1. The distance in between 2 points differing in phase by 60° having wave velocity 360 m/s and frequency of wave 500 Hz is:[BP 2013, BP 2012]
(a) Δx = (Δφ/360°) × (v/f) = (60/360) × (360/500) = 0.12 m 2. Sound travels fastest in
(a) Sound travels fastest in solids with high elasticity. Iron has higher speed (~5130 m/s) than other options. 3. Oxygen is 16 times heavier than H₂. Equal volumes of hydrogen and oxygen are mixed, the ratio of velocity of sound in mixture to O₂ is[BP 2012]
(b) v ∝ 1/√ρ. For mixture: ρ_mix = (ρ_H₂ + ρ_O₂)/2 = (1+16)/2 = 8.5; v_mix/v_O₂ = √(16/8.5) ≈ √2 4. The relation between phase difference and path difference is[BP 2011]
- Δφ = kΔx
- Δφ = 2πνΔx
- Δφ = (2π/λ)Δx
- Δφ = (1/λ)Δx
(c) Phase difference Δφ = (2π/λ) × path difference Δx (standard wave relation). 5. Phenomenon associated with transverse wave only is[BP 2010]
- Diffraction
- Polarization
- Interference
- Refraction
(b) Polarization only occurs in transverse waves as it requires perpendicular oscillation direction. 6. The frequency of sound audible to humans is[BP 2010]
- <20 Hz and > 20000 Hz
- >20 Hz and <20000 Hz
- <20 Hz and >200000 Hz
- <20 Hz and <20000 Hz
(b) Standard human hearing range is 20 Hz to 20 kHz. 7. The intensity of sound at night increases because of:[IOM 2013]
- Low temperature
- Increase in density
- Decrease in density
- Calmness
(a) Lower temperature increases air density, enhancing sound propagation. 8. Sound waves in rocks are[IOM 2009]
- Longitudinal stationary
- Transverse stationary
- Longitudinal and transverse wave
- Wave in rock does not propagate
(c) Rocks support both longitudinal (P-waves) and transverse (S-waves). 9. Velocity of sound in air at STP is 330 m/s. The distance covered by sound in 2 seconds when atmospheric temperature is 30°C will be:[MOE 2013]
(b) v = 330 + (0.6×30) = 348 m/s; Distance = 348×2 = 696 m ≈ 0.7 km 10. If R is the radius of a resonance tube, the end correction to be applied is[MOE 2013]
(d) End correction e ≈ 0.6R (standard acoustic correction for open tubes). 11. If distance between source of sound and a cliff is s. If the velocity of sound is V, the time taken to hear the 2nd echo is[MOE 2012]
(c) 2nd echo travels 4s distance (to cliff and back twice); Time = 4s/V 12. The velocity of sound in water is 1400 m/s. The density of water is 1000 kg/m³. Bulk modulus of elasticity of water is[MOE 2012]
- 5 × 10⁹ N/m²
- 1.96 × 10⁹ N/m²
- 4 × 10⁹ N/m²
- 1.96 × 10¹⁰ N/m²
(b) B = v²ρ = (1400)² × 1000 = 1.96 × 10⁹ N/m² 13. Laplace's correction in the expression for velocity of sound is needed because sound waves[MOE 2010]
- Are longitudinal
- Propagate adiabatically
- Propagate isothermally
- Are of long wavelengths
(b) Sound propagation is adiabatic (not isothermal) for high-frequency waves. 14. Angle between particle and wave velocity in transverse wave is:[MOE 2010]
(c) In transverse waves, particle motion is perpendicular to wave propagation. 15. The intensity of sound gets reduced by 10% on passing through a block. If it passes through two such blocks, then intensity of outgoing sound is:[MOE 2009]
(c) After two blocks: I = I₀ × (0.9)² = 0.81I₀ (81%) 16. Sound waves are travelling in a medium whose adiabatic elasticity is Eₐ and isothermal elasticity is Eᵢ. The velocity of sound is proportional to[MOE 2009]
(d) v ∝ √(Eₐ/ρ) (Laplace correction for adiabatic conditions). 17. A change in temperature affects which property of sound?[KU 2014]
- Frequency
- Amplitude
- Wavelength
- Loudness
(c) v ∝ √T ⇒ λ = v/f changes with temperature. 18. Which of the following is a mechanical wave?[KU 2012]
- Radio waves
- Light waves
- X-rays
- Sound waves
(d) Sound requires material medium (mechanical wave). 19. A radio station has a band of 30 m. The frequency of electromagnetic waves from this station will be:[IOM 2012]
- 10 MHz
- 3 × 10¹¹ Hz
- 10 kHz
- 11 × 10¹⁰ Hz
(a) f = c/λ = (3×10⁸)/30 = 10 MHz 20. The increase in velocity of sound for 10°C rise of temperature is (v₀ = 332 m/s)
- 0.16 m/s
- 0.61 m/s
- 6.1 m/s
- 61 m/s
(c) Δv = 0.6 × ΔT = 0.6 × 10 = 6 m/s (closest to 6.1 m/s) 21. If 4 g of Helium under STP has a volume of 22.4 litre, the speed of sound waves in an atmosphere of Helium at 0°C and 1 atm pressure is
- 973 m/s
- 963 m/s
- 953 m/s
- 943 m/s
(a) v = √(γRT/M) = √((5/3 × 8.314 × 273)/0.004) ≈ 973 m/s 22. If x = a sin(ωt + π/6) and x' = a cosωt, then what is the phase difference between the two waves
(b) cosωt = sin(ωt + π/2); Phase difference = (π/2 - π/6) = π/3 23. Two sound waves are given by y = a sin(ωt - kx) and y' = b cos(ωt - kx). The phase difference between the two waves is
(c) cos(ωt - kx) = sin(ωt - kx + π/2) ⇒ π/2 phase difference 24. Two waves are represented by y₁ = a sin(ωt + π/6) and y₂ = a cosωt. Their resultant amplitude is
(c) A_res = √(a² + a² + 2a²cos(π/2 - π/6)) = √3a 25. Equation of progressive wave is given by y = 4 sin[π(t/5 - x/3) + π/6]. Which is correct?
- v = 5 cm/s
- λ = 18 m
- a = 0.04 cm
- f = 50 Hz
(b) Comparing with y = A sin(2π(t/T - x/λ) + φ): λ = 6π/(π/3) = 18 m 26. If the pressure amplitude in a sound wave is tripled, then by what factor the intensity of the sound wave is increased
(c) I ∝ P₀² ⇒ (3)² = 9 times 27. The velocity of sound in air is independent of change in[IOM 2001]
- Pressure
- Density
- Temperature
- Humidity
(a) At constant temperature, v is independent of pressure (Boyle's Law). 28. Sound waves having which frequency are audible to humans?
- 5 Hz
- 27000 Hz
- 5000 Hz
- 50000 Hz
(c) Human range: 20 Hz - 20 kHz (5000 Hz is within this range). 29. The temperature at which the speed of sound in air becomes double of its value at 27°C is
(b) v ∝ √T ⇒ T₂ = 4T₁ = 4 × 300K = 1200K (927°C) 30. The speed of sound in air at NTP is 300 m/s. If air pressure becomes four times, then the speed will be
- 150 m/s
- 300 m/s
- 600 m/s
- 1200 m/s
(b) Speed depends only on temperature (constant here). 31. Velocity of sound is measured in hydrogen and oxygen gases at given temperature. The ratio (V_H/V_O) will be
(b) v ∝ 1/√M ⇒ V_H/V_O = √(32/2) = 4:1 32. The speed of sound in hydrogen at NTP is 1270 m/s. Speed in a mixture of H₂ and O₂ (4:1 by volume) will be
- 317 m/s
- 635 m/s
- 830 m/s
- 950 m/s
(b) ρ_mix = (4×2 + 1×32)/5 = 8; v = 1270 × √(2/8) ≈ 635 m/s 33. What must be the minimum distance of reflecting boundary to hear distinct echo? (v = 330 m/s)
(b) Minimum time = 0.1s ⇒ d = (330×0.1)/2 = 16.5 m 34. A string of length 'L' and mass 'M' hangs freely. The velocity at distance 'x' from free end is
(b) Tension at x = μxg ⇒ v = √(T/μ) = √(gx) 35. A wave y = A sin(ωt - kx) reflected from rigid boundary becomes
- y = A' sin(ωt + kx)
- y = -A' sin(ωt + kx)
- y = A' sin(ωt - kx)
- y = -A' sin(ωt - kx)
(b) Rigid boundary causes π phase change and direction reversal. 36. Two sound waves with 60° phase difference have path difference of
(b) Δx = (Δφ/360°)λ = (60/360)λ = λ/6 37. A pulse reaching fixed end reflects with
- Same phase, velocity reversed
- Same phase, no velocity change
- 180° phase change, no velocity reversal
- 180° phase change with velocity reversal
(d) Fixed end causes π phase change and velocity reversal. 38. A man between two cliffs hears echoes at 1s intervals (v = 340 m/s). Distance between cliffs is
(c) Time difference = 2d/v ⇒ d = (340×1)/2 = 170 m (total distance = 3×170 = 510 m) 39. When sound goes from one medium to another, unchanged quantity is
- Speed
- Amplitude
- Frequency
- Wavelength
(c) Frequency remains constant (determined by source). 40. Speed of sound is maximum in
- Monoatomic gas
- Diatomic gas
- Polyatomic gas
- Equal in all
(a) v ∝ √(γRT/M); γ is highest (5/3) for monoatomic gases. 41. The ratio of intensities of waves y₁ = 0.06 sin 2π(0.04t + 0.1x) and y₂ = 0.03 sin 2π(0.08t + 0.2x) is
(c) I ∝ A²ω² ⇒ (0.06²×0.04²)/(0.03²×0.08²) = 4:1 42. A man hears thunder 6s after lightning (T = 27°C). Distance is (v₀ = 332 m/s)[IOM 2001]
(c) v = 332 + 0.6×27 ≈ 348 m/s; d = 348×6 ≈ 2088 m 43. A man hears echo on 3rd clap (2 claps/s). Distance if v = 320 m/s is[IOM 1998]
(a) Time between claps = 0.5s; Echo heard at t = 1s ⇒ d = (320×1)/2 = 160 m (but options may need recheck) 44. Two uniform wires with diameter ratio 1:2 under same tension have frequency ratio[MOE 2066]
(b) f ∝ 1/d (since μ ∝ d²) ⇒ f₁/f₂ = 2/1 45. Phase difference between y₁ = a sinωt and y₂ = a cosωt is[MOE 2008]
(c) cosωt = sin(ωt + π/2) ⇒ π/2 difference 46. The equation y = 10 sinπ(0.01x - 2t) has frequency[MOE 2063]
(c) Comparing with y = A sin(kx - ωt): f = ω/2π = 2π/2π = 1 Hz 47. Frequency of 15 MHz radio waves has wavelength[MOE 2056]
(a) λ = c/f = (3×10⁸)/(15×10⁶) = 20 m 48. Ratio of intensities of y₁ = 20sin8π and y₂ = 40sin100π is[MOE 2063]
(a) I ∝ A² ⇒ (20²)/(40²) = 1:4 49. For sound wave (f = 500 Hz, v = 350 m/s), distance between particles with 60° phase difference is
(b) Δx = (60/360) × (350/500) ≈ 0.116 m ≈ 12 cm 50. In y = Y₀ sin2π(ft - x/λ), if v_max = 4v_wave then
- λ = πY₀
- λ = πY₀/2
- λ = 2πY₀
- λ = 4πY₀
(b) v_max = ωY₀ = 4v ⇒ 2πfY₀ = 4fλ ⇒ λ = πY₀/2 51. For y = 6cos(1800t - 60x), ratio of maximum particle velocity to wave velocity is
- 3.6×10⁶
- 3.6×10⁴
- 3.6×10⁻⁴
- 360
(c) v_max = 6×10⁻6 × 1800; v_wave = 1800/60 = 30 ⇒ Ratio = 0.0108/30 = 3.6×10⁻⁴ 52. A string (L=0.6m, μ=0.2 kg/m) vibrates in 3 segments (A=0.5cm) at T=80N. Particle velocity amplitude is
- 1.57 m/s
- 3.14 m/s
- 6.28 m/s
- 9.42 m/s
(b) v_particle = ωA = 2πf × 0.005; f = (3/2L)√(T/μ) ≈ 100 Hz ⇒ v ≈ 3.14 m/s 53. Speed of sound in moist H₂ vs dry H₂ is
- More in dry H₂
- More in moist H₂
- Same
- None
(c) Speed is nearly same (minor density differences). 54. String stretched by 20% has wave speed v. If stretched by 16%, new speed is[MOE]
(b) v ∝ √T ∝ √ΔL ⇒ v'/v = √(16/20) ≈ 0.89 (No exact match, but closest to v/√2) 55. Time for sound to travel 1m in steel rod (Y=2×10¹¹ N/m², ρ=8000 kg/m³) is
- 1×10⁻⁴ s
- 2×10⁻⁴ s
- 4×10⁻⁴ s
- 16×10⁻⁴ s
(b) v = √(Y/ρ) ≈ 5000 m/s ⇒ t = 1/5000 = 2×10⁻⁴ s 56. Depth of sea if echo returns in 2s (B=2.3×10⁹ N/m², ρ=1.1 g/cc) is
(b) v = √(B/ρ) ≈ 1445 m/s ⇒ d = (1445×2)/2 ≈ 1410 m 57. Percentage change in speed of sound when T increases from 300K to 301K is
(a) Δv/v = 0.5 × ΔT/T = 0.5 × 1/300 ≈ 0.167% 58. Amplitude at 40cm compared to 10cm is
(b) Assuming spherical waves: A ∝ 1/r ⇒ A'/A = 10/40 = 1/4 (but options may vary) 59. Ratio of amplitudes at 9m and 25m from point source is
(c) A ∝ 1/r ⇒ A₁/A₂ = 25/9 60. If pressure amplitude increases by 1%, intensity increases by
(b) I ∝ P₀² ⇒ 2 × 1% = 2% increase 61. Intensity of sound in air (ρ=1.3 kg/m³, v=330 m/s, P=1.01×10⁵ Pa) is
- 1.6×10⁵ W/m²
- 16×10⁵ W/m²
- 160×10⁵ W/m²
- 1600×10⁵ W/m²
(a) I = P²/(2ρv) ≈ (1.01×10⁵)²/(2×1.3×330) ≈ 1.6×10⁵ W/m² 62. For y = 5/[5 + (x+30)²], maximum displacement is
(d) At x=-30, y_max = 5/5 = 1 cm 63. For y = 2/[2 + (x+20)²], wave velocity is
(a) Form suggests stationary pulse (no velocity implied in given equation) 64. For y = 5/[5 + (4x+50)²]
- Travels in +x direction
- Amplitude = 0.16 cm
- Travels 25 cm in 2s
- Both b and c
(d) Amplitude = 5/5² = 0.2 cm (closest to 0.16); Velocity = 50/2 = 25 cm/s 65. Displacement changes from y = 1 + x² at t=0 to y = 1 + (x-1)² at t=2s. Velocity is
(a) Pulse moved 1m in 2s ⇒ v = 0.5 m/s 66. Ratio of sound speed in H₂ (γ=7/5) to He (γ=5/3) at same T is
(a) v_H/v_He = √[(7/5×4)/(5/3×4)] = √(21/25) (No exact match) 67. Jet plane sound heard at 30° when overhead has velocity (sound speed = v)
(d) tan30° = v/v_jet ⇒ v_jet = v/tan30° = √3v 68. Ripple tank pulses at 0.1s intervals create 30mm spacing. New spacing at 0.5s intervals is
(c) v = 30/0.1 = 300 mm/s ⇒ New spacing = 300×0.5 = 150 mm 69. 1000 Hz wave travels 600m in 2s. Number of wavelengths in this distance is
(d) λ = v/f = (600/2)/1000 = 0.3m ⇒ N = 600/0.3 = 2000 70. Cloud at 60° elevation produces thunder after 8s (v=300 m/s). Height is
- 8×300 m
- 8×300 sin60° m
- 8×300 cos60° m
- 8×300 tan60° m
(c) Horizontal distance = v×t; Height = d×tan60° = 300×8×cos60°×tan60° (simplifies to 8×300 sin60°) 71. If note B has 1/8th frequency of A with equal energy, amplitude of B is
(b) E ∝ A²f² ⇒ A_B = A_A × √(f_A/f_B) = 2A_A 72. For gas with sound speeds v₁,v₂ at T₁,T₂ and rms speeds v₁',v₂'
- v₂' = v₁' √(v₂/v₁)
- v₂' = v₁' (v₂/v₁)
- v₂' = v₁' (v₂/v₁)²
- v₂' = v₁' (v₂/v₁)^(3/2)
(a) v_sound ∝ √T; v_rms ∝ √T ⇒ v_rms ratio matches √(v_sound ratio) 73. Two wires with density ratio 1:3 have frequency ratio[IOM 2017]
(b) f ∝ 1/√μ ⇒ f₁/f₂ = √(3/1) = √3:1 74. Velocity of sound in steel (Y=2×10¹¹ N/m², ρ=78×10³ kg/m³) is[IOM 2017]
- 340 m/s
- 900 m/s
- 3×10⁸ m/s
- 5060 m/s
(d) v = √(Y/ρ) ≈ √(2×10¹¹/7800) ≈ 5060 m/s 