the phase of transmitted wave always remains unchanged

the amplitude of transmitted wave does not depend upon the velocity of wave in media

the amplitude of reflected wave and transmitted wave are same to each other for a given incident wave

the amplitude of rerflected wave is equal to the amplitude of incident wave

MCQ Questions for Class 11 Engineering Physics Waves Quiz 3

The relation between frequency

$\upsilon$ , wavelength

$\lambda$ and velocity of propagation of wave

$\nu$ is

0%
$\upsilon = \dfrac{\lambda}{\nu}$
0%
$\upsilon = \lambda \nu$
0%
$\upsilon = \dfrac{\nu}{\lambda}$
0%
None of these

Explanation

Velocity of the wave,

$\nu = \upsilon \lambda$ or

$\upsilon = \dfrac{\nu}{\lambda}$

Two identical sinusoidal waves each of amplitude

$10 mm$ with a phase difference of

$90^o$ are travelling in the same direction in string. The amplitude of the resultant wave is :

0%
$5 mm$
0%
$10 \sqrt{2} mm$
0%
$15 mm$
0%
$20 mm$

Explanation

Since the two waves are identical, they have same amplitudes with phase difference of

$90^o$ .
The amplitude of the resultant wave is,

$A=\sqrt { a^2+b^2 } =\sqrt { a^2+a^2 } =a\sqrt { 2 }$

Here,

$a=10 mm, A= 10 \sqrt{2} mm$

Which of the following waves is udsed in sonography?

0%
Radio waves
0%
X-rays
0%
Ultrasonic waves
0%
Gamma rays

Which of the following waves does not travel in vacuum?

0%
Seismic waves
0%
X-rays
0%
Light
0%
Radio waves

Ultrasonic waves produced by a vibrating quartz crystal are:

0%
only longitudinal
0%
only transverse
0%
both longitudinal and transverse
0%
neither longitudinal nor transverse

Regarding open organ pipe, which of following is correct?

0%
Both ends are pressure antinodes
0%
Both ends are displacement nodes
0%
Both ends are pressure nodes
0%
Both (1)and (2)

Explanation

Regarding open pipe both ends are pressure nodes.

So, option

$C$ is correct.

A simple harmoinic motion is represented by

$F(t) = 10 \, \sin(10\tau + 0.5)$ . The amplitude of the S.H.M. is

0%
$a = 30$
0%
$a = 20$
0%
$a = 10$
0%
$a = 5$

Explanation

We know that SHM is represented by

$F(x,t)=Asin(wt-kx)$

Here,

$A=amplitute$

$w=frequency$

Given,

$F(t)=10sin(10t+0.5)$

By comparing both the equation we get

$A=10$

Correct option is C.

In the equation

$y=A\sin{(kx-\omega t+{\phi}_{0})}$ the term phase is defined as:

0%
${\phi}_{0}$
0%
${\phi}_{0}-\omega t$
0%
$kx+{\phi}_{0}$
0%
$kx-\omega t-{\phi}_{0}$

Explanation

Given,

$y=Asin(kx-\omega t+\phi_0)$

In the equation, the term phase is defined as

$\phi_0$ .

The correct option is A.

Displacement time equation of a particle executing SHM is x = A sin(wt + /6). Time taken by the particle to go directly from x = - A/2 to x = +A/2 is

0%
$\pi/\omega$
0%
$\pi/3\omega$
0%
$3\pi/\omega$
0%
$\pi/2\omega$

Explanation

At t =0, the particle was at x = A sin (0 + /6) = A/2.

The particle reaches equilibrium at $t= t_1$ . Thus $0 = A sin (wt_1 + /6) or t_1 = -T/12$ . Since time taken cannot be negative, $t_1 = T/12$ .

Similarly same time will be used to travel from mean position to +A/2. Thus, total time taken is $2(T/12) = T/6 = \dfrac{\pi} {3w}$

The correct option is (c)

The equation of a simple harmonic motion is given by,

$x=8 \,sin (8 \pi t)+6 cos (8 \pi t)$ , the initial phase angle is

0%
$tan^{-1} (4/3)$
0%
$tan^{-1} (3/4)$
0%
$tan^{-1} (2/3)$
0%
$tan^{-1} (5/8)$

Two wave have amplitude ratio 10 : 1 then

$v_{1}/v_{2}$ is (Approx) -

0%
10 : 1
0%
1 : 10
0%
2 : 3
0%
3 : 2

A particle starts oscillating from position equal to half the amplitude. What is its initial phase?

0%
$\pi/6$
0%
$\pi/3$
0%
zero
0%
none of these

A single wave is called

0%
a wave pulse
0%
a crest
0%
a trough
0%
a crest and a trough

A particle is executing SHM with an amplitude 4 cm. What time does it take to reach the distance equal to amplitude, if the particle starts from rest at t = 0 from the mean position

0%
T
0%
T/2
0%
T/4
0%
3T/4

Phase of a particle gives us

0%
the velocity of the particle
0%
the direction of motion of the particle
0%
the acceleration of the particle
0%
the displacement of the particle

Which of the following functions represent a wave

0%
$(x-vt)^{2}$
0%
$\log (x+vt)$
0%
$e^{-(x-vt)^{2}}$
0%
$\dfrac{1}{x+vt}$

Explanation

Although all the four function are written in the form

$f\left(ax \pm bt\right),$ only option

$(c)$ among the four function is finite everywhere at all times. Hence only option

$(c)$ represents a wave.

The equation of a wave traveling in a string can be written as

$y=3 \sin\pi(100t-x)$ cm. Its wavelength is:

0%
3 cm
0%
100 cm
0%
5 cm
0%
2 cm

Three waves

$y _ { 1 } = 2 A \sin \left( \omega t - \dfrac { \pi } { 3 } \right)$

$y _ { 2 } = 2 A \sin \left( \omega t + \dfrac { \pi } { 3 } \right)$ and

$y _ { 3 } = - A \sin ( \omega t )$ interfere each other. The amplitude of the resultant wave is

0%
A
0%
zero
0%
2 A
0%
3 A

Explanation

Amplitude of resultant wave

$y=y_1+y_2+y_3$

$y=2A\sin \left(\omega t-\dfrac{\pi}{3}\right)+2A\sin\left(\omega t+\dfrac{\pi}{3}\right)-A\sin\omega t$

$y=2A\sin\omega t\cos\dfrac{\pi}{3}-2A\cos \omega t\sin\dfrac{\pi}{3}+2A\sin\omega t\cos \dfrac{\pi}{3}+2A\cos \omega t\sin\dfrac{\pi}{3}-A\sin\omega t$

$y=4A\sin\omega t\cos \dfrac{\pi}{3}-A\sin\omega t$

$y=2A\sin\omega t-A\sin\omega t$

$y=A\sin\omega t$

Thus, the resultant amplitude is A.

1197304_1400021_ans_173daf6bee194939aa44ee8bcc235cd7.png

If two waves of same frequency and amplitude superpose to produce resultant of the amplitude of either wave, then their phase difference is-

0%
$\pi$
0%
$2\pi/3$
0%
$\pi/3$
0%
$Zero$

Explanation

Let,

$\begin{array}{l} { x_{ 1 } }=A\sin wt \\ { x_{ 2 } }=A\sin \left( { wt+\phi } \right) \\ x={ x_{ 1 } }+{ x_{ 2 } } \end{array}$

$\therefore$ For resultant amplitude to be equal to

$\phi$ must be

$\dfrac{{2\pi }}{3}$ .

$\therefore$ phase difference is

$\dfrac{{2\pi }}{3}$

$\therefore$ Option

$B$ is correct answer.

1207848_1284508_ans_3fef86ab8a384168b02c4074e1bcd0ca.JPG

The motion of a particle is said to be simple harmonic
It it moves to and fro about a fixed point and given that driving force is towards mean position

0%
True
0%
False

The phase difference between the displacement and acceleration of particle executing S.H.M. radian is:

0%
$\pi / 4$
0%
$\pi / 2$
0%
$\pi$
0%
$2 \pi$

Explanation

$va\sin(\omega t+\phi)$

Differentiating

$v=a\omega\cos(\omega t+\phi)$

Differentiating again,

$a=-a\omega^2\sin(\omega+\phi)$

Now taking minus sign

$a=a\omega^2\sin(\omega t+\phi+\pi)$

So, phase difference between displacement and acceleration will be:-

$=\omega t+\phi+\pi-{\omega t+\phi}\\=\omega t+\phi+\pi-\omega t -\phi\\=\pi$

The displacement of a particle executing SMH is given by

$x=0.01{\,\,}sin{\,\,}100\pi(t+0.05).$ The time period is

0%
$0.01$ sec
0%
$0.02$ sec
0%
$0.1$ sec
0%
$0.2$ sec

Explanation

Given, Angular frequency

$(\omega) = 100\pi$

So, Time period

$= \dfrac{2\pi}{\omega} = \dfrac{2 \pi}{100 \pi} = 0.02 \;sec$

Therefore, B is correct option.

The two waves represented by

$y=a \sin (\omega t)$ and

$y_2=b \cos (\omega t)$ have a phase difference of

0%
$0$
0%
$\dfrac{\pi}{2}$
0%
$\pi$
0%
$\dfrac{\pi}{4}$

Explanation

$y_1=a \sin \omega t$ , and

$y_2=b \cos \omega t=b \sin \left(\omega t+\dfrac{\pi}{2}\right)$
So phase difference

$\phi=\pi/2$

The displacement of a particle is given by

$x = 3 sin(5\pi t) + 4 cos 5(5 \pi t)$ The amplitude of the particle is [MP PMT 1999]

0%
3
0%
4
0%
5
0%
7

Explanation

For the given imposing imposing waves

$a_1 = 5, a_2 = 4$ and phase difference

$\phi = \dfrac{\pi}{2}$

$\Rightarrow A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 cos \pi /2} = \sqrt{(3)^2 + (4)^2} = 5$

The frequency of light ray having the wavelength

$3000\ A^o$ is

0%
$9 \times 10^{13}\ cycles/sec$
0%
$10^{15}\ cycles/sec$
0%
$90 \ cycles/sec$
0%
$3000 \ cycles/sec$

Explanation

The frequency of the light ray is given as:

$\nu=\dfrac{c}{\lambda}$

$=\dfrac{3 \times 10^8}{3000 \times 10^{-10}}$

$=10^{15}\ cycles/sec$

A water wave in a shallow tank passes through a gap in a barrier. What happens to the speed and what happens to the wavelength of the wave as it passes through the gap?

0%
speed - decreases ; wavelength - decreases
0%
speed - decreases ; wavelength - remains constant
0%
speed - remains constant ; wavelength - decreases
0%
speed - remains constant ; wavelength - remains constant

Wavelength of light of frequency

$100\ Hz$

0%
$2 \times 10^6\ m$
0%
$3 \times 10^6\ m$
0%
$4 \times 10^6\ m$
0%
$5 \times 10^6\ m$

Explanation

$\lambda=\dfrac{c}{v}$

$=\dfrac{3 \times 10^8}{100}=3 \times 10^6\ m$

In how many parts radio wave frequency is divided

0%
$2$
0%
$3$
0%
$4$
0%
$5$

A source of sound of frequency 600 Hz is placed inside water. The speed in water is 1500 m/s and in air it is 300 m/s. The frequency of sound recorded by an observer standing in air is

0%
200 Hz
0%
3000 Hz
0%
120 Hz
0%
600 Hz

The principle of superposition of waves can not be used to explain wave phenomena like

0%
Interference
0%
Diffraction
0%
Stationary Waves and beats
0%
Dispersion

Transverse waves are produced in a long string by attaching its free end to a vibrating tuning fork. Figure shows the shape of a part of the string. The points in phase are

0%
A and D
0%
B and E
0%
C and F
0%
A and G

Explanation

Points in phase will have a phase difference of

$0 \ or \ 2\pi$

$\Rightarrow$ A and G are the points in same phase.

Human audible frequency range is 20Hz to 20KHz. If velocity of sound in air is 340m/s, the minimum wavelength of audible sound wave is

0%
0.17mm
0%
1.77mm
0%
17mm
0%
170mm

Explanation

Using the well known relation between wavelength frequency and velocity of a wave,

$= \nu \lambda$

$\dfrac{340}{20}$ >

$\lambda > \dfrac{340}{20000}$

Thus,

$\lambda = [0.017,17]$ in meters

Two light waves are represented by

$y_{1} =3\sin\omega t$ and

$y_{2}=4\sin( \omega t+\dfrac{\pi }{3} )$ . The resultant amplitude due to interference will be

0%
$\sqrt{21}$
0%
$\sqrt{26}$
0%
$\sqrt{37}$
0%
$\sqrt{41}$

Explanation

The resultant wave

$y=y_{1}+y_{2}$

$y=3\sin \omega t+4\sin( \omega t+\pi /3)$

$=3\sin \omega t+2\sin \omega t+2\sqrt{3}\cos \omega t$

$=5\sin \omega t+2\sqrt{3}\cos \omega t$

Amplitude of the resultant wave is given by

$=\sqrt{(5)^{2}+(2\sqrt{3})^{2}}$

$=\sqrt{25+12}$

$=\sqrt{37}$

During propagation of longitudinal plane wave in a medium, two particles separated by a distance equivalent to one wavelength at an instant will be/have

0%
In phase, same displacement
0%
In phase, different displacement
0%
Different phase, same displacement
0%
Different phase, different displacement

A plane progressive wave has frequency

$25 \ Hz$ and amplitude

$2.5 \times 10^{-5}$ m and initial phase zero, propagates along the negative x-direction with a velocity of

$300 m/s$ . The phase difference between the oscillations at two points

$6m$ apart along the line of propagation is :

0%
$\pi$
0%
$\displaystyle \dfrac{\pi}{2}$
0%
$2\pi$
0%
$\displaystyle \dfrac{\pi}{4}$

Explanation

$t=\dfrac{d}{V}$

$=\dfrac{6}{300}$

$\Rightarrow t=2\times 10^{-2}$

$T=\dfrac{1}{25}$

$=4\times 10^{-2}$
For

$2\pi$ time taken to

$T$ and phase difference in

$t$

$=\dfrac{2\pi}{T}\times t$

$=\dfrac{2\pi}{4\times 10^{-2}}\times 2\times 10^{-2}$

$=\pi$

The equations of displacement of two waves are given as

$Y_{1}=10 \sin(\pi t+\pi /3)$

$Y_{2}=5(\sin3\pi t +\sqrt{3}\cos3\pi t$ ), then the ratio of their amplitude is

0%
$1:2$
0%
$2:1$
0%
$1:1$
0%
$3:1$

Explanation

Amplitude of

${ Y }_{ 1 }=10\sin { \left( m\pi t+\pi /3 \right) }$ is

${ A }_{ 1 }=10$

Ampltude of

${ Y }_{ 2 }=10\left( \dfrac { 1 }{ 2 } \sin { 3\pi t } +\dfrac { \sqrt { 3 } }{ 2 } \cos { 3\pi t } \right) =10\sin { \left( 3\pi t+\pi /3 \right) }$ is

${ A }_{ 2}=10$

$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =1:1$

The frequency of vibration of a rod is

$200Hz$ . If the velocity of sound in air is

$340m/s$ , the wave length of the sound produced is

0%
$1.7m$
0%
$6.8 m$
0%
$3.4 m$
0%
$13.6 m$

Explanation

$V=f\lambda$ where

$V =$ velocity,

$\lambda$ = wavelength and

$f =$ frequency

$\lambda=\dfrac{V}{f}$

$=\dfrac{340}{200}$

$\lambda=1.7 m$

Mark correct option:

0%
the phase of transmitted wave always remains unchanged
0%
the amplitude of transmitted wave does not depend upon the velocity of wave in media
0%
the amplitude of reflected wave and transmitted wave are same to each other for a given incident wave
0%
the amplitude of rerflected wave is equal to the amplitude of incident wave

Three waves of equal frequency having amplitudes

$10 \ mm, 4 \ mm$ and

$7\ mm$ arrive at a given point with successive phase difference

$\dfrac \pi 2$ . The amplitude of the resulting wave (in mm) is given by:

0%
$7$
0%
$6$
0%
$5$
0%
$4$

Explanation

It is given that the

$3$ waves arrive at a point with a successive difference of

$90^o$ .

Therefore, the waves with amplitude

$10 mm$ and

$7 mm$ will be completely out of phase and interfere destructively.

Hence, the amplitude of the resultant wave is

$10 - 7 = 3mm$

Now, the waves of

$3 \ mm$ and

$4 \ mm$ are out of phase by

$90^o$

Therefore, applying Pythagorus theorem, they will interfere with each other causing the new amplitude of the wave to be,

$A^2 = 3^2 + 4^2$

$A^2 = 25$ or

$A = 5 \ mm$

When a transverse plane wave traverses a medium, individual particles execute periodic motion given by the equation

$y=0.25\cos(2\pi t-\pi x)$ . The phase difference for two positions of same particle which are occupied by time intervals

$0.4 second$ apart is

0%
$144^{o}$
0%
$135^{o}$
0%
$72^{o}$
0%
$108^{o}$

Explanation

$\omega =\dfrac{2\pi}{T}$

$T=\dfrac{2\pi}{\omega}$

$T=1sec$

$\therefore 2\pi \ rad \ in \ 1 sec$

$in \ 0.4 sec \ 2\pi\times 0.9$

$=144^o$

The equation of a wave pulse is given as

$y=\displaystyle \frac{0.8}{(4x+5t)+4}$ , the amplitude of the pulse is:

0%
$0.2 \ units$
0%
$0.4\ units$
0%
$0.6 \ units$
0%
$0.8 \ units$

Explanation

The amplitude of wave is given by

$y$ , at

$x=0$ and

$t=0$ .

So amplitude is

$y = 0.8/(4\times0 + 5\times0 + 4) = 0.2\ units$

The displacement of a particle varies according to the relation

$X=4(\cos\pi t+\sin\pi t)$
The amplitude of the particle is

0%
$-4$
0%
$4$
0%
$4\sqrt{2}$
0%
$8$

Explanation

$x=4(\cos\ \pi t+\sin\ \pi t)$

$= 4 \sqrt2(\cos \ \pi t \sin \pi /4 + \sin \ \pi t \cos \ \pi /4)$

$=4\sqrt{2}\sin (\pi t+\ \dfrac{\pi }{4})$
So, amplitude

$=4\sqrt{2}$

A Sound wave with an amplitude of

$3 cm$ starts moving towards right from origin and gets reflected at a rigid wall after a second. If the velocity of the wave is

$340 ms^{-1}$ and it has a wavelength of

$2 m$ , the equations of incident and reflected waves are

0%
$y = 3\times 10^{-2}\sin \pi (340 t - x), y = -3\times 10^{-2}\sin\pi(340t + x)$ towards left
0%
$y = 4\times 10^{-2} \sin \pi (340 t + x), y = -4\times 10^{-2}\sin\pi(340t + x)$ towards left
0%
$y = 5\times 10^{-2} \sin \pi (340 t - x), y = -5\times 10^{-2}\sin\pi(340t - x)$ towards left
0%
$y = 6\times10^{-2}\sin \pi(340 t - x), y = -6\times 10^{-2}\sin\pi(340t + x)$ towards left

Explanation

$A=3cm$

$y=A \sin (Vt-x)$

$V= \lambda / t$

$f=170 sec$

$y_{1}=A \sin (V t-x)$

$y_{2}=A \sin (V t+x)$

$\therefore y_{1}=3\times 10^{-2} \sin k (340 t-x),$

$y_{2}=3\times 10^{-2} \sin k(340 t +x)$

$k=\dfrac{2\pi}{\lambda}$

$k=\dfrac{2\pi}{2}$

$k=\pi$

$\therefore y_{1}=3\times 10^{-2} \sin \pi(340t-x)$

$y_{2}=3\times 10^{-2} \sin \pi (340t + x)$

A man is travelling towards east at 20ms

$^{-1}$ and sound waves are moving towards east at 340ms

$^{-1}$ . He finds that

$500$ waves cross him in

$2$ seconds, the wavelength of the sound wave received by the man is :

0%
$1.28 m$
0%
$2.28 m$
0%
$3.28 m$
0%
$4.28 m$

Explanation

Speed of sound relate to man is

$=340-20$

$=320\ m/s$

$f=\dfrac{number\ of\ waves\ crossing }{time}$

$=\dfrac{500}{2}$

$=250\ Hz$

$f=\dfrac{u}{\lambda }$

$\lambda=\dfrac{u}{f}$

$=\dfrac{320}{250}$

$=1.28m$

The equation

$y=4+2\sin(6t-3x)$ represents a wave motion with amplitude of

0%
$6 units$
0%
$2 units$
0%
$20 units$
0%
$8 units$

Explanation

A wave can be represented by:

$y=C+a\sin(\omega t \pm kx)$ , where a is the amplitude of the wave. Comparing with the equation given in the question,

$a=2$ .

The equation of a wave travelling on a stretched string is

$y=4\displaystyle \sin 2\pi(\frac{t}{0.02}-\frac{x}{100})$ here x, y are in cm and t in seconds. The relative deformation amplitude of medium is

0%
0.02 $\pi$
0%
0.08 $\pi$
0%
0.06 $\pi$
0%
0.01 $\pi$

Explanation

$y=4sin\ 2\pi \left( \dfrac{t}{0.02}-\dfrac {x}{100} \right)$

$\dfrac{dy}{dx}=\dfrac{8\pi}{100}\ \ \ cos\ 2\pi\left( \dfrac{t}{0.02}-\dfrac{x}{100} \right)$
Deformation amplitude

$A=\left | \dfrac{dy}{dx} \right |=\dfrac{8\pi}{100}$

$=0.08\pi$

A wave of length

$2m$ is superposed on its reflected wave to form a stationary wave. A node is located at

$x=3m$ The next node will be located at

$x=$

0%
$4m$
0%
$3.75m$
0%
$3.50m$
0%
$3.25m$

Explanation

Since wave length is

$2m$ , half of wavelength is

$1m$ . Node forms after each half of wave length. So, node will be formed at each

$1 m$
So, as node is formed at

$3 m$
next node will be formed at

$3+1=4\ m.$

The equation of a plane progressive wave is

$y = 0.9 \sin 4\pi [t-\displaystyle \dfrac{x}{2}]$ , when it is reflected at a rigid support, its amplitude becomes

$2/3$ of its previous value. The equation
of the reflected wave is :

0%
$y=0.9\displaystyle \sin 4\pi[t+\dfrac{x}{2}]$
0%
$y=-0.6\displaystyle \sin 4\pi[t+\dfrac{x}{2}]$
0%
${y}=0.9\displaystyle \sin 8\pi[t-\dfrac{x}{2}]$
0%
${y}=0.6\displaystyle \sin 4\pi[t+\dfrac{x}{2}]$

Explanation

$y=0.9\ \sin\ 4\ \pi (t-\dfrac{x}{2})$
Reflected wave equation will be opposite in direction and of amplitude

$\dfrac{2}{3}\times 0.9=0.6$ and amplitude will be negative of incident wave.
So,

$y_{reflected}=-0.6\ \sin\ 4\pi \left ( t+\dfrac{x}{2} \right ).$

The equation of a progressive wave is

$y = 0.5 \sin \ 2\pi (\dfrac{t}{0.004} - \dfrac{x}{50})$ , where

$x$ is in

$cm$ . The phase difference between two points separated by a distance

$5 \ cm$ at any instant is :

0%
$\dfrac \pi 5$
0%
$\dfrac \pi 3$
0%
$\dfrac \pi 2$
0%
$\pi$

Explanation

$Wavelength(\lambda)=\dfrac{2\pi}{k}= \dfrac{2\pi \times 50}{2\pi}=50\: cm$

We know that path difference of

$\lambda$ gives rise to phase difference of

$2\pi$ . So path difference of

$5\: cm$ gives rise to phase difference of

$\dfrac{2\pi}{\lambda} \times 5=\dfrac{\pi}{5}$

In a stationary wave that forms as a result of reflection of waves from an obstacle, the ratio of the amplitude at an antinode to the amplitude at node is

$6$ . The percentage of energy transmitted is

0%
$25\%$
0%
$49\%$
0%
$58\%$
0%
$72\%$

Explanation

the ratio of the amplitude at an antinode to the amplitude at node is 6, so

$\dfrac{a_i+a_r}{a_i-a_r}=6$
or,

$\dfrac{a_r}{a_i}=5/7$
Now, the energy is proportional to the square of the amplitudes.
So,

$\dfrac{E_r}{E_i}=25/49=51\%$
So the rest will be the transmitted energy, i.e.

$49\%$ .

CBSE Questions for Class 11 Engineering Physics Waves Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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