Sound
(a) Longitudinal ✅
(b) Transverse
(c) Standing only
(d) Random
Explanation: In a longitudinal wave, particles oscillate parallel to the direction of propagation, creating compressions and rarefactions. In gases and liquids, restoring forces due to pressure and inertia support compressional (longitudinal) waves. Transverse shear waves require rigidity (shear modulus), which fluids lack, so transverse sound cannot propagate in fluids.
Q2. The physical mechanism that allows sound to travel in air is mainly:
(a) Pressure and density fluctuations ✅
(b) Temperature gradients only
(c) Electric field variations
(d) Magnetic field variations
Explanation: A vibrating source produces alternating high-pressure (compression) and low-pressure (rarefaction) regions. These propagate as density fluctuations with the local speed of sound determined by thermodynamic properties of air. Temperature gradients can affect speed, but the fundamental propagation is via pressure/density oscillations governed by the acoustic wave equation.
Q3. Sound needs a medium to travel because:
(a) It is a mechanical wave ✅
(b) It is an electromagnetic wave
(c) It is a quantum field only
(d) None
Explanation: Mechanical waves require material particles to oscillate and transfer energy. Unlike light (EM waves) which can travel in vacuum, sound cannot propagate without a medium since the restoring and inertial properties (bulk modulus and density) of the medium are essential to support oscillations.
Q4. The regions of high and low pressure in a sound wave are called:
(a) Compressions and rarefactions ✅
(b) Nodes and antinodes
(c) Crests and troughs
(d) Beats
Explanation: Compressions are regions with higher-than-average pressure; rarefactions are lower-than-average pressure areas. In longitudinal standing waves you will also see nodes (no displacement) and antinodes (max displacement), but traveling sound waves are described primarily by compressions and rarefactions.
Q5. Displacement of particles in a longitudinal sound wave is:
(a) Parallel to wave propagation ✅
(b) Perpendicular to propagation
(c) Zero
(d) Random
Explanation: Longitudinal waves involve particle motion in the same direction as energy transport. This parallel oscillation produces periodic pressure changes that define the acoustic wave.
Q6. Which statement is true about sound in solids?
(a) Both longitudinal and transverse sound can propagate ✅
(b) Only longitudinal sound propagates
(c) Only transverse sound propagates
(d) No sound can propagate
Explanation: Solids possess bulk modulus (resistance to compression) and shear modulus (resistance to shape change), enabling both compressional (P-waves) and shear (S-waves) acoustic modes. Their speeds depend on elastic moduli and density.
Q7. The quantity that primarily determines the pitch of a sound is:
(a) Frequency ✅
(b) Amplitude
(c) Phase
(d) Speed
Explanation: Pitch is the perceptual correlate of frequency; higher frequency sounds are heard as higher pitch. Amplitude affects loudness, not pitch; phase is perceptually less salient; speed depends on medium and does not directly set pitch.
Q8. Loudness is most closely related to:
(a) Intensity/amplitude ✅
(b) Frequency only
(c) Wavelength only
(d) Phase
Explanation: Loudness (a subjective measure) increases with sound intensity, which is proportional to the square of amplitude. Frequency influences pitch, not loudness, though human hearing sensitivity varies with frequency.
Q9. Timbre (quality) of sound is determined by:
(a) Harmonic content and envelope ✅
(b) Speed of sound only
(c) Amplitude only
(d) Phase only
Explanation: Timbre distinguishes sounds with the same pitch and loudness. It depends on the spectrum (relative strengths of harmonics/overtones), transient behavior (attack/decay), and time envelope. This is why the same note sounds different on a flute vs. violin.
Q10. The audible frequency range for a typical human is approximately:
(a) 20 Hz to 20 kHz ✅
(b) 2 Hz to 2 kHz
(c) 200 Hz to 200 kHz
(d) 20 kHz to 200 kHz
Explanation: Healthy young humans generally perceive frequencies from about 20 Hz up to ~20,000 Hz. Upper limits decrease with age or noise exposure; lower limits can be felt as vibration at very high amplitudes but not heard.
Q11. Frequencies below the audible range are called:
(a) Infrasonic ✅
(b) Ultrasonic
(c) Hypersonic
(d) Supersonic
Explanation: Infrasonic refers to frequencies lower than ~20 Hz. Ultrasonic is above ~20 kHz. Supersonic and hypersonic describe speeds greater than sound, not frequency bands.
Q12. Frequencies above the audible range are called:
(a) Ultrasonic ✅
(b) Infrasonic
(c) Hypersonic
(d) Supersonic
Explanation: Ultrasonic sound exceeds ~20,000 Hz and is widely used in imaging (ultrasound), cleaning, and non-destructive testing due to its short wavelength and directionality.
Q13. Which statement about wavelength of a sound is correct?
(a) Wavelength depends on the medium’s sound speed and the wave’s frequency ✅
(b) Wavelength is fixed for all media
(c) Wavelength depends only on amplitude
(d) Wavelength is independent of frequency
Explanation: λ = v/f. For a given frequency, a higher sound speed (e.g., in solids vs air) yields a longer wavelength. Amplitude does not affect wavelength in linear acoustics.
Q14. The relation between pressure amplitude and displacement amplitude in sound depends on:
(a) Medium’s bulk modulus and density ✅
(b) Gravitational acceleration only
(c) Electrical conductivity
(d) Magnetic permeability
Explanation: Acoustic pressure p and particle displacement ξ are linked via the medium’s acoustic impedance Z = ρv and compressibility. Bulk modulus (K) and density (ρ) determine wave speed v = √(K/ρ), shaping how displacement translates into pressure.
Q15. The typical speed of sound in dry air at room temperature (~20°C) is closest to:
(a) 343 m/s ✅
(b) 300 m/s
(c) 400 m/s
(d) 500 m/s
Explanation: v_air ≈ 343 m/s at 20°C, increasing with temperature. The exact value depends on temperature, humidity, and gas composition, but 343 m/s is a widely used reference.
Q16. Which medium generally has the highest sound speed among the options?
(a) Steel ✅
(b) Air
(c) Water
(d) Wood
Explanation: Sound travels fastest in solids due to higher elastic moduli and typically moderate densities. Steel’s longitudinal sound speed is ~5,000–6,000 m/s, water ~1,480 m/s, air ~343 m/s (at 20°C).
Q17. The speed of sound in air increases primarily with:
(a) Temperature ✅
(b) Amplitude
(c) Frequency
(d) Pressure at constant temperature
Explanation: For ideal gases at constant composition, v ≈ √(γRT/M). As temperature (T) rises, v increases. Amplitude and frequency do not change v in linear acoustics; pressure changes at constant T have negligible effect on v for ideal gases.
Q18. The parameter γ in the gas sound speed formula v = √(γRT/M) is:
(a) Ratio of specific heats (Cp/Cv) ✅
(b) Gas constant
(c) Molecular mass
(d) Bulk modulus
Explanation: γ (adiabatic index) is Cp/Cv, reflecting thermodynamic behavior of the gas. R is the universal gas constant, T is absolute temperature, and M is molar mass.
Q19. Comparing wavelengths in the same medium, a higher-frequency sound has:
(a) Shorter wavelength ✅
(b) Longer wavelength
(c) The same wavelength
(d) Zero wavelength
Explanation: For fixed v, λ = v/f. Increasing frequency reduces wavelength proportionally, affecting diffraction and directionality (higher frequencies diffract less and are more directional).
Q20. Infrasonic waves are useful in:
(a) Seismology and animal communication (e.g., elephants) ✅
(b) Medical imaging
(c) SONAR only
(d) Microwave heating
Explanation: Low-frequency waves travel long distances and penetrate structures, aiding earthquake monitoring and long-range biological communication. Ultrasonics, not infrasonics, are used in medical imaging.
Q21. Ultrasonic waves are widely used because they:
(a) Have short wavelengths and can be focused precisely ✅
(b) Cannot penetrate biological tissue
(c) Are inaudible and thus useless
(d) Are unaffected by medium properties
Explanation: Short λ improves resolution in imaging (ultrasound), enables precise cleaning, drilling, and non-destructive testing. Attenuation and tissue properties must be considered; they do penetrate tissue within limits.
Q22. The amplitude of a sound wave doubles. What happens to intensity (in linear acoustics)?
(a) It quadruples ✅
(b) It doubles
(c) It halves
(d) It remains unchanged
Explanation: Intensity I ∝ A² (A is amplitude). Doubling A makes I → (2A)² = 4A², i.e., four times. This underlies the logarithmic nature of loudness scales like decibels.
Q23. The decibel (dB) scale relates intensity via:
(a) L = 10 log10(I/I0) ✅
(b) L = 20 log10(I/I0)
(c) L = ln(I/I0)
(d) L = I/I0
Explanation: Sound intensity level L is defined using a base-10 logarithm with a factor of 10. For pressure amplitude ratios, the relation uses 20 log10(p/p0) because intensity ∝ p².
Q24. If intensity increases by a factor of 100, the sound level increases by:
(a) 20 dB ✅
(b) 10 dB
(c) 3 dB
(d) 30 dB
Explanation: ΔL = 10 log10(100) = 10 × 2 = 20 dB. This reflects the logarithmic nature of perceived loudness changes relative to physical intensity changes.
Q25. Which statement best explains why high-frequency sounds are more directional?
(a) Shorter wavelengths diffract less around obstacles ✅
(b) Higher amplitudes push air more strongly
(c) High frequencies travel faster in air
(d) Low frequencies are absorbed more
Explanation: Diffraction depends on the ratio of obstacle/aperture size to wavelength. Short-wavelength (high-frequency) waves bend less, making them more beam-like; this is exploited in ultrasound imaging and high-frequency tweeters.
Q26. Speed of sound in air at 0°C is approximately:
(a) 331 m/s ✅
(b) 343 m/s
(c) 300 m/s
(d) 350 m/s
Explanation: v = 331 m/s at 0°C. Formula: v ≈ 331 + 0.6T (T in °C). At 20°C, v ≈ 343 m/s.
Q27. Speed of sound in water at room temperature is about:
(a) 1480 m/s ✅
(b) 1200 m/s
(c) 1600 m/s
(d) 343 m/s
Explanation: Water’s high bulk modulus and density yield ~1480 m/s, much faster than air.
Q28. Speed of sound in steel is about:
(a) 5000 m/s ✅
(b) 343 m/s
(c) 1500 m/s
(d) 2000 m/s
Explanation: Solids transmit sound fastest due to high elasticity. Steel ~5000–6000 m/s.
Q29. Speed of sound in air increases with:
(a) Temperature ✅
(b) Pressure
(c) Amplitude
(d) Frequency
Explanation: v ∝ √T. Pressure has negligible effect at constant temperature.
Q30. Formula for speed of sound in ideal gas:
(a) v = √(γRT/M) ✅
(b) v = √(RT)
(c) v = √(γM/R)
(d) None
Explanation: γ = Cp/Cv, R = gas constant, T = temperature, M = molar mass.
Q31. Speed of sound in helium is higher than in air because:
(a) Lower molar mass ✅
(b) Higher density
(c) Higher pressure
(d) None
Explanation: v ∝ 1/√M. Helium’s small M increases speed (~1000 m/s).
Q32. Intensity of sound is defined as:
(a) Power per unit area ✅
(b) Energy per unit volume
(c) Pressure per unit length
(d) None
Explanation: I = P/A. Units: W/m².
Q33. Intensity is proportional to:
(a) Amplitude² ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: I ∝ A². Doubling amplitude quadruples intensity.
Q34. Sound level in decibels is:
(a) L = 10 log10(I/I0) ✅
(b) L = 20 log10(I/I0)
(c) L = ln(I/I0)
(d) None
Explanation: I0 = 10⁻¹² W/m² (threshold of hearing).
Q35. If intensity increases 10 times, sound level increases by:
(a) 10 dB ✅
(b) 20 dB
(c) 30 dB
(d) None
Explanation: ΔL = 10 log10(10) = 10 dB.
Q36. If intensity increases 100 times, sound level increases by:
(a) 20 dB ✅
(b) 10 dB
(c) 30 dB
(d) None
Explanation: ΔL = 10 log10(100) = 20 dB.
Q37. If intensity increases 1000 times, sound level increases by:
(a) 30 dB ✅
(b) 20 dB
(c) 40 dB
(d) None
Explanation: ΔL = 10 log10(1000) = 30 dB.
Q38. Threshold of hearing corresponds to intensity:
(a) 10⁻¹² W/m² ✅
(b) 10⁻⁶ W/m²
(c) 10⁻³ W/m²
(d) None
Explanation: This is the reference intensity I0 used in decibel calculations.
Q39. Threshold of pain corresponds to intensity:
(a) 1 W/m² ✅
(b) 10⁻¹² W/m²
(c) 0.01 W/m²
(d) None
Explanation: Very loud sounds (~120 dB) correspond to ~1 W/m².
Q40. Echo is heard when reflected sound reaches after:
(a) 0.1 s ✅
(b) 0.01 s
(c) 1 s
(d) None
Explanation: Human ear distinguishes two sounds if separated by ≥0.1 s.
Q41. Minimum distance for echo in air (v = 343 m/s):
(a) 17.2 m ✅
(b) 10 m
(c) 20 m
(d) None
Explanation: d = v×t/2 = 343×0.1/2 ≈ 17.2 m.
Q42. Reverberation is:
(a) Persistence of sound due to multiple reflections ✅
(b) Echo only
(c) Silence
(d) None
Explanation: Reverberation occurs in halls when reflected sounds overlap with direct sound.
Q43. Reverberation time is:
(a) Time for sound intensity to drop to 1/10⁶ of initial ✅
(b) Time for echo
(c) Time for silence
(d) None
Explanation: Defined as time for sound to decay by 60 dB.
Q44. Reverberation is desirable in:
(a) Concert halls ✅
(b) Classrooms
(c) Studios
(d) None
Explanation: Moderate reverberation enriches music. Excess causes muddiness.
Q45. Reverberation is undesirable in:
(a) Recording studios ✅
(b) Concert halls
(c) Auditoriums
(d) None
Explanation: Studios require clarity, so reverberation must be minimized.
Q46. In architectural acoustics, excessive reverberation causes:
(a) Loss of speech clarity ✅
(b) Increase in loudness
(c) Better music quality
(d) None
Explanation: Too much reverberation overlaps syllables, reducing intelligibility in classrooms and auditoriums.
Q47. SONAR uses:
(a) Ultrasonic echoes ✅
(b) Infrasonic waves
(c) Visible light
(d) None
Explanation: SONAR measures depth/distance using reflected ultrasonic pulses.
Q48. Medical ultrasound imaging uses:
(a) Ultrasonic echoes ✅
(b) Infrasonic waves
(c) Audible sound
(d) None
Explanation: Ultrasonic pulses reflect from tissues, producing images.
Q49. Whales communicate using:
(a) Infrasonic waves ✅
(b) Ultrasonic waves
(c) Audible sound only
(d) None
Explanation: Whale calls are infrasonic, traveling long distances underwater.
Q50. Echo depth calculation: SONAR pulse returns in 2 s, v = 1500 m/s. Depth =
(a) 1500 m ✅
(b) 3000 m
(c) 750 m
(d) None
Explanation: Distance = v×t/2 = 1500×2/2 = 1500 m.
Q51. In a closed pipe, the fundamental frequency occurs when:
(a) Length = λ/4 ✅
(b) Length = λ/2
(c) Length = λ
(d) None
Explanation: Closed pipe supports odd harmonics only. The first resonance occurs at L = λ/4.
Q52. In an open pipe, the fundamental frequency occurs when:
(a) Length = λ/2 ✅
(b) Length = λ/4
(c) Length = λ
(d) None
Explanation: Open pipe supports all harmonics. First resonance at L = λ/2.
Q53. In a closed pipe, next resonance after fundamental occurs at:
(a) 3λ/4 ✅
(b) λ/2
(c) λ
(d) None
Explanation: Closed pipe harmonics: L = (2n−1)λ/4. Next after λ/4 is 3λ/4.
Q54. In an open pipe, second resonance occurs at:
(a) λ ✅
(b) λ/2
(c) 3λ/4
(d) None
Explanation: Open pipe harmonics: L = nλ/2. Second resonance at L = λ.
Q55. Closed pipe supports:
(a) Odd harmonics only ✅
(b) Even harmonics only
(c) All harmonics
(d) None
Explanation: Because one end is node, other antinode → only odd multiples.
Q56. Open pipe supports:
(a) All harmonics ✅
(b) Odd harmonics only
(c) Even harmonics only
(d) None
Explanation: Both ends antinodes → all harmonics possible.
Q57. Fundamental frequency of closed pipe of length 34 cm, v = 340 m/s:
(a) 250 Hz ✅
(b) 500 Hz
(c) 125 Hz
(d) None
Explanation: f = v/4L = 340/(4×0.34) ≈ 250 Hz.
Q58. Fundamental frequency of open pipe of length 34 cm, v = 340 m/s:
(a) 500 Hz ✅
(b) 250 Hz
(c) 125 Hz
(d) None
Explanation: f = v/2L = 340/(0.68) ≈ 500 Hz.
Q59. Resonance tube experiment measures:
(a) Speed of sound ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Known tuning fork frequency, measure resonance length → calculate v.
Q60. Resonance occurs when:
(a) Natural frequency = driving frequency ✅
(b) Natural frequency ≠ driving frequency
(c) Amplitude = zero
(d) None
Explanation: Resonance amplifies oscillations when frequencies match.
Q61. In resonance tube, first resonance length L1 = 20 cm, second L2 = 62 cm. Wavelength =
(a) 84 cm ✅
(b) 42 cm
(c) 62 cm
(d) None
Explanation: Difference L2 − L1 = λ/2 = 42 cm → λ = 84 cm.
Q62. Frequency of tuning fork if λ = 84 cm, v = 336 m/s:
(a) 400 Hz ✅
(b) 336 Hz
(c) 420 Hz
(d) None
Explanation: f = v/λ = 336/0.84 = 400 Hz.
Q63. In closed pipe, third resonance corresponds to:
(a) 5λ/4 ✅
(b) λ
(c) 3λ/2
(d) None
Explanation: Closed pipe resonances: L = (2n−1)λ/4. For n=3 → 5λ/4.
Q64. In open pipe, third resonance corresponds to:
(a) 3λ/2 ✅
(b) λ
(c) 5λ/4
(d) None
Explanation: Open pipe resonances: L = nλ/2. For n=3 → 3λ/2.
Q65. End correction in resonance tube accounts for:
(a) Antinode slightly beyond open end ✅
(b) Node inside tube
(c) Pressure variation
(d) None
Explanation: Effective length is slightly longer than physical length due to end effect.
Q66. End correction is approximately:
(a) 0.6r ✅
(b) r
(c) 0.1r
(d) None
Explanation: r = radius of tube. Effective length L = physical length + 0.6r.
Q67. Doppler effect formula for observer moving towards source:
(a) f' = f(1 + v₀/v) ✅
(b) f' = f(v/(v−vs))
(c) f' = f(1 − v₀/v)
(d) None
Explanation: Observer motion increases relative frequency.
Q68. Doppler effect formula for source moving towards observer:
(a) f' = f(v/(v−vs)) ✅
(b) f' = f(1 + v₀/v)
(c) f' = f(v/(v+vs))
(d) None
Explanation: Source motion compresses wavefronts, increasing frequency.
Q69. Doppler effect formula for source moving away:
(a) f' = f(v/(v+vs)) ✅
(b) f' = f(v/(v−vs))
(c) f' = f(1 − v₀/v)
(d) None
Explanation: Source motion stretches wavefronts, decreasing frequency.
Q70. Doppler effect formula for observer moving away:
(a) f' = f(1 − v₀/v) ✅
(b) f' = f(1 + v₀/v)
(c) f' = f(v/(v+vs))
(d) None
Explanation: Observer motion reduces relative frequency.
Q71. Doppler effect is used in:
(a) Radar, astronomy, medical imaging ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Applications include speed detection, star velocity, blood flow.
Q72. If two tuning forks produce 6 beats per second, their frequencies differ by:
(a) 6 Hz ✅
(b) 12 Hz
(c) 3 Hz
(d) None
Explanation: Beat frequency equals the difference between the two frequencies.
Q73. Two tuning forks 256 Hz and 260 Hz produce beat frequency:
(a) 4 Hz ✅
(b) 2 Hz
(c) 6 Hz
(d) None
Explanation: |260 − 256| = 4 Hz.
Q74. Beats are heard when:
(a) Frequencies are close but not equal ✅
(b) Frequencies are equal
(c) Frequencies are very different
(d) None
Explanation: Beat phenomenon requires small frequency difference.
Q75. Maximum beat frequency audible is about:
(a) 10 Hz ✅
(b) 100 Hz
(c) 1 Hz
(d) None
Explanation: Human ear perceives beats up to ~10 Hz. Beyond that, heard as separate tones.
Q76. In a guitar string fixed at both ends, fundamental frequency occurs when:
(a) Length = λ/2 ✅
(b) Length = λ/4
(c) Length = λ
(d) None
Explanation: Both ends are nodes → first mode has half a wavelength fitting in the length.
Q77. In a flute (open pipe), fundamental frequency occurs when:
(a) Length = λ/2 ✅
(b) Length = λ/4
(c) Length = λ
(d) None
Explanation: Both ends are antinodes → first mode is λ/2.
Q78. In a clarinet (closed pipe), fundamental frequency occurs when:
(a) Length = λ/4 ✅
(b) Length = λ/2
(c) Length = λ
(d) None
Explanation: One end node, other antinode → first mode is λ/4.
Q79. Harmonics in open pipe are:
(a) All harmonics ✅
(b) Odd only
(c) Even only
(d) None
Explanation: Open pipe supports all harmonics.
Q80. Harmonics in closed pipe are:
(a) Odd only ✅
(b) Even only
(c) All
(d) None
Explanation: Closed pipe supports odd harmonics only.
Q81. Frequency of fundamental in string length L, tension T, mass/length μ:
(a) f = (1/2L)√(T/μ) ✅
(b) f = (1/L)√(T/μ)
(c) f = (1/4L)√(T/μ)
(d) None
Explanation: Derived from wave speed v = √(T/μ), fundamental λ = 2L.
Q82. If tension quadruples, frequency:
(a) Doubles ✅
(b) Halves
(c) Same
(d) None
Explanation: f ∝ √T. Quadruple T → f doubles.
Q83. If length doubles, frequency:
(a) Halves ✅
(b) Doubles
(c) Same
(d) None
Explanation: f ∝ 1/L. Doubling L halves f.
Q84. Doppler effect: source 500 Hz, moving towards observer at 30 m/s, v = 340 m/s. Observed f =
(a) 546 Hz ✅
(b) 540 Hz
(c) 550 Hz
(d) None
Explanation: f' = f(v/(v−vs)) = 500(340/310) ≈ 546 Hz.
Q85. Doppler effect: observer moves towards source at 30 m/s, f = 500 Hz, v = 340 m/s. Observed f =
(a) 544 Hz ✅
(b) 540 Hz
(c) 550 Hz
(d) None
Explanation: f' = f(1+v0/v) = 500(1+30/340) ≈ 544 Hz.
Q86. Doppler effect: source moves away at 30 m/s, f = 500 Hz, v = 340 m/s. Observed f =
(a) 456 Hz ✅
(b) 460 Hz
(c) 450 Hz
(d) None
Explanation: f' = f(v/(v+vs)) = 500(340/370) ≈ 456 Hz.
Q87. Doppler effect: observer moves away at 30 m/s, f = 500 Hz, v = 340 m/s. Observed f =
(a) 456 Hz ✅
(b) 460 Hz
(c) 450 Hz
(d) None
Explanation: f' = f(1−v0/v) = 500(1−30/340) ≈ 456 Hz.
Q88. SONAR pulse returns in 3 s, v = 1500 m/s. Depth =
(a) 2250 m ✅
(b) 3000 m
(c) 1500 m
(d) None
Explanation: Distance = v×t/2 = 1500×3/2 = 2250 m.
Q89. SONAR pulse returns in 4 s, v = 1500 m/s. Depth =
(a) 3000 m ✅
(b) 1500 m
(c) 6000 m
(d) None
Explanation: Distance = v×t/2 = 1500×4/2 = 3000 m.
Q90. Ultrasonic waves are used in:
(a) Medical imaging, cleaning, SONAR ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Applications exploit short wavelength and high frequency.
Q91. Ultrasound imaging resolution improves with:
(a) Higher frequency ✅
(b) Lower frequency
(c) Higher amplitude
(d) None
Explanation: Shorter wavelength increases resolution, but penetration decreases.
Q92. Elephants communicate using:
(a) Infrasonic waves ✅
(b) Ultrasonic waves
(c) Audible sound only
(d) None
Explanation: Low-frequency calls travel kilometers through ground and air.
Q93. Dogs can hear up to:
(a) 45 kHz ✅
(b) 20 kHz
(c) 25 kHz
(d) None
Explanation: Dogs detect ultrasonic frequencies beyond human hearing.
Q94. Resonance tube experiment measures:
(a) Speed of sound ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Known tuning fork frequency, measure resonance length → calculate v.
Q95. Beats disappear when:
(a) Frequencies become equal ✅
(b) Amplitude increases
(c) Wavelength decreases
(d) None
Explanation: No difference in frequency → no beats.
Q96. Two tuning forks 512 Hz and 520 Hz produce beat frequency:
(a) 8 Hz ✅
(b) 6 Hz
(c) 10 Hz
(d) None
Explanation: |520−512| = 8 Hz.
Q97. Beats are audible when frequency difference is:
(a) ≤10 Hz ✅
(b) ≥100 Hz
(c) ≥50 Hz
(d) None
Explanation: Human ear perceives beats up to ~10 Hz.
Q98. Quality (timbre) of sound depends on:
(a) Harmonic content ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Harmonics distinguish instruments playing same note.
Q99. Loudness depends on:
(a) Intensity ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: Loudness ∝ intensity, which ∝ amplitude².
Q100. Pitch depends on:
(a) Frequency ✅
(b) Amplitude
(c) Intensity
(d) None
Explanation: Pitch is perceptual correlate of frequency.
Q101. Sound intensity level at threshold of hearing (I0 = 10⁻¹² W/m²):
(a) 0 dB ✅
(b) 10 dB
(c) 20 dB
(d) None
Explanation: L = 10 log10(I/I0). At I = I0, L = 0 dB.
Q102. Sound intensity level at threshold of pain (I = 1 W/m²):
(a) 120 dB ✅
(b) 100 dB
(c) 110 dB
(d) None
Explanation: L = 10 log10(1/10⁻¹²) = 10×12 = 120 dB.
Q103. If intensity increases 1000 times, sound level increases by:
(a) 30 dB ✅
(b) 20 dB
(c) 10 dB
(d) None
Explanation: ΔL = 10 log10(1000) = 30 dB.
Q104. If sound level increases by 20 dB, intensity increases by:
(a) 100 times ✅
(b) 10 times
(c) 1000 times
(d) None
Explanation: ΔL = 10 log10(I2/I1). 20 = 10 log10(100).
Q105. If sound level increases by 30 dB, intensity increases by:
(a) 1000 times ✅
(b) 100 times
(c) 10 times
(d) None
Explanation: ΔL = 10 log10(1000).
Q106. Reverberation time depends on:
(a) Volume and absorption of hall ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Sabine’s formula: T = 0.161 V/A, where V = volume, A = absorption.
Q107. Reverberation time is desirable in:
(a) Concert halls ✅
(b) Studios
(c) Classrooms
(d) None
Explanation: Moderate reverberation enriches music.
Q108. Reverberation time is undesirable in:
(a) Studios ✅
(b) Concert halls
(c) Auditoriums
(d) None
Explanation: Studios require clarity, so reverberation must be minimized.
Q109. Reverberation time in a hall is reduced by:
(a) Adding carpets and curtains ✅
(b) Increasing hall volume
(c) Removing absorptive materials
(d) None
Explanation: Soft materials absorb sound, reducing reflections and reverberation time.
Q110. SONAR uses:
(a) Ultrasonic echoes ✅
(b) Infrasonic waves
(c) Visible light
(d) None
Explanation: SONAR measures depth/distance using reflected ultrasonic pulses.
Q111. SONAR depth calculation: Pulse returns in 2 s, v = 1500 m/s. Depth =
(a) 1500 m ✅
(b) 3000 m
(c) 750 m
(d) None
Explanation: Distance = v×t/2 = 1500×2/2 = 1500 m.
Q112. SONAR depth calculation: Pulse returns in 4 s, v = 1500 m/s. Depth =
(a) 3000 m ✅
(b) 1500 m
(c) 6000 m
(d) None
Explanation: Distance = v×t/2 = 1500×4/2 = 3000 m.
Q113. Medical ultrasound imaging uses:
(a) Ultrasonic echoes ✅
(b) Infrasonic waves
(c) Audible sound
(d) None
Explanation: Ultrasonic pulses reflect from tissues, producing images.
Q114. Ultrasound in pregnancy is used to:
(a) Monitor fetal growth ✅
(b) Measure bone density
(c) Detect earthquakes
(d) None
Explanation: Safe imaging technique for fetus.
Q115. Medical Doppler ultrasound measures:
(a) Blood flow velocity ✅
(b) Bone density
(c) Muscle strength
(d) None
Explanation: Frequency shift in reflected waves indicates blood speed.
Q116. SONAR is used in:
(a) Submarine navigation ✅
(b) Cooking
(c) Heating
(d) None
Explanation: SONAR detects depth and obstacles underwater.
Q117. Communication systems use sound waves to:
(a) Transmit information ✅
(b) Transmit matter
(c) Transmit heat
(d) None
Explanation: Signals encoded in waves carry information.
Q118. Bandwidth is:
(a) Range of frequencies transmitted ✅
(b) Single frequency
(c) Amplitude range
(d) None
Explanation: Determines data capacity.
Q119. Higher bandwidth means:
(a) Higher data rate ✅
(b) Lower data rate
(c) Same data rate
(d) None
Explanation: More frequencies → more information per second.
Q120. Modulation is:
(a) Superimposing information signal on carrier wave ✅
(b) Removing carrier
(c) Amplifying only
(d) None
Explanation: Enables long-distance transmission.
Q121. Amplitude modulation varies:
(a) Amplitude of carrier ✅
(b) Frequency
(c) Phase
(d) None
Explanation: Carrier amplitude changes with signal.
Q122. Frequency modulation varies:
(a) Frequency of carrier ✅
(b) Amplitude
(c) Phase
(d) None
Explanation: Carrier frequency changes with signal.
Q123. Phase modulation varies:
(a) Phase of carrier ✅
(b) Amplitude
(c) Frequency
(d) None
Explanation: Carrier phase changes with signal.
Q124. Advantage of FM over AM:
(a) Less noise ✅
(b) More noise
(c) Lower fidelity
(d) None
Explanation: Frequency is less affected by amplitude noise.
Q125. Optical fiber losses are mainly due to:
(a) Scattering and absorption ✅
(b) Reflection only
(c) Refraction only
(d) None
Explanation: Imperfections cause attenuation of signal.
Q126. Dispersion in optical fibers causes:
(a) Pulse broadening ✅
(b) Pulse narrowing
(c) No effect
(d) None
Explanation: Different wavelengths travel at different speeds.
Q127. Optical fiber communication advantage:
(a) High bandwidth and low loss ✅
(b) Low bandwidth
(c) High loss
(d) None
Explanation: Enables fast internet.
Q128. Optical fiber core is made of:
(a) Glass or plastic ✅
(b) Metal
(c) Wood
(d) None
Explanation: Transparent medium guides light.
Q129. Optical fiber cladding has:
(a) Lower refractive index than core ✅
(b) Higher refractive index
(c) Same index
(d) None
Explanation: Ensures total internal reflection.
Q130. Optical fibers are used in:
(a) Communication, medical imaging ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Applications include internet cables, endoscopy.
Q131. Dispersion in optical fibers causes:
(a) Pulse broadening ✅
(b) Pulse narrowing
(c) No effect
(d) None
Explanation: Different wavelengths travel at different speeds.
Q132. Coaxial cables transmit signals by:
(a) Electrical conduction ✅
(b) Optical reflection
(c) Magnetic induction
(d) None
Explanation: Inner conductor carries signal, outer shields.
Q133. Satellite communication uses:
(a) Microwaves ✅
(b) Infrared
(c) Visible light
(d) None
Explanation: Microwaves penetrate atmosphere and reach satellites.
Q134. Frequency range for satellite communication:
(a) GHz range ✅
(b) MHz range
(c) kHz range
(d) None
Explanation: Typically 1–40 GHz.
Q135. Mobile communication uses:
(a) Radio waves ✅
(b) Microwaves only
(c) Infrared
(d) None
Explanation: Radio waves in MHz–GHz range.
Q136. Internet via optical fiber uses:
(a) Light waves ✅
(b) Radio waves
(c) Sound waves
(d) None
Explanation: High-speed data transmitted using light pulses.
Q137. Doppler effect in astronomy measures:
(a) Velocity of stars/galaxies ✅
(b) Mass
(c) Temperature
(d) None
Explanation: Red/blue shift indicates motion.
Q138. Radar uses Doppler effect to measure:
(a) Speed of vehicles ✅
(b) Distance only
(c) Temperature
(d) None
Explanation: Frequency shift gives velocity.
Q139. Medical ultrasound uses Doppler effect to measure:
(a) Blood flow velocity ✅
(b) Bone density
(c) Muscle strength
(d) None
Explanation: Frequency shift in reflected waves.
Q140. Beats are used to:
(a) Tune musical instruments ✅
(b) Measure amplitude
(c) Measure wavelength
(d) None
Explanation: Musicians adjust until beats vanish, indicating equal frequencies.
Q141. Two tuning forks 256 Hz and 260 Hz produce beat frequency:
(a) 4 Hz ✅
(b) 2 Hz
(c) 6 Hz
(d) None
Explanation: |260 − 256| = 4 Hz.
Q142. Beats are heard when:
(a) Frequencies are close but not equal ✅
(b) Frequencies are equal
(c) Frequencies are very different
(d) None
Explanation: Beat phenomenon requires small frequency difference.
Q143. Maximum beat frequency audible is about:
(a) 10 Hz ✅
(b) 100 Hz
(c) 1 Hz
(d) None
Explanation: Human ear perceives beats up to ~10 Hz.
Q144. Musical instruments produce different timbre because:
(a) Harmonic content varies ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Harmonics distinguish instruments playing same note.
Q145. Loudness depends on:
(a) Intensity ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: Loudness ∝ intensity, which ∝ amplitude².
Q146. Pitch depends on:
(a) Frequency ✅
(b) Amplitude
(c) Intensity
(d) None
Explanation: Pitch is perceptual correlate of frequency.
Q147. Quality (timbre) depends on:
(a) Harmonic content ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Harmonics distinguish instruments.
Q148. Resonance occurs when:
(a) Natural frequency = driving frequency ✅
(b) Natural frequency ≠ driving frequency
(c) Amplitude = zero
(d) None
Explanation: Resonance amplifies oscillations when frequencies match.
Q149. Resonance tube experiment measures:
(a) Speed of sound ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Known tuning fork frequency, measure resonance length → calculate v.
Q150. End correction in resonance tube accounts for:
(a) Antinode slightly beyond open end ✅
(b) Node inside tube
(c) Pressure variation
(d) None
Explanation: Effective length is slightly longer than physical length due to end effect.
Q151. A closed pipe of length 25 cm resonates with tuning fork of 340 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/4L → v = 4Lf = 4×0.25×340 = 340 m/s.
Q152. An open pipe of length 50 cm resonates with tuning fork of 340 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/2L → v = 2Lf = 2×0.5×340 = 340 m/s.
Q153. A string length 1 m, tension 100 N, μ = 0.01 kg/m. Fundamental frequency =
(a) 50 Hz ✅
(b) 100 Hz
(c) 25 Hz
(d) None
Explanation: f = (1/2L)√(T/μ) = (1/2)√(100/0.01) = 50 Hz.
Q154. If tension increases 4 times, frequency:
(a) Doubles ✅
(b) Halves
(c) Same
(d) None
Explanation: f ∝ √T. Quadruple T → f doubles.
Q155. If length halves, frequency:
(a) Doubles ✅
(b) Halves
(c) Same
(d) None
Explanation: f ∝ 1/L. Halving L doubles f.
Q156. Doppler effect: source 500 Hz, moving towards observer at 20 m/s, v = 340 m/s. Observed f =
(a) 531 Hz ✅
(b) 520 Hz
(c) 540 Hz
(d) None
Explanation: f' = f(v/(v−vs)) = 500(340/320) ≈ 531 Hz.
Q157. Doppler effect: observer moves towards source at 20 m/s, f = 500 Hz, v = 340 m/s. Observed f =
(a) 529 Hz ✅
(b) 520 Hz
(c) 540 Hz
(d) None
Explanation: f' = f(1+v0/v) = 500(1+20/340) ≈ 529 Hz.
Q158. Doppler effect: source moves away at 20 m/s, f = 500 Hz, v = 340 m/s. Observed f =
(a) 471 Hz ✅
(b) 480 Hz
(c) 460 Hz
(d) None
Explanation: f' = f(v/(v+vs)) = 500(340/360) ≈ 471 Hz.
Q159. Doppler effect: observer moves away at 20 m/s, f = 500 Hz, v = 340 m/s. Observed f =
(a) 471 Hz ✅
(b) 480 Hz
(c) 460 Hz
(d) None
Explanation: f' = f(1−v0/v) = 500(1−20/340) ≈ 471 Hz.
Q160. Beat frequency: two tuning forks 256 Hz and 258 Hz →
(a) 2 Hz ✅
(b) 4 Hz
(c) 6 Hz
(d) None
Explanation: |258−256| = 2 Hz.
Q161. Beat frequency: two tuning forks 512 Hz and 520 Hz →
(a) 8 Hz ✅
(b) 6 Hz
(c) 10 Hz
(d) None
Explanation: |520−512| = 8 Hz.
Q162. Beats are audible when frequency difference is:
(a) ≤10 Hz ✅
(b) ≥100 Hz
(c) ≥50 Hz
(d) None
Explanation: Human ear perceives beats up to ~10 Hz.
Q163. Reverberation time depends on:
(a) Volume and absorption of hall ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Sabine’s formula: T = 0.161 V/A.
Q164. Reverberation time desirable in:
(a) Concert halls ✅
(b) Studios
(c) Classrooms
(d) None
Explanation: Moderate reverberation enriches music.
Q165. Reverberation time undesirable in:
(a) Studios ✅
(b) Concert halls
(c) Auditoriums
(d) None
Explanation: Studios require clarity.
Q166. Optimum reverberation time for a classroom is:
(a) 0.5–1.0 s ✅
(b) 2–3 s
(c) 4–5 s
(d) None
Explanation: Short reverberation ensures speech clarity; longer times are suitable for music halls.
Q167. SONAR pulse returns in 5 s, v = 1500 m/s. Depth =
(a) 3750 m ✅
(b) 3000 m
(c) 1500 m
(d) None
Explanation: Distance = v×t/2 = 1500×5/2 = 3750 m.
Q168. SONAR pulse returns in 6 s, v = 1500 m/s. Depth =
(a) 4500 m ✅
(b) 3000 m
(c) 1500 m
(d) None
Explanation: Distance = v×t/2 = 1500×6/2 = 4500 m.
Q169. Ultrasound therapy is used to:
(a) Heal tissues ✅
(b) Cook food
(c) Measure velocity
(d) None
Explanation: Ultrasonic waves stimulate healing in physiotherapy.
Q170. Radar differs from SONAR because it uses:
(a) Radio waves ✅
(b) Sound waves
(c) Light waves
(d) None
Explanation: Radar uses electromagnetic waves, SONAR uses sound.
Q171. Echocardiography uses:
(a) Ultrasound ✅
(b) Infrasound
(c) Audible sound
(d) None
Explanation: Ultrasound images heart structure and motion.
Q172. Communication systems use sound waves to:
(a) Transmit information ✅
(b) Transmit matter
(c) Transmit heat
(d) None
Explanation: Signals encoded in waves carry information.
Q173. Bandwidth is:
(a) Range of frequencies transmitted ✅
(b) Single frequency
(c) Amplitude range
(d) None
Explanation: Determines data capacity.
Q174. Higher bandwidth means:
(a) Higher data rate ✅
(b) Lower data rate
(c) Same data rate
(d) None
Explanation: More frequencies → more information per second.
Q175. Modulation is:
(a) Superimposing information signal on carrier wave ✅
(b) Removing carrier
(c) Amplifying only
(d) None
Explanation: Enables long-distance transmission.
Q176. A closed pipe of length 34 cm resonates with tuning fork of 250 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/4L → v = 4Lf = 4×0.34×250 = 340 m/s.
Q177. An open pipe of length 34 cm resonates with tuning fork of 500 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/2L → v = 2Lf = 2×0.34×500 = 340 m/s.
Q178. A string length 1 m, tension 400 N, μ = 0.01 kg/m. Fundamental frequency =
(a) 100 Hz ✅
(b) 200 Hz
(c) 50 Hz
(d) None
Explanation: f = (1/2L)√(T/μ) = (1/2)√(400/0.01) = 100 Hz.
Q179. If tension increases 9 times, frequency:
(a) Triples ✅
(b) Doubles
(c) Halves
(d) None
Explanation: f ∝ √T. Nine times T → f triples.
Q180. Doppler effect: source 600 Hz, moving towards observer at 20 m/s, v = 340 m/s. Observed f =
(a) 635 Hz ✅
(b) 640 Hz
(c) 630 Hz
(d) None
Explanation: f' = f(v/(v−vs)) = 600(340/320) ≈ 635 Hz.
Q181. Doppler effect: observer moves towards source at 20 m/s, f = 600 Hz, v = 340 m/s. Observed f =
(a) 635 Hz ✅
(b) 640 Hz
(c) 630 Hz
(d) None
Explanation: f' = f(1+v0/v) = 600(1+20/340) ≈ 635 Hz.
Q182. Doppler effect: source moves away at 20 m/s, f = 600 Hz, v = 340 m/s. Observed f =
(a) 566 Hz ✅
(b) 570 Hz
(c) 560 Hz
(d) None
Explanation: f' = f(v/(v+vs)) = 600(340/360) ≈ 566 Hz.
Q183. Doppler effect: observer moves away at 20 m/s, f = 600 Hz, v = 340 m/s. Observed f =
(a) 566 Hz ✅
(b) 570 Hz
(c) 560 Hz
(d) None
Explanation: f' = f(1−v0/v) = 600(1−20/340) ≈ 566 Hz.
Q184. SONAR pulse returns in 8 s, v = 1500 m/s. Depth =
(a) 6000 m ✅
(b) 7500 m
(c) 3000 m
(d) None
Explanation: Distance = v×t/2 = 1500×8/2 = 6000 m.
Q185. SONAR pulse returns in 10 s, v = 1500 m/s. Depth =
(a) 7500 m ✅
(b) 6000 m
(c) 3000 m
(d) None
Explanation: Distance = v×t/2 = 1500×10/2 = 7500 m.
Q186. Ultrasonic flaw detection is used in:
(a) Engineering materials ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Detects cracks in metals and composites.
Q187. Ultrasonic cleaning works because:
(a) Cavitation bubbles remove dirt ✅
(b) Heat melts dirt
(c) Pressure dissolves dirt
(d) None
Explanation: Ultrasonic vibrations create bubbles that scrub surfaces.
Q188. Ultrasonic welding is used in:
(a) Plastics and metals ✅
(b) Cooking
(c) Heating
(d) None
Explanation: High-frequency vibrations fuse materials without melting.
Q189. Communication systems use sound waves to:
(a) Transmit information ✅
(b) Transmit matter
(c) Transmit heat
(d) None
Explanation: Signals encoded in waves carry information.
Q190. Bandwidth is:
(a) Range of frequencies transmitted ✅
(b) Single frequency
(c) Amplitude range
(d) None
Explanation: Determines data capacity.
Q191. Higher bandwidth means:
(a) Higher data rate ✅
(b) Lower data rate
(c) Same data rate
(d) None
Explanation: More frequencies → more information per second.
Q192. Modulation is:
(a) Superimposing information signal on carrier wave ✅
(b) Removing carrier
(c) Amplifying only
(d) None
Explanation: Enables long-distance transmission.
Q193. Amplitude modulation varies:
(a) Amplitude of carrier ✅
(b) Frequency
(c) Phase
(d) None
Explanation: Carrier amplitude changes with signal.
Q194. Frequency modulation varies:
(a) Frequency of carrier ✅
(b) Amplitude
(c) Phase
(d) None
Explanation: Carrier frequency changes with signal.
Q195. Phase modulation varies:
(a) Phase of carrier ✅
(b) Amplitude
(c) Frequency
(d) None
Explanation: Carrier phase changes with signal.
Q196. Advantage of FM over AM:
(a) Less noise ✅
(b) More noise
(c) Lower fidelity
(d) None
Explanation: Frequency is less affected by amplitude noise.
Q197. Optical fiber used in medicine for:
(a) Endoscopy ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fiber optics transmit light inside body.
Q198. Optical fiber used in communication for:
(a) Internet cables ✅
(b) Radio only
(c) Sound waves
(d) None
Explanation: High-speed data transmission.
Q199. Optical fiber used in sensors for:
(a) Measuring strain and temperature ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fiber sensors detect physical changes.
Q200. Optical fiber used in defense for:
(a) Secure communication ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Immune to electromagnetic interference.
Q201. A closed pipe of length 17 cm resonates with tuning fork of 500 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/4L → v = 4Lf = 4×0.17×500 = 340 m/s.
Q202. An open pipe of length 34 cm resonates with tuning fork of 500 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/2L → v = 2Lf = 2×0.34×500 = 340 m/s.
Q203. A string length 2 m, tension 400 N, μ = 0.04 kg/m. Fundamental frequency =
(a) 50 Hz ✅
(b) 100 Hz
(c) 25 Hz
(d) None
Explanation: f = (1/2L)√(T/μ) = (1/4)√(400/0.04) = 50 Hz.
Q204. If tension increases 16 times, frequency:
(a) 4 times ✅
(b) 2 times
(c) Same
(d) None
Explanation: f ∝ √T. Sixteen times T → f quadruples.
Q205. Doppler effect: source 400 Hz, moving towards observer at 40 m/s, v = 340 m/s. Observed f =
(a) 448 Hz ✅
(b) 440 Hz
(c) 450 Hz
(d) None
Explanation: f' = f(v/(v−vs)) = 400(340/300) ≈ 448 Hz.
Q206. Doppler effect: observer moves towards source at 40 m/s, f = 400 Hz, v = 340 m/s. Observed f =
(a) 447 Hz ✅
(b) 440 Hz
(c) 450 Hz
(d) None
Explanation: f' = f(1+v0/v) = 400(1+40/340) ≈ 447 Hz.
Q207. Doppler effect: source moves away at 40 m/s, f = 400 Hz, v = 340 m/s. Observed f =
(a) 362 Hz ✅
(b) 370 Hz
(c) 360 Hz
(d) None
Explanation: f' = f(v/(v+vs)) = 400(340/380) ≈ 362 Hz.
Q208. Doppler effect: observer moves away at 40 m/s, f = 400 Hz, v = 340 m/s. Observed f =
(a) 353 Hz ✅
(b) 360 Hz
(c) 370 Hz
(d) None
Explanation: f' = f(1−v0/v) = 400(1−40/340) ≈ 353 Hz.
Q209. SONAR pulse returns in 12 s, v = 1500 m/s. Depth =
(a) 9000 m ✅
(b) 7500 m
(c) 6000 m
(d) None
Explanation: Distance = v×t/2 = 1500×12/2 = 9000 m.
Q210. SONAR pulse returns in 14 s, v = 1500 m/s. Depth =
(a) 10,500 m ✅
(b) 9000 m
(c) 7500 m
(d) None
Explanation: Distance = v×t/2 = 1500×14/2 = 10,500 m.
Q211. Ultrasonic waves in industry are used for:
(a) Cleaning delicate instruments ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Cavitation bubbles clean surgical tools and electronics.
Q212. Infrasonic waves are used in:
(a) Earthquake monitoring ✅
(b) Medical imaging
(c) SONAR
(d) None
Explanation: Seismology relies on infrasonic waves.
Q213. Ultrasonic waves are used in:
(a) Non-destructive testing ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Detects cracks and flaws in materials.
Q214. Communication systems use sound waves to:
(a) Transmit information ✅
(b) Transmit matter
(c) Transmit heat
(d) None
Explanation: Signals encoded in waves carry information.
Q215. Bandwidth is:
(a) Range of frequencies transmitted ✅
(b) Single frequency
(c) Amplitude range
(d) None
Explanation: Determines data capacity.
Q216. Higher bandwidth means:
(a) Higher data rate ✅
(b) Lower data rate
(c) Same data rate
(d) None
Explanation: More frequencies → more information per second.
Q217. Modulation is:
(a) Superimposing information signal on carrier wave ✅
(b) Removing carrier
(c) Amplifying only
(d) None
Explanation: Enables long-distance transmission.
Q218. Amplitude modulation varies:
(a) Amplitude of carrier ✅
(b) Frequency
(c) Phase
(d) None
Explanation: Carrier amplitude changes with signal.
Q219. Frequency modulation varies:
(a) Frequency of carrier ✅
(b) Amplitude
(c) Phase
(d) None
Explanation: Carrier frequency changes with signal.
Q220. Phase modulation varies:
(a) Phase of carrier ✅
(b) Amplitude
(c) Frequency
(d) None
Explanation: Carrier phase changes with signal.
Q221. Advantage of FM over AM:
(a) Less noise ✅
(b) More noise
(c) Lower fidelity
(d) None
Explanation: Frequency is less affected by amplitude noise.
Q222. Optical fiber used in surgery for:
(a) Laser delivery ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fibers guide laser beams.
Q223. Optical fiber used in imaging for:
(a) Fiber scopes ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Flexible fibers transmit images.
Q224. Optical fiber used in industry for:
(a) Monitoring machines ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fiber sensors detect vibrations.
Q225. Optical fiber used in research for:
(a) Particle detection ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fibers measure signals in experiments such as particle physics and advanced sensing.
Q226. A closed pipe of length 20 cm resonates with tuning fork of 425 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/4L → v = 4Lf = 4×0.20×425 = 340 m/s.
Q227. An open pipe of length 40 cm resonates with tuning fork of 425 Hz. Speed of sound =
(a) 340 m/s ✅
(b) 300 m/s
(c) 360 m/s
(d) None
Explanation: f = v/2L → v = 2Lf = 2×0.40×425 = 340 m/s.
Q228. A string length 1.5 m, tension 225 N, μ = 0.09 kg/m. Fundamental frequency =
(a) 25 Hz ✅
(b) 50 Hz
(c) 75 Hz
(d) None
Explanation: f = (1/2L)√(T/μ) = (1/3)√(225/0.09) = 25 Hz.
Q229. If tension increases 25 times, frequency:
(a) 5 times ✅
(b) 2 times
(c) Same
(d) None
Explanation: f ∝ √T. Twenty‑five times T → f increases 5 times.
Q230. Doppler effect: source 300 Hz, moving towards observer at 30 m/s, v = 340 m/s. Observed f =
(a) 327 Hz ✅
(b) 330 Hz
(c) 325 Hz
(d) None
Explanation: f' = f(v/(v−vs)) = 300(340/310) ≈ 327 Hz.
Q231. Doppler effect: observer moves towards source at 30 m/s, f = 300 Hz, v = 340 m/s. Observed f =
(a) 326 Hz ✅
(b) 330 Hz
(c) 325 Hz
(d) None
Explanation: f' = f(1+v0/v) = 300(1+30/340) ≈ 326 Hz.
Q232. Doppler effect: source moves away at 30 m/s, f = 300 Hz, v = 340 m/s. Observed f =
(a) 274 Hz ✅
(b) 280 Hz
(c) 270 Hz
(d) None
Explanation: f' = f(v/(v+vs)) = 300(340/370) ≈ 274 Hz.
Q233. Doppler effect: observer moves away at 30 m/s, f = 300 Hz, v = 340 m/s. Observed f =
(a) 274 Hz ✅
(b) 280 Hz
(c) 270 Hz
(d) None
Explanation: f' = f(1−v0/v) = 300(1−30/340) ≈ 274 Hz.
Q234. SONAR pulse returns in 16 s, v = 1500 m/s. Depth =
(a) 12,000 m ✅
(b) 10,500 m
(c) 9000 m
(d) None
Explanation: Distance = v×t/2 = 1500×16/2 = 12,000 m.
Q235. SONAR pulse returns in 18 s, v = 1500 m/s. Depth =
(a) 13,500 m ✅
(b) 12,000 m
(c) 10,500 m
(d) None
Explanation: Distance = v×t/2 = 1500×18/2 = 13,500 m.
Q236. Reverberation time in large auditorium depends on:
(a) Volume and absorption coefficient ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Sabine’s formula T = 0.161 V/A.
Q237. Reverberation time desirable in:
(a) Opera houses ✅
(b) Studios
(c) Classrooms
(d) None
Explanation: Moderate reverberation enriches music.
Q238. Reverberation time undesirable in:
(a) Recording studios ✅
(b) Opera houses
(c) Auditoriums
(d) None
Explanation: Studios require clarity.
Q239. Optimum reverberation time for a concert hall is:
(a) 1.5–2.5 s ✅
(b) 0.5–1.0 s
(c) 3–4 s
(d) None
Explanation: Longer reverberation enriches musical tones, but too long causes muddiness.
Q240. Maximum beat frequency audible to humans is about:
(a) 10 Hz ✅
(b) 100 Hz
(c) 50 Hz
(d) None
Explanation: Beyond ~10 Hz, beats merge into separate tones.
Q241. Two tuning forks 400 Hz and 404 Hz produce beat frequency:
(a) 4 Hz ✅
(b) 2 Hz
(c) 6 Hz
(d) None
Explanation: |404−400| = 4 Hz.
Q242. Beats are audible when frequency difference is:
(a) ≤10 Hz ✅
(b) ≥100 Hz
(c) ≥50 Hz
(d) None
Explanation: Human ear perceives beats up to ~10 Hz.
Q243. Musical instruments produce different timbre because:
(a) Harmonic content varies ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Harmonics distinguish instruments playing same note.
Q244. Loudness depends on:
(a) Intensity ✅
(b) Frequency
(c) Wavelength
(d) None
Explanation: Loudness ∝ intensity, which ∝ amplitude².
Q245. Pitch depends on:
(a) Frequency ✅
(b) Amplitude
(c) Intensity
(d) None
Explanation: Pitch is perceptual correlate of frequency.
Q246. Quality (timbre) depends on:
(a) Harmonic content ✅
(b) Frequency only
(c) Amplitude only
(d) None
Explanation: Harmonics distinguish instruments.
Q247. Resonance occurs when:
(a) Natural frequency = driving frequency ✅
(b) Natural frequency ≠ driving frequency
(c) Amplitude = zero
(d) None
Explanation: Resonance amplifies oscillations when frequencies match.
Q248. Resonance tube experiment measures:
(a) Speed of sound ✅
(b) Frequency
(c) Amplitude
(d) None
Explanation: Known tuning fork frequency, measure resonance length → calculate v.
Q249. Optical fiber used in environment monitoring for:
(a) Detecting pollutants ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fiber sensors measure chemical levels and pollutants in air and water.
Q250. Optical fiber used in aviation for:
(a) Aircraft communication ✅
(b) Cooking
(c) Heating
(d) None
Explanation: Fibers transmit secure signals in aircraft systems, immune to electromagnetic interference.
Post a Comment
Post a Comment