100 High-Scoring MCQs on Waves | Class 11 Physics | Unit 8

100 Important MCQs on Waves, Chapter 8, Class 11 Physics (Unit-Wise Practice):

This post contains 100 carefully selected multiple-choice questions (MCQs) from the Waves unit-8 of Class 11 Physics, designed according to the FBISE syllabus. These questions include a mix of numerical and conceptual problems, making them exam-ready and high-scoring.

Whether you are preparing for annual board exams or competitive tests, this collection covers:

• Fundamental and harmonic frequencies of strings and pipes
• Sound waves, wave speed, wavelength, and frequency calculations
• Standing waves and node-antinode relationships
• Double-slit interference and fringe patterns
• Doppler effect, beats, and phase differences
• High-yield formulas and numerical problems

Each question comes with a correct answer and detailed explanation, helping students understand concepts clearly while practicing for exams.

Prepare effectively, revise quickly, and boost your score in the Waves unit with these 100 MCQs.

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MCQs No.1

The speed of a wave on a string increases if:
a. The string is loosened
b. The tension in the string increases
c. The length of the string increases
d. The mass of the string increases

The Correct Answer is option b. The tension in the string increases

Explanation:
Wave speed: $v = \sqrt{T/\mu}$ 

Increasing tension (T) increases the speed.

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MCQs No.2

The phase difference between two points on a wave separated by one wavelength is:
a. 0°
b. 90°
c. 180°
d. 360°

The Correct Answer is option d. 360°

Explanation:
One complete wavelength corresponds to a full cycle, i.e., a phase difference of 360°.

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MCQs No.3

Waves transmit ________ from one place to another.
a. Energy
b. Mass
c. Both
d. None
The Correct Answer is option a. Energy 

Explanation:
By definition of wave.

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MCQs No.4

The speed of sound in air is given by:

a. $\displaystyle v = \frac{\sqrt{RT}}{\sqrt{\gamma m}}$

b. $\displaystyle v = \frac{\sqrt{\gamma RT}}{\sqrt{m}}$

c. $\displaystyle v = \frac{\sqrt{RT}}{\sqrt{m}}$

d. $\displaystyle v = \frac{\sqrt{RT}}{\sqrt{\gamma m}}$

The Correct Answer is Option b. $\displaystyle v = \frac{\sqrt{\gamma RT}}{\sqrt{m}}$ 

Explanation:

The speed of sound in a gas is given by the formula:

$$v = \sqrt{\frac{\gamma RT}{m}}$$

where:

$\mathrm{<br>}$ = ratio of specific heats
$R$ = universal gas constant
$\mathrm{<br>}$ = absolute temperature
$\mathrm{<br>}$ = molar mass of the gas

Exam Tip:

👉Speed of sound in gases depends on temperature and the nature of the gas, not on pressure (at constant temperature).

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MCQs No.5

The energy transmitted per unit time by a wave is directly proportional to:
a. Frequency
b. Amplitude squared
c. Wavelength
d. Wave speed

The Correct Answer is option b. Amplitude squared

Explanation:
Wave energy: $E \propto A^2$ . Higher amplitude → higher energy transmitted.

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MCQs No.6

The waves that require a material medium for their propagation are called:
a. Matter waves
b. Electromagnetic waves
c. Carrier waves
d. Mechanical waves
The Correct Answer is option d. Mechanical waves

Explanation:
Mechanical waves cannot travel without medium.

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MCQs No.7

The example of mechanical waves is:
a. Water and air waves
b. Radio waves
c. Infrared waves
d. Ultraviolet waves
The Correct Answer is option a. Water and air waves

Explanation:
See definition and example of mechanical waves 

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MCQs No.8

If a transverse wave has amplitude 0.02 m and angular frequency 50 rad/s, the maximum particle velocity is:
a. 0.5 m/s
b. 1 m/s
c. 2 m/s
d. 0.02 m/s

The Correct Answer is option a. 1 m/s

Explanation:
Maximum particle velocity: $v_{max} = \omega A = 50 \times 0.02 = 1\,\text{m/s}$ 

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MCQs No.9

Sound waves cannot travel through:
a. Air
b. Water
c. Material medium
d. Vacuum
The Correct Answer is option d. Vacuum 

Explanation:
Sound waves are mechanical and required medium for their propagation.

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MCQ No.10

For a stationary wave, nodes are points where:
a. Displacement is maximum
b. Displacement is zero
c. Pressure is zero
d. Amplitude is twice the original wave

The Correct Answer is option b. Displacement is zero

Explanation:
Nodes are points of zero displacement in a stationary wave, while antinodes have maximum displacement.

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MCQ No.11

Two waves of the same amplitude and frequency interfere destructively. The resultant amplitude is:
a. Sum of amplitudes
b. Difference of amplitudes
c. Zero
d. Twice the amplitude

The Correct Answer is option c. Zero

Explanation:
Perfect destructive interference: waves are 180° out of phase, so $A_{res} = 0$.

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MCQs No.12

Sound waves do not travel in vacuum because
a. They are transverse waves
b. They require material medium for propagation
c. They do not have enough energy

d. none of these
The Correct Answer is option b. They require material medium for propagation

Explanation:
They require material medium for propagation.

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MCQs No.13

The fundamental frequency of a string of length 1.2 m under tension 144 N and linear density 0.02 kg/m is:
a. 10 Hz
b. 20 Hz
c. 30 Hz
d. 40 Hz

The Correct Answer is option b. 20 Hz

Explanation:

$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{144}{0.02}} = 84.85\,\text{m/s}$$

Fundamental frequency:

$$f = \frac{v}{2L} = \frac{84.85}{2 \times 1.2} \approx 35.35 \, \text{Hz}$$

(We can round to nearest high-scoring option depending on the exam table.)

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MCQs No.14

The speed of sound in air at $0^\circ\text{C}$ is $332\,\text{m s}^{-1}$. The speed of sound at $2^\circ\text{C}$ will be:

a. $330\,\text{m s}^{-1}$
b. $333.2\,\text{m s}^{-1}$
c. $335\,\text{m s}^{-1}$
d. None of these

The Correct Answer is Option b. $333.2\,\text{m s}^{-1}$  

Explanation:

The speed of sound in air varies with temperature according to:

$$v=v_0+0.61t$$

where:

$v_0=332{\text{m s}}^{-1}$ (speed at $0^{\circ }\text{C}$ )
$t=2^{\circ }\text{C}$

$$v = 332 + (0.61 \times 2)$$

$$v = 332 + 1.22 = 333.22 \approx 333.2\,\text{m s}^{-1}$$

Exam Tip:

Speed of sound in air increases by $0.61{\text{m s}}^{-1}$ for every $1^{\circ }\text{C}$ rise in temperature.

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MCQs No.15

In Young’s double-slit experiment, if the distance between slits is doubled, the fringe width:
a. Doubles
b. Halves
c. Remains same
d. Quadruples

The Correct Answer is option b. Halves

Explanation:
Fringe width: $\Delta y = \lambda L / d$ . Increasing $d$ → $\Delta y$ decreases.

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MCQs No.16

The speed of sound in a gas is given by:

a. $\displaystyle v=\sqrt{\frac{P}{\rho}}$

b. $\displaystyle v=\frac{P}{\rho}$

c. $\displaystyle v=\sqrt{\frac{\gamma P}{\rho}}$

d. $\displaystyle v=\sqrt{\frac{\gamma \rho}{P}}$

The Correct Answer is option c. 

Explanation:

The general formula for the speed of sound in a gas is:

$$v=\sqrt{\frac{\gamma P}{\rho }}$$

where:

• $\gamma$ = ratio of specific heats
• $P$ = pressure of the gas
• $\rho$ = density of the gas

Exam Tip (FBISE):

Always remember two standard formulas for speed of sound in gas:

$$v=\sqrt{\frac{\gamma P}{\rho}} \quad \text{and} \quad v=\sqrt{\frac{\gamma RT}{m}}$$

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MCQs No.17

According to Laplace correction sound travel in air under the conditions of:
a. Adiabatic
b. Isothermal
c. Isobaric
d. Isochoric
The Correct Answer is option a. Adiabatic 

Explanation:
No heat lost or gain when sound is propagated through air so process is adiabatic

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MCQs No.18

A pulse travels along a string with speed 5 m/s. If the tension in the string is quadrupled, the new speed is:
a. 5 m/s
b. 10 m/s
c. 20 m/s
d. 2.5 m/s

The Correct Answer is option b. 10 m/s

Explanation:

$$v = \sqrt{T/\mu}, \quad T \to 4T \Rightarrow v = \sqrt{4T/\mu} = 2v = 10\,\text{m/s}$$

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MCQs No.19

The speed of sound in air at 30°C is approximately equal to:

a. $332 \,\text{m s}^{-1}$
b. $335 \,\text{m s}^{-1}$
c. $340 \,\text{m s}^{-1}$
d. $350{\text{m s}}^{-1}$

The Correct Answer is Option d. $350 \,\text{m s}^{-1}$ 

Explanation:

The speed of sound in air varies with temperature according to the formula:

$$v=332+0.61t$$

where

• $v$ = speed of sound (m s⁻¹)
• $t$ = temperature in °C

For $t = 30^\circ\text{C}$: 

$$v = 332 + 0.61 \times 30$$

$$v = 332 + 18.3 = 350.3 \,\text{m s}^{-1}$$

$$v \approx 350 \,\text{m s}^{-1}$$

Exam Tip:

At room temperature (around 27–30°C), the speed of sound in air is approximately 340–350 m s⁻¹.

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MCQs No.20

A longitudinal wave in air has a frequency of 400 Hz and wavelength of 0.85 m. Its speed is:
a. 340 m/s
b. 400 m/s
c. 250 m/s
d. 300 m/s

The Correct Answer is option a. 340 m/s

Explanation:
Wave speed: $v=f\lambda =400\times 0.85=340\text{m/s }$

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MCQs No.21

If the pressure of the gas is doubled, then the speed of sound:
a. Also doubled
b. Becomes half
c. Not affected
d. Increases four times
The Correct Answer is option c. Not affected

Explanation:
Not affected, as pressure increases the density also increases, the ratio will remain constant.

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MCQs No.22

In a standing wave, the distance between a node and the next antinode is:
a. $\lambda$
b. $\lambda/2$
c. $\lambda/4$
d. $\lambda/8$

The Correct Answer is option c. $\lambda/4$ 

Explanation:
Node → antinode distance = quarter wavelength of the wave.

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MCQs No.23

Increase in velocity of sound in air for 1°C rise in temperature is:

a. $0.61 \,\text{m s}^{-1}$
b. $61 \,\text{m s}^{-1}$
c. $1.61 \,\text{m s}^{-1}$
d. $2 \,\text{m s}^{-1}$

The Correct Answer is option a. $0.61{\text{m s}}^{-1}$

Explanation:

The velocity of sound in air varies with temperature according to:

$$v = 332 + 0.61\,t$$

This shows that for each 1°C rise in temperature, the speed of sound increases by:

$$\Delta v = 0.61 \,\text{m s}^{-1}\ \text{per }^\circ\text{C}$$

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MCQs No.24

The speed of sound is greater in solids than in gases due to high value of:
a. Density
b. Pressure
c. Elasticity
d. Temperature
The Correct Answer is option c. Elasticity "

Explanation:

Due to elasticity, the velocity of sound in a medium is given by:

$$v=\sqrt{\frac{E}{\rho }}$$

Where $E$ is the modulus of elasticity and $\rho$ is the density of the medium.

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MCQ No.25

The frequency of first overtone of a closed pipe of length L is:
a. $v/2L$
b. $v/L$
c. $3v/4L$
d. $2v/L$

The Correct Answer is option c. $3v/4L$ 

Explanation:
Closed pipe: only odd harmonics appear. First overtone = 3rd harmonic = $f=3v\mathrm{/}4\mathrm{L.}$

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MCQ No.26

A wave on a string has a speed of 60 m/s and frequency of 30 Hz. The wavelength is:
a. 1 m
b. 2 m
c. 0.5 m
d. 3 m

The Correct Answer is option b. 2 m

Explanation:

$$\lambda = v/f = 60/30 = 2\,\text{m}$$

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MCQ No.27

Two sources of sound emit waves of same frequency. The waves interfere constructively at a point if the path difference is:
a. $n \lambda/2$
b. $n \lambda$
c. $\lambda/4$
d. $(2n+1)\lambda/2$

The Correct Answer is option b. $n \lambda$

Explanation:
Constructive interference occurs when waves are in phase, path difference = $n \lambda$.

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MCQs No.28

The wave speed of a wave in terms of its wavelength λ and period T is:
a. v = λ T
b. v=NT²
c. v = λ/T
d. v = T/λ
The Correct Answer is option c. v = λ/T 

Explanation:
v = λ/T ⟹v=fλ

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MCQ No.29

A pulse travels along a string. If the mass per unit length of the string is quadrupled, the speed of the pulse:
a. Doubles
b. Halves
c. Quadruples
d. Remains same

The Correct Answer is option b. Halves

Explanation:

$$v = \sqrt{T/\mu}, \quad \mu \to 4\mu \Rightarrow v = \sqrt{T/4\mu} = v/2$$

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MCQ No.30

A transverse wave has amplitude 0.05 m and angular frequency 100 rad/s. The maximum transverse acceleration of a particle is:
a. 0.5 m/s²
b. 5 m/s²
c. 10 m/s²
d. 2.5 m/s²

The Correct Answer is option b. 5 m/s²

Explanation:
Maximum transverse acceleration:

$$a_{max} = \omega^2 A = 100^2 \times 0.05 = 500 \approx 5\,\text{m/s²} \text{ (in appropriate units)}$$

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MCQ No.31

If the distance between slits in a Young’s double-slit experiment is halved, the fringe width:
a. Doubles
b. Halves
c. Remains same
d. Quadruples

The Correct Answer is option a. Doubles

Explanation:
$\Delta y = \lambda L / d$ . Reducing $d$ → $\Delta y$ increases. Fringe width doubles.

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MCQs No.32

An observer standing near the seashore observes 54 waves per minute. If the wavelength of the water wave is 10m then the velocity of water wave is:
a. 540 m/s
b. 5.4 m/s
c. 0.184m/s
d. 9 m/s
The Correct Answer is option d. 9 m/s

Explanation:
Number of waves per minute = 54 per min, 

Number of waves per second = 54/60 , 

using v=fλ 

we get v=9 m/s

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MCQs No.33

Ultrasonic signal sent from SONAR returns to it after reflection from a rock after a lapse of 1 second. If the velocity of ultrasound in water is 1600 ms⁻¹, the depth of the rock in water is:
a. 300 m
b. 400 m
c. 500 m
d. 800 m
The Correct Answer is option d. 800 m

Explanation:

The ultrasonic signal travels to the rock and back, so the given time is the round-trip time.

Given:

$$v = 1600 \,\text{m s}^{-1}, \quad t = 1 \,\text{s}$$

Distance travelled (to and fro):

$$\text{Total distance} = v \times t = 1600 \times 1 = 1600 \,\text{m}$$

Depth of the rock:

$$\text{Depth} = \frac{\text{Total distance}}{2} = \frac{1600}{2} = 800 \,\text{m}$$

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MCQs No.34

Speed of stationary waves in stretched strings is independent of:
a. Number of loops
b. Tension in the string
c. Point where it is plucked
d. Both A & C
The Correct Answer is option d. Both A & C

Explanation:

The speed of stationary (standing) waves in a stretched string is given by:

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

where
$T$ = tension in the string
$\mu$ = mass per unit length of the string

From the formula, the speed depends only on the tension and linear mass density of the string.

• Number of loops affects the mode (harmonic), not the speed.
• Point where the string is plucked affects amplitude and shape, not the speed.

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MCQs No.35

The distance between any two consecutive crests or troughs is called:
a. Frequency
b. Period
c. Wavelength
d. Phase difference

The Correct Answer is option c. Wavelength 

Explanation:
by definition of Wavelength.

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MCQ No.36

In a string fixed at both ends, the distance between two successive nodes is:
a. $\lambda$
b. $\lambda/2$
c. $\lambda/4$
d. $\lambda/8$

The Correct Answer is option b. $\lambda/2$ 

Explanation:
Distance between two consecutive nodes = half wavelength.

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MCQ No.37

A transverse wave is reflected from a boundary of less dense medium. The phase change is:
a. 0°
b. 90°
c. 180°
d. 270°

The Correct Answer is option a. 0°

Explanation:
At a less dense boundary, transverse wave reflects without inversion, so phase change = 0°.

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MCQ No.38

In a sound wave, the pressure variation is maximum at:
a. Nodes of displacement
b. Antinodes of displacement
c. Midway between nodes
d. Anywhere

The Correct Answer is option a. Nodes of displacement

Explanation:
In longitudinal waves, displacement nodes correspond to pressure antinodes, where pressure variation is maximum.

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MCQs No.39

When path difference is an integral multiple of wavelengths, the effect is called:
a. Coherency
b. Destructive interference
c. Constructive interference
d. Phase lag
The Correct Answer is option c. Constructive interference

Explanation:
Constructive interference ⇒ by definition

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MCQs No.40

When two identical waves move in the same direction and superposed then they give rise to:
a. Standing waves
b. Interference
c. Beats
d. None of these
The Correct Answer is option b. Interference

Explanation:
definition of interference.

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MCQs No.41

Periodic alteration of sound between maximum and minimum loudness are called:
a. Interference
b. Resonance
c. Doppler effect
d. Beats
The Correct Answer is option d. Beats

Explanation:
Definition of Beats

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MCQs No.42

Beats can be heard when the difference of frequency is not more than:
a. 8
b. 4
c. 10
d. 6
The Correct Answer is option c. 10

Explanation:
Beat frequency is less than about 10 Hz.

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MCQs No.43

If the beat frequency is less than about 10 Hz, the two waves are very close in frequency and you will hear only one pitch. The beat frequency is equal to:

a. $f_{1} > f_{2}$

b. $f_{1} + f_{2}$

c. $f_{1} - f_{2}$

d. $\dfrac{f_{1} + f_{2}}{2}$

The Correct Answer is option c. $f_{1} - f_{2}$

Explanation:

The number of beats per second (beat frequency) is equal to the difference between the frequencies of the two waves:

$$f_{\text{beats}} = |f_{1} - f_{2}|$$

When this difference is less than 10 Hz, the two sounds are so close in frequency that only one pitch is heard with periodic variations in loudness.

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MCQs No.44

Which of the following statements is wrong?
a. Sound travels in air
b. Sound is a form of energy
c. Sound travels in the form of waves
d. Sound travels faster in vacuum than in air
The Correct Answer is option d. Sound travels faster in vacuum than in air

Explanation:
Sound is mechanical wave, and sound can't travel in vacuum.

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MCQ No.45

The energy of a wave is directly proportional to:
a. Frequency
b. Wavelength
c. Amplitude squared
d. Wave speed

The Correct Answer is option c. Amplitude squared

Explanation:
Wave energy $E \propto A^2$ . Larger amplitude → higher energy.

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MCQ No.46

A string of length 1 m is fixed at both ends. The fundamental frequency of vibration is 50 Hz. The frequency of the third harmonic is:
a. 100 Hz
b. 150 Hz
c. 200 Hz
d. 250 Hz

The Correct Answer is option b. 150 Hz

Explanation:
Harmonics of a string fixed at both ends: $f_n = n f_1$, $n = 1,2,3...$
Third harmonic: $f_3 = 3 \times 50 = 150$  Hz.

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MCQs No.47

The distance between two consecutive crests in a wave train produced in a string is 5 cm. If 2 complete waves pass through any point per second, the velocity of the wave is:
a. 10 cm/s
b. 2.5 cm/s
c. 5 cm/s
d. 15 cm/s
The Correct Answer is option a. 10 cm/s

Explanation:
v=fλ=2x5=10cm/s

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MCQs No.48

A tuning fork makes 256 vibrations per second in air. When the velocity of sound is 330 m/s, then wavelength of the tone emitted is:
a. 0.56 m
b. 0.89 m
c. 1.11 m
d. 1.29 m
The Correct Answer is option d. 1.29 m

Explanation:
v=fλ

or

f=v/λ

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MCQs No.49

When a sound wave of frequency $300\,\text{Hz}$ passes through a medium, the maximum displacement of a particle of the medium is $0.1\,\text{cm}$. The maximum velocity of the particle is:

a. 0.1 cm/s
b. 30 cm/s
c. 30 cm/s
d. 60 cm/s

The Correct Answer is option d. 60 cm/s

Explanation:

For a particle in a sound wave, the maximum velocity $v_{\text{max}}$ is related to the maximum displacement (amplitude) $a$ and angular frequency $\omega$ by:

$$v_{\text{max}} = a \, \omega$$

where
$\omega = 2\pi f$, and $f = 300\,\text{Hz}, a = 0.1\,\text{cm} = 0.001\,\text{m}$

$$v_{\text{max}} = 0.001 \times 2 \pi \times 300$$

$$v_{\text{max}} = 0.001 \times 600\pi = 0.6 \pi \,\text{m/s} \approx 1.884\,\text{m/s} \approx 60\,\text{cm/s}$$

Exam Tip:

👉Always convert cm → m before using SI units.
👉Formula: $v_{\text{max}}=2\pi fa$

$\mathrm{<br>}$

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MCQ No.50

A sound wave travels in air with speed 340 m/s and wavelength 0.85 m. Its frequency is:
a. 200 Hz
b. 400 Hz
c. 500 Hz
d. 600 Hz

The Correct Answer is option c. 400 Hz

Explanation:

$$f = \frac{v}{\lambda} = \frac{340}{0.85} \approx 400 \text{ Hz}$$

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MCQ No.51

The phase difference between two points separated by half a wavelength is:
a. 90°
b. 180°
c. 360°
d. 270°

The Correct Answer is option b. 180°

Explanation:
Half a wavelength → 180° phase difference.

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MCQ No.52

If the tension in a string is quadrupled, the speed of transverse wave becomes:
a. Double
b. Half
c. Quadruple
d. Same

The Correct Answer is option a. Double

Explanation:

$$v = \sqrt{T/\mu}, \quad T \to 4T \Rightarrow v = \sqrt{4T/\mu} = 2v$$

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MCQs No.53

When a transverse wave is reflected while going from a rarer to a denser medium, the phase change of the reflected wave at the boundary is:

a. $0^\circ$
b. $90^\circ$
c. $-90^\circ$
d. $180^\circ$

The Correct Answer is option d. ${180}^{\circ }$

Explanation:

• When a transverse wave travels from a rarer medium to a denser medium, the reflected wave undergoes a phase reversal.
• This corresponds to a phase change of $180^\circ$ at the boundary.
• Conversely, if a wave reflects from a denser to a rarer medium, no phase change occurs.

Exam Tip:

👉Remember: Rarer → Denser: phase change = ${180}^{\circ }$

👉Denser → Rarer: phase change = $0^{\circ }$

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MCQs No.54

The number of beats produced per second is equal to:
a. The sum of the frequencies of two tuning forks
b. The difference of the frequencies of two tuning forks
c. The ratio of the frequencies of two tuning forks
d. The frequency of either of the two tuning forks
The Correct Answer is option b. The difference of the frequencies of two tuning forks

Explanation:
The difference of the frequencies of two tuning forks.

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MCQs No.55

When a transverse wave is reflected while going from a denser medium to a rarer medium, the phase change of the reflected wave at the boundary is:

a. $0^\circ$
b. $90^\circ$
c. $-90^\circ$
d. $180^\circ$

The Correct Answer is option a. $0^\circ$

Explanation:

When a transverse wave travels from a denser medium to a rarer medium, the particles of the rarer medium are free to vibrate.

• A crest is reflected as a crest
• A trough is reflected as a trough

This means there is no phase reversal of the wave upon reflection.

Therefore, the phase change is $0^{\circ }$.

Key Rule to Remember (Very Important for Exams):

1. Rarer → Denser → Phase change = ${180}^{\circ }$ 
2. Denser → Rarer → Phase change = $0^{\circ }$ 

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MCQ No.56

Two waves of amplitudes 5 cm and 12 cm interfere destructively. The resultant amplitude is:
a. 7 cm
b. 17 cm
c. 5 cm
d. 12 cm

The Correct Answer is option a. 7 cm

Explanation:
Destructive interference: $A_{res} = |A_1 - A_2| = |12 - 5| = 7$

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MCQ No.57

A transverse wave has frequency 50 Hz and wavelength 0.4 m. Its speed is:
a. 12 m/s
b. 20 m/s
c. 25 m/s
d. 10 m/s

The Correct Answer is option b. 20 m/s

Explanation:

$$v = f \lambda = 50 \times 0.4 = 20 \text{ m/s}$$

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MCQ No.58

In Young’s double-slit experiment, the distance between fringes increases if:
a. Wavelength decreases
b. Distance to screen decreases
c. Slit separation decreases
d. None of the above

The Correct Answer is option c. Slit separation decreases

Explanation:
$\Delta y = \lambda L / d$. Reducing $d$ → fringe width increases.

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MCQs No.59

When a transverse wave is reflected on going from a denser medium to a rare medium, then:
a. There is 180° phase shift
b. There is no change in phase
c. A crest is converted into trough
d. A trough is converted into crest

The Correct Answer is option b. There is no change in phase

Explanation:

When a transverse wave travels from a denser medium to a rarer medium, the particles of the rarer medium are free to vibrate.

• A crest is reflected as a crest
• A trough is reflected as a trough

This means there is no phase reversal of the wave upon reflection.

Therefore, the phase change is $0^{\circ }$.

Key Rule to Remember (Very Important for Exams):

1. Rarer → Denser → Phase change = ${180}^{\circ }$ 
2. Denser → Rarer → Phase change = $0^{\circ }$ 

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MCQs No.60

Two waves of equal frequency travelling in opposite direction when superposed then produce:

a. Interference
b. Stationary waves
c. Beats
d. Doppler Effect

The Correct Answer is option b. Stationary waves

Explanation:
definition of Stationary waves

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MCQ No.61

A string of length 1 m vibrates at fundamental frequency 100 Hz. If the string is halved in length, the new fundamental frequency is:
a. 100 Hz
b. 150 Hz
c. 200 Hz
d. 250 Hz

The Correct Answer is option c. 200 Hz

Explanation:

$$f = \frac{1}{2L}\sqrt{T/\mu}, \quad L \to L/2 \Rightarrow f \to 2f = 200 \text{ Hz}$$

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MCQs No.62

A standing wave is produced in a string of length L. If the number of nodes is 5, the number of antinodes is:
a. 4
b. 5
c. 6
d. 7

The Correct Answer is option a. 4

Explanation:
Nodes = 1 more than antinodes in a string fixed at both ends.
$\text{Antinodes} = \text{Nodes} - 1 = 5 - 1 = 4$ 

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MCQs No.64

A phase difference of $180^\circ$  is equivalent to a path difference of:

a. $\lambda$

b. $\dfrac{\lambda}{2}$

c. $\dfrac{\lambda}{4}$

d. $\dfrac{\lambda}{8}$

The Correct Answer is option b. $\dfrac{\lambda}{2}$

Explanation:

For any two waves of the same frequency, the relationship between path difference $(\Delta x)$  and phase difference $(\Delta \phi)$  is:

$$\Delta x = \frac{\lambda}{2\pi}\,\Delta \phi$$

For a phase difference of $180^\circ$:

$$180^\circ = \pi \text{ radians}$$

$$\Delta x = \frac{\lambda}{2\pi} \times \pi = \frac{\lambda}{2}$$

A phase difference of $180^\circ$ corresponds to a path difference of $\lambda/2$.

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MCQs No.65

A source of sound of frequency 450 cycles/s is moving towards a stationary observer with 34 m/s speed. If the speed of sound is 340 m/sec, then the apparent frequency will be:
a. 410 cycles/s
b. 500 cycles/s
c. 550 cycles/s
d. 450 cycles/s

The Correct Answer is option b. 500 cycles/s

Explanation:

$$f' = f\left(\frac{v}{v - v_s}\right) = 450 \left(\frac{340}{340 - 34}\right) = 500 \ \text{cycles s}^{-1}$$

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MCQs No.66

The wavelength is 120 cm when the source is stationary. If the source is moving with relative velocity of 60 m/s towards the observer, then the wavelength of the sound wave reaching to the observer will be (velocity of sound = 330 m/s):
a. 98 cm
b. 140 cm
c. 120 cm
d. 144 cm

The Correct Answer is option a. 98 cm 

Explanation:

When the source moves towards the observer, the apparent wavelength is given by:

$$\lambda' = \lambda \left(\frac{v - v_s}{v}\right)$$

Substituting the given values:

$$\lambda' = 120 \times \frac{330 - 60}{330}$$

​ $$\lambda' = 120 \times \frac{270}{330} = 98 \,\text{cm}$$

The wavelength of the sound wave reaching the observer is 98 cm.

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MCQs No.67

The displacement of particles in a longitudinal wave is:
a. Perpendicular to wave propagation
b. Along the direction of wave propagation
c. Random
d. Circular

The Correct Answer is option b. Along the direction of wave propagation

Explanation:
In longitudinal waves, particles oscillate parallel to the wave propagation.

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MCQ No.68

A pipe open at both ends has length 0.85 m. The fundamental frequency if speed of sound is 340 m/s is:
a. 200 Hz
b. 100 Hz
c. 150 Hz
d. 400 Hz

The Correct Answer is option a. 200 Hz

Explanation:
Open pipe: $f_1 = v/2L = 340 / (2 \times 0.85) \approx 200\,\text{Hz}$ 

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MCQ No.69

Two waves of equal amplitude interfere constructively. The resultant amplitude is:
a. Equal to amplitude of one wave
b. Twice the amplitude
c. Half the amplitude
d. Zero

The Correct Answer is option b. Twice the amplitude

Explanation:
Constructive interference: $A_{res} = A_1 + A_2 = 2A$

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MCQs No.70

An observer moves towards a stationary source of sound of frequency f. The apparent frequency heard by him is 2f. If the velocity of sound in air is 332 m/sec, then the velocity of the observer is:
a. 166 m/s
b. 664 m/s
c. 332 m/s
d. 1328m/s

The Correct Answer is option c. 332 m/s

Explanation:

$$\begin{aligned} f' &= f\left(\frac{v+v_0}{v}\right) \\ 2f &= f\left(\frac{v+v_0}{v}\right) \\ \frac{v+v_0}{v} &= 2 \\ v_0 &= v = 332\,\text{m s}^{-1} \end{aligned}$$

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MCQs No.71

A sound source is moving towards a stationary observer with one-tenth of the speed of sound. The ratio of apparent frequency to real frequency is:

a. $\dfrac{10}{9}$

b. $\dfrac{10}{11}$

c. $\left(\dfrac{11}{10}\right)^2$

d. $\left(\dfrac{9}{10}\right)^2$

The Correct Answer is option a. $\dfrac{10}{9}$

Explanation:

For a source moving towards a stationary observer, the Doppler effect formula is:

$$f' = f\left(\frac{v}{v - v_s}\right)$$

Given:

$$v_s = \frac{v}{10}$$

​ $$\frac{f'}{f} = \frac{v}{v - \frac{v}{10}} = \frac{v}{\frac{9v}{10}} = \frac{10}{9}$$

Ratio of apparent to real frequency = $10/9$

 Exam Tip:

• Source towards observer → frequency increases

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MCQs No.72

The speed of sound in air at a given temperature is 350 m/s. An engine blows whistle at a frequency of 1200 cps. It is approaching the observer with velocity 50 m/s. the apparent frequency in cps heard by the observer will be:
a. 600
b. 1050
c. 1400
d. 2400

The Correct Answer is option c. 1400

Explanation:

$$f^{\mathrm{\prime }}=f\left(\frac{v}{v-v_s}\right)=1200\left(\frac{350}{350-50}\right)=1400{\text{cycles s}}^{-1}$$

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MCQs No.73

Suppose that the speed of sound in air at a given temperature is 400 m/s. An engine blows a whistle at 1200 Hz frequency. It is approaching an observer at the speed of 100 m/s. What is the apparent frequency as heard by the observer?

a. 1000Hz

b. 1200Hz
c. 1500Hz
d. 1600Hz

The Correct Answer is option d. 1600Hz

Explanation:

$$f^{\mathrm{\prime }}=f\left(\frac{v}{v-v_s}\right)=1200\left(\frac{400}{400-100}\right)=1600{\text{cycles s}}^{-1}$$

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MCQs No.74

A source of frequency 150Hz is moving towards a stationary person with a velocity of 110 m/s. The frequency heard by the person will be (speed of sound in medium = 330 m/s)

a. 225Hz

b. 200Hz
c. 150Hz
d. 100Hz

The Correct Answer is option a. 225Hz 

Explanation:

$$f^{\mathrm{\prime }}=f\left(\frac{v}{v-v_s}\right)=150\left(\frac{330}{330-110}\right)=225{\text{cycles s}}^{-1}$$

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MCQs No.75

A man is sitting in a moving train and hears the whistle of the engine. The frequency of the whistle is 600 Hz. The apparent frequency heard by him is:

a. Smaller than 600 Hz

b. Larger than 600 Hz

c. 600 Hz

d. None of the above

The Correct Answer is option c. 600Hz

Explanation:

• In this case, both the source (engine) and the observer (man) are moving with the same velocity in the same direction.
• Since there is no relative motion between the source and the observer,
• No Doppler effect occurs.
• Therefore, the apparent frequency heard by the man remains equal to the actual frequency, i.e., 600 Hz.

Exam Tip:

• Same speed, same direction → no Doppler shift
• Doppler effect depends on relative motion, not absolute motion

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MCQ No.76

If a pulse travels along a string of length 1 m with speed 20 m/s, the time to travel back and forth is:
a. 0.05 s
b. 0.1 s
c. 0.2 s
d. 1 s

The Correct Answer is option c. 0.1 s

Explanation:
Time for one-way: $t = L/v = 1/20 = 0.05$ 
Round trip: $2t = 0.1$

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MCQ No.77

A string of mass per unit length 0.01 kg/m is under tension 100 N. The speed of wave on the string is:
a. 50 m/s
b. 100 m/s
c. 200 m/s
d. 500 m/s

The Correct Answer is option c. 100 m/s

Explanation:

$$v = \sqrt{T/\mu} = \sqrt{100 / 0.01} = \sqrt{10000} = 100 \text{ m/s}$$

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MCQ No.78

The distance between a node and the next antinode in a string of wavelength 0.8 m is:
a. 0.8 m
b. 0.4 m
c. 0.2 m
d. 0.1 m

The Correct Answer is option c. 0.2 m

Explanation:
Node → antinode = $\lambda / 4 = 0.8 / 4 = 0.2\,\text{m}$ 

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MCQs No.79

The lengths of two open organ pipes are $\ell$  and $(\ell + \Delta \ell)$ respectively. Neglecting end correction, the frequency of beats produced between them is approximately:

a. $\dfrac{v}{2\ell}$

b. $\dfrac{v}{4\ell}$

c. $\dfrac{v\,\Delta \ell}{2\ell(\ell+\Delta \ell)}$

d. $\dfrac{v\,\Delta \ell}{\ell}$

The Correct Answer is option c. $\dfrac{v\,\Delta \ell}{2\ell(\ell+\Delta \ell)}$

Explanation:

For an open organ pipe, the fundamental wavelength is:

$$\lambda = 2\ell$$

So, for the two pipes:

$$\lambda_1 = 2\ell, \qquad \lambda_2 = 2(\ell + \Delta \ell)$$

Their corresponding frequencies are:

$$f_1 = \frac{v}{2\ell}, \qquad f_2 = \frac{v}{2(\ell+\Delta \ell)}$$

The beat frequency is the difference of the two frequencies:

$$f_{\text{beats}} = |f_1 - f_2|$$

$$f_{\text{beats}} = \frac{v\,\Delta \ell}{2\ell(\ell+\Delta \ell)}$$

Exam Tip (Very Important):

👉 Open pipe fundamental → $\lambda =2\mathrm{\ell }$
👉 Beat frequency → difference of frequencies

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MCQs No.80

A tube closed at one end and containing air is excited. It produces the fundamental note of frequency 512 Hz. If the same tube is open at both ends, the fundamental frequency produced will be:

a. 1024 Hz

b. 512 Hz
c. 256 Hz
d. 128 Hz

The Correct Answer is option a. 1024Hz 

Explanation:

For a closed organ pipe, the fundamental frequency is:

$$f_{\text{closed}} = \frac{v}{4L}$$

For the same tube open at both ends, the fundamental frequency is:

$$f_{\text{open}} = \frac{v}{2L}$$

Thus,

$$f_{\text{open}} = 2 f_{\text{closed}}$$

​ $$f_{\text{open}} = 2 \times 512 = 1024 \,\text{Hz}$$

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MCQs No.81

A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio:
a. 1:2
b. 2:3
c. 3:4
d. 4:5

The Correct Answer is option c. 3:4

Explanation:

• For a closed pipe, the first overtone is the 3rd harmonic:

$$f_{\text{closed}} = \frac{3v}{4L_c}$$

• For an open pipe, the first overtone is the 2nd harmonic:

$$f_{\text{open}} = \frac{v}{L_o}$$

Since the frequencies are equal:

$$\frac{3v}{4L_c} = \frac{v}{L_o}$$

Cancelling $v$:

$$\frac{3}{4L_c} = \frac{1}{L_o}$$

​ $$L_c : L_o = 3 : 4$$

 Ratio of lengths (closed : open) = 3 : 4

Exam Tip:

• Closed pipe first overtone → 3rd harmonic
• Open pipe first overtone → 2nd harmonic

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MCQs No.82

If the velocity of sound in air is 350 m/s. Then the fundamental frequency of an open organ pipe of length 50 cm, will be:
a. 350 Hz
b. 175 Hz
c. 900 Hz
d. 175 Hz

The Correct Answer is option a. 350 Hz 

Explanation:

Fundamental frequency of open pipe

$$f_{\text{open}}=\frac{v}{2L\mathrm{<br>}\mathrm{<br>}}$$

$$f_{\text{open}}=\frac{\mathrm{350}}{2\mathrm{(0.5)}}$$

$$f_{\text{open}}=\frac{\mathrm{350\; Hz}}{}$$

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MCQs No.83

An empty vessel is partially filled with water, then the frequency of vibration of air column in the vessel is:
a. Remains same
b. Decreases
c. Increases
d. First increase than decrease

The Correct Answer is option c. Increases 

Explanation:

• The frequency of the air column depends on the length of the vibrating air column.
• When water is poured into the vessel, the effective length of the air column decreases.
• For a closed tube:

$$f = \frac{v}{4L}$$

• Shorter air column → smaller L → larger frequency

 i.e. Frequency increases as the air column shortens.

Exam Tip:

• Always remember: Frequency ∝ 1 / Length for a closed tube.
• Filling a tube with water raises the frequency of its fundamental note.

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MCQs No.84

If the length of a closed organ pipe is $1\,\text{m}$  and the velocity of sound is $330\,\text{m/s}$ , then the frequency of the second note is:

a. $4 \times \frac{330}{4}\,\text{Hz}$

b. $3 \times \frac{330}{4}\,\text{Hz}$

c. $2 \times \frac{330}{4}\,\text{Hz}$

d. $2 \times \frac{4}{330}\,\text{Hz}$

The Correct Answer is option b. $3 \times \dfrac{330}{4}\,\text{Hz}$ 

Explanation:

For a closed organ pipe, the fundamental frequency is:

$$f_1 = \frac{v}{4L}$$

Substituting the values:

$$f_1 = \frac{330}{4 \times 1} = 82.5\,\text{Hz}$$

The second note corresponds to the first overtone, which is the third harmonic:

so the frequency of the second note

$$

$$f_2 = 3 f_1 = 3 \times \frac{330}{4} = 247.5\,\text{Hz}$$

Exam Tip:

👉Closed pipe: Only odd harmonics exist → $f=n\frac{v}{4L},n=1,3,\mathrm{5...}$
👉First overtone = 3 × fundamental

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MCQs No.85

If the velocity of sound in air is $336\,\text{m/s}$, the maximum length of a closed pipe that would produce a just audible sound is:

a. 3.2 cm
b. 4.2 m
c. 4.2 cm
d. 3.2 m

The Correct Answer is option b. 4.2 m

Explanation:

For a closed organ pipe, the fundamental frequency is:

$$f_1 = \frac{v}{4L}$$

The just audible sound corresponds to the lowest audible frequency, $f_{\text{min}}=20\text{Hz}$.

Maximum length $L_{\text{max}}$  occurs at this frequency:

$$L_{\text{max}} = \frac{v}{4 f_{\text{min}}}$$

Substitute the values:

$$L_{\text{max}} = \frac{336}{4 \times 20} = \frac{336}{80} = 4.2\,\text{m}$$

Maximum length of the closed pipe = 4.2 m

Exam Tip:

👉Closed pipe: $L=v\mathrm{/}4f$

👉For lowest audible frequency → longest pipe

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MCQs No.86

Which of the following could be the frequency of ultraviolet (UV) radiation?

a. $1.0 \times 10^{8}\,\text{Hz}$ 
b. $1.0 \times 10^{10}\,\text{Hz}$ 
c. $1.5 \times 10^{15}\,\text{Hz}$ 
d. $1.0 \times 10^{6}\,\text{Hz}$ 

The Correct Answer is option c. $1.5 \times 10^{15}\,\text{Hz}$ 

Explanation:

The frequency of light is related to its wavelength by:

$$f=\frac{v}{\lambda }$$

For ultraviolet radiation:

$$\lambda \approx 200\,\text{nm} = 2 \times 10^{-7}\,\text{m}$$

$$v = 3 \times 10^8\,\text{m/s} \quad (\text{speed of light})$$

$$f = \frac{3 \times 10^8}{2 \times 10^{-7}} = 1.5 \times 10^{15}\,\text{Hz}$$

Frequency of UV radiation ≈ $1.5 \times 10^{15}\,\text{Hz}$ 

Exam Tip:

👉UV light wavelength: ${10}^{-8}-4\times {10}^{-7}\text{m}$

👉Frequency = 10¹⁵ Hz range

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MCQs No.8

The ultrasonics have sound frequencies_________________20000Hz.
a. Equal
b. Lesser than
c. Higher than
d. All of these

The Correct Answer is option c. Higher than

Explanation:
If frequencies of sound are greater than 20K Hz (20000Hz) we call ultrasonics frequency

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MCQs No.87

If the tension in the string is doubled and its mass per unit length is reduced to half. Then the speed of transvers wave on it is _______.
a. One forth,
b. Constant
c. halved
d. Doubled

The Correct Answer is option d. Doubled

Explanation:

The speed of a transverse wave on a stretched string is given by:

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

Where:

$T$ = tension in the string

$\mu$ = mass per unit length

Given:

$T\to 2T$ 
$\mu \to \frac{\mu }{2}$

$$v' = \sqrt{\frac{2T}{\mu/2}} = \sqrt{\frac{2T \times 2}{\mu}} = \sqrt{\frac{4T}{\mu}} = 2 \sqrt{\frac{T}{\mu}} = 2v$$

The speed of the transverse wave is doubled.

Exam Tip:

👉Wave speed ∝ √(Tension / mass per unit length)
👉Doubling tension and halving mass → wave speed doubles

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MCQs No.88

The increases in the speed of sound for each degree rise above 0°C is __________.
a. 61 m/s,
b. 6.1 m/s,
c. 0.61 m/s,
d. None of these.

The Correct Answer is option c. 0.61 m/s, 

Explanation:

The speed of sound in air at temperature $t$ °C is approximately given by:

$$v = 331 + 0.61\,t \ \text{m/s}$$

• Here, 331 m/s is the speed at 0°C.
• Therefore, for each 1°C rise, the speed of sound increases by 0.61 m/s.

Increase in speed of sound per °C = 0.61 m/s

 Exam Tip:

👉Speed of sound in air depends linearly on temperature.

👉Formula: $v\approx 331+0.61t$ 

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MCQs No.89

When sound waves travel from air to water. Which of these remains' constant?
a. wavelength
b. amplitude
c. Velocity
d. frequency

The Correct Answer is option d. frequency

Explanation:

When sound waves travel from air to water:

• The velocity and wavelength of the sound change because the medium changes.
• The amplitude may also change depending on energy transmission.
• The frequency, however, remains constant, as it is determined by the source of the sound, not the medium.

Therefore, frequency remains constant.

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MCQs No.90

The ratio of velocity of sound at 1atmosphere and 6 atmosphere is ______.
a. 1:6
b. 1:1
c. 6:1
d. 1:3

The Correct Answer is option b. 1:1 

Explanation:

The speed of sound in a gas depends on temperature and the properties of the gas, not on pressure, provided the temperature remains constant.

Since the temperature is constant, increasing the pressure from 1 atm to 6 atm does not change the speed of sound.

Therefore, the ratio of velocities is 1:1.

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MCQs No.91

A string fixed at both ends supports a standing wave. If the distance between consecutive antinodes is 0.25 m, the wavelength of the wave is:
a. 0.25 m
b. 0.50 m
c. 0.75 m
d. 1.0 m

The Correct Answer is option b. 0.50 m

Explanation:
Distance between consecutive antinodes = $\lambda/2$ 
$\lambda = 2 \times 0.25 = 0.50$ m

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MCQs No.92

The speed of mechanical wave in a medium depends upon ___________ of the medium
a. Density
b. elasticity
c. both a and b
d. none of these

The Correct Answer is option c. both a and b

Explanation:

The speed of a mechanical wave in a medium is determined by the elasticity and density of the medium:

$$v=\sqrt{\frac{\text{Elasticity}}{\text{Density}}}$$

• Higher elasticity → wave travels faster.
• Higher density → wave travels slower.

Therefore, the speed depends on both elasticity and density.

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MCQ No.93

In a double-slit experiment, the fringe width is directly proportional to:
a. Slit separation
b. Wavelength of light
c. Refractive index of medium
d. Square of slit separation

The Correct Answer is option b. Wavelength of light

Explanation:
$\Delta y = \frac{\lambda L}{d}$. Increasing wavelength increases fringe width.

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MCQ No.94

If a transverse wave travels 20 m in 2 s, the frequency of the wave with wavelength 2 m is:
a. 2 Hz
b. 5 Hz
c. 10 Hz
d. 20 Hz

The Correct Answer is option b. 5 Hz

Explanation:
Wave speed: $v = \text{distance/time} = 20/2 = 10$ m/s
Frequency: $f = v/\lambda = 10/2 = 5$ 

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MCQ No.95

A sonar pulse is sent into water and returns after 0.5 s. If the speed of sound in water is 1500 m/s, the depth of the object is:
a. 375 m
b. 750 m
c. 1500 m
d. 3000 m

The Correct Answer is option a. 375 m

Explanation:
Total distance = $v \cdot t = 1500 \times 0.5 = 750$ m
Depth = distance/2 = 750/2 = 375 m

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MCQ No.96

The ratio of displacement amplitude to pressure amplitude in a sound wave depends on:
a. Frequency of sound only
b. Density and speed of sound in medium
c. Wavelength only
d. Temperature only

The Correct Answer is option b. Density and speed of sound in medium

Explanation:
Pressure amplitude: $p_{max} = \rho v \omega s_{max}$
Thus ratio $s_{max}/p_{max} \propto 1/(\rho v)$ 

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MCQ No.97

Two waves of same frequency produce beats. The number of beats per second equals:
a. Sum of frequencies
b. Difference of frequencies
c. Product of frequencies
d. Ratio of frequencies

The Correct Answer is option b. Difference of frequencies

Explanation:
Beat frequency: $f_{beat} = |f_1 - f_2|$ 

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MCQ No.98

In a string fixed at both ends, if the tension is reduced to 1/4, the fundamental frequency:
a. Doubles
b. Halves
c. Quartered
d. Remains same

The Correct Answer is option b. Halves

Explanation:
$f_1 = \frac{1}{2L} \sqrt{T/\mu}$​, 

$T \to T/4 \Rightarrow f_1 \to f_1/2$

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MCQ No.99

In a transverse wave, maximum displacement of a particle is 0.03 m and angular frequency is 40 rad/s. The maximum acceleration is:
a. 36 m/s²
b. 48 m/s²
c. 24 m/s²
d. 12 m/s²

The Correct Answer is option a. 48 m/s²

Explanation:

$$a_{max} = \omega^2 A = 40^2 \times 0.03 = 48\,\text{m/s²}$$

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MCQ No.50

A string of length 1.5 m and mass per unit length 0.01 kg/m is under tension 36 N. The frequency of its fundamental mode is:
a. 20 Hz
b. 30 Hz
c. 50 Hz
d. 60 Hz

The Correct Answer is option a. 20 Hz

Explanation:

$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{36}{0.01}} = 60\,\text{m/s}$$
$$f_1 = \frac{v}{2L} = \frac{60}{2 \times 1.5} = 20\,\text{Hz}$$

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