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Sound — Homi Bhabha Class 9 Physics MCQs with Solutions
Free Homi Bhabha Class 9 Physics Sound MCQs with step-by-step solutions covering Nature & Propagation of Sound, Frequency & Amplitude, Echo & Reverberation, Human Ear Mechanism, Ultrasound Applications. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Nature & Propagation of Sound · easy · theory
Through which of the following can sound NOT travel?
A. A perfect vacuum ✓ Correct
B. Air in a room
C. A solid iron bar
D. Water in a tank
Solution: Sound is a mechanical wave and needs a material medium (solid, liquid or gas) to carry the vibrations. In a vacuum there are no particles to vibrate, so sound cannot travel.
Q2 — Nature & Propagation of Sound · easy · theory
When sound travels through air, the air particles vibrate back and forth along the same direction in which the wave moves. Such a wave is called a
A. Electromagnetic wave
B. Water surface wave
C. Longitudinal wave ✓ Correct
D. Transverse wave
Solution: In sound, particles of the medium oscillate parallel to the direction of wave travel, creating compressions and rarefactions. This makes sound a longitudinal wave.
Q3 — Frequency & Amplitude · easy · theory
The loudness of a sound depends mainly on which property of the sound wave?
A. Its frequency
B. Its speed
C. Its wavelength
D. Its amplitude ✓ Correct
Solution: A larger amplitude means the particles vibrate with more energy, which the ear senses as a louder sound. Frequency controls pitch, not loudness.
Q4 — Echo & Reverberation · easy · theory
An echo is produced because of which property of sound?
A. Reflection of sound ✓ Correct
B. Diffraction of sound
C. Absorption of sound
D. Refraction of sound
Solution: An echo is simply the original sound heard again after it bounces back from a distant surface. This bouncing back is reflection of sound.
Q5 — Human Ear Mechanism · easy · theory
Which part of the ear changes sound vibrations into electrical signals that are sent to the brain?
A. The pinna
B. The ear canal
C. The cochlea ✓ Correct
D. The eardrum
Solution: The fluid-filled cochlea in the inner ear contains tiny hair cells that convert the vibrations into electrical nerve impulses, which the auditory nerve carries to the brain.
Q6 — Human Ear Mechanism · easy · theory
The range of frequencies that a normal, healthy human ear can hear is about
A. 20 Hz to 20,000 Hz ✓ Correct
B. 1 Hz to 20 Hz
C. 20,000 Hz to 40,000 Hz
D. 0 Hz to 100 Hz
Solution: The human audible range runs roughly from 20 Hz (lowest) to 20,000 Hz (highest). Below this is infrasound and above it is ultrasound, both inaudible to us.
Q7 — Ultrasound Applications · easy · theory
Ultrasound is the name given to sound whose frequency is
A. above 20,000 Hz ✓ Correct
B. exactly 20 Hz
C. between 20 Hz and 20,000 Hz
D. below 20 Hz
Solution: Ultrasound has frequencies above 20,000 Hz, which is beyond the upper limit of human hearing. Sound below 20 Hz is called infrasound.
Q8 — Nature & Propagation of Sound · hard · theory
An electric bell ringing inside a sealed glass jar becomes inaudible as air is pumped out. This shows that sound:
A. is an electromagnetic wave
B. needs light to travel
C. cannot travel through vacuum ✓ Correct
D. travels faster in vacuum
Solution: With no air (medium) there are no particles to carry the compressions and rarefactions, so sound cannot propagate.
Q9 — Nature & Propagation of Sound · hard · theory
Sound travels fastest in:
A. solids ✓ Correct
B. liquids
C. vacuum
D. gases
Solution: Particles are most tightly packed in solids, so sound travels fastest in solids and slowest in gases (and not at all in vacuum).
Q10 — Nature & Propagation of Sound · hard · numerical
Sound takes 3 s to travel from a source to a listener 1020 m away. The speed of sound in that air is:
A. 3060 m/s
B. 306 m/s
C. 1020 m/s
D. 340 m/s ✓ Correct
Solution: Speed $= \dfrac{1020}{3} = 340$ m/s.
Q11 — Frequency & Amplitude · hard · numerical
A sound wave has a speed of 340 m/s and a frequency of 170 Hz. Its wavelength is:
A. 0.5 m
B. 510 m
C. 2.5 m
D. 2 m ✓ Correct
Solution: Wavelength $= \dfrac{v}{f} = \dfrac{340}{170} = 2$ m.
Q12 — Frequency & Amplitude · hard · numerical
The time period of a wave of frequency 250 Hz is:
A. 250 s
B. 0.04 s
C. 0.004 s ✓ Correct
D. 0.25 s
Solution: Time period $= \dfrac{1}{f} = \dfrac{1}{250} = 0.004$ s.
Q13 — Frequency & Amplitude · hard · theory
Two sounds have the same loudness, but one is shriller than the other. The shriller sound has a greater:
A. frequency ✓ Correct
B. speed
C. time period
D. amplitude
Solution: Shrillness (pitch) is decided by frequency; equal loudness means equal amplitude.
Q14 — Echo & Reverberation · hard · numerical
A person shouts towards a cliff and hears the echo after 3 s. If the speed of sound is 340 m/s, the distance of the cliff is:
A. 113 m
B. 1020 m
C. 340 m
D. 510 m ✓ Correct
Solution: Sound travels to the cliff and back: total distance $= 340 \times 3 = 1020$ m, so the cliff is $\dfrac{1020}{2} = 510$ m away.
Q15 — Echo & Reverberation · hard · theory
To hear a distinct echo, the reflecting surface must be at least about 17 m away because:
A. echoes need bright light
B. the surface must be very hard
C. sound travels very slowly
D. the ear can distinguish two sounds separated by at least 0.1 s ✓ Correct
Solution: The ear resolves sounds ~0.1 s apart; in 0.1 s sound travels 34 m (there and back), so the surface must be at least ~17 m away.
Q16 — Echo & Reverberation · hard · numerical
A ship sends an ultrasound pulse to the sea bed and receives the echo after 2 s. If the speed of sound in water is 1500 m/s, the depth of the sea is:
A. 1500 m ✓ Correct
B. 1000 m
C. 750 m
D. 3000 m
Solution: Total path $= 1500 \times 2 = 3000$ m; depth $= \dfrac{3000}{2} = 1500$ m (this method is SONAR).
Q17 — Human Ear Mechanism · hard · theory
The three tiny bones (ossicles) of the middle ear function to:
A. convert light to sound
B. produce sound
C. store sound
D. amplify and transmit the vibrations to the inner ear ✓ Correct
Solution: The hammer, anvil and stirrup amplify the eardrum's vibrations and pass them to the cochlea.
Q18 — Human Ear Mechanism · hard · theory
The correct path of sound through the ear is:
A. pinna → cochlea → eardrum → ossicles
B. cochlea → eardrum → pinna → nerve
C. eardrum → pinna → cochlea → ossicles
D. pinna → ear canal → eardrum → ossicles → cochlea → auditory nerve ✓ Correct
Solution: Sound is collected by the pinna, travels down the ear canal to the eardrum, is amplified by the ossicles, converted in the cochlea, and carried by the auditory nerve.
Q19 — Ultrasound Applications · hard · theory
Ultrasound is used to detect cracks inside metal blocks because it:
A. melts the metal
B. is audible to workers
C. passes straight through cracks unchanged
D. is reflected back from a crack (a change in medium) ✓ Correct
Solution: A crack is a gap that reflects the ultrasound, revealing the flaw that a solid block would not.
Q20 — Ultrasound Applications · hard · theory
Bats and dolphins navigate and locate prey mainly using:
A. infrasound only
B. their sense of smell
C. visible light
D. ultrasound echoes (echolocation) ✓ Correct
Solution: They emit ultrasonic pulses and interpret the echoes to locate obstacles and prey — echolocation.
Q21 — Nature & Propagation of Sound · hard · theory
A person strikes a long steel rail at one end. A listener standing at the other end hears the sound twice, once through the rail and once through the air. Which is heard first and why?
A. Through the rail first, because sound travels faster in solids than in gases ✓ Correct
B. Through the air first, because air is lighter than steel
C. Through the rail first, because the rail is longer than the air path
D. Both are heard at exactly the same instant
Solution: Solids are far more rigid (elastic) than gases and their particles are packed close together, so vibrations are passed on much faster. Sound therefore reaches the listener through the steel rail well before it arrives through the air.
Q22 — Frequency & Amplitude · hard · numerical
A sound wave in air has a speed of 340 m/s and a wavelength of 0.5 m. If, in the same air, the frequency of the source is doubled, what is the new wavelength?
A. 0.5 m
B. 0.125 m
C. 0.25 m ✓ Correct
D. 1.0 m
Solution: Speed depends only on the medium, so it stays 340 m/s. Since speed = frequency x wavelength is constant, doubling the frequency halves the wavelength: 0.5 / 2 = 0.25 m.
Q23 — Echo & Reverberation · hard · numerical
The ear can tell an echo apart from the original sound only if they are separated by at least 0.1 s. At a sound speed of 340 m/s, what is the minimum distance of a reflecting wall for a distinct echo to be heard?
A. 17 m ✓ Correct
B. 8.5 m
C. 3.4 m
D. 34 m
Solution: In 0.1 s sound covers 340 x 0.1 = 34 m, but this is the to-and-fro path. The wall must therefore be at least 34 / 2 = 17 m away.
Q24 — Echo & Reverberation · hard · theory
In a large empty hall, sound seems to persist for a while even after the source stops. Hanging thick curtains and laying carpets reduces this effect. The persistence of sound is called ___ and the curtains help by ___.
A. resonance; amplifying the sound
B. refraction; bending the sound
C. reverberation; absorbing the sound ✓ Correct
D. echo; reflecting the sound better
Solution: Repeated reflections from bare hard walls make sound linger; this is reverberation. Soft porous materials such as curtains and carpets absorb sound energy, cutting down the reflections and the persistence.
Q25 — Human Ear Mechanism · hard · theory
The eardrum has a much larger area than the small oval window that leads into the inner ear. This difference in area mainly helps to
A. change the frequency of the sound
B. block out all loud sounds completely
C. reduce the loudness of every sound
D. increase the pressure of the vibrations reaching the inner ear ✓ Correct
Solution: The same vibration force is collected over the large eardrum and delivered to the tiny oval window. Because pressure = force / area, focusing the force onto a smaller area raises the pressure, so the faint vibrations are strong enough to move the inner-ear fluid.
Q26 — Ultrasound Applications · hard · theory
Ultrasound is preferred over ordinary audible sound for medical imaging and for detecting tiny flaws inside metals mainly because
A. it does not need any medium to travel through
B. it travels faster than audible sound in every medium
C. it can have a higher frequency and shorter wavelength, giving sharper detail and better directionality ✓ Correct
D. it is much louder and so can be heard clearly
Solution: Its high frequency means a short wavelength, so ultrasound can travel in a narrow directed beam and resolve very small features. This sharp detail and straight-line travel is what makes it ideal for scanning bodies and finding hidden cracks.
Q27 — Nature & Propagation of Sound · medium · theory
Sound waves are:
A. electromagnetic waves
B. longitudinal mechanical waves that need a material medium ✓ Correct
C. transverse waves that travel through vacuum
D. waves that need no medium
Solution: Sound is a longitudinal mechanical wave; it needs particles of a medium (solid, liquid or gas) to travel.
Q28 — Nature & Propagation of Sound · medium · numerical
Sound travels 1500 m in water in 1 s. The speed of sound in water is:
A. 150 m/s
B. 750 m/s
C. 15000 m/s
D. 1500 m/s ✓ Correct
Solution: Speed $= \dfrac{\text{distance}}{\text{time}} = \dfrac{1500}{1} = 1500$ m/s.
Q29 — Nature & Propagation of Sound · medium · theory
The regions of high pressure/density in a sound wave travelling through air are called:
A. crests
B. compressions ✓ Correct
C. rarefactions
D. troughs
Solution: In a longitudinal sound wave, compressions are high-density regions and rarefactions are low-density regions.
Q30 — Frequency & Amplitude · medium · theory
The pitch (shrillness) of a sound depends on its:
A. amplitude
B. wavelength only
C. speed
D. frequency ✓ Correct
Solution: Higher frequency gives a higher pitch; loudness depends on amplitude.