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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
|
(a) $100 \mathrm{~J}$ (b) $80 \mathrm{~J}$ (c) $40 \mathrm{~J}$ (d) $20 \mathrm{~J}$
|
(c) $40 \mathrm{~J}$
|
$\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$ $(\Delta \mathrm{U}){\mathrm{ACB}}=(\Delta \mathrm{U}){\mathrm{ADB}}$ $60-30=\Delta \mathrm{Q}-10$ $\Delta \mathrm{Q}=40 \mathrm{~J}$
|
Heat Transfer
|
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)
|
(a) $(2 \mathrm{K} / 3) \Delta \mathrm{T}$ (b) $1 / 2 R \Delta T$ (c) $(3 / 2) \mathrm{R} \Delta \mathrm{T}$ (d) $(1 / 2) \mathrm{KR} \Delta \mathrm{T}$
|
(b) $1 / 2 R \Delta T$
|
$\mathrm{VT}=\mathrm{K}$, $\mathrm{V}(\mathrm{PV} / \mathrm{R})=\mathrm{K}$ $\mathrm{PV}^{2}=$ constant For a polytropic process $\mathrm{C}=(\mathrm{R} / 1-\mathrm{x})+\mathrm{C}_{\mathrm{v}}=(\mathrm{R} / 1-2)+3 \mathrm{R} / 2=\mathrm{R} / 2$ $\Delta \mathrm{Q}=\mathrm{n} \mathrm{C} T=(1 / 2) \mathrm{R} \Delta \mathrm{T}$
|
Heat Transfer
|
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 $R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{K}^{-1}$ ]
|
(a) $350 \mathrm{~J}$ (b) $700 \mathrm{~J}$ (c) $748 \mathrm{~J}$ (d) 374 J
|
(c) 748 J
|
Number of moles of gas, $\mathrm{n}=(67.2 / 22.4)=3 \mathrm{~mol}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=3 imes(3 / 2) \mathrm{R} imes \Delta \mathrm{T}=3 imes(3 / 2) imes(8.31) imes 20=747.9=748 \mathrm{~J}$
|
Heat Transfer
|
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 $=4184 \mathrm{~J} / \mathrm{kg} / \mathrm{K}$ )
|
(a) $167.4 \mathrm{~kJ}$ (b) $8.4 \mathrm{~kJ}$ (c) $4.2 \mathrm{~kJ}$ (d) $16.7 \mathrm{~kJ}$
|
(d) 16.7 kJ
|
For isochoric process, $\Delta \mathrm{U}=\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}$ Here, $\mathrm{m}=200 \mathrm{~g}=0.2 \mathrm{~kg}, \mathrm{~s}=4184 \mathrm{~J} / \mathrm{kg} / \mathrm{K}$ $\Delta \mathrm{T}=60^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}=20^{\circ} \mathrm{C}$ $\Delta \mathrm{U}=0.2 imes 4184 imes 20=16736 \mathrm{~J}=16.7 \mathrm{~kJ}$
|
Heat Transfer
|
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?
|
(a) $25 \mathrm{~J}$ (b) $30 \mathrm{~J}$ (c) $35 \mathrm{~J}$ (d) $40 \mathrm{~J}$
|
(c) 35 J
|
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) $\Delta \mathrm{Q}=35 \mathrm{~J}$
|
Heat Transfer
|
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
|
(1) $90 \mathrm{Wm}^{-2}$ (2) $120 \mathrm{Wm}^{-2}$ (3) $65 \mathrm{Wm}^{-2}$ (4) $200 \mathrm{Wm}^{-2}$
|
(1) $90 \mathbf{Wm}^{-2}$
|
\\section*\{Temp. of \\ heat source $(\mathrm{dQ} / \mathrm{dt})=(\mathrm{kA} \Delta \mathrm{T}) / \mathrm{l}$ Energy flux, $\frac{1}{A}\left(\frac{d Q}{d t}\right)=\frac{k \Delta T}{l}$ $=(0.1)(900) / 1=90 \mathrm{~W} / \mathrm{m}^{2}$
|
Heat Transfer
|
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
|
(1) $\left(\mathrm{K}{1}+\mathrm{K}{2}
ight) / 2$ (2) $\mathrm{K}{1}+\mathrm{K}{2}$ (3) $\left(2 \mathrm{~K}{1}+3 \mathrm{~K}{2}
ight) / 5$ (4) $\left(\mathrm{K}{1}+3 \mathrm{~K}{2}
ight) / 4$
|
(4) $\left(\mathrm{K}{1}+3 \mathrm{~K}{2}
ight) / 4$
|
$\mathrm{K}{\mathrm{eq}}=\left(\mathrm{K}{1} \mathrm{A}{1}+\mathrm{K}{2} \mathrm{A}{2}
ight) /\left(\mathrm{A}{1}+\mathrm{A}{2}
ight)$ $\mathrm{K}{\mathrm{eq}}=\mathrm{K}{1} \pi \mathrm{R}^{2}+\mathrm{K}{2}\left[\pi(2 \mathrm{R})^{2}-\pi \mathrm{R}^{2}
ight] / \pi(2 \mathrm{R})^{2}$ $=\mathrm{K}{1} \pi \mathrm{R}^{2}+\mathrm{K}{2}\left[3 \pi \mathrm{R}^{2}
ight] / \pi 4 \mathrm{R}^{2}$ $=\left(\mathrm{K}{1}+3 \mathrm{~K}{2}
ight) / 4$
|
Heat Transfer
|
Q11: Heat given to a body which raises its temperature by $1^{\circ} \mathrm{C}$ is
|
(a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient
|
(b) thermal capacity
|
Heat Transfer
|
|
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 $T_{1}\left(T_{2}>T_{1}
ight)$. The rate of heat transfer through the slab, in a steady-state, is $\left[A\left(T_{2}-T_{1}
ight) K / x
ight] f$, with $f$ equal to
|
(1) 1 (2) $1 / 2$ (3) $2 / 3$ (4) $1 / 3$
|
(4) $1 / 3$
|
For the first surface, $\mathrm{Q}{1}=\mathrm{KA}\left(\mathrm{T}{2}-\mathrm{T}
ight) \mathrm{t} / \mathrm{x}$ For second surface, $\mathrm{Q}{2}=(2 \mathrm{~K}) \mathrm{A}\left(\mathrm{T}-\mathrm{T}{1}
ight) \mathrm{t} /(4 \mathrm{x})$ At steady state, $\mathrm{Q}{2}=\mathrm{Q}{1} \Rightarrow \mathrm{KA}\left(\mathrm{T}{2}-\mathrm{T}
ight) \mathrm{t} / \mathrm{x}=(2 \mathrm{~K}) \mathrm{A}\left(\mathrm{T}-\mathrm{T}{1}
ight) \mathrm{t} /(4 \mathrm{x})$ Or $2\left(\mathrm{T}{2}-\mathrm{T}
ight)=\left(\mathrm{T}-\mathrm{T}{1}
ight)$ $\mathrm{T}=\left(2 \mathrm{T}{2}+\mathrm{T}{1}
ight) / 3$ $\mathrm{Q}{1}=\mathrm{KA}\left(\mathrm{T}{2}-\left[\left(2 \mathrm{T}{2}+\mathrm{T}{1}
ight) / 3
ight]
ight) t / \mathrm{x}$ $\mathrm{H}=\mathrm{Q}{1} / \mathrm{t}=\mathrm{KA}\left(\mathrm{T}{2}-\left[\left(2 \mathrm{T}{2}+\mathrm{T}{1}
ight) / 3
ight]
ight) / \mathrm{x}$\ $=\mathrm{KA}\left(\mathrm{T}{2-} \mathrm{T}{1}
ight) / 3 \mathrm{x}$ $\left[ rac{A\left(T_{2}-T_{1}
ight) K}{x}
ight] f= rac{K A}{x}\left[ rac{T_{2}-T_{1}}{3}
ight]$ $\mathrm{f}=1 / 3$
|
Heat Transfer
|
Q13: According to Newton's law of cooling, the rate of cooling of a body is proportional to $(\Delta \Theta)^{ ext {n }}$, where is the difference of the temperature of the body and the surroundings, and $\mathbf{n}$ is equal to
|
(a) two (b) three (c) four (d) one
|
(d) one
|
According to Newton's law of cooling, the rate of cooling is proportional to $\Delta \Theta(\Delta \Theta)^{n}=(\Delta \Theta)$ or $n=1$.
|
Heat Transfer
|
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
|
(a) $2 / 5$ (b) $3 / 5$ (c) $3 / 7$ (d) $5 / 7$
|
(d) $5 / 7$
|
$\Delta \mathrm{U}=\mathrm{nC}{\mathrm{v}} \Delta \mathrm{T}$ $\Delta \mathrm{Q}=\mathrm{nC}{\mathrm{p}} \Delta \mathrm{T}$ Therefore, $\Delta \mathrm{U} / \Delta \mathrm{Q}=\mathrm{nC}{\mathrm{v}} \Delta \mathrm{T} / \mathrm{nC}{\mathrm{p}} \Delta \mathrm{T}=\mathrm{C}{\mathrm{v}} / \mathrm{C}{\mathrm{p}}=1 / \gamma=5 / 7$
|
Heat Transfer
|
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 $\left(30^{\circ} \mathrm{C}\ text { to } \left.35^{\circ} \mathrm{C}
ight)$ at constant volume is
|
(1) 30 (2) 50 (3) 70 (4) 90
|
(2) 50
|
$\mathrm{Q}=\mathrm{nC}{\mathrm{p}} \mathrm{dT}$ $\mathrm{C}{\mathrm{p}}=\mathrm{Q} / \mathrm{ndT}=70 / 2 \mathrm{x}(35-30)$ $\mathrm{C}{\mathrm{p}}=70 /(2 imes 5)$ $\mathrm{C}{\mathrm{p}}=7 \mathrm{cal} / \mathrm{mol} imes \mathrm{K}$ Now, $\mathrm{C}{\mathrm{v}}=\mathrm{C}{\mathrm{p}}-\mathrm{R}$ $\mathrm{C}{\mathrm{v}}=7-2=5 \mathrm{cal} / \mathrm{mol} imes \mathrm{K}$ $\mathrm{Q}^{\prime}=\mathrm{nC}{\mathrm{v}} \mathrm{dT}=2 imes 5 imes 5=50 \mathrm{cal}$
|
Heat Transfer
|
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
|
(a) $y=\operatorname{asin}(\omega t+k x)$
(b) $y=-\operatorname{asin}(\omega t+k x)$
(c) $\mathrm{y}=\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx})$
|
(b) $y=-\operatorname{asin}(\omega t+k x)$
|
Consider option (a)
Stationary wave: $\mathrm{Y}=\operatorname{asin}(\omega \mathrm{t}+\mathrm{kx})+\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx})$
when, $\mathrm{x}=0, \mathrm{Y}$ is not zero.
The option is not acceptable.
Consider option (b)
Stationary wave: $\mathrm{Y}=\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx})-\operatorname{asin}(\omega \mathrm{t}+\mathrm{kx})$ At $\mathrm{x}=0, \mathrm{Y}=\mathrm{a} \sin \omega \mathrm{t}-\mathrm{a} \sin \omega \mathrm{t}=\mathrm{zero}$
This option holds good.
Option (c) gives $\mathrm{Y}=\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx})+\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx})$
$\mathrm{Y}=2 \mathrm{asin} \omega \mathrm{t}($ At $\mathrm{x}=0, \mathrm{Y}$ is not zero)
Hence only option (b) holds good
|
Waves on String
|
Q2: The displacement $y$ of a wave travelling in the $x$-direction is given by
$y=10^{-4} \sin (600 t-2 x+\pi / 3)$ metre, where $x$ is expressed in metre and $t$ in second. The speed of the wave motion, in $\mathrm{m} \mathrm{s}^{-1}$ is
|
(a) 300
(b) 600
(c) 1200
(d) 200
|
(a) 300
|
Given wave equation: $\mathrm{y}=10^{-4} \sin (600 \mathrm{t}-2 \mathrm{x}+\pi / 3) \mathrm{m}$
Standard wave equation: $\mathrm{y}=\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx}+\Phi)$
Compare them
Angular speed, $\omega=600 \sec ^{-1}$
Propagation constant, $\mathrm{k}=2 \mathrm{~m}^{-1}$
$\omega / \mathrm{k}=(2 \pi \mathrm{f}) /(2 \pi / \lambda)=\mathrm{f} \lambda=$ velocity
Since velocity $=\omega / \mathrm{k}=600 / 2=300 \mathrm{~m} / \mathrm{sec}$
|
Waves on String
|
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
|
(a) $10.5 \mathrm{~Hz}$
(b) $105 \mathrm{~Hz}$
(c) $1.05 \mathrm{~Hz}$
(d) $1050 \mathrm{~Hz}$.
|
(b) $105 \mathrm{~Hz}$
|
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.
$315=$ nv/2L--------(1)
$420=(n+1) v / 2 L--(2)$
Dividing equa (1) by equa (2) we get
$\mathrm{n}=3$
From equa (1), lowest resonant frequency
$\mathrm{f}_{0}=\mathrm{v} / 2 \mathrm{~L}=315 / 3=105 \mathrm{~Hz}$
|
Waves on String
|
Q4: The transverse displacement $y(x, t)$ of a wave on a string is given by
$$
y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}
$$
\section*{This represents a}
|
(a) wave moving in $+\mathrm{x}$-direction with speed $\sqrt{\frac{a}{b}}$
(b) wave moving in $-\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
(c) standing wave of frequency $\sqrt{b}$
(d) standing wave of frequency $1 / \sqrt{ } b$
|
(b) wave moving in $-\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
|
Given
$$
\begin{aligned}
& y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)} \end{aligned}
$$
Comparing equation (1) with standard equation $y(x, t)=f(a x+b t)$
As there is a positive sign between $\mathrm{x}$ and $\mathrm{t}$ terms, hence wave travel in $-\mathrm{x}$ direction.
Wave speed $=$ Coefficient of $\mathrm{t} /$ Coefficient of $\mathrm{x}=\sqrt{\frac{b}{a}}$
|
Waves on String
|
Q5: The equation of a wave on a string of linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}$ is given by
$$
y=0.02(m) \sin 2 \pi\left(\frac{t}{0.04(s)}-\frac{x}{0.50(m)}\right)
$$
The tension in the string is
|
(a) $6.25 \mathrm{~N}$
(b) $4.0 \mathrm{~N}$
(c) $12.5 \mathrm{~N}$
(d) $0.5 \mathrm{~N}$
|
(a) $6.25 \mathrm{~N}$
|
Here, linear mass density $\mu=0.04 \mathrm{~kg} \mathrm{~m}^{-1}$
The given equation of a wave is
$$
y=0.02(m) \sin \left[2 \pi\left(\frac{t}{0.04}-\frac{x}{0.50}\right)\right]
$$
Compare it with the standard wave equation $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$
we get, $\omega=2 \pi / 0.04 \mathrm{rad} \mathrm{s}^{-1}, \mathrm{k}=2 \pi / 0.5 \mathrm{rad} \mathrm{m}^{-1}$
Wave velocity, $\mathrm{v}=\omega / \mathrm{k}=(2 \pi / 0.04) /(2 \pi / 0.5) \mathrm{ms}^{-1}$
$v=\sqrt{\frac{T}{\mu}}$
Also,
Here, $\mathrm{T}$ is the tension in the string and $\mu$ is the linear mass density
Equating equations (1) and (2), we get
$$
\frac{\omega}{k}=\sqrt{\frac{T}{\mu}}
$$
$\mathrm{T}=\mu \omega^{2} / \mathrm{k}^{2}$
$\mathrm{T}=\left[0.04 \times(2 \pi / 0.04)^{2}\right] /(2 \pi / 0.5)^{2}=6.25 \mathrm{~N}$
|
Waves on String
|
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
|
(a) \alpha=12.50 \pi, \beta=\pi / 2.0
(b) \alpha=25.00 \pi, \beta=\pi
(c) \alpha=0.08 / \pi, \beta=2.0 / \pi
(d) \alpha=0.04 / \pi, \beta=1.0 / \pi
|
(b) \alpha=25.00 \pi, \beta=\pi
|
The wave travelling along the $\mathrm{x}$-axis is given by
y(x, t)=0.005 \cos (\alpha x-\beta t)
Therefore, $\alpha=\mathrm{k}=2 \pi / \lambda$ As $\lambda=0.08 \mathrm{~m}$
$\alpha=2 \pi / 0.08=\pi / 0.04 \Rightarrow \alpha=(\pi / 4) \times 100=25.00 \pi$
$\omega=\beta \Rightarrow 2 \pi / 2=\pi$
$\therefore \alpha=25.00 \pi$
$\beta=\pi$
|
Waves on String
|
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 $\mathbf{g}=\mathbf{1 0} \mathbf{m ~ s}^{-2}$ )
|
(a) $5 \pi \sqrt{ } 5 \mathrm{~s}$
(b) $2 \mathrm{~s}$
(c) $2 \sqrt{ } 2 \mathrm{~s}$
(d) $\sqrt{ } 5$
|
(c) $2 \sqrt{2} s$
|
A uniform string of length $20 \mathrm{~m}$ is suspended from a rigid support.
As $\mu$ is mass per unit length of the rope, then $\mu=\mathrm{m} / \mathrm{L}$
$v=\sqrt{\frac{T}{\mu}}$
$\frac{d x}{d t}=\sqrt{\frac{m g x / L}{m / L}}=\sqrt{g} \sqrt{x}$
$\frac{d x}{\sqrt{x}}=\sqrt{g} d t$
Integrating, on both sides, we get
$\int_{0}^{L} x^{-1 / 2} d x=\sqrt{g} \int_{0}^{L} d t$
$t=2 \sqrt{L} / \sqrt{g}=2 \times \sqrt{\frac{20}{10}}=2 \sqrt{2} s$
|
Waves on String
|
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?
|
(a) $770 \mathrm{~Hz}$
(b) $188.5 \mathrm{~Hz}$
(c) $178.2 \mathrm{~Hz}$
(d) 200.5
|
(c) $178.2 \mathrm{~Hz}$
|
Frequency, $\mathrm{f}=\mathrm{v} / 21$
$=\frac{1}{2 I} \sqrt{\frac{T}{\mu}}=\frac{1}{2 I} \sqrt{\frac{T}{A d}}$
Where, $\mathrm{T}$ is tension in the wire and $\mu$ is the mass per unit length of wire.
Also, Young's modulus, $\mathrm{Y}=\mathrm{Tl} / \mathrm{A} \Delta 1$
$\mathrm{T} / \mathrm{A}=\mathrm{Y} \Delta \mathrm{I} / 1$
Substituting this value in frequency expression, we get
$f=\frac{1}{2 l} \sqrt{\frac{y \Delta l}{l d}}$
Given, $1=1.5 \mathrm{~m}, \Delta \mathrm{l} / 1=0.01$
$\mathrm{d}=7.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$\mathrm{Y}=2.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
Substituting these values we have
$f=\frac{1}{2 l} \sqrt{\frac{2.2 \times 10^{11} \times 0.01}{7.7 \times 10^{3}}}$
$\mathrm{f}=178.2 \mathrm{~Hz}$
|
Waves on String
|
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
|
(a) $428 \mathrm{~Hz}$
(b) $430 \mathrm{~Hz}$
(c) $420 \mathrm{~Hz}$
(d) $422 \mathrm{~Hz}$
|
(c) $420 \mathrm{~Hz}$
|
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 $420 \mathrm{~Hz}$.
|
Waves on String
|
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)
|
(a) $180 \mathrm{~m} / \mathrm{s}$
(b) $160 \mathrm{~m} / \mathrm{s}$
(c) $120 \mathrm{~m} / \mathrm{s}$
(d) $90 \mathrm{~m} / \mathrm{s}$
|
(b) $160 \mathrm{~m} / \mathrm{s}$
|
Given $\mathrm{y}(\mathrm{x}, \mathrm{t})=0.5 \sin [(5 \pi / 4) \mathrm{x}] \cos (200 \pi \mathrm{t})$
Comparing 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, $\mathrm{k}=5 \pi / 4$ $\mathrm{rad} / \mathrm{m}$, $\omega=200 \mathrm{rad} / \mathrm{s}$
Speed of the travelling wave, $v=\omega / \mathrm{k}=200 \pi /(5 \pi / 4)=160 \mathrm{~m} / \mathrm{s}$
|
Waves on String
|
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
|
(a) 20
(b) 80
(c) 40
(d) 120
|
(b) 80
|
$\lambda_{\max } / 2=40 \Rightarrow \lambda_{\max }=80 \mathrm{~cm}$
|
Waves on String
|
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.
Statement-2: Intensity of waves of given frequency in the same medium is proportional to square of amplitude only.
|
(a) Statement-1 is false, statement-2 is true
(b) Statement-1 is true, statement-2 is false
(c) Statement-1 is true, statement-2 true; statement-2 is the correct explanation of statement-1
(d) Statement-1 is true, statement-2 is true; statement-2 is not correct explanation of statement-1
|
(b) Statement-1 is true, statement-2 is false
|
$\mathrm{y}_{1}(\mathrm{x}, \mathrm{t})=2 \mathrm{asin}(\omega \mathrm{t}-\mathrm{kx})$
$\mathrm{y}_{2}(\mathrm{x}, \mathrm{t})=2 \mathrm{asin}(2 \omega \mathrm{t}-2 \mathrm{kx})$
But Intensity, $I=1 / 2\left(\rho \omega^{2} \mathrm{~A}^{2} v\right)$
Here, $\rho=$ density of the medium,
$\mathrm{A}=$ amplitude
$\mathrm{v}=$ velocity of the wave
Intensity depends upon amplitude, frequency, and velocity of the wave.
Also, $\mathrm{I}_{1}=\mathrm{I}_{2}$
|
Waves on String
|
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
|
(a) $\mathrm{f}$
(b) $\mathrm{f} / 2$
(c) $3 \mathrm{f} / 4$
(d) $2 \mathrm{f}$
|
(a) f
|
When the pipe is open at both ends
$\lambda / 2=1$
$\lambda=21$
$\mathrm{v}=\mathrm{f} \lambda$
$\mathrm{f}=\mathrm{v} / \lambda=\mathrm{v} / 21$
When the pipe is dipped vertically in the water so that half of the pipe is in water
$\lambda / 4=1 / 2$
$\lambda=21 \Rightarrow \mathrm{v}=\mathrm{f}^{\prime} \lambda$
$\mathrm{f}^{\prime}=\mathrm{v} / \lambda=\mathrm{v} / 21=\mathrm{f}--$
Thus, the fundamental frequency of the air column is now,
$\mathrm{f}=\mathrm{f}^{\prime}$
|
Waves on String
|
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
|
(a) $10 \mathrm{~cm}$
(b) $33.3 \mathrm{~cm}$
(c) $16.6 \mathrm{~cm}$
(d) $20.0 \mathrm{~cm}$
|
(d) $20.0 \mathrm{~cm}$
|
Velocity of the wave on the string
$$
\begin{aligned}
& V=
& \sqrt{\frac{t}{\mu}}
& =
& \sqrt{\frac{8}{5} \times 1000}
\end{aligned}
$$
$\mathrm{V}=40 \mathrm{~m} / \mathrm{s}$
Here, $\mathrm{T}=$ tension and $\mu=$ mass/length
Wavelength of the wave $\lambda=\mathrm{v} / \mathrm{n}=40 / 100$
Separation between successive nodes,
$\lambda / 2=40 /(2 \mathrm{x} 100)=20 \mathrm{~cm}$
|
Waves on String
|
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
|
(a) $10 \mathrm{~N}$
(b) $7.5 \mathrm{~N}$
(c) $12.5 \mathrm{~N}$
(d) $5 \mathrm{~N}$
|
(c) $12.5 \mathrm{~N}$
|
We have given,
y=0.03 \sin (450 t-9 x)
Comparing it with standard equation of wave, we get
\omega=450 \mathrm{k}=9
\mathrm{v}=\omega / \mathrm{k}=450 / 9=50 \mathrm{~m} / \mathrm{s}
Velocity of the travelling wave on a string is given by
$$
\begin{aligned}
& \mathrm{v}=
& \sqrt{\frac{t}{\mu}}
\end{aligned}
$$
$$
50=\sqrt{\frac{T}{50 \times 10^{-3}}}
$$
\mu=$ linear mass density
\mathrm{T}=2500 \times 5 \times 10^{-3}
\mathrm{T}=12.5 \mathrm{~N}$
|
Waves on String
|
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
|
(a) $3.39 \times 10^{3} \mathrm{~J}$
(b) $5.65 \times 10^{2} \mathrm{~J}$
(c) $2.26 \times 10^{3} \mathrm{~J}$
(d) $5.17 \times 10^{2} \mathrm{~J}$
|
(d) $5.17 \times 10^{2} \mathrm{~J}$
|
Impedance, $\mathrm{Z}=\left(\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}\right)^{1 / 2}$
$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=(2 \pi \mathrm{vL})$
$\mathrm{X}_{\mathrm{L}}=6.28 \times 50 \times 20 \times 10^{-3}=6.28 \Omega$
$\mathrm{X}_{\mathrm{C}}=(1 / \omega \mathrm{C})=1 /(2 \pi \mathrm{vC})=1 /\left(6.28 \times 120 \times 10^{-6} \times 50\right)=26.54 \Omega$
$\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}=26.54-6.28=20.26$
$\mathrm{Z}=\left((60)^{2}+(20.26)^{2}\right)^{1 / 2}$
$\mathrm{Z}^{2}=4010 \Omega^{2}$
Average power dissipated, $\operatorname{Pav}=\varepsilon_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \Phi$
$\mathrm{P}_{\mathrm{av}}=\varepsilon_{\text {rms }} \times\left(\varepsilon_{\text {rms }} / \mathrm{Z}\right) \times(\mathrm{R} / \mathrm{Z})$
$\mathrm{P}_{\mathrm{av}}=\left(\varepsilon_{\text {rms }}{ }^{2} / \mathrm{Z}^{2}\right) \times \mathrm{R}=\left[(24)^{2} / 4010\right] \times 60 \mathrm{~W}=8.62 \mathrm{~W}$
Energy dissipated in $60 \mathrm{~S}=8.62 \times 60=5.17 \times 10^{2} \mathrm{~J}$
|
Alternating Current
|
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
|
(a) NAB
(b) NABR
(c) NAB \omega
(d) NABRc
|
(c) In an a.c. generator, maximum emf $=\mathbf{N A B} \omega$
|
Alternating Current
|
|
Q3: The phase difference between the alternating current and emf is \(\pi / 2\). Which of the following cannot be the constituent of the circuit?
|
(a) LC
(b) L alone
(c) C alone
(d) R, L
|
(d) R, L
|
$\mathrm{R}$ and $\mathrm{L}$ cause phase difference to lie between 0 and $\pi / 2$ but never 0 and $\pi / 2$ at extremities
|
Alternating Current
|
Q4: Alternating current cannot be measured by D.C ammeter because
|
(a) A.C cannot pass through D.C ammeter
(b) A.C changes direction
(c) The average value of current for the complete cycle is zero
(d) D.C. ammeter will get damaged
|
(c) The average value of current for the complete cycle is zero
|
The average value of A.C for the complete cycle is zero. Hence A.C cannot be measured by D.C ammeter
|
Alternating Current
|
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
|
(a) $25 \mathrm{~A}$
(b) $50 \mathrm{~A}$
(c) $45 \mathrm{~A}$
(d) $35 \mathrm{~A}$
|
(c) 45 A
|
Given $\epsilon_{\mathrm{p}}=2300 \mathrm{~V}, \mathrm{~N}_{\mathrm{p}}=4000$
$\epsilon_{\mathrm{s}}=230 \mathrm{~V}$,
$\mathrm{I}_{\mathrm{p}}=5 \mathrm{~A}$,
$\eta=90 \%=0.9$
$\eta=\mathrm{P}_{\mathrm{o}} / \mathrm{P}_{\mathrm{i}}=\left(\epsilon_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}\right) /\left(\epsilon_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}\right)$
$\mathrm{I}_{\mathrm{s}}=\eta \epsilon_{\mathrm{p}} \mathrm{I}_{\mathrm{p}} / \epsilon_{\mathrm{s}}=(0.9 \times 2300 \times 5) / 230=45 \mathrm{~A}$
|
Alternating Current
|
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?
|
(a) $\mathrm{RC}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{C}=10 \mu \mathrm{F}$
(b) $\mathrm{RL}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{L}=10 \mathrm{mH}$
(c) $\mathrm{RC}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{C}=1 \mu \mathrm{F}$
(d) $\mathrm{RL}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{L}=1 \mathrm{mH}$
|
(a) $RC$ circuit with $R=1 \mathrm{k} \Omega$ and $C=10 \mu \mathrm{F}$
|
$\mathrm{X}_{\mathrm{c}}=\mathrm{R}$
$1 / \omega \mathrm{C}=\mathrm{R}$
$\omega=100 \mathrm{rad} / \mathrm{s}$
Therefore, $1 / 100=\mathrm{RC}$
Substituting values from option (a)
$\mathrm{R}=1 \mathrm{k} \Omega=10^{3} \Omega$
$\mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$
$\mathrm{RC}=10^{3} \times 10 \times 10^{-6}=1 / 100$
Therefore, option (a) is correct
|
Alternating Current
|
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
|
(a) 50,10
(b) $1000 / \sqrt{2}, 10$
(c) $50 / \sqrt{2},0$
(d) 50,0
|
(b) 1000/ $\sqrt{2}, 10$
|
$\mathrm{P}_{\mathrm{avg}}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \theta$
$\mathrm{P}_{\text {avg }}=\left(\mathrm{V}_{0} / \sqrt{2}\right)\left(\mathrm{I}_{0} / \sqrt{2}\right) \cos \theta$
$=(100 / \sqrt{2})(20 / \sqrt{2}) \cos 45^{\circ}$
$P_{\text {avg }}=1000 / \sqrt{2}$ watt
Wattless current $=\mathrm{I}_{\mathrm{rms}} \sin \theta$
Wattless current $=\left(\mathrm{I}_{0} / \sqrt{2}\right) \sin \theta$
$=(20 / \sqrt{2}) \sin 45^{0}$
$=10 \mathrm{amp}$
|
Alternating Current
|
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
|
(a) $-\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / 5 \mathrm{Rnt}$
(b) $-\mathrm{n}\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / 5 \mathrm{Rt}$
(c) $-\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / \mathrm{Rnt}$
(d) $-\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / 5 \mathrm{Rnt}$
|
(b) $-\mathbf{n}\left(\mathbf{W}_{2}-\mathbf{W}_{1}\right) / 5 R t$
|
The emf induced in the coil is $\mathrm{e}=-\mathrm{n}(\mathrm{d} \Phi / \mathrm{dt})$
Induced current, $\mathrm{I}=\mathrm{e} / \mathrm{R}^{\prime}=-\left(\mathrm{n} / \mathrm{R}^{\prime}\right)(\mathrm{d} \Phi / \mathrm{dt})$
Given, $\mathrm{R}^{\prime}=\mathrm{R}+4 \mathrm{R}=5 \mathrm{R}$
$\mathrm{d} \Phi=\mathrm{W}_{2}-\mathrm{W}_{1}$
$\mathrm{dt}=\mathrm{t}$
(here, $\mathrm{W}_{1}$ and $\mathrm{W}_{2}$ are flux associated with one turn)
Substituting the given values in equa(1) we get
$\mathrm{I}=(-\mathrm{n} / 5 \mathrm{R})(\mathrm{W}_{2}-\mathrm{W}_{1} / \mathrm{t})$
|
Alternating Current
|
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
|
(a) $5 \mathrm{~ms}$
(b) $2.2 \mathrm{~ms}$
(c) $7.2 \mathrm{~ms}$
(d) $3.3 \mathrm{~ms}$
|
(d) $3.3 \mathrm{~ms}$
|
As $v(t)=220 \sin 100 \pi t$
So $\mathrm{I}(\mathrm{t})=(220 / 50) \sin 100 \pi \mathrm{t}$
I.e., $\mathrm{I}=\mathrm{I}_{\mathrm{m}}=\sin 100 \pi \mathrm{t}$
For $\mathrm{I}=\mathrm{I}_{\mathrm{m}}$
$\mathrm{t}_{1}=(\pi / 2) \times(1 / 100 \pi)=(1 / 200) \mathrm{sec}$,
And for $\mathrm{I}=\mathrm{I}_{\mathrm{m}} / 2$
$\Rightarrow(\mathrm{I}_{\mathrm{m}} / 2)=\mathrm{I}_{\mathrm{m}} \sin \left(100 \pi \mathrm{t}_{2}\right)$
$\Rightarrow(\pi / 6)=100 \pi \mathrm{t}_{2}$
$\Rightarrow \mathrm{t}_{2}=(1 / 600) \mathrm{s}$
$\therefore \mathrm{t}_{\text {req }}=(1 / 200)-(1 / 600)=(2 / 600)=(1 / 300) \mathrm{s}=3.3 \mathrm{~ms}$
|
Alternating Current
|
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
|
(a) $0.065 \mathrm{H}$
(b) $80 \mathrm{H}$
(c) $0.08 \mathrm{H}$
(d) $0.044 \mathrm{H}$
|
(a) $0.065 \mathrm{H}$
|
$\mathrm{I}=10 \mathrm{~A}$
$\mathrm{V}=80 \mathrm{~V}$
$\mathrm{R}=8 \Omega$
$10=220 /\left(8^{2}+\mathrm{X}_{\mathrm{L}}{ }^{2}\right)^{1 / 2}$
$64+\mathrm{X}_{\mathrm{L}}{ }^{2}=484$
$X_{L}=\sqrt{ } 420$
$2 \pi \times 50 \mathrm{~L}=\sqrt{ } 420$
$\mathrm{L}=\sqrt{ } 420 / 100 \pi$
$\mathrm{L}=0.065 \mathrm{H}$
|
Alternating Current
|
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
|
(a) $242 \mathrm{~W}$
(b) $305 \mathrm{~W}$
(c) $210 \mathrm{~W}$
(d) Zero
|
(a) $242 \mathrm{~W}$
|
Tan $\Phi=\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{c}}\right) / \mathrm{R}$
$\operatorname{Tan} 30^{\circ}=\mathrm{X}_{\mathrm{c}} / \mathrm{R}=\mathrm{X}_{\mathrm{c}}=\mathrm{R} / \sqrt{3}$
$\operatorname{Tan} 30^{\circ}=\mathrm{X}_{\mathrm{L}} / \mathrm{R}=\mathrm{X}_{\mathrm{L}}=\mathrm{R} / \sqrt{3}$
$\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{c}} \Rightarrow$ Condition for resonance
So $\Phi=0^{0}$\n$\mathrm{P}=\mathrm{VI} \cos 0^{0}$
$\mathrm{P}=\mathrm{V}^{2} / \mathrm{R}=(220)^{2} / 200=242 \mathrm{~W}$
|
Alternating Current
|
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
|
(a) $8.85 \times 10^{-10} \mathrm{C}$
(b) $7.85 \times 10^{-10} \mathrm{C}$
(c) $9.85 \times 10^{-10} \mathrm{C}$
(d) $6.85 \times 10^{-10} \mathrm{C}$
|
(a) $8.85 \times 10^{-10} \mathrm{C}$
|
The electric field between two plates is
$\mathrm{E}=\sigma / \varepsilon_{0}=\mathrm{q} / \mathrm{A} \varepsilon_{0} \Rightarrow \mathrm{q}=\mathrm{EA} \varepsilon_{0}$
$\mathrm{q}=(100)(1)\left(8.85 \times 10^{-12}\right)=8.85 \times 10^{-10} \mathrm{C}$
|
Capacitor
|
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
|
(a) $\mathrm{V}$
(b) $(\mathrm{n}+1) \mathrm{V} /(\mathrm{K}+\mathrm{n})$
(c) $(\mathrm{nV}) /(\mathrm{K}+\mathrm{n})$
(d) $\mathrm{V} /(\mathrm{K}+\mathrm{n})$
|
(b) $(\mathrm{n}+1) \mathrm{V} /(\mathrm{K}+\mathrm{n})$
|
For parallel combination, $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}+\mathrm{nC}=\mathrm{C}(\mathrm{n}+1)$
Charge on capacitor, $\mathrm{q}=\mathrm{C}_{\mathrm{eq}} \mathrm{V}=\mathrm{CV}(\mathrm{n}+1)$
Now, after removing the battery, dielectric material is placed.
Then, $\mathrm{C}_{\mathrm{eq}}=\mathrm{KC}+\mathrm{nC}=\mathrm{C}(\mathrm{K}+\mathrm{n})$
New potential difference, $\mathrm{V}^{\prime}=\mathrm{q} / \mathrm{C}_{\mathrm{eq}}=\mathrm{CV}(\mathrm{n}+1) / \mathrm{C}(\mathrm{K}+\mathrm{n})=(\mathrm{n}+1) \mathrm{V} /(\mathrm{K}+\mathrm{n})$
|
Capacitor
|
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
|
(a) $\mathrm{C}$
(b) $\mathrm{nC}$
(c) $(n-1) C$
(d) $(\mathrm{n}+1) \mathrm{C}$
|
(c) $(n-1) C$
|
$\mathrm{n}$ plates connected alternately give rise to $(\mathrm{n}-1$ ) capacitors connected in parallel
Resultant capacitance $=(n-1) C$.
|
Capacitor
|
Q4: Capacitance (in $F$ ) of a spherical conductor with radius $1 \mathrm{~m}$ is
|
(a) $1.1 \times 10^{-10}$
(b) $10^{-6}$
(c) $9 \times 10^{-9}$
(d) $10^{-3}$
|
(a) $1.1 \times 10^{-10}$
|
$\mathrm{C}=4 \pi \varepsilon_{0} \mathrm{R}=1 / 9 \times 10^{9}=1.1 \times 10^{-10} \mathrm{~F}$
|
Capacitor
|
Q5: If there are $\mathbf{n}$ capacitors in parallel connected to $\mathbf{V}$ volt source, then the energy stored is equal to
|
(a) $\mathrm{CV}$
(b) $(1 / 2) \mathrm{nCV}^{2}$
(c) $\mathrm{CV}^{2}$
(d) $(1 / 2 \mathrm{n}) \mathrm{CV}^{2}$
|
(b) $(1 / 2) n C V^{2}$
|
Total capacity $=\mathrm{nC}$
$\therefore$ Energy $=1 / 2 \mathrm{nCV}{ }^{2}$
|
Capacitor
|
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
|
(a) $20.25 \mathrm{pF}$
(b) $1.8 \mathrm{pF}$
(c) $45 \mathrm{pF}$
(d) $40.5 \mathrm{pF}$
|
(d) $40.5 \mathrm{pF}$
|
$$\mathrm{K}_{1}=3 \quad \mathrm{~K}_{2}=3$$
$$\mathrm{C}=\varepsilon_{0} \mathrm{~A} / \mathrm{d}=9 \times 10^{-12} \mathrm{~F}$$
With dielectric, $$\mathrm{C}=\varepsilon_{0} \mathrm{Ak} / \mathrm{d}$$
$$\mathrm{C}_{1}=\varepsilon_{0} \mathrm{~A} 3 /(\mathrm{d} / 3)=9 \mathrm{C}$$
$$\mathrm{C}_{2}=\varepsilon_{0} \mathrm{~A} 6 /(2 \mathrm{~d} / 3)=9 \mathrm{C}$$
$$\mathrm{C}_{\text {total }}=\mathrm{C}_{1} \mathrm{C}_{2} /(\mathrm{C}_{1}+\mathrm{C}_{2})$$ as they are in series
$$\mathrm{C}_{\text {total }}=(9 \mathrm{C} \times 9 \mathrm{C}) / 18 \mathrm{C}=(9 / 2) \mathrm{C}=(9 / 2)(9 \times 10^{-12} \mathrm{~F})$$
$$\mathrm{C}_{\text {total }}=40.5 \mathrm{pF}$$
|
Capacitor
|
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
|
(a) zero
(b) $1 / 2(\mathrm{~K}-1) \mathrm{CV}^{2}$
(c) $\mathrm{CV}^{2}(\mathrm{~K}-1) / \mathrm{K}$
(d) $(\mathrm{K}-1) \mathrm{CV}^{2}$
|
(a) zero
|
The potential energy of a charged capacitor
$$\mathrm{U}_{\mathrm{i}}=\mathrm{q}^{2} / 2 \mathrm{C}$$\where $$\mathrm{U}_{\mathrm{i}}$$ is the initial potential energy.
If a dielectric slab is slowly introduced, the energy $$=\mathrm{q}^{2} / 2 \mathrm{KC}$$
Once 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.
Total work done $$=$$ zero
Answer: (a) zero
|
Capacitor
|
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
|
(a) $1 / 2$
(b) 1
(c) 2
(d) $1 / 4$
|
(a) $1 / 2$
|
Let $$\mathrm{E}$$ be emf of the battery
Work done by the battery $$\mathrm{W}=\mathrm{CE}^{2}$$
Energy stored in the capacitor
$$\mathrm{U}=1 / 2 \mathrm{CE}^{2}$$
$$\mathrm{U} / \mathrm{W}=1 / 2 \mathrm{CE}^{2} / \mathrm{CE}^{2}=1 / 2$$
Answer: (a) $1 / 2$
|
Capacitor
|
Q9: A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor
|
(a) decreases
(b) remains unchanged
(c) becomes infinite
(d) increases
|
(b) remains unchanged
|
With air as dielectric, $$\mathrm{C}=\varepsilon_{0} \mathrm{~A} / \mathrm{d}$$
With space partially filled, $$\mathrm{C}^{\prime}=\varepsilon_{0} \mathrm{~A} /(\mathrm{d}-\mathrm{t})=\varepsilon_{0} \mathrm{~A} / \mathrm{d}=\mathrm{C}$$
Aluminium is a good conductor. Its sheet introduced between the plates of a capacitor is of negligible thickness. Therefore, capacity remains unchanged.
Answer: (b) remains unchanged
|
Capacitor
|
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
|
(a) $(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
(b) $(\Phi_{1}+\Phi_{2}) / \varepsilon_{0}$
(c) $(\Phi_{2}-\Phi_{1}) / \varepsilon_{0}$
(d) $(\Phi_{1}+\Phi_{2}) \varepsilon_{0}$
|
(a) $(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
|
According to Gauss theorem,
$(\Phi_{2}-\Phi_{1})=\mathrm{Q} / \varepsilon_{0}$
$\Rightarrow \mathrm{Q}=(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
The 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 $\Phi_{1}>\Phi_{2}, \Phi_{2}>\Phi_{1}, \mathrm{Q}=\mathrm{q} 1 \sim \mathrm{q} 2$
Answer: (a) $(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
|
Capacitor
|
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 $\varepsilon_{0}=8.86 \times 10^{-12} C^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )
|
(a) 3.8
(b) 8.5
(c) 6.2
(d) 4.5
|
(b) 8.5
|
$\mathrm{C}=\mathrm{K} \varepsilon_{0} \mathrm{~A} / \mathrm{d}$ or $\mathrm{K}=\mathrm{CV} / \varepsilon_{0} \mathrm{AE}_{\max }$
$\mathrm{K}=(15 \times 10^{-12} \times 500) /(8.86 \times 10^{-12} \times 10^{-4} \times 10^{6})=8.5$
Answer: (b) 8.5
|
Capacitor
|
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
(take $1 / 4 \pi \varepsilon_{0}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$ )
(a) $3.0 \times 10^{4} \mathrm{~m} / \mathrm{s}$
(b) $1.0 \mathrm{~m} / \mathrm{s}$
(c) $1.5 \times 10^{2} \mathrm{~m} / \mathrm{s}$
(d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
\section*{Solution}
Using work energy theorem
$\mathrm{W}_{\mathrm{E}}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=(1 / 2) \mathrm{mv}^{2}$ or $\mathrm{kq}_{1} \mathrm{q}_{2}[\left(1 / \mathrm{r}_{1}\right)-\left(1 / \mathrm{r}_{2}\right)]=(1 / 2) \mathrm{mv}^{2}$
$$
\begin{aligned}
& \mathrm{W}_{\mathrm{E}}=\left(9 \times 10^{9}\right) \times\left(1 \times 10^{-6}\right)^{2}\left\{\left(1 / 10^{-3}\right)-\left(1 /\left(9 \times 10^{-3}\right)\right\}=(1 / 2) \mathrm{mv}^{2}\right. \\
& \Rightarrow\left(9 \times 10^{9} \times 2 \times 10^{-12}\right) /\left(4 \times 10^{-6}\right)=\mathrm{v}^{2} \Rightarrow \mathrm{v}=2.0 \times 10^{3} \mathrm{~ms}^{-1}
\end{aligned}
$$
Answer: (d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
|
(d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
|
Using work energy theorem
$\mathrm{W}_{\mathrm{E}}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=(1 / 2) \mathrm{mv}^{2}$ or $\mathrm{kq}_{1} \mathrm{q}_{2}[\left(1 / \mathrm{r}_{1}\right)-\left(1 / \mathrm{r}_{2}\right)]=(1 / 2) \mathrm{mv}^{2}$
$$
\begin{aligned}
& \mathrm{W}_{\mathrm{E}}=\left(9 \times 10^{9}\right) \times\left(1 \times 10^{-6}\right)^{2}\left\{\left(1 / 10^{-3}\right)-\left(1 /\left(9 \times 10^{-3}\right)\right\}=(1 / 2) \mathrm{mv}^{2}\right. \\
& \Rightarrow\left(9 \times 10^{9} \times 2 \times 10^{-12}\right) /\left(4 \times 10^{-6}\right)=\mathrm{v}^{2} \Rightarrow \mathrm{v}=2.0 \times 10^{3} \mathrm{~ms}^{-1}
\end{aligned}
$$
Answer: (d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
|
Capacitor
|
|
Q1: The wavelength of the carrier waves in a modern optical fibre communication network is close to
|
(a) 600 nm
(b) 2400 nm
(c) 1500 nm
(d) 900 nm
|
Answer: (c) 1500 nm
|
Fibre optics communication is mainly conducted in a wavelength range from 1260 nm to 1625 nm
|
Communication System
|
Q2: The physical sizes of the transmitter and receiver antenna in a communication system are
|
(a) inversely proportional to the modulation frequency
(b) proportional to the carrier frequency
(c) independent of both carrier and modulation frequency
(d) inversely proportional to the carrier frequency
|
Answer: (d) inversely proportional to the carrier frequency
|
The physical size of the transmitter and receiver antenna is inversely proportional to the carrier frequency.
|
Communication System
|
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 ?
|
(a) 2 × 10^3
(b) 2 × 10^4
(c) 2 × 10^5
(d) 2 × 10^6
|
Answer: (c) 2 × 10^5
|
Frequency of carrier wave = 10 × 10^9 Hz
Available bandwidth 10% of 10 × 10^9 Hz = 10^9 Hz
Bandwidth for each telephonic channel 5 kHz = 5 × 10^3 Hz
Number of channels = 10^9 / (5 × 10^3) = 2 × 10^5
|
Communication System
|
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 × 10^6 m )
|
(a) 65 km
(b) 48 km
(c) 40 km
(d) 80 km
|
Answer (a) 65 km
|
Maximum distance upto which signal can be broadcasted is
d_max = √(2 R h_T) + √(2 R h_R)
where h_T and h_R are heights of transmitter tower and height of receiver respectively.
Putting all values, we get
d_max = √(2 × 6.4 × 10^6) [√140 + √40]
d_max = 65 km
|
Communication System
|
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
|
(a) 2005 kHz, 2000 kHz, and 1995 kHz
(b) 2000 kHz and 1995 kHz
(c) 2 MHz only
(d) 2005 kHz and 1995 kHz
|
Answer: (a) 2005 kHz, 2000 kHz, and 1995 kHz
|
Given, f_m = 5 kHz, f_c = 2 MHz = 2000 kHz
The 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
|
Communication System
|
Q6: Consider telecommunication through optical fibres. Which of the following statements is not true?
|
(a) Optical fibres can be of graded refractive index
(b) Optical fibres are subject to electromagnetic interference from outside
(c) Optical fibres have extremely low transmission loss
(d) Optical fibres may have a homogeneous core with suitable cladding
|
Answer: (b) Optical fibres are not subject to electromagnetic interference from outside
|
Communication System
|
|
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?
|
(a) 2750 kHz
(b) 2900 kHz
(c) 2250 kHz
(d) 2000 kHz
|
Answer: (d) 2000 kHz
|
10% of f_c = 250 kHz
Hence, range of signal = (2500 ± 250 kHz) = 2250 kHz to 2750 kHz
10% of 2000 kHz = 200 kHz
Range is 1800 kHz to 2200 kHz
Hence, allocated broadcast frequency will be 2000 kHz
|
Communication System
|
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=3×10^8 m/s, h=6.6×10^(-34) Js)
|
(a) 3.75 × 10^6
(b) 4.87 × 10^5
(c) 6.25 × 10^5
(d) 3.86 × 10^6
|
Answer: (c) 6.25 × 10^5
|
f = C / λ = (3 × 10^8) / (8 × 10^-7) = 3.75 × 10^14 Hz
1% of f = 3.75 × 10^17 Hz = 3.75 × 10^6 MHz
Number of channels = (3.75 × 10^6) / 6
Number of channels = 6.25 × 10^5
|
Communication System
|
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.
|
(a) Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz
(b) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz
(c) Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz
(d) Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz
|
Answer: (b) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz
|
Modulation index, m = (V_m / V_c) = (5 / 25) = 0.2
Frequency of carrier wave, f_c = 1.2 × 10^3 kHz = 1200 kHz
Frequency of modulate wave = 20 kHz
f1 = f_c - f_m = 1200 - 20 = 1180 kHz
f2 = f_c + f_m = 1200 + 20 = 1220 kHz
|
Communication System
|
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
|
(a) 2
(b) 5
(c) 6
(d) 3
|
Answer: (c) 6
|
Band width for both signals = 15200 Hz - 200 Hz = 15000 Hz
Band width for human speed 2700 Hz - 200 Hz = 2500 Hz
The ratio = 15000 / 2500 = 6
|
Communication System
|
Q11: In an amplitude modulator circuit, the carrier wave is given by, C(t)=4 sin(20000πt), while modulating signal is given by, m(t)=2 sin(2000πt). The values of modulation index and lower sideband frequency are
|
(a) 0.4 and 10 kHz
(b) 0.5 and 9 kHz
(c) 0.3 and 9 kHz
(d) 0.5 and 10 kHz
|
Answer: (b) 0.5 and 9 kHz
|
Given C(t) = 4 sin(20000πt) ⇒ A_c = 4
m(t) = 2
|
Communication System
|
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 $lpha$ is its coefficient of linear expansion. Then, $F$ is equal to
|
a. $\mathrm{L}^{2} \mathrm{Y} lpha \Delta \mathrm{T}$
b. $\mathrm{AY} / lpha \Delta \mathrm{T}$
c. AY $lpha \Delta \mathrm{T}$
d. LAY $lpha \Delta \mathrm{T}$
|
c
|
Thermal expansion, $\Delta \mathrm{L}=\mathrm{L} lpha \Delta \mathrm{T}$\ 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)\ $\mathrm{F}=\mathrm{YAx}(\mathrm{L} lpha \Delta \mathrm{T}) / \mathrm{L}=\mathrm{YA} lpha \Delta \mathrm{T}$
|
Elasticity
|
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
|
a. $0.2 \mathrm{~J}$
b. $10 \mathrm{~J}$
c. $20 \mathrm{~J}$
d. $0.1 \mathrm{~J}$
|
d
|
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\ $=1 / 2 imes F / A imes(\Delta L / L) imes(A L)$\ $=1 / 2 imes$ F $\Delta \mathrm{L}$\ $=1 / 2 imes 200 imes 10^{-3}$\ Elastic Energy $=0.1 \mathrm{~J}$
|
Elasticity
|
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 $lpha$. 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
|
a.F/A $lpha \Delta \mathrm{T}$
b.F/A $lpha(\Delta \mathrm{T}-273)$
c. F/2A $lpha \Delta$
d. $2 \mathrm{~F} / \mathrm{A} lpha \Delta \mathrm{T}$
|
a
|
Young's Modulus $\mathrm{Y}=$ stress/strain $=(\mathrm{F} / \mathrm{A}) /(\Delta l / l)$\ Substituting the coefficient of linear expansion\ $lpha=\Delta l /(l \Delta \mathrm{T})$\ $\Delta l / l=lpha \Delta \mathrm{T}$\ $\mathrm{Y}=(\mathrm{F} / \mathrm{A} lpha \Delta \mathrm{T})$
|
Elasticity
|
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 $t$
|
a. $1.5 \mathrm{~mm}$
b. $1.9 \mathrm{~mm}$
c. $1.7 \mathrm{~mm}$
d. $1.3 \mathrm{~mm}$
|
c
|
$\Delta_{1}=\Delta_{2}$\ $\left(\mathrm{Fl}{1} / \pi \mathrm{r}{1}^{2} \mathrm{y}{1}
ight)=\left(\mathrm{Fl}{2} / \pi \mathrm{r}{2}^{2} \mathrm{y}{2}
ight)$\ $2 /(\mathrm{R}^{2} imes 7)=1.5 /(2^{2} imes 4)$\ $\mathrm{R}=1.75 \mathrm{~mm}$
|
Elasticity
|
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?
|
a. $1 \mathrm{~mm}$
b. $1.15 \mathrm{~mm}$
c. $0.90 \mathrm{~mm}$
d. $1.36 \mathrm{~mm}$
|
b
|
Stress $=\mathrm{F} / \mathrm{A}$\ Stress $=400 imes 4 / \pi \mathrm{d}^{2}$\ $=379 imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$\ $\mathrm{d}^{2}=(400 imes 4) /(379 imes 10^{6} \pi)$\ $\mathrm{d}=1.15 \mathrm{~mm}$
|
Elasticity
|
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
|
a. $9 \mathrm{~F} /(\pi \mathrm{r}^{2} \mathrm{YT})$
b. $6 \mathrm{~F} /(\pi \mathrm{r}^{2} \mathrm{YT})$
c. $3 \mathrm{~F} /(\pi \mathrm{r}^{2} \mathrm{YT})$
d. $\mathrm{F} /(3 \pi \mathrm{r}^{2} \mathrm{YT})$
|
c
|
$\mathrm{Y}=\left(\mathrm{F} / \pi \mathrm{r}^{2}
ight) imes \mathrm{L} / \Delta \mathrm{L}$\ $\Delta \mathrm{L}=\mathrm{F} / / \pi \mathrm{r}^{2} \mathrm{Y}-------(1)$\ Change in length due to temperature change\ $\Delta \mathrm{L}=\mathrm{L} lpha \Delta \mathrm{T}$\ From equa (1) and (2)\ $\mathrm{L} lpha \Delta \mathrm{T}=\mathrm{FL} / \mathrm{AY}$\ $lpha=\mathrm{F} / \mathrm{AY} \Delta \mathrm{T}$\ $lpha=\mathrm{F} / \pi \mathrm{r}^{2} \mathrm{YT}$\ Coefficient of volume expansion\ $3 \propto=3 \mathrm{~F} / \pi \mathrm{r}^{2} \mathrm{YT}$
|
Elasticity
|
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?
|
(a) length $=200 \mathrm{~cm}$, diameter $=2 \mathrm{~mm}$
(b) length $=300 \mathrm{~cm}$, diameter $=3 \mathrm{~mm}$
(c) length $=50 \mathrm{~cm}$, diameter $=0.5 \mathrm{~mm}$
(d) length $=100 \mathrm{~cm}$, diameter $=1 \mathrm{~mm}$
|
c
|
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) $\mathrm{L} / \mathrm{D}^{2}=200 /(0.2)^{2}=5 imes 10^{3} \mathrm{~cm}^{-1}$\ In (b) $\mathrm{L} / \mathrm{D}^{2}=300 /(0.3)^{2}=3.3 imes 10^{3} \mathrm{~cm}^{-1}$\ In (c) $\mathrm{L} / \mathrm{D}^{2}=50 /(0.5)^{2}=20 imes 10^{3} \mathrm{~cm}^{-1}$\ In (d) $\mathrm{L} / \mathrm{D}^{2}=100 /(0.1)^{2}=10 imes 10^{3} \mathrm{~cm}^{-1}$
|
Elasticity
|
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
|
(a) $1 / 9$
(b) 81
(c) $1 / 81$
(d) 9
|
d
|
Stress $=$ Force $/$ Area\ Stress $=$ Force $/ L^{2}$\ Now, dimensions increases by a factor of 9\ Now, $\mathrm{S}=$ (volume $\mathrm{x}$ density) $\mathrm{xg} / \mathrm{L}^{2}$\ $\mathrm{S}=\mathrm{L}^{3} imes
ho \mathrm{g} / \mathrm{L}^{2}=\mathrm{L}
ho \mathrm{g}$\ Stress S $\propto \mathrm{L}$\ $\mathrm{S}{2} / \mathrm{S}{1}=\mathrm{L}{2} \mathrm{~L}{1}=9 \mathrm{~L}{1} / \mathrm{L}{1}=9$
|
Elasticity
|
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 $\mathrm{Mg} / lpha \mathrm{AB}$. Find the value of $lpha$.
|
a. 4
b. 5
c. 3
d. 2
|
c
|
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 $lpha=3$
|
Elasticity
|
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?
|
a. $4.8 imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
b. $3.1 imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
c. $6.2 imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
d. $5.2 imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
|
b
|
Tensile stress $=$ Force/Area\ Tensile stress $=(4)(3.1 \pi) / \pi\left(2 imes 10^{-3}
ight)^{2}$\ Tensile stress $=3.1 imes 10^{6} \mathrm{Nm}^{-2}$
|
Elasticity
|
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 $1.2 imes 10^{-5} \mathrm{~K}^{-1}$ and $2 imes 10^{11} \mathrm{Nm}^{-2}$ respectively, the force developed in the rail is approximately
|
a. $2 imes 10^{9} \mathrm{~N}$
b. $3 imes 10^{-5} \mathrm{~N}$
c. $2 imes 10^{7} \mathrm{~N}$
d. $1 imes 10^{5} \mathrm{~N}$
|
d
|
$\mathrm{A}=40 \mathrm{~cm}^{2}=4 imes 10^{-3} \mathrm{~m}^{2}$\ $\Delta \mathrm{T}=10^{\circ} \mathrm{C}$\ $\mathrm{Y}=2 imes 10^{11} \mathrm{Nm}^{-2}$\ $lpha=1.2 imes 10^{-5} \mathrm{~K}^{-1}$\ Force $=\mathrm{YA} lpha \Delta \mathrm{T}$\ Force $=\left(2 imes 10^{11}
ight)\left(4 imes 10^{-3}
ight)\left(1.2 imes 10^{-5}
ight)(10)=9.6 imes 10^{4} \mathrm{~N}$\ Force $pprox 1 imes 10^{5} \mathrm{~N}$
|
Elasticity
|
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
|
a. $2 \mathrm{Y} / \mathrm{S}$
b. $\mathrm{S} / 2 \mathrm{Y}$
c. $2 S^{2} Y$
d. $\mathrm{S}^{2} / 2 \mathrm{Y}$
|
d
|
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)
|
Elasticity
|
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?
|
a. F
b. $4 F$
c. $6 F$
d. $9 F$
|
d
|
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 $3 A$\ $V=V_{1}=V_{2}$\ $V=A imes L_{1}=3 A imes L_{2} \Rightarrow L_{2}=L_{1} / 3$\ $Y=(F / A) /(\Delta L / L)$\ $F_{1}=YA(\Delta L_{1} / L_{1})$\ $F_{2}=Y 3 A(\Delta L_{2}, L_{2})$\ Given $\Delta L_{1}=\Delta L_{2}=\Delta x$ (for the same extension)\ $F_{2}=Y 3 A(\Delta x /(L_{1} / 3))=9 (YA \Delta x / L_{1})=9 F_{1}=9 F$
|
Elasticity
|
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 $\mathrm{mm}$ )
|
a. $l / 2$
b. l
c. 2
d. Zero
|
b
|
$Y=( ext{Force} imes L) /(A imes l)=WL / Al$\ $l=WL / AY$\ 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) $l$
|
Elasticity
|
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
|
a. $10^{-2} \mathrm{~T}$
b. $1 \mathrm{~T}$
c. $10^{-4} \mathrm{~T}$
d. $10^{2} \mathrm{~T}$
|
c
|
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) $10^{-4} \mathrm{~T}$
|
Electromagnetic Waves
|
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
|
a. $6 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{x}$-direction
b. $2 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{y}$-direction
c. $2 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{z}$-direction
d. $6 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{z}$-direction
|
c
|
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
|
Electromagnetic Waves
|
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 $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right.$ )
|
a. $20 \times 10^{-8} \mathrm{~N}$
b. $10 \times 10^{-8} \mathrm{~N}$
c. $35 \times 10^{-8} \mathrm{~N}$
d. $15 \times 10^{-8} \mathrm{~N}$
|
a
|
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) $20 \times 10^{-8} \mathrm{~N}$
|
Electromagnetic Waves
|
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
|
a. $5.48 \mathrm{~V} / \mathrm{m}$
b. $7.75 \mathrm{~V} / \mathrm{m}$
c. $1.73 \mathrm{~V} / \mathrm{m}$
d. $2.45 \mathrm{~V} / \mathrm{m}$
|
d
|
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) $2.45 \mathrm{~V} / \mathrm{m}$
|
Electromagnetic Waves
|
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
|
a. (i) (ii) (iii) (iv)
b. (iv) (iii) (ii) (i)
c. (i) (ii) (iv) (iii)
d. (iii) (ii) (i) (iv)
|
a
|
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)
|
Electromagnetic Waves
|
Q6: During the propagation of electromagnetic waves in a medium
|
a. both electric and magnetic energy densities are zero
b. electric energy density is double of the magnetic energy density
c. electric energy density is half of the magnetic energy density
d. electric energy density is equal to the magnetic energy density
|
d
|
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
|
Electromagnetic Waves
|
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
|
a. $12 \mathrm{~V} / \mathrm{m}$
b. $3 \mathrm{~V} / \mathrm{m}$
c. $6 \mathrm{~V} / \mathrm{m}$
d. $9 \mathrm{~V} / \mathrm{m}$
|
c
|
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
|
Electromagnetic Waves
|
Q8: An electromagnetic wave of frequency $=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with permittivity $=4.0$. Then
|
a. wavelength is doubled and the frequency remains unchanged
b. wavelength is doubled and frequency becomes half
c. wavelength is halved and frequency remains unchanged
d. wavelength and frequency both remain unchanged
|
c
|
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
|
Electromagnetic Waves
|
Q9: Electromagnetic waves are transverse in nature is evident by
|
a. polarization
b. interference
c. reflection
d. diffraction
|
a
|
Answer: (a) Polarization proves the transverse nature of electromagnetic waves.
|
Electromagnetic Waves
|
Q10: The energy associated with electric field is $\left(U_{E}
ight)$ and with magnetic field is $\left(U_{B}
ight)$ for an electromagnetic wave in free space. Then
|
a. $U_{E}>U_{B}$
b. $U_{E}=U_{B} / 2$
c. $U_{E}=U_{B}$
d. $U_{E}<U_{B}$
|
c
|
Answer: (c) $\mathbf{U}_{\mathrm{E}}=\mathbf{U}_{\mathrm{B}}$
|
Electromagnetic Waves
|
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 $\mathbf{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
|
a. $2 \mathrm{kV} / \mathrm{m}$
b. $0.7 \mathrm{kV} / \mathrm{m}$
c. $1 \mathrm{kV} / \mathrm{m}$
d. $1.4 \mathrm{kV} / \mathrm{m}$
|
d
|
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) $1.4 \mathrm{kV} / \mathrm{m}$}
|
Electromagnetic Waves
|
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
|
a) $3.3 \times 10^{-3} \mathrm{~J} / \mathrm{m}^{3}$
b) $4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}$
c) $6.37 \times 10^{-9} \mathrm{~J} / \mathrm{m}^{3}$
d) $81.35 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}$
|
b
|
$\mathrm{U}=\left(8.85 \times 10^{-12}\right) \times(720)^{2}=4.58 \times 10^{-6} \mathrm{Jm}^{-1}$\ \section*{Answer (b) $4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}$}
|
Electromagnetic Waves
|
Q13: Which of the following are not electromagnetic waves?
|
a) cosmic rays
b) gamma rays
c) $\beta$-rays
d) X-rays
|
c
|
Answer: (c) $\beta$-rays are not electromagnetic waves
|
Electromagnetic Waves
|
Q14: If a source of power $4 \mathrm{~kW}$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called
|
a) microwaves
b) y-rays
c) X-rays
d) ultraviolet rays
|
c
|
Solution\ \section*{Answer: (c) X-rays}
|
Electromagnetic Waves
|
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?
|
a) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=4$
b) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=2$
c) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 4$
d) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 2$
|
c
|
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) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 4$}
|
Electromagnetic Waves
|
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.
|
due north
30° east of north
30° west of north
60° east of north
|
(1) due north
|
Let the time taken to cross the river be t
t=d/V_s cosθ
For the time t to be minimum, cosθ should be maximum
θ should be 0
Therefore swimmer should swim due north
|
Kinematics 2D
|
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.
|
t_p < t_Q
t_p > t_Q
t_p = t_Q
t_p / t_Q = (length of arc ABC / length of chord AB )
|
(1) t_P < t_Q
|
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.
|
Kinematics 2D
|
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° 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/s², is
|
a = 5 m/s²
|
R = 1 / 2 at² + 1.15
u² sin 2θ / g = 1 / 2 a (2 u sin θ / g)² + 1.15
100 × √3 / 2 / 10 = 1 / 2 a (4 × 100 × 3 / 4 / 100) + 1.15
5 √3 = 3 a / 2 + 1.15
2 × 5 √3 = 3 a + 2.30
10 √3 = 3 a + 2.30
10 √3 - 2.30 = 3 a
10 × 1.73 - 2.3 = 3 a
17.3 - 2.3 = 3 a
15 / 3 = a
a = 5 m/s²
|
Kinematics 2D
|
|
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
|
1
3
4
√41
|
(2) 3 km/h
|
Time t = 15 min = 15 / 60 = 1 / 4 hour
Velocity = distance / time
V_b cos θ = 1 / (1 / 4) = 4 km/h
5 cos θ = 4 km/h
cos θ = 4 / 5
sin θ = 3 / 5
Consider the triangle of velocity ABC,
V_r / V_b = 3 / 5
V_r / 5 = 3 / 5
Or V_r = 3 km/h
|
Kinematics 2D
|
Q8: A ball is projected from the ground at an angle of 45° 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° with the horizontal surface. The maximum height it reaches after the bounce, in metres, is
|
H₂ = 30 m
|
H₁ = u² sin² 45 / 2g = 120
u² / 2g = 120
When half of the kinetic energy is lost v = u / √2
H₂ = (u / √2)² sin² 30 / 2g = u² / 16g
From (1) and (2)
H₂ = H₁ / 4 = 30 m
|
Kinematics 2D
|
|
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?
|
True
|
$v^2=u^2-2gH$
As the body falls back to the point of projection, $H=0$
$$v^2=u^2-0=u^2$$
$$\operatorname{Or} v=u$$
Hence the statement is true
|
Kinematics 2D
|
|
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?
|
True
|
The horizontal velocity of the projectile $=$ $u \cos \theta$
It remains constant throughout the motion
The vertical velocity becomes zero at the top of the path
Therefore the velocity is minimum at the top
Hence the statement is true
|
Kinematics 2D
|
|
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?
|
False
|
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.
|
Kinematics 2D
|
|
Q13: A particle moves in a circle of radius R. In half the period of revolution its displacement is and distance covered is
|
Displacement $AB=2r$
Distance $ACB=2\pi r / 2=\pi r$
|
Displacement $AB=2r$
Distance $ACB=2\pi r / 2=\pi r$
|
Kinematics 2D
|
|
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
|
Time Taken $=(d / \sqrt{2}) / (v / \sqrt{2})=d / v$
|
It is inferred on the basis of symmetry that all four persons will meet at $O$, the centre of the square.
At any instant, the component velocity along
$$KO=v \cos 45^\circ=v / \sqrt{2}$$
Distance $KO=d \cos 45^\circ=d / \sqrt{2}$
Time taken $=$ Distance $KO$ / velocity along $KO$
Time Taken $=(d / \sqrt{2}) / (v / \sqrt{2})=d / v$
|
Kinematics 2D
|
End of preview. Expand
in Data Studio
PhyscisQA dataset comprises 370 carefully selected high school physics questions sourced from online resources. These questions are notably complex, often requiring the application of multiple concepts, intricate computations, and multihop reasoning. Each question is paired with a comprehensive, step-by-step solution, to support the evaluation of LLMs for physics reasoning.PhysicsQA offers a more robust evaluation and analysis of LLM performance by encompassing a diverse range of questions, both in terms of complexity and the topics covered.
Paper: Improving Physics Reasoning in Large Language Models Using Mixture of Refinement Agents [AAAI 2026 TrustAgent]
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