Waves Motion MCQs (Level 2) – 100 Advanced & Numerical MCQs with Answers

Waves Motion MCQs (Level 2) – 100 Advanced & Numerical MCQs with Answers

100 Advanced & Numerical MCQs (Level -2) on Waves Motion, Physics (Unit-Wise MCQs Practice): 



Whether you are preparing for board examinations, chapter tests, college assessments, or competitive entrance exams (MDCAT, ECAT, NUST, PIEAS, GIKI, UET, FAST, and other engineering or medical admission tests), this comprehensive Wave Motion MCQ Collection is designed to help you master every important concept of wave motion and sound. The questions are arranged progressively—from basic concepts to advanced numerical problems and higher-order thinking—ensuring complete and systematic preparation for every type of examination.

This chapter-wise MCQ collection includes:

  • 100 Basic MCQs Level-1 (1–100) – Covering the fundamental concepts of wave motion, types of waves, wave characteristics, sound waves, wave propagation, reflection, refraction, diffraction, interference, standing waves, Doppler Effect, ultrasound, and SONAR.
  • 100 Advanced & Numerical MCQs (101–200) – Focusing on wave equations, speed of sound calculations, beats, harmonics, organ pipes, vibrating strings, Doppler Effect numericals, ultrasound, SONAR, echo calculations, and practical problem-solving.
  • 50 Higher-Order Thinking Skills (HOTS) MCQs (201–250) – Designed to strengthen analytical reasoning, conceptual understanding, assertion–reason questions, experimental analysis, real-life applications, and multi-concept problem-solving skills.
  • 50 Challenging MCQs Quiz with Answers – A carefully selected mix of conceptual, numerical, and HOTS questions designed for quick revision, self-assessment, and complete exam preparation.

This MCQ collection covers:

  • Fundamentals of Wave Motion
  • Mechanical and Electromagnetic Waves
  • Transverse and Longitudinal Waves
  • Wave Characteristics (Amplitude, Wavelength, Frequency, Time Period, Wave Speed)
  • Wave Equation (v=fλ)(v=f\lambda)
  • Propagation of Sound Waves
  • Speed of Sound in Solids, Liquids, and Gases
  • Effect of Temperature on the Speed of Sound
  • Newton's Formula and Laplace Correction
  • Reflection, Refraction, and Diffraction of Waves
  • Principle of Superposition
  • Interference of Waves
  • Constructive and Destructive Interference
  • Beats and Beat Frequency
  • Standing Waves
  • Nodes and Antinodes
  • Vibrating Strings
  • Harmonics and Overtones
  • Open and Closed Organ Pipes
  • Doppler Effect
  • Ultrasound and Piezoelectric Effect
  • Medical Applications of Ultrasound
  • SONAR and Echo Sounding
  • Echolocation in Bats and Dolphins
  • Real-Life Applications of Wave Motion

Every MCQ includes the correct answer along with a clear, concept-based explanation to strengthen understanding, improve problem-solving skills, and reinforce important physics concepts.

This question bank helps students to:

  • Build a strong conceptual foundation in Wave Motion
  • Master wave properties and sound propagation
  • Improve numerical and analytical problem-solving skills
  • Understand interference, standing waves, harmonics, and resonance
  • Strengthen concepts of the Doppler Effect, ultrasound, and SONAR
  • Avoid common examination mistakes
  • Increase speed, accuracy, and confidence in objective-type questions
  • Prepare effectively for both board examinations and competitive entrance tests

With 250 carefully selected MCQs arranged into 100 Basic, 100 Advanced & Numerical, and 50 Higher-Order Thinking (HOTS) questions, along with a Top 50 Challenging MCQs Quiz, this all-in-one Wave Motion MCQ Bank provides complete chapter preparation. It is an excellent study resource for strengthening concepts, improving exam performance, and achieving success in both board examinations and competitive physics entrance tests.


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Waves Motion MCQs (Level 2) – 100 Advanced & Numerical MCQs (MCQs 101–200)


MCQs 101–125: 

Topics Covered:

  • Wave Equation
  • Frequency, Time Period & Wavelength
  • Wave Speed
  • Numerical Applications
  • Conceptual Reasoning

MCQ No. 101

A wave has a frequency of 25 Hz and a wavelength of 8 m. Its speed is:

a) 100 m/s

b) 150 m/s

c) 200 m/s

d) 250 m/s

Correct Answer: c) 200 m/s

Explanation: Using v=fλv=f\lambda:

v=25×8=200 m/sv=25\times8=200\ \text{m/s}

MCQ No. 102

A sound wave travels at 330 m/s and has a frequency of 110 Hz. Its wavelength is:

a) 2 m

b) 3 m

c) 4 m

d) 5 m

Correct Answer: b) 3 m

Explanation:

λ=vf=330110=3 m\lambda=\frac{v}{f}=\frac{330}{110}=3\ \text{m}

MCQ No. 103

A wave has a wavelength of 0.5 m and travels with a speed of 150 m/s. The frequency is:

a) 75 Hz

b) 150 Hz

c) 300 Hz

d) 450 Hz

Correct Answer: c) 300 Hz

Explanation:

f=vλ=1500.5=300 Hzf=\frac{v}{\lambda}=\frac{150}{0.5}=300\ \text{Hz}

MCQ No. 104

The time period of a wave having a frequency of 40 Hz is:

a) 0.20 s

b) 0.025 s

c) 0.04 s

d) 0.40 s

Correct Answer: b) 0.025 s

Explanation:

T=140=0.025 sT=\frac{1}{40}=0.025\ \text{s}

MCQ No. 105

A wave completes one oscillation in 0.01 s. Its frequency is:

a) 10 Hz

b) 50 Hz

c) 100 Hz

d) 200 Hz

Correct Answer: c) 100 Hz

Explanation:

f=1T=10.01=100 Hzf=\frac1T=\frac1{0.01}=100\ \text{Hz}

MCQ No. 106

The wavelength of a wave is doubled while its frequency remains constant. The wave speed becomes:

a) Half

b) Double

c) Four times

d) Unchanged

Correct Answer: b) Double

Explanation: Since v=fλv=f\lambda, doubling wavelength doubles the speed when frequency is constant.


MCQ No. 107

The frequency of a wave is doubled while its wavelength remains constant. The speed becomes:

a) Half

b) Double

c) Unchanged

d) One-fourth

Correct Answer: b) Double

Explanation: Wave speed is directly proportional to frequency if wavelength remains constant.


MCQ No. 108

A sound wave has a wavelength of 1.5 m and a frequency of 240 Hz. Its speed is:

a) 180 m/s

b) 240 m/s

c) 300 m/s

d) 360 m/s

Correct Answer: d) 360 m/s

Explanation:

v=240×1.5=360 m/sv=240\times1.5=360\ \text{m/s}

MCQ No. 109

A wave travels 1200 m in 4 s. If its frequency is 100 Hz, the wavelength is:

a) 1 m

b) 3 m

c) 6 m

d) 12 m

Correct Answer: b) 3 m

Explanation:

v=12004=300 m/sv=\frac{1200}{4}=300\ \text{m/s}
λ=300100=3 m\lambda=\frac{300}{100}=3\ \text{m}

MCQ No. 110

If the wavelength decreases by 20% while frequency remains constant, the wave speed:

a) Increases by 20%

b) Decreases by 20%

c) Remains constant

d) Doubles

Correct Answer: b) Decreases by 20%

Explanation: Speed is directly proportional to wavelength when frequency remains unchanged.


MCQ No. 111

A tuning fork of 512 Hz produces sound in air moving at 340 m/s. The wavelength is closest to:

a) 0.33 m

b) 0.50 m

c) 0.66 m

d) 1.00 m

Correct Answer: c) 0.66 m

Explanation:

λ=3405120.66 m\lambda=\frac{340}{512}\approx0.66\ \text{m}

MCQ No. 112

The frequency of a wave depends upon:

a) The source of vibration

b) The medium

c) The amplitude

d) The wave speed

Correct Answer: a) The source of vibration

Explanation: The frequency is determined by the vibrating source and does not change when the wave enters another medium.


MCQ No. 113

A wave enters another medium. Its speed decreases by 25%. The frequency remains constant. The wavelength:

a) Increases by 25%

b) Decreases by 25%

c) Remains constant

d) Doubles

Correct Answer: b) Decreases by 25%

Explanation: Since v=fλv=f\lambda , wavelength changes in the same ratio as speed.


MCQ No. 114

The time period of a wave is 0.005 s. Its wavelength is 2 m. The speed is:

a) 200 m/s

b) 300 m/s

c) 400 m/s

d) 500 m/s

Correct Answer: c) 400 m/s

Explanation:

f=10.005=200 Hzf=\frac1{0.005}=200\ \text{Hz}
v=200×2=400 m/sv=200\times2=400\ \text{m/s}

MCQ No. 115

Which quantity remains unchanged when a wave is reflected?

a) Frequency

b) Direction

c) Phase

d) Position

Correct Answer: a) Frequency

Explanation: Reflection changes direction but not the frequency determined by the source.


MCQ No. 116

A wave travels 600 m in 2 s. If its wavelength is 4 m, the frequency is:

a) 50 Hz

b) 75 Hz

c) 100 Hz

d) 150 Hz

Correct Answer: b) 75 Hz

Explanation:

v=6002=300 m/sv=\frac{600}{2}=300\ \text{m/s}
f=3004=75 Hzf=\frac{300}{4}=75\ \text{Hz}

MCQ No. 117

The wavelength of a 500 Hz sound wave travelling at 350 m/s is:

a) 0.50 m

b) 0.60 m

c) 0.70 m

d) 0.80 m

Correct Answer: c) 0.70 m

Explanation:

λ=350500=0.70 m\lambda=\frac{350}{500}=0.70\ \text{m}

MCQ No. 118

Which graph is used to determine the wavelength directly?

a) Velocity-Time graph

b) Displacement-Distance graph

c) Force-Time graph

d) Pressure-Time graph

Correct Answer: b) Displacement-Distance graph

Explanation: Wavelength is measured as the distance between two successive points in the same phase.


MCQ No. 119

A wave has a frequency of 60 Hz and a period of:

a) 0.60 s

b) 0.06 s

c) 0.0167 s

d) 0.00167 s

Correct Answer: c) 0.0167 s

Explanation:

T=160=0.0167 sT=\frac1{60}=0.0167\ \text{s}

MCQ No. 120

A wave has a speed of 500 m/s and a wavelength of 10 m. Its frequency is:

a) 25 Hz

b) 40 Hz

c) 50 Hz

d) 60 Hz

Correct Answer: c) 50 Hz

Explanation:

f=50010=50 Hzf=\frac{500}{10}=50\ \text{Hz}

MCQ No. 121

If both the frequency and wavelength of a wave are doubled, the speed becomes:

a) Half

b) Double

c) Four times

d) Unchanged

Correct Answer: c) Four times

Explanation: Since v=fλv=f\lambda , doubling both quantities makes the speed four times greater.


MCQ No. 122

A sound wave has a wavelength of 0.8 m and frequency 425 Hz. The speed is:

a) 300 m/s

b) 320 m/s

c) 340 m/s

d) 360 m/s

Correct Answer: c) 340 m/s

Explanation:

v=425×0.8=340 m/sv=425\times0.8=340\ \text{m/s}

MCQ No. 123

A wave with frequency 250 Hz travels at 500 m/s. How many wavelengths are contained in 20 m?

a) 5

b) 8

c) 10

d) 12

Correct Answer: c) 10

Explanation:

λ=500250=2 m\lambda=\frac{500}{250}=2\ \text{m}
202=10\frac{20}{2}=10

MCQ No. 124

Which quantity is inversely proportional to the time period?

a) Wavelength

b) Frequency

c) Speed

d) Amplitude

Correct Answer: b) Frequency

Explanation:

f=1Tf=\frac1T

MCQ No. 125

A wave completes 800 oscillations in 4 seconds. If its wavelength is 1.5 m, the wave speed is:

a) 150 m/s

b) 250 m/s

c) 300 m/s

d) 450 m/s

Correct Answer: c) 300 m/s

Explanation:

f=8004=200 Hzf=\frac{800}{4}=200\ \text{Hz}
v=200×1.5=300 m/sv=200\times1.5=300\ \text{m/s}

MCQs 126–150

Topics Covered:

  • Speed of Sound
  • Newton's Formula
  • Laplace Correction
  • Effect of Temperature
  • Pressure & Density
  • Mixed Numerical Problems

MCQ No. 126

The speed of sound in air at 0°C is approximately:

a) 300 m/s

b) 331 m/s

c) 343 m/s

d) 360 m/s

Correct Answer: b) 331 m/s

Explanation: The speed of sound in dry air at 0°C is approximately 331 m/s.


MCQ No. 127

The speed of sound increases by approximately ______ for every 1°C rise in temperature.

a) 0.16 m/s

b) 0.34 m/s

c) 0.61 m/s

d) 1.61 m/s

Correct Answer: c) 0.61 m/s

Explanation: The speed of sound in air increases by about 0.61 m/s per °C.


MCQ No. 128

What is the approximate speed of sound at 20°C?

a) 331 m/s

b) 337 m/s

c) 343 m/s

d) 350 m/s

Correct Answer: c) 343 m/s

Explanation:

v=331+(0.61×20)v = 331 + (0.61 \times 20)
v=343.2343 m/sv = 343.2 \approx 343\ \text{m/s}

MCQ No. 129

The speed of sound at 30°C is closest to:

a) 331 m/s

b) 343 m/s

c) 349 m/s

d) 355 m/s

Correct Answer: c) 349 m/s

Explanation:

331+(0.61×30)=349.3 m/s331 + (0.61 \times 30)=349.3\ \text{m/s}

MCQ No. 130

At 10°C, the speed of sound is approximately:

a) 331 m/s

b) 337 m/s

c) 343 m/s

d) 349 m/s

Correct Answer: b) 337 m/s

Explanation:

331+(0.61×10)=337.1 m/s331 + (0.61 \times10)=337.1\ \text{m/s}

MCQ No. 131

According to Newton's formula, the speed of sound is:

a) γPρ\sqrt{\frac{\gamma P}{\rho}}

b) Pρ\sqrt{\frac{P}{\rho}}

c) fλf\lambda

d) λT\frac{\lambda}{T}

Correct Answer: b) Pρ\sqrt{\frac{P}{\rho}}

Explanation: Newton assumed sound propagation to be an isothermal process.


MCQ No. 132

Newton's formula gives a value lower than the experimental value because it assumes the process is:

a) Adiabatic

b) Isothermal

c) Isobaric

d) Isochoric

Correct Answer: b) Isothermal

Explanation: This incorrect assumption caused Newton's formula to underestimate the speed of sound.


MCQ No. 133

Laplace corrected Newton's formula by introducing:

a) Frequency

b) Amplitude

c) Ratio of specific heats (γ)

d) Wavelength

Correct Answer: c) Ratio of specific heats (γ)

Explanation: Laplace assumed sound propagation is adiabatic and multiplied the bulk modulus by γ.


MCQ No. 134

The corrected formula for the speed of sound in gases is:

a) v=fλv=f\lambda

b) v=Pρv=\sqrt{\frac{P}{\rho}}

c) v=γPρv=\sqrt{\frac{\gamma P}{\rho}}

d) v=λTv=\frac{\lambda}{T}

Correct Answer: c) v=γPρv=\sqrt{\frac{\gamma P}{\rho}}

Explanation: This is the modern expression for the speed of sound in gases.


MCQ No. 135

For air, the value of γ is approximately:

a) 1.00

b) 1.20

c) 1.40

d) 1.67

Correct Answer: c) 1.40

Explanation: Air is mainly diatomic, so its ratio of specific heats is about 1.4.


MCQ No. 136

If the absolute temperature of a gas becomes four times its original value, the speed of sound becomes:

a) Half

b) Double

c) Four times

d) Unchanged

Correct Answer: b) Double

Explanation:
Since

vTv\propto\sqrt{T}
4=2\sqrt4=2

MCQ No. 137

The speed of sound in a gas is proportional to:

a) Temperature

b) Square of temperature

c) Square root of absolute temperature

d) Density only

Correct Answer: c) Square root of absolute temperature

Explanation:

vTv\propto\sqrt{T}

MCQ No. 138

At constant temperature, increasing the pressure of an ideal gas causes the speed of sound to:

a) Increase

b) Decrease

c) Remain unchanged

d) Become zero

Correct Answer: c) Remain unchanged

Explanation: Pressure and density increase proportionally, so P/ρP/\rho remains constant.


MCQ No. 139

The speed of sound is greatest in:

a) Hydrogen

b) Air

c) Water

d) Iron

Correct Answer: d) Iron

Explanation: Solids possess much greater elasticity than gases and liquids.


MCQ No. 140

A sound wave travels 686 m in 2 seconds. Its speed is:

a) 331 m/s

b) 340 m/s

c) 343 m/s

d) 350 m/s

Correct Answer: c) 343 m/s

Explanation:

v=6862=343 m/sv=\frac{686}{2}=343\ \text{m/s}

MCQ No. 141

A sound wave has a speed of 343 m/s and a frequency of 490 Hz. Its wavelength is:

a) 0.50 m

b) 0.60 m

c) 0.70 m

d) 0.80 m

Correct Answer: c) 0.70 m

Explanation:

λ=343490=0.70 m\lambda=\frac{343}{490}=0.70\ \text{m}

MCQ No. 142

If the wavelength of a sound wave is 1.4 m and the speed is 350 m/s, its frequency is:

a) 200 Hz

b) 225 Hz

c) 250 Hz

d) 275 Hz

Correct Answer: c) 250 Hz

Explanation:

f=3501.4=250 Hzf=\frac{350}{1.4}=250\ \text{Hz}

MCQ No. 143

A sound wave of frequency 500 Hz travels with a speed of 340 m/s. The period is:

a) 0.02 s

b) 0.002 s

c) 0.20 s

d) 2 s

Correct Answer: b) 0.002 s

Explanation:

T=1500=0.002 sT=\frac1{500}=0.002\ \text{s}

MCQ No. 144

If the frequency of a sound wave is doubled while the speed remains constant, the wavelength becomes:

a) Double

b) Half

c) Four times

d) Unchanged

Correct Answer: b) Half

Explanation:
Since

λ=vf\lambda=\frac{v}{f}

MCQ No. 145

The speed of sound in air changes mainly because of:

a) Loudness

b) Temperature

c) Pitch

d) Amplitude

Correct Answer: b) Temperature

Explanation: Temperature has the greatest effect on the speed of sound in gases.


MCQ No. 146

Which medium has the lowest speed of sound?

a) Steel

b) Water

c) Air

d) Vacuum

Correct Answer: d) Vacuum

Explanation: Sound cannot propagate through a vacuum because there are no particles to transmit vibrations.


MCQ No. 147

A sound wave takes 5 s to travel 1700 m. Its speed is:

a) 300 m/s

b) 320 m/s

c) 340 m/s

d) 360 m/s

Correct Answer: c) 340 m/s

Explanation:

v=17005=340 m/sv=\frac{1700}{5}=340\ \text{m/s}

MCQ No. 148

The theoretical speed of sound agrees with the experimental value after applying:

a) Einstein's correction

b) Maxwell's correction

c) Laplace's correction

d) Faraday's correction

Correct Answer: c) Laplace's correction

Explanation: Laplace's adiabatic correction brought theory into close agreement with experiment.


MCQ No. 149

Which equation correctly expresses the relationship between wave speed and temperature?

a) vTv\propto T

b) vT2v\propto T^2

c) vTv\propto\sqrt{T}

d) v1Tv\propto\frac1T

Correct Answer: c) vTv\propto\sqrt{T}

Explanation: The speed of sound in gases varies as the square root of the absolute temperature.


MCQ No. 150

A sound wave has a frequency of 680 Hz and wavelength 0.5 m. Its speed is:

a) 300 m/s

b) 320 m/s

c) 340 m/s

d) 360 m/s

Correct Answer: c) 340 m/s

Explanation:

v=fλ=680×0.5=340 m/sv=f\lambda=680\times0.5=340\ \text{m/s}

MCQs 151–175

Topics Covered:

  • Principle of Superposition
  • Interference of Waves
  • Beats
  • Standing Waves
  • Vibrating Strings
  • Harmonics & Overtones
  • Open & Closed Organ Pipes

MCQ No. 151

Two coherent waves each having an amplitude of 4 cm interfere constructively. The resultant amplitude is:

a) 2 cm

b) 4 cm

c) 8 cm

d) 16 cm

Correct Answer: c) 8 cm

Explanation: In constructive interference, amplitudes add directly.

A=4+4=8 cmA=4+4=8\ \text{cm}

MCQ No. 152

Two waves of equal amplitude 6 cm interfere destructively. The resultant amplitude is:

a) 12 cm

b) 6 cm

c) 3 cm

d) 0 cm

Correct Answer: d) 0 cm

Explanation: Equal amplitudes with a phase difference of 180° completely cancel each other.


MCQ No. 153

The path difference required for the second bright fringe is:

a) λ

b) 2λ

c) λ/2

d) 3λ/2

Correct Answer: b) 2λ

Explanation: Bright fringes occur at path differences of nλn\lambda. For the second bright fringe, n=2n=2.


MCQ No. 154

The path difference for the first dark fringe is:

a) λ

b) λ/4

c) λ/2

d) 2λ

Correct Answer: c) λ/2

Explanation: The first destructive interference occurs at a path difference of λ2\frac{\lambda}{2}.


MCQ No. 155

Two tuning forks have frequencies 256 Hz and 262 Hz. The beat frequency is:

a) 2 Hz

b) 4 Hz

c) 6 Hz

d) 8 Hz

Correct Answer: c) 6 Hz

Explanation:

fb=262256=6 Hzf_b=|262-256|=6\ \text{Hz}

MCQ No. 156

A beat frequency of 8 Hz is produced with one tuning fork of 512 Hz. The frequency of the other fork may be:

a) 500 Hz

b) 504 Hz

c) 520 Hz

d) Both 504 Hz and 520 Hz

Correct Answer: d) Both 504 Hz and 520 Hz

Explanation: The second frequency can be either 5128=504512-8=504 Hz or 512+8=520512+8=520 Hz.


MCQ No. 157

If two tuning forks have identical frequencies, the beat frequency will be:

a) 1 Hz

b) 2 Hz

c) Zero

d) Infinite

Correct Answer: c) Zero

Explanation: Beats disappear when both frequencies are exactly equal.


MCQ No. 158

Two waves have frequencies 300 Hz and 294 Hz. The number of beats heard in 10 seconds is:

a) 6

b) 30

c) 60

d) 300

Correct Answer: c) 60

Explanation:
Beat frequency

=300294=6 Hz=300-294=6\ \text{Hz}

Beats in 10 s

=6×10=60=6\times10=60

MCQ No. 159

The distance between two consecutive nodes in a standing wave is 40 cm. The wavelength is:

a) 20 cm

b) 40 cm

c) 80 cm

d) 160 cm

Correct Answer: c) 80 cm

Explanation:

Distance between nodes=λ2\text{Distance between nodes}=\frac{\lambda}{2}

MCQ No. 160

The distance between a node and the nearest antinode is 15 cm. The wavelength is:

a) 30 cm

b) 45 cm

c) 60 cm

d) 75 cm

Correct Answer: c) 60 cm

Explanation:

λ4=15\frac{\lambda}{4}=15
λ=60 cm\lambda=60\ \text{cm}

MCQ No. 161

A stretched string has a fundamental frequency of 250 Hz. Its second harmonic is:

a) 250 Hz

b) 375 Hz

c) 500 Hz

d) 750 Hz

Correct Answer: c) 500 Hz

Explanation:

f2=2f1f_2=2f_1

MCQ No. 162

A string vibrates with a fundamental frequency of 120 Hz. The third harmonic is:

a) 240 Hz

b) 300 Hz

c) 360 Hz

d) 480 Hz

Correct Answer: c) 360 Hz

Explanation:

3×120=360 Hz3\times120=360\ \text{Hz}

MCQ No. 163

If the length of a stretched string is doubled while tension remains constant, the fundamental frequency becomes:

a) Double

b) Half

c) Four times

d) Unchanged

Correct Answer: b) Half

Explanation: Fundamental frequency is inversely proportional to the length of the string.


MCQ No. 164

If the tension in a stretched string becomes four times, the fundamental frequency becomes:

a) Half

b) Double

c) Four times

d) Unchanged

Correct Answer: b) Double

Explanation:

fTf\propto\sqrt{T}
4=2\sqrt4=2

MCQ No. 165

If the linear density of a string becomes four times, the fundamental frequency becomes:

a) Double

b) Half

c) Four times

d) Unchanged

Correct Answer: b) Half

Explanation:

f1μf\propto\frac1{\sqrt{\mu}}

MCQ No. 166

An open organ pipe has a fundamental frequency of 300 Hz. Its second harmonic is:

a) 150 Hz

b) 300 Hz

c) 600 Hz

d) 900 Hz

Correct Answer: c) 600 Hz

Explanation: Open organ pipes support all harmonics.


MCQ No. 167

An open organ pipe has a fundamental frequency of 200 Hz. The third harmonic is:

a) 400 Hz

b) 500 Hz

c) 600 Hz

d) 800 Hz

Correct Answer: c) 600 Hz

Explanation:

3×200=600 Hz3\times200=600\ \text{Hz}

MCQ No. 168

A closed organ pipe has a fundamental frequency of 250 Hz. Its first overtone is:

a) 500 Hz

b) 750 Hz

c) 1000 Hz

d) 1250 Hz

Correct Answer: b) 750 Hz

Explanation: In a closed pipe, the first overtone is the third harmonic.

3×250=750 Hz3\times250=750\ \text{Hz}

MCQ No. 169

A closed organ pipe has a fundamental frequency of 150 Hz. The next possible harmonic is:

a) 300 Hz

b) 450 Hz

c) 600 Hz

d) 750 Hz

Correct Answer: b) 450 Hz

Explanation: Closed organ pipes produce only odd harmonics.


MCQ No. 170

An open organ pipe of length 0.50 m produces its fundamental note. If the speed of sound is 340 m/s, the frequency is:

a) 170 Hz

b) 340 Hz

c) 510 Hz

d) 680 Hz

Correct Answer: b) 340 Hz

Explanation:

f=v2L=3402×0.5=340 Hzf=\frac{v}{2L} =\frac{340}{2\times0.5} =340\ \text{Hz}

MCQ No. 171

A closed organ pipe is 0.25 m long. The speed of sound is 340 m/s. Its fundamental frequency is:

a) 170 Hz

b) 250 Hz

c) 340 Hz

d) 680 Hz

Correct Answer: c) 340 Hz

Explanation:

f=v4L=3404×0.25=340 Hzf=\frac{v}{4L} =\frac{340}{4\times0.25} =340\ \text{Hz}

MCQ No. 172

The first overtone in an open organ pipe is:

a) First harmonic

b) Second harmonic

c) Third harmonic

d) Fifth harmonic

Correct Answer: b) Second harmonic

Explanation: In an open pipe, the first overtone corresponds to the second harmonic.


MCQ No. 173

The first overtone in a closed organ pipe corresponds to:

a) Second harmonic

b) Third harmonic

c) Fourth harmonic

d) Fifth harmonic

Correct Answer: b) Third harmonic

Explanation: Closed organ pipes support only odd harmonics.


MCQ No. 174

A string fixed at both ends vibrates in its third harmonic. The number of loops formed is:

a) One

b) Two

c) Three

d) Four

Correct Answer: c) Three

Explanation: In the nth harmonic, a stretched string forms n loops (antinodes).


MCQ No. 175

A standing wave on a string has 6 loops. The harmonic produced is:

a) Third

b) Fourth

c) Fifth

d) Sixth

Correct Answer: d) Sixth

Explanation: The number of loops (antinodes) in a standing wave equals the harmonic number.


MCQs 176–200

Topics Covered:

  • Doppler Effect
  • Ultrasound
  • SONAR
  • Medical Applications
  • Mixed Numerical & Conceptual Problems
  • ECAT, MDCAT, NUST, PIEAS & GIKI Level Questions

MCQ No. 176

A stationary observer hears a higher frequency when the sound source is:

a) Moving away from the observer

b) Moving towards the observer

c) Stationary

d) Moving perpendicular to the observer

Correct Answer: b) Moving towards the observer

Explanation: As the source approaches, successive wavefronts become closer together, increasing the observed frequency.


MCQ No. 177

An ambulance siren emits a frequency of 600 Hz while approaching a stationary observer. The observed frequency will be:

a) Less than 600 Hz

b) Equal to 600 Hz

c) Greater than 600 Hz

d) Zero

Correct Answer: c) Greater than 600 Hz

Explanation: Due to the Doppler Effect, the observed frequency increases when the source approaches.


MCQ No. 178

A sound source moves away from a stationary observer. The observer will hear:

a) Higher pitch

b) Lower pitch

c) No sound

d) Constant amplitude only

Correct Answer: b) Lower pitch

Explanation: The frequency decreases when the source moves away, resulting in a lower pitch.


MCQ No. 179

Which quantity changes due to the Doppler Effect?

a) Actual frequency of the source

b) Speed of sound

c) Observed frequency

d) Amplitude of sound

Correct Answer: c) Observed frequency

Explanation: The source frequency remains constant; only the observed frequency changes.


MCQ No. 180

An observer hears a train whistle at 520 Hz while approaching and 480 Hz while receding. The actual frequency of the whistle is closest to:

a) 480 Hz

b) 500 Hz

c) 520 Hz

d) 540 Hz

Correct Answer: b) 500 Hz

Explanation: The true frequency lies between the approaching and receding observed frequencies.


MCQ No. 181

Ultrasound waves have frequencies:

a) Below 20 Hz

b) Between 20 Hz and 20 kHz

c) Above 20 kHz

d) Above 200 MHz

Correct Answer: c) Above 20 kHz

Explanation: Ultrasound consists of sound waves with frequencies greater than 20,000 Hz.


MCQ No. 182

The main advantage of ultrasound in medical diagnosis is that it:

a) Uses ionizing radiation

b) Is non-ionizing and safe for imaging

c) Produces X-rays

d) Destroys body tissues

Correct Answer: b) Is non-ionizing and safe for imaging

Explanation: Ultrasound provides images of internal organs without exposing patients to ionizing radiation.


MCQ No. 183

SONAR is based on the principle of:

a) Diffraction

b) Refraction

c) Reflection of sound

d) Polarization

Correct Answer: c) Reflection of sound

Explanation: SONAR detects reflected ultrasonic waves (echoes) from underwater objects.


MCQ No. 184

A SONAR pulse returns after 4 s. If the speed of sound in seawater is 1500 m/s, the depth of the sea is:

a) 1500 m

b) 3000 m

c) 6000 m

d) 750 m

Correct Answer: b) 3000 m

Explanation:
The sound travels to the seabed and back.

d=vt2=1500×42=3000 md=\frac{vt}{2} =\frac{1500\times4}{2} =3000\ \text{m}

MCQ No. 185

An echo is heard 3 seconds after a shout. If the speed of sound is 340 m/s, the reflecting surface is:

a) 340 m away

b) 510 m away

c) 680 m away

d) 1020 m away

Correct Answer: b) 510 m away

Explanation:

d=340×32=510 md=\frac{340\times3}{2}=510\ \text{m}

MCQ No. 186

A ship sends an ultrasonic pulse that returns after 2 s. The speed of sound in seawater is 1500 m/s. The depth is:

a) 750 m

b) 1000 m

c) 1500 m

d) 3000 m

Correct Answer: c) 1500 m

Explanation:

d=1500×22=1500 md=\frac{1500\times2}{2}=1500\ \text{m}

MCQ No. 187

Ultrasonic waves are commonly generated using:

a) Piezoelectric crystals

b) Electric motors

c) Permanent magnets

d) Galvanometers

Correct Answer: a) Piezoelectric crystals

Explanation: Piezoelectric crystals vibrate rapidly to produce ultrasonic waves.


MCQ No. 188

Which medical instrument commonly uses the Doppler Effect?

a) Stethoscope

b) Doppler ultrasound scanner

c) Thermometer

d) ECG machine

Correct Answer: b) Doppler ultrasound scanner

Explanation: Doppler ultrasound measures blood flow by detecting frequency shifts in reflected ultrasonic waves.


MCQ No. 189

Bats detect obstacles using:

a) Infrared radiation

b) Radar

c) Ultrasonic echoes

d) Radio waves

Correct Answer: c) Ultrasonic echoes

Explanation: Bats use echolocation by emitting ultrasonic waves and analyzing their echoes.


MCQ No. 190

Which marine animal uses echolocation similar to SONAR?

a) Shark

b) Dolphin

c) Whale shark

d) Octopus

Correct Answer: b) Dolphin

Explanation: Dolphins emit ultrasonic clicks and detect echoes to locate prey and navigate.


MCQ No. 191

The audible frequency range of a healthy human ear is approximately:

a) 2 Hz – 200 Hz

b) 20 Hz – 20 kHz

c) 200 Hz – 200 kHz

d) 2 kHz – 200 kHz

Correct Answer: b) 20 Hz – 20 kHz

Explanation: Humans can generally hear sounds between 20 Hz and 20,000 Hz.


MCQ No. 192

If the speed of sound in water is 1500 m/s and the frequency is 750 kHz, the wavelength is:

a) 1 mm

b) 2 mm

c) 4 mm

d) 8 mm

Correct Answer: b) 2 mm

Explanation:

λ=1500750000=0.002 m=2 mm\lambda=\frac{1500}{750000}=0.002\ \text{m}=2\ \text{mm}

MCQ No. 193

Which application uses high-frequency sound waves to detect internal body structures?

a) MRI

b) CT Scan

c) Ultrasonography

d) X-ray radiography

Correct Answer: c) Ultrasonography

Explanation: Ultrasonography uses reflected ultrasonic waves to produce images of internal organs.


MCQ No. 194

If an echo is received after 0.5 s and the speed of sound is 340 m/s, the obstacle is:

a) 85 m away

b) 170 m away

c) 255 m away

d) 340 m away

Correct Answer: a) 85 m away

Explanation:

d=340×0.52=85 md=\frac{340\times0.5}{2}=85\ \text{m}

MCQ No. 195

Which wave property is used in echo sounding?

a) Polarization

b) Reflection

c) Dispersion

d) Diffraction

Correct Answer: b) Reflection

Explanation: Echo sounding measures the time taken by reflected sound waves.


MCQ No. 196

A submarine uses SONAR to determine:

a) Atmospheric pressure

b) Ocean depth and underwater objects

c) Earth's magnetic field

d) Wind speed

Correct Answer: b) Ocean depth and underwater objects

Explanation: SONAR determines the distance to underwater objects using reflected ultrasonic waves.


MCQ No. 197

Which statement about ultrasound is correct?

a) It can travel only in a vacuum.

b) It has frequencies below the audible range.

c) It has frequencies above 20 kHz.

d) It is an electromagnetic wave.

Correct Answer: c) It has frequencies above 20 kHz.

Explanation: Ultrasound is a mechanical sound wave with frequencies greater than 20,000 Hz.


MCQ No. 198

The Doppler Effect is observed because of:

a) Relative motion between the source and the observer

b) Change in wave speed

c) Reflection only

d) Interference only

Correct Answer: a) Relative motion between the source and the observer

Explanation: Relative motion changes the spacing of wavefronts reaching the observer, altering the observed frequency.


MCQ No. 199

A SONAR signal takes 6 s to return from the seabed. If the speed of sound in seawater is 1500 m/s, the depth is:

a) 3000 m

b) 4000 m

c) 4500 m

d) 6000 m

Correct Answer: c) 4500 m

Explanation:

d=1500×62=4500 md=\frac{1500\times6}{2}=4500\ \text{m}

MCQ No. 200

A wave has a frequency of 1 MHz and travels through tissue at 1540 m/s. Its wavelength is approximately:

a) 0.154 mm

b) 1.54 mm

c) 15.4 mm

d) 154 mm

Correct Answer: b) 1.54 mm

Explanation:

λ=15401,000,000=0.00154 m=1.54 mm\lambda=\frac{1540}{1,000,000}=0.00154\ \text{m}=1.54\ \text{mm}

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