The ratio $\\mathrm{C}_{\\mathrm{P}} / \\mathrm{C}_{V}$ of the mixture is",
        "input": "(a) 1.4\n(b) 1.54\n(c) 1.59\n(d) 1.62",
        "answer": "(d) 1.62",
        "solution": "For $16 \\mathrm{~g}$ of helium, $\\mathrm{n}_{1}=16 / 4=4$\nFor $16 \\mathrm{~g}$ of oxygen, $\\mathrm{n}_{2}=16 / 32=1 / 2$\nFor a mixture of gases\n$\\mathrm{C}_{\\mathrm{v}}=\\left(\\mathrm{n}_{1} \\mathrm{Cv}_{1}+\\mathrm{n}_{2} \\mathrm{Cv}_{2}\\right) /\\left(\\mathrm{n}_{1}+\\mathrm{n}_{2}\\right)$ where $\\mathrm{C}_{\\mathrm{v}}=(\\mathrm{f} / 2) \\mathrm{R}$\n$C_{p}=\\left(n_{1} C_{1}+n_{2} C p_{2}\\right) /\\left(n_{1}+n_{2}\\right)$ where $C_{p}=([f / 2]+1) R$\nFor helium, $f=3, n_{1}=4$\nFor oxygen, $f=5, n_{2}=1 / 2$\n$$\n\\frac{C p}{C v}=\\frac{\\left(4 \\times \\frac{5}{2} R\\right)+\\left(\\frac{1}{2} \\times \\frac{7}{2} R\\right)}{\\left(4 \\times \\frac{3}{2} R\\right)+\\left(\\frac{1}{2} \\times \\frac{5}{2} R\\right)}=\\frac{47}{29}=1.62\n$$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q9: Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will",
        "input": "(a) increase\n(b) decrease\n(c) remain the same\nd) decrease for some, while the increase for others.",
        "answer": "(c) remain the same",
        "solution": "It is the relative velocities between molecules that are important. Root mean square velocities are different from lateral translation.",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q10: At what temperature is the root mean square velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\\circ} \\mathrm{C}$ ?",
        "input": "(a) $80 \\mathrm{~K}$\n(b) $-73 K$\n(c) $3 \\mathrm{~K}$\n(d) $20 \\mathrm{~K}$.",
        "answer": "(d) $20 \\mathrm{~K}$",
        "solution": "$$\nv_{r m s}=\\sqrt{R T / M}\\left(\\mathrm{~V}_{\\mathrm{rms}}\\right) \\mathrm{O}_{2}=\\left(\\mathrm{V}_{\\mathrm{rms}}\\right) \\mathrm{H}_{2}\n$$\n$$\\sqrt{\\frac{273+47}{32}}=\\sqrt{\\frac{T}{2}}$$\n$\\mathrm{T}=20 \\mathrm{~K}$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q11:The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to [Boltzmann constant, $\\mathrm{k}_{\\mathrm{B}}=1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K}$, Avogadro Number, $\\mathrm{N}_{\\mathrm{A}}=6.02 \\times 10^{26} / \\mathrm{kg}$, Radius of Earth: $6.4 \\times 10^{6} \\mathrm{~m}$, Gravitational acceleration on Earth $=10 \\mathrm{~m} \\mathrm{~s}^{-2}$ ]",
        "input": "(a) $10^{4} \\mathrm{~K}$\n(b) $800 \\mathrm{~K}$\n(c) $650 \\mathrm{~K}$\n(d) $3 \\times 10^{5} \\mathrm{~K}$",
        "answer": "(a) $10^{4} \\mathrm{~K}$",
        "solution": "The escape speed of the molecule, $v_{e}=\\sqrt{2 g R}$\nRoot mean square velocity, $v_{r m s}=\\sqrt{\\frac{\\left.3\\left(k_{B} N\\right) T\\right)}{m}}$\nSo, for $v_{e}=v_{r m s} \\Rightarrow 2 g R=3\\left(k_{B} \\mathrm{~N}\\right) \\mathrm{T} / \\mathrm{m}$\n$\\mathrm{T}=2 \\mathrm{gRm} / 3 \\mathrm{k}_{\\mathrm{B}} \\mathrm{N}=\\left(2 \\times 10 \\times 6.4 \\times 10^{6} \\times 2 \\times 10^{-3}\\right) /\\left(3 \\times 1.38 \\times 10^{-23} \\times 6.02 \\times 10^{23}\\right)=10^{4} \\mathrm{~K}$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q13: The value closest to the thermal velocity of a Helium atom at room temperature ( $300 \\mathrm{~K}$ ) in $\\mathrm{m} \\mathrm{s}$ ${ }^{1}$ is $\\left[k_{B}=1.4 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K} ; \\mathrm{m}_{\\mathrm{He}}=7 \\times 10^{-27} \\mathrm{~kg}\\right]$",
        "input": "(a) $1.3 \\times 10^{3}$\n(b) $1.3 \\times 10^{5}$\n(c) $1.3 \\times 10^{2}$\n(d) $1.3 \\times 10^{4}$",
        "answer": "(a) $1.3 \\times 10^{3}$",
        "solution": "$$\n\\begin{aligned}\n& \\text { (3/2) } \\mathrm{k}_{\\mathrm{B}} \\mathrm{T}=1 / 2 \\mathrm{mv} \\\\\n& v=\\sqrt{\\frac{3 k_{B} T}{m}}=\\sqrt{\\frac{3 \\times 1.4 \\times 10^{-23} \\times 300}{7 \\times 10^{-27}}} \\\\\n& =1.3 \\times 10^{3} \\mathrm{~ms}^{-1}\n\\end{aligned}\n$$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q14: An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure ( $C_{P}$ ) and at constant volume ( $C_{V}$ ) is",
        "input": "(a) 6\n(b) $7 / 2$\n(c) $5 / 2$\n(d) $7 / 5$",
        "answer": "(d) $7 / 5$",
        "solution": "An ideal gas has molecules with 5 degrees of freedom, then\n$\\mathrm{Cv}=(5 / 2) \\mathrm{R}$ and $\\mathrm{Cp}=(7 / 2) \\mathrm{R}$\n$\\mathrm{Cp} / \\mathrm{Cv}=(7 / 2) \\mathrm{R} /(5 / 2) \\mathrm{R}$\n$\\mathrm{Cp} / \\mathrm{Cv}=7 / 5$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "Q15: $\\mathbf{N}$ moles of a diatomic gas in a cylinder is at a temperature $\\mathrm{T}$. Heat is supplied to the cylinder such that the temperature remains constant but $n$ moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas?",
        "input": "(a) 0\n(b) $(5 / 2) n R T$\n(c) $\\left(\\frac{1}{2}\\right) \\mathrm{nRT}$\n(d) $(3 / 2) n R T$",
        "answer": "(c) $(1 / 2)$ nRT",
        "solution": "Initial kinetic energy of the system $K_{i}=(5 / 2)$ RTN\nFinal kinetic energy of the system $K_{f}=(5 / 2) R T(N-n)+(3 / 2) R T(2 n)$\n$K_{f}-K_{i}=\\Delta K=n R T(3-5 / 2)=(1 / 2) n R T$",
        "topic": "Kinetic Theory of Gases"
    },
    {
        "question": "A paramagnetic material has $10^{28}$ atoms $/ \\mathrm{m}^{3}$. It's magnetic susceptibility at temperature $350 \\mathrm{~K}$ is $2.8 \\times 10^{-4}$. Its susceptibility at $300 \\mathrm{~K}$ is",
        "input": "(a) $3.267 \\times 10^{-4}$, (b) $3.672 \\times 10^{-4}$, (c) $2.672 \\times 10^{-4}$, (d) $3.726 \\times 10^{-4}$",
        "answer": "(a) $3.267 \\times 10^{-4}$",
        "solution": "For a paramagnetic material, magnetic susceptibility $\\chi \\propto 1 / T$ $\\chi_{2}=\\chi_{1}\\left(T_{1} / T_{2}\\right)=2.8 \\times 10^{-4} \\times(350 / 300)=3.267 \\times 10^{-4}$",
        "topic": "Magnetism"
    },
    {
        "question": "A magnetic compass needle oscillates 30 times per minute at a place where the dip is $45^{\\circ}$, and 40 times per minute where the dip is $30^{\\circ}$. If $B_{1}$ and $B_{2}$ are respectively the total magnetic field due to the earth at the two places, then the ratio $B_{1} / B_{2}$ is best given by",
        "input": "(a) 0.7, (b) 3.6, (c) 1.8, (d) 2.2",
        "answer": "(a) 0.7",
        "solution": "Frequency of oscillation $=\\sqrt{\\text { magnetic field }}$ $\\frac{30}{40}=\\sqrt{\\frac{B_{1} \\cos 45^{\\circ}}{B_{2} \\cos 30^{\\circ}}}$ (Horizontal component of magnetic field $=\\mathrm{B} \\cos \\delta$ ) $\\left(B_{1} / B_{2}\\right)=0.7$",
        "topic": "Magnetism"
    },
    {
        "question": "A magnetic needle of magnetic moment $6.7 \\times 10^{-2} \\mathrm{~A} \\mathrm{~m}^{2}$ and moment of inertia $7.5 \\times 10^{-6} \\mathrm{~kg} \\mathrm{~m}^{2}$ is performing simple harmonic oscillations in a magnetic field of $0.01 \\mathrm{~T}$. Time taken for 10 complete oscillations is",
        "input": "(a) $6.65 \\mathrm{~s}$, (b) $8.89 \\mathrm{~s}$, (c) $6.98 \\mathrm{~s}$, (d) $8.76 \\mathrm{~s}$",
        "answer": "(a) $6.65 \\mathrm{~s}$",
        "solution": "Time period of magnetic needle oscillating simple harmonically is given by $T=2 \\pi \\sqrt{\\frac{I}{M B}} T=2 \\pi \\sqrt{\\frac{7.5 \\times 10^{-6}}{6.7 \\times 10^{-2} \\times 0.01}}$ $\\mathrm{T}=(2 \\pi / 10) \\times 1.05 \\mathrm{~s}$ For 10 oscillations, total time taken $\\mathrm{T}^{\\prime}=10 \\mathrm{~T}=2 \\pi \\times 1.05=6.65 \\mathrm{~s}$",
        "topic": "Magnetism"
    },
    {
        "question": "A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\\circ}$. One of the fields has a magnitude of $15 \\mathrm{mT}$. The dipole attains stable equilibrium at an angle of $30^{\\circ}$ with this field. The magnitude of the other field (in $\\mathrm{mT}$ ) is close to",
        "input": "(a) 1, (b) 11, (c) 36, (d) 1060",
        "answer": "(b) 11",
        "solution": "The magnetic dipole attains stable equilibrium under the influence of these two fields making an angle $\\theta_{1}=30^{\\circ}$ with $B_{1}$ and $\\theta_{2}=75^{\\circ}-30^{\\circ}=45^{\\circ}$ with $B_{2}$. For stable equilibrium, net torque acting on dipole must be zero, I.e., $\\tau_{1}+\\tau_{2}=0$ $\\tau_{1}=\\tau_{2}$ $m B_{1} \\sin \\theta_{1}=m B_{2} \\sin \\theta_{2}$ $B_{2}=B_{1}\\left(\\sin \\theta_{1} / \\sin \\theta_{2}\\right)=15 \\mathrm{mT} \\times\\left(\\sin 30^{\\circ} / \\sin 45^{\\circ}\\right)$ $=15 \\mathrm{mT} \\times(1 / 2) \\times \\sqrt{2}=10.6 \\mathrm{mT}=$",
        "topic": "Magnetism"
    },
    {
        "question": "A short bar magnet is placed in the magnetic meridian of the earth with a north pole pointing north. Neutral points are found at a distance of $30 \\mathrm{~cm}$ from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $\\mathrm{Am}^{2}$ is close to (Given $\\left(\\mu_{0} / 4 \\pi\\right)=10^{-7}$ in SI units and $B_{H}=$ Horizontal component of earth's magnetic field $=3.6 \\times 10^{-}$ ${ }^{5}$ Tesla.)",
        "input": "(a) 9.7, (b) 4.9, (c) 19.4, (d) 14.6",
        "answer": "(a) 9.7",
        "solution": "$\\left(\\mu_{0} / 4 \\pi\\right)\\left(M / r^{3}\\right)=3.6 \\times 10^{-5}$ $M=\\left[\\left(3.6 \\times 10^{-5}\\right) /\\left(10^{-7}\\right)\\right](0.3)^{3}$ $M=9.7 \\mathrm{Am}^{2}$",
        "topic": "Magnetism"
    },
    {
        "question": "Needles $\\mathrm{N}_{1}, \\mathrm{~N}_{2}$ and $\\mathrm{N}_{3}$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will",
        "input": "(a) attract all three of them, (b) attract $\\mathrm{N}_{1}$ and $\\mathrm{N}_{2}$ strongly but repel $\\mathrm{N}_{3}$, (c) attract $\\mathrm{N}_{1}$ strongly, $\\mathrm{N}_{2}$ weakly and repel $\\mathrm{N}_{3}$ weakly, (d) attract $\\mathrm{N}_{1}$ strongly, but repel $\\mathrm{N}_{2}$ and $\\mathrm{N}_{3}$ weakly",
        "answer": "(c) Magnet will attract $\\mathrm{N}_{1}$ strongly, $\\mathrm{N}_{2}$ weakly and repel $\\mathrm{N}_{3}$ weakly",
        "solution": "",
        "topic": "Magnetism"
    },
    {
        "question": "A magnetic needle is kept in a non-uniform magnetic field. It experiences",
        "input": "(a) a force and a torque, (b) a force but not a torque, (c) a torque but not a force, (d) neither a force nor a torque",
        "answer": "(a) A force and a torque act on a magnetic needle kept in a non-uniform magnetic field",
        "solution": "",
        "topic": "Magnetism"
    },
    {
        "question": "The magnetic lines of force inside a bar magnet",
        "input": "(a) are from north-pole to south-pole of the magnet, (b) do not exist, (c) depend upon the area of cross-section of the bar magnet, (d) are from south-pole to north-pole of the magnet",
        "answer": "(b) Materials of low retentivity and low coercivity are suitable for making electromagnets",
        "solution": "",
        "topic": "Magnetism"
    },
    {
        "question": "Curie temperature is the temperature above which",
        "input": "(a) a ferromagnetic material becomes paramagnetic, (b) a paramagnetic material becomes diamagnetic, (c) a ferromagnetic material becomes diamagnetic, (d) a paramagnetic material becomes ferromagnetic",
        "answer": "(a) a ferromagnetic material becomes paramagnetic",
        "solution": "",
        "topic": "Magnetism"
    },
    {
        "question": "The materials suitable for making electromagnets should have",
        "input": "(a) high retentivity and high coercivity, (b) low retentivity and low coercivity, (c) high retentivity and low coercivity, (d) low retentivity and high coercivity",
        "answer": "(b) Materials of low retentivity and low coercivity are suitable for making electromagnets.",
        "solution": "",
        "topic": "Magnetism"
    },
    {
        "question": "Relative permittivity and permeability of a material are $\\epsilon_{r}$ and $\\mu_{r}$, respectively. Which of the following values of these quantities are allowed for a diamagnetic material?",
        "input": "(a) $\\epsilon_{r}=1.5, \\mu_{r}=1.5$, (b) $\\epsilon_{r}=0.5, \\mu_{r}=1.5$, (c) $\\epsilon_{r}=1.5, \\mu_{r}=0.5$, (d) $\\epsilon_{r}=0.5, \\mu_{r}=0.5$",
        "answer": "(c) $\\epsilon_{r}=1.5, \\mu_{r}=0.5$",
        "solution": "The values of relative permeability of diamagnetic materials are slightly less than 1 and $\\varepsilon_{r}$ is quite high. Therefore $\\varepsilon_{\\mathrm{r}}=1.5$ and $\\mu_{\\mathrm{r}}=0.5$ could be the allowed values",
        "topic": "Magnetism"
    },
    {
        "question": "In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a",
        "input": "(a) circle, (b) helix, (c) straight line, (d) ellipse",
        "answer": "(c) straight line",
        "solution": "Magnetic field exerts a force $=$ Bevsin0 $=0$ Electric field exerts force along a straight line The path of the charged particle will be a straight line",
        "topic": "Magnetism"
    },
    {
        "question": "A charged particle moves through a magnetic field perpendicular to its direction. Then",
        "input": "(a) kinetic energy changes but the momentum is constant, (b) the momentum changes but the kinetic energy is constant, (c) both momentum and kinetic energy of the particle are not constant, (d) both momentum and kinetic energy of the particle are constant.",
        "answer": "(b) the momentum changes but the kinetic energy is constant",
        "solution": "The magnetic force will act perpendicular to velocity at every instant, the path will be circular. Due to change in direction momentum will change but the total energy will remain the same.",
        "topic": "Magnetism"
    },
    {
        "question": "A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to",
        "input": "(a) induction of electrical charge on the plate, (b) shielding of magnetic lines of force as aluminium is a paramagnetic material, (c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping, (d) development of air current when the plate is placed",
        "answer": "(c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.",
        "solution": "",
        "topic": "Magnetism"
    },
    {
        "question": "An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8: 27 . The ratio of the radii of the nuclei (assumed to be spherical) is",
        "input": "(a) $3: 2$\n(b) $2: 3$\n(c) 4: 9\n(d) 8: 27",
        "answer": "(a) 3: 2",
        "solution": "$\\operatorname{As}\\left(v_{1} / v_{2}\\right)=8 / 27 ;\\left(r_{1} / r_{2}\\right)=$ ?\nUsing law of conservation of linear momentum, $0=m_{1} \\mathrm{~V}_{1}-\\mathrm{m}_{2} \\mathrm{~V}_{2}$\n(As both are moving in opposite directions)\nOr $m_{1} / m_{2}=v_{2} / v_{1}=27 / 8$\n$$\n\\frac{\\rho\\left(\\frac{4}{3}\\right) \\pi r_{1}^{3}}{\\rho\\left(\\frac{4}{3}\\right) \\pi r_{2}^{3}}=\\frac{27}{8}\n$$\nTherefore, $\\left(r_{1} / r_{2}\\right)=3 / 2$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "According to Bohr's theory, the time-averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the $\\mathrm{n}^{\\text {th }}$ orbit is proportional to ( $\\mathrm{n}=$ principal quantum number)",
        "input": "(a) $\\mathrm{n}^{-2}$\n(b) $\\mathrm{n}^{-3}$\n(c) $\\mathrm{n}^{-4}$\n(d) $n^{-5}$",
        "answer": "(d) $\\mathrm{n}^{-5}$",
        "solution": "Magnetic field at the centre, $B_{n}=\\mu_{0} / / 2 r_{n}$\nFor a hydrogen atom, the radius of $\\mathrm{n}^{\\text {th }}$ orbit is given by\n$r_{n}=\\left(n^{2} / m\\right)(h / 2 \\pi)^{2}\\left(4 \\pi \\varepsilon_{0} / e^{2}\\right)$\n$r_{n} \\propto n^{2}$\n$\\mathrm{I}=\\mathrm{e} / \\mathrm{T}=\\mathrm{e} /\\left(2 \\pi r_{n} / v_{n}\\right)=e v_{n} / 2 \\pi r_{n}$\nAlso, $\\mathrm{v}_{\\mathrm{n}} \\propto \\mathrm{n}^{-1} \\therefore 1 \\propto \\mathrm{n}^{-3}$\nHence, $B_{n} \\propto n^{-5}$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is (given binding energy per nucleon for deuteron $=1.1 \\mathrm{MeV}$ and for helium $=7.0 \\mathrm{MeV}$ )",
        "input": "(a) $25.8 \\mathrm{MeV}$\n(b) $32.4 \\mathrm{MeV}$\n(c) $30.2 \\mathrm{MeV}$\n(d) $23.6 \\mathrm{MeV}$",
        "answer": "(d) 23.6 MeV",
        "solution": "${ }_{1} \\mathrm{H}^{2}+{ }_{1} \\mathrm{H}^{2} \\rightarrow{ }_{2} \\mathrm{He}^{4}$\nEnergy released $=4\\left(\\right.$ Binding Energy of $\\left.\\left(1_{1}^{2} \\mathrm{H}\\right)\\right)-4\\left(\\right.$ Binding Energy of $\\left.\\left(2{ }^{4} \\mathrm{He}\\right)\\right)=4 \\times 7-4 \\times 1.1=23.6 \\mathrm{MeV}$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion",
        "input": "(a) kinetic energy decreases, potential energy increases but the total energy remains the same\n(b) kinetic energy and total energy decrease but potential energy increases\n(c) its kinetic energy increases but potential energy and total energy decrease\n(d) kinetic energy, potential energy and total energy decrease.",
        "answer": "(c) its kinetic energy increases but potential energy and total energy decrease",
        "solution": "For an electron in $\\mathrm{n}^{\\text {th }}$ excited state of hydrogen atom,\nkinetic energy $=\\mathrm{e}^{2} /\\left(8 \\pi \\epsilon_{0} \\mathrm{n}^{2} \\mathrm{a}_{0}\\right)$\npotential energy $=-e^{2} /\\left(4 \\pi \\epsilon_{0} n^{2} a_{0}\\right)$ and total energy $=-e^{2} /\\left(8 \\pi \\epsilon_{0} n^{2} a_{0}\\right)$\nwhere ao is Bohr radius.\nAs an electron makes a transition from an excited state to the ground state, $\\mathrm{n}$ decreases. Therefore kinetic energy increases but potential energy and total energy decrease.",
        "topic": "Nuclear Physics"
    },
    {
        "question": "Energy required for the electron excitation in $\\mathrm{Li}^{++}$from the first to the third Bohr orbit is",
        "input": "(a) $12.1 \\mathrm{eV}$\n(b) $36.3 \\mathrm{eV}$\n(c) $108.8 \\mathrm{eV}$\n(d) $122.4 \\mathrm{eV}$",
        "answer": "(c) $108.8 \\mathrm{eV}$",
        "solution": "Using, $E_{n}=\\left(13.6 Z^{2} / n^{2}\\right) \\mathrm{eV}$\nHere, $\\mathrm{Z}=3$ (For $\\mathrm{Li}^{++}$)\nTherefore, $\\mathrm{E}_{1}=\\left(13.6(3)^{2} / 1^{2}\\right)=-122.4 \\mathrm{eV}$\nand $E_{3}=\\left(13.6(3)^{2} / 3^{2}\\right)=-13.6 \\mathrm{eV}$\n$\\Delta \\mathrm{E}=\\mathrm{E}_{3}-\\mathrm{E}_{1}=-13.6+122.4=108.8 \\mathrm{eV}$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "If the binding energy per nucleon in ${ }_{3} \\mathrm{Li}^{7}$ and ${ }_{2} \\mathrm{He}^{4}$ nuclei are $5.60 \\mathrm{MeV}$ and $7.06 \\mathrm{MeV}$ respectively, then in the reaction",
        "input": "(a) $39.2 \\mathrm{MeV}$\n(b) $28.24 \\mathrm{MeV}$\n(c) $17.28 \\mathrm{MeV}$\n(d) $1.46 \\mathrm{MeV}$",
        "answer": "(c) 17.28 MeV",
        "solution": "Binding energy of $3 \\mathrm{Li}^{7}=7 \\times 5.60=39.2 \\mathrm{MeV}$\nBinding energy of ${ }_{2} \\mathrm{He}^{4}=4 \\times 7.06=28.24 \\mathrm{MeV}$\nEnergy of proton $=$ Energy of $\\left[2\\left(2 \\mathrm{He}^{4}\\right)-3 \\mathrm{Li}^{7}\\right]$\n$=2 \\times 28.24-39.2=17.28 \\mathrm{MeV}$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be",
        "input": "(a) $2^{1 / 3}: 1$\n(b) $1: 3^{1 / 2}$\n(c) $3^{1 / 2}: 1$\n(d) $1: 2^{1 / 3}$",
        "answer": "(d) $1: 2^{1 / 3}$",
        "solution": "Momentum is conserved during disintegration $\\mathrm{m}_{1} \\mathrm{v}_{1}=\\mathrm{m}_{2} \\mathrm{v}_{2}--$-(1)\nFor an atom $R=R_{0} A^{1 / 3}$\n$\\left(R_{1} / R_{2}\\right)=\\left(A_{1} / A_{2}\\right)^{1 / 3}=\\left(m_{1} / m_{2}\\right)^{1 / 3}=\\left(v_{2} / v_{1}\\right)^{1 / 3}$ from (1)\n$\\left(R_{1} / R_{2}\\right)=(1 / 2)^{1 / 3}=1 / 2^{1 / 3}$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "A nuclear transformation is denoted by $X(n, \\alpha)_{3} L i^{7}$. Which of the following is the nucleus of element $X$ ?",
        "input": "(a) ${ }_{5} \\mathrm{~B}^{9}$\n(b) ${ }_{4} \\mathrm{Be}^{11}$\n(c) ${ }_{6} \\mathrm{C}^{12}$\n(d) ${ }_{5} \\mathrm{~B}^{10}$",
        "answer": "(d) $5^{10}$",
        "solution": "The nuclear transformation is given by\n${ }_{2} X^{A}+{ }_{o n}{ }^{1} \\rightarrow{ }_{2} \\mathrm{He}^{4}+{ }_{3} \\mathrm{Li}^{7}$\nAccording to conservation of mass number\n$A+1=4+7$ or $A=10$\nAccording to conservation of charge number\n$Z+0 \\rightarrow 2+3$ or $Z=5$\nSo the nucleus of the element be ${ }_{5} \\mathrm{~B}^{10}$.",
        "topic": "Nuclear Physics"
    },
    {
        "question": "Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At $t=0$ it was 1600 counts per second and at $t=8$ seconds it was 100 counts per second. The count rate observed, as counts per second, at $t=6$ seconds is close to",
        "input": "(a) 200\n(b) 360\n(c) 150\n(d) 400",
        "answer": "(a) 200",
        "solution": "According to law of radioactivity, the count rate at $\\mathrm{t}=8$ seconds is\n$\\mathrm{N}_{1}=\\mathrm{N}_{0} \\mathrm{e}^{-\\lambda \\mathrm{t}}$\n$\\mathrm{dN} / \\mathrm{dt}=\\lambda N=\\lambda N_{0} \\mathrm{e}^{-\\lambda t}$\n$1600=\\lambda N_{0} e^{0}=\\lambda N_{0} \\Rightarrow 100=\\lambda N_{0} e^{-8 \\lambda}=1600 e^{-8 \\lambda}$\nAt $t=6 \\mathrm{sec}$\n$d N / d t=\\lambda N_{0} e^{-6 \\lambda}=1600 \\times\\left(e^{-2 \\lambda}\\right)^{3}=1600 \\times(18)=200$",
        "topic": "Nuclear Physics"
    },
    {
        "question": "Heat cannot by itself flow from a body at a lower temperature to a body at a higher temperature is a statement or consequence of",
        "input": "(a) The second law of thermodynamics\n(b) conservation of momentum\n(c) conservation of mass\n(d) The first law of thermodynamics",
        "answer": "Second law of thermodynamics.",
        "solution": "",
        "topic": "Thermodynamics"
    },
    {
        "question": "Which of the following is incorrect regarding the first law of thermodynamics?",
        "input": "(a) It introduces the concept of internal energy\n(b) It introduces the concept of entropy\n(c) It is not applicable to any cyclic process\n(d) It is a restatement of the principle of conservation of energy",
        "answer": "Statements (b) and (c) are incorrect regarding the first law of thermodynamics.",
        "solution": "",
        "topic": "Thermodynamics"
    },
    {
        "question": "Which of the following statements is correct for any thermodynamic system?",
        "input": "(a) The internal energy changes in all processes\n(b) Internal energy and entropy are state functions\n(c) The change in entropy can never be zero\n(d) The work done in an adiabatic process is always zero",
        "answer": "Internal energy and entropy are state functions",
        "solution": "",
        "topic": "Thermodynamics"
    },
    {
        "question": "Which of the following parameters does not characterize the thermodynamic state of matter?",
        "input": "(a) temperature\n(b) pressure\n(c) work\n(d) volume",
        "answer": "The work does not characterize the thermodynamic state of matter",
        "solution": "",
        "topic": "Thermodynamics"
    },
    {
        "question": "Even a Carnot engine cannot give 100 % efficiency because we cannot",
        "input": "(a) prevent radiation\n(b) find ideal sources\n(c) reach absolute zero temperature\n(d) eliminate friction.",
        "answer": "We cannot reach absolute zero temperature",
        "solution": "",
        "topic": "Thermodynamics"
    },
    {
        "question": "Which statement is incorrect?",
        "input": "(a) All reversible cycles have the same efficiency\n(b) The reversible cycle has more efficiency than an irreversible one\n(c) Carnot cycle is a reversible one\n(d) Carnot cycle has the maximum efficiency in all cycles",
        "answer": "All reversible cycles do not have the same efficiency.",
        "solution": "",
        "topic": "Thermodynamics"
    },
    {
        "question": "A Carnot engine operating between temperatures T1 and T2 has efficiency 1 / 6 When T2 is lowered by 62 K, its efficiency increases to 1 / 3. Then T1 and T2 are, respectively",
        "input": "(a) 372 K and 310 K\n(b) 372 K and 330 K\n(c) 330 K and 268 K\n(d) 310 K and 248 K",
        "answer": "372 K and 310 K",
        "solution": "The efficiency of Carnot engine, \u03b7=1\u2212(T2/T1) \u03b7=1/6 T2/T1=5/6 T1=6 T2/5 As per the question, when T2 is lowered by 62 K, then its efficiency becomes 1/3 1/3=[1\u2212(T2\u221262/T1)] [T2\u221262/T1]=1\u2212(1/3) (Using equa (1)) 5(T2\u221262)/6 T2=2/3 5 T2\u2212310=4 T2 \u21d2 T2=310 K From equation (1) T1=(6\u00d7310)/5=372 K",
        "topic": "Thermodynamics"
    },
    {
        "question": "100 g of water is heated from 30\u2218 C to 50\u2218 C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 Jkg\u22121 K\u22121 )",
        "input": "(a) 4.2 kJ\n(b) 8.4 kJ\n(c) 84 kJ\n(d) 2.1 kJ",
        "answer": "8.4 kJ",
        "solution": "\u0394 Q=ms \u0394 T Here m=100 g=100 \u00d710\u22123 Kg S=4184 J kg\u22121 K\u22121 and \u0394 T=(50\u221230)=20\u2218 C \u0394 Q=100 \u00d710\u22123 \u00d74184 \u00d720=8.4 \u00d7103 J \u0394 Q=\u0394 U+\u0394 W Change in internal energy \u0394 U=\u0394 Q=8.4 \u00d7103 J=8.4 kJ",
        "topic": "Thermodynamics"
    },
    {
        "question": "200 g water is heated from 40\u2218 C to 60\u2218 C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water =4184 J / kg / K )",
        "input": "(a) 167.4 kJ\n(b) 8.4 kJ\n(c) 4.2 kJ\n(d) 16.7 kJ",
        "answer": "16.7 kJ",
        "solution": "For isobaric process, \u0394 U=Q=ms \u0394 T Here, m=200 g=0.2 Kg, s=4184 J / Kg / K \u0394 T=60\u2218 C\u221240\u2218 C=20\u2218 C \u0394 U=(0.2)(4184)(20)=16736 J=16.7 kJ",
        "topic": "Thermodynamics"
    },
    {
        "question": "The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7\u2218 C. The gas is (R=8.3 J mol\u22121 K\u22121)",
        "input": "(a) monoatomic\n(b) diatomic\n(c) triatomic\n(d) a mixture of monoatomic and diatomic",
        "answer": "diatomic",
        "solution": "According to the first law of thermodynamics \u0394 Q=\u0394 U+\u0394 W For an adiabatic process, \u0394 Q=0 0=\u0394 U+\u0394 W \u0394 U=\u2212\u0394 W nC v \u0394 T=\u2212\u0394 W C v=\u2212\u0394 W /n \u0394 T=\u2212(\u2212146)\u00d7103 /[(1 \u00d7103) \u00d77]=20.8 Jmol\u22121 K\u22121 For diatomic gas, C v=(5/2) R=(5/2) \u00d78.3=20.8 Jmol\u22121 K\u22121 Hence, the gas is diatomic",
        "topic": "Thermodynamics"
    },
    {
        "question": "From the following statements, concerning ideal gas at any given temperature T, select the correct.",
        "input": "(a)The coefficient of volume expansion at constant pressure is the same for all ideal gases\n(b)The average translational kinetic energy per molecule of oxygen gas is 3 kT, k being Boltzmann constant\n(c)The mean-free path of molecules increases with an increase in the pressure\n(d)In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different",
        "answer": "The coefficient of volume expansion at constant pressure is the same for all ideal gases",
        "solution": "Y=d V /(V 0 \u00d7d T) at a constant temperature Y=1/V 0(d V/d T) p since P V=R T PdV=RdT or (dV/dT)=R/P 0 Therefore, \u03d5=(1/V 0)(R/P 0)=R/R T 0 \u03b3=1/T 0 \u03b3=1/273",
        "topic": "Thermodynamics"
    },
    {
        "question": "Calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 \u2218 C and it is defined under which of the following conditions?",
        "input": "(a) From 14.5\u2218 C to 15.5\u2218 C at 760 mm of Hg\n(b) From 98.5\u2218 C to 99.5\u2218 C at 760 mm of Hg\n(c) From 13.5\u2218 C to 14.5\u2218 C at 76 mm of Hg\n(d) From 3.5\u2218 C to 4.5\u2218 C at 76 mm of Hg",
        "answer": "From 14.5\u2218 C to 15.5\u2218 C at 760 mm of Hg",
        "solution": "1 calorie is the amount of heat required to raise the temperature of 1 gm of water from 14.5\u2218 C to 15.5 \u2218 C at 760 mm of Hg",
        "topic": "Thermodynamics"
    },
    {
        "question": "The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N 2 (molar mass 28) molecules in eV at the same temperature is",
        "input": "(a) 0.0015\n(b) 0.003\n(c) 0.048\n(d) 0.768",
        "answer": "0.048",
        "solution": "Average Kinetic Energy =(3/2) KT It depends on temperature and does not depend on molar mass For both the gases, average translational kinetic energy will be the same ie., 0.048 eV",
        "topic": "Thermodynamics"
    },
    {
        "question": "One mole of a monoatomic gas is heated at a constant pressure of 1 atmosphere from 0 K to 100 K. If the gas constant R=8.32 J / mol K, the change in internal energy of the gas is approximately [1998]",
        "input": "(a) 2.3 J\n(b) 46 J\n(c) 8.67 \u00d7103 J\n(d) 1.25 \u00d7103 J",
        "answer": "1.25 \u00d7103 J",
        "solution": "\u0394 U=C v dT=(3 R/2) \u0394 T \u0394 U=(3/2) \u00d7(8.3) \u00d7(100)=1.25 \u00d7103 J",
        "topic": "Thermodynamics"
    },
    {
        "question": "An ideal gas heat engine is operating between 227\u2218 C and 127\u2218 C. It absorbs 104 J of heat at the higher temperature. The amount of heat converted into work is",
        "input": "(a) 2000 J\n(b) 4000 J\n(c) 8000 J\n(d) 5600 J",
        "answer": "2000 J",
        "solution": "\u03b7=1\u2212(T 2/T 1) \u03b7=1\u2212(127+273)/(227+273)=1\u2212(400/500)=1/5 W=\u03b7 Q 1=1/5\u00d710 4=2000 J",
        "topic": "Thermodynamics"
    },
    {
        "question": "Two masses of $1 \\mathrm{~g}$ and $4 \\mathrm{~g}$ are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is",
        "input": "(a) $4: 1$ (b) $\\sqrt{ } 2: 1$ (c) $1: 2$ (d) $1: 16$",
        "answer": "(c) $1: 2$",
        "solution": "$p=\\sqrt{2 E m}$ Where $\\mathrm{p}$ denotes momentum, $\\mathrm{E}$ denotes kinetic energy and $m$ denotes mass. $$ \\frac{p_{1}}{p_{2}}=\\sqrt{\\frac{m_{1}}{m_{2}}} \\frac{p_{1}}{p_{2}}=\\sqrt{\\frac{1}{4}}=\\frac{1}{2} $$",
        "topic": "Work Power Energy"
    },
    {
        "question": "If a machine is lubricated with oil",
        "input": "(a) the mechanical advantage of the machine increases (b) the mechanical efficiency of the machine increases (c) both its mechanical advantage and efficiency increase (d) its efficiency increases, but its mechanical advantage decreases.",
        "answer": "(b) When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.",
        "solution": "When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.",
        "topic": "Work Power Energy"
    },
    {
        "question": "The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$ U(x)=\\frac{a}{x^{12}}-\\frac{b}{x^{6}} $$ between the atoms. if the dissociation energy of the molecule is $(U(x=\\infty)-U$ at equilibrium), $D$ is",
        "input": "(a) $b^{2} / 6 a$ (b) $b^{2} / 2 a$ (c) $b^{2} / 12 a$ (d) $b^{2} / 4 a$",
        "answer": "(d) $b^{2} / 4 a$",
        "solution": "$\\mathrm{U}(\\mathrm{x}=\\infty)=0$, at equilibrium $\\mathrm{dU} / \\mathrm{dx}=0$ $U(x)=\\frac{a}{x^{12}}-\\frac{b}{x^{6}}--(1)$ $-12 a x^{-13}+6 b x^{-7}=0$ $-12 a / x^{13}=6 b / x^{7}$ $x^{6}=2 a / b--(2)$ From 1 and 2 we have $$ U=\\frac{a-b(2 a / b)}{(2 a / b)^{2}} $$ $U=-b^{2} / 4 a$ $D=\\left(0+b^{2} / 4 a\\right)$ $\\mathrm{D}=\\mathrm{b}^{2} / 4 \\mathrm{a}$",
        "topic": "Work Power Energy"
    },
    {
        "question": "At time $t=0$ s particle starts moving along the $x$-axis. If its kinetic energy increases uniformly with time ' $t$ ', the net force acting on it must be proportional to",
        "input": "(a) $\\sqrt{ } \\mathrm{t}$ (b) constant (c) $t$ (d) $1 / \\sqrt{ } \\mathrm{V}$",
        "answer": "(d) $1 / \\sqrt{t}$",
        "solution": "Linear dependency with initial Kinetic Energy as zero is given as $\\mathrm{KE}=\\mathrm{kt}$, where $\\mathrm{k}$ is the proportionality constant. Kinetic energy can be written as $\\mathrm{KE}=1 / 2 \\mathrm{mv}^{2}$ so we can write $1 / 2 m v^{2}=k t$ $$ v=\\sqrt{\\frac{2 k t}{m}} \\frac{d v}{d t}=\\sqrt{\\frac{k}{2 m t}} $$ $F=m . d v / d t$ $$ F=m \\sqrt{\\frac{k}{2 m t}} $$ F $\\infty 1 / \\sqrt{t}$",
        "topic": "Work Power Energy"
    },
    {
        "question": "This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. If two springs $S_{1}$ and $S_{2}$ of force constants $k_{1}$ and $k_{2}$, respectively, are stretched by the same force, it is found that more work is done on spring $\\mathrm{S}_{1}$ than on spring $\\mathrm{S}_{2}$. Statement-1: If stretched by the same amount, work done on $\\mathrm{S}_{1}$, will be more than that on $\\mathrm{S}_{2}$ Statement-2 $: k_{1}<k_{2}$",
        "input": "(a) Statement-1 is True, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1. (b) Statement-1 is False, Statement-2 is true (c) Statement-1 is True, Statement-2 is false (d) Statement-1 is True, Statement-2 is true and Statement-2 is the correct explanation of statement-1.",
        "answer": "(b) Statement-1 is False, Statement-2 is true",
        "solution": "Given same force $F=k_{1} x_{1}=k_{2} x_{2}$ $k_{1} / k_{2}=x_{1} / x_{2}$ $W_{1}=1 / 2 k_{1} x_{1}^{2}$ and $W_{2}=1 / 2 k_{2} x_{2}^{2}$ As $W_{1} / W_{2}>1$ so $1 / 2 k_{1} x_{1}{ }^{2} 11 / 2 k_{2} x_{2}{ }^{2}>1$ $\\mathrm{Fx}_{1} / \\mathrm{Fx}_{2}>1 \\Rightarrow \\mathrm{k}_{2} / \\mathrm{k}_{1}>1$ Therefore $k_{2}>k_{1}$ Statement- 2 is true Or if $x_{1}=x_{2}=x$ $W_{1} / W_{2}=1 / 2 k_{1} x^{2} / 1 / 2 k_{2} x^{2}$ $W_{1} / W_{2}=k_{1} / k_{2}<1$\\ $\\mathrm{W}_{1}<\\mathrm{W}_{2}$, Statement-1 is False",
        "topic": "Work Power Energy"
    },
    {
        "question": "A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration ac is varying with time ' $t$ ' as ac $=k^{2} r t^{2}$ where ' $k$ ' is a constant. The power delivered to the particle by the force acting on it is",
        "input": "(a) $2 \\pi \\mathrm{mk}^{2} \\mathrm{r}^{2} \\mathrm{~s}$ (b) $m^{2} r^{2} t$ (c) $\\mathrm{mk}^{4} \\mathrm{r}^{2} \\mathrm{t} / 3$ (d) zero",
        "answer": "(b) $\\mathrm{mk}^{2} \\mathrm{r}^{2} \\mathrm{t}$",
        "solution": "Centripetal acceleration $\\mathrm{a}_{\\mathrm{c}}=\\mathrm{k}^{2} \\mathrm{rt}^{2}$ $\\mathrm{V}^{2} / \\mathrm{r}=\\mathrm{k}^{2} \\mathrm{t}^{2}$ $\\mathrm{V}^{2}=\\mathrm{k}^{2} \\mathrm{r}^{2} \\mathrm{t}^{2}$ $\\mathrm{mv}^{2} / 2=\\mathrm{m} \\mathrm{k}^{2} \\mathrm{r}^{2} \\mathrm{t} 2 / 2$ Kinetic energy $=\\mathrm{m}^{2} \\mathrm{r}^{2} \\mathrm{t}^{2} / 2$ $$ \\frac{d}{d t}(K \\cdot E)=m k^{2} r^{2} t $$ Power $=\\mathrm{m} \\mathrm{k}^{2} \\mathrm{r}^{2} \\mathrm{t}$",
        "topic": "Work Power Energy"
    },
    {
        "question": "A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is",
        "input": "(a) $\\sqrt{u^{2}-2 g L}$ (b) $\\sqrt{u^{2}-g L}$ (c) $\\sqrt{2\\left(u^{2}-g L\\right)}$ (d) $\\sqrt{2 g L}$",
        "answer": "(c) $\\sqrt{2\\left(u^{2}-g L\\right)}$",
        "solution": "At lowest position B, Potential Energy $=0$ Kinetic Energy $=1 / 2$ mu Total Energy $=1 / 2 \\mathrm{mu}$ At $\\mathrm{C}$, when string is horizontal, Potential Energy $=\\mathrm{mgL}$ Kinetic Energy $=1 / 2 \\mathrm{mv}^{2}$ Total Energy $=1 / 2 \\mathrm{mv}^{2}+\\mathrm{mgL}$ Since energy is conserved, $1 / 2 m v^{2}+m g L=1 / 2 m u^{2}$ $v^{2}=u^{2}-2 g L$ Since $v$ is in vertical direction and $u$ is in horizontal direction, they are mutually perpendicular to each other. Change in velocity $=\\sqrt{u^{2}+v^{2}}|\\Delta \\vec{v}|=$ $$ \\sqrt{u^{2}+\\left(u^{2}-2 g L\\right)}|\\Delta \\vec{v}|=\\sqrt{\\left.2\\left(u^{2}-g L\\right)\\right)} $$",
        "topic": "Work Power Energy"
    },
    {
        "question": "When a rubber-band is stretched by a distance $x$, it exerts a restoring force of magnitude $F=a x+$ $b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is",
        "input": "(a) $a L^{2} / 2+b L^{3} / 3$ (b) $1 / 2\\left(a L^{2} / 2+b L^{3} / 3\\right)$ (c) $a L^{2}+b L^{3}$ (d) $1 / 2\\left(a L^{2}+b L^{3}\\right)$",
        "answer": "(a) $a L^{2} / 2+b L^{3} / 3$",
        "solution": "Work done by a variable force $$ W=\\int \\vec{F} \\cdot \\overrightarrow{d s} $$ Wherein $\\operatorname{Vvec}\\{\\mathrm{F}\\} F$ is the variable force and $\\backslash \\mathrm{vec}\\{\\mathrm{ds}\\} d s$ is small displacement $F=a x+b x^{2}$ Work done in displacing rubber through $\\mathrm{dx}=\\mathrm{Fdx}$ $$ W=\\int_{0}^{L}\\left(a x+b x^{2}\\right) d x $$ $\\mathrm{W}=\\mathrm{aL}^{2} / 2+\\mathrm{bL} 3 / 3$",
        "topic": "Work Power Energy"
    },
    {
        "question": "A person trying to lose weight by burning fat lifts a mass of $10 \\mathrm{~kg}$ up to a height of $1 \\mathrm{~m} 1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \\mathrm{X}$ $10^{7} \\mathrm{~J}$ of energy per kg which is converted to mechanical energy with a $20 \\%$ efficiency rate. Take $\\mathrm{g}=$ $9.8 \\mathrm{~m} / \\mathrm{s}^{2}$",
        "input": "(a) $12.89 \\times 10^{-3} \\mathrm{~kg}$ (b) $2.45 \\times 10^{-3} \\mathrm{~kg}$ (c) $6.45 \\times 10^{3} \\mathrm{~kg}$ (d) $9.89 \\times 10^{-3} \\mathrm{~kg}$",
        "answer": "(a) $12.89 \\times 10^{-3} \\mathrm{Kg}$",
        "solution": "Work done against gravity $=(\\mathrm{mgh}) 1000$ in lifting 1000 times $=10 \\times 9.8 \\times 10^{3}$ $=9.8 \\times 10^{4}$ Joule $20 \\%$ of the efficiency is used to convert fat into energy $\\left(20 \\%\\right.$ of $\\left.3.8 \\times 10^{7} \\mathrm{~J}\\right) \\times \\mathrm{m}=9.8 \\times 10^{4}$ Where $m$ is mass $\\mathrm{m}=12.89 \\times 10^{-3} \\mathrm{Kg}$",
        "topic": "Work Power Energy"
    },
    {
        "question": "A body of mass $\\mathrm{m}=10^{-2} \\mathrm{~kg}$ is moving in a medium and experiences a frictional force $\\mathrm{F}=-\\mathrm{KV}^{2}$. Its initial speed is $v_{0}=10 \\mathrm{~ms}^{-1}$. If, after $10 \\mathrm{~s}$, its energy is, the value of $k$ will be",
        "input": "(a) $10^{-4} \\mathrm{~kg} \\mathrm{~m}^{-1}$ (b) $10^{-1} \\mathrm{~kg} \\mathrm{~m}^{-1} \\mathrm{~s}^{-1}$ (c) $10^{-3} \\mathrm{~kg} \\mathrm{~m}^{-1}$ (d) $10^{-3} \\mathrm{~kg} \\mathrm{~s}^{-1}$",
        "answer": "(a) $10^{-4} \\mathrm{~kg} \\mathrm{~m}^{-1}$",
        "solution": "$1 / 2 m v_{t}^{2}=1 / 8 m v_{0}{ }^{2}$ $V_{\\mathrm{f}}=\\mathrm{V}_{0} / 2=5 \\mathrm{~m} / \\mathrm{s}$ $V_{\\mathrm{f}}=\\mathrm{v}_{\\mathrm{o}} / 2=5 \\mathrm{~m} / \\mathrm{s}$ $\\left(10^{-2}\\right) \\mathrm{dV} / \\mathrm{dt}=-\\mathrm{kV}^{2}$ $\\int_{10}^{5} \\frac{d v}{v^{2}}=-100 k \\int_{0}^{10} d t$ $1 / 5-1 / 10=100 k(10)$ $\\mathrm{k}=10^{-4} \\mathrm{~kg} \\mathrm{~m}^{-1}$",
        "topic": "Work Power Energy"
    },
    {
        "question": "A time-dependent force $F=6 t$ acts on a particle of mass $1 \\mathrm{~kg}$. If the particle starts from rest, the work done by the force during the first $1 \\mathrm{sec}$. will be :",
        "input": "(a) $9 \\mathrm{~J}$ (b) $18 \\mathrm{~J}$ (c) $4.5 \\mathrm{~J}$ (d) $22 \\mathrm{~J}$",
        "answer": "(c) $4.5 \\mathrm{~J}$",
        "solution": "$F=6 t, m=1 \\mathrm{Kg}, u=0$ Now, $F=m a=m(d v / d t)=1 x(d v / d t)$ $\\mathrm{dv} / \\mathrm{dt}=6 \\mathrm{t}$ $\\int_{0}^{v} d v=\\int_{0}^{1} 6 t d t$ $v=3\\left(1^{2}-0\\right)=3 \\mathrm{~m} / \\mathrm{s}$ From work-energy theorem $W=\\Delta K E=1 / 2 m\\left(v^{2}-u^{2}\\right)$ $W=1 / 2 \\times 1\\left(3^{2}-0\\right)=4.5 \\mathrm{~J}$",
        "topic": "Work Power Energy"
    },
    {
        "question": "A particle is moving in a circular path of radius a under the action of an attractive potential $\\mathrm{U}=$ $\\mathrm{k} / 2 \\mathrm{r}^{2}$.Its total energy is",
        "input": "(a) $k / 2 a^{2}$ (b) Zero (c) $-3 / 2\\left(\\mathrm{k} / \\mathrm{a}^{2}\\right)$ (d) $-k / 4 a^{2}$",
        "answer": "(b) Zero",
        "solution": "$\\mathrm{F}=\\mathrm{mv} v^{2}=\\mathrm{K} / \\mathrm{r}^{2}$ $K E=1 / 2 m v^{2}=K / 2 r^{2}$ Total energy $=$ P.E + K.E $-K / 2 r^{2}+K / 2 r^{2}=$ Zero",
        "topic": "Work Power Energy"
    },
    {
        "question": "A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of",
        "input": "(a) $(2 / 3) \\mathrm{k}$ (b) $(3 / 2) \\mathrm{k}$ (c) $3 \\mathrm{k}$ (d) $6 \\mathrm{k}$",
        "answer": "(b) (3/2) k",
        "solution": "For a spring, $(K \\times L)$ is a constant Therefore, $\\mathrm{k} \\times \\mathrm{L}=\\mathrm{k} \\times \\mathbf{x}(2 \\mathrm{~L} / 3$ ) Where $2 L / 3=$ length of longer piece Or k'=(3/2)k",
        "topic": "Work Power Energy"
    },
    {
        "question": "A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to",
        "input": "(a) $v$ (b) $v^{2}$ (c) $v^{3}$ (d) $v^{4}$",
        "answer": "(c) $v^{3}$",
        "solution": "Force $=v \\times d m / d t$ $$ \\begin{aligned} & \\text { Force }=\\frac{d}{d t}(\\text { volume } x \\text { density })=v \\frac{d}{d t}(\\text { Axp }) \\\\ & \\text { Force }=v \\text { Ap } \\frac{d x}{d t} \\\\ & \\text { Power }=\\text { Force } x \\text { velocity } \\\\ & \\left(\\text { Apv }{ }^{2}\\right) v=\\mathrm{Apv}^{3} \\\\ & \\text { Power } \\propto \\mathrm{V}^{3} \\\\ & \\text { Answer: (c) } \\mathbf{v}^{3} \\end{aligned} $$",
        "topic": "Work Power Energy"
    },
    {
        "question": "Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is",
        "input": "(a) 30 m/s\n(b) 20 m/s\n(c) 10 m/s\n(d) 5 m/s",
        "answer": "(c) 10 m/s",
        "solution": "Just after collision\nv_c = m1v1 + m2v2 / (m1 + m2)\nv_c = (10 * 14 + 4 * 0) / (10 + 4)\nv_c = 10 m/s",
        "topic": "Centre of Mass"
    },
    {
        "question": "Two particles A and B, initially at rest, move towards each other under the mutual force of attraction. At the instant when the speed of A is v and the speed of B is 2v, the speed of the centre of mass of the system is",
        "input": "(a) 3v\n(b) v\n(c) 1.5v\n(d) zero",
        "answer": "(d) zero",
        "solution": "FA is the force on particle A\nFA = mA * aA = mA * v / t\nFB is the force on particle B\nFB = mB * aB = mB * 2v / t\nSince FA = FB\nmA * v / t = mB * 2v / t\nSo mA = 2mB\nFor the centre of mass of the system\nv = (mA * vA + mB * vB) / (mA + mB)\nv = (2mB * v - mB * 2v) / (2mB + mB) = 0\nThe negative sign is used because the particles are travelling in the opposite directions",
        "topic": "Centre of Mass"
    },
    {
        "question": "A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 m/s. Which of the following statements(s) are correct for the system of these two masses",
        "input": "(a) The total momentum of the system is 3 kg m/s\n(b) The momentum of 5 kg mass after the collision is 4 kg m/s\n(c) The kinetic energy of the centre of mass is 0.75 J\n(d) The total kinetic energy of the system is 4 J",
        "answer": "(a) and (c)",
        "solution": "If the velocity of 1 kg mass before the collision is u and velocity of 5 kg mass after the collision is v then\nFrom the conservation of linear momentum\n1 * u + 5 * 0 = 1 * (-2) + 5 * v\nu = -2 + 5v\nSince the collision is elastic, total kinetic energy will be equal\nTherefore,\n1/2 * 1 * u^2 = 1/2 * 1 * 2^2 + 1/2 * 5 * v^2\nu^2 = 4 + 5v^2\n(-2 + 5v)^2 = 4 + 5v^2\n4 - 20v + 25v^2 = 0\n-20v + 20v^2 = 0\n-20v(1 - v) = 0\nv = 1 m/s\nThus, u = 3 m/s\nThe velocity of the centres of mass of the combined system\nV_cm = (m1v1 + m2v2) / (m1 + m2)\nV_cm = (1 * (-2) + 5 * 1) / (1 + 5)\nV_cm = 3 / 6\nV_cm = 1 / 2\nThe combined Kinetic Energy of the system\nKinetic Energy = 1/2(1 + 5) * (1 / 2)^2\nK.E = 6 / 8\nKinetic Energy = 0.75 J\nTotal momentum of the system\n= (m1 + m2) * v_cm\n= (1 + 5) * 1 / 2\n= 3 kg-m/s",
        "topic": "Centre of Mass"
    },
    {
        "question": "Consider a rubber ball freely falling from a height h=4.9 m onto a horizontal elastic plate. Assume that the duration of a collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be",
        "input": "",
        "answer": "The velocity of the ball is reversed when it strikes the surface",
        "solution": "The velocity of the ball is reversed when it strikes the surface",
        "topic": "Centre of Mass"
    },
    {
        "question": "Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.\n\nStatement-2: Principle of conservation of momentum holds true for all kinds of collisions.",
        "input": "(a) Statement-1 is true, Statement-2 is false\n(b) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1\n(c) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1\n(d) Statement-1 is false, Statement-2 is true",
        "answer": "(b) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1",
        "solution": "In a completely inelastic collision then m1 v1 + m2 v2 = m1 v + m2 v\nv = (m1 v1 + m2 v2) / (m1 + m2)\nKinetic Energy = p1^2 / 2 m1 + p2^2 / 2 m2\nAs p1 and p2 both cannot be zero simultaneously therefore total kinetic energy cannot be lost.",
        "topic": "Centre of Mass"
    },
    {
        "question": "A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collisions are perfectly inelastic, the percentage loss in the energy during the collision is close to",
        "input": "(a) 56%\n(b) 62%\n(c) 44%\n(d) 50%",
        "answer": "(a) 56%",
        "solution": "Conservation of linear momentum in the x-direction\n(pi)x = (pf)x or 2mv = (2m + m)Vx\nvx = 2/3 V\nConservation of linear momentum in y-direction\n(pi)y = (pf)y or 2mv = (2m + m) Vy\nvy = 2/3 V\nThe initial kinetic energy of the two particles system is\nEi = 1/2 m (2v)^2 + 1/2 (2m) v^2\nEi = 1/2 * 4mv + 1/2 * 2mv^2\nEi = 2mv^2 + mv^2 = 3mv^2\nFinal energy of the combined two particles system is\nEf = 1/2 (3m) (vx^2 + vy^2)\nEf = 1/2 (3m) [4v^2 / 9 + 4v^2 / 9]\nEf = 4mv^2 / 3\nLoss of energy \u0394E = Ei - Ef\n\u0394E = mv^2 (3 - 4/3) = (5/3) mv^2\nPercentage loss in the energy during the collision\n\u0394E / Ei * 100 = [(5/3) mv^2 / 3mv^2] * 100\n= (5/9) * 100 \u2248 56%",
        "topic": "Centre of Mass"
    },
    {
        "question": "The distance of the centre of mass of a solid uniform cone from its vertex is Z0. If the radius of its base is R and its height is h then Z0 is equal to",
        "input": "(a) 3h / 4\n(b) h^2 / 4R\n(c) 5h / 8\n(d) 3h^2 / 8R",
        "answer": "(a) 3h / 4",
        "solution": "For a solid cone centre of mass from the base is h / 4\nPosition of centre of mass of a solid cone from the vertex = h - h / 4 = 3h / 4",
        "topic": "Centre of Mass"
    },
    {
        "question": "It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively :",
        "input": "(a) (0.28,0.89)\n(b) (0,0)\n(c) (0,1)\n(d) (0.89,0.28)",
        "answer": "(a) (0.28,0.89)",
        "solution": "Let the initial speed of neutron is v0 and kinetic energy is K\nFirst collision:\nBy momentum conservation\nmv0 = mv1 + 2mv2 \u21d2 v0 = v1 + 2v2\nBy e = 1 v2 - v1 = v0\nv2 = 2v0 / 3\nv1 = -v0 / 3\nFrictional loss = [1/2 mv0^2 - 1/2 m (v0 / 3)^2] / 1/2 mv0^2\nPd = 8/9 \u2248 0.89\nSecond Collision:\nBy momentum conservation\nmv0 = mv1 + 12mv2\nv1 + 12v2 = v0\nBy e = 1 v2 - v1 = v0\nv2 = 2v0 / 13\nv1 = -11v0 / 13\nNow fraction loss of energy\nPc = [1/2 mv0^2 - 1/2 m (11v0 / 13)^2] / 1/2 mv0^2\nPc = 48/169 = 0.28",
        "topic": "Centre of Mass"
    },
    {
        "question": "In a collinear collision, a particle with an initial speed V0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is",
        "input": "(a) \u221a2 v0\n(b) v0 / 2\n(c) v0 \u221a2\n(d) v0 / 4",
        "answer": "(a) \u221a2 v0",
        "solution": "Total kinetic energy after the collision = 1/2 mv1^2 + 1/2 mv2^2 = 3/2 (1/2 mv0^2)\nv1^2 + v2^2 = (3/2) v0^2\nBy momentum conservation\nmv0 = m(v1 + v2)\n(v1 + v2)^2 = v0^2\nv1^2 + v2^2 + 2v1v2 = v0^2\n2v1v2 = -v0^2 / 2\n(v1 - v2)^2 = v1^2 + v2^2 - 2v1v2 = (3/2) v0^2 + v0^2 / 2\n(v1 - v2) = \u221a2 v0",
        "topic": "Centre of Mass"
    },
    {
        "question": "Q1: Three charges $+Q, q,+Q$ are placed respectively, at distance $0, d / 2$ and $d$ from the origin, on the $x$-axis. If the net force experienced by $+Q$ placed at $x=0$ is zero, then value of $q$ is",
        "input": "(a) $+\\mathrm{Q} / 4$ (b) $-\\mathrm{Q} / 2$ (c) $+\\mathrm{Q} / 2$ (d) $-\\mathrm{Q} / 4$",
        "answer": "(d) $-\\mathbf{Q} / 4$",
        "solution": "$\\mathrm{QQ} / \\mathrm{d}^{2}+\\mathrm{Qq} /(\\mathrm{d} / 2)^{2}=0$ $Q+4 q=0$ or $\\mathrm{q}=-\\mathrm{Q} / 4$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q2: A parallel plate capacitor having capacitance $12 \\mathrm{pF}$ is charged by a battery to a potential difference of $10 \\mathrm{~V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is",
        "input": "(a) $508 \\mathrm{pJ}$ (b) $692 \\mathrm{pJ}$ (c) $560 \\mathrm{pJ}$ (d) $600 \\mathrm{pJ}$",
        "answer": "(a) 508 pJ",
        "solution": "Initial Energy of the capacitor, $\\mathrm{U}_{\\mathrm{i}}=(1 / 2) \\mathrm{CV}^{2}$ $=(1 / 2) \\times 12 \\mathrm{pF} \\times 10 \\times 10$ $=600 \\mathrm{pJ}$ After the slab, the energy of the slab, $\\mathrm{U}_{\\mathrm{f}}=(1 / 2) \\mathrm{Q}^{2} / \\mathrm{C}^{\\prime}$ $\\mathrm{Q}=\\mathrm{CV}=(12 \\mathrm{pF})(10 \\mathrm{~V})=120 \\mathrm{pC}$ $\\mathrm{C}^{\\prime}=\\mathrm{kC}=6.5 \\times 120 \\times 10^{-12} \\mathrm{~F}$ Therefore, $\\mathrm{U}_{\\mathrm{f}}=\\left[(1 / 2)\\left(120 \\times 10^{-12}\\right)^{2}\\right] /\\left[6.5 \\times 120 \\times 10^{-12}\\right]$ $\\mathrm{U}_{\\mathrm{f}}=92 \\mathrm{pJ}$ $\\mathrm{W}+\\mathrm{U}_{\\mathrm{f}}=\\mathrm{U}_{\\mathrm{i}}$ $\\Rightarrow \\mathrm{W}=\\mathrm{U}_{\\mathrm{i}}-\\mathrm{U}_{\\mathrm{f}}$ $=600 \\mathrm{pJ}-92 \\mathrm{pJ}$ $=508 \\mathrm{pJ}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q3: An electric field of $1000 \\mathrm{~V} / \\mathrm{m}$ is applied to an electric dipole at an angle of $45^{\\circ}$. The value of the electric dipole moment is $10^{-29} \\mathrm{Cm}$. What is the potential energy of the electric dipole?",
        "input": "(a) $-10 \\times 10^{-29} \\mathrm{~J}$ (b) $-7 \\times 10^{-27} \\mathrm{~J}$ (c) $-20 \\times 10^{-18} \\mathrm{~J}$ (d) $-9 \\times 10^{-20} \\mathrm{~J}$",
        "answer": "(b) $-\\mathbf{7} \\times 10^{-27} \\mathrm{~J}$",
        "solution": "$\\mathrm{E}=1000 \\mathrm{~V} / \\mathrm{m}, \\mathrm{p}=10^{-29} \\mathrm{~cm}, \\theta=45^{\\circ}$ Potential energy stored in the dipole, $U=-p . E \\cos \\theta=-10^{-29} \\times 1000 \\times \\cos 45^{\\circ}$ $\\mathrm{U}=-0.707 \\times 10^{-26} \\mathrm{~J}=-7 \\times 10^{-27} \\mathrm{~J}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q4: A solid conducting sphere, having a charge $Q$, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$. If the shell is now given a charge of $-4 Q$, the new potential difference between the same two surfaces is",
        "input": "(a) $4 \\mathrm{~V}$ (b) $\\mathrm{V}$ (c) $2 \\mathrm{~V}$ (d) $-2 \\mathrm{~V}$",
        "answer": "(b) V",
        "solution": "$\\mathrm{V}_{\\mathrm{a}}-\\mathrm{V}_{\\mathrm{b}}=\\mathrm{k}\\{(\\mathrm{Q} / \\mathrm{a})-(\\mathrm{Q} / \\mathrm{b})\\}$ $\\mathrm{V}_{\\mathrm{a}}{ }^{\\prime}=\\mathrm{kQ} / \\mathrm{a}+\\mathrm{k}(-4 \\mathrm{Q}) / \\mathrm{b}$ $\\mathrm{V}_{\\mathrm{b}}{ }^{\\prime}=\\mathrm{kQ} / \\mathrm{b}+\\mathrm{k}(-4 \\mathrm{Q}) / \\mathrm{b}$ $\\mathrm{V}_{\\mathrm{a}}^{\\prime}-\\mathrm{V}_{\\mathrm{b}}^{\\prime}=\\mathrm{kQ} / \\mathrm{a}-\\mathrm{kQ} / \\mathrm{b}=\\mathrm{V}_{\\mathrm{a}}-\\mathrm{V}_{\\mathrm{b}}=\\mathrm{V}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q5: Voltage rating of a parallel plate capacitor is $500 \\mathrm{~V}$. Its dielectric can withstand a maximum electric field of $106 \\mathrm{~V} \\mathrm{~m}^{-1}$. The plate area is $10^{-4} \\mathrm{~m}^{2}$. What is the dielectric constant if the capacitance is $15 \\mathrm{pF}$ ? (given $\\varepsilon_{0}=8.86 \\times 10^{-12} \\mathrm{C}^{2} \\mathrm{~N}^{-1} \\mathrm{~m}^{-2}$ )",
        "input": "(a) 3.8 (b) 8.5 (c) 6.2 (d) 4.5",
        "answer": "(b) 8.5",
        "solution": "$\\mathrm{C}=\\mathrm{K} \\varepsilon_{0} \\mathrm{~A} / \\mathrm{d}$ and $\\mathrm{V}=\\mathrm{Ed}$ Or $\\mathrm{K}=\\mathrm{CV} / \\varepsilon_{0} \\mathrm{AE}_{\\max }$ $\\mathrm{K}=\\left(15 \\times 10^{-12} \\times 500\\right) /\\left(8.86 \\times 10^{-12} \\times 10^{-4} \\times 10^{6}\\right)=8.5$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q6: The bob of a simple pendulum has a mass of $2 \\mathrm{~g}$ and a charge of $5.0 \\mathrm{C}$. It is at rest in a uniform horizontal electric field of intensity $2000 \\mathrm{~V} \\mathrm{~m}^{-1}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g=10 \\mathrm{~m} \\mathrm{~s}^{-2}$ )",
        "input": "(a) $\\tan ^{-1}(0.2)$ (b) $\\tan ^{-1}(0.5)$ (c) $\\tan ^{-1}(2.0)$ (d) $\\tan ^{-1}(5.0)$",
        "answer": "(b) $\\tan ^{-1}(0.5)$",
        "solution": "The forces acting on the bob are its weight and the force due to field. At equilibrium, $\\mathrm{T} \\cos \\theta=\\mathrm{mg}$ $\\mathrm{T} \\sin \\theta=\\mathrm{qE}$ Dividing (2) by (1) $\\tan \\theta=\\mathrm{qE} / \\mathrm{mg}$ $\\theta=\\tan ^{-1}\\left(\\left(5 \\times 10^{-6} \\times 2 \\times 10^{3}\\right) /\\left(2 \\times 10^{-3} \\times 10\\right)\\right)=\\tan ^{-1}(0.5)$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q7: A parallel plate capacitor has $1 \\mu \\mathrm{F}$ capacitance. One of its two plates is given $+2 \\mu \\mathrm{C}$ charge and the other plate, $+4 \\mu \\mathrm{C}$ charge. The potential difference developed across the capacitor is",
        "input": "(a) $3 \\mathrm{~V}$ (b) $2 \\mathrm{~V}$ (c) $5 \\mathrm{~V}$ (d) $1 \\mathrm{~V}$",
        "answer": "(d) $1 \\mathrm{~V}$",
        "solution": "Potential difference $V_{1}-V_{2}=\\left(E_{1}-E_{2}\\right) d$ $\\mathrm{V}_{1}-\\mathrm{V}_{2}=\\left[\\left(\\sigma_{1} / 2 \\varepsilon_{0}\\right)-\\left(\\sigma_{2} / 2 \\varepsilon_{0}\\right)\\right] \\mathrm{d}$ $\\mathrm{V}_{1}-\\mathrm{V}_{2}=\\left(\\mathrm{q}_{1} \\mathrm{~d} / 2 \\mathrm{~A} \\varepsilon_{0}\\right)-\\left(\\mathrm{q}_{2} \\mathrm{~d} / 2 \\mathrm{~A} \\varepsilon_{0}\\right)=(4-2) /(2 \\times 1)=1 \\mathrm{~V}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q8: A capacitor with a capacitance $5 \\mu \\mathrm{F}$ is charged to $5 \\mu \\mathrm{C}$. If the plates are pulled apart to reduce the capacitance to $2 \\mu \\mathrm{F}$, how much work is done?",
        "input": "(a) $6.25 \\times 10^{-6} \\mathrm{~J}$ (b) $3.75 \\times 10^{-6} \\mathrm{~J}$ (c) $2.16 \\times 10^{-6} \\mathrm{~J}$ (d) $2.55 \\times 10^{-6} \\mathrm{~J}$",
        "answer": "(b) $3.75 \\times 10^{-6} \\mathrm{~J}$",
        "solution": "Work done $=\\mathrm{U}_{\\mathrm{f}}-\\mathrm{U}_{\\mathrm{i}}=(1 / 2) \\mathrm{q}^{2} / \\mathrm{C}_{\\mathrm{f}}-(1 / 2) \\mathrm{q}^{2} / \\mathrm{C}_{\\mathrm{i}}$ Work done $=\\mathrm{q}^{2} / 2\\left[1 / \\mathrm{C}_{\\mathrm{f}}-1 / \\mathrm{C}_{\\mathrm{i}}\\right]$ Work done $=\\left[\\left(5 \\times 10^{-6}\\right)^{2} / 2\\right]\\left[\\left(1 /\\left(2 \\times 10^{-6}\\right)\\right)-\\left(1 /\\left(5 \\times 10^{-6}\\right)\\right)\\right]$ Work done $=3.75 \\times 10^{-6} \\mathrm{~J}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q9: A parallel plate capacitor of capacitance $90 \\mathrm{pF}$ is connected to a battery of emf $20 \\mathrm{~V}$. If a dielectric material of dielectric constant $K=5 / 3$ is inserted between the plates, the magnitude of the induced charge will be",
        "input": "(a) $1.2 \\mathrm{nC}$ (b) $0.3 \\mathrm{nC}$ (c) $2.4 \\mathrm{nC}$ (d) $0.9 \\mathrm{nC}$",
        "answer": "(a) $1.2 \\mathrm{nC}$",
        "solution": "Induced charge on dielectric, $\\mathrm{Q}_{\\text {ind }}=\\mathrm{Q}(1-1 / \\mathrm{K})$ Final charge on capacitor, $\\mathrm{Q}=\\mathrm{K} \\mathrm{C}_{0} \\mathrm{~V}$ $\\mathrm{Q}=(5 / 3) \\times 90 \\times 10^{-12} \\times 20=3 \\times 10^{-9} \\mathrm{C}=3 \\mathrm{nC}$ $\\mathrm{Q}_{\\text {ind }}=3(1-3 / 5)=3 \\times 2 / 5=1.2 \\mathrm{nC}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q10: The energy stored in the electric field produced by a metal sphere is $4.5 \\mathrm{~J}$. If the sphere contains $4 \\mu \\mathrm{C}$ charges, its radius will be [Take: $\\left(1 / 4 \\pi \\varepsilon_{0}\\right)=9 \\times 10^{9} \\mathrm{Nm}^{2} / \\mathrm{C}^{2}$ ]",
        "input": "(a) $32 \\mathrm{~mm}$ (b) $20 \\mathrm{~mm}$ (c) $16 \\mathrm{~mm}$ (d) $28 \\mathrm{~mm}$",
        "answer": "(c) 16 mm",
        "solution": "The energy stored in the electric field produced by a metal sphere $=4.5 \\mathrm{~J}$ $\\Rightarrow \\mathrm{Q}^{2} / 2 \\mathrm{C}=4.5$ or $\\mathrm{C}=\\mathrm{Q}^{2} / 2 \\times 4.5$ Capacitance of spherical conductor $=4 \\pi \\varepsilon_{0} \\mathrm{R}$ $4 \\pi \\varepsilon_{0} \\mathrm{R}=\\mathrm{Q}^{2} /(2 \\times 4.5)$ $\\mathrm{R}=\\left(1 / 4 \\pi \\varepsilon_{0}\\right) \\times\\left[\\left(4 \\times 10^{-6}\\right)^{2} /(2 \\times 4.5)\\right]=9 \\times 10^{9} \\times(16 / 9) \\times 10^{-12}=16 \\times 10^{-3} \\mathrm{~m}=16 \\mathrm{~mm}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q11: There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $P$, in the region, is found to vary between the limits $589.0 \\mathrm{~V}$ to $589.8 \\mathrm{~V}$. What is the potential at a point on the sphere whose radius vector makes an angle of $60^{\\circ}$ with the direction of the field?",
        "input": "(a) $589.2 \\mathrm{~V}$ (b) $589.6 \\mathrm{~V}$ (c) $589.5 \\mathrm{~V}$ (d) $589.4 \\mathrm{~V}$",
        "answer": "(d) 589.4 V",
        "solution": "$\\Delta \\mathrm{V}=\\mathrm{E} . \\mathrm{d}$ $\\Delta \\mathrm{V}=\\mathrm{Ed} \\cos \\theta=0.8 \\times \\cos 60^{\\circ}$ $\\Delta \\mathrm{V}=0.4$ Hence the new potential at the point on the sphere is $589.0+0.4=589.4 \\mathrm{~V}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q12: Two identical conducting spheres $A$ and $B$, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is $\\mathbf{F}$. A third identical conducting sphere, $\\mathbf{C}$, is uncharged. Sphere $C$ is first touched to $A$, then to $B$, and then removed. As a result, the force between $A$ and $B$ would be equal to",
        "input": "(a) $3 \\mathrm{~F} / 8$ (b) F/2 (c) $3 \\mathrm{~F} / 4$ (d) F",
        "answer": "(a) $3 \\mathrm{~F} / 8$",
        "solution": "Initially force between spheres $\\mathrm{A}$ and $\\mathrm{B}, \\mathrm{F}=\\mathrm{kq}^{2} / \\mathrm{r}$ When A and C are touched, charge on both will be $\\mathrm{q} / 2$ Again $\\mathrm{C}$ is touched with $\\mathrm{B}$ the charge on $\\mathrm{B}$ is given by $\\mathrm{q}_{\\mathrm{B}}=((\\mathrm{q} / 2)+\\mathrm{q}) / 2=3 \\mathrm{q} / 4$ Required force between spheres A and B is given by $\\mathrm{F}^{\\prime}=\\mathrm{kq}_{\\mathrm{A}} \\mathrm{q}_{\\mathrm{B}} / \\mathrm{r}^{2}=[\\mathrm{k} \\times(\\mathrm{q} / 2) \\times(3 \\mathrm{q} / 4)] / \\mathrm{r}^{2}=(3 / 8)\\left(\\mathrm{kq}^{2} / \\mathrm{r}^{2}\\right)=3 / 8 \\mathrm{~F}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q13: A parallel plate capacitor is made of two circular plates separated by a distance of $5 \\mathrm{~mm}$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \\times 10^{4} \\mathrm{~V} / \\mathrm{m}$, the charge density of the positive plate will be close to",
        "input": "(a) $6 \\times 10^{4} \\mathrm{C} / \\mathrm{m}^{2}$ (b) $6 \\times 10^{-7} \\mathrm{C} / \\mathrm{m}^{2}$ (c) $3 \\times 10^{-7} \\mathrm{C} / \\mathrm{m}^{2}$ (d) $3 \\times 10^{4} \\mathrm{C} / \\mathrm{m}^{2}$",
        "answer": "(b) $6 \\times 10^{-7} \\mathbf{C m}^{-2}$",
        "solution": "Here, $\\mathrm{K}=2.2, \\mathrm{E}=3 \\times 10^{4} \\mathrm{Vm}^{-1}$ Electric field between the parallel plate capacitor with dielectric, $\\mathrm{E}=\\sigma / \\mathrm{K} \\varepsilon_{0} \\Rightarrow \\sigma=\\mathrm{K} \\varepsilon_{0} \\mathrm{E}=2.2 \\times 8.85 \\times 10^{-12} \\times 3 \\times 10^{4}$ $\\mathrm{E}=6 \\times 10^{-7} \\mathrm{Cm}^{-2}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q14: Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 \\mathrm{~V}$ and $200 \\mathrm{~V}$, respectively. It is found that by connecting them together the potential on each one can be made zero. Then",
        "input": "(a) $9 \\mathrm{C}_{1}=4 \\mathrm{C}_{2}$ (b) $5 \\mathrm{C}_{1}=3 \\mathrm{C}_{2}$ (c) $3 \\mathrm{C}_{1}=5 \\mathrm{C}_{2}$ (d) $3 \\mathrm{C}_{1}+5 \\mathrm{C}_{2}=0$",
        "answer": "(c) $3 \\mathrm{C}_{1}=5 \\mathrm{C}_{2}$",
        "solution": "For potential to be made zero, after connection $120 \\mathrm{C}_{1}=200 \\mathrm{C}_{2}$ $6 \\mathrm{C}_{1}=10 \\mathrm{C}_{2}$ $3 \\mathrm{C}_{1}=5 \\mathrm{C}_{2}$",
        "topic": "Electrostatics"
    },
    {
        "question": "Q15: An electric dipole is placed at an angle of $30^{\\circ}$ to a non-uniform electric field. The dipole will experience",
        "input": "(a) a torque only (b) a translational force only in the direction of the field (c) a translational force only in a direction normal to the direction of the field (d) a torque as well as a translational force",
        "answer": "(d) a torque as well as a translational force",
        "solution": "In a non-uniform electric field, the dipole will experience torque as well as a translational force.",
        "topic": "Electrostatics"
    },
    {
        "question": "Q1: A sample of radioactive material $\\mathrm{A}$, that has an activity of $10 \\mathrm{mCi}\\left(1 \\mathbf{C i}=3.7 \\times 10^{10}\\right.$ decays/s) has twice the number of nuclei as another sample of a different radioactive material $B$, which has an activity of 20 $\\mathrm{mCi}$. The correct choices for half-lives of $A$ and $B$ would then be respectively",
        "input": "(a) 20 days and 5 days\n(b) 10 days and 40 days\n(c) 20 days and 10 days\n(d) 5 days and 10 days",
        "answer": "(a) 20 days and 5 days",
        "solution": "$\\mathrm{R}_{\\mathrm{A}}=10 \\mathrm{mCi}, \\mathrm{R}_{\\mathrm{B}}=20 \\mathrm{mCi}, \\mathrm{N}_{\\mathrm{A}}=2 \\mathrm{~N}_{\\mathrm{B}}$\n\n$\\mathrm{R}_{\\mathrm{A}} / \\mathrm{R}_{\\mathrm{B}}=\\lambda_{\\mathrm{A}} \\mathrm{N}_{\\mathrm{A}} / \\lambda_{\\mathrm{B}} \\mathrm{N}_{\\mathrm{B}}=\\left[\\left(\\mathrm{T}_{1 / 2}\\right)_{\\mathrm{B}} /\\left(\\mathrm{T}_{1 / 2}\\right)_{\\mathrm{A}}\\right] \\times\\left[\\mathrm{N}_{\\mathrm{A}} / \\mathrm{N}_{\\mathrm{B}}\\right]$\n\n$(1 / 2)=\\left[\\left(\\mathrm{T}_{1 / 2}\\right)_{\\mathrm{B}} /\\left(\\mathrm{T}_{1 / 2}\\right)_{\\mathrm{A}}\\right] \\times 2 \\Rightarrow\\left(\\mathrm{T}_{1 / 2}\\right)_{\\mathrm{A}}=4\\left(\\mathrm{~T}_{1 / 2}\\right)_{\\mathrm{B}}$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q2: Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At $t$ $=0$ it was 1600 counts per second and at $t=8$ seconds it was 100 counts per second. The count rate observed, as counts per second, at $t=6$ seconds is close to",
        "input": "(a) 200\n(b) 360\n(c) 150\n(d) 400",
        "answer": "(a) 200",
        "solution": "According to the law of radioactivity, the count rate at $\\mathrm{t}=8$ seconds is\n\n$\\mathrm{N}_{1}=\\mathrm{N}_{0} \\mathrm{e}^{-\\lambda \\mathrm{t}}$\n\n$\\mathrm{dN} / \\mathrm{dt}=\\lambda \\mathrm{N}=\\lambda \\mathrm{N}_{0} \\mathrm{e}^{-\\lambda \\mathrm{t}}$\n\nAt $\\mathrm{t}=0,1600=\\lambda \\mathrm{N}_{0} \\mathrm{e}^{0}=\\lambda \\mathrm{N}_{0}-----(1)$\n\nAt $\\mathrm{t}=8 \\mathrm{~s}, 100=\\lambda \\mathrm{N}_{0} \\mathrm{e}^{-8 \\lambda}$\n\n$\\Rightarrow 100=1600 \\mathrm{e}^{-8 \\lambda}$\n\n$\\mathrm{e}^{8 \\lambda}=16$\n\nTherefore half life is $\\mathrm{t} 1 / 2=2 \\mathrm{sec}$\n\nAt $t=6 \\mathrm{sec}$\n\n$(\\mathrm{dN} / \\mathrm{dt})=\\lambda \\mathrm{N}_{0} \\mathrm{e}^{-6 \\lambda}=1600 \\times\\left(\\mathrm{e}^{-2 \\lambda}\\right)^{3}=1600 \\mathrm{x}(1 / 8)=200$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q3: Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on $H e+$ ions in $n=1$ and $n=2$ states. The possible transition of helium ions as they absorb energy from the radiation?",
        "input": "",
        "answer": "$\\mathbf{E}_{3}$ and $\\mathbf{E}_{4}$ is possible",
        "solution": "$\\mathrm{E}=13.6(1 / 1-1 / 4)=13.6 \\times(3 / 4)=10.2 \\mathrm{eV}$\n\nLet us check the transitions possible on $\\mathrm{He}$\n\n$\\mathrm{n}=1$ or $\\mathrm{n}=2$\n\n$\\mathrm{E}_{1}=4 \\times 13.6(1-1 / 4)=40.8 \\mathrm{eV}\\left[\\mathrm{E}_{1}>\\mathrm{E}\\right.$, hence not possible $]$\n\n$\\mathrm{n}=1$ or $\\mathrm{n}=3$\n\n$\\mathrm{E}_{2}=4 \\times 13.6(1-(1 / 9))=48.3 \\mathrm{eV}\\left[\\mathrm{E}_{2}>\\mathrm{E}\\right.$, hence not possible $]$\n\n$\\mathrm{n}=2$ or $\\mathrm{n}=3$\n\n$\\mathrm{E}_{3}=4 \\times 13.6((1 / 4)-(1 / 9))=7.56 \\mathrm{eV}\\left[\\mathrm{E}_{3}<\\mathrm{E}\\right.$, hence it is possible $]$\n\n$\\mathrm{n}=2$ or $\\mathrm{n}=4$\n\n$\\mathrm{E}_{4}=4 \\times 13.6((1 / 4)-(1 / 6))=10.2 \\mathrm{eV}\\left[\\mathrm{E}_{4}=\\mathrm{E}\\right.$, hence it is possible $]$\n\nHence $\\mathrm{E}_{3}$ and $\\mathrm{E}_{4}$ can be possible",
        "topic": "Radioactivity"
    },
    {
        "question": "Q4: Two radioactive materials $A$ and $B$ have decay constants $10 \\lambda$ and $\\lambda$, respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that of $B$ will be $1 / e$ after a time",
        "input": "(a) $1 / 9 \\lambda$\n(b) $11 / 10 \\lambda$\n(c) $1 / 10 \\lambda$\n(d) $1 / 11 \\lambda$",
        "answer": "(a) $1 / 9 \\lambda$",
        "solution": "$\\mathrm{N}=\\mathrm{N}_{0} \\mathrm{e}^{-\\lambda \\mathrm{t}}$\n\nSo, $\\mathrm{N}_{1}=\\mathrm{N}_{0} \\mathrm{e}^{-10 \\lambda t}$ and $\\mathrm{N}_{2}=\\mathrm{N}_{0} \\mathrm{e}^{-\\lambda t}$\n\n$\\Rightarrow(1 / \\mathrm{e})=\\left(\\mathrm{N}_{1} / \\mathrm{N}_{2}\\right)=\\left(\\mathrm{N}_{0} \\mathrm{e}^{-10 \\lambda t}\\right) /\\left(\\mathrm{N}_{0} \\mathrm{e}^{-2 \\mathrm{t}}\\right)$\n\n$\\Rightarrow(1 / \\mathrm{e})=\\mathrm{e}^{-9 \\lambda \\mathrm{t}}=\\mathrm{e}^{-1}=\\mathrm{e}^{-9 \\lambda \\mathrm{t}}$\n\n$\\Rightarrow 1=9 \\lambda \\mathrm{t} \\Rightarrow \\mathrm{t}=1 / 9 \\lambda$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q5: Half-lives of two radioactive nuclei $A$ and $B$ are 10 minutes and 20 minutes, respectively. If, initially a sample has an equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei $A$ and B will be",
        "input": "(a) $3: 8$\n(b) $1: 8$\n(c) $9: 8$\n(d) $8: 1$",
        "answer": "(c) 9: 8",
        "solution": "By the law of radioactivity $\\mathrm{N}=\\mathrm{N}_{0} \\mathrm{e}^{-\\lambda \\lambda}$\n\nFor nuclei A,\n\n$\\mathrm{N}_{\\mathrm{A}}=\\mathrm{N}_{0 \\mathrm{~A}} \\mathrm{e}^{-2 \\mathrm{t}}$\n\n$\\operatorname{Or}\\left(\\mathrm{N}_{\\mathrm{A}} / \\mathrm{N}_{0 \\mathrm{~A}}\\right)=(1 / 2)^{\\mathrm{n}}=(1 / 2)^{t / 10}=(1 / 2)^{6}$\n\n$\\mathrm{N}_{\\mathrm{A}}=\\mathrm{N}_{0 \\mathrm{~A}} / 2^{6}$\n\nFor nuclei B,\n\n$\\left(\\mathrm{N}_{\\mathrm{B}} / \\mathrm{N}_{0 \\mathrm{~B}}\\right)=(1 / 2)^{\\mathrm{n}}=(1 / 2)^{t 20}=(1 / 2)^{3}$\n\n$\\Rightarrow \\mathrm{N}_{\\mathrm{B}}=\\left(\\mathrm{N}_{0 \\mathrm{~B}}\\right) / 2^{3}$\n\nRatio of nuclei decayed will be\n\n$\\left(\\mathrm{N}^{\\prime}{ }_{A} / \\mathrm{N}^{\\prime}{ }_{\\mathrm{B}}\\right)=\\left(\\mathrm{N}_{0 A}-\\mathrm{N}_{\\mathrm{A}}\\right) /\\left(\\mathrm{N}_{0 B}-\\mathrm{N}_{\\mathrm{B}}\\right)=\\left(\\mathrm{N}_{0 A} / \\mathrm{N}_{0 B}\\right)\\left[1-(1 / 2)^{6} / 1-(1 / 2)^{3}\\right]=9 / 8$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q6: Two radioactive substances $A$ and $B$ have decay constant $5 \\lambda$ and $\\lambda$ respectively. At $t=0$, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(1 / e)^{2}$ will be",
        "input": "(a) $2 / \\lambda$\n(b) $1 / \\lambda$\n(c) $1 / 4 \\lambda$\n(d) $1 / 2 \\lambda$",
        "answer": "(d) $1 / 2 \\lambda$",
        "solution": "The number of undecayed nuclei at any time $t$,\n\n$\\mathrm{N}=\\mathrm{N}_{0} \\mathrm{e}^{-\\lambda t}$\n\nAs $\\mathrm{N}_{0 \\mathrm{~A}}=\\mathrm{N}_{0 \\mathrm{~B}}$ (given)\n\nSo, for nuclei A and B\n\n$\\left.\\left(\\mathrm{N}_{\\mathrm{A}} / \\mathrm{N}_{\\mathrm{B}}\\right)=\\mathrm{e}^{\\left(-\\lambda_{A}\\right.}+\\lambda_{B}\\right) \\mathrm{t}$\n\n$\\mathrm{t}=\\left[1 /\\left(\\lambda_{\\mathrm{B}}-\\lambda_{\\mathrm{A}}\\right)\\right] \\operatorname{In}\\left(\\mathrm{N}_{\\mathrm{A}} / \\mathrm{N}_{\\mathrm{B}}\\right)=1 /(\\lambda-5 \\lambda) \\operatorname{In}\\left(1 / \\mathrm{e}^{2}\\right)=1 / 2 \\lambda$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q7: The radiation corresponding to $3 \\rightarrow 2$ transitions of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3 \\times 10^{-4} \\mathrm{~T}$. If the radius of the largest circular path followed by these electrons is $10.0 \\mathrm{~mm}$, the work function of the metal is close to",
        "input": "(a) $1.6 \\mathrm{eV}$\n(b) $1.8 \\mathrm{eV}$\n(c) $1.1 \\mathrm{eV}$\n(d) $0.8 \\mathrm{eV}$",
        "answer": "(c) 1.1 eV",
        "solution": "Radius of a charged particle moving in a constant magnetic field is given by\n\n$\\mathrm{R}=(\\mathrm{mv} / \\mathrm{qB})$ or $\\mathrm{R}^{2}=\\mathrm{m}^{2} \\mathrm{v}^{2} / \\mathrm{q}^{2} \\mathrm{~B}^{2}$\n\n$\\mathrm{R}^{2}=\\left[2 \\mathrm{~m}\\left((1 / 2) \\mathrm{mv} v^{2}\\right)\\right] / \\mathrm{q}^{2} \\mathrm{~B}^{2}$\n\n$\\mathrm{R}^{2}=2 \\mathrm{~m}(\\mathrm{~K} . \\mathrm{E}) / \\mathrm{q}^{2} \\mathrm{~B}^{2}$\n\n$\\Rightarrow K . E=\\left(q^{2} B^{2} R^{2}\\right) / 2 m \\Rightarrow K . E_{\\max }=\\left(q^{2} B^{2} R^{2}{ }_{\\max }\\right) / 2 m=0.80 \\mathrm{eV}$\n\nEnergy of photon corresponding transition from orbit $3 \\rightarrow 2$ in hydrogen atom.\n\n$\\left.\\mathrm{E}=13.6\\left(1 / 2^{2}\\right)-\\left(1 / 3^{2}\\right)\\right)=1.89 \\mathrm{eV}$\n\nUsing Einstein photoelectric equation.\n\n$\\mathrm{E}=\\mathrm{K} . \\mathrm{E}_{\\text {max }}+\\Phi$\n\n$\\Rightarrow 1.89=0.8+\\Phi$\n\n$\\Rightarrow \\Phi=1.09 \\approx 1.1 \\mathrm{eV}$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q8: Assume that a neutron breaks into a proton and an electron. The energy released during this process is",
        "input": "(Mass of neutron $=1.6725 \\times 10^{-27} \\mathrm{~kg}$\nMass of proton $=1.6725 \\times 10^{-27} \\mathrm{~kg}$\nMass of electron $=9 \\times 10^{-31} \\mathrm{~kg}$ )\n\n(a) $7.10 \\mathrm{MeV}$\n(b) $6.30 \\mathrm{MeV}$\n(c) $5.4 \\mathrm{MeV}$\n(d) $0.73 \\mathrm{MeV}$",
        "answer": "None of the given options are correct",
        "solution": "Mass defect, $\\Delta \\mathrm{m}=\\mathrm{m}_{\\mathrm{p}}+\\mathrm{m}_{\\mathrm{e}}-\\mathrm{m}_{\\mathrm{n}}$\n\n$\\Delta \\mathrm{m}=\\left(1.6725 \\times 10^{-27}\\right)+\\left(9 \\times 10^{-31}\\right)-\\left(1.6725 \\times 10^{-27}\\right) \\mathrm{Kg}$\n\n$\\Delta \\mathrm{m}=9 \\times 10^{-31} \\mathrm{~kg}$\n\nEnergy released $=\\Delta \\mathrm{mc}^{2}$\n\nEnergy released $=\\left(9 \\times 10^{-31}\\right) \\times\\left(3 \\times 10^{8}\\right)^{2} \\mathrm{~J}$\n\nEnergy released $=\\left[\\left(9 \\times 10^{-31}\\right) \\times\\left(9 \\times 10^{16}\\right)\\right] /\\left[1.6 \\times 10^{-13}\\right] \\mathrm{Mev}=0.51 \\mathrm{MeV}$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q9: The half-life period of a radioactive element $X$ is the same as the mean lifetime of another radioactive element Y. Initially, they have the same number of atoms. Then",
        "input": "(a) $\\mathrm{X}$ and $\\mathrm{Y}$ decay at the same rate always\n(b) $\\mathrm{X}$ will decay faster than $\\mathrm{Y}$\n(c) $\\mathrm{Y}$ will decay faster than $\\mathrm{X}$\n(d) $\\mathrm{X}$ and $\\mathrm{Y}$ have the same decay rate initially",
        "answer": "(c) Y will decay faster than $\\mathbf{X}$",
        "solution": "$\\mathrm{T}_{1 / 2}$, half-life of $\\mathrm{X}=\\mathrm{T}_{\\text {mean }}$, mean life of $\\mathrm{y}$\n\nOr $0.693 / \\lambda_{\\mathrm{x}}=1 / \\lambda_{\\mathrm{y}}$\n\n$\\lambda_{\\mathrm{x}}=0.693 \\lambda_{\\mathrm{y}}$\n\n$\\lambda_{\\mathrm{x}}<\\lambda_{\\mathrm{y}}$\n\nRate of decay $=\\lambda \\mathrm{N}$\n\nInitially, number of atoms $(\\mathrm{N})$ of both are equal but since $\\lambda_{\\mathrm{x}}<\\lambda_{\\mathrm{y}}$, therefore $\\mathrm{y}$ will decay at a faster rate than $\\mathrm{x}$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q10: If the binding energy of the electron in a hydrogen atom is $13.6 \\mathrm{eV}$, the energy required to remove the electron from the first excited state of $\\mathrm{Li}^{\\text {t+ }}$ is",
        "input": "(a) $30.6 \\mathrm{eV}$\n(b) $13.6 \\mathrm{eV}$\n(c) $3.4 \\mathrm{eV}$\n(d) $122.4 \\mathrm{eV}$",
        "answer": "(a) $30.6 \\mathrm{eV}$",
        "solution": "$\\mathrm{E}_{2}=\\left(-\\mathrm{Z}^{2} \\mathrm{E}_{0}\\right) / \\mathrm{n}^{2}$\n\n$\\mathrm{E}_{2}=\\left(-(3)^{2} \\times 13.6\\right) /(2)^{2}$\n\n$=-30.6 \\mathrm{eV}$\n\nEnergy required $=30.6 \\mathrm{eV}$",
        "topic": "Radioactivity"
    },
    {
        "question": "Q1: A convex lens is put $10 \\mathrm{~cm}$ from a light source and it makes a sharp image on a screen, kept $10 \\mathrm{~cm}$ from the lens. Now a glass block (refractive index 1.5 ) of $1.5 \\mathrm{~cm}$ thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. Then $d$ is",
        "input": "(a) $0.55 \\mathrm{~cm}$ towards the lens\n(b) 0\n(c) $1.1 \\mathrm{~cm}$ away from the lens\n(d) $0.55 \\mathrm{~cm}$ away from the lens",
        "answer": "(d) $0.55 \\mathrm{~cm}$ away from the lens",
        "solution": "Case I: $u=-10 \\mathrm{~cm}, \\mathrm{v}=10 \\mathrm{~cm}, \\mathrm{f}=$ ?\n\nUsing lens formula,\n\n$(1 / \\mathrm{f})=[(1 / \\mathrm{v})-(1 / \\mathrm{u})]$\n\n$1 / \\mathrm{f}=[(1 / 10)-(1 /-10)]$\n\n$\\mathrm{f}=5 \\mathrm{~cm}$\n\nCase II: Due to introduction of slab, shift in the source is $=\\mathrm{t}[1-(1 / \\mu)]=1.5[1-(2 / 3)]=0.5$\n\nNow, $u=-9.5 \\mathrm{~cm}, \\mathrm{v}=10.55 \\mathrm{~cm} \\mathrm{~d}=10.55-10=0.55 \\mathrm{~cm}$ away from the lens.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q2: The speed of light in the medium is",
        "input": "(a) maximum on the axis of the beam\n(b) minimum on the axis of the beam\n(c) the same everywhere in the beam\n(d) directly proportional to the intensity I",
        "answer": "(b) minimum on the axis of the beam",
        "solution": "Given $\\mu=\\mu_{0}+\\mu_{2} \\mathrm{I}$\n\nAs $\\mu=$ Speed of light in vacuum/Speed of light in the medium\n\n$\\mu=\\mathrm{c} / \\mathrm{v}$\n\\$mathrm{v}=\\mathrm{c} / \\mu=\\mathrm{c} /\\left(\\mu_{0}+\\mu_{2} \\mathrm{I}\\right)$\n\nAs the intensity is maximum on the axis of the beam, therefore $v$ is minimum on the axis of the beam.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q3: As the beam enters the medium, it will",
        "input": "(a) travel as a cylindrical beam\n(b) diverge\n(c) converge\n(d) diverge near the axis and converge near the periphery",
        "answer": "(c) converge",
        "solution": "As the beam enters the medium, it will converge.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q4: The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \\mathrm{~mm}$ ). This surface separates two media of refractive indices 1 and 1.34 . Calculate the distance from the refracting surface at which a parallel beam of light will come to focus",
        "input": "(a) $4.0 \\mathrm{~cm}$\n(b) $1 \\mathrm{~cm}$\n(c) $3.1 \\mathrm{~cm}$\n(d) $2 \\mathrm{~cm}$",
        "answer": "(c) $3.1 \\mathrm{~cm}$",
        "solution": "$(1.34 / v)-(1 / \\infty)=(1.34-1) / 7.8$\n\n$1.34 / \\mathrm{v}=0.34 / 7.8$\n\n$\\mathrm{v}=(1.34 \\times 7.8) / 0.34$\n\n$\\mathrm{v}=30.7 \\mathrm{~mm} \\approx 3.1 \\mathrm{~cm}$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q5: In an interference experiment the ratio of amplitudes of coherent waves is $\\left(a_{1} / a_{2}\\right)=1 / 3$. The ratio of maximum and minimum intensities of fringes will be",
        "input": "(a) 4\n(b) 9\n(c) 2\n(d) 18",
        "answer": "(a) 4",
        "solution": "$\\left(\\mathrm{I}_{\\max } / \\mathrm{I}_{\\text {min }}\\right)=\\left(\\mathrm{a}_{1}+\\mathrm{a}_{2}\\right)^{2} /\\left(\\mathrm{a}_{1}-\\mathrm{a}_{2}\\right)^{2}=(1+3)^{2} /(1-3)^{2}=16 / 4=4$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q6: The image formed by an objective of a compound microscope is",
        "input": "(a) virtual and diminished\n(b) real and diminished\n(c) real and enlarged\n(d) virtual and enlarged",
        "answer": "(c) The objective of a compound microscope forms a real and enlarged image.",
        "solution": "The image formed by an objective of a compound microscope is real and enlarged.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q7: Calculate the limit of resolution of a telescope objective having a diameter of $200 \\mathrm{~cm}$, if it has to detect light of wavelength $500 \\mathrm{~nm}$ coming from a star.",
        "input": "(a) $610 \\times 10^{-9}$ radian\n(b) $152.5 \\times 10^{-9}$ radian\n(c) $457.5 \\times 10^{-9}$ radian\n(d) $305 \\times 10^{-9}$ radian",
        "answer": "(d) $305 \\times 10^{-9}$ radian",
        "solution": "The limit of resolution,\n\n$\\Delta \\theta=1.22 \\lambda / \\mathrm{a}=\\left(1.22 \\times 500 \\times 10^{-9} / 200 \\times 10^{-2}\\right)=3.05 \\times 10^{-7} \\mathrm{radian}$\n\n$\\Delta \\theta=305 \\times 10^{-9}$ radian",
        "topic": "Ray Optics"
    },
    {
        "question": "Q8: A concave mirror for face viewing has a focal length of $0.4 \\mathrm{~m}$. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is",
        "input": "(a) $0.16 \\mathrm{~m}$\n(b) $1.60 \\mathrm{~m}$\n(c) $0.32 \\mathrm{~m}$\n(d) $0.24 \\mathrm{~m}$",
        "answer": "(c) $0.32 \\mathrm{~m}$",
        "solution": "Given, $\\mathrm{f}=-0.4 \\mathrm{~m}, \\mathrm{~m}=5=(-\\mathrm{v} / \\mathrm{u})=(\\mathrm{y} /-\\mathrm{x}) \\Rightarrow \\mathrm{y}=5 \\mathrm{x}$\n\n$(1 / \\mathrm{f})=(1 / \\mathrm{v})+(1 / \\mathrm{u})$\n\n$\\Rightarrow(1 /-0.4)=(1 / 5 x)+(1 /-x)=4 /-5 x$\n\n$\\Rightarrow \\mathrm{x}=-0.32 \\mathrm{~m}$ so $\\mathrm{u}=0.32 \\mathrm{~m}$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q9: The value of the numerical aperture of the objective lens of a microscope is 1.25. If the light of wavelength $5000 \\AA$ is used, the minimum separation between two points, to be seen as distinct, will be",
        "input": "(a) $0.48 \\mathrm{~m}$\n(b) $0.12 \\mathrm{~m}$\n(c) $0.38 \\mathrm{~m}$\n(d) $0.24 \\mathrm{~m}$",
        "answer": "(d) $0.24 \\mathrm{~m}$",
        "solution": "Numerical aperture of the objective lens of microscope $=0.61 \\lambda / \\mathrm{d}$\n\nMinimum separation between two points $\\mathrm{d}$ to be seen clearly,\n\n$\\mathrm{d}=0.61 \\lambda /$ Numerical aperture\n\n$\\mathrm{d}=\\left(0.61 \\times 5000 \\times 10^{-10}\\right) / 1.25$\n\n$\\mathrm{d}=0.24 \\mathrm{~m}$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q10: A convergent doublet of separated lenses, corrected for spherical aberration, has a resultant focal length of $10 \\mathrm{~cm}$. The separation between the two lenses is $2 \\mathrm{~cm}$. The focal lengths of the component lenses are",
        "input": "(a) $18 \\mathrm{~cm}, 20 \\mathrm{~cm}$\n(b) $12 \\mathrm{~cm}, 14 \\mathrm{~cm}$\n(c) $16 \\mathrm{~cm}, 18 \\mathrm{~cm}$\n(d) $10 \\mathrm{~cm}, 12 \\mathrm{~cm}$",
        "answer": "(a) $18 \\mathrm{~cm}, 20 \\mathrm{~cm}$",
        "solution": "Since the doublet is corrected for the spherical aberration, it satisfies the following condition\n\n$\\mathrm{f}_{1}-\\mathrm{f}_{2}=\\mathrm{d}=2 \\mathrm{~cm}$\n\n$\\mathrm{f}_{1}=\\mathrm{f}_{2}+2$\n\nEquivalent focal length $=\\mathrm{F}$\n\n$\\mathrm{F}=\\left(\\mathrm{f}_{1} \\mathrm{f}_{2}\\right) /\\left(\\mathrm{f}_{1}+\\mathrm{f}_{2}-\\mathrm{d}\\right)=10 \\mathrm{~cm}$\n\nSolving it, we get\n\n$\\mathrm{f}_{1}=20 \\mathrm{~cm}$\n\n$\\mathrm{f}_{2}=18 \\mathrm{~cm}$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q11: A single slit of width $b$ is illuminated by coherent monochromatic light of wavelength $\\lambda$. If the second and fourth minima in the diffraction pattern at a distance $1 \\mathrm{~m}$ from the slit are at $3 \\mathrm{~cm}$ and $6 \\mathrm{~cm}$, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)",
        "input": "(a) $6.0 \\mathrm{~cm}$\n(b) $1.5 \\mathrm{~cm}$\n(c) $4.5 \\mathrm{~cm}$\n(d) $3.0 \\mathrm{~cm}$",
        "answer": "(d) $3.0 \\mathrm{~cm}$",
        "solution": "For single slit diffraction, $\\sin \\theta=\\mathrm{n} \\lambda / \\mathrm{b}$\n\nPosition of $\\mathrm{n}^{\\text {th }}$ minima from central maxima $=\\mathrm{n} \\lambda_{\\mathrm{D}} / \\mathrm{b}$\n\nWhen $\\mathrm{n}=2$, then $\\mathrm{x}_{2}=2 \\lambda_{\\mathrm{D}} / \\mathrm{b}=0.03$\n\nWhen $\\mathrm{n}=4$, then $\\mathrm{x}_{4}=4 \\lambda_{\\mathrm{D}} / \\mathrm{b}=0.06$\n\nEqn. (2) - Eqn. (1)\n\n$\\mathrm{x}_{4}-\\mathrm{x}_{2}=\\left(4 \\lambda_{\\mathrm{D}} / \\mathrm{b}\\right)-\\left(2 \\lambda_{\\mathrm{D}} / \\mathrm{b}\\right)=0.03$ or\n\nthen width of central maximum $=2 \\lambda_{\\mathrm{D}} / \\mathrm{b}=2 \\times(0.03 / 2)=0.03 \\mathrm{~m}=3 \\mathrm{~cm}$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q12: An observer looks at a distant tree of height $10 \\mathrm{~m}$ with a telescope of the magnifying power of 20 . To the observer, the tree appears",
        "input": "(a) 10 times taller\n(b) 10 times nearer\n(c) 20 times taller\n(d) 20 times nearer",
        "answer": "(d) 20 times nearer",
        "solution": "The telescope resolves and brings the objects closer, which are far away from the telescope. Hence, for a telescope with magnifying power 20 , the tree appears 20 times nearer.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q13: To determine the refractive index of a glass slab using a travelling microscope, the minimum number of readings required are",
        "input": "(a) Two\n(b) Four\n(c) Three\n(d) Five",
        "answer": "(c) Three",
        "solution": "To determine the refractive index of a glass slab using a travelling microscope, the minimum number of readings required are three.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q14: You are asked to design a shaving mirror assuming that a person keeps it $10 \\mathrm{~cm}$ from his face and views the magnified image of the face at the closest comfortable distance of $25 \\mathrm{~cm}$. The radius of curvature of the mirror would then be",
        "input": "(a) $30 \\mathrm{~cm}$\n(b) $24 \\mathrm{~cm}$\n(c) $60 \\mathrm{~cm}$\n(d) $-24 \\mathrm{~cm}$",
        "answer": "(c) $60 \\mathrm{~cm}$",
        "solution": "$(1 / 15)+(1 /-10)=1 / \\mathrm{f}$\n\n$\\mathrm{f}=-30 \\mathrm{~cm}$\n\n$R=2 f=2(-30)=-60 \\mathrm{~cm}$",
        "topic": "Ray Optics"
    },
    {
        "question": "Q15: A green light is incident from the water to the air-water interface at the critical angle ( $\\boldsymbol{\\theta}$ ). Select the",
        "input": "(a) The entire spectrum of visible light will come out of the water at various angles to the normal\n(b) The entire spectrum of visible light will come out of the water at an angle of $90^{\\circ}$ to the normal\n(c) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium\n(d) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium",
        "answer": "(c) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium",
        "solution": "As, $\\sin \\theta=1 / \\mu$\n\nAlso, the refractive index $(\\mu)$ of the medium depends on the wavelength of the light. $\\mu$ is less for the greater wavelength (i.e. lesser frequency).\n\nSo, $\\theta$ will be more for a lesser frequency of light. Hence, the spectrum of visible light whose frequency is less than that of green light will come out to the air medium.",
        "topic": "Ray Optics"
    },
    {
        "question": "Q1: Two-point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod which is of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of",
        "input": "1. 0.42 m from the mass of 0.3 kg\n2. 0.70 m from the mass of 0.7 kg\n3. 0.98 m from the mass of 0.3 kg\n4. 0.98 m from the mass of 0.7 kg",
        "answer": "(3) 0.98 m from the mass of 0.3 kg",
        "solution": "The moment of inertia of the system about the axis of rotation O is I = I1 + I2 = (0.3)x^2 + 0.7(1.4 - x)^2\nI = 0.3 x^2 + 0.7(1.96 + x^2 - 2.8 x)\n= 0.3 x^2 + 1.372 + 0.7 x^2 - 1.96 x\n= x^2 + 1.372 - 1.96 x\nThe work done in rotating the rod is converted into its rotational kinetic energy.\nW = 1 / 2 I \u03c9^2\n= 1 / 2 [x^2 + 1.372 - 1.96 x] \u03c9^2\nFor the work done to be minimum\ndW / dx = 0 \u21d2 2 x - 1.96 = 0\nx = 1.96 / 2 = 0.98 m",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q2: A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?",
        "input": "1. centre of the circle\n2. on the circumference of the circle\n3. inside the circle\n4. outside the circle",
        "answer": "(1) centre of the circle",
        "solution": "The force will pass through the centre of the circle. Therefore, the angular momentum will remain conserved at the centre of the circle.",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q3: A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is",
        "input": "1. 2 K\n2. K / 2\n3. K / 4\n4. 4 K",
        "answer": "(2) K / 2",
        "solution": "According to the conservation of angular momentum\nI \u03c90 = I1 \u03c91\nSo we have I1 = 2 I\n\u03c91 = \u03c90 / 2\nInitial kinetic energy = I \u03c90^2 / 2 = K\nFinal Kinetic energy = I1 \u03c91^2 / 2 = 2 I (\u03c90 / 2)^2 / 2 = K / 2",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q4: A particle is confined to rotate in a circular path with decreasing linear speed. Then which of the following is correct?",
        "input": "1. angular momentum is conserved about the centre\n2. only the direction of L is conserved\n3. it spirals towards the centre\n4. its acceleration is towards the centre",
        "answer": "(2) only the direction of L is conserved",
        "solution": "L is not conserved in magnitude since v is changing (decreasing). It is given that a particle is confined to rotate in a circular path, it cannot have a spiral path. Since the particle has two accelerations ac and at, therefore the net acceleration is not towards the centre. The direction of L remains the same even when the speed decreases.",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q5: A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach point B. The angle \u03b8 at which the speed of the bob is half of that at A satisfies",
        "input": "1. \u03b8 = \u03c0 / 2\n2. \u03c0 / 4 < \u03b8 < \u03c0 / 2\n3. \u03c0 / 2 < \u03b8 < 3 \u03c0 / 4\n4. 3 \u03c0 / 2 < \u03b8 < \u03c0",
        "answer": "(4) 3 \u03c0 / 2 < \u03b8 < \u03c0",
        "solution": "This is the case of vertical motion when the body just completes the circle. Here\nv = \u221a(5 g L)\nApplying energy conservation,\n1 / 2 mv0^2 = 1 / 2 mv^2 + mgl(1 - cos \u03b8)\nwhere v0 is the horizontal velocity at the bottom point, v is the velocity of bob where the bob inclined \u03b8 with vertical.\nAlso, we know the relation between the velocity at the topmost and velocity at the bottom point.\nmg(2l) = 1 / 2 mv0^2 - 1 / 2 mvtop^2\nSince v0 is just sufficient\nmvtop^2 / l = T + mg\nT = 0\nvtop = \u221a(g l)\nThen equation 2 becomes\nv0 = \u221a(5 g l)\nAccording to the question v = v0 / 2\nSo from equation (1)\n1 / 2 m(5 gl) = 1 / 2 m(5 gl / 4) + mgl(1 - cos \u03b8)\n(20 mgl - 5 mgl) / 8 = mgl(1 - cos \u03b8)\n(1 - cos \u03b8) = 15 / 8\ncos \u03b8 = 7 / 8\nHence, 3 \u03c0 / 2 < \u03b8 < \u03c0",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q6: A ball of mass (m) 0.5 kg is attached to the end of a string having a length (L) 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in radian/s) is",
        "input": "1. 9\n2. 18\n3. 27\n4. 36",
        "answer": "(4) 36",
        "solution": "m \u03c9^2 L sin \u03b8 cos \u03b8 = mg sin \u03b8\ncos \u03b8 = g / \u03c9^2 L\nsin \u03b8 = 1 / \u03c9^2 L\nsin \u03b8 = 1 / (\u03c9^2 L) \u221a((\u03c9^2 L)^2 - g^2)\nT = mg cos \u03b8 + m \u03c9^2 L sin^2 \u03b8\nT = mg (g / \u03c9^2 L) + (m \u03c9^2 / \u03c9^2 L)^2 ((\u03c9^2 L)^2 - g^2)\nT = mg / \u03c9^2 L [ g^2 + (\u03c9^2 L)^2 - g^2]\nT = m \u03c9^2 L = 324 (given)\n\u03c9 = \u221a(324 / 0.5 * 0.5)\n\u03c9 = 36 rad / s",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q8: A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m / s. A small ball of mass 0.1 kg, moving velocity 20 m / s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m / s. Immediately after the collision.",
        "input": "1. The ring has pure rotation about its stationary CM\n2. The ring comes to a complete stop\n3. Friction between the ring and the ground is to the left\n4. There is no friction between the ring and the ground",
        "answer": "(1) The ring has pure rotation about its stationary CM\n(3) Friction between the ring and the ground is to the left",
        "solution": "Let's assume that friction between the ground and the ring gives no impulse during the collision with the ball. Using conservation of momentum along the x-axis we get that the CM of the ring will come to rest. Thus option A is correct.\nSecondly, the question tells us that the ball gets a velocity in the vertical direction, hence there must be an impulse in the vertical direction. There will be horizontal and a vertical impulse on the ring at the point of contact. These will have components along the tangent of the ring, which will provide angular impulses.\nUsing angular impulse = change in angular momentum, we get\n= 2 cos 30\u00b0(1 / 2) - 1 sin 30\u00b0(1 / 2) = 2(1 / 4)(\u03c92 - \u03c91)\nnote that we have assumed that the direction of angular velocities is the same before and after and since LHS of the above equation is positive \u03c92 > \u03c91 so the ring must be slipping to right and hence the friction will be to the left as it will be opposite to the direction of motion. Thus option C is correct",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q9: A carpet of mass M, made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on the rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R / 2.",
        "input": "1. \u221a(14 / 3 g R)\n2. \u221a(7 / 3 g R)\n3. \u221a(2 g R)\n4. \u221a(g R)",
        "answer": "(1) \u221a(14 / 3 g R)",
        "solution": "The radius is reduced to R / 2, the mass of the rolled carpet = (M / \u03c0 R^2 1) \u00d7 (R / 2)^2 1 = M / 4\nRelease of potential energy = MgR - (M / 4) gR / 2 = 7 / 8 MgR\nThe kinetic energy at this instant is given by = 1 / 2 Mv^2 + 1 / 2 I \u03c9^2 = 1 / 2(M / 4) v^2 + 1 / 2(MR^2 / 32)(2v / R)^2 = (3 / 16) Mv^2\n(3 / 16) Mv^2 = 7 / 8 MgR\nv = \u221a(14 / 3 g R)",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q10: A bob of mass m attached to an inextensible string of length is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed \u03c9 rad / s about the vertical. About the point of suspension",
        "input": "1. Angular momentum changes in direction but not in magnitude\n2. Angular momentum changes both in direction and magnitude\n3. Angular momentum is conserved\n4. Angular momentum changes in magnitude but not in direction.",
        "answer": "(1) Angular momentum changes in direction but not in magnitude",
        "solution": "",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q14: Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is",
        "input": "1. (55 / 2) MR^2\n2. (73 / 2) MR^2\n3. (181 / 2) MR^2\n4. (19 / 2) MR^2",
        "answer": "(3) (181 / 2) MR^2",
        "solution": "I0 = Icm + md^2\nI0 = (7 MR^2 / 2) + 6 (M * (2 R)^2) = 55 MR^2 / 2\nIp = I0 + md^2\nIp = 55 MR^2 / 2 + 7 M (3 R)^2 = (181 / 2) MR^2",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q15: A pulley of radius 2 m is rotated about its axis by a force F = 20 t - 5 t^2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kgm^2, the number of rotations made by the pulley before its direction of motion it reversed, is",
        "input": "1. more than 6 but less than 9\n2. more than 9\n3. less than 3\n4. more than 3 but less than 6",
        "answer": "(4) more than 3 but less than 6",
        "solution": "Torque is given by \u03c4 = FR\nOr \u03b1 = FR / I\nGiven F = 20 t - 5 t^2\nR = 2 m,\nI = 10 kgm^2\n\u03b1 = [(20 t - 5 t^2) * 2] / 10\n\u03b1 = 4 t - t^2\n\u03c9 = \u222b0^t \u03b1 dt = 2 t^2 - t^3 / 3\nat \u03c9 = 0 \u21d2 2 t^2 - t^3 / 3 = 0\nt^3 = 6 t^2\nt = 6\n\u03b8 = \u222b0^6 \u03c9 dt = \u222b0^6 (2 t^2 - t^3 / 3) dt\n\u03b8 = 36 / 2 \u03c0 < 6",
        "topic": "Rotational Motion"
    },
    {
        "question": "Q1: The energy band gap is maximum in",
        "input": "(a) metals\n(b) superconductors\n(c) insulators\n(d) semiconductors",
        "answer": "(c) The energy band gap is maximum in insulators",
        "solution": "",
        "topic": "Semiconductors"
    },
    {
        "question": "Q2: A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of",
        "input": "(a) each of them increases\n(b) each of them decreases\n(c) copper decreases and germanium increases\n(d) copper increases and germanium decreases",
        "answer": "(c) copper decreases and germanium increases",
        "solution": "Copper is a conductor. Germanium is a semiconductor. When cooled, the resistance of copper decreases and that of germanium increases.",
        "topic": "Semiconductors"
    },
    {
        "question": "Q3: In the common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor ( \u03b2 ) will be",
        "input": "(a) 48\n(b) 49\n(c) 50\n(d) 51",
        "answer": "(b) 49",
        "solution": "\u03b2=Ic / Ib=Ic /(Ie-Ic)=5.488 /(5.60-5.488)=5.488 / 0.112=49",
        "topic": "Semiconductors"
    },
    {
        "question": "Q4: Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?",
        "input": "(a) The number of free electrons for conduction is significant only in Si and Ge but small in C\n(b) The number of free conduction electrons is significant in C but small in Si and Ge\n(c) The number of free conduction electrons is negligibly small in all the three\n(d) The number of free electrons for conduction is significant in all three",
        "answer": "(a) The number of free electrons for conduction is significant only in Si and Ge but small in C",
        "solution": "Carbon (C), silicon (Si) and germanium (Ge) have the same lattice structure and their valence electrons are 4. For C, these electrons are in the second orbit, for Si it is third and for germanium, it is the fourth orbit. In solid-state, the higher the orbit, the greater the possibility of overlapping of energy bands. Ionization energies are also less therefore Ge has more conductivity compared to Si. Both are semiconductors. Carbon is an insulator.",
        "topic": "Semiconductors"
    },
    {
        "question": "Q5: In the ratio of the concentration of electrons that of holes in a semiconductor is 7 / 5 and the ratio of currents is 7 / 4 then what is the ratio of their drift velocities?",
        "input": "(a) 4 / 7\n(b) 5 / 8\n(c) 4 / 5\n(d) 5 / 4",
        "answer": "(d) 5 / 4",
        "solution": "Drift velocity, Vd=I / nAe\n(vd)electron /(vd)hole=(Ie / Ih)(nh / ne)=(7 / 4) \u00d7(5 / 7)=5 / 4",
        "topic": "Semiconductors"
    },
    {
        "question": "Q6: At absolute zero, silicon (Si) acts as",
        "input": "(a) non-metal\n(b) metal\n(c) insulator\n(d) none of these",
        "answer": "(c) insulator",
        "solution": "Semiconductors like silicon (Si) and germanium (Ge) act as insulators at low temperature.",
        "topic": "Semiconductors"
    },
    {
        "question": "Q7: Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an n-type semiconductor, the density of electrons is 10^19 m^-3 and their mobility is 1.6 m^2 /(V-s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to",
        "input": "(a) 0.2 m\n(b) 4 m\n(c) 2 m\n(d) 0.4 m",
        "answer": "(d) 0.4 m",
        "solution": "J=neVd\nResistivity, \u03c1=E / j=E / neVd=1 / ne (vd / E)=1 / ne \u03bce\nResistivity, 1 /(10^19 \u00d7 1.6 \u00d7 10^-19 \u00d7 1.6)=0.39 \u03a9 m=0.4 \u03a9 m",
        "topic": "Semiconductors"
    },
    {
        "question": "Q8: The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480 nm is incident on it. The bandgap in (eV) for the semiconductor is",
        "input": "(a) 0.5 eV\n(b) 0.7 eV\n(c) 1.1 eV\n(d) 2.5 eV",
        "answer": "(a) 0.5 eV",
        "solution": "Band gap = Energy of photon of =2480 nm\nEnergy =(hc / \u03bb) J=(hc / \u03bbe) e V\nBand gap =((6.63 \u00d7 10^-34) \u00d7(3 \u00d7 10^8)) /((2480 \u00d7 10^-9) \u00d7(1.6 \u00d7 10^-19))=0.5 eV",
        "topic": "Semiconductors"
    },
    {
        "question": "Q9: The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the",
        "input": "(a) crystal structure\n(b) variation of the number of charge carriers with temperature\n(c) type of bonding\n(d) variation of scattering mechanism with temperature",
        "answer": "(b) variation of the number of charge carriers with temperature",
        "solution": "",
        "topic": "Semiconductors"
    },
    {
        "question": "Q10: A strip of copper and another germanium are cooled from room temperature to 80 K. The resistance of",
        "input": "(a) each of these decreases\n(b) copper strip increases and that of germanium decreases\n(c) copper strip decreases and that of germanium increases\n(d) each of these increases",
        "answer": "(c) copper strip decreases and that of germanium increases",
        "solution": "Copper is conductor and germanium is a semiconductor. When cooled, the resistance of copper strip decreases and that of germanium increases.",
        "topic": "Semiconductors"
    },
    {
        "question": "Q1. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in second is",
        "input": "(a) 8\u03c0 / 3\n(b) 4\u03c0 / 3\n(c) 3\u03c0 / 8\n(d) 7\u03c0 / 3",
        "answer": "Answer: (a) 8\u03c0 / 3",
        "solution": "In SHM, speed v = \u03c9 \u221a(A^2 - x^2), at x = 4, v = \u03c9 \u221a((5)^2 - (4)^2) = 3\u03c9\nAcceleration a = -\u03c9^2 x\nat x = 4, a = -4\u03c9^2\nAs |v| = |a|\n\u03c9 = 3 / 4 \u21d2 T = 2\u03c0 / \u03c9 = 8\u03c0 / 3",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q2: A cylindrical plastic bottle of negligible mass is filled with 310 mL of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency. If the radius of the bottle is 2.5 cm then \u03c9 is close to (density of water = 10^3 kg/m^3 )",
        "input": "(a) 2.50 rad/s\n(b) 3.75 rad/s\n(c) 5.00 rad/s\n(d) 1.25 rad/s",
        "answer": "None of the options given are correct",
        "solution": "Restoring force due to pressing the bottle with amount x,\nF = - (\u03c1Ax)g\na = - (\u03c1Ag/m)x\nTherefore, \u03c9^2 = \u03c1Ag/m = [\u03c1(\u03c0r^2)g] / m\n\u03c9 = \u221a(10^3 * \u03c0 * (2.5 * 10^-2)^2 * 10 / 310 * 10^-3)",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q3: A particle undergoing simple harmonic motion has time-dependent displacement given by x(t) = Asin(\u03c0t / 90). The ratio of kinetic to the potential energy of this particle at t = 210 s will be",
        "input": "(a) 1 / 3\n(b) 2\n(c) 1\n(d) 3",
        "answer": "Answer: (a) 1 / 3",
        "solution": "The maximum kinetic energy of the particle, K.E = (1 / 2)mA^2\u03c9^2cos^2(\u03c9t)\nThe potential energy of the particle at any time t, U = (1 / 2)mA^2\u03c9^2sin^2(\u03c9t)\nThe ratio of kinetic energy to potential energy, r = K.E/U = [(1 / 2)mA^2\u03c9^2cos^2(\u03c9t)] / [(1 / 2)mA^2\u03c9^2sin^2(\u03c9t)]\nr = cos^2(\u03c9t) / sin^2(\u03c9t)\nr = cot^2(\u03c9t)\nThe angular frequency is given as \u03c9 = \u03c0 / 90\nTime, t = 210 s\nTherefore, r = cot^2(\u03c0 / 90) * 210\nr = cot^2(7\u03c0 / 3)\nr = cot^2(2\u03c0 + \u03c0 / 3)\nWe know,\ncot(2n\u03c0 + \u03b8) = cot\u03b8\nTherefore, r = cot^2(\u03c0 / 3)\nr = (1 / \u221a3)^2\nr = 1 / 3",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q4: A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then",
        "input": "(a) K2 = K1\n(b) K2 = K1 / 2\n(c) K2 = 2K1\n(d) K2 = K1 / 4",
        "answer": "Answer: (c) K2 = 2K1",
        "solution": "Maximum kinetic energy = (1 / 2)m\u03c9^2A^2\n\u03c9 = \u221a(g / L)\nA = L\u03b8\nK.E = (1 / 2)mgL\u03b8^2\nK1 = (1 / 2)mgL\u03b8^2\nIf the length is doubled\nK2 = (1 / 2)mg2L\u03b8^2\nFrom (1) and (2)\n(K1 / K2) = (1 / 2)mgL\u03b8^2 / (1 / 2)mg2L\u03b8^2\n= 1 / 2\nK2 = 2K1",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q5: A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^-2 m. The relative change in the angular frequency of the pendulum is best given by",
        "input": "(a) 10^-5 rad/s\n(b) 10^-1 rad/s\n(c) 1 rad/s\n(d) 10^-3 rad/s",
        "answer": "Answer: (d) 10^-3 rad/s",
        "solution": "Angular frequency of the pendulum\n\u03c9 = \u221a(g / l)\n(\u0394\u03c9 / \u03c9) = (1 / 2)(\u0394geff / geff)\n(\u0394\u03c9 / \u03c9) = (1 / 2)(2A\u03c9^2 / g)\n\u0394\u03c9 / \u03c9 = (A\u03c9^2 / g)\n= (1^2 * 10^-2 / 10) = 10^-3",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q6: A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 / 1000 of the original amplitude is close to",
        "input": "(a) 100 s\n(b) 10 s\n(c) 20 s\n(d) 50 s",
        "answer": "Answer: (c) 20 s",
        "solution": "Frequency of damped oscillation, f = 5 Hz\nFor A = A / 2\nt1 = 2 s\nAlso, A = A0 e^-(b / 2m)t or 1 / 2 = e^-(b / 2m)2\n(b / m) = ln2\nFor A = A / 1000, t2 = ?\n1 / 1000 = e^-(b / 2m)t2\nOr 10^-3 = e^-(b / 2m)t2\n(b / 2m)t2 = 3ln10 or t2 = 6ln10 / ln2\nt2 = 20 s",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q7: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to",
        "input": "(a) 335 m/s\n(b) 322 m/s\n(c) 328 m/s\n(d) 341 m/s",
        "answer": "Answer: (c) 328 m/s",
        "solution": "Due to the jagged end\n(\u03bb1 / 4) = 11 cm, so v / (512 * 4) = 11 cm------(1)\n(\u03bb2 / 4) = 27 cm, so v / (256 * 4) = 27 cm------(2)\n(2) - (1)\nv / (512 * 4) = 0.16\nv = 328 m/s",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q9: In an engine, the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston is close to",
        "input": "(a) 0.7 Hz\n(b) 1.9 Hz\n(c) 1.2 Hz\n(d) 0.1 Hz",
        "answer": "Answer: (b) 1.9 Hz",
        "solution": "The amplitude of S.H.M., A = 7 cm = 0.07 m\nAs the washer does not stay in contact with the piston, at some particular frequency i.e. normal force on the washer = 0\nMaximum acceleration of washer = A\u03c9^2 = g\n\u03c9 = \u221a(g / A) = \u221a(10 / 0.07) = \u221a(1000 / 7)\nFrequency of the piston, v = \u03c9 / 2\u03c0\n= 1.9 Hz",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q10: A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to",
        "input": "(a) 680 Hz\n(b) 510 Hz\n(c) 340 Hz\n(d) 170 Hz",
        "answer": "Answer: (d) 170 Hz",
        "solution": "vair = 340 m/s, v = 5 m/s\nf1 - f2 = 5 beats per second\nApparently frequency heard by the observer on reflection from the wall,\nf1 = [v / (v - vs)]f = [340 / (340 - 5)]f = (340 / 335)f\nf2 = [v / (v + vs)]f = [340 / (340 + 5)]f = (340 / 345)f\nSince f1 - f2 = 5\n[(340 / 335) - (340 / 345)]f = 5 \u21d2 f = (5 / 340)([335 * 345] / 10) = 169.96 Hz = 170 Hz",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q11: A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^12 s^-1. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver 108 and Avogadro number 6.02 * 10^23 gm mole^-1 )",
        "input": "(a) 6.4 N m^-1\n(b) 7.1 N m^-1\n(c) 2.2 N m^-1\n(d) 5.5 N m^-1",
        "answer": "Answer: (b) 7.1 N m^-1",
        "solution": "Frequency of a particle executing SHM,\nf = 1 / (2\u03c0) \u221a(k / m)\nk = 4\u03c0^2 * f^2 * m\nHere, f = 10^12 s^-1, m = [108 / (6.02 * 10^23)] * 10^-3 Kg\nk = 4 * (3.14)^2 * (10^12)^2 * [108 * 10^-3 / 6.02 * 10^23] = 7.1 Nm^-1",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q12: The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s^-1. At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is \u03c0 / 4",
        "input": "(a) 500 \u221a2 m/s^2\n(b) 500 m/s^2\n(c) 750 \u221a2 m/s^2\n(d) 750 m/s^2",
        "answer": "Answer: (a) 500 \u221a2 m/s^2",
        "solution": "For simple harmonic motion,\nMaximum acceleration/Maximum velocity = 10 \u21d2 \u03c9^2a / \u03c9a = 10 or \u03c9 = 10\nAt t = 0; displacement, x = 5\nx = asin(\u03c9t + \u03a6)\n5 = asin(0 + \u03c0 / 4) or 5 = asin(\u03c0 / 4) or a = 5\u221a2 m\nMaximum acceleration = \u03c9^2a = 10^2 * 5\u221a2 = 500\u221a2 ms^-2",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q13: A child swinging on a swing in sitting position, stands up, then the time period of the swing will",
        "input": "(a) increase\n(b) decrease\n(c) remains the same\n(d) increases if the child is long and decreases if the child is short",
        "answer": "Answer: (b) decrease",
        "solution": "The time period will decrease.\nWhen the child stands up, the centre of gravity is shifted upwards and so the length of swing decreases. T = 2\u03c0\u221a(l / g)",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q1. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in second is",
        "input": "(a) 8\u03c0 / 3\n(b) 4\u03c0 / 3\n(c) 3\u03c0 / 8\n(d) 7\u03c0 / 3",
        "answer": "Answer: (a) 8\u03c0 / 3",
        "solution": "In SHM, speed v = \u03c9 \u221a(A^2 - x^2), at x = 4, v = \u03c9 \u221a((5)^2 - (4)^2) = 3\u03c9\nAcceleration a = -\u03c9^2 x\nat x = 4, a = -4\u03c9^2\nAs |v| = |a|\n\u03c9 = 3 / 4 \u21d2 T = 2\u03c0 / \u03c9 = 8\u03c0 / 3",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q2: A cylindrical plastic bottle of negligible mass is filled with 310 mL of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency. If the radius of the bottle is 2.5 cm then \u03c9 is close to (density of water = 10^3 kg/m^3 )",
        "input": "(a) 2.50 rad/s\n(b) 3.75 rad/s\n(c) 5.00 rad/s\n(d) 1.25 rad/s",
        "answer": "None of the options given are correct",
        "solution": "Restoring force due to pressing the bottle with amount x,\nF = - (\u03c1Ax)g\na = - (\u03c1Ag/m)x\nTherefore, \u03c9^2 = \u03c1Ag/m = [\u03c1(\u03c0r^2)g] / m\n\u03c9 = \u221a(10^3 * \u03c0 * (2.5 * 10^-2)^2 * 10 / 310 * 10^-3)",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q3: A particle undergoing simple harmonic motion has time-dependent displacement given by x(t) = Asin(\u03c0t / 90). The ratio of kinetic to the potential energy of this particle at t = 210 s will be",
        "input": "(a) 1 / 3\n(b) 2\n(c) 1\n(d) 3",
        "answer": "Answer: (a) 1 / 3",
        "solution": "The maximum kinetic energy of the particle, K.E = (1 / 2)mA^2\u03c9^2cos^2(\u03c9t)\nThe potential energy of the particle at any time t, U = (1 / 2)mA^2\u03c9^2sin^2(\u03c9t)\nThe ratio of kinetic energy to potential energy, r = K.E/U = [(1 / 2)mA^2\u03c9^2cos^2(\u03c9t)] / [(1 / 2)mA^2\u03c9^2sin^2(\u03c9t)]\nr = cos^2(\u03c9t) / sin^2(\u03c9t)\nr = cot^2(\u03c9t)\nThe angular frequency is given as \u03c9 = \u03c0 / 90\nTime, t = 210 s\nTherefore, r = cot^2(\u03c0 / 90) * 210\nr = cot^2(7\u03c0 / 3)\nr = cot^2(2\u03c0 + \u03c0 / 3)\nWe know,\ncot(2n\u03c0 + \u03b8) = cot\u03b8\nTherefore, r = cot^2(\u03c0 / 3)\nr = (1 / \u221a3)^2\nr = 1 / 3",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q4: A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then",
        "input": "(a) K2 = K1\n(b) K2 = K1 / 2\n(c) K2 = 2K1\n(d) K2 = K1 / 4",
        "answer": "Answer: (c) K2 = 2K1",
        "solution": "Maximum kinetic energy = (1 / 2)m\u03c9^2A^2\n\u03c9 = \u221a(g / L)\nA = L\u03b8\nK.E = (1 / 2)mgL\u03b8^2\nK1 = (1 / 2)mgL\u03b8^2\nIf the length is doubled\nK2 = (1 / 2)mg2L\u03b8^2\nFrom (1) and (2)\n(K1 / K2) = (1 / 2)mgL\u03b8^2 / (1 / 2)mg2L\u03b8^2\n= 1 / 2\nK2 = 2K1",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q5: A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^-2 m. The relative change in the angular frequency of the pendulum is best given by",
        "input": "(a) 10^-5 rad/s\n(b) 10^-1 rad/s\n(c) 1 rad/s\n(d) 10^-3 rad/s",
        "answer": "Answer: (d) 10^-3 rad/s",
        "solution": "Angular frequency of the pendulum\n\u03c9 = \u221a(g / l)\n(\u0394\u03c9 / \u03c9) = (1 / 2)(\u0394geff / geff)\n(\u0394\u03c9 / \u03c9) = (1 / 2)(2A\u03c9^2 / g)\n\u0394\u03c9 / \u03c9 = (A\u03c9^2 / g)\n= (1^2 * 10^-2 / 10) = 10^-3",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q6: A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 / 1000 of the original amplitude is close to",
        "input": "(a) 100 s\n(b) 10 s\n(c) 20 s\n(d) 50 s",
        "answer": "Answer: (c) 20 s",
        "solution": "Frequency of damped oscillation, f = 5 Hz\nFor A = A / 2\nt1 = 2 s\nAlso, A = A0 e^-(b / 2m)t or 1 / 2 = e^-(b / 2m)2\n(b / m) = ln2\nFor A = A / 1000, t2 = ?\n1 / 1000 = e^-(b / 2m)t2\nOr 10^-3 = e^-(b / 2m)t2\n(b / 2m)t2 = 3ln10 or t2 = 6ln10 / ln2\nt2 = 20 s",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q7: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to",
        "input": "(a) 335 m/s\n(b) 322 m/s\n(c) 328 m/s\n(d) 341 m/s",
        "answer": "Answer: (c) 328 m/s",
        "solution": "Due to the jagged end\n(\u03bb1 / 4) = 11 cm, so v / (512 * 4) = 11 cm------(1)\n(\u03bb2 / 4) = 27 cm, so v / (256 * 4) = 27 cm------(2)\n(2) - (1)\nv / (512 * 4) = 0.16\nv = 328 m/s",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q9: In an engine, the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston is close to",
        "input": "(a) 0.7 Hz\n(b) 1.9 Hz\n(c) 1.2 Hz\n(d) 0.1 Hz",
        "answer": "Answer: (b) 1.9 Hz",
        "solution": "The amplitude of S.H.M., A = 7 cm = 0.07 m\nAs the washer does not stay in contact with the piston, at some particular frequency i.e. normal force on the washer = 0\nMaximum acceleration of washer = A\u03c9^2 = g\n\u03c9 = \u221a(g / A) = \u221a(10 / 0.07) = \u221a(1000 / 7)\nFrequency of the piston, v = \u03c9 / 2\u03c0\n= 1.9 Hz",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q10: A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to",
        "input": "(a) 680 Hz\n(b) 510 Hz\n(c) 340 Hz\n(d) 170 Hz",
        "answer": "Answer: (d) 170 Hz",
        "solution": "vair = 340 m/s, v = 5 m/s\nf1 - f2 = 5 beats per second\nApparently frequency heard by the observer on reflection from the wall,\nf1 = [v / (v - vs)]f = [340 / (340 - 5)]f = (340 / 335)f\nf2 = [v / (v + vs)]f = [340 / (340 + 5)]f = (340 / 345)f\nSince f1 - f2 = 5\n[(340 / 335) - (340 / 345)]f = 5 \u21d2 f = (5 / 340)([335 * 345] / 10) = 169.96 Hz = 170 Hz",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q11: A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^12 s^-1. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver 108 and Avogadro number 6.02 * 10^23 gm mole^-1 )",
        "input": "(a) 6.4 N m^-1\n(b) 7.1 N m^-1\n(c) 2.2 N m^-1\n(d) 5.5 N m^-1",
        "answer": "Answer: (b) 7.1 N m^-1",
        "solution": "Frequency of a particle executing SHM,\nf = 1 / (2\u03c0) \u221a(k / m)\nk = 4\u03c0^2 * f^2 * m\nHere, f = 10^12 s^-1, m = [108 / (6.02 * 10^23)] * 10^-3 Kg\nk = 4 * (3.14)^2 * (10^12)^2 * [108 * 10^-3 / 6.02 * 10^23] = 7.1 Nm^-1",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q12: The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s^-1. At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is \u03c0 / 4",
        "input": "(a) 500 \u221a2 m/s^2\n(b) 500 m/s^2\n(c) 750 \u221a2 m/s^2\n(d) 750 m/s^2",
        "answer": "Answer: (a) 500 \u221a2 m/s^2",
        "solution": "For simple harmonic motion,\nMaximum acceleration/Maximum velocity = 10 \u21d2 \u03c9^2a / \u03c9a = 10 or \u03c9 = 10\nAt t = 0; displacement, x = 5\nx = asin(\u03c9t + \u03a6)\n5 = asin(0 + \u03c0 / 4) or 5 = asin(\u03c0 / 4) or a = 5\u221a2 m\nMaximum acceleration = \u03c9^2a = 10^2 * 5\u221a2 = 500\u221a2 ms^-2",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q13: A child swinging on a swing in sitting position, stands up, then the time period of the swing will",
        "input": "(a) increase\n(b) decrease\n(c) remains the same\n(d) increases if the child is long and decreases if the child is short",
        "answer": "Answer: (b) decrease",
        "solution": "The time period will decrease.\nWhen the child stands up, the centre of gravity is shifted upwards and so the length of swing decreases. T = 2\u03c0\u221a(l / g)",
        "topic": "Simple Harmonic Motion"
    },
    {
        "question": "Q1: A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to",
        "input": "666 Hz\n753 Hz\n500 Hz\n333 Hz",
        "answer": "666 Hz",
        "solution": "Frequency of the sound produced by the open flute\nf=2(v / 21)=(2 x 330) /(2 x 0.5)=660 Hz\nVelocity of observer, v0=10 x(5 / 18)=(25 / 9) m/s\nAs the source is moving towards the observer, according to the Doppler effect.\nFrequency detected by observer\nf'={((v+v0) / v)} f={((25 / 9)+330) / 330} 660\n=2((25 / 9)+330)\nf'=665.55 \u2248 666 Hz",
        "topic": "Sound Waves"
    },
    {
        "question": "Q2: A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source when it is moving away from the observer after crossing him? (Take the velocity of sound in air as 350 m/s)",
        "input": "750 Hz\n857 Hz\n1143 Hz\n807 Hz",
        "answer": "750 Hz",
        "solution": "When source is moving towards a stationary observer,\nfapp= fsource {(V-0) / (V-50)}\n1000=fsource (350 / 300)\nWhen source is moving away from observer\nf'=fsource {350 / (350+50)}\nf'={(1000 x 300) / 350} x(350 / 400)\nf'=750 Hz",
        "topic": "Sound Waves"
    },
    {
        "question": "Q3: Two sources of sound S1 and S2 produce sound waves of the same frequency 660 Hz. A listener is moving from source S1 towards S2 with constant speed u m/s and he hears 10 beats. The velocity of sound is 330 m/s. Then u is equals",
        "input": "5.5 m/s\n15.0 m/s\n2.5 m/s\n10.0 m/s",
        "answer": "2.5 m/s",
        "solution": "f1=f[(v-v0) / v]\nf2=f[(v+v0) / v]\nFrequency f2-f1=fx(2v0 / v)\n10=660 x(2u / 330)\nu=2.5 m/s",
        "topic": "Sound Waves"
    },
    {
        "question": "Q4: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to",
        "input": "322 ms^-1\n341 ms^-1\n335 ms^-1\n328 ms^-1",
        "answer": "328 ms^-1",
        "solution": "(\u03bb / 4)=0.11+e\n{V /(512) 4}=0.11+e\n{V /(256) 4}=0.27+e\nAfter solving (1) and (2) we get\nV=328 ms^-1",
        "topic": "Sound Waves"
    },
    {
        "question": "Q5: A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? (Given reference intensity of sound is 10^-12 W/m^2)",
        "input": "40 cm\n20 cm\n10 cm\n30 cm",
        "answer": "40 cm",
        "solution": "Using \u03b2=10 log10(I / I0)\nOr 120=10 log 10(I / 10^-12)\nI / 10^-12=10^12\nI=1\nAlso I=1=P / 4\u03c0r^2=2 / 4\u03c0r^2\nr^2=2 / 4\u03c0I=2 / 4\u03c0\nr^2=2 / 4(3.14)\nOn solving the above equations, we get\nr=40 cm",
        "topic": "Sound Waves"
    },
    {
        "question": "Q6: Two cars A and B are moving away from each other in opposite directions. Both the cars are moving at a speed of 20 m/s with respect to the ground. If an observer in car A detects a frequency of 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B ? (speed of sound in air = 340 m/s)",
        "input": "2250 Hz\n200 Hz\n2300 Hz\n2150 Hz",
        "answer": "2250 Hz",
        "solution": "f'=f{(v-v0) / (v+vs)}\n2000=f{(340-20) / (340+20)}\nf=2250 Hz",
        "topic": "Sound Waves"
    },
    {
        "question": "Q7: A person standing on an open ground hears the sound of a jet aeroplane, coming from the north at an angle 60 degrees with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, the speed of the plane is",
        "input": "(\u221a3 / 2) v\n2v / \u221a3\nV\nv / 2",
        "answer": "v / 2",
        "solution": "Distance, PQ=vp xt (Distance = speed x time)\nDistance, QR=V . t\nCos 60=PQ / QR\n1 / 2=(vp x t) / (v.t)\nvp=v / 2",
        "topic": "Sound Waves"
    },
    {
        "question": "Q8: Three sound waves of equal amplitudes have frequencies(f-1, f, f+1). They superpose to give beats. The number of beats produced per second is",
        "input": "2\n1\n4\n3",
        "answer": "2",
        "solution": "Beat produced between(f-1) and f is 1 .\nBeat produced between f and f+1 is 1 .\nBeat produced between(f-1) and (f+1) is 2\nTherefore No of beats produced per second will be 2",
        "topic": "Sound Waves"
    },
    {
        "question": "Q9: A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in the air is 340 m/s.",
        "input": "6\n4\n12\n8",
        "answer": "6",
        "solution": "Fundamental frequency of the closed organ pipe is\nf=v / 4 L\nf=340 /(4 x 0.85)=100 Hz\nThe natural frequencies of the closed organ pipe is\nfn=(2n-1) f=f, 3f, 5f, 7f, 9f, 11f, 13f ...\nSo the possible frequencies below 1250 Hz are\nfn=100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz\nNumber of frequencies =6",
        "topic": "Sound Waves"
    },
    {
        "question": "Q10: An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light =3 x 10^8 m/s)",
        "input": "17.3 GHz\n15.3 GHz\n10.1 GHz\n12.1 GHz",
        "answer": "17.3 GHz",
        "solution": "Doppler effect in light (speed of observer is not very small compared to speed of light)\nf'=f[(1+v / c) / (1-v / c)]^{1 / 2}\nHere, frequency (v / c)=1 / 2\nSo, f'=f [(3 / 2) / (1 / 2)]^{1 / 2}\nf'=10 x \u221a3=17.3 GHz",
        "topic": "Sound Waves"
    },
    {
        "question": "Q1: In Young's double-slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 1/40 rad by using the light of wavelength \u03bb1. When the light of wavelength \u03bb2 is used a bright fringe is seen at the same angle in the same setup. Given that 1 and 2 are in the visible range (380 nm to 740 nm), their values are",
        "input": "(a) 400 nm, 500 nm\n(b) 625 nm, 500 nm\n(c) 380 nm, 525 nm\n(d) 380 nm, 500 nm",
        "answer": "(b) 625 nm, 500 nm",
        "solution": "Path difference = dsin\u03b8 = d \u00d7 \u03b8 = (0.1 mm)(1 / 40) = 2.5 \u00d7 10^{-3} mm = 2500 nm\nFor bright fringes, path difference = n\u03bb\nSo, 2500 = n\u03bb1 = m\u03bb2\nn = 4, m = 5\nor \u03bb1 = 2500 / 4 = 625 nm\n\u03bb2 = 2500 / 5 = 500 nm",
        "topic": "Wave Optics"
    },
    {
        "question": "Q2: In Young's double-slit experiment, the path difference, at a certain point on the screen, between two interfering waves is (1/8)th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to",
        "input": "(a) 0.80\n(b) 0.94\n(c) 0.85\n(d) 0.74",
        "answer": "(c) 0.85",
        "solution": "The phase difference between two waves is given as\n(\u0394x) x (2\u03c0/\u03bb) = (\u03bb/8) x (2\u03c0/\u03bb) = \u03c0/4\nSo, the intensity at this point is\nI = I0 cos\u00b2(\u03a6/2)\nI = I0 cos\u00b2(\u03c0/8)\nI = I0[(1 + cos(\u03c0/4)) / 2] = I0[(1 + (1/\u221a2)) / 2] = 0.85 I0",
        "topic": "Wave Optics"
    },
    {
        "question": "Q3: In a double-slit experiment, green light (5303 \u00c5) falls on a double slit having a separation of 19.44 m and a width of 4.05 m. The number of bright fringes between the first and the second diffraction minima is",
        "input": "(a) 10\n(b) 04\n(c) 05\n(d) 09",
        "answer": "(c) 05",
        "solution": "\u03bbg = 5303 \u00c5, d = 19.44 m, a = 4.05 m\nFor diffraction location of first minima and second minima\ny1 = D\u03bb/a, y2 = 2D\u03bb/a\ny2 - y1 = (2D\u03bb/a) - (D\u03bb/a) = D\u03bb/a\n\u03b2 = D\u03bb/d\nNumber of bright fringes = (y2 - y1) / \u03b2\n= (D\u03bb/a) x (d/D\u03bb)\n= d/a\n= 19.44 / 4.05 = 5",
        "topic": "Wave Optics"
    },
    {
        "question": "Q4: In an interference experiment the ratio of amplitudes of coherent waves is (a1 / a2) = (1 / 3). The ratio of maximum and minimum intensities of fringes will be",
        "input": "(a) 4\n(b) 9\n(c) 2\n(d) 18",
        "answer": "(a) 4",
        "solution": "(a1 / a2) = (1 / 3)\nImax = a1 + a2\nImin = a1 - a2\n(Imax / Imin) = [(a1 + a2)\u00b2 / (a1 - a2)\u00b2] = [(1 + 1/3)\u00b2 / (1 - 1/3)\u00b2] = (4/2)\u00b2 = 4",
        "topic": "Wave Optics"
    },
    {
        "question": "Q5: Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.",
        "input": "(a) 610 x 10^{-9} radian\n(b) 152.5 x 10^{-9} radian\n(c) 457.5 x 10^{-9} radian\n(d) 305 x 10^{-9} radian",
        "answer": "(d) 305 x 10^{-9} radian",
        "solution": "The limit of resolution,\n\u03b8\u0394 = 1.22\u03bb/a = (1.22 x 500 x 10^{-9}) / (200 x 10^{-2}) = 3.05 x 10^{-7} radian = 305 x 10^{-9} radian",
        "topic": "Wave Optics"
    },
    {
        "question": "Q6: In Young's double-slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be",
        "input": "(a) (\u221a3 + 1)\u00b2 : 16\n(b) 9 : 1\n(c) 25 : 9\n(d) 4 : 1",
        "answer": "(b) 9 : 1",
        "solution": "I1 = 4I0\nI2 = I0\nImax = (\u221aI1 + \u221aI2)\u00b2\n= (2\u221aI0 + \u221aI0)\u00b2 = 9I0\nImin = (\u221aI1 - \u221aI2)\u00b2\n= (2\u221aI0 - \u221aI0)\u00b2 = I0\n(Imax / Imin) = 9 / 1",
        "topic": "Wave Optics"
    },
    {
        "question": "Q7: In Young's double-slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide are",
        "input": "(a) 1.56 mm\n(b) 7.8 mm\n(c) 9.75 mm\n(d) 15.6 mm",
        "answer": "(b) 7.8 mm",
        "solution": "Let y be the distance from the central maximum to the point where the bright fringes due to both the wavelengths coincides.\nNow, for \u03bb1, y = m\u03bb1D/d\nFor \u03bb2, y = n\u03bb2D/d\nm\u03bb1 = n\u03bb2\n(m/n) = \u03bb2/\u03bb1 = 520/650 = 4/5\ni.e. with respect to central maximum 4th bright fringe of \u03bb1 coincides with 5th bright fringe of \u03bb2\nNow, y = (4 x 650 x 10^{-9} x 1.5) / (0.5 x 10^{-3}) m\n y = 7.8 x 10^{-3} m or y = 7.8 mm",
        "topic": "Wave Optics"
    },
    {
        "question": "Q8: A single slit of width b is illuminated by coherent monochromatic light of wavelength \u03bb. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)",
        "input": "(a) 6.0 cm\n(b) 1.5 cm\n(c) 4.5 cm\n(d) 3.0 cm",
        "answer": "(b) 1.5 cm",
        "solution": "For single slit diffraction, sin\u03b8 = n\u03bb/b (n = 1,2,3----)\nsin\u03b8 \u2248 \u03b8 \u2248 tan\u03b8\nb tan\u03b8 = n\u03bb\nb tan\u03b81 = 2\u03bb\nb(y1/D) = 2\u03bb\ntan\u03b82 = 2\u03bb\nb(y2/D) = 2\u03bb\n(y2 - y1) b / D = 2\u03bb\n(6 - 3) b / D = 2\u03bb\n3b / D = 2\u03bb\n\u03bbD / b = 3/2 = 1.5 cm",
        "topic": "Wave Optics"
    },
    {
        "question": "Q9: Unpolarized light of intensity I0 is incident on the surface of a block of glass at Brewster's angle. In that case, which one of the following statements is true?",
        "input": "(a) Transmitted light is partially polarized with intensity I0 / 2\n(b) Transmitted light is completely polarized with intensity less than I0 / 2\n(c) The reflected light is completely polarized with intensity less than I0 / 2\n(d) The reflected light is partially polarized with intensity I0 / 2",
        "answer": "(c) The reflected light is completely polarized with intensity less than I0 / 2",
        "solution": "At Brewster's angle, i = tan\u207b\u00b9(\u03bc), the reflected light is completely polarized, whereas refracted light is partially polarized. Thus, the reflected ray will have lesser intensity compared to the refracted ray.",
        "topic": "Wave Optics"
    },
    {
        "question": "Q10: A beam of unpolarized light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45\u00b0 relative to that of A. The intensity of the emergent light is",
        "input": "(a) I0 / 8\n(b) I0\n(c) I0 / 2\n(d) I0 / 4",
        "answer": "(d) I0 / 4",
        "solution": "The intensity of light after passing polaroid A is\nI1 = I0 / 2\nNow this light will pass through the second polaroid B whose axis is inclined at an angle of 45\u00b0 to the axis of polaroid A. So in accordance with Malus law, the intensity of light emerging from polaroid B is\nI2 = I1 cos\u00b2 45 = (I0 / 2)(1/\u221a2)\u00b2 = I0 / 4",
        "topic": "Wave Optics"
    },
    {
        "question": "Q1: When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coil is",
        "input": "(a) 0.67 H\n(b) 1.67 H\n(c) 3 H\n(d) 6 H",
        "answer": "(b) 1.67 H",
        "solution": "I1=5 A, I2=2 A, \u0394I=2-5=-3 A, \u0394t=0.1, S=50 V, As, \u03b5=-L(\u0394I / \u0394t), 50=-L(-3 / 0.1), 50=30 L, L=5 / 3=1.67 H",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q2: Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A= 10 cm^2 and length =20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (\u03bc0=4 \u03c0 \u00d7 10^-7 T mA^-1)",
        "input": "(a) 2.4 \u03c0 \u00d7 10^-4 H\n(b) 2.4 \u03c0 \u00d7 10^-5 H\n(c) 4.8 \u03c0 \u00d7 10^-4 H\n(d) 4.8 \u03c0 \u00d7 10^-5 H",
        "answer": "(a) 2.4 \u03c0 \u00d7 10^-4 H",
        "solution": "A=\u03c0 r1^2=10 cm^2, l=20 cm, N1=300, N2=400, M=\u03bc0 N1 N2 / l, M=(4 \u03c0 \u00d7 10^-7 \u00d7 300 \u00d7 400 \u00d7 10 \u00d7 10^-4) / 0.20=2.4 \u03c0 \u00d7 10^-4 H",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q3: The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of",
        "input": "(a) 1 \u03bcF\n(b) 2 \u03bcF\n(c) 4 \u03bcF\n(d) 8 \u03bcF",
        "answer": "(a) 1 \u03bcF",
        "solution": "For maximum power, L \u03c9=1 / C \u03c9, C=1 / L \u03c9^2, C=1 /[10 \u00d7(2 \u03c0 \u00d7 50)^2]=1 /(10 \u00d7 10^4 \u00d7(\u03c0)^2)=10^-6 F=1 \u03bcF (Taking \u03c0^2=10)",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q4: A coil of inductance 300 mH and resistance 2 \u03a9 is connected to a source of voltage 2 V. The current reaches half of its steady state value in",
        "input": "(a) 0.15 s\n(b) 0.3 s\n(c) 0.05 s\n(d) 0.1 s",
        "answer": "(d) 0.1 s",
        "solution": "During growth of charge in an inductance, I=I0(1-e^(-Rt / L)), or I0 / 2=I0(1-e^(-Rt / L)), e^(-Rt / L)=1 / 2=2^-1, or Rt / L=ln 2 \u21d2 t=(L / R) ln 2, t=[(300 \u00d7 10^-3) / 2] \u00d7(0.693), t=0.1 sec",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q5: In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to",
        "input": "(a) 4 L\n(b) 2 L\n(c) L / 2\n(d) L / 4",
        "answer": "(c) L / 2",
        "solution": "At resonance, \u03c9=1 / \u221a(L C) when \u03c9 is constant, (1 / L1 C1)=(1 / L2 C2)=(1 / LC)=1 / L2(2 C)=1 / 2 L2 C, L2=L / 2",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q6: Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon",
        "input": "(a) the rates at which currents are changing in the two coils\n(b) relative position and orientation of the two coils\n(c) the materials of the wires of the coils\n(d) the currents in the two coils",
        "answer": "(c) the materials of the wires of the coils",
        "solution": "Mutual inductance between two coils depends on the materials of the wires of the coils",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q7: The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity \u03c9 is",
        "input": "(a) R / \u03c9 L\n(b) R / (R^2+\u03c9^2 L^2)^(1 / 2)\n(c) \u03c9 L / R\n(d) R / (R^2-\u03c9^2 L^2)^(1 / 2)",
        "answer": "(b) R / (R^2+\u03c9^2 L^2)^(1 / 2)",
        "solution": "Power Factor, cos \u03c6=1 / (1+tan^2 \u03c6)^(1 / 2), cos \u03c6=1 / (1+(\u03c9 L / R)^2)^(1 / 2)(because tan \u03c6=\u03c9 L / R), cos \u03c6=R / (R^2+\u03c9^2 L^2)^(1 / 2)",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q8: An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to",
        "input": "(a) 80 H\n(b) 0.08 H\n(c) 0.044 H\n(d) 0.065 H",
        "answer": "(d) 0.065 H",
        "solution": "Given I=10 A, V=80 V, R=V / I=80 / 10=8 \u03a9 and \u03c9=50 Hz, I=V / (8^2+X_L^2)^(1 / 2), 10=220 / (64+X_L^2)^(1 / 2), (64+X_L^2)^(1 / 2)=22, Squaring on both sides, we get 64+X_L^2=484, X_L^2=484-64=420, X_L=(420)^(1 / 2) \u21d2 2 \u03c0 \u00d7 \u03c9 L=(420)^(1 / 2), Series inductor on an arc lamp, L=(420)^(1 / 2) / (2 \u03c0 \u00d7 50)=0.065 H",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q9: A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90 %, the output current would be",
        "input": "(a) 50 A\n(b) 25 A\n(c) 45 A\n(d) 20 A",
        "answer": "(c) 45 A",
        "solution": "Efficiency \u03b7=0.9=Ps / Pp, Vs Is=0.9 \u00d7 Vp Ip, Is=(0.9 \u00d7 2300 \u00d7 5) / 230=45 A",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q10: A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity \u03c9, the maximum e.m.f. induced in the coil will be",
        "input": "(a) (3 / 2) nBA \u03c9\n(b) nBA\n(c) 3 nBA \u03c9\n(d) (1 / 2) nBA \u03c9",
        "answer": "(b) nBA",
        "solution": "Emf induced in the coil is given by \u03b5=BAn \u03c9 sin \u03c9 t, \u03b5max=BA (The induced emf is maximum when sin \u03c9 t=1)",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q11: A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth's magnetic field of 0.3 \u00d7 10^-4 Wb/m^2. The value of the induced emf in wire is",
        "input": "(a) 0.3 \u00d7 10^-3 V\n(b) 2.5 \u00d7 10^-3 V\n(c) 1.5 \u00d7 10^-3 V\n(d) 1.1 \u00d7 10^-3 V",
        "answer": "(c) 1.5 \u00d7 10^-3 V",
        "solution": "The motional emf is given as \u03b5=|v(l \u00d7 B)| =5 \u00d7(0.3 \u00d7 10^-4)(10) sin 90\u00b0=1.5 \u00d7 10^-3 V",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q12: A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil",
        "input": "(a) increases by a factor of 27\n(b) decreases by a factor of 3\n(c) increases by a factor of 3\n(d) decreases by a factor of 9",
        "answer": "(c) increases by a factor of 3",
        "solution": "",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q13: The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is",
        "input": "(a) 637.5 J\n(b) 540 J\n(c) 437.5 J\n(d) 740 J",
        "answer": "(c) 437.5 J",
        "solution": "\u03b5=25 V, I1=10 A, I2=25 A, t=1 s, \u0394E=?, \u03b5=L(\u0394I / \u0394t), 25=L[(25-10) / 1], L=25 / 15=(5 / 3) H, \u0394E=1 / 2 L(I2^2-I1^2)=(1 / 2) \u00d7(5 / 3) \u00d7(25^2-10^2)=(5 / 3) \u00d7 525=437.5 J",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q14: There are two long co-axial solenoids of the same length l. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self inductance of the inner coil is",
        "input": "(a) n2 / n1\n(b) (n2 / n1)(r2^2 / r1^2)\n(c) (n2 / n1)(r1 / r2)\n(d) n1 / n2",
        "answer": "(a) n2 / n1",
        "solution": "The mutual inductance of the inner coil M=\u03bc0 n1 n2 r1^2, Self inductance (L) of the inner coil is L=\u03bc0 n1^2 r1^2, Using (1) and (2), M / L=n2 / n1",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q15: A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is",
        "input": "(a) 2 mV\n(b) 6 mV\n(c) 1 mV\n(d) 12 mV",
        "answer": "(d) 12 mV",
        "solution": "Emf developed across the given edges, \u03b5=Blv=0.1 \u00d7 0.02 \u00d7 6=12 \u00d7 10^-3 V=12 mV, As all the edges are parallel between the faces perpendicular to the x-axis, hence required potential difference is 12 mV",
        "topic": "Electromagnetic Induction"
    },
    {
        "question": "Q1: If the distance between the earth and the sun were half its present value, the number of days in a year would have been",
        "input": "(a) 64.5\n(b) 129\n(c) 182.5\n(d) 730",
        "answer": "(b) 129 days",
        "solution": "From Kepler's law, T\u00b2 \u221d R\u00b3\nTherefore, (T\u2082/T\u2081)\u00b2 = (R\u2082/R\u2081)\u00b3\nT\u2081 = 365 days, R\u2081 = R, R\u2082 = R/2\n(T\u2082/365)\u00b2 = (R/2/R)\u00b3\nT\u2082\u00b2 = (365)\u00b2/8\nT\u2082\u00b2 = 16,653\nT\u2082 = 129 days",
        "topic": "Gravitation"
    },
    {
        "question": "Q2: The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g' and 'R' (radius of the earth) is 10 m/s\u00b2 and 6400 km respectively. The required energy for this work will be",
        "input": "(a) 6.4 \u00d7 10\u00b9\u2070 Joules\n(b) 6.4 \u00d7 10\u00b9\u00b9 Joules\n(c) 6.4 \u00d7 10\u2078 Joules\n(d) 6.4 \u00d7 10\u2070 Joules",
        "answer": "(a) 6.4 \u00d7 10\u00b9\u2070 Joules",
        "solution": "The energy required is given by = GMm/R = gR\u00b2 \u00d7 m/R (because g = GM/R\u00b2) = mgR = 1000 \u00d7 10 \u00d7 6400 \u00d7 10\u00b3 = 64 \u00d7 10\u2079 J = 6.4 \u00d7 10\u00b9\u2070 J",
        "topic": "Gravitation"
    },
    {
        "question": "Q3: An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) energy E\u2080. Its potential energy is",
        "input": "(a) -E\u2080\n(b) 1.5E\u2080\n(c) 2E\u2080\n(d) E\u2080",
        "answer": "(c) 2E\u2080",
        "solution": "Total Energy, E\u2080 = -GMm/2r\nPotential Energy, U = -GMm/r = 2E\u2080",
        "topic": "Gravitation"
    },
    {
        "question": "Q4: The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is",
        "input": "(a) R/2\n(b) R/3\n(c) 2R\n(d) 3R",
        "answer": "(c) 2R",
        "solution": "Acceleration due to gravity at a height \"h\" is given by g' = g(R/R + h)\u00b2\nHere, g is the acceleration due to gravity on the surface\nR is the radius of the earth\nAs g' is given as g/9, we get g/9 = g(R/R + h)\u00b2\n1/3 = R/R + h\nh = 2R",
        "topic": "Gravitation"
    },
    {
        "question": "Q5: A simple pendulum has a time period T\u2081 when on the earth's surface, and T\u2082 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T\u2082/T\u2081 is",
        "input": "(a) 1\n(b) 3\n(c) 4\n(d) 2",
        "answer": "(d) 2",
        "solution": "2\u03c0\u221a(l/g)\nThe time period of a simple pendulum =\nOn the surface of earth g = GM/R\u00b2\nAt a height R above the earth g = GM/(2R)\u00b2\nThe time period on the surface of the earth T\u2081 = 2\u03c0\u221a(lR\u00b2/GM)\nThe time period on the surface of the earth T\u2082 = 2\u03c0\u221a(l(2R)\u00b2/GM)\nT\u2082/T\u2081 = 2",
        "topic": "Gravitation"
    },
    {
        "question": "Q6: A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (R = 6,400 km) will approximately be",
        "input": "(a) 1/2 hr\n(b) 1 hr\n(c) 2 hr\n(d) 4 hr",
        "answer": "(c) 2 hr",
        "solution": "According to Kepler's law, T\u00b2 \u221d R\u00b3\nTherefore, (T\u2082/T\u2081)\u00b2 = (R\u2082/R\u2081)\u00b3\nFor a spy satellite time period is given by T\u2081\nR\u2081 = 6400 km\nFor a geostationary satellite, T\u2082 = 24 hours\nR\u2082 = 36,000 km\n(24/T\u2081)\u00b2 = (36000/6400)\u00b3\n(24/T\u2081)\u00b2 = 178\n(24/T\u2081) = 13.34\nT\u2081 = 24/13.34 \u2248 2 hr",
        "topic": "Gravitation"
    },
    {
        "question": "Q7: Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is",
        "input": "(a) -6 Gm/r\n(b) -9 Gm/r\n(c) Zero\n(d) -4 Gm/r",
        "answer": "(b) -9 Gm/r",
        "solution": "P is the point where the field is zero, and a unit mass is placed at P.\nApplying Newton's law of gravitation,\n(Gm \u00d7 1)/x\u00b2 = (G 4m \u00d7 1)/(r - x)\u00b2\nx\u00b2/(r - x)\u00b2 = 1/4\nx/(r - x) = 1/2\n2x = r - x\nx = r/3\nPotential at the point P, V = -Gm/x - G(4m)/(r - x)",
        "topic": "Gravitation"
    },
    {
        "question": "Q8: A Binary star system consists of two stars A and B which have time periods T\u2090 and T\u1d66, radii R\u2090 and R\u1d66 and masses M\u2090 and M\u1d66. Then",
        "input": "(a) T\u2090 > T\u1d66 then R\u2090 > R\u1d66\n(b) T\u2090 > T\u1d66 then M\u2090 > M\u1d66\n(c) (T\u2090/T\u1d66)\u00b2 = (R\u2090/R\u1d66)\u00b2\n(d) T\u2090 = T\u1d66",
        "answer": "(d) T\u2090 = T\u1d66",
        "solution": "Angular velocity of binary stars are the equal \u03c9\u2090 = \u03c9\u1d66\n2\u03c0/T\u2090 = 2\u03c0/T\u1d66\n\u2234 T\u2090 = T\u1d66",
        "topic": "Gravitation"
    },
    {
        "question": "Q9: Two particles of equal mass \"m\" go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is",
        "input": "(a) \u221a(Gm/R)\n(b) \u221a(Gm/4R)\n(c) \u221a(Gm/3R)\n(d) \u221a(Gm/2R)",
        "answer": "(b) \u221a(Gm/4R)",
        "solution": "Gm\u00b2/4R\u00b2 = mv\u00b2/R\nV = \u221a(Gm/4R)",
        "topic": "Gravitation"
    },
    {
        "question": "Q10: A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass \"m\" is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is",
        "input": "(a) 1/2 mV\u00b2\n(b) mV\u00b2\n(c) 3/2 mV\u00b2\n(d) 2 mV\u00b2",
        "answer": "(b) mV\u00b2",
        "solution": "Kinetic energy at the time of ejection = 1/2 mv\u2091\u00b2\nV\u2091 is the escape velocity = \u221a2 \u00d7 orbital velocity\nV\u2091 is the escape velocity = \u221a2 v\nKinetic energy at the time of ejection = 1/2 m(\u221a2 v)\u00b2 = mv\u00b2",
        "topic": "Gravitation"
    },
    {
        "question": "Q11: Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. (F = Gm\u2081m\u2082/r\u00b2)",
        "input": "",
        "answer": "V = \u221a(Gm/a) and T = 2\u03c0 \u221a(a\u00b3/3 Gm)",
        "solution": "ABC is an equilateral triangle of side a\nEach particle moves in a circle of radius r\nAD\u00b2 = a\u00b2 - a\u00b2/4, r = 2/3 AD\nr = 2/3 \u00d7 \u221a(a\u00b2 - a\u00b2/4) r = a \u221a3---(1)\nTo find v\nLet v = Initial velocity given to each particle.\nFor circular motion, centripetal force should be provided. It is provided by the gravitational force between two masses. Let F denote this force.\nF = Gm\u00b2/a\u00b2\nResultant force = \u221a(F\u00b2 + F\u00b2 + 2 F\u00b2 cos 60\u00b0)\nResultant force = \u221a3 F\nResultant force = Centripetal force\n\u221a3 F = mv\u00b2/r\nV\u00b2 = \u221a3 F r/m = \u221a3/m \u00d7 (Gm\u00b2/a\u00b2) (a/\u221a3) = Gm/a\nV = \u221a(Gm/a)\nTo find time period of circular motion(T)\nT = 2\u03c0r/V = 2\u03c0(a/\u221a3) \u00d7 \u221a(a/ Gm) = 2\u03c0 \u221a(a\u00b3/3 Gm)\ntherefore, T = 2\u03c0 \u221a(a\u00b3/3 Gm)",
        "topic": "Gravitation"
    },
    {
        "question": "Q12: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?",
        "input": "(a) 5 GmM/6 R\n(b) 2 GmM/3 R\n(c) GmM/2 R\n(d) GmM/3 R",
        "answer": "(a) 5 GmM/6 R",
        "solution": "Energy of the satellite on the surface of the Earth E\u2081 = -GMm/R\nEnergy at a distance 2R is given by E\u2082 = -GMm/3R + 1/2 mv\u2080\u00b2\nE\u2082 = -GMm/3R + 1/2 m[GM/3R]\nE\u2082 = -GmM/6R\nE\u2082 - E\u2081 = (-GmM/6R) - (-GMm/R) = 5GmM/6R",
        "topic": "Gravitation"
    },
    {
        "question": "Q13: Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours, respectively. The radius of the orbit of S1 is 10\u2074 km. When S2 is closest to S1, find the angular speed of S2 as actually observed by an astronaut in S1",
        "input": "",
        "answer": "Angular speed = 7\u03c0/8",
        "solution": "Angular velocity \u03c9 = 2\u03c0/T\nHere T = 1 hour for S\u2081, \u03c9\u2081 = 2\u03c0 rad/hr\nSimilarly, for S\u2082, \u03c9\u2082 = 2\u03c0\nGiven that they are rotating in the same sense. So, relative angular velocity = 2\u03c0 - 2\u03c0/8 = 7\u03c0/8",
        "topic": "Gravitation"
    },
    {
        "question": "Q14: Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence",
        "input": "(a) There will be no change in weight anywhere on the earth\n(b) Weight of the object, everywhere on the earth, will increase\n(c) Except at poles, weight of the object on the earth will decrease\n(d) Weight of the object, everywhere on the earth, will decrease.",
        "answer": "(c) Except at poles, weight of the object on the earth will decrease",
        "solution": "The effect of rotation of earth on acceleration due to gravity is given by g' = g - \u03c9\u00b2R cos 2\u03a6\nWhere \u03a6 is latitude. There will be no change in gravity at poles as \u03a6 = 90\u00b0, at all points as \u03c9 increases g' will decrease.",
        "topic": "Gravitation"
    },
    {
        "question": "Q15: A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n\u1d57\u02b0 power of R. If the period of rotation of the particle is T, then",
        "input": "(a) T \u221d R\u00b3/\u2082\n(b) T \u221d R(m 2)+1\n(c) T \u221d R(n+1)/2\n(d) T \u221d R\u207f/2",
        "answer": "(c) T \u221d R(n+1)/2",
        "solution": "According to the question, central force is given by F\u1d9c \u221d 1/R\u207f\nF\u1d9c = k(1/R\u207f)\nm\u03c9\u00b2R = k(1/R\u207f)\nm(2\u03c0)\u00b2/T\u00b2 = k(1/R\u207f+\u00b9)\nOr T\u00b2 \u221d R\u207f+\u00b9\nT \u221d R(n+1)/2",
        "topic": "Gravitation"
    },
    {
        "question": "Q1: A solid sphere of radius $R$ acquires a terminal velocity $v_{1}$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\\eta$. The sphere is broken into 27 identical spheres. If each of these acquires a terminal velocity $v_{2}$, when falling through the same fluid, the ratio ( $\\left.v_{1} / v_{2}\right)$ equals",
        "input": "(a) 9\n(b) $1 / 27$\n(c) $1 / 9$\n(d) 27",
        "answer": "(a) 9",
        "solution": "$27 \\mathrm{x}(4 / 3) \\pi \\mathrm{r}^{3}=(4 / 3) \\pi \\mathrm{R}^{3}$\n$\\operatorname{Or} \\mathrm{r}=\\mathrm{R} / 3$\nTerminal velocity, $\\mathrm{v} \\propto \\mathrm{r}^{3}$\nTherefore, $\\left(\\mathrm{v}_{1} / \\mathrm{v}_{2}\right)=\\left(\\mathrm{R}^{2} / \\mathrm{r}^{2}\right)$\n$\\mathrm{v}_{1} \\mathrm{v}_{2}=[\\mathrm{R} /(\\mathrm{R} / 3)]^{2}=9$\n$\\left(\\mathrm{v}_{1} / \\mathrm{v}_{2}\right)=9$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q2: Spherical balls of radius $R$ are falling in a viscous fluid of viscosity with a velocity $v$. The retarding viscous force acting on the spherical ball is",
        "input": "(a) directly proportional to $\\mathrm{R}$ but inversely proportional to $\\mathrm{v}$\n(b) directly proportional to both radius $\\mathrm{R}$ and velocity $\\mathrm{v}$\n(c) inversely proportional to both radius $\\mathrm{R}$ and velocity $\\mathrm{v}$\n(d) inversely proportional to $\\mathrm{R}$ but directly proportional to velocity $\\mathrm{v}$",
        "answer": "(b) directly proportional to both radius $R$ and velocity $v$",
        "solution": "Retarding viscous force $=6 \\pi \\eta R v$\nobviously option (b) holds goods",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q3: A long cylindrical vessel is half-filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of the vessel is $5 \\mathrm{~cm}$ and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in $\\mathrm{cm}$, will be",
        "input": "(a) 0.4\n(b) 2.0\n(c) 0.1\n(d) 1.2",
        "answer": "(b) 2.0",
        "solution": "The linear speed of the liquid at the sides is $\\mathrm{r} \\omega$. So, the difference in height is given as follows\n$2 \\mathrm{gh}=\\omega^{2} \\mathrm{r}^{2}$\n$\\mathrm{h}=\\omega^{2} \\mathrm{r}^{2} / 2 \\mathrm{~g}$\nhere $\\omega=2 \\pi \\mathrm{f}$\nTherefore, $\\mathrm{h}=\\left[(2 \times 2 \\pi)^{2}\\left(5 \times 10^{-2}\right)^{2}\right] /(2 \times 10)=2 \\mathrm{~cm}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q4: Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3} \\mathrm{~m}$. The water velocity as it leaves the tap is $0.4 \\mathrm{~ms}^{-1}$. The diameter of the water stream at a distance $2 \times 10^{-1} \\mathrm{~m}$ below the tap is close to",
        "input": "(a) $5.0 \times 10^{-3} \\mathrm{~m}$\n(b) $7.5 \times 10^{-3} \\mathrm{~m}$\n(c) $9.6 \times 10^{-3} \\mathrm{~m}$\n(d) $3.6 \times 10^{-3} \\mathrm{~m}$",
        "answer": "(d) $3.6 \times 10^{-3} \\mathrm{~m}$",
        "solution": "Here, $\\mathrm{d}_{1}=8 \times 10^{-3} \\mathrm{~m}$\n$\\mathrm{v}_{1}=0.4 \\mathrm{~m} \\mathrm{~s}^{-1}, \\mathrm{~h}=0.2 \\mathrm{~m}$\nAccording to equation of motion,\n$$\n\\begin{aligned}\n& \\quad v_{2}=\\sqrt{v_{1}^{2}+2 g h}=\\sqrt{(0.4)^{2}+2+10 \times 0.2} \\n& =2 \\mathrm{~m} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\nAccording to equation of continuity $\\mathrm{a}_{1} \\mathrm{v}_{1}=\\mathrm{a}_{2} \\mathrm{v}_{2}$\nAccording to equation of continuity $\\mathrm{a}_{1} \\mathrm{v}_{1}=\\mathrm{a}_{2} \\mathrm{v}_{2}$\n$\\left(\\pi \\mathrm{D}_{1}{ }^{2} / 4\right) \\mathrm{v}_{1}=\\left(\\pi \\mathrm{D}_{2}{ }^{2} / 4\right) \\mathrm{v}_{2}$\n$\\mathrm{D}_{2}{ }^{2}=\\left(\\mathrm{v}_{1} / \\mathrm{v}_{2}\right) \\mathrm{D}_{1}{ }^{2}$\n$\\mathrm{D}_{2}=\\left[\\sqrt{ }\\left(\\mathrm{v}_{1} / \\mathrm{v}_{2}\right)\right] \\mathrm{D}_{1}$\n$=[\\sqrt{ }(0.4 / 2)] \times 8 \times 10^{-3} \\mathrm{~m}$\n$\\mathrm{D}_{2}=3.6 \times 10^{-3} \\mathrm{~m}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q5: A $20 \\mathrm{~cm}$ long capillary tube is dipped in water. The water rises up to $8 \\mathrm{~cm}$. If the entire arrangement is put in a freely falling elevator the length of the water column in the capillary tube will be",
        "input": "(a) $4 \\mathrm{~cm}$\n(b) $20 \\mathrm{~cm}$\n(c) $8 \\mathrm{~cm}$\n(d) $10 \\mathrm{~cm}$",
        "answer": "(b) $20 \\mathrm{~cm}$",
        "solution": "In a freely falling elevator, $\\mathrm{g}=0$ Water will rise to the full length i.e., $20 \\mathrm{~cm}$ to tube",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q6: A spherical solid ball of volume $V$ is made of a material of density $\rho_{1}$. It is falling through a liquid of density $\rho_{2}\\left(\rho_{2}<1\right)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{\text {viscous }}=-\\mathbf{k v}^{2}(k>0)$. The terminal speed of the ball is",
        "input": "(a) $\\operatorname{Vg}\\left(\rho_{1}-\rho_{2}\right)$\n(b) \\[ \\sqrt{ \\frac{V g\\left(\\rho_{1}-\\rho_{2}\\right)}{k}} \\]\n(c) $\\operatorname{Vg} \rho_{1} / \\mathrm{k}$\n(d) $\\sqrt{\\frac{V g \rho_{1}}{k}}$",
        "answer": "(b) \\[ \\sqrt{ \\frac{V g\\left(\\rho_{1}-\\rho_{2}\\right)}{k}} \\]",
        "solution": "The forces acting on the solid ball when it is falling through a liquid is \"mg\" downwards, thrust by Archimedes principle upwards and the force due to the force of friction also acting upwards. The viscous force rapidly increases with velocity, attaining a maximum when the ball reaches the terminal velocity.\nThen the acceleration is zero\n$\\mathrm{mg}-\\mathrm{V} \rho_{2} \\mathrm{~g}-\\mathrm{kv}_{\\mathrm{t}}^{2}=$ ma where $\\mathrm{V}$ is volume,\n$\\mathrm{v}_{\\mathrm{t}}$ is the terminal velocity\nWhen the ball is moving with terminal velocity, $\\mathrm{a}=0$\nTherefore $\\mathrm{V} \rho_{1} \\mathrm{~g}-\\mathrm{V} \rho_{2} \\mathrm{~g}-\\mathrm{kv}_{\\mathrm{t}}^{2}=0$\n$\\mathrm{v}_{\\mathrm{t}}=\\sqrt{\\frac{V g\\left(\\rho_{1}-\\rho_{2}\\right)}{k}}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q7: Water flows into a large tank with a flat bottom at the rate of $10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. Water is also leaking out of a hole of area $1 \\mathrm{~cm}^{2}$ at its button. If the height of the water in the tank remains steady, then this height is",
        "input": "(a) $5 \\mathrm{~cm}$\n(b) $7 \\mathrm{~cm}$\n(c) $4 \\mathrm{~cm}$\n(d) $9 \\mathrm{~cm}$",
        "answer": "(a) $5 \\mathrm{~cm}$",
        "solution": "Since the height of the water column is constant\nWater inflow rate $\\left(\\mathrm{Q}_{\text {in }}\right)=$ Water outflow rate $\\left(\\mathrm{Q}_{\text {out }}\right)$\n$\\mathrm{Q}_{\text {in }}=10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$\n$\\mathrm{Q}_{\text {out }}=10^{-4} \\mathrm{x} \\sqrt{ }(2 \\mathrm{gh})$\n$10^{-4}=10^{-4} \\mathrm{x} \\sqrt{ } 20 \\mathrm{xh}$\n$\\mathrm{h}=(1 / 20) \\mathrm{m}=5 \\mathrm{~cm}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q8: A submarine experiences a pressure of $5.05 \times 10^{6} \\mathrm{~Pa}$ at depth of $d_{1}$ in a sea. When it goes further to a depth of $d_{2}$, it experiences a pressure of $8.08 \times 10^{6} \\mathrm{~Pa}$. Then $d_{1}-d_{2}$ is approximately (density of water $=10^{3}$ $\\mathrm{ms}^{-2}$ and acceleration due to gravity $=10 \\mathrm{~ms}^{-2}$ )",
        "input": "(a) $300 \\mathrm{~m}$\n(b) $400 \\mathrm{~m}$\n(c) $600 \\mathrm{~m}$\n(d) $500 \\mathrm{~m}$",
        "answer": "(a) $300 \\mathrm{~m}$",
        "solution": "$\\mathrm{P}_{1}=\\mathrm{P}_{0}+\rho \\mathrm{gd}_{1}$\n$\\mathrm{P}_{2}=\\mathrm{P}_{0}+\rho \\mathrm{gd}_{2}$\n$\\Delta \\mathrm{P}=\\mathrm{P}_{2}-\\mathrm{P}_{1}=\rho g \\Delta \\mathrm{d}$\n$\\left(8.08 \times 10^{6}-5.05 \times 10^{6}\right)=10^{3} \times 10 \times \\Delta \\mathrm{d}$\n$3.03 \times 10^{6}=10^{3} \times 10 \times \\Delta \\mathrm{d}$\n$\\Delta \\mathrm{d}=303 \\mathrm{~m} \u0007pprox 300 \\mathrm{~m}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q9: Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is $5 \\mathrm{~cm}$, the Reynolds number for the flow is of the order (density of water $=1000 \\mathrm{~kg} / \\mathrm{m}^{3}$, coefficient of viscosity of water $=1 \\mathrm{mPa} \\mathrm{s}$ )",
        "input": "(a) $10^{3}$\n(b) $10^{4}$\n(c) $10^{2}$\n(d) $10^{6}$",
        "answer": "(b) $10^{4}$",
        "solution": "Reynolds number $=\rho v d / \\eta$\nVolume flow rate $=\\mathrm{vx} \\pi \\mathrm{r}^{2}$\n$\\mathrm{v}=\\left(100 \times 10^{-3} / 60\right) \times\\left(1 / \\pi \times 25 \times 10^{-4}\right)$\n$\\mathrm{v}=(2 / 3 \\pi) \\mathrm{m} / \\mathrm{s}$\nReynolds number $=\\left\\{\\left(10^{3} \times 2 \times 10 \times 10^{-2}\right) /\\left(10^{-3} \times 3 \\pi\\right)\\right\\}$\n$\\simeq 2 \times 10^{4}$\nOrder of $10^{4}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q10: The top of a water tank is open to the air and its water level is maintained. It is giving out $0.74 \\mathrm{~m}^{3}$ water per minute through a circular opening of $2 \\mathrm{~cm}$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to",
        "input": "(a) $6.0 \\mathrm{~m}$\n(b) $4.8 \\mathrm{~m}$\n(c) $9.6 \\mathrm{~m}$\n(d) $2.9 \\mathrm{~m}$",
        "answer": "(b) $4.8 \\mathrm{~m}$",
        "solution": "Here, volumetric flow rate $=(0.74 / 60)=\\pi r^{2} v=\\left(\\pi \times 4 \times 10^{-4}\right) \times \\sqrt{ } 2$ gh\n$\\Rightarrow \\sqrt{ } 2 \\mathrm{gh}=[(74 \times 100) / 240 \\pi)]$\n$\\Rightarrow \\sqrt{ } 2 \\mathrm{gh}=740 / 24 \\pi$\n$2 \\mathrm{gh}=(740 / 24 \\pi)^{2}$\n$\\mathrm{h}=[(740 \times 740) / 24 \times 24 \times 10)]\\left(\\right.$ since $\\left.\\pi^{2}=10\\right)$\n$\\mathrm{h} \u0007pprox 4.8 \\mathrm{~m}$",
        "topic": "Fluid Mechanics"
    },
    {
        "question": "Q1: During peddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts",
        "input": "(a) in the backward direction on the front wheel and in the forward direction on the rear wheel\n(b) in the forward direction on the front wheel and in the backward direction on the rear wheel\n(c) in the backward direction on both, the front and the rear wheels\n(d) in the forward direction on both, the front and the rear wheels.",
        "answer": "Answer: (a) Due to peddling, the point of contact of the rear wheel has a tendency to move backwards. So frictional force opposes the backwards tendency i.e., the frictional force acts in the forward direction. But the back wheel accelerates the front wheel in the forward direction. To oppose this frictional force acts in the backward direction on the front wheel.",
        "solution": "",
        "topic": "Friction"
    },
    {
        "question": "Q2: A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5 , the magnitude of the frictional force acting on the block is",
        "input": "(a) 2.5 N\n(b) 0.98 N\n(c) 4.9 N\n(d) 0.49 N",
        "answer": "Answer: (a) 2.5 N",
        "solution": "Solution: Consider the forces, acting on the block in the vertical direction\nForce of friction f=\u03bcR\nf=0.5\u00d75\nf=2.5 N",
        "topic": "Friction"
    },
    {
        "question": "Q3: A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2 . The weight of the block is-",
        "input": "(a) 20 N\n(b) 50 N\n(c) 100 N\n(d) 2 N",
        "answer": "Answer: (d) 2N",
        "solution": "Solution: Frictional force balances the weight of the body\nFrictional force f=\u03bcN=mg\nf=0.2X 10=2 N\nTherefore, weight of the block mg=2 N",
        "topic": "Friction"
    },
    {
        "question": "Q4: A marble block of mass 2 kg lying on ice when given a velocity of 6 m / s is stopped by friction in 10 s. Then the coefficient of friction is (consider g=10 m / s)",
        "input": "(a) 0.02\n(b) 0.03\n(c) 0.06\n(d) 0.01",
        "answer": "Answer: (c) 0.06",
        "solution": "Solution: u=6 m / s\nv=0\nt=10 s\na=-f / m=-\u03bcmg / m=-\u03bcg=-10 \u03bc\nSubstituting values in v=u+at\n0=6-10 \u03bc\u00d7 10\nTherefore, \u03bc=0.06",
        "topic": "Friction"
    },
    {
        "question": "Q5: A block P of mass m is placed on a horizontal frictionless plane. The second block of the same mass m is placed on it and is connected to a spring of spring constant k. The two blocks are pulled by a distance A. Block Q oscillates without slipping. What is the maximum value of the frictional force between the two block.",
        "input": "(a)kA/2\n(b) kA\n(c) \u03bcs mg\n(d) zero",
        "answer": "Answer: (a) fm=k A / 2",
        "solution": "Solution: Block Q oscillates but does not slip on P. It means that acceleration is the same for Q and P both. There is a force of friction between the two blocks while the horizontal plane is frictionless. The spring is connected to the upper block. The (P-Q) system oscillates with angular frequency \u03c9. The spring is stretched by A.\nTherefore, \u03c9=\u221ak / 2 m\nTherefore, Maximum acceleration in SHM=\u03c9^2 A\na_m=kA / 2 m\nNow consider the lower block.\nLet the maximum force of friction =f_m\nf_m=ma_m or f_m=mx(kA / 2 m)\nf_m=kA / 2",
        "topic": "Friction"
    },
    {
        "question": "Q6: Statement 1: A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30\u00b0 with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in mechanical energy in the second situation is smaller than that in the first situation.",
        "input": "Statement-2: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.\n(a) Statement-1 and 2 are true and statement-2 is a correct explanation for statement-1\n(b) Statement-1 and 2 are true and statement-2 is not a correct explanation for statement-1\n(c) Statement-1 is true, statement-2 is false\n(d) Statement-1 is false, statement-2 is true.",
        "answer": "Answer: (c) Statement-1 is true, statement-2 is false",
        "solution": "",
        "topic": "Friction"
    },
    {
        "question": "Q7: A block rests on a rough inclined plane making an angle of 30\u00b0 with the horizontal. The coefficient of static friction between the block and the plane is 0.8 . If the frictional force on the block is 10 N, the mass of the block (in kg) is : (taken g=10 m / s\u00b2 )",
        "input": "(a) 2.0\n(b) 4.0\n(c) 1.6\n(d) 2.5",
        "answer": "Answer: (a) 2",
        "solution": "Solution: \u03bc=0.8\nf=10 N\nmg sin 30=10\nmx 10\u00d7(1 / 2)=10\nm=2 kg",
        "topic": "Friction"
    },
    {
        "question": "Q8: A smooth block is released at rest on a 45\u00b0 incline and then slides a distance d. The time taken to slide is n times as much to slide on a rough incline than on a smooth incline. The coefficient of friction is-",
        "input": "(a) \u03bc k=1-1 / n\u00b2\n(b) \u03bc k=\u221a(1-1 / n\u00b2)\n(c) \u03bc s=1-1 / n\u00b2\n(d) \u03bc s=\u221a(1-1 / n\u00b2)",
        "answer": "Answer: (a) \u03bc k=1-1 / n\u00b2",
        "solution": "Solution: On a smooth plane\nd=1 / 2(g sin \u03b8) t_1\nt_1=\u221a(2 d / g sin \u03b8)\nOn a rough surface\nd=1 / 2(g sin \u03b8-\u03bc g cos \u03b8) t_2\nt_2=\u221a(2 d / g sin \u03b8-\u03bc g cos \u03b8)\nFrom the question, we know\nt_2=nt_1\n\u221a(2 d / g sin \u03b8-\u03bc g cos \u03b8)\nn\u221a(2 d / g sin \u03b8)\nn=1 / \u221a(1-\u03bc_k)\n(since cos 45\u00b0=sin 45\u00b0\n1 / \u221a2)\nn\u00b2=1 / 1-\u03bc_k\n1-\u03bc_k=1 / n\u00b2\n\u03bc_k=1-1 / n\u00b2",
        "topic": "Friction"
    },
    {
        "question": "Q9: The upper half of an inclined plane with inclination \u03a6 is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by",
        "input": "(a) 2 sin \u03a6\n(b) 2 cos \u03a6\n(c) 2 tan \u03a6\n(d) tan \u03a6",
        "answer": "Answer: (c) 2 tan \u03a6",
        "solution": "Solution: Suppose the length of the plane is L. When the block slides down the plane, the increase in the K.E will be equal to the decrease in the P.E\nWork done, W= Change in K.E(\u0394K)=(1 / 2) mu\u00b2-(1 / 2) mv\u00b2=0\nWork done by friction (W_f)+ Work done by gravity (W_g)=0\n-\u03bc g cos \u03a6(L / 2)+mgLsin \u03a6=0\n(\u03bc / 2) cos \u03a6=sin \u03a6\n\u03bc=2 tan \u03a6",
        "topic": "Friction"
    },
    {
        "question": "Q10: Consider a car moving on a straight road with a speed of 100 m / s. The distance at which a car can be stopped is : (\u03bc_k=0.5)",
        "input": "(a) 800 m\n(b) 1000 m\n(c) 100 m\n(d) 400 m",
        "answer": "Answer: (b) 1000",
        "solution": "Solution: When the car is stopped by friction then its retarding force is\nma=\u03bc R\nma=\u03bc mg\na=\u03bc g\nConsider the equation v\u00b2=u\u00b2-2 as (the car is retarding so a is negative)\n0=u\u00b2-2 as\n2as =u\u00b2\ns=u\u00b2 / 2 a\ns=u\u00b2 / 2 \u03bc g\ns=(100)\u00b2 / 2\u00d7 0.5\u00d7 10\ns=1000 m",
        "topic": "Friction"
    },
    {
        "question": "Q11: The minimum force required to start pushing a body up a rough (frictional coefficient \u03bc ) inclined plane F\u2081 while the minimum force needed to prevent it from sliding down is F\u2082. If the inclined plane makes an angle \u03b8 from the horizontal such that tan \u03b8=2 \u03bc then the ratio F\u2081 / F\u2082 is",
        "input": "(a) 4\n(b) 1\n(c) 2\n(d) 3",
        "answer": "Answer: (d) 3",
        "solution": "Solution: We have to work against the sliding force due to gravity and frictional force to push upwards. Therefore, the force F\u2081=mg sin \u03b8+\u03bc mg cos \u03b8.\nTo stop the body from sliding work is done against sliding force but frictional force stops the body from sliding.\nTherefore, the force F\u2082=mg sin \u03b8-\u03bc mg cos \u03b8.\nF\u2081 / F\u2082=(mg sin \u03b8+\u03bc mg cos \u03b8) /(mg sin \u03b8-\u03bc mg cos \u03b8)\nF\u2081 / F\u2082=(tan \u03b8+\u03bc) /(tan \u03b8-\u03bc)\n(since tan \u03b8=2 \u03bc )\nF\u2081 / F\u2082=(2 \u03bc+\u03bc) /(2 \u03bc-\u03bc)\nF\u2081 / F\u2082=3",
        "topic": "Friction"
    },
    {
        "question": "Q12: A body of mass m=10\u207b\u00b2 kg is moving in a medium and experiences a frictional force F=-kv\u00b2. Its initial speed is v\u2080=10 ms\u207b\u00b9. If, after 10 s, its energy is 1 / 8 mv\u2080\u00b2, the value of k will be",
        "input": "(a) 10\u207b\u00b3 kg m\u207b\u00b9\n(b) 10\u207b\u00b3 kg s\u207b\u00b9\n(c) 10\u207b\u2074 kg m\u207b\u00b9\n(d) 10\u207b\u00b9 kg m\u207b\u00b9 s\u207b\u00b9",
        "answer": "Answer: (c) 10\u207b\u2074 kg m\u207b\u00b9",
        "solution": "Solution: Mass m=10\u00b2 Kg\nInitial velocity v\u2080=10 ms\u207b\u00b9\nTime t=10 s\nFrictional Force F=-kv\u00b2\n1 / 2 mv_f\u00b2=1 / 8 mv\u2080\u00b2\nv_f=v\u2080 / 2\nv_f=10 / 2=5 m / s\nNow, the force is\n(10\u207b\u00b2) dv / dt=-kv\u00b2\n\u222b_{10}^{5} (dv / v\u00b2)=-100 k \u222b_{0}^{10} (dt)\n1 / 5=1 / 10=100 kx 10\nk=10\u207b\u2074",
        "topic": "Friction"
    },
    {
        "question": "Q13: A block of mass m is placed on a surface with a vertical cross-section given by y=x\u00b3 / 6. If the coefficient of friction is 0.5 , the maximum height above the ground at which the block can be placed without slipping is",
        "input": "(a) 1 / 3 m\n(b) 1 / 2 m\n(c) 1 / 6 m\n(d) 2 / 3 m",
        "answer": "Answer: (c) 1 / 6",
        "solution": "Solution: For equilibrium under limiting friction\nmg sin \u03b8=\u03bc mg cos \u03b8\ntan \u03b8=\u03bc\nThe equation of the surface is given by y=x\u00b3 / 6\nSlope dy / dx=x\u00b2 / 2\nFrom (1) and (2) we get\n\u03bc=x\u00b2 / 2=0.5\nx=1\nSo, y=1 / 6",
        "topic": "Friction"
    },
    {
        "question": "Q2: The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is",
        "input": "(a) 2 / 5\n(b) 3 / 2\n(c) 3 / 5\n(d) 2 / 3",
        "answer": "(a) 2 / 5",
        "solution": "For an ideal gas in an isobaric process, Heat supplied, Q=nCp \u0394T Work done, W=P \u0394V=nR \u0394T Required Ratio =W / Q Required Ratio =(nR \u0394T)/(nCp \u0394T) R / Cp=R/(\u03b3R / \u03b3-1) [since \u03b3=5 / 3, for a monoatomic gas ] =2 / 5",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q3: The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7\u00b0C. The gas is (R=8.3 J mol\u22121 K\u22121)",
        "input": "(a) monoatomic\n(b) diatomic\n(c) triatomic\n(d) a mixture of monoatomic and diatomic",
        "answer": "(b) diatomic",
        "solution": "Increase in temperature, \u0394T=T2\u2212T1=7 K (Since it is change in temperature so its value will be same in Kelvin and degree celsius) Work done on the system, W=146 KJ=146 \u00d7 1000 J Number of moles of gas, n=1000 Work done in an adiabatic process, W=nR(T2\u2212T1)/\u03b3-1 put the values, 146000=1000 \u00d7 8.3 \u00d7 7 /(\u03b3-1) \u03b3-1=0.39 \u03b3\u22481.4 Hence the gas is diatomic",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q4: Two moles of an ideal monatomic gas occupies a volume V at 27\u00b0C. The gas expands adiabatically to a volume 2 V. Calculate (i) the final temperature of the gas and (ii) change in its internal energy.",
        "input": "(a) (i) 198 K (ii) 2.7 kJ\n(b) (i) 195 K (ii) 2.7 kJ\n(c) (i) 189 K (ii) 2.7 kJ\n(d) (i) 195 K (ii) 2.7 kJ",
        "answer": "(c) (i) 189 K (ii) 2.7 kJ",
        "solution": "For an adiabatic process, PV^\u03b3=constant (nRT/V)V^\u03b3=constant or TV^(\u03b3-1)=constant T1 V1^(\u03b3-1)=T2 V2^(\u03b3-1) T2=T1[V1 / V2]^(\u03b3-1) Here, T1=27\u00b0C=300 K, V1=V, V2=2 V, \u03b3=5 / 3 T2=300(V / 2 V)(5 / 3-1)=300(1 / 2)^(2 / 3)=189 K Change in internal energy, \u0394U=nCv \u0394T \u0394U=n[(f / 2)R](T2\u2212T1)=2 \u00d7(3 / 2) \u00d7(25 / 3)(189-300) \u0394U=-2.7 kJ",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q6: The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 1000\u00b0C is (For steel Young's modulus is 2 \u00d7 10^11 Nm\u22122 and coefficient of thermal expansion is 1.1 \u00d7 10\u22125 K\u22121)",
        "input": "(a) 2.2 \u00d7 10^7 Pa\n(b) 2.2 \u00d7 10^6 Pa\n(c) 2.2 \u00d7 10^8 Pa\n(d) 2.2 \u00d7 10^9 Pa",
        "answer": "(c) 2.2 \u00d7 10^8 Pa",
        "solution": "Given, \u0394T=100\u00b0C, Y=2 \u00d7 10^11 N m\u22122 \u03b1=1.1 \u00d7 10\u22125 K\u22121 Young's modulus, Y= stress/strain =(F / A) /\u0394L / L Therefore, Y=P /\u0394L / L Y=P / \u03b1 \u0394T [ since \u0394L=L \u03b1 \u0394T] Thermal stress in a rod is the pressure due to the thermal strain. Required pressure P=Y \u03b1 \u0394T=2 \u00d7 10^11 \u00d7 1.1 \u00d7 10\u22125 \u00d7 100=2.2 \u00d7 10^8 Pa",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q7: Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by \u0394x on applying force F, how much force is needed to stretch wire 2 by the same amount?",
        "input": "(a) 6 F\n(b) 9 F\n(c) F\n(d) 4 F",
        "answer": "(d) 9 F",
        "solution": "For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire 11 is A and that of 12 is 3 A. V=V1=V2 V=A \u00d7 11=3 A \u00d7 12 12=11 / 3 Y=(F / A) /\u03941 / l \u21d2F1=YA(\u039411 / 11) F2=Y 3 A(\u0394l2 / l2) Given \u0394l1=\u0394l2=\u0394x (for the same extension) F2=Y 3 A(\u0394x / 11 / 3)=9(YA \u0394x / 11)=9 F1 or 9 F",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q8: A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to",
        "input": "(a) 9 F / (\u03c0 r^2 YT)\n(b) F / (3 \u03c0 r^2 YT)\n(c) 6 F / (\u03c0 r^2 YT)\n(d) 3 F / (\u03c0 r^2 YT)",
        "answer": "(d) 3 F / (\u03c0 r^2 YT)",
        "solution": "Given, Length =L Longitudinal force =F Radius =r Young's modulus =Y Temperature =T Since length remains same (Stress)Compressive =( Stress ) Thermal F / A=Y \u03b1 T \u21d2F / \u03c0 r^2=Y \u03b1 T (where \u03b1= coeficient of linear expansion) \u03b1=F / \u03c0 r 2 YT Coefficient of volume expansion, \u03b3=3 \u03b1 \u03b3=3 F / \u03c0 r^2 YT",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q9: During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP / CV for the gas is",
        "input": "(a) 4 / 3\n(b) 2\n(c) 5 / 3\n(d) 3 / 2",
        "answer": "(d) 3 / 2",
        "solution": "In an adiabatic process, T^\u03b3= (constant) P^\u03b3-1 T^\u03b3 / \u03b3-1= (constant)P Given T^3= (constant) P \u03b3 / \u03b3-1=3 \u21d2 3 \u03b3-3=\u03b3 \u21d2 \u03b3=3 / 2",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q10: An external pressure P is applied on a cube at 0\u00b0C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by",
        "input": "(a) 3 / P \u03b1 K\n(b) P / 3 \u03b1 K\n(c) 3 \u03b1 / PK\n(d) 3 PK \u03b1",
        "answer": "(b) P / 3 \u03b1 K",
        "solution": "The bulk modulus of the gas is given by K=-P /\u0394V / V 0 (Here negative sign indicates the decrease in volume with pressure) \u0394V / V0=P / K [in magnitude] Also, V=V0(1+\u03b3 \u0394T) or \u0394V / V0=\u03b3 \u0394T Comparing eq. (1) and (2), we get P / K=\u03b3 \u0394T \u0394T=P / 3 \u03b1 K (since \u03b3=3 \u03b1 )",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q11: Steel wire of length ' L ' at 40\u00b0C is suspended from the ceiling and then a mass ' m ' is hung from its free end. The wire is cooled down from 40\u00b0C to 30\u00b0C to regain its original length ' L '. The coefficient of linear thermal expansion of the steel is 10\u22125 /\u00b0C, Young's modulus of steel is 10^11 N / m^2 and radius of the wire is 1 mm. Assume that L > diameter of the wire. Then the value of ' m ' in kg is nearly",
        "input": "",
        "answer": "3",
        "solution": "E=(F / A) /\u0394L / L =(F \u00d7 L)/(A \u00d7 \u0394L) \u21d2(\u0394L / L)=F / AE Since, (\u0394L / L)=\u03b1 \u0394t \u03b1 \u0394t=F / AE \u03b1 \u0394t=mg / AE m=[(\u03b1 \u0394t) / g] AE m=3.14 \u2248 3",
        "topic": "Thermal Expansion"
    },
    {
        "question": "Q12: A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?",
        "input": "(a) 25 J\n(b) 35 J\n(c) 30 J\n(d) 40 J",
        "answer": "(b) 35 J",
        "solution": "The value of Cp for a diatomic gas is Cp=(7 / 2) R-(1) It is given that work done during expansion at constant pressure is dW=10 J=nR \u0394T Since the gas undergoes an isobaric process \u21d2 \u0394Q=n(7 / 2) R \u0394T=7 / 2(nR \u0394T)=(7 / 2)(10)=35 J",
        "topic": "Thermal Expansion"
    }
]