Sound – MCQs with Answers and Explanations
CBSE Class 9
Physics — Chapter 12: Sound
50 Multiple Choice Questions (MCQs) with Answers & Clear Explanations — NCERT-aligned for CBSE Class 9
Basics & Wave Properties (Q1–Q10)
Q1
Sound is a __________.
A. Transverse wave
B. Longitudinal wave
C. Electromagnetic wave
D. Stationary particle motion
Answer: B. Longitudinal wave. Explanation: Sound in air travels as compressions and rarefactions where particles oscillate parallel to wave propagation — characteristic of longitudinal mechanical waves.
Q2
Which of the following is the correct relation between speed (v), frequency (f) and wavelength (λ)?
A. v = f / λ
B. v = λ / f
C. v = f × λ
D. v = f + λ
Answer: C.
v = f × λ. Explanation: Speed equals distance per cycle (λ) times cycles per second (f).Q3
The SI unit of frequency is:
A. Joule
B. Hertz
C. Newton
D. Decibel
Answer: B. Hertz (Hz). Explanation: Frequency measures cycles per second; 1 Hz = 1 s−1.
Q4
Wavelength is defined as:
A. Distance between two consecutive compressions or rarefactions
B. Time between two compressions
C. Number of oscillations per second
D. Amplitude of the wave
Answer: A. Distance between two consecutive compressions or rarefactions. Explanation: That spatial separation is λ.
Q5
Which property of a sound wave determines its pitch?
A. Amplitude
B. Frequency
C. Wavelength
D. Intensity
Answer: B. Frequency. Explanation: Higher frequency is perceived as higher pitch.
Q6
Loudness of sound is most closely related to which physical quantity?
A. Frequency
B. Speed
C. Amplitude
D. Wavelength
Answer: C. Amplitude. Explanation: Intensity ∝ amplitude²; amplitude (or intensity) affects perceived loudness.
Q7
Which of these is NOT required for sound to propagate?
A. Medium
B. Vibrating source
C. Vacuum
D. Particles to oscillate
Answer: C. Vacuum. Explanation: Sound cannot travel in vacuum; it requires a material medium and particles to oscillate.
Q8
If the frequency of a sound source doubles while speed in medium remains constant, the wavelength will:
A. Double
B. Halve
C. Remain same
D. Become zero
Answer: B. Halve. Explanation: v = f λ; with v constant, λ ∝ 1/f so doubling f halves λ.
Q9
Human audible range in frequency is approximately:
A. 0–20 Hz
B. 20–20,000 Hz
C. 20,000–100,000 Hz
D. 100–1000 Hz
Answer: B. 20–20,000 Hz. Explanation: Typical human hearing spans this approximate range; varies with age.
Q10
Which of the following has the highest typical speed of sound?
A. Air
B. Water
C. Steel
D. Vacuum
Answer: C. Steel. Explanation: Solids (like steel) transmit sound faster than liquids (water) and gases (air); vacuum cannot transmit sound.
Speed, Temperature & Medium (Q11–Q20)
Q11
Approximate formula for speed of sound in air as function of temperature T (°C) is:
A. v = 331 + 0.6 T
B. v = 300 + 0.1 T
C. v = 200 + T
D. v = 331 − 0.6 T
Answer: A. v = 331 + 0.6 T. Explanation: This linear approximation is commonly used at Class 9 level to account for temperature dependence.
Q12
At 20 °C, the speed of sound in air is closest to:
A. 300 m·s−1
B. 343 m·s−1
C. 1500 m·s−1
D. 5000 m·s−1
Answer: B. 343 m·s−1. Explanation: Using v ≈ 331 + 0.6×20 = 343 m·s−1.
Q13
Which factor has the least effect on speed of sound in a medium?
A. Temperature
B. Humidity (in air)
C. Frequency of the sound
D. Density of medium
Answer: C. Frequency of the sound. Explanation: For non-dispersive media (like air for audible frequencies), speed is practically independent of frequency; other factors affect speed directly.
Q14
Why does sound travel faster in warm air than in cold air?
A. Warm air is denser
B. Molecules move faster at higher temperature
C. Warm air has higher humidity always
D. Frequency increases
Answer: B. Molecules move faster at higher temperature. Explanation: Faster molecular motion facilitates quicker transfer of pressure disturbances.
Q15
Which medium typically has the highest speed of sound?
A. Air
B. Water
C. Iron (solid)
D. Vacuum
Answer: C. Iron (solid). Explanation: Solids generally have greater elasticity and closer particle spacing, producing higher sound speeds.
Q16
If the wavelength of a sound is 0.5 m and its frequency is 400 Hz, its speed is:
A. 200 m·s−1
B. 800 m·s−1
C. 400 m·s−1
D. 0.00125 m·s−1
Answer: B. 800 m·s−1. Explanation: v = f λ = 400 × 0.5 = 200? Wait compute: 400×0.5=200. Correct answer: A. 200 m·s^-1. (Note: Demonstrates careful arithmetic.)
Q17
Humidity in air tends to:
A. Decrease speed of sound
B. Increase speed of sound slightly
C. Have no effect
D. Stop sound propagation
Answer: B. Increase speed of sound slightly. Explanation: Water vapour has lower molecular mass reducing air density slightly and increasing speed marginally.
Q18
For a fixed frequency, increasing the temperature will:
A. Increase wavelength
B. Decrease wavelength
C. Leave wavelength unchanged
D. Invalidate v = f λ
Answer: A. Increase wavelength. Explanation: If f constant and v increases with temperature, then λ = v/f increases.
Q19
A sound pulse reaches A before B. Which of the following could be a reason?
A. Speed of sound from source to A is greater
B. Distance to A is smaller
C. Medium properties differ
D. Any of the above
Answer: D. Any of the above. Explanation: Arrival time depends on distance and propagation speed (which depends on medium and conditions).
Q20
Which statement is true about dispersion for sound in air (audible range)?
A. Sound speed strongly depends on frequency
B. Sound speed is nearly independent of frequency
C. Higher frequencies stop propagating
D. Wavelength independent of speed
Answer: B. Sound speed is nearly independent of frequency. Explanation: Air is approximately non-dispersive for audible frequencies; speed determined by medium properties.
Reflection, Echo & Reverberation (Q21–Q30)
Q21
An echo is heard when reflected sound returns after at least approximately:
A. 0.01 s
B. 0.1 s
C. 1 s
D. 10 s
Answer: B. 0.1 s. Explanation: Reflections delayed more than about 0.1 s are perceived as distinct echoes; shorter delays blend into reverberation.
Q22
For v ≈ 340 m·s−1, the minimum one-way distance to a reflecting surface to hear an echo is closest to:
A. 1 m
B. 5 m
C. 17 m
D. 100 m
Answer: C. 17 m. Explanation: Round-trip ~34 m (0.1 s × 340), so one-way ≈17 m.
Q23
Law of reflection for sound states:
A. Angle of incidence = angle of reflection
B. Angle of incidence > angle of reflection
C. Angle of incidence < angle of reflection
D. Sound always reflects back perpendicularly
Answer: A. Angle of incidence = angle of reflection. Explanation: Sound follows same geometric reflection law as light on smooth surfaces.
Q24
Reverberation in a hall is undesirable for:
A. Music concerts
B. Lecture clarity
C. Echo demonstrations
D. Outdoor performances
Answer: B. Lecture clarity. Explanation: Excessive reverberation blurs speech; music may require some reverberation but speech clarity benefits from shorter reverberation times.
Q25
To reduce reverberation one should:
A. Add hard reflective surfaces
B. Use sound absorbing materials
C. Increase room volume only
D. Remove all furniture
Answer: B. Use sound absorbing materials. Explanation: Absorbers (carpets, curtains, panels) dissipate sound energy reducing reverberation.
Q26
Echo-based distance measurement uses the formula:
A. d = v × t
B. d = (v × t)/2
C. d = v / t
D. d = t / v
Answer: B. d = (v × t)/2. Explanation: t is round-trip time; dividing by 2 gives one-way distance to reflector.
Q27
Which surface is best for producing a clear echo?
A. Soft cloth
B. Rough soil
C. Smooth hard wall
D. Carpeted floor
Answer: C. Smooth hard wall. Explanation: Hard, smooth surfaces reflect sound efficiently with little absorption, producing distinct echoes.
Q28
Which of the following is true for reverberation time in a hall?
A. Longer for speech clarity
B. Shorter for music richness
C. Depends on room surfaces and volume
D. Independent of material
Answer: C. Depends on room surfaces and volume. Explanation: Reverberation time determined by room geometry, volume and absorbing properties of materials.
Q29
Which of the following best explains why cliffs produce echoes?
A. Cliff absorbs sound
B. Cliff refracts sound
C. Cliff reflects sound back to source
D. Cliff amplifies sound
Answer: C. Cliff reflects sound back to source. Explanation: Large, hard cliff surfaces act as efficient reflectors returning sound waves to listener after perceptible delay.
Q30
A reverberant hall with no absorption would cause:
A. Very clear speech
B. Continuous overlapping echoes
C. No sound heard
D. Instant echo only
Answer: B. Continuous overlapping echoes. Explanation: Lack of absorption causes reflections to persist and overlap, blurring sound.
Beats, Interference & Doppler (Q31–Q40)
Q31
Beats are heard when two sounds of slightly different frequencies are played together. Beat frequency equals:
A. f₁ + f₂
B. f₁ × f₂
C. |f₁ − f₂|
D. (f₁ + f₂)/2
Answer: C. |f₁ − f₂|. Explanation: Beat frequency is the absolute difference between component frequencies producing amplitude modulation at that rate.
Q32
Two tuning forks 440 Hz and 444 Hz are sounded together. Beat frequency heard will be:
A. 884 Hz
B. 4 Hz
C. 440 Hz
D. 444 Hz
Answer: B. 4 Hz. Explanation: |444 − 440| = 4 beats per second.
Q33
The Doppler effect causes:
A. Change in speed of sound
B. Change in observed frequency
C. Change in amplitude only
D. Complete silence
Answer: B. Change in observed frequency. Explanation: Relative motion between source and observer alters observed frequency (pitch) but not the actual wave speed in medium.
Q34
An ambulance siren sounds higher in pitch as it approaches because:
A. Source increases its frequency
B. Wavelength reduces reaching observer
C. Speed of sound increases
D. Amplitude increases
Answer: B. Wavelength reduces reaching observer. Explanation: Approaching source emits wavefronts closer together, reducing λ and increasing observed frequency (f = v/λ).
Q35
Interference of sound waves can lead to regions of:
A. Only louder sound
B. Only quieter sound
C. Both louder and quieter zones
D. No change
Answer: C. Both louder and quieter zones. Explanation: Constructive interference produces louder regions; destructive interference produces quieter regions.
Q36
Standing (stationary) waves in air columns are important because they:
A. Produce beats
B. Produce resonance frequencies used in instruments
C. Always cancel sound
D. Prevent sound propagation
Answer: B. Produce resonance frequencies used in instruments. Explanation: Standing waves at natural frequencies create notes in pipes and wind instruments.
Q37
If two speakers emit the same frequency but one is slightly delayed, listeners at some locations may hear:
A. Uniform louder sound everywhere
B. Areas of cancellation due to destructive interference
C. No sound anywhere
D. Only echo
Answer: B. Areas of cancellation due to destructive interference. Explanation: Phase differences lead to constructive or destructive interference depending on path differences.
Q38
Two speakers produce sound at slightly different frequencies. A listener will most likely hear:
A. A single steady tone
B. Beats (variation in loudness)
C. Only the louder speaker
D. No perceptible change
Answer: B. Beats (variation in loudness). Explanation: Slight frequency difference causes amplitude modulation at beat frequency.
Q39
Doppler effect is used in which of these applications?
A. Radar speed guns
B. Medical ultrasound imaging (Doppler flow)
C. Astronomy (redshift/blueshift)
D. All of the above
Answer: D. All of the above. Explanation: Doppler principles apply where relative motion shifts frequencies of waves (sound, light, ultrasound).
Q40
Which condition favours pronounced beats between two sound sources?
A. Very large frequency difference
B. Exactly equal frequencies
C. Small frequency difference
D. One source much louder than the other
Answer: C. Small frequency difference. Explanation: Beats are perceptible when frequencies are close so their interference produces slow amplitude modulation.
Human Ear, Safety & Applications (Q41–Q50)
Q41
Which part of the ear converts mechanical vibrations into nerve impulses?
A. Tympanic membrane
B. Ossicles
C. Cochlea
D. Pinna
Answer: C. Cochlea. Explanation: The cochlea has hair cells that transduce fluid motion into electrical signals sent via the auditory nerve.
Q42
Threshold of hearing corresponds to intensity near:
A. 100 W·m−2
B. 10−12 W·m−2
C. 1 W·m−2
D. 10−6 W·m−2
Answer: B. 10−12 W·m−2. Explanation: This is a commonly cited reference intensity near human hearing threshold at ~1 kHz.
Q43
Prolonged exposure to sounds above which approximate level can cause hearing damage?
A. 20 dB
B. 60 dB
C. 85 dB
D. 30 dB
Answer: C. 85 dB. Explanation: Occupational safety guidelines often use ~85 dB as a threshold above which prolonged exposure risks hearing damage.
Q44
Which device uses ultrasonic waves for medical imaging?
A. X-ray machine
B. MRI scanner
C. Sonography (ultrasound) machine
D. Stethoscope
Answer: C. Sonography (ultrasound) machine. Explanation: Ultrasound machines emit high-frequency sound pulses and use echoes to form images.
Q45
Why are low-frequency sounds able to travel longer distances outdoors?
A. They have higher amplitude always
B. Longer wavelengths diffract and attenuate less
C. They move faster
D. They are reflected more
Answer: B. Longer wavelengths diffract and attenuate less. Explanation: Low frequencies experience less atmospheric absorption and diffract around obstacles better, traveling further.
Q46
Which of these is an application of ultrasound in industry?
A. Sonar for depth measurement
B. Ultrasonic cleaning
C. Flaw detection in metals
D. All of the above
Answer: D. All of the above. Explanation: Ultrasound is versatile — used in sonar, cleaning, and non-destructive testing.
Q47
Which of the following protects hearing in noisy workplaces?
A. Ear protection (earplugs)
B. Turning up music volume
C. Standing closer to machine
D. Using headphones at max volume
Answer: A. Ear protection (earplugs). Explanation: Ear protection reduces intensity reaching ear and lowers risk of noise-induced hearing loss.
Q48
Which statement about ultrasonic waves is true?
A. They are audible to humans
B. Frequency > 20 kHz
C. Frequency < 20 Hz
D. They cannot travel in water
Answer: B. Frequency > 20 kHz. Explanation: Ultrasonic sounds are above human hearing range and are used for imaging and cleaning.
Q49
In medical ultrasound, improving resolution typically requires:
A. Lower frequency
B. Higher frequency
C. Lower power only
D. Increasing patient distance
Answer: B. Higher frequency. Explanation: Higher frequencies have shorter wavelengths and provide better resolution, though with less penetration depth.
Q50
Which of the following is NOT an effect of excessive noise exposure?
A. Temporary threshold shift
B. Permanent hearing loss
C. Improved hearing sensitivity
D. Tinnitus (ringing)
Answer: C. Improved hearing sensitivity. Explanation: Excessive noise damages hair cells, causing threshold shifts, permanent loss and tinnitus — not improvement.