Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Heat Transfer
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
Options: (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
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)
Options: (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
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 $R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{K}^{-1}$ ]
Options: (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, $\mathrm{n}=(67.2 / 22.4)=3 \mathrm{~mol}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=3 	imes(3 / 2) \mathrm{R} 	imes \Delta \mathrm{T}=3 	imes(3 / 2) 	imes(8.31) 	imes 20=747.9=748 \mathrm{~J}$
Topic: 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 $=4184 \mathrm{~J} / \mathrm{kg} / \mathrm{K}$ )
Options: (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, $\mathrm{m}=200 \mathrm{~g}=0.2 \mathrm{~kg}, \mathrm{~s}=4184 \mathrm{~J} / \mathrm{kg} / \mathrm{K}$ $\Delta \mathrm{T}=60^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}=20^{\circ} \mathrm{C}$ $\Delta \mathrm{U}=0.2 	imes 4184 	imes 20=16736 \mathrm{~J}=16.7 \mathrm{~kJ}$
Topic: 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?
Options: (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
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
Options: (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
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
Options: (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) $\left(\mathrm{K}{1}+3 \mathrm{~K}{2}ight) / 4$
Answer: (4) $\left(\mathrm{K}{1}+3 \mathrm{~K}{2}ight) / 4$
Solution: $\mathrm{K}{\mathrm{eq}}=\left(\mathrm{K}{1} \mathrm{A}{1}+\mathrm{K}{2} \mathrm{A}{2}ight) /\left(\mathrm{A}{1}+\mathrm{A}{2}ight)$ $\mathrm{K}{\mathrm{eq}}=\mathrm{K}{1} \pi \mathrm{R}^{2}+\mathrm{K}{2}\left[\pi(2 \mathrm{R})^{2}-\pi \mathrm{R}^{2}ight] / \pi(2 \mathrm{R})^{2}$ $=\mathrm{K}{1} \pi \mathrm{R}^{2}+\mathrm{K}{2}\left[3 \pi \mathrm{R}^{2}ight] / \pi 4 \mathrm{R}^{2}$ $=\left(\mathrm{K}{1}+3 \mathrm{~K}{2}ight) / 4$
Topic: Heat Transfer
Q11: Heat given to a body which raises its temperature by $1^{\circ} \mathrm{C}$ is
Options: (a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient
Answer: (b) thermal capacity
Topic: 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
Options: (1) 1 (2) $1 / 2$ (3) $2 / 3$ (4) $1 / 3$
Answer: (4) $1 / 3$
Solution: For the first surface, $\mathrm{Q}{1}=\mathrm{KA}\left(\mathrm{T}{2}-\mathrm{T}ight) \mathrm{t} / \mathrm{x}$ 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 $2\left(\mathrm{T}{2}-\mathrm{T}ight)=\left(\mathrm{T}-\mathrm{T}{1}ight)$ $\mathrm{T}=\left(2 \mathrm{T}{2}+\mathrm{T}{1}ight) / 3$ $\mathrm{Q}{1}=\mathrm{KA}\left(\mathrm{T}{2}-\left[\left(2 \mathrm{T}{2}+\mathrm{T}{1}ight) / 3ight]ight) t / \mathrm{x}$ $\mathrm{H}=\mathrm{Q}{1} / \mathrm{t}=\mathrm{KA}\left(\mathrm{T}{2}-\left[\left(2 \mathrm{T}{2}+\mathrm{T}{1}ight) / 3ight]ight) / \mathrm{x}$\ $=\mathrm{KA}\left(\mathrm{T}{2-} \mathrm{T}{1}ight) / 3 \mathrm{x}$ $\left[rac{A\left(T_{2}-T_{1}ight) K}{x}ight] f=rac{K A}{x}\left[rac{T_{2}-T_{1}}{3}ight]$ $\mathrm{f}=1 / 3$
Topic: 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
Options: (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
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
Options: (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
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
Options: (1) 30 (2) 50 (3) 70 (4) 90
Answer: (2) 50
Solution: $\mathrm{Q}=\mathrm{nC}{\mathrm{p}} \mathrm{dT}$ $\mathrm{C}{\mathrm{p}}=\mathrm{Q} / \mathrm{ndT}=70 / 2 \mathrm{x}(35-30)$ $\mathrm{C}{\mathrm{p}}=70 /(2 	imes 5)$ $\mathrm{C}{\mathrm{p}}=7 \mathrm{cal} / \mathrm{mol} 	imes \mathrm{K}$ Now, $\mathrm{C}{\mathrm{v}}=\mathrm{C}{\mathrm{p}}-\mathrm{R}$ $\mathrm{C}{\mathrm{v}}=7-2=5 \mathrm{cal} / \mathrm{mol} 	imes \mathrm{K}$ $\mathrm{Q}^{\prime}=\mathrm{nC}{\mathrm{v}} \mathrm{dT}=2 	imes 5 	imes 5=50 \mathrm{cal}$
Topic: Waves on String
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
Options: (a) $y=\operatorname{asin}(\omega t+k x)$

(b) $y=-\operatorname{asin}(\omega t+k x)$

(c) $\mathrm{y}=\operatorname{asin}(\omega \mathrm{t}-\mathrm{kx})$
Answer: (b) $y=-\operatorname{asin}(\omega t+k x)$
Solution: 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
Topic: 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
Options: (a) 300

(b) 600

(c) 1200

(d) 200
Answer: (a) 300
Solution: 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 $=\omega / \mathrm{k}=600 / 2=300 \mathrm{~m} / \mathrm{sec}$
Topic: 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
Options: (a) $10.5 \mathrm{~Hz}$

(b) $105 \mathrm{~Hz}$

(c) $1.05 \mathrm{~Hz}$

(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.

$315=$ 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

$\mathrm{f}_{0}=\mathrm{v} / 2 \mathrm{~L}=315 / 3=105 \mathrm{~Hz}$
Topic: 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}
Options: (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 $1 / \sqrt{ } b$
Answer: (b) wave moving in $-\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
Solution: 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 $\mathrm{x}=\sqrt{\frac{b}{a}}$
Topic: 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
Options: (a) $6.25 \mathrm{~N}$

(b) $4.0 \mathrm{~N}$

(c) $12.5 \mathrm{~N}$

(d) $0.5 \mathrm{~N}$
Answer: (a) $6.25 \mathrm{~N}$
Solution: Here, linear mass density $\mu=0.04 \mathrm{~kg} \mathrm{~m}^{-1}$

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

$$
\frac{\omega}{k}=\sqrt{\frac{T}{\mu}}
$$

$\mathrm{T}=\mu \omega^{2} / \mathrm{k}^{2}$

$\mathrm{T}=\left[0.04 \times(2 \pi / 0.04)^{2}\right] /(2 \pi / 0.5)^{2}=6.25 \mathrm{~N}$
Topic: 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
Options: (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
Answer: (b) \alpha=25.00 \pi, \beta=\pi
Solution: 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 $\lambda=0.08 \mathrm{~m}$
$\alpha=2 \pi / 0.08=\pi / 0.04 \Rightarrow \alpha=(\pi / 4) \times 100=25.00 \pi$

$\omega=\beta \Rightarrow 2 \pi / 2=\pi$

$\therefore \alpha=25.00 \pi$

$\beta=\pi$
Topic: 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 $\mathbf{g}=\mathbf{1 0} \mathbf{m ~ s}^{-2}$ )
Options: (a) $5 \pi \sqrt{ } 5 \mathrm{~s}$
(b) $2 \mathrm{~s}$
(c) $2 \sqrt{ } 2 \mathrm{~s}$
(d) $\sqrt{ } 5$
Answer: (c) $2 \sqrt{2} s$
Solution: 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

$\int_{0}^{L} x^{-1 / 2} d x=\sqrt{g} \int_{0}^{L} d t$

$t=2 \sqrt{L} / \sqrt{g}=2 \times \sqrt{\frac{20}{10}}=2 \sqrt{2} s$
Topic: 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?
Options: (a) $770 \mathrm{~Hz}$
(b) $188.5 \mathrm{~Hz}$
(c) $178.2 \mathrm{~Hz}$
(d) 200.5
Answer: (c) $178.2 \mathrm{~Hz}$
Solution: 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

$f=\frac{1}{2 l} \sqrt{\frac{2.2 \times 10^{11} \times 0.01}{7.7 \times 10^{3}}}$

$\mathrm{f}=178.2 \mathrm{~Hz}$
Topic: 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
Options: (a) $428 \mathrm{~Hz}$
(b) $430 \mathrm{~Hz}$
(c) $420 \mathrm{~Hz}$
(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
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)
Options: (a) $180 \mathrm{~m} / \mathrm{s}$
(b) $160 \mathrm{~m} / \mathrm{s}$
(c) $120 \mathrm{~m} / \mathrm{s}$
(d) $90 \mathrm{~m} / \mathrm{s}$
Answer: (b) $160 \mathrm{~m} / \mathrm{s}$
Solution: Given $\mathrm{y}(\mathrm{x}, \mathrm{t})=0.5 \sin [(5 \pi / 4) \mathrm{x}] \cos (200 \pi \mathrm{t})$

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, $v=\omega / \mathrm{k}=200 \pi /(5 \pi / 4)=160 \mathrm{~m} / \mathrm{s}$
Topic: 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
Options: (a) 20
(b) 80
(c) 40
(d) 120
Answer: (b) 80
Solution: $\lambda_{\max } / 2=40 \Rightarrow \lambda_{\max }=80 \mathrm{~cm}$
Topic: 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.
Options: (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
Answer: (b) Statement-1 is true, statement-2 is false
Solution: $\mathrm{y}_{1}(\mathrm{x}, \mathrm{t})=2 \mathrm{asin}(\omega \mathrm{t}-\mathrm{kx})$

$\mathrm{y}_{2}(\mathrm{x}, \mathrm{t})=2 \mathrm{asin}(2 \omega \mathrm{t}-2 \mathrm{kx})$

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, $\mathrm{I}_{1}=\mathrm{I}_{2}$
Topic: 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
Options: (a) $\mathrm{f}$
(b) $\mathrm{f} / 2$
(c) $3 \mathrm{f} / 4$
(d) $2 \mathrm{f}$
Answer: (a) f
Solution: 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,

$\mathrm{f}=\mathrm{f}^{\prime}$
Topic: 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
Options: (a) $10 \mathrm{~cm}$
(b) $33.3 \mathrm{~cm}$
(c) $16.6 \mathrm{~cm}$
(d) $20.0 \mathrm{~cm}$
Answer: (d) $20.0 \mathrm{~cm}$
Solution: 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,

$\lambda / 2=40 /(2 \mathrm{x} 100)=20 \mathrm{~cm}$
Topic: 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
Options: (a) $10 \mathrm{~N}$
(b) $7.5 \mathrm{~N}$
(c) $12.5 \mathrm{~N}$
(d) $5 \mathrm{~N}$
Answer: (c) $12.5 \mathrm{~N}$
Solution: 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}$