JEE Physics Waves MCQs Set 12

Multiple Choice Questions (MCQs) for JEE Physics: Waves

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Multiple Choice Questions

Question 1: A source and an observer move away from each other with a velocity of 15 m/s each with respect to the ground. If the observer finds the frequency of sound coming from the source to be 1950 Hz, the actual frequency of the source is (v = 340 m/s):
(a) 1785 Hz
(b) 1968 Hz
(c) 1950 Hz
(d) 2130 Hz
Show Answer & Explanation

Answer: (d) 2130 Hz

Explanation:
1. f′ = f(v − v0)/(v + vs) = f × 325/355.
2. f = 1950 × 355/325 = 2130 Hz.

Teacher's Note:
1. Both motions reduce the observed frequency.
2. Observer moving away: v − v0. Source moving away: v + vs.

Question 2: A sound source of frequency 170 Hz is placed near a wall. A man walking from the source towards the wall finds that there is a periodic rise and fall of sound intensity. If the speed of sound in air is 340 m/s, the distance between two adjacent minima is:
(a) 3/2 m
(b) 2 m
(c) 1 m
(d) 1/2 m
Show Answer & Explanation

Answer: (c) 1 m

Explanation:
1. The direct and reflected waves form a standing wave.
2. λ = v/f = 340/170 = 2 m, and adjacent minima (nodes) are λ/2 = 1 m apart.

Teacher's Note:
1. Nodes in a standing wave are λ/2 apart.
2. Compute λ from v/f first.

Question 3: A whistle revolves in a circle with angular speed ω = 20 rad/s using a string of length 50 cm. If the frequency of the sound from the whistle is 385 Hz, the minimum frequency heard by an observer far away from the centre is (v = 340 m/s):
(a) 333 Hz
(b) 374 Hz
(c) 385 Hz
(d) 394 Hz
Show Answer & Explanation

Answer: (b) 374 Hz

Explanation:
1. Speed of the whistle = ωr = 20 × 0.5 = 10 m/s.
2. Minimum frequency is when the source moves directly away: f′ = 385 × 340/(340 + 10) = 374 Hz.

Teacher's Note:
1. Speed of source = ωr.
2. The minimum occurs when the source recedes.

Question 4: A source of frequency 200 Hz moves towards an observer with a velocity equal to the speed V of the observer, who is moving away from the source. The apparent frequency heard by the observer is:
(a) 200 Hz
(b) 50 Hz
(c) 160 Hz
(d) 150 Hz
Show Answer & Explanation

Answer: (a) 200 Hz

Explanation:
1. f′ = f(v − V)/(v − V) = f, since the observer's recession and the source's approach have equal speeds.
2. f′ = 200 Hz.

Teacher's Note:
1. Write both factors with the same speed V.
2. The numerator and denominator are equal.

Question 5: A bus moves with a velocity of 5 m/s towards a huge wall. The driver sounds a horn of frequency 165 Hz. If the speed of sound in air is 335 m/s, the number of beats heard per second by a passenger on the bus is:
(a) 6
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. The passenger hears the direct sound at 165 Hz (no relative motion with the horn).
2. The wall receives 165 × 335/(335 − 5) and reflects it; the moving passenger hears f′ = 165 × (335 + 5)/(335 − 5) = 170 Hz.
3. Beats = 170 − 165 = 5 per second.

Teacher's Note:
1. The source and passenger move together.
2. Direct sound is unshifted; the echo is raised twice.

Question 6: A person walks away from a wall towards an observer at a speed of 1 m/s and blows a whistle of frequency 680 Hz. The number of beats heard by the stationary observer per second is (v = 340 m/s):
(a) 2
(b) zero
(c) 4
(d) 8
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Direct sound: source moving towards the observer, f1 = 680 × 340/339.
2. Echo: source moving away from the wall, f2 = 680 × 340/341.
3. f1 − f2 = 680 × 340 × (2/(339 × 341)) ≈ 4 per second.

Teacher's Note:
1. One sound is shifted up, the other shifted down.
2. Compute the two shifts separately.

Question 7: A siren emitting a sound of frequency 900 Hz moves away from an observer towards a cliff at a speed of 30 m/s. The frequency of the echo heard by the observer is (v = 330 m/s):
(a) 1000 Hz
(b) 990 Hz
(c) 930 Hz
(d) 960 Hz
Show Answer & Explanation

Answer: (b) 990 Hz

Explanation:
1. The cliff receives sound from a source approaching it: f′ = 900 × 330/(330 − 30) = 990 Hz.
2. The cliff reflects 990 Hz; the stationary observer hears 990 Hz.

Teacher's Note:
1. Treat the cliff as a stationary source of the received frequency.
2. The observer is at rest.

Question 8: The wavelength of the light received from a galaxy is 0.4% greater than the wavelength measured on Earth. The velocity of the galaxy relative to the Earth is:
(a) 1.2 × 10⁶ m/s
(b) 1.2 × 10⁷ m/s
(c) 1.2 × 10⁵ m/s
(d) 1.2 × 10⁴ m/s
Show Answer & Explanation

Answer: (a) 1.2 × 10⁶ m/s

Explanation:
1. Δλ/λ = v/c = 0.004.
2. v = 0.004 × 3 × 10⁸ = 1.2 × 10⁶ m/s (the galaxy is receding).

Teacher's Note:
1. Red shift: Δλ/λ = v/c.
2. 0.4% = 0.004.

Question 9: The Doppler effect for sound depends on the relative motion of the source and the listener and also on which of them is in motion, whereas for light it depends only on their relative motion. The reason is:
(a) Einstein's theory of relativity
(b) Einstein's mass-energy relation
(c) the photoelectric effect
(d) none of these
Show Answer & Explanation

Answer: (a) Einstein's theory of relativity

Explanation:
1. Sound needs a medium, so motion relative to the medium matters.
2. Light needs no medium and its speed is the same for all observers (relativity), so only relative motion matters.

Teacher's Note:
1. The constancy of the speed of light is the key.
2. Sound has a preferred frame, the medium.

Question 10: The Doppler effect for light differs from that for sound in that:
(a) the relative frequency shift is smaller for light than for sound
(b) the velocity of light is very large compared to sound
(c) light waves are electromagnetic but sound waves are mechanical
(d) the velocity addition valid for sound is not true for light waves
Show Answer & Explanation

Answer: (d) the velocity addition valid for sound is not true for light waves

Explanation:
1. For sound, velocities simply add relative to the medium.
2. For light the speed is c in every frame, so the classical addition does not apply and the relativistic formula is needed.

Teacher's Note:
1. Relativity replaces simple velocity addition for light.
2. The other options are true statements but not the reason.

Question 11: An observer moves towards a stationary source of sound with a speed 1/5 of the speed of sound. The wavelength and frequency of the source are λ and f. The apparent frequency and wavelength recorded by the observer are respectively:
(a) f and 1.2λ
(b) 0.8f and 0.8λ
(c) 1.2f and λ
(d) 1.2f and 1.2λ
Show Answer & Explanation

Answer: (c) 1.2f and λ

Explanation:
1. f′ = f(v + v/5)/v = 1.2f.
2. The wavelength in the medium is unchanged because the source is at rest: λ.

Teacher's Note:
1. Moving observer: frequency shift only.
2. The wave in the air is unchanged.

Question 12: An astronomical object moves with such a speed that a red shift of 1 nm is observed in a wavelength of 600 nm of the wave received from it. The speed of the object is:
(a) 2 × 10⁵ m/s
(b) 5 × 10⁵ m/s
(c) 4 × 10⁵ m/s
(d) 3 × 10⁵ m/s
Show Answer & Explanation

Answer: (b) 5 × 10⁵ m/s

Explanation:
1. Δλ/λ = v/c, so v = 3 × 10⁸ × (1/600) = 5 × 10⁵ m/s.

Teacher's Note:
1. Use Δλ/λ = v/c.
2. Both lengths in nm.

Question 13: A progressive wave of frequency 500 Hz travels with a velocity of 360 m/s. Two points on the wave that are 60° out of phase are separated by:
(a) 0.12 m
(b) 0.72 m
(c) 0.24 m
(d) 0.06 m
Show Answer & Explanation

Answer: (a) 0.12 m

Explanation:
1. λ = v/f = 360/500 = 0.72 m.
2. Path difference = (λ/2π) × Δφ = (0.72/2π) × (π/3) = 0.12 m.

Teacher's Note:
1. Δx = λΔφ/2π.
2. 60° = π/3 rad.

Question 14: A man generates a symmetrical pulse in a string by moving his hand up and down. At t = 0 the point in his hand moves downward. The pulse travels at 3 m/s and his hand passes 6 times per second through the mean position. The point on the string at a distance 3 m reaches its upper extreme for the first time at t =
(a) 1 s
(b) 0.25 s
(c) 1.5 s
(d) 1.25 s
Show Answer & Explanation

Answer: (d) 1.25 s

Explanation:
1. Passing the mean position 6 times per second means 3 oscillations per second: T = 1/3 s.
2. Time for the pulse to travel 3 m is 3/3 = 1 s.
3. The point then moves downward first and reaches its upper extreme after 3T/4 = 0.25 s. Total = 1.25 s.

Teacher's Note:
1. The point starts moving down, as the source did.
2. From mean position going down to the top takes 3T/4.

Question 15: The equation of a wave is y = 0.05 sin[(π/2)(10x − 40t) − π/4] m. Its wavelength, frequency and speed are respectively:
(a) 0.2 m, 10 Hz, 2 m/s
(b) 0.8 m, 5 Hz, 4 m/s
(c) 0.4 m, 10 Hz, 4 m/s
(d) 0.4 m, 20 Hz, 8 m/s
Show Answer & Explanation

Answer: (c) 0.4 m, 10 Hz, 4 m/s

Explanation:
1. Rewrite as sin(5πx − 20πt − π/4): k = 5π, ω = 20π.
2. λ = 2π/k = 0.4 m, f = ω/2π = 10 Hz.
3. v = fλ = 4 m/s.

Teacher's Note:
1. Expand the bracket first.
2. v = ω/k.

Question 16: For the wave y = 0.05 sin[(π/2)(10x − 40t) − π/4] m, the particle velocity at x = 0.5 m and t = 0.05 s is about:
(a) 0.7 m/s
(b) 2.22 m/s
(c) −2.22 m/s
(d) 7.0 m/s
Show Answer & Explanation

Answer: (b) 2.22 m/s

Explanation:
1. vp = ∂y/∂t = −0.05 × 20π cos(5πx − 20πt − π/4).
2. Phase = 2.5π − π − π/4 = 1.25π, and cos 1.25π = −0.707.
3. vp = −π × (−0.707) = +2.22 m/s.

Teacher's Note:
1. Differentiate with respect to t at fixed x.
2. Evaluate the phase carefully.

Question 17: Six litres of alcohol (speed of sound 1280 m/s, density 0.81 g/cm³) are subjected to a pressure decrease from 200 cm of Hg to 75 cm of Hg (ρHg = 13.6 g/cm³, g = 9.81 m/s²). The change in volume is about:
(a) 0.75 cm³
(b) 7.5 cm³
(c) 0.075 cm³
(d) 75 cm³
Show Answer & Explanation

Answer: (a) 0.75 cm³

Explanation:
1. B = ρv², and B = −VΔP/ΔV, so |ΔV| = V|ΔP|/(ρv²).
2. ΔP = 125 × 13.6 × 981 = 1.667 × 10⁶ dyn/cm²; V = 6000 cm³; ρv² = 0.81 × (1.28 × 10⁵)² dyn/cm².
3. ΔV = 6000 × 1.667 × 10⁶/(0.81 × 1.638 × 10¹⁰) = 0.75 cm³ (an increase, as pressure falls).

Teacher's Note:
1. Bulk modulus links speed and compressibility.
2. Use CGS units throughout.

Question 18: The speed of sound in air at NTP is 332 m/s. If the density of hydrogen at NTP is 1/16 that of air, the speed of sound in hydrogen at NTP is:
(a) 83 m/s
(b) 664 m/s
(c) 5312 m/s
(d) 1328 m/s
Show Answer & Explanation

Answer: (d) 1328 m/s

Explanation:
1. At the same pressure and γ, v ∝ 1/√ρ.
2. v(H) = 332 × √16 = 1328 m/s.

Teacher's Note:
1. Lighter gas, faster sound.
2. √16 = 4.

Question 19: Neon is monatomic (γ = 5/3, molar mass 20.2 g/mol) and water vapour is polyatomic (γ = 4/3, molar mass 18 g/mol). The ratio of the speed of sound in neon to that in water vapour at the same temperature is about:
(a) 1.25
(b) 0.80
(c) 1.055
(d) 0.948
Show Answer & Explanation

Answer: (c) 1.055

Explanation:
1. v = √(γRT/M), so vNe/vW = √[(γNe/γW)(MW/MNe)].
2. = √[(5/4) × (18/20.2)] = √1.114 = 1.055.

Teacher's Note:
1. T is the same, so it cancels.
2. Use molar masses in the same units.

Question 20: If the degree of interference pattern, (Imax − Imin)/(Imax + Imin), is 60%, the ratio of the intensities of the two interfering waves is:
(a) 25 : 4
(b) 9 : 1
(c) 3 : 1
(d) 4 : 1
Show Answer & Explanation

Answer: (b) 9 : 1

Explanation:
1. (Imax − Imin)/(Imax + Imin) = 3/5, so Imax/Imin = 4.
2. (a1 + a2)/(a1 − a2) = 2, so a1/a2 = 3.
3. I1/I2 = 9.

Teacher's Note:
1. Imax/Imin = ((a1 + a2)/(a1 − a2))².
2. Intensity ratio = (amplitude ratio)².

Question 21: A tuning fork of frequency 300 Hz produces 5 beats per second with an unknown fork. When some wax is added to the arm of the known fork, the number of beats decreases. The frequency of the unknown fork is:
(a) 295 Hz
(b) 305 Hz
(c) 300 Hz
(d) 290 Hz
Show Answer & Explanation

Answer: (a) 295 Hz

Explanation:
1. The unknown fork is 295 Hz or 305 Hz.
2. Wax lowers the known fork below 300 Hz. If the unknown were 305 Hz the beats would increase.
3. The beats decrease, so the unknown fork is 295 Hz.

Teacher's Note:
1. Waxing lowers the frequency of the known fork.
2. Compare the new difference with 5.

Question 22: A tuning fork of frequency 158 Hz produces 3 beats per second with an unknown fork. After the arm of the unknown fork is filed, the beats become 7 per second. The frequency of the unknown fork may be:
(a) 155 Hz or 161 Hz
(b) 155 Hz only
(c) 161 Hz only
(d) 151 Hz or 165 Hz
Show Answer & Explanation

Answer: (a) 155 Hz or 161 Hz

Explanation:
1. Before filing the unknown fork is 155 Hz or 161 Hz.
2. Filing raises its frequency. From 155 Hz or 161 Hz it can reach 165 Hz, giving 7 beats with 158 Hz.
3. Both are consistent with the data.

Teacher's Note:
1. Filing raises the frequency.
2. Test each candidate: 165 − 158 = 7.

Question 23: 41 tuning forks are arranged in series so that each fork produces 3 beats per second with its neighbour. If the frequency of the last fork is 3 times that of the first, the frequencies of the 1st, 11th, 16th, 21st and last forks are respectively (in Hz):
(a) 30, 60, 75, 90, 150
(b) 60, 93, 108, 123, 183
(c) 40, 70, 85, 100, 160
(d) 60, 90, 105, 120, 180
Show Answer & Explanation

Answer: (d) 60, 90, 105, 120, 180

Explanation:
1. Frequencies are in AP with common difference 3: n41 = n1 + 40 × 3.
2. n1 + 120 = 3n1, so n1 = 60 Hz.
3. n11 = 60 + 30 = 90, n16 = 60 + 45 = 105, n21 = 60 + 60 = 120, n41 = 180.

Teacher's Note:
1. Forty gaps between forty-one forks.
2. nth term = n1 + (n − 1)d.

Question 24: Two loudspeakers 3 m apart, vibrating in phase, emit sound of equal amplitude. A listener stands 4.0 m in front of one of the speakers (on the line perpendicular to the line joining the speakers). The speed of sound is 330 m/s. The lowest frequency for which the listener hears a minimum is:
(a) 495 Hz
(b) 82.5 Hz
(c) 165 Hz
(d) 330 Hz
Show Answer & Explanation

Answer: (c) 165 Hz

Explanation:
1. Distance to the farther speaker = √(3² + 4²) = 5 m, so the path difference is 1 m.
2. Destructive interference: 1 = (2m + 1)λ/2, so f = 165(2m + 1) Hz.
3. The lowest is 165 Hz (then 495 Hz, 825 Hz, ...).

Teacher's Note:
1. Path difference = odd multiple of λ/2.
2. Use the 3-4-5 triangle.

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