MCQ Questions for Class 11 Engineering Physics Waves Quiz 1

A transverse wave travels on a taut steel wire with a velocity of

$v$ when the tension in it is

$2.06\times { 10 }^{ 4 }N$ . When the tension is changed to

$T$ , the velocity changed to

$v/2$ . The value of

$T$ is closet to:

0%
$2.50\times { 10 }^{ 4 }N$
0%
$10.2\times { 10 }^{ 2 }N$
0%
$30.5\times { 10 }^{ 4 }N\quad$
0%
$5.15\times { 10 }^{ 3 }N$

Explanation

Velocity of sound in a medium

$v = \sqrt{\dfrac{T}{\mu}}$

$v \propto \sqrt{T}$

$\Rightarrow \dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}$

Given

$T_1 = 2.06 \times 10^4N$

$v_2 = \dfrac{v}{2}$

$v_1 = v$

$\dfrac{v}{v/2} = \sqrt{\dfrac{2.06\times 10^4}{T}}$

$4 = \dfrac{2.06\times 10^4}{T}$

$T = .515 \times 10^4 N$

$T = 5.15 \times 10^3N$

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

When the displacement ,

$x=a\cos wt+b\sin wt$ , the amplitude will be

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

Here,

$a=4, b=4$ so

$A=\sqrt{4^2+4^2}=4\sqrt 2$

Amplitude of a wave is represented by

$A = \dfrac {c}{a + b - c}$ . Then resonance will occur when

0%
$b = -c/ 2$
0%
$b = 0$ and $a = c$
0%
$b = -a/2$
0%
None of the above

Explanation

$A = \dfrac {c}{a + b - c}$ : when

$b = 0, a = c$
Amplitude

$A \rightarrow \infty$ . This correspondence to resonance.

A simple wave motion represented by

$y=5(\sin 4\pi t+\sqrt 3 \cos 4\pi t)$ . Its amplitude is:

0%
$5$
0%
$5\sqrt 3$
0%
$10\sqrt 3$
0%
$10$

Explanation

$y=5(sin4\pi t +\sqrt{3}cos4\pi t)$

$=5\times 2(\dfrac{1}{2}sin4\pi t+\dfrac{\sqrt{3}}{2}cos4\pi t)$

$=10(cos\dfrac{\pi}{3}sin4\pi t+sin\dfrac{\pi}{3}cos4\pi t)$

$=10sin(4\pi t+\dfrac{\pi}{3})$

Hence the amplitude of the wave is

$10$ .

The phase change between incident and reflected sound wave from a free end is

0%
$0$
0%
$\pi$
0%
$3\pi$
0%
$2\pi$

Explanation

At a free and, the wave is reflected as it is with just as change in its direction of propagation

$y= A \sin ( \omega t+kx)$

$y_r= A \sin (\omega t-kx)$
phase difference in time domain

$= 0$

Which of the following is conserved when light waves interfere?

0%
momentum
0%
amplitude
0%
energy
0%
intensity

Phase difference between a particle at a compression and a particle at the next rarefaction is

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

Explanation

Phase difference between two successive compression of rarefaction is

$2\pi$
As rarefaction appears between two compression, phase difference is

$\pi$ .

Thickness of very thin films can be found by the technique of

0%
Dispersion
0%
Interference
0%
polarization
0%
Diffraction

Two light waves are represented by

$y_{1}=4\sin \omega t$ and

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

0%
$5cm$
0%
$7cm$
0%
$1cm$
0%
$0$

Explanation

The two waves are

$y_{1}=4\sin \omega t$ and

$y_{2}=3\cos \omega t$ , the resultant wave

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

$=4\sin \omega t + 3\cos \omega t$ .
The resultant amplitude

$=\sqrt{4^{2}+3^{2}}$

$=\sqrt{25}=5$

For the sustained interference of light, the necessary condition is that the two sources should

0%
have constant phase difference only
0%
be narrow
0%
be close to each other
0%
of same amplitude with constant phase difference

The necessary condition for an interference by two sources of light is that:

0%
two light sources must have the same wavelength
0%
two point sources should have the same amplitude and same wavelength
0%
two sources should have the same wavelength, nearly the same amplitude and have a constant phase angle difference
0%
the two point sources should have a randomly varying phase difference

Four different independent waves are represented by
a)

$y_{1}=a_{1} \sin \omega _{1}t$
b)

$y_{2}=a_{2} \sin \omega _{2}t$
c)

$y_{3}=a_{3} \sin \omega _{3}t$
d)

$y_{4}=a_{4} \sin (\omega _{4} t + \pi /3)$
The sustained interference is possible due to

0%
a & c
0%
a & d
0%
c & d
0%
Not possible with any combination

Explanation

Interference takes place during the interaction of waves that have, nearly the same frequency. Four different independent waves have frequencies

$w_{1}$ ,

$w_{2}$ ,

$w_{3}$ and

$w_{4}$ respectively. As these frequencies are different, the interference does not take place between any combinations.
Hence, option D is correct.

The phase difference between the particle at one compression and another particle in third compression is

0%
$\pi$ radians
0%
$2\pi$ radians
0%
$3\pi$ radians
0%
$4\pi$ radians

Explanation

phase difference between two successive compression is

$2\pi$

$\therefore$ phase difference between a particle at one compression and in third compression is

$2(2\pi)= 4\pi$

When viewed in white light, soap bubble show colors because of

0%
Interference
0%
Scattering
0%
Diffraction
0%
Dispersion

If a wave completes 24 cycles in 0.8 s, then frequency of the wave is

0%
30 Hz
0%
8 Hz
0%
24 Hz
0%
12 Hz

Explanation

The frequency of a wave is the number of full cycle per second. Here the frequency of the wave will be:

$24/0.8=30$ Hz.

For superposition of two waves, which of the following is correct

0%
They must have the same frequency and wavelength
0%
They must have equal frequencies but may have unequal wavelengths
0%
They must have the same wave-length, but may have different frequencies
0%
They may have different wavelength and different frequencies

Explanation

Principle of superposition states that the displacement of two waves

$y_1= f_1 (t-\dfrac{x}{v}) \ and \ y_2= f_2(t+\dfrac{x}{v})$
is

$y=y_1+y_2$
Wavelength and frequency depend on the function

$f_1 \ and \ f_2$

$\therefore$ They can have any wavelength and any frequency

If the frequency of a wave is doubled, then its wavelength

0%
Becomes doubled
0%
Remains same
0%
Becomes half of the original
0%
None of these

A wave creates disturbance and transmits matter from one place to another.

0%
True
0%
False

Speed of sound wave in a given medium is

0%
Directly proportional to its frequency
0%
Inversely proportional to its frequency
0%
Directly proportional to the square of frequency
0%
Independent of its frequency

Explanation

The relationship of the velocity (v) of sound wave in a medium with its frequency and wavelength is :

$v=\nu \times \lambda$
So,

$v \propto \nu$ .

The interference phenomenon can take place

0%
in transverse wave
0%
in longitudinal wave
0%
in electromagnetic waves
0%
in all waves

The disturbance caused in a medium is called

0%
Straight line
0%
Wave
0%
Disturbance
0%
None of these

The product of the time period of a wave and its frequency is

0%
Infinite
0%
Zero
0%
More than unity but less than infinity
0%
Unity

Explanation

The relation between frequency (f) and time period(T) is

$f=1/T or f \times T=1$

Which of the following quantity decrease as sound wave travels through a medium :

0%
Amplitude
0%
Frequency
0%
Velocity
0%
Wavelength

Two waves ${y_1} = {A_1}\sin (\omega t - {\beta _1}){y_2} = {A_2}\sin (\omega t - {\beta _2})$ superimpose to form a resultant wave whose amplitude is

0%
$\sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos ({\beta _1} - {\beta _2})}$
0%
$\sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\sin ({\beta _1} - {\beta _2})}$
0%
$A_1 + A_2$
0%
$|A_1 + A_2|$

Explanation

$y=y_1 +y_2$

$y=A Sin (wt - \beta)$

$A=\sqrt{A_1^2 + A_2^2 + 2A_1 A_2 Cos (\beta_1- \beta_2)}$

When two plane progressive waves travelling in same direction superimpose over each other, the velocity of the resultant wave will

0%
increase
0%
remain unchanged
0%
be zero
0%
decrease

What will be the wave velocity, if the radar gives 54 waves per min and wavelength of the given wave is 10 m?

0%
4 m s $^{-1}$
0%
6 m s $^{-1}$
0%
9 m s $^{-1}$
0%
5 m s $^{-1}$

Explanation

$v = v \displaystyle \lambda = \frac{54}{60}H_z \times 10 m = 9 m s^{-1}$

The particle of a medium vibrates about their mean position whenever a wave travels through that medium. The phase difference between the vibrations of two such particles

0%
varies with time only
0%
varies with distance separating them only
0%
varies with time as well as distance
0%
is always zero

Explanation

The phase difference between the vibrations of two particles of the medium is given by :

$\Delta \phi=\dfrac{2\pi}{\lambda}\Delta x$

it is clear that phase difference varies as the path difference between the particles varies, which is the distance, separating the particles.

An ultrasonic source emits sound of frequency 220 kHz in air. If this sound meets a water surface, what is the wavelength of the reflected sound?

0%
$1.6 \times 10^{-3} m$
0%
$2.6 \times 10^{-3} m$
0%
$3.6 \times 10^{-3} m$
0%
$4.6 \times 10^{-3} m$

Explanation

Here,

$v=220 kHz = 220 \times 10^3$

$=2.2 \times 10^5 Hz$ ;
Speed of sound in air,

$v_a = 352 m s^{-1}$
Speed of sound in water,

$v_w = 1.496 m s^{-1}$
The reflected sound: After reflection, the ultrasonic sound continues to travel in air. If

$\lambda_a$ is wave length in air, then

$\lambda_a = \displaystyle \frac{v_a}{v} = \frac{352}{2.2 \times 10^5} = 1.6 \times 10^{-3} m$

A piezo - electric quartz plate is vibrating to produce wavelength of

$10^{-1}$ m. Find the frequency if for quartz

$Y=8 \times 10^{10} N m^{-2}$ and

$\rho = 2.65 \times 10^3 kg m^{-3}$

0%
350 kHz
0%
450 kHz
0%
550 kHz
0%
650 kHz

Explanation

We know that for longitudinal waves in solids,

$v = \displaystyle \sqrt{\frac{Y}{\rho}}$ , so,

$v = \displaystyle \sqrt{\frac{8 \times 10^{10}}{2.65 \times 10^3}}$

$= 5.5 \times 10^3 m/s$
So,

$10^{-2}m$
But as

$v=f \lambda$ , i.e.,

$f = (\frac{v}{\lambda})$
so f

$[5.5 \times 10^3 / 10^{-2}] = 5.5 \times 10^5 Hz$

$=550 kHz$

The frequency of a man's voice is

$300\space Hz$ . If the velocity of sound waves is

$336\space ms^{-1}$ , the wavelength of the sound is

0%
$1.12\space m$
0%
$300\times336\space m$
0%
$330/336\space m$
0%
None of these

Explanation

We are given, frequency

$f=300 Hz$ , and velocity

$v=336 ms^{-1}$ ,

From the relation

$v=f \times \lambda$ ,

$\lambda = v/f =336/300=1.12 m$

Option "A" is correct.

In ordinary talk, the amplitude of vibration is approximately:

0%
$10^{-12} m$
0%
$10^{-11} m$
0%
$10^{-8} m$
0%
$10^{-7} m$

Explanation

In ordinary talk, the amplitude of vibration is about

$10^{-8}$

$m$ .

The frequency, wavelength and speed of a sound wave are related as :

$v=velocity\ of\ wave,\ u=frequency,\ \lambda=wavelength$

0%
$\lambda=vT$
0%
$\lambda=v/u$
0%
$u=v/\lambda$
0%
$\lambda=uv$

Explanation

We know that,

$v=u\lambda$ , where

$v$ is the speed of sound wave,

$u$ is the frequency and

$\lambda$ is the wavelength.

Also,

$u=1/T$ , where

$T$ is the Time Period.

Now using these two equations and comparing them, we will get our answer as option

$A,B$ and

$C$

The phenomenon of interference is shown by

0%
longitudinal mechanical waves only
0%
transverse mechanical waves only
0%
non mechanical transverse waves only
0%
all the above types of waves

Two waves represented by

$y_1= a sin\omega t$ and

$y_2 = a sin(\omega t+ \phi)$ and

$\phi =\dfrac{\pi}{2}$ are superposed at any point at a particular instant. The resultant amplitude is

0%
$a$
0%
$4a$
0%
$\sqrt{2} a$
0%
$zero$

Explanation

Since the angle between them is

$90$ , so

$cos\ 90^0=0$

So,

$R=\sqrt {a^2+a^2}=\sqrt 2a$

The picture shows a wave generated in a laboratory. What is the amplitude of the wave shown ?

0%
0.85 cm
0%
1.7 cm
0%
2.7 cm
0%
There is not enough information to determine the answer.

The velocity of sound in a gas is 4 times that in air at the same temperature. When a tuning fork is sounded in a wave of frequency 480 and wavelength

$\lambda_1$ is produced. The same fork is sounded in the gas and of

$\lambda_2$ is the wavelength of the wave, then

$\lambda_2\,/\, \lambda_1$ is :

0%
1
0%
2
0%
4
0%
1/2

Explanation

We have ,

$v=f\lambda$ ,

when tuning fork is sounded in air ,

$v_{a}=f\lambda_{1}$ , ........................eq1

when tuning fork is sounded in gas ,

$v_{g}=f\lambda_{2}$ , .......................eq2 ,

frequency will remain same as the source of sound is same ,

dividing eq2 by eq1 ,

$v_{g}/v_{a}=\lambda_{2}/\lambda_{1}$ ,....................eq3

now given ,

$v_{g}=4v_{a}$ ,

putting ,

$v_{g}=4v_{a}$ in eq3,

we get ,

$4v_{a}/v_{a}=\lambda_{2}/\lambda _{1}$ ,

$4=\lambda_{2}/\lambda _{1}$

In the following properties of a wave, the one which is independent of the other is

0%
amplitude
0%
velocity
0%
wavelength
0%
frequency

Explanation

We know that velocity of wave

$(v)=$ frequency

$(f) \times$ wavelength

$(\lambda)$ .

Thus, we can say that the amplitude (A) is independent of other means it is not related to the other quantities.

'Which is not true for a wave ?

0%
Wave velocity v = n $\lambda$
0%
Energy is transferred during wave motion
0%
All waves can pass through vacuum
0%
Unit of wave velocity is $m/s$

Explanation

(A) In a wave , frequency and wavelength are related by ,

$v=n\lambda$ .

(B) During wave motion , a disturbance (energy) is transferred by medium particles in a wave .

(C) Mechanical waves cannot travel through vacuum , because they are dependent on the properties of medium , so it is a wrong statement .

(D) Wave velocity is given by ,

$v=n(s^{-1})\lambda(m)=m/s$

Time period of a sound wave having the wavelength

$0.2 m$ and frequency

$10$ Hz will be

0%
$2s$
0%
$0.2s$
0%
$0.1s$
0%
$0.02s$

Explanation

Given -

$n=10Hz$ ,

we have , time period ,

$T=1/n=1/10=0.1s$

The stethoscope used by doctors works on the principle of

0%
Refraction of sound
0%
Reflection of sound
0%
Interference of sound
0%
Absorption of sound

The SI unit of amplitude of oscillation is :

0%
$cm$
0%
$m$
0%
$km$
0%
$\mu m$

A source of frequency

$500 Hz$ emits waves of wavelength

$0.2 \ m$ . The time the wave takes to travel a distance of

$300\ m$ is :

0%
75 seconds
0%
60 seconds
0%
12 seconds
0%
3 seconds

Explanation

We have , speed of wave ,

$v=n\lambda$ ,

where

$n=$ frequency of wave ,

$\lambda=$ wavelength of wave ,

given

$n=500Hz , \lambda=0.2m , d=300m$ ,

hence

$v=500\times0.2=100m/s$ ,

therefore , time taken to cover a distance of 300m ,

$t=d/v=300/100=3s$

The maximum displacement of the particles of a medium in which a wave is travelling is called _________ .

0%
amplitude
0%
frequency
0%
beat
0%
phase

Explanation

A wave has basically three components which is used to define all of its properties,

$\rightarrow$ Amplitude

$\rightarrow$ Wavelength/ frequency

$\rightarrow$ Phase

Amplitude of the wave is defined as the maximum displacement of the particle from its mean or initial position. For instance,

In a single or solitude wave we can see maximum displacement of the particle two times

$(+k,-k)$ , where k represents the maximum displacement or Amplitude.

$\therefore$ Option (A) is the correct answer.

Two particles having the same phase must be at

0%
one crest and one trough
0%
one crest and the mean position
0%
one trough and the mean position
0%
two consecutive crests

Light : Ray : : Sound : ?

0%
Hear
0%
Wave
0%
Audio
0%
Pitch

Explanation

The Light has particle nature

These particles (i.e photons) propagate

in from of Rays. Whereas in case

of sound, it propagates in the

from of pressure waves

Option

$-B$ is correct.

1104597_320961_ans_cf32c1021c574275a9f06d3b68f35689.jpg

The relation between frequency

$(n)$ and wavelength

$(\lambda)$ of a sound wave is given by

(

$v$ is velocity,

$n$ is frequency and

$T$ is time-period)

0%
$v=n\lambda$
0%
$n=-v\lambda$
0%
$v=\displaystyle\frac{n}{2\lambda}$
0%
$n=\displaystyle\frac{T}{\lambda }$

Explanation

Time period

$(T)$ of a wave is defined as the time taken to produce one complete wave, and frequency

$(n)$ of a wave is defined as the number of complete waves produced in unit time.

Thus they are related as,

$n=\dfrac{1}{T}$ or

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

In one time period (T), a wave advances by a distance equal to one

wavelength. Thus if

$v$ is the velocity of the given wave, it is given by,

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

$n=\dfrac{1}{T}$ , we may write,

$v=n\lambda$

Two particles having the same phase must be at

0%
a crest and a trough respectively at the same time
0%
a crest and at the mean position respectively at the same time
0%
a trough and at the mean position respectively at the same time
0%
Two consecutive crests respectively at the same time

Explanation

For same phase angle, the two consecutive crest of wave 1 and wave 2 should be occurring at the same time period i.e. when they are in the same phase.Also same phase implies same frequency as well as same wavelength which will happen only if

$A_B = A_D$ and

$A_A = A_C$

If the maximum acceleration of the body executing S.H.M. is a

$_0$ and maximum velocity is V

$_0$ the amplitude is given by

0%
$\displaystyle \frac{1}{a_0 v_0}$
0%
$\displaystyle \frac{a_0}{v_0}$
0%
$v_0 a_0$
0%
$\displaystyle \frac{v_0^2}{a_0}$

Explanation

Max. acceleration in S.H.M,

$a_0 = \omega^2 a$ .....(i) [a = amplitude]
max. velocity =

$\omega$ a

$v_0 = \omega a$

$\Rightarrow v_0^2 = \omega^2 a^2$ ........ (ii)

$\therefore \displaystyle \frac{(ii)}{(i)} \Rightarrow a = \frac{v_0^2}{a_0}$

What type of wave travels the faster?

0%
Sound
0%
Earthquake
0%
Ocean
0%
Light

Frequency

$(v)$ and time period

$(T)$ of a sound wave are related as :

0%
$v\times T=1$
0%
$\displaystyle\frac{v}{T}=1$
0%
$v\times T^2$
0%
$v=T^{-2}$

Explanation

We have the relation,

$v=\frac{1}{T}$

where,

$v$ is the frequency

$T$ is the time period

Then, the product of the frequency anf time period of the sound wave is unity.

$v \times T = \frac{1}{T} \times T = 1$

CBSE Questions for Class 11 Engineering Physics Waves Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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