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Sound and Waves PYQs – Physics, SSC & Railway

123 Sound and Waves (Physics) questions asked in recent SSC and Railway exams, newest first, with exam, date & shift, answers and solutions.

📚 123 questionsSSC 30 · Railway 93
12
RRB Group D 2025-26 – 25 August 2026 Shift 1 · Loudness, pitch, quality and decibel

Q1. Which of the following examples best illustrates a high pitch sound in everyday life?

  1. (a) The hum of a refrigerator
  2. (b) A child blowing a whistle
  3. (c) The ticking of a large wall clock
  4. (d) The beat of a large drum
Answer & solution

Answer: (b) A child blowing a whistle

Correct Answer:
Correct Answer: Option 2 – A child blowing a whistle
Explanation:
Pitchof a sound depends on itsfrequency— higher frequency means higher pitch. A whistle produces a sound of very high frequency, resulting in a shrill, high-pitched sound. In contrast, a refrigerator's hum, a clock's ticking, and a drum beat are all examples of low-pitched (low frequency) sounds.
Important Facts:
• Pitch should not be confused with loudness, which depends on amplitude.
• A large drum produces a low-frequency (bass), low-pitch sound due to its large vibrating surface.
• Human audible range is roughly 20 Hz to 20,000 Hz.
RRB Group D 2025-26 – 25 August 2026 Shift 1 · Wave characteristics (frequency, wavelength, amplitude, time period)

Q2. If the frequency of a sound wave increases while its speed in air remains constant, what happens to its wavelength?

  1. (a) Its wavelength decreases
  2. (b) Its wavelength fluctuates randomly
  3. (c) Its wavelength increases
  4. (d) Its wavelength remains unchanged
Answer & solution

Answer: (a) Its wavelength decreases

Correct Answer:
Correct Answer: Option 1 – Its wavelength decreases
Explanation:
The wave equation isv = f × λ(speed = frequency × wavelength). If the speed (v) in air remains constant and the frequency (f) increases, then the wavelength (λ) mustdecreaseproportionally to keep the product v constant (λ = v/f).
Important Facts:
• Frequency and wavelength are inversely proportional for a given wave speed.
• Speed of sound in air depends mainly on the medium's properties (temperature, elasticity, density), not on frequency.
• This relationship applies to all wave types, including sound and light.
RRB Group D 2025-26 – 25 August 2026 Shift 2 · Human ear and sound production

Q3. Which of the following situations best demonstrates that a vibrating object is necessary to produce sound?

  1. (a) Plucking a taut guitar string and hearing a tone
  2. (b) Holding a metal rod steady/still and hearing a hum
  3. (c) Shining a torch light on the wall in a dark room
  4. (d) Looking at a glass bottle placed on the table without disturbing it
Answer & solution

Answer: (a) Plucking a taut guitar string and hearing a tone

Correct Answer:
Correct Answer: Option 1 – Plucking a taut guitar string and hearing a tone
Explanation:
Sound is produced when an objectvibrates. When a guitar string is plucked (displaced and released), it vibrates back and forth, and these vibrations are transmitted through the air as sound waves, producing an audible tone — directly demonstrating that vibration is necessary for sound production. Holding a metal rod steady (not letting it vibrate) would not naturally produce a hum; shining a torch and looking at an undisturbed bottle involve no vibration or sound at all (light, not sound, in one case; complete stillness in the other).
RRB Group D 2025-26 – 25 August 2026 Shift 3 · Applications of ultrasound and SONAR

Q4. Which of the following statements best describes how ultrasound is utilized in medical imaging?

  1. (a) Ultrasound waves ionize tissues to produce visible images.
  2. (b) Ultrasound waves reflect off tissues to create internal images.
  3. (c) Ultrasound waves pass through blood without interaction to form images.
  4. (d) Ultrasound waves are absorbed by bones to visualize soft tissue.
Answer & solution

Answer: (b) Ultrasound waves reflect off tissues to create internal images.

Correct Answer:
Correct Answer: Option 2 – Ultrasound waves reflect off tissues to create internal images.
Explanation:
In ultrasound imaging (sonography), high-frequency sound waves are sent into the body and reflect differently off tissues and organs of different densities. These reflected echoes are captured and processed to construct an image of internal structures, without ionizing radiation.
Important Facts:
• Ultrasound frequency is above 20 kHz, beyond human hearing range.
• Used commonly for imaging organs, foetal scanning, and detecting kidney stones/gallstones.
• Unlike X-rays, ultrasound does not ionize tissue.
RRB Group D 2025-26 – 21 August 2026 Shift 2 · Range of hearing, infrasonic and ultrasonic sound

Q5. A sound wave with a frequency of 15 Hz would be best categorized as which type?

  1. (a) Ultrasonic sound
  2. (b) Infrasonic sound
  3. (c) Audible sound
  4. (d) Hypersonic sound
Answer & solution

Answer: (b) Infrasonic sound

Correct Answer:
Correct Answer: Option 2 – Infrasonic sound
Explanation:
The audible range for humans is about 20 Hz to 20,000 Hz. A frequency of 15 Hz lies below this range, so it is classified as infrasonic sound.
RRB Group D 2025-26 – 21 August 2026 Shift 3 · Reflection of sound, echo and reverberation

Q6. How does a stethoscope use the principle of reflection of sound?

  1. (a) By directing sound outward
  2. (b) By filtering sound by frequency
  3. (c) By absorbing sound waves
  4. (d) Through multiple reflections within its tube to transmit heartbeat sounds
Answer & solution

Answer: (d) Through multiple reflections within its tube to transmit heartbeat sounds

Correct Answer:
Correct Answer: Option 4 – Through multiple reflections within its tube to transmit heartbeat sounds
Explanation:
A stethoscope works on the principle ofmultiple reflection of sound. Sound produced by the heartbeat undergoes repeated reflections along the tube's walls, reaching the doctor's ears with negligible loss of energy/intensity.
Important Facts:
• This is a classic example of sound reflection used in NCERT Class 9/10 science.
RRB Group D 2025-26 – 20 August 2026 Shift 1 · Range of hearing, infrasonic and ultrasonic sound

Q7. How does the human ear respond to frequencies below 20 Hz?

  1. (a) The human ear amplifies only high frequencies
  2. (b) The human ear converts low frequencies into high frequencies
  3. (c) The human ear cannot hear frequencies below 20 Hz
  4. (d) The human ear can hear all frequencies below 20 Hz
Answer & solution

Answer: (c) The human ear cannot hear frequencies below 20 Hz

Correct Answer:
Correct Answer: Option 3 – The human ear cannot hear frequencies below 20 Hz
Explanation:
The audible range for the human ear is approximately20 Hz to 20,000 Hz. Sound waves with frequency below 20 Hz are calledinfrasonicwaves and cannot be perceived by the human ear.
Important Facts:
• Frequencies above 20,000 Hz are called ultrasonic waves.
• Bats and dolphins can perceive ultrasonic frequencies.
• Elephants and whales can produce/detect infrasonic frequencies.
RRB Group D 2025-26 – 20 August 2026 Shift 3 · Numericals on sound and waves

Q8. A sound wave has a frequency of 500 Hz and travels at 340 m/s. What is its wavelength?

  1. (a) Its wavelength is 170 meters
  2. (b) Its wavelength is 0.68 meters
  3. (c) Its wavelength is 0.34 meters
  4. (d) Its wavelength is 1.47 meters
Answer & solution

Answer: (b) Its wavelength is 0.68 meters

Correct Answer:
Option 2 – Its wavelength is 0.68 meters
Explanation:
Wavelength (λ) = velocity (v) / frequency (f) = 340 / 500 =0.68 m.
RRB Group D 2025-26 – 20 August 2026 Shift 3 · Reflection of sound, echo and reverberation

Q9. Which of the following best explains why an echo is heard when a sound is produced near a large, flat wall?

  1. (a) Sound is absorbed by the wall and then emitted slowly.
  2. (b) Sound waves pass directly through the wall and reappear on the other side.
  3. (c) The wall vibrates and produces a new sound identical to the original.
  4. (d) Sound waves reflect off the wall and return to the listener after a short delay.
Answer & solution

Answer: (d) Sound waves reflect off the wall and return to the listener after a short delay.

Correct Answer:
Option 4 – Sound waves reflect off the wall and return to the listener after a short delay.
Explanation:
An echo is heard because sound waves, upon striking a large, hard, flat surface, getreflectedback and reach the listener's ear after a time delay sufficient for the brain to perceive it as a distinct repeated sound (requires at least ~17 m distance from the reflecting surface, giving a delay of at least 0.1 s).
RRB Group D 2025-26 – 20 August 2026 Shift 3 · Loudness, pitch, quality and decibel

Q10. Which statement best explains the difference between a low pitch sound and a high pitch sound?

  1. (a) A low pitch sound is always louder than a high pitch sound.
  2. (b) A low pitch sound has a lower frequency compared to a high pitch sound.
  3. (c) A low pitch sound has a higher frequency than a high pitch sound.
  4. (d) A low pitch sound and a high pitch sound have the same frequency.
Answer & solution

Answer: (b) A low pitch sound has a lower frequency compared to a high pitch sound.

Correct Answer:
Option 2 – A low pitch sound has a lower frequency compared to a high pitch sound.
Explanation:
Pitch is determined by frequency: alower frequencyproduces a low-pitched sound, while a higher frequency produces a high-pitched sound. Pitch is independent of loudness (which depends on amplitude).
RRB Group D 2025-26 – 19 August 2026 Shift 1 · Reflection of sound, echo and reverberation

Q11. A group of engineers has been tasked with designing a lecture hall for maximum speech intelligibility. Which combination of features should they prioritize to minimize reverberation?

  1. (a) Soft wall coverings, acoustic ceiling tiles, and cushioned seating
  2. (b) High ceiling with reflective panels
  3. (c) Polished stone floors and glass walls
  4. (d) Cushioned seating with bare walls and floors
Answer & solution

Answer: (a) Soft wall coverings, acoustic ceiling tiles, and cushioned seating

Correct Answer:
Correct Answer: Option 1 – Soft wall coverings, acoustic ceiling tiles, and cushioned seating
Explanation:
Reverberation (persistence of sound due to multiple reflections) is minimized by usingsound-absorbing materials.Soft wall coverings, acoustic ceiling tiles, and cushioned/upholstered seatingall absorb sound energy rather than reflecting it, reducing echo and improving speech clarity.
Important Facts:
• Hard, smooth, reflective surfaces (glass, polished stone, bare walls) increase reverberation.
• Auditoriums and concert halls use acoustic tiles, carpets, and curtains to control reverberation time.
• Excessive reverberation reduces speech intelligibility by overlapping successive sounds.
RRB Group D 2025-26 – 19 August 2026 Shift 1 · Wave characteristics (frequency, wavelength, amplitude, time period)

Q12. A wave has a frequency of 200 Hz and a wavelength of 1.7 m. What does their product represent?

  1. (a) The period of the wave
  2. (b) The amplitude of the wave
  3. (c) The speed of the wave
  4. (d) The loudness of the wave
Answer & solution

Answer: (c) The speed of the wave

Correct Answer:
Correct Answer: Option 3 – The speed of the wave
Explanation:
The wave equation states thatwave speed (v) = frequency (f) × wavelength (λ). Thus, the product of frequency (200 Hz) and wavelength (1.7 m) gives thespeed of the wave(= 340 m/s, approximately the speed of sound in air).
Important Facts:
• v = f × λ is a fundamental relation for all types of waves.
• The period (T) is the reciprocal of frequency: T = 1/f.
RRB Group D 2025-26 – 19 August 2026 Shift 2 · Nature and types of waves (longitudinal, transverse)

Q13. When a bell rings inside a jar and the air is slowly removed, what happens to the sound?

  1. (a) The sound becomes louder
  2. (b) The sound becomes fainter as air is removed
  3. (c) The pitch of the sound increases
  4. (d) The bell stops vibrating immediately as air is removed
Answer & solution

Answer: (b) The sound becomes fainter as air is removed

Correct Answer:
Correct Answer: Option 2 – The sound becomes fainter as air is removed
Explanation:
Sound is a mechanical wave and requires a material medium (like air) to travel. As air is gradually pumped out of the jar, fewer air molecules remain to transmit the vibrations from the bell, so the sound becomes progressivelyfainter, eventually becoming inaudible in a near-vacuum, even though the bell continues to vibrate.
Important Facts:
• This is a classic demonstration proving that sound cannot travel through a vacuum, unlike light.
• The bell's hammer mechanism can still be seen striking (if visible) even when no sound is heard.
RRB Group D 2025-26 – 19 August 2026 Shift 3 · Loudness, pitch, quality and decibel

Q14. Which pair of changes will increase both the loudness and the pitch of a sound produced by a guitar string?

  1. (a) Decreasing amplitude and increasing frequency
  2. (b) Increasing amplitude and decreasing frequency
  3. (c) Increasing amplitude and increasing frequency
  4. (d) Decreasing amplitude and decreasing frequency
Answer & solution

Answer: (c) Increasing amplitude and increasing frequency

Correct Answer:
Option 3 – Increasing amplitude and increasing frequency
Explanation:
Loudnessof a sound depends on theamplitudeof the vibration — higher amplitude means louder sound.Pitchdepends on thefrequencyof vibration — higher frequency means higher pitch. To increase both loudness and pitch simultaneously, both amplitude and frequency must be increased.
Important Facts:
• Loudness is measured in decibels (dB).
• Pitch is related to frequency, measured in hertz (Hz).
Correct Answer: Option 3 – Increasing amplitude and increasing frequency
RRB Group D 2025-26 – 18 August 2026 Shift 1 · Reflection of sound, echo and reverberation

Q15. Which of the following best describes the effect of excessive reverberation in a large auditorium during a speech?

  1. (a) It improves the quality of music by making it sharper and more distinct.
  2. (b) It causes spoken words to overlap and become unclear to the audience.
  3. (c) It allows the audience to hear sound for a shorter duration than intended.
  4. (d) It increases the volume of the speaker's voice for everyone present.
Answer & solution

Answer: (b) It causes spoken words to overlap and become unclear to the audience.

Excessive reverberation
Reverberation is the persistence of sound due to repeated reflections. When reverberation is excessive (as in a poorly designed large auditorium), successive spoken syllables/words overlap with reflected sound of previous ones before dying out.
This overlapping makes speech unclear and difficult for the audience to understand, which is why auditoriums use sound-absorbing materials to control reverberation time.
Correct Answer: It causes spoken words to overlap and become unclear to the audience.
RRB Group D 2025-26 – 18 August 2026 Shift 3 · Human ear and sound production

Q16. Why does touching a vibrating tuning fork to water cause ripples?

  1. (a) Vibrations from the fork transfer energy, moving water
  2. (b) Water forms ripples without vibration
  3. (c) The fork heats the water and makes ripples
  4. (d) The fork makes sound but does not move water
Answer & solution

Answer: (a) Vibrations from the fork transfer energy, moving water

Concept: Sound and Vibration
A tuning fork produces sound by vibrating rapidly. When its vibrating prongs touch the surface of water, the mechanical vibrations (energy) are transferred directly to the water molecules, causing the water surface to move up and down and produce ripples. This is a simple, classic demonstration that sound is caused by vibration and that vibrating bodies transfer energy to whatever they touch.
Correct Answer: Vibrations from the fork transfer energy, moving water
RRB Group D 2025-26 – 18 August 2026 Shift 3 · Numericals on sound and waves

Q17. A student shouts facing a building 220 meters away. If the speed of sound is 330 m/s, how long after the shout will the echo be heard?

  1. (a) 0.44 s
  2. (b) 0.67 s
  3. (c) 2.20 s
  4. (d) 1.33 s
Answer & solution

Answer: (d) 1.33 s

Calculation: Echo Time
The sound must travel to the building (220 m) and then reflect back to the student (another 220 m), so the total distance travelled by the sound wave is:
• Total distance = 2 × 220 m = 440 m
• Speed of sound = 330 m/s
• Time = Distance / Speed = 440 / 330 = 1.333 s ≈ 1.33 s
Correct Answer: 1.33 s
RRB Group D 2025-26 – 17 August 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q18. Which of the following statements best explains why sound can NOT travel through a vacuum?

  1. (a) Sound travels faster in a vacuum because there are no obstacles.
  2. (b) Sound waves in a vacuum change their pitch to inaudible levels.
  3. (c) Sound needs light to travel, and a vacuum lacks light.
  4. (d) Sound requires particles to transfer energy, but a vacuum has no particles.
Answer & solution

Answer: (d) Sound requires particles to transfer energy, but a vacuum has no particles.

Sound Needs a Material Medium
Sound is amechanical wave— it propagates by the vibration of particles of the medium (solid, liquid, or gas), transferring energy from one particle to the next through collisions/oscillations.
Avacuum has no particles, so there is nothing to vibrate and carry the sound energy forward. Hence sound cannot travel through a vacuum. (This is unlike light, which is an electromagnetic wave and travels through vacuum.)
Correct Answer: Sound requires particles to transfer energy, but a vacuum has no particles.
RRB Group D 2025-26 – 17 August 2026 Shift 1 · Reflection of sound, echo and reverberation

Q19. Which of the following actions would most effectively reduce reverberation in a large auditorium?

  1. (a) Removing some of the seats in the auditorium
  2. (b) Painting the walls with regular paint
  3. (c) Adding more loudspeakers to the auditorium
  4. (d) Installing curtains and carpets throughout the space
Answer & solution

Answer: (d) Installing curtains and carpets throughout the space

Reducing Reverberation
Reverberation is caused by multiple reflections of sound off hard surfaces, causing prolonged persistence of sound. It is reduced by usingsound-absorbing materialson walls, floors and ceilings.
Installingcurtains and carpetsthroughout the auditorium increases sound absorption significantly, cutting down reflected sound and reducing reverberation time. Regular paint and loudspeakers do not absorb sound, and removing seats would actually reduce absorption (seats/audience absorb sound).
Correct Answer: Installing curtains and carpets throughout the space
RRB Group D 2025-26 – 17 August 2026 Shift 2 · Reflection of sound, echo and reverberation

Q20. Which of the following conditions is necessary to hear an echo clearly?

  1. (a) Time interval must be at least 0.1 s
  2. (b) Source of sound must be very strong
  3. (c) The reflecting surface must be soft
  4. (d) Listener must be moving
Answer & solution

Answer: (a) Time interval must be at least 0.1 s

Condition for a Clear Echo
The human ear retains the sensation of sound for about 0.1 second (persistence of hearing). For an echo to be heard as distinct from the original sound, the time gap between the original sound and its reflection must beat least 0.1 s, which corresponds to the reflecting surface being at least about 17 m away (using speed of sound ≈ 340 m/s).
• A very strong source, soft reflecting surface, or a moving listener are not the defining conditions for hearing a clear echo — the minimum time gap is the key requirement.
• In fact, hard, large surfaces reflect sound better than soft ones.
Correct Answer: Time interval must be at least 0.1 s
RRB Group D 2025-26 – 17 August 2026 Shift 2 · Nature and types of waves (longitudinal, transverse)

Q21. When a tuning fork vibrates and produces sound, what type of waves is created in the surrounding air?

  1. (a) A longitudinal wave with compressions and rarefactions.
  2. (b) A surface wave moving in circles.
  3. (c) A transverse wave with crests and troughs.
  4. (d) A wave where energy moves but particles do not.
Answer & solution

Answer: (a) A longitudinal wave with compressions and rarefactions.

Nature of Sound Waves
Sound waves travel through air aslongitudinal waves, in which particles of the medium vibrate back and forth parallel to the direction of wave propagation, creating regions of compression (high pressure) and rarefaction (low pressure).
• Transverse waves (crests and troughs) occur in media like strings or water surfaces and in light waves, not in sound waves through air.
• Sound cannot travel as a surface wave through air produced by a tuning fork.
Correct Answer: A longitudinal wave with compressions and rarefactions.
RRB Group D 2025-26 – 14 August 2026 Shift 1 · Range of hearing, infrasonic and ultrasonic sound

Q22. Why can humans NOT hear the sound produced by a bat?

  1. (a) The frequency of bat sounds is above 20,000 Hz.
  2. (b) The frequency of bat sounds is below 20 Hz.
  3. (c) The sound travels only in water.
  4. (d) The sound is too loud for humans.
Answer & solution

Answer: (a) The frequency of bat sounds is above 20,000 Hz.

Answer: Option 1 – The frequency of bat sounds is above 20,000 Hz.
• Bats produceultrasonic sound (> 20,000 Hz), beyond the human hearing range (20–20,000 Hz).
RRB Group D 2025-26 – 14 August 2026 Shift 1 · Reflection of sound, echo and reverberation

Q23. Which of the following changes will most likely reduce the clarity of an echo in a large room?

  1. (a) Painting the walls a light colour
  2. (b) Adding thick curtains to the walls
  3. (c) Installing smooth ceramic tiles
  4. (d) Making the room larger
Answer & solution

Answer: (b) Adding thick curtains to the walls

Answer: Option 2 – Adding thick curtains to the walls.
• Soft, thick curtainsabsorb sound, reducing reflection and hence the echo.
RRB Group D 2025-26 – 14 August 2026 Shift 3 · Loudness, pitch, quality and decibel

Q24. If two sounds have the same frequency but different amplitudes, what will be different?

  1. (a) Their pitch will be different
  2. (b) Their speed will be different
  3. (c) Their loudness will be different
  4. (d) Their direction will be different
Answer & solution

Answer: (c) Their loudness will be different

Answer: Option 3 – Their loudness will be different.
• Same frequency ⇒ same pitch; different amplitude ⇒ differentloudness.
RRB Group D 2025-26 – 12 August 2026 Shift 1 · Speed of sound in different media

Q25. Which of the following statements best describes how sound travels through different materials?

  1. (a) Sound travels at the same speed in all materials.
  2. (b) Sound travels fastest in liquids and slowest in gases.
  3. (c) Sound travels faster in solids than in liquids and gases.
  4. (d) Sound travels fastest in gases and slowest in solids.
Answer & solution

Answer: (c) Sound travels faster in solids than in liquids and gases.

Answer: Option 3 – Sound travels faster in solids than in liquids and gases.
• Sound needs a medium and travels fastest where particles are closest, i.e.solids > liquids > gases.
RRB Group D 2025-26 – 12 August 2026 Shift 1 · Loudness, pitch, quality and decibel

Q26. Which statement is true about two sounds with the same amplitude but different frequencies?

  1. (a) They have the same frequency but different amplitudes
  2. (b) They have the same loudness but different pitch
  3. (c) They travel at different speeds in the same medium
  4. (d) They have the same pitch but different loudness
Answer & solution

Answer: (b) They have the same loudness but different pitch

Answer: Option 2 – They have the same loudness but different pitch.
• Amplitude decides loudness (same here); frequency decides pitch (different) ⇒ same loudness, different pitch.
RRB Group D 2025-26 – 12 August 2026 Shift 2 · Reflection of sound, echo and reverberation

Q27. Which of the following best describes an echo in the context of sound propagation and reflection?

  1. (a) An echo is a continuous sound produced by a vibrating object.
  2. (b) An echo is a change in the pitch of a sound due to distance from the source.
  3. (c) An echo is a reflected sound heard distinctly after the original sound.
  4. (d) An echo is a louder sound produced when sound reflects from a wall.
Answer & solution

Answer: (c) An echo is a reflected sound heard distinctly after the original sound.

Answer: Option 3 – A reflected sound heard distinctly after the original sound.
• Anechois a reflected sound heard separately after the direct sound (needs a time gap ≥ 0.1 s).
RRB Group D 2025-26 – 12 August 2026 Shift 2 · Numericals on sound and waves

Q28. A certain sound wave has a frequency of 400 Hz. What is the time period for this wave?

  1. (a) 0.25 seconds
  2. (b) 0.0025 seconds
  3. (c) 0.025 seconds
  4. (d) 0.05 seconds
Answer & solution

Answer: (b) 0.0025 seconds

Answer: Option 2 – 0.0025 seconds.
• T = 1/f = 1/400 =0.0025 s.
RRB Group D 2025-26 – 12 August 2026 Shift 3 · Numericals on sound and waves

Q29. A student shouts facing a tall building 176 meters away and hears the echo after 1 second. What is the speed of sound in air based on this observation?

  1. (a) 400 m/s
  2. (b) 330 m/s
  3. (c) 352 m/s
  4. (d) 176 m/s
Answer & solution

Answer: (c) 352 m/s

Answer: Option 3 – 352 m/s.
• Echo travels to the wall and back = 2×176 = 352 m in 1 s ⇒ speed = 352/1 =352 m/s.
RRB Group D 2025-26 – 12 August 2026 Shift 3 · Speed of sound in different media

Q30. Which of the following statements best describes how the speed of sound changes as it moves from air into water?

  1. (a) The speed of sound decreases when moving from air to water.
  2. (b) The speed of sound remains the same in both air and water.
  3. (c) The speed of sound increases when moving from air to water.
  4. (d) The speed of sound is slower in water than in air.
Answer & solution

Answer: (c) The speed of sound increases when moving from air to water.

Answer: Option 3 – The speed of sound increases when moving from air to water.
• Sound travelsfaster in water(denser, more elastic medium) than in air.
RRB Group D 2025-26 – 11 August 2026 Shift 2 · Loudness, pitch, quality and decibel

Q31. If a musical instrument produces a sound with a frequency of 200 Hz, which of the following frequencies will produce a sound of higher pitch than this instrument?

  1. (a) 120 Hz
  2. (b) 180 Hz
  3. (c) 150 Hz
  4. (d) 350 Hz
Answer & solution

Answer: (d) 350 Hz

Answer: Option 4 – 350 Hz.
• Higher frequency = higher pitch. Only350 Hzexceeds 200 Hz.
RRB Group D 2025-26 – 11 August 2026 Shift 3 · Nature and types of waves (longitudinal, transverse)

Q32. Which of the following features is unique to longitudinal sound waves?

  1. (a) They move as crests and troughs through a medium.
  2. (b) They travel as side-to-side pulses in a medium.
  3. (c) They vibrate only up and down in a medium.
  4. (d) They form regions of compression and rarefaction.
Answer & solution

Answer: (d) They form regions of compression and rarefaction.

Answer: Option 4 – They form regions of compression and rarefaction.
•Longitudinal waves(like sound) travel as alternatingcompressions and rarefactionsalong the direction of motion.
RRB Group D 2025-26 – 11 August 2026 Shift 3 · Loudness, pitch, quality and decibel

Q33. If two notes are played, one at 300 Hz and the other at 600 Hz, how do their pitches compare?

  1. (a) The 300 Hz note has a higher pitch
  2. (b) Both notes have the same pitch
  3. (c) The 600 Hz note has a higher pitch
  4. (d) The 300 Hz note is louder than the 600 Hz note
Answer & solution

Answer: (c) The 600 Hz note has a higher pitch

Answer: Option 3 – The 600 Hz note has a higher pitch.
• Higher frequency = higher pitch, so600 Hzhas the higher pitch.
RRB Group D 2025-26 – 10 August 2026 Shift 1 · Reflection of sound, echo and reverberation

Q34. Which of the following methods is commonly used to measure reverberation in a room?

  1. (a) Calculating the frequency of the sound wave
  2. (b) Counting the number of echoes heard
  3. (c) Measuring the speed of sound in the room
  4. (d) Timing how long sound persists after the source stops
Answer & solution

Answer: (d) Timing how long sound persists after the source stops

Answer: Option 4 — Timing how long sound persists after the source stops
Reverberation time = how long soundpersists after the source stops→ Option 4.
RRB Group D 2025-26 – 10 August 2026 Shift 3 · Loudness, pitch, quality and decibel

Q35. Why does a tightly stretched drum skin make a different sound than a loose one?

  1. (a) A tightly stretched drum skin is heavier than a loose one
  2. (b) A tightly stretched drum skin contains more air than a loose one
  3. (c) A tightly stretched drum skin is thicker than a loose one
  4. (d) A tightly stretched drum skin vibrates faster than a loose one
Answer & solution

Answer: (d) A tightly stretched drum skin vibrates faster than a loose one

Answer: Option 4 – A tightly stretched drum skin vibrates faster than a loose one.
• Greater tension raises thefrequencyof vibration, giving a higher-pitched sound. A loose skin vibrates slowly ⇒ lower pitch.
RRB Group D 2025-26 – 10 August 2026 Shift 3 · Nature and types of waves (longitudinal, transverse)

Q36. Which of the following statements best explains why striking a bell in a vacuum does NOT produce sound we can hear?

  1. (a) Sound needs a medium to travel from the bell to our ears
  2. (b) The bell does not vibrate in a vacuum
  3. (c) Light from the bell reaches our ears instead
  4. (d) The bell’s material prevents sound from traveling
Answer & solution

Answer: (a) Sound needs a medium to travel from the bell to our ears

Answer: Option 1 – Sound needs a medium to travel.
• Sound is a mechanical wave; it requires amaterial medium. In a vacuum there is no medium, so the sound cannot reach our ears.
RRB Group D 2025-26 – 9 August 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q37. What forms the compressions and rarefactions in a sound wave traveling through air?

  1. (a) Particles in the air permanently collide and stick together
  2. (b) Particles in the air bunch together and spread apart alternately
  3. (c) Particles in the air move in a spiral pattern as the wave passes
  4. (d) Particles in the air maintain equal spacing at all times
Answer & solution

Answer: (b) Particles in the air bunch together and spread apart alternately

Answer: Option 2 — Particles bunch together and spread apart alternately
Compressions (bunched) and rarefactions (spread) form as air particlesoscillate→ Option 2.
RRB Group D 2025-26 – 9 August 2026 Shift 2 · Numericals on sound and waves

Q38. If a sound wave has a wavelength of 0.5 m and travels at 250 m/s, what is its frequency?

  1. (a) 250 Hz
  2. (b) 500 Hz
  3. (c) 1000 Hz
  4. (d) 125 Hz
Answer & solution

Answer: (b) 500 Hz

Answer: Option 2 — 500 Hz
f = v/λ = 250/0.5 =500 Hz→ Option 2.
RRB Group D 2025-26 – 6 August 2026 Shift 1 · Applications of ultrasound and SONAR

Q39. Why are ultrasound waves used for cleaning delicate objects such as jewelry?

  1. (a) They heat the jewelry to remove stains
  2. (b) They dissolve dirt using special chemicals
  3. (c) They attract dirt particles with magnetism
  4. (d) They cause tiny vibrations that dislodge dirt
Answer & solution

Answer: (d) They cause tiny vibrations that dislodge dirt

Answer: Option 4 — They cause tiny vibrations that dislodge dirt
Ultrasonic cleaning creates vibrating cavitation bubbles thatdislodge dirtfrom delicate objects → Option 4.
RRB Group D 2025-26 – 6 August 2026 Shift 3 · Loudness, pitch, quality and decibel

Q40. A student beats a large drum and blows a high-pitched plastic whistle at the same time. Which property of the sound waves is the main reason why the whistle sounds so different from the drum?

  1. (a) The loudness of the sound waves is always the same for both
  2. (b) The speed of the sound waves through the air is different for each
  3. (c) The frequency of the sound waves is different for each instrument
  4. (d) The type of air in the room changes when each instrument is played
Answer & solution

Answer: (c) The frequency of the sound waves is different for each instrument

Answer: Option 3 — The frequency of the sound waves is different
Pitch depends onfrequency; the whistle (high pitch) has a higher frequency than the drum → Option 3.
RRB Group D 2025-26 – 5 August 2026 Shift 1 · Applications of ultrasound and SONAR

Q41. Which of the following best explains how ultrasound is used in medical imaging?

  1. (a) Ultrasound waves pass through tissues without any reflection for imaging
  2. (b) Ultrasound is used to measure temperature changes in body tissues
  3. (c) Ultrasound uses low-frequency sound waves for medical imaging
  4. (d) Ultrasound waves reflect off tissues to create real-time images on a monitor
Answer & solution

Answer: (d) Ultrasound waves reflect off tissues to create real-time images on a monitor

Answer: Option 4 — Ultrasound reflects off tissues to create real-time images
Medical ultrasound sends high-frequency waves thatreflect off tissues (echoes)to form real-time images → Option 4.
RRB Group D 2025-26 – 4 August 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q42. What is the role of a medium in the propagation of sound?

  1. (a) A medium provides particles that carry sound vibrations
  2. (b) A medium blocks sound from moving between places
  3. (c) A medium is not needed for sound to move from one point to another
  4. (d) A medium changes sound into light energy for travel
Answer & solution

Answer: (a) A medium provides particles that carry sound vibrations

Answer: Option 1 — A medium provides particles that carry sound vibrations
Sound is a mechanical wave; it needs amaterial mediumwhose particles vibrate to carry it → Option 1.
RRB Group D 2025-26 – 4 August 2026 Shift 2 · Applications of ultrasound and SONAR

Q43. How can ultrasound be used to detect flaws in a metal block?

  1. (a) By measuring the color of the metal
  2. (b) By heating the block with the ultrasound
  3. (c) By measuring the reflection of waves at internal cracks
  4. (d) By observing visible movement inside the block
Answer & solution

Answer: (c) By measuring the reflection of waves at internal cracks

Answer: Option 3 — By measuring the reflection of waves at internal cracks
Ultrasound wavesreflect backfrom internal cracks/flaws; detecting these echoes reveals the defects → Option 3.
RRB Group D 2025-26 – 3 August 2026 Shift 1 · Applications of ultrasound and SONAR

Q44. What is the key reason for ultrasound to be used in cleaning delicate electronic components?

  1. (a) It removes dirt without causing physical damage
  2. (b) It scratches the surface to remove debris
  3. (c) It chemically alters the material for cleaning
  4. (d) It melts the components for better cleaning
Answer & solution

Answer: (a) It removes dirt without causing physical damage

Answer: Option 1 — It removes dirt without causing physical damage
Concept:Ultrasonic cleaning uses high-frequency sound waves that create tiny vibrating bubbles (cavitation) in the cleaning liquid. These dislodge dirt from fine grooves and delicate partswithout mechanical scratching, making it ideal for electronic components → Option 1.
RRB Group D 2025-26 – 3 August 2026 Shift 2 · Speed of sound in different media

Q45. Why can you hear a train approaching when you put your ear on the railway track?

  1. (a) Vibrations travel only along the surface of the rails
  2. (b) Rails vibrate more loudly than air does
  3. (c) Sound travels efficiently through solid steel rails to your ear
  4. (d) Sound moves through the air above the track
Answer & solution

Answer: (c) Sound travels efficiently through solid steel rails to your ear

Answer: Option 3 — Sound travels efficiently through solid steel rails to your ear
Sound travelsfaster and more efficiently through solids(steel) than through air, so you hear the approaching train earlier through the rail → Option 3.
RRB NTPC UG CBT-1 2025-26 – 18 June 2026 Shift 2 · Nature and types of waves (longitudinal, transverse)

Q46. You can often hear a person speaking from behind a tall fence even if you cannot see them. Which property of sound waves explains this phenomenon?

  1. (a) Diffraction
  2. (b) Reflection
  3. (c) Refraction
  4. (d) Polarization
Answer & solution

Answer: (a) Diffraction

Topic: Properties of Sound Waves
Diffractionis the bending of waves around obstacles or through openings.
Sound waves have long wavelengths (cm to meters range).
When sound encounters a fence, it bends around the edges and spreads into the shadow region behind it.
This allows us to hear someone speaking even when we cannot see them directly.
Answer: Option 1 → Diffraction
RRB NTPC UG CBT-1 2025-26 – 18 June 2026 Shift 3 · Reflection of sound, echo and reverberation

Q47. What do we call the lasting of sound caused by repeated reflections?

  1. (a) Resonance
  2. (b) Quality
  3. (c) Reverberation
  4. (d) Echo
Answer & solution

Answer: (c) Reverberation

Topic: Sound Reflection Phenomena
Reverberation:Persistence of sound after the source stops due to multiple reflections from surfaces (walls, ceiling, floor).
Echo:Single distinct reflection heard separately (requires minimum distance ~17m).
Resonance:Vibration of object at its natural frequency when exposed to matching external frequency.
Quality/Timbre:Characteristic that distinguishes sounds of same pitch and loudness.
Answer: Option 3 → Reverberation
RRB NTPC UG CBT-1 2025-26 – 16 June 2026 Shift 1 · Reflection of sound, echo and reverberation

Q48. What minimum distance should the reflecting surface be for an echo to be heard?

  1. (a) 10 m
  2. (b) 17.2 m
  3. (c) 5 m
  4. (d) 34.2 m
Answer & solution

Answer: (b) 17.2 m

Topic:Echo and Sound Reflection
For echo to be heard distinctly, time gap ≥ 0.1 s
Speed of sound in air = 344 m/s (approx)
Distance = Speed × Time
Total distance = 344 × 0.1 = 34.4 m
This is round trip (to surface and back)
Minimum distance to surface = 34.4/2 = 17.2 m
Answer: Option 2 → 17.2 m
RRB NTPC UG CBT-1 2025-26 – 16 June 2026 Shift 3 · Loudness, pitch, quality and decibel

Q49. Which term describes sound energy received per second per unit area?

  1. (a) Pitch
  2. (b) Frequency
  3. (c) Echo
  4. (d) Intensity
Answer & solution

Answer: (d) Intensity

Topic: Sound Energy and Intensity
Intensityis defined as sound energy received per second per unit area.
Formula:I = P/A(Power per unit area)
Unit: W/m² (Watt per square meter)
Other terms:
• Pitch - Related to frequency of sound
• Frequency - Number of vibrations per second
• Echo - Reflection of sound waves
Answer: Option 4 → Intensity
RRB NTPC UG CBT-1 2025-26 – 14 June 2026 Shift 2 · Speed of sound in different media

Q50. In which of the following media does sound move the fastest at a particular temperature?

  1. (a) Water
  2. (b) Air
  3. (c) Vaccum
  4. (d) Iron
Answer & solution

Answer: (d) Iron

Topic: Speed of Sound in Different Media
Speed of sound depends onelasticity and densityof medium.
Speed hierarchy:Solids > Liquids > Gases
Soundcannot travel in vacuum(no medium).
At room temperature:
Air ≈ 343 m/s
Water ≈ 1480 m/s
Iron ≈ 5130 m/s
Iron (solid) has highest speed among given options.
Answer: Option 4 → Iron
RRB NTPC UG CBT-1 2025-26 – 14 June 2026 Shift 2 · Speed of sound in different media

Q51. What is the approximate speed of sound in air at room temperature (approximately 20°C)?

  1. (a) 5000 m/s
  2. (b) 1500 m/s
  3. (c) 343 m/s
  4. (d) 300 m/s
Answer & solution

Answer: (c) 343 m/s

Topic:Speed of Sound in Air
At room temperature (20°C), sound travels through air
at approximately343 m/s.
Key Facts:
• Speed varies with temperature
• Formula: v ≈ 331 + 0.6T (where T in °C)
• At 20°C: v ≈ 331 + 0.6(20) = 343 m/s
Answer: Option 3 → 343 m/s
SSC GD 2026 – 27 May 2026 Shift 2 · Range of hearing, infrasonic and ultrasonic sound

Q52. Which of these animal groups produces ultrasound?

  1. (a) Dolphins, bats and porpoises
  2. (b) Whales, elephants and giraffes
  3. (c) Dogs, cats and lions
  4. (d) Rhinos, moths and tigers
Answer & solution

Answer: (a) Dolphins, bats and porpoises

Correct Answer: Dolphins, bats and porpoises — Option (a)
Dolphins, bats, and porpoises produce ultrasound for echolocation. About Ultrasound: Definition: Sound with frequency above 20,000 Hz (20 kHz) — beyond the upper limit of human hearing; Infrasound: Below 20 Hz; Audible sound: 20 Hz to 20,000 Hz; Animals that produce/use ultrasound: Bats ( Chiroptera ): Use echolocation (biosonar) to navigate in darkness and locate prey; Emit ultrasonic pulses (20,000 to 100,000 Hz) from their larynx or nose; Interpret echoes to detect obstacles and insects; Dolphins and Porpoises (Cetaceans): Use echolocation for navigation and hunting in water; Produce clicks and squeaks in the ultrasonic range; Dolphins can detect objects as small as a golf ball from 70 meters away; Porpoises ( शिशुमार ): A type of small cetacean (like small dolphins) — also use echolocation; Other ultrasound producers: Some whales (sperm whales), shrews, some mice and rats (for communication), some insects (for communication/defense); About echolocation: The animal emits ultrasonic pulses; The sound bounces off objects and returns as echoes; The animal interprets the time delay and frequency shifts of echoes; This gives precise information about location, size, shape, and movement of objects; Applications of ultrasound: Medical imaging (sonography/ultrasonography); Industrial testing (detecting flaws in metals); Cleaning delicate instruments; Sonar in submarines.
Option (b) Whales, elephants and giraffes — Incorrect: While some whales (like sperm whales) do produce ultrasound for echolocation, elephants and giraffes do NOT produce ultrasound; elephants actually use INFRASOUND (below 20 Hz) for long-distance communication; giraffes produce very low frequency sounds; this combination is incorrect.
Option (c) Dogs, cats and lions — Incorrect: Dogs, cats, and lions do NOT produce ultrasound for echolocation; dogs can HEAR ultrasound (up to 65,000 Hz) and cats can hear up to 79,000 Hz but they do not PRODUCE ultrasound for navigation; lions communicate through roars in the audible frequency range.
Option (d) Rhinos, moths and tigers — Incorrect: Rhinos, moths, and tigers do NOT produce ultrasound for echolocation; some moths can DETECT ultrasound emitted by hunting bats (as a defense mechanism) but they don't produce it for navigation; rhinos use infrasound; tigers roar in audible frequencies.
SSC GD 2026 – 22 May 2026 Shift 3 · Range of hearing, infrasonic and ultrasonic sound

Q53. What is the audible range of sound for humans?

  1. (a) 2 Hz to 2000 kHz
  2. (b) 20 Hz to 200 kHz
  3. (c) 20 Hz to 20,000 Hz
  4. (d) 20 Hz to 2000 Hz
Answer & solution

Answer: (c) 20 Hz to 20,000 Hz

Correct Answer: 20 Hz to 20,000 Hz — Option (c)
The audible range of sound for humans is 20 Hz to 20,000 Hz (20 kHz). Classification of sound by frequency: Infrasound: Below 20 Hz; Humans cannot hear these frequencies; Produced by earthquakes, volcanoes, waterfalls, large animals (elephants, whales); Some animals (elephants, rhinoceroses) use infrasound for long-distance communication; Audible sound (for humans): 20 Hz to 20,000 Hz (20 kHz); This is the normal human hearing range; The range narrows with age — older adults typically cannot hear frequencies above 15,000-16,000 Hz; Ultrasound: Above 20,000 Hz (20 kHz); Humans cannot hear these frequencies; Bats and dolphins use ultrasound for echolocation (navigation and hunting); Applications: medical ultrasound imaging, sonar, industrial testing (non-destructive testing); Dog hearing: 40 Hz to 65,000 Hz (hence dog whistles work at frequencies humans cannot hear); Bat hearing: 20 Hz to 120,000 Hz; Dolphin hearing: up to 150,000 Hz; About sound: Sound is a mechanical wave that requires a medium (solid, liquid, or gas) to propagate; Speed of sound in air at 20°C: approximately 343 m/s; Speed increases in denser media: faster in water (1,480 m/s) and even faster in solids (steel: ~5,000 m/s).
Option (a) 2 Hz to 2000 kHz — Incorrect: The human audible range starts at 20 Hz (not 2 Hz) and ends at 20,000 Hz = 20 kHz (not 2000 kHz); 2000 kHz = 2,000,000 Hz is far beyond any biological hearing range.
Option (b) 20 Hz to 200 kHz — Incorrect: The human audible range is 20 Hz to 20,000 Hz (20 kHz) — NOT up to 200 kHz; 200 kHz = 200,000 Hz is well into the ultrasound range; some bats and marine mammals can hear up to such frequencies, but not humans.
Option (d) 20 Hz to 2000 Hz — Incorrect: The human audible range extends up to 20,000 Hz (20 kHz) — NOT just 2,000 Hz; 2,000 Hz is only 2 kHz, which is a mid-frequency sound; limiting human hearing to 2,000 Hz is incorrect. The candidate selected this incorrect option.
SSC GD 2026 – 19 May 2026 Shift 2 · Reflection of sound, echo and reverberation

Q54. For approximately how long does the experience of sound persist in our brain?

  1. (a) 0.01 s
  2. (b) 0.1 s
  3. (c) 1 s
  4. (d) 10 s
Answer & solution

Answer: (b) 0.1 s

Correct Answer: 0.1 s — Option (b)
The persistence of hearing (also called 'auditory persistence') is approximately 0.1 seconds (100 milliseconds). This is analogous to the persistence of vision for light, which is also about 0.1 seconds. This means that after a sound stimulus ends, our brain continues to perceive it for about 0.1 seconds. This principle is important in music and speech perception — it allows us to connect individual sounds into a continuous stream. This is why we can hear musical notes as connected melodies rather than isolated sounds, and why speech sounds blend into understandable words. The 0.1-second persistence is used in the design of audio systems and has applications in hearing aids and cochlear implants.
Option (a) 0.01 s — Incorrect: 0.01 seconds (10 milliseconds) is too short for auditory persistence; the brain retains sound experience for approximately 0.1 seconds.
Option (c) 1 s — Incorrect: 1 second is too long for the basic persistence of a sound experience; the standard value of auditory persistence is 0.1 seconds.
Option (d) 10 s — Incorrect: 10 seconds refers to short-term memory retention, not the basic neurological persistence of sound experience, which is about 0.1 seconds.
SSC GD 2026 – 18 May 2026 Shift 3 · Speed of sound in different media

Q55. What is the speed of sound in air at 0°C?

  1. (a) 344 m/s
  2. (b) 331 m/s
  3. (c) 316 m/s
  4. (d) 346 m/s
Answer & solution

Answer: (b) 331 m/s

Correct Answer: 331 m/s — Option (b)
The speed of sound in air at 0°C (273 K) is approximately 331 m/s. The speed of sound in air depends on temperature according to the formula: v = 331 + 0.6T m/s (where T is temperature in °C). At 0°C: v = 331 + 0.6(0) = 331 m/s. As temperature increases, air molecules move faster and collide more frequently, transmitting sound waves more quickly. At 25°C (room temperature), the speed is approximately 346 m/s. Sound travels faster in denser or more elastic media: faster in liquids than gases, and fastest in solids.
Option (a) 344 m/s — Incorrect: 344 m/s is approximately the speed of sound at around 20°C (room temperature), not at 0°C.
Option (c) 316 m/s — Incorrect: 316 m/s is below the actual speed at 0°C; this value does not correspond to any standard atmospheric temperature condition.
Option (d) 346 m/s — Incorrect: 346 m/s corresponds to the speed of sound at approximately 25°C, not at 0°C.
SSC GD 2026 – 9 May 2026 Shift 2 · Applications of ultrasound and SONAR

Q56. Which of the following organs of the human body CANNOT be
imaged using ultrasound waves?
I. Liver II. Gallbladder

  1. (a) Only I
  2. (b) Only II
  3. (c) Both I and II
  4. (d) Neither I nor II
Answer & solution

Answer: (d) Neither I nor II

Correct Answer: Neither I nor II — Option (d)
Both the Liver (Statement I) and Gallbladder (Statement II) CAN be effectively imaged using ultrasound (sonography). Ultrasound imaging (sonography) uses high-frequency sound waves to create images of structures inside the body and is routinely used to image: the Liver (to detect liver diseases, tumours , cirrhosis); the Gallbladder (to detect gallstones — ultrasound is the most preferred investigation for gallstones); kidneys, uterus, ovaries, thyroid, and other soft tissue organs. Since both Liver and Gallbladder can be imaged using ultrasound, 'Neither I nor II' is the correct answer — meaning neither is correct as an organ that CANNOT be imaged by ultrasound.
Option (a) Only I — Incorrect: The Liver CAN be imaged by ultrasound; it is routinely used for liver assessment.
Option (b) Only II — Incorrect: The Gallbladder CAN be imaged by ultrasound; abdominal ultrasound is the gold standard investigation for gallstones.
Option (c) Both I and II — Incorrect: Both organs CAN be imaged by ultrasound; neither the Liver nor the Gallbladder is excluded from ultrasound imaging.
RRB NTPC UG CBT-1 2025-26 – 9 May 2026 Shift 2 · Speed of sound in different media

Q57. What happens to the speed of sound as the temperature of the air increases?

  1. (a) It decreases, because the air becomes less dense
  2. (b) It decreases rapidly because the air molecules move extremely slower
  3. (c) It remains the same, as the speed is constant in a given medium
  4. (d) It increases, because the air molecules move faster
Answer & solution

Answer: (d) It increases, because the air molecules move faster

Topic:Speed of Sound and Temperature
Speed of sound in air: v ∝ √T
Key Facts:
1. Higher temperature → faster molecular motion
2. Faster molecules → quicker sound wave propagation
3. Speed increases by ~0.6 m/s per °C rise
Answer: Option 4 → It increases, because the air molecules move faster
SSC GD 2026 – 8 May 2026 Shift 3 · Human ear and sound production

Q58. What is the primary function of a hearing aid?

  1. (a) To reduce noise
  2. (b) To convert sound waves into light signals
  3. (c) To amplify sound for clear hearing
  4. (d) To change the frequency of sound
Answer & solution

Answer: (c) To amplify sound for clear hearing

Correct Answer: To amplify sound for clear hearing — Option (c)
A hearing aid is an electroacoustic device designed primarily to amplify sound for people with hearing loss (partial deafness). It works by: (1) A microphone picks up environmental sounds and converts them to electrical signals; (2) An amplifier increases the strength of these electrical signals; (3) A speaker (receiver) converts the amplified electrical signals back to sound waves and delivers them to the ear canal. Modern digital hearing aids can also filter background noise, but their PRIMARY function is amplification — making sounds louder and clearer for the user.
Option (a) To reduce noise — Incorrect: While some advanced hearing aids have noise-reduction features, this is a secondary function; the primary purpose is amplification of sound.
Option (b) To convert sound waves into light signals — Incorrect: Converting sound to light is not the function of a hearing aid; it converts sound to electrical signals and back to amplified sound.
Option (d) To change the frequency of sound — Incorrect: Hearing aids do not primarily change the frequency (pitch) of sounds; they amplify sound intensity to compensate for hearing loss.
RRB NTPC UG CBT-1 2025-26 – 8 May 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q59. In a sound wave traveling through air, which of the following terms refers to a region where the particles are crowded together, resulting in high pressure and high density?

  1. (a) Crest
  2. (b) Wavelength
  3. (c) Rarefaction
  4. (d) Compression
Answer & solution

Answer: (d) Compression

Sound Wave Characteristics
Sound waves arelongitudinal waveswhere particles oscillate parallel to wave direction.
Compression:Region where particles are crowded together → high pressure, high density.
Rarefaction:Region where particles are spread apart → low pressure, low density.
Crest/Trough:Terms used for transverse waves (like light), not sound.
Answer: Option 4 → Compression
RRB NTPC UG CBT-1 2025-26 – 8 May 2026 Shift 3 · Range of hearing, infrasonic and ultrasonic sound

Q60. Frequencies higher than the human audible limit of 20,000 Hz are classified as _____.

  1. (a) Infrasound
  2. (b) Supersonic
  3. (c) Tones
  4. (d) Ultrasound
Answer & solution

Answer: (d) Ultrasound

Sound Frequency Classification
Human audible range:20 Hz to 20,000 Hz
Below 20 Hz →Infrasound(earthquakes, elephants)
Above 20,000 Hz →Ultrasound(medical imaging, bats)
Supersonic refers to speed > speed of sound, not frequency
Answer: Option 4 → Ultrasound
SSC GD 2026 – 7 May 2026 Shift 2 · Applications of ultrasound and SONAR

Q61. When ultrasonic waves are sent into a cleaning solution,
what happens to particles of dust, grease, and dirt?

  1. (a) They get evaporated
  2. (b) They get dissolved
  3. (c) They separate and fall down
  4. (d) They get burnt
Answer & solution

Answer: (c) They separate and fall down

Correct Answer: They separate and fall down — Option (c)
When ultrasonic waves (high-frequency sound waves above 20,000 Hz) are passed through a cleaning solution, the process of ultrasonic cleaning occurs. The waves create rapid alternating compression and rarefaction cycles in the liquid, generating millions of tiny cavitation bubbles that implode violently near surfaces. This cavitation effect dislodges and detaches particles of dust, grease, and dirt from the surfaces being cleaned — the particles separate and fall down into the solution. This method is used to clean delicate items like jewellery , surgical instruments, optical lenses, and electronic components.
Option (a) They get evaporated — Incorrect: Dirt particles do not evaporate in the cleaning solution; they are physically dislodged by cavitation.
Option (b) They get dissolved — Incorrect: The cleaning action of ultrasound is mechanical (cavitation), not chemical dissolution; the particles separate from surfaces but do not dissolve.
Option (d) They get burnt — Incorrect: Ultrasonic waves are sound waves (mechanical energy), not heat; they do not burn the particles.
RRB NTPC UG CBT-1 2025-26 – 7 May 2026 Shift 1 · Loudness, pitch, quality and decibel

Q62. How does increasing the Amplitude of a sound wave affect the sound produced?

  1. (a) The speed of the sound increases
  2. (b) The frequency decreases
  3. (c) The pitch increases
  4. (d) The loudness increases
Answer & solution

Answer: (d) The loudness increases

Sound Wave Properties:
Amplitude→ Determines the energy and intensity of sound
Higher amplitude = More energy = Louder sound
Lower amplitude = Less energy = Softer sound
Note:Amplitude does NOT affect:
• Speed (depends on medium)
• Frequency/Pitch (independent property)
Answer: Option 4 → The loudness increases
SSC GD 2026 – 6 May 2026 Shift 1 · Range of hearing, infrasonic and ultrasonic sound

Q63. What is the frequency range of sound that can be heard by the human ear?

  1. (a) 2 Hz to 200 Hz
  2. (b) 20 Hz to 20,000 Hz
  3. (c) 200 Hz to 2000 Hz
  4. (d) 2000 Hz to 20,000 Hz
Answer & solution

Answer: (b) 20 Hz to 20,000 Hz

Correct Answer: 20 Hz to 20,000 Hz — Option (b)
The human ear can detect sound waves in the frequency range of 20 Hz to 20,000 Hz (20 kHz). This range is called the audible range or sonic range. Sound below 20 Hz is called infrasound (e.g., produced by earthquakes, large animals like elephants), and sound above 20,000 Hz is called ultrasound (e.g., used in medical imaging, by bats and dolphins). The human ear is most sensitive to frequencies between 1,000 Hz and 4,000 Hz.
Option (a) 2 Hz to 200 Hz — Incorrect: 2 Hz falls in the infrasound range (below human hearing); 200 Hz is within the audible range but the upper limit of human hearing is 20,000 Hz, not 200 Hz.
Option (c) 200 Hz to 2000 Hz — Incorrect: While this range falls within human hearing, it does not represent the full audible range, which begins at 20 Hz (not 200 Hz).
Option (d) 2000 Hz to 20,000 Hz — Incorrect: This range misses the lower portion of human hearing (20 Hz to 2000 Hz), which is where most speech and low-pitched sounds are heard.
SSC GD 2026 – 30 April 2026 Shift 1 · Reflection of sound, echo and reverberation

Q64. What is the minimum distance of the obstacle from the source of sound to hear a distinct echo?

  1. (a) 34.4 m
  2. (b) 22.5 m
  3. (c) 17.2 m
  4. (d) 10.6 m
Answer & solution

Answer: (c) 17.2 m

Correct Answer: 17.2 m — Option (c)
The minimum distance of the obstacle from the source of sound to hear a distinct echo is 17.2 metres . Scientific explanation: For a person to hear a distinct echo, the reflected sound must reach their ear at least 1/10th of a second (0.1 second) after the original sound, because the human ear can distinguish two sounds only if there is a time gap of at least 0.1 seconds between them; Speed of sound at 20°C (approximately room temperature) = 344 m/s; Total distance travelled by sound (to obstacle and back) = Speed × Time = 344 × 0.1 = 34.4 m; Since sound has to travel to the obstacle AND back, the minimum distance of the obstacle from the source = 34.4/2 = 17.2 metres . Therefore, for a distinct echo to be heard, the reflecting surface must be at least 17.2 metres away from the source of sound. Note: The speed of sound changes with temperature (approximately 331 m/s at 0°C, 344 m/s at 20°C). At the commonly used standard value of 344 m/s, the minimum distance for echo = 17.2 m.
Option (a) 34.4 m — Incorrect: 34.4 metres is the total distance that sound must travel (to the obstacle and back) for the time gap to be 0.1 seconds; the minimum DISTANCE OF THE OBSTACLE from the source is half of this, i.e., 17.2 m — not 34.4 m.
Option (b) 22.5 m — Incorrect: 22.5 metres is not the minimum distance for a distinct echo; this value does not correspond to any standard calculation for echo distance; the correct minimum distance is 17.2 m (based on speed of sound 344 m/s and minimum time gap 0.1 s). The candidate selected this incorrect option.
Option (d) 10.6 m — Incorrect: 10.6 metres is less than the required minimum distance; at this distance, the echo would return in 10.6×2/344 = 0.0616 seconds, which is less than 0.1 seconds — insufficient for the human ear to distinguish the echo from the original sound.
SSC GD 2026 – 30 April 2026 Shift 3 · Reflection of sound, echo and reverberation

Q65. The rolling of thunder is mainly due to:

  1. (a) Resonance of clouds
  2. (b) Successive reflections of sound
  3. (c) Lightning effect
  4. (d) Interference of sound waves
Answer & solution

Answer: (b) Successive reflections of sound

Correct Answer: Successive reflections of sound — Option (b)
Thunder is the sound produced by the rapid expansion of air heated by lightning. The 'rolling' or rumbling effect of thunder occurs because the sound waves from different parts of the lightning bolt (which can be several kilometres long) reach the listener at different times, and the sound is also successively reflected from clouds, mountains, and other surfaces. This prolonged, rolling effect is due to successive reflections of sound.
Option (a) Resonance of clouds — Incorrect: Clouds do not resonate like a musical instrument to produce the rolling effect.
Option (c) Lightning effect — Incorrect: Lightning causes thunder, but the 'rolling' character is due to reflections, not the lightning itself.
Option (d) Interference of sound waves — Incorrect: Interference produces patterns of amplification and cancellation, not the prolonged rolling of thunder.
RRB NTPC GL CBT-1 2025-26 – 22 March 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q66. In which type of wave do particles of the medium move parallel to the direction of wave propagation?

  1. (a) Longitudinal wave
  2. (b) Electromagnetic wave
  3. (c) Transverse wave
  4. (d) Surface wave
Answer & solution

Answer: (a) Longitudinal wave

Wave Motion Types:
Longitudinal Wave:Particles oscillate parallel to wave direction.
Examples: Sound waves, compression waves in springs.
Transverse Wave:Particles oscillate perpendicular to wave direction.
Examples: Light waves, waves on strings.
Surface Wave:Combination of both motions.
In longitudinal waves, compressions and rarefactions travel in the same direction as particle displacement.
Answer: Option 1 → Longitudinal wave
RRB NTPC GL CBT-1 2025-26 – 22 March 2026 Shift 3 · Nature and types of waves (longitudinal, transverse)

Q67. The interference of two identical waves moving in opposite directions produces:

  1. (a) Longitudinal Waves
  2. (b) Standing Waves
  3. (c) Incident Waves
  4. (d) Transversal waves
Answer & solution

Answer: (b) Standing Waves

Topic: Wave Interference
When two identical waves travel in opposite directions,
they superpose and create a pattern where certain points
remain stationary (nodes) and others oscillate with maximum
amplitude (antinodes).
This phenomenon is calledStanding Wavesor
Stationary Waves.
Examples: Vibrating string, organ pipes, resonance tubes.
Answer: Option 2 → Standing Waves
RRB NTPC GL CBT-1 2025-26 – 18 March 2026 Shift 3 · Speed of sound in different media

Q68. Which of the following options is closest to the speed of sound in air?

  1. (a) 340 m/s
  2. (b) 3 x 108 m/s
  3. (c) 3400 km/s
  4. (d) 12000 m/s
Answer & solution

Answer: (a) 340 m/s

Speed of sound in air at 20°C ≈343 m/s
Check other options:
Option 2: 3×10⁸ m/s = Speed of light (not sound)
Option 3: 3400 km/s = 3,400,000 m/s (too high)
Option 4: 12000 m/s (too high, supersonic jets ~1000 m/s)
Answer: Option 1 → 340 m/s
RRB NTPC GL CBT-1 2025-26 – 17 March 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q69. The waves which can propagate in gases are:

  1. (a) Stationary Waves
  2. (b) Gravity Waves
  3. (c) Transverse Waves
  4. (d) Longitudinal Waves
Answer & solution

Answer: (d) Longitudinal Waves

Topic: Wave Propagation in Gases
Gases have no shear modulus (no rigidity).
Only compression and rarefaction can propagate.
Longitudinal waves→ particles vibrate parallel to wave direction
Examples: Sound waves in air
Transverse waves→ need rigid medium (solids only)
Cannot propagate in gases/liquids
Answer: Option 4 → Longitudinal Waves
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Numericals on sound and waves

Q70. A sound wave has a frequency of 2 kHz and a wavelength of 45 cm. How much time will it take to cover a distance of 1.8 km?

  1. (a) 1 second
  2. (b) 4 seconds
  3. (c) 2 seconds
  4. (d) 0.5 seconds
Answer & solution

Answer: (c) 2 seconds

Frequency (f) = 2 kHz = 2000 Hz
Wavelength (λ) = 45 cm = 0.45 m
Speed of sound (v) = Frequency × Wavelength
v = 2000 × 0.45 = 900 m/s
Distance to cover = 1.8 km = 1800 m
Time (t) = Distance / Speed
t = 1800 / 900 = 2 seconds
But the closest option given is 4 seconds, check again as speed calculation is correct.
Re-calculating can confirm:
Speed = 2 kHz × 0.45 m = 900 m/s
Time = 1800 m ÷ 900 m/s = 2 seconds
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Speed of sound in different media

Q71. Which of the following statements is/are true regarding the speed of sound?
(i) The speed of sound decreases with an increase in the temperature of the medium.
(ii) The speed of sound in solid state is greater than the speed of sound in gaseous state under the same external conditions.
(iii) The speed of sound is independent of the properties of the medium through which it travels.

  1. (a) Both (i) and (iii)
  2. (b) Both (i) and (ii)
  3. (c) Only (iii)
  4. (d) Only (ii)
Answer & solution

Answer: (d) Only (ii)

Statement (i) is incorrect: Speed of sound increases with an increase in temperature, because warmer air has molecules moving faster, aiding wave propagation.
Statement (ii) is correct: Speed of sound is greatest in solids, slower in liquids, and slowest in gases because particles are closer and more tightly bonded in solids.
Statement (iii) is incorrect: Speed of sound depends strongly on properties of the medium like density and elasticity.
Hence, only statement (ii) is true.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Numericals on sound and waves

Q72. An echo is heard in 6 s. If the distance of the reflecting surface from the source is 990 m, then how much is the speed of sound?

  1. (a) 165 m/s
  2. (b) 1320 m/s
  3. (c) 660 m/s
  4. (d) 330 m/s
Answer & solution

Answer: (d) 330 m/s

Echo time = 6 seconds is the total time for the sound to travel to the reflecting surface and back.
Distance to reflecting surface = 990 m.
Total distance travelled by sound = 2 × 990 = 1980 m.
Speed of sound = Total distance / Time = 1980 m / 6 s = 330 m/s.
Hence, the speed of sound is 330 m/s.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Numericals on sound and waves

Q73. An ultrasonic wave has a frequency of 33 kHz. What will be the speed of this wave if its wavelength is 0.01 m?

  1. (a) 330 m/s
  2. (b) 0.33 m/s
  3. (c) 3.3 m/s
  4. (d) 33 m/s
Answer & solution

Answer: (a) 330 m/s

Speed of wave (v) is given by the formula: v = frequency × wavelength.
Frequency (f) = 33 kHz = 33,000 Hz.
Wavelength (λ) = 0.01 m.
Calculate speed: v = 33,000 × 0.01 = 330 m/s.
Therefore, the speed of the ultrasonic wave is 330 m/s.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Reflection of sound, echo and reverberation

Q74. Read the given assertion and reason and select the correct option based on the analysis of these statements.
Assertion: Stethoscope uses the concept of multiple reflection of sound waves for its working.
Reason: Sound waves require a material medium for their propagation.

  1. (a) Both assertion and reason are true and the reason is the correct explanation of the assertion.
  2. (b) Assertion is true but reason is false.
  3. (c) Assertion is false and reason is true.
  4. (d) Both assertion and reason are true and the reason is not the correct explanation of the assertion.
Answer & solution

Answer: (d) Both assertion and reason are true and the reason is not the correct explanation of the assertion.

Assertion is true: Stethoscope works by multiple reflection of sound waves inside its tubes to amplify sounds.
Reason is true: Sound waves need a material medium (like air or body tissues) for propagation.
However, the reason is not the correct explanation for multiple reflections in a stethoscope.
Multiple reflection in a stethoscope happens to enhance sound transmission, not just because sound requires a medium.
Therefore, both statements are true but the reason does not explain the assertion correctly.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Speed of sound in different media

Q75. The speed of sound depends primarily on the nature and ____________ of the medium.

  1. (a) Temperature
  2. (b) Amplitude
  3. (c) Volume
  4. (d) Pressure
Answer & solution

Answer: (a) Temperature

The speed of sound is affected mainly by the properties of the medium it travels through.
Temperature changes the medium's elasticity and density, altering sound speed.
Amplitude, volume, and pressure have little direct effect on speed.
Thus, temperature is the primary factor influencing sound speed in a medium.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Human ear and sound production

Q76. Vibration, which produces sound, is defined as a kind of rapid _____________.

  1. (a) To and fro motion of an object
  2. (b) Linear forward motion
  3. (c) Slow lateral motion
  4. (d) Single directional displacement
Answer & solution

Answer: (a) To and fro motion of an object

The correct option is 1.
Vibration that produces sound is the rapid to and fro motion of an object.
This motion causes waves in the surrounding medium which we perceive as sound.
Other options like linear forward motion or slow lateral motion do not define vibration producing sound.
Hence, option 1 is correct for the definition of vibration that produces sound.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Range of hearing, infrasonic and ultrasonic sound

Q77. What is the audible frequency range for human beings?

  1. (a) 200 Hz to 200,000 Hz
  2. (b) 2 Hz to 200 Hz
  3. (c) 20 Hz to 20,000 Hz
  4. (d) 20 kHz to 200 kHz
Answer & solution

Answer: (c) 20 Hz to 20,000 Hz

The audible frequency range for human beings is the range of sound frequencies that the average human ear can hear.
This range typically starts at about 20 Hz and goes up to about 20,000 Hz (20 kHz).
Frequencies below 20 Hz are called infrasound and those above 20,000 Hz are ultrasound, which humans cannot hear.
Therefore, the audible frequency range is 20 Hz to 20,000 Hz.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Nature and types of waves (longitudinal, transverse)

Q78. When sound propagates through air, the region of high pressure created by the vibrating object pushing air forward is called a __________.

  1. (a) Medium boundary
  2. (b) Compression (C)
  3. (c) Transverse wave
  4. (d) Rarefaction (R)
Answer & solution

Answer: (b) Compression (C)

When a sound wave travels through air, it is a longitudinal wave.
The vibrating object pushes air molecules forward, creating regions of high pressure.
These high-pressure regions are called compressions.
Hence, the correct term for the region of high pressure is Compression.
Option 2 is correct.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Loudness, pitch, quality and decibel

Q79. Which of the following statements is/are INCORRECT about the loudness and softness of a sound wave?
(i) A loud sound has more amplitude as compared to a soft sound.
(ii) A loud sound has lesser energy associated with it as compared to a soft sound.
(iii) Loudness of sound wave is the measure of the response of our heart to the sound.

  1. (a) Both (ii) and (iii)
  2. (b) Both (i) and (iii)
  3. (c) Only (i)
  4. (d) Only (ii)
Answer & solution

Answer: (a) Both (ii) and (iii)

Statement (i) is correct: Loud sounds have greater amplitude than soft sounds.
Statement (ii) is incorrect: A loud sound has more energy, not less, compared to a soft sound.
Statement (iii) is incorrect: Loudness is related to the intensity of sound perceived by the ear, not the heart's response.
Hence, statements (ii) and (iii) are incorrect.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Numericals on sound and waves

Q80. A sound source has a frequency of 50 Hz. How many vibrations does it complete in one minute?

  1. (a) 600
  2. (b) 3000
  3. (c) 5000
  4. (d) 50
Answer & solution

Answer: (b) 3000

Frequency = 50 Hz means 50 vibrations per second.
Number of vibrations in one minute = 50 vibrations/second × 60 seconds = 3000 vibrations.
Hence, the sound source completes 3000 vibrations in one minute.
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Reflection of sound, echo and reverberation

Q81. Multiple reflection of sound can be used to __________.

  1. (a) Hear sounds clearly over long distances
  2. (b) Reduce loudness
  3. (c) Increase the frequency of sound
  4. (d) Change the pitch of sound
Answer & solution

Answer: (a) Hear sounds clearly over long distances

Multiple reflection of sound helps sound waves to travel longer distances by reflecting several times from surfaces.
This allows sounds to be heard clearly even over long distances as the sound energy is sustained.
Hence, multiple reflection enhances sound clarity and audibility over larger areas.
Therefore, the correct option is that multiple reflections help to hear sounds clearly over long distances.
Final answer: Option 1
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Applications of ultrasound and SONAR

Q82. SONAR technology uses ultrasound to ________.

  1. (a) Produce high temperatures
  2. (b) Measure atmospheric pressure
  3. (c) Measure electric current
  4. (d) Locate underwater objects
Answer & solution

Answer: (d) Locate underwater objects

Sonar (Sound Navigation and Ranging) uses ultrasound waves.
These ultrasound waves are emitted and their echoes are detected after bouncing from objects underwater.
This helps to locate objects under water, such as submarines, shipwrecks, or the seabed.
So, SONAR technology is used to locate underwater objects.
Final answer: Option 4
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Reflection of sound, echo and reverberation

Q83. What is the primary reason curved ceilings are used in concert halls?

  1. (a) To absorb sound
  2. (b) To block noise
  3. (c) To decorate the hall
  4. (d) To reflect sound evenly
Answer & solution

Answer: (d) To reflect sound evenly

Correct option: 4
Curved ceilings in concert halls are designed to reflect sound evenly throughout the hall.
This helps in distributing sound waves uniformly so every audience member hears clear and balanced sound.
The curvature prevents sound from focusing on one spot and reduces echoes and dead spots.
Hence, the main purpose is sound reflection, not absorption or noise blocking.
Decoration is secondary, not the primary reason.
Closing the answer.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Numericals on sound and waves

Q84. What will be the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium?

  1. (a) 2 m
  2. (b) 440 m
  3. (c) 220 m
  4. (d) 0.5 m
Answer & solution

Answer: (a) 2 m

The wavelength (λ) of a wave is calculated using the formula: λ = v / f
Here, v = 440 m/s (speed of sound), f = 220 Hz (frequency)
So, λ = 440 / 220 = 2 metres
Therefore, the wavelength of the sound wave is 2 metres.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 1 · Speed of sound in different media

Q85. A sound wave travels from air into water. How does its frequency, speed, and wavelength change as it crosses the boundary?

  1. (a) Its frequency decreases, speed decreases, and wavelength increases
  2. (b) Its frequency remains constant, speed decreases, and wavelength decreases
  3. (c) Its frequency increases, speed remains the same, and wavelength decreases
  4. (d) Its frequency remains constant, speed increases, and wavelength increases
Answer & solution

Answer: (d) Its frequency remains constant, speed increases, and wavelength increases

When a sound wave travels from air to water, its frequency remains constant because frequency depends on the source, not the medium.
The speed of sound increases in water since it is denser and more elastic than air.
Wavelength changes because it is related to speed and frequency by the equation: speed = frequency × wavelength.
Since frequency is constant and speed increases, wavelength must also increase in water.
Final answer: Option 4
SSC MTS 2025 – 19 February 2026 Shift 2 · Loudness, pitch, quality and decibel

Q86. What term describes a measure of the response of the ear to sound?

  1. (a) Frequency
  2. (b) Loudness
  3. (c) Pitch
  4. (d) Intensity
Answer & solution

Answer: (b) Loudness

The correct answer is option 2.
Loudness is the subjective measure of how the human ear perceives the strength or magnitude of a sound. It is the ear's response to sound intensity — a physiological and psychological measure. Loudness is measured in phons or sones and depends on both the intensity and frequency of sound.
Option 1 is incorrect: Frequency refers to the number of vibrations per second (measured in Hz); it determines the pitch of sound, not the ear's response.
Option 3 is incorrect: Pitch is the perceived highness or lowness of a sound, related to frequency; it is not the term for the ear's response to sound strength.
Option 4 is incorrect: Intensity is the objective, physical measure of sound energy per unit area (measured in W/m² or decibels). Loudness is the subjective ear response to intensity.
SSC MTS 2025 – 18 February 2026 Shift 2 · Reflection of sound, echo and reverberation

Q87. Why does the rolling of thunder occur?

  1. (a) Due to a single reflection of sound waves
  2. (b) Due to the absorption of sound waves by clouds
  3. (c) Due to successive reflections of sound from a number of reflecting surfaces
  4. (d) Due to the scattering of light waves
Answer & solution

Answer: (c) Due to successive reflections of sound from a number of reflecting surfaces

The correct answer is option 3.
The prolonged rolling sound of thunder is caused by successive reflections of sound waves from multiple surfaces — such as clouds, mountains, and the ground. The original sound reaches the listener at different time intervals after bouncing off these various surfaces, creating a continuous rumbling effect rather than a single sharp bang.
Option 1 is incorrect: A single reflection would produce one sharp echo, not a prolonged rolling sound.
Option 2 is incorrect: Clouds do absorb some sound energy, but absorption does not explain the rolling or prolonged nature of thunder.
Option 4 is incorrect: Scattering of light waves is an optical phenomenon related to sky colour ; it has no connection to the sound of thunder.
SSC MTS 2025 – 17 February 2026 Shift 2 · Reflection of sound, echo and reverberation

Q88. To hear a distinct echo, the time interval between the original sound and the reflected sound must be at least _____ seconds.

  1. (a) 0.01
  2. (b) 0.1
  3. (c) 1
  4. (d) 10
Answer & solution

Answer: (b) 0.1

To hear a distinct echo, the time interval between the original and reflected sound must be at least 0.1 seconds. This is because the human ear can distinguish two sounds only if they are separated by at least 0.1 seconds (persistence of hearing = 1/10 second). At the speed of sound (~340 m/s), this requires the reflecting surface to be at least 17 metres away.
SSC MTS 2025 – 17 February 2026 Shift 2 · Human ear and sound production

Q89. What component of a hearing aid receives sound?

  1. (a) Microphone
  2. (b) Amplifier
  3. (c) Speaker
  4. (d) Battery
Answer & solution

Answer: (a) Microphone

The Microphone is the component of a hearing aid that receives (picks up) sound from the environment and converts it into electrical signals. The Amplifier then increases the strength of these signals; the Speaker (receiver) converts them back to sound; the Battery powers the device.
SSC MTS 2025 – 16 February 2026 Shift 2 · Reflection of sound, echo and reverberation

Q90. What is reverberation?

  1. (a) The initial sound heard before reflection
  2. (b) The persistence of sound due to repeated reflections from walls
  3. (c) The absorption of sound by materials
  4. (d) The amplification of sound in a small space
Answer & solution

Answer: (b) The persistence of sound due to repeated reflections from walls

Reverberation is the phenomenon of persistence (continuation) of sound in an enclosed space after the original sound source has stopped, caused by multiple reflections of sound waves from the walls, ceiling, and floor. Concert halls and auditoriums are designed to control reverberation time for optimal sound quality.
Why other options are incorrect:
- The initial sound before reflection: This describes the direct sound, not reverberation.
- Absorption of sound by materials: Acoustic absorption reduces reverberation; it is the opposite of the reverberation phenomenon.
- Amplification of sound in a small space: Sound amplification involves increasing sound intensity; reverberation is about persistence of sound due to reflections.
SSC MTS 2025 – 11 February 2026 Shift 1 · Nature and types of waves (longitudinal, transverse)

Q91. Which of the following is correct regarding sound waves?
I. Compression – Region of high pressure
II. Rarefaction – Region of low pressure

  1. (a) Only I
  2. (b) Only II
  3. (c) Both I and II
  4. (d) Neither I nor II
Answer & solution

Answer: (c) Both I and II

The correct answer is Both I and II (Option 3). In sound waves (longitudinal waves): Compression is the region where particles are pushed together, creating a region of high pressure. Rarefaction is the region where particles are spread apart, creating a region of low pressure. Both statements I and II are scientifically correct. Options 1 and 2 are incomplete. Option 4 is incorrect.
SSC MTS 2025 – 11 February 2026 Shift 2 · Numericals on sound and waves

Q92. A sound wave has a frequency of 5 kHz and wave length 50 cm. How long (in seconds) will it take to travel 5 km?

  1. (a) 2
  2. (b) 3
  3. (c) 4
  4. (d) 5
Answer & solution

Answer: (a) 2

The correct answer is 2 seconds (Option 1). Speed of sound = frequency x wavelength = 5000 Hz x 0.50 m = 2500 m/s. Time = Distance / Speed = 5000 m / 2500 m/s = 2 seconds. Options 2 (3 seconds), 3 (4 seconds), and 4 (5 seconds) are incorrect based on this calculation.
SSC MTS 2025 – 9 February 2026 Shift 2 · Applications of ultrasound and SONAR

Q93. What technique uses ultrasonic waves to form images of the heart?

  1. (a) Electrocardiography
  2. (b) Sonography
  3. (c) Echocardiography
  4. (d) Radiography
Answer & solution

Answer: (c) Echocardiography

Correct Answer: 3 – Echocardiography.
Echocardiography (Echo) uses high-frequency ultrasonic waves (sound waves) to create detailed images of the heart's structure and function. It is used to assess the pumping function, valves, and walls of the heart.
Electrocardiography (ECG) records the electrical activity of the heart using electrodes – it does not use ultrasound.
Sonography is a general term for ultrasound imaging of various body parts, not specific to the heart.
Radiography uses X-rays to form images, not ultrasonic waves.
SSC MTS 2025 – 5 February 2026 Shift 2 · Reflection of sound, echo and reverberation

Q94. Why are the ceilings of concert halls, conference halls, and cinema halls generally curved?

  1. (a) To ensure sound after reflection reaches all corners of the hall
  2. (b) To stop sound waves
  3. (c) To focus sound waves in one direction
  4. (d) To absorb sound waves
Answer & solution

Answer: (a) To ensure sound after reflection reaches all corners of the hall

Correct Answer: To ensure sound after reflection reaches all corners of the hall
Curved ceilings in concert halls, conference halls, and cinema halls act as concave reflectors. They reflect sound waves uniformly so that reflected sound reaches all corners and areas of the hall evenly. This ensures that the audience in every seat experiences good sound quality without dead zones or echoes.
Why other options are wrong:
To stop sound waves - Curved ceilings reflect sound, not stop it.
To focus in one direction - Curved ceilings distribute sound widely, not in one direction.
To absorb sound waves - Sound absorption is done by soft materials like foam panels, not curved ceilings.
SSC MTS 2025 – 4 February 2026 Shift 3 · Nature and types of waves (longitudinal, transverse)

Q95. A _____ is a disturbance that moves through a medium when the particles of the medium set neighbouring particles into motion.

  1. (a) Frequency
  2. (b) Wave
  3. (c) Phase
  4. (d) Tone
Answer & solution

Answer: (b) Wave

Correct Answer: Wave
A Wave is a disturbance that propagates through a medium by setting neighbouring particles into motion, thereby transferring energy without bulk movement of matter. Examples include sound waves, water waves, and seismic waves.
Why other options are wrong:
Frequency - A property of a wave (number of cycles per second); not a disturbance itself.
Phase - Describes the position of a point in the wave cycle; not a disturbance.
Tone - Refers to the quality or pitch of sound; not a type of disturbance.
Delhi Police Head Constable (Ministerial) 2025-26 – 7 January 2026 Shift 1 · Doppler effect, resonance and shock waves

Q96. Read the following and choose the correct option:
Assertion (A): An observer on the ground hears a sonic boom only after a supersonic aircraft has passed over them.
Reason (R): The sonic boom is caused by a conical shock wave formed when the aircraft travels faster than the speed of sound, and this shock wave continuously trails behind the aircraft as it moves.

  1. (a) Both A and R are true and R is the correct explanation of A.
  2. (b) Both A and R are true but R is not the correct explanation of A.
  3. (c) A is true but R is false.
  4. (d) A is false but R is true.
Answer & solution

Answer: (a) Both A and R are true and R is the correct explanation of A.

Correct Answer: Option 1 – Both A and R are true and R is the correct explanation of A.
Assertion (A) is TRUE: An observer on the ground hears the sonic boom only AFTER the supersonic aircraft has passed over them because the shock wave cone trails behind the aircraft. The boom reaches the observer after the aircraft has already moved past.
Reason (R) is TRUE and correctly explains A: When an aircraft exceeds the speed of sound, it produces a conical (Mach cone) shock wave that continuously trails behind it. As this cone sweeps over the ground observer, they hear the sonic boom — but this happens only after the plane has passed overhead.
R correctly explains why A is true: the trailing nature of the shock wave cone is the direct reason the boom is heard after the aircraft has passed.
SSC Stenographer 2025 – 8 December 2025 Shift 3 · Speed of sound in different media

Q97. Read the following and choose the correct option:
Assertion (A): Sound waves travel faster in steel than in air at the same temperature.
Reason (R): The particles in solids (like steel) are more closely packed and experience stronger intermolecular forces compared to particles in gases (like air), allowing vibrations to be transmitted more efficiently and rapidly.

  1. (a) Both A and R are true but R is not the correct explanation of A.
  2. (b) A is true but R is false.
  3. (c) A is false but R is true.
  4. (d) Both A and R are true and R is the correct explanation of A.
Answer & solution

Answer: (d) Both A and R are true and R is the correct explanation of A.

The correct answer is Both A and R are true and R is the correct explanation of A . Sound waves need a medium to travel, and their speed depends on the density and elasticity of the medium. In solids like steel, particles are packed closely together and bound by strong intermolecular forces. This allows vibrations (sound energy) to be transmitted faster and more efficiently compared to gases like air, where particles are far apart and the intermolecular forces are weaker. Therefore, sound travels faster in steel than in air, and the given Reason (R) correctly explains Assertion (A).
SSC Stenographer 2025 – 6 December 2025 Shift 1 · Loudness, pitch, quality and decibel

Q98. Which property of sound determines its loudness?

  1. (a) Pitch
  2. (b) Wavelength
  3. (c) Frequency
  4. (d) Amplitude
Answer & solution

Answer: (d) Amplitude

The correct answer is Amplitude . Loudness of sound depends on the amplitude of the sound wave. A sound wave with larger amplitude carries more energy and is perceived as louder, while a wave with smaller amplitude is softer. Pitch is determined by frequency, and wavelength is related to frequency and velocity, but loudness is directly controlled by amplitude.
SSC CHSL Tier-1 2025 – 25 November 2025 Shift 2 · Loudness, pitch, quality and decibel

Q99. What is the term used to describe how the brain interprets the frequency of an emitted sound?

  1. (a) Tone
  2. (b) Wavelength
  3. (c) Frequency
  4. (d) Pitch
Answer & solution

Answer: (d) Pitch

Pitch is the term used to describe how the brain perceives and interprets the frequency of a sound. A higher frequency sound is perceived as a higher pitch; a lower frequency sound as a lower pitch.
Tone refers to the quality or character of a sound; Wavelength is a physical property (not how it is perceived); Frequency is the physical measurement (cycles per second/Hz), not the perceptual interpretation.
SSC CGL Tier-1 2025 – 23 September 2025 Shift 1 · Doppler effect, resonance and shock waves

Q100. What happens to the perceived frequency when both the source and the listener move with the same speed in the same direction?

  1. (a) It increases
  2. (b) It decreases
  3. (c) It remains unchanged
  4. (d) It fluctuates rapidly
Answer & solution

Answer: (c) It remains unchanged

Doppler Effect: if source and observer have the same velocity in the same direction, relative velocity = 0. No change in perceived frequency.
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