JEE Physics Waves MCQs Set 28

Physics Objective Questions and Answers: Waves

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Multiple Choice Questions

Question 1: A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if the density and elasticity of steel are 7.7 × 10³ kg/m³ and 2.2 × 10¹¹ N/m² respectively?
(a) 188.5 Hz
(b) 178.2 Hz
(c) 200.5 Hz
(d) 770 Hz
Show Answer & Explanation

Answer: (b) 178.2 Hz

Explanation:
1. Stress = Y × strain = 2.2 × 10⁹ N/m².
2. v = √(stress/ρ) = 534.5 m/s, and f = v/2L = 534.5/3 ≈ 178.2 Hz.

Teacher's Note:
1. v = √(T/μ) = √(stress/ρ).

Question 2:
Assertion (A): It is not possible to have interference between the waves produced by two independent sources of the same frequency.
Reason (R): For interference of two waves, the phase difference between the waves must remain constant.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Independent sources do not keep a constant phase difference.
2. Without a constant phase relation there is no stable interference pattern.

Teacher's Note:
1. Coherence is required.

Question 3: A balloon is filled with hydrogen. For sound waves, this balloon behaves like
(a) a converging lens
(b) a diverging lens
(c) a concave mirror
(d) nothing can be said
Show Answer & Explanation

Answer: (b) a diverging lens

Explanation:
1. Sound travels faster in hydrogen than in air.
2. A medium in which the wave is faster bends the waves away from the axis, like a diverging lens.

Teacher's Note:
1. Faster medium diverges.

Question 4: A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency 513 Hz. The unknown frequency is
(a) 246 Hz
(b) 240 Hz
(c) 260 Hz
(d) 254 Hz
Show Answer & Explanation

Answer: (d) 254 Hz

Explanation:
1. The unknown frequency is 246 or 254 Hz, so its second harmonic is 492 or 508 Hz.
2. Only 508 Hz gives 5 beats with 513 Hz, so the frequency is 254 Hz.

Teacher's Note:
1. Test both possibilities.

Question 5: There is no transmission of energy in
(a) electromagnetic waves
(b) transverse progressive waves
(c) longitudinal progressive waves
(d) stationary waves
Show Answer & Explanation

Answer: (d) stationary waves

Explanation:
1. In a stationary wave the energy is confined between the nodes.
2. There is no net flow of energy along the medium.

Teacher's Note:
1. Standing waves do not transport energy.

Question 6: The fifth harmonic of a stretched string fixed at both ends is set up. How many nodes (including the two fixed ends) are present?
(a) 4
(b) 6
(c) 5
(d) 10
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. The nth harmonic has n loops, so it has n + 1 nodes.
2. For n = 5 there are 6 nodes.

Teacher's Note:
1. Nodes = n + 1.

Question 7: The length of a sonometer wire between two fixed ends is 110 cm. If the fundamental frequencies of three segments are in the ratio 1 : 2 : 3, what is the ratio of the lengths of these segments of the wire?
(a) 3 : 2 : 1
(b) 6 : 3 : 2
(c) 6 : 2 : 3
(d) 2 : 3 : 6
Show Answer & Explanation

Answer: (b) 6 : 3 : 2

Explanation:
1. f ∝ 1/L, so L ∝ 1/f.
2. The lengths are in the ratio 1 : 1/2 : 1/3 = 6 : 3 : 2.

Teacher's Note:
1. L ∝ 1/f.

Question 8: Which of the following phenomena cannot be observed for sound waves?
(a) refraction
(b) interference
(c) diffraction
(d) polarisation
Show Answer & Explanation

Answer: (d) polarisation

Explanation:
1. Polarisation is possible only for transverse waves.
2. Sound in air is longitudinal, so it cannot be polarised.

Teacher's Note:
1. Only transverse waves polarise.

Question 9: Doppler phenomenon is related with
(a) pitch (frequency)
(b) loudness
(c) quality
(d) reflection
Show Answer & Explanation

Answer: (a) pitch (frequency)

Explanation:
1. The Doppler effect changes the apparent frequency, that is the pitch.

Teacher's Note:
1. Apparent pitch changes.

Question 10:
Assertion (A): The principle of superposition can be used for any physical quantity.
Reason (R): The principle of superposition can be used only when the amplitude of the quantity is small.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Superposition holds only for linear systems, not for every quantity, so the assertion is false.
2. The reason is true: it holds when the amplitude is small enough for the response to be linear.

Teacher's Note:
1. Superposition needs linearity.

Question 11: The speed of a progressive wave is given by
(a) \( v=n\lambda \)
(b) \( v=n/\lambda \)
(c) \( v=\lambda/n \)
(d) \( v=T/\lambda \)
Show Answer & Explanation

Answer: (a) \( v=n\lambda \)

Explanation:
1. Wave speed = frequency × wavelength.

Teacher's Note:
1. v = fλ.

Question 12: The rate of transfer of energy in a wave depends
(a) directly on the square of the wave amplitude and square of the wave frequency
(b) directly on the square of the wave amplitude and root of the wave frequency
(c) directly on the wave amplitude and square of the wave frequency
(d) on none of these
Show Answer & Explanation

Answer: (a) directly on the square of the wave amplitude and square of the wave frequency

Explanation:
1. The energy density is proportional to ω²A².
2. The power transmitted is therefore proportional to A² and f².

Teacher's Note:
1. P ∝ A²ω².

Question 13: A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has frequency 2f. If v is the velocity of sound, the velocity of the car is
(a) \( v/2 \)
(b) \( v/\sqrt2 \)
(c) \( v/3 \)
(d) \( v/4 \)
Show Answer & Explanation

Answer: (c) \( v/3 \)

Explanation:
1. The echo frequency is f(v + u)/(v − u) = 2f.
2. Solving, v + u = 2v − 2u, so u = v/3.

Teacher's Note:
1. Two Doppler shifts.

Question 14: When the temperature increases, the frequency of a tuning fork
(a) increases
(b) decreases
(c) remains the same
(d) increases or decreases depending on the material
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. The prongs expand and the elastic modulus decreases with temperature.
2. The frequency of the fork therefore decreases.

Teacher's Note:
1. Heating lowers the pitch.

Question 15: A source of sound S emits waves of frequency 100 Hz and an observer O is located at some distance from it. The source moves at 19.4 m/s in a direction making an angle of 60° with the line from S to O. The observer is at rest. The apparent frequency heard is (speed of sound = 330 m/s)
(a) 103 Hz
(b) 106 Hz
(c) 97 Hz
(d) 100 Hz
Show Answer & Explanation

Answer: (a) 103 Hz

Explanation:
1. Only the component of the source velocity along SO matters: 19.4 cos 60° = 9.7 m/s towards O.
2. f′ = 100 × 330/(330 − 9.7) ≈ 103 Hz.

Teacher's Note:
1. Use the radial component.

Question 16: The relation between the velocity of sound in a gas (v) and the r.m.s. velocity of the molecules of the gas (v_rms) is
(a) \( v=v_{rms}\,(\gamma/3)^{1/2} \)
(b) \( v_{rms}=v\,(2/3)^{1/2} \)
(c) \( v=v_{rms} \)
(d) \( v_{rms}=v\,(3/\gamma)^{1/2} \)
Show Answer & Explanation

Answer: (a) \( v=v_{rms}\,(\gamma/3)^{1/2} \)

Explanation:
1. v = √(γP/ρ) and v_rms = √(3P/ρ).
2. So v = v_rms √(γ/3).

Teacher's Note:
1. Divide the two formulas.

Question 17: Ultrasonic, infrasonic and audible waves travel through a medium with speeds v_u, v_i and v_a respectively. Then
(a) v_u, v_i and v_a are nearly equal
(b) \( v_u\ge v_a\ge v_i \)
(c) \( v_u\le v_a\le v_i \)
(d) \( v_a\le v_u\text{ and }v_u\approx v_i \)
Show Answer & Explanation

Answer: (a) v_u, v_i and v_a are nearly equal

Explanation:
1. The speed of sound in a medium does not depend on its frequency.
2. So all three speeds are nearly equal.

Teacher's Note:
1. Speed is independent of frequency.

Question 18: A wave travelling on a string is described by y(x, t) = 0.005 sin(80.0x − 3.0t). The period of the wave is
(a) 3.00 s
(b) 2.09 s
(c) 0.48 s
(d) 0.05 s
Show Answer & Explanation

Answer: (b) 2.09 s

Explanation:
1. ω = 3.0 rad/s.
2. T = 2π/ω = 2.09 s.

Teacher's Note:
1. T = 2π/ω.

Question 19: A source of sound is moving with a uniform speed along a circle. The frequency of the sound as heard by a listener stationed at the centre of the path
(a) increases
(b) decreases
(c) remains the same
(d) may increase and decrease alternately
Show Answer & Explanation

Answer: (c) remains the same

Explanation:
1. The source velocity is always perpendicular to the line joining it to the listener.
2. There is no radial component, so the frequency heard stays the same.

Teacher's Note:
1. No radial velocity.

Question 20:
Assertion (A): Stringed instruments are provided with hollow boxes.
Reason (R): The hollow box increases the surface area of vibration, which in turn increases the loudness of the sound.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. The box forces a larger volume of air into vibration (resonance).
2. A larger vibrating surface radiates more sound, so the sound is louder.

Teacher's Note:
1. Sounding box amplifies.

Question 21: When a sound wave goes from one medium to another, the quantity that remains unchanged is
(a) frequency
(b) amplitude
(c) wavelength
(d) speed
Show Answer & Explanation

Answer: (a) frequency

Explanation:
1. The frequency is fixed by the source.
2. Speed and wavelength change with the medium.

Teacher's Note:
1. f depends on the source.

Question 22: The Doppler effect in sound takes place when the source and the observer are
(a) stationary
(b) moving with the same velocity
(c) in relative motion
(d) none of these
Show Answer & Explanation

Answer: (c) in relative motion

Explanation:
1. A shift in frequency occurs only if there is relative motion between them.

Teacher's Note:
1. Relative motion is required.

Question 23: Choose the incorrect statement(s). I. Mechanical waves transfer energy and matter both from one point to another. II. Mechanical waves transfer only energy from one point to another. III. Mechanical waves transfer only matter from one point to another.
(a) I only
(b) II only
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (c) I and III

Explanation:
1. Mechanical waves transfer energy but not matter.
2. So I and III are incorrect and II is correct.

Teacher's Note:
1. Energy only.

Question 24: The equation of a wave on a string of linear mass density 0.04 kg/m is y = 0.02 sin[2π(t/0.04 − x/0.50)] (SI units). The tension in the string is
(a) 4.0 N
(b) 12.5 N
(c) 0.5 N
(d) 6.25 N
Show Answer & Explanation

Answer: (d) 6.25 N

Explanation:
1. T = 0.04 s and λ = 0.50 m, so v = λ/T = 12.5 m/s.
2. Tension = μv² = 0.04 × 156.25 = 6.25 N.

Teacher's Note:
1. T = μv².

Question 25:
Assertion (A): The velocity of the particles while crossing the mean position in a stationary wave varies from a maximum at antinodes to zero at nodes.
Reason (R): The amplitude of vibration at antinodes is maximum and at nodes the amplitude is zero, and all particles between two successive nodes cross the mean position together.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. The maximum particle speed is ω times the local amplitude.
2. The amplitude is largest at antinodes and zero at nodes, which explains the variation.

Teacher's Note:
1. v_max = ω × amplitude.

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