CBSE Class 10 Physics Waves and Sound MCQs Set 37

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Multiple Choice Questions

Question 1: The intensity of a sound wave at a distance of 40 m from the source is I. What is the intensity, approximately, at a distance of 90 m from the source?
(a) I/5
(b) I/2
(c) I/3
(d) I/7
Show Answer & Explanation

Answer: (a) I/5

Explanation:
1. Intensity is inversely proportional to the square of the distance.
2. I' = I × (40/90)2 = 16I/81 ≈ I/5.

Teacher's Note:
1. Intensity falls as 1/r2 for a point source.

Question 2: The velocity of sound in O2 is 340 m s-1 at a certain pressure and temperature. What is the velocity of sound in CO2 under the same conditions? (molar masses: O2 = 32, CO2 = 44)
(a) \( \frac{680}{7} \) m s-1
(b) \( 340\sqrt{2} \) m s-1
(c) \( 340\sqrt{\frac{8}{11}} \) m s-1
(d) \( \frac{340}{\sqrt{2}} \) m s-1
Show Answer & Explanation

Answer: (c) \( 340\sqrt{\frac{8}{11}} \) m s-1

Explanation:
1. At the same temperature and pressure v ∝ 1/√M (M = molar mass).
2. vCO2 = 340 × √(32/44) = \( 340\sqrt{8/11} \) m s-1.

Teacher's Note:
1. Heavier gases carry sound more slowly.

Question 3: A scooterist moves away from a wall at 54 km h-1. He blows the horn when he is 100 m from the wall. After what time does he hear the echo? (speed of sound = 340 m s-1)
(a) 0.6 s
(b) 0.9 s
(c) 2.5 s
(d) 3.45 s
Show Answer & Explanation

Answer: (a) 0.6 s

Explanation:
1. 54 km h-1 = 15 m s-1.
2. The sound travels 100 m to the wall and back to the scooterist, who has moved 15t further: 100 + (100 + 15t) = 340t.
3. t = 200/325 = 0.6 s.

Teacher's Note:
1. Add the distance the scooter moves while the sound is travelling.

Question 4: One end of a string of length 2 m and mass 50 g is fixed and a 2 kg mass hangs from the other end. What is the velocity of the wave on the string? (g = 10 m s-2)
(a) 10 m s-1
(b) 20 m s-1
(c) \( 20\sqrt{2} \) m s-1
(d) \( 30\sqrt{3} \) m s-1
Show Answer & Explanation

Answer: (c) \( 20\sqrt{2} \) m s-1

Explanation:
1. Tension T = mg = 2 × 10 = 20 N and the linear density μ = 0.05/2 = 0.025 kg m-1.
2. v = \( \sqrt{T/\mu} \) = \( \sqrt{800} \) = \( 20\sqrt{2} \) m s-1.

Teacher's Note:
1. v = √(T/μ) for waves on a stretched string.

Question 5: Which of the following is different from the others?
(a) motion of the Earth about its own axis
(b) motion of the Earth around the Sun
(c) motion of the needle of a sewing machine
(d) motion of the Moon around the Earth
Show Answer & Explanation

Answer: (c) motion of the needle of a sewing machine

Explanation:
1. The needle moves to and fro about a mean position, which is an oscillatory (vibratory) motion.
2. The other three are rotational or orbital periodic motions.

Teacher's Note:
1. Oscillatory motion is back-and-forth about a mean position.

Question 6: The first and second resonances in a resonating air column occur at 21 cm and 61 cm. If the frequency of vibration is 400 Hz, what is the velocity of sound?
(a) 300 m s-1
(b) 260 m s-1
(c) 320 m s-1
(d) 340 m s-1
Show Answer & Explanation

Answer: (c) 320 m s-1

Explanation:
1. The difference between successive resonance lengths = λ/2 = 40 cm, so λ = 0.8 m.
2. v = fλ = 400 × 0.8 = 320 m s-1.

Teacher's Note:
1. Successive resonances in an air column differ by half a wavelength.

Question 7: A string is in its first overtone when vibrated with a tuning fork of frequency 50 Hz. If the length of the string is 20 cm, what is the velocity of the wave on the string?
(a) 10 m s-1
(b) 20 m s-1
(c) 15 m s-1
(d) 30 m s-1
Show Answer & Explanation

Answer: (a) 10 m s-1

Explanation:
1. The first overtone is the second harmonic, for which the string length equals one wavelength: λ = 0.2 m.
2. v = fλ = 50 × 0.2 = 10 m s-1.

Teacher's Note:
1. First overtone = second harmonic = L = λ.

Question 8: Arrange the steps to determine the velocity of sound in air by the resonating air column experiment.
(A) Note the first resonating length ℓ1.
(B) Note the second resonating length ℓ2.
(C) Determine the velocity of sound using v = 2f(ℓ2 - ℓ1).
(D) Find the difference in the resonating lengths.
(a) A C B D
(b) A B D C
(c) C B D A
(d) B D C A
Show Answer & Explanation

Answer: (b) A B D C

Explanation:
1. Note ℓ1 (A) and ℓ2 (B), then find ℓ2 - ℓ1 (D) and calculate v = 2f(ℓ2 - ℓ1) (C).
2. The order is A, B, D, C.

Teacher's Note:
1. ℓ2 - ℓ1 = λ/2, so v = fλ = 2f(ℓ2 - ℓ1).

Question 9: Match Column A with Column B.
(P) Velocity of sound in air (Q) Velocity of sound in solids (R) Fundamental frequency of a string
(p) \( \sqrt{Y/d} \) (q) \( \frac{1}{2\ell}\sqrt{T/m} \) (r) \( \sqrt{\gamma P/\rho} \)
(a) P → r; Q → p; R → q
(b) P → r; Q → q; R → p
(c) P → p; Q → q; R → q
(d) P → q; Q → r; R → p
Show Answer & Explanation

Answer: (a) P → r; Q → p; R → q

Explanation:
1. In a gas v = \( \sqrt{\gamma P/\rho} \) (r).
2. In a solid v = \( \sqrt{Y/d} \), with Y the Young's modulus and d the density (p).
3. The fundamental frequency of a string of length ℓ is (1/2ℓ)\( \sqrt{T/m} \) (q).

Teacher's Note:
1. Speed = √(elastic property / inertial property).

Question 10: Assertion (A): The velocity of sound in iron is greater than that in water.
Reason (R): The velocity of sound in solids is greater than that in liquids.
(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) Both A and R are false.
Show Answer & Explanation

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

Explanation:
1. Solids are much more elastic than liquids, so sound travels fastest in solids; A is true and R is true.
2. R explains A.

Teacher's Note:
1. Order of speeds: solids > liquids > gases.

Question 11: Which of the following has no unit?
(a) amplitude
(b) pitch
(c) frequency
(d) time period
Show Answer & Explanation

Answer: (b) pitch

Explanation:
1. Amplitude is a length, frequency is in Hz and time period is in seconds.
2. Pitch is only a sensation by which we judge a sound as shrill or flat, so it has no unit.

Teacher's Note:
1. Pitch depends on frequency but is itself not a unit-carrying quantity.

Question 12: The apparent frequency heard by an observer changes from 250 Hz to 280 Hz when a source moving towards the observer changes its speed from 10 m s-1 to x m s-1. Find x. (speed of sound in air = 350 m s-1)
(a) 15 m s-1
(b) 30 m s-1
(c) 46 m s-1
(d) 46 km s-1
Show Answer & Explanation

Answer: (c) 46 m s-1

Explanation:
1. f' = f v/(v - vs): 250 = f × 350/340, so f = 242.86 Hz.
2. 280 = 242.86 × 350/(350 - x) gives 350 - x = 303.6, so x = 46 m s-1.

Teacher's Note:
1. For an approaching source f' = f v/(v - vs).

Question 13: A string of length 8 m fixed at both ends resonates with a tuning fork of frequency 420 Hz in its 7th overtone. What is the distance between the third and the fifth antinodes?
(a) 7 m
(b) 1 m
(c) 2 m
(d) 5 m
Show Answer & Explanation

Answer: (c) 2 m

Explanation:
1. The 7th overtone is the 8th harmonic, so 8(λ/2) = 8 m and λ = 2 m.
2. Adjacent antinodes are λ/2 apart, so the third and fifth antinodes are λ = 2 m apart.

Teacher's Note:
1. Distance between consecutive antinodes is λ/2.

Question 14: Which of the following is in resonance with a tuning fork of frequency 110 Hz? (speed of sound = 330 m s-1)
(a) a vibrating string of length 2 m and mass 30 g with tension 90 N
(b) an open pipe of length 3 m
(c) a closed organ pipe of length 3 m
(d) both (a) and (b)
Show Answer & Explanation

Answer: (b) an open pipe of length 3 m

Explanation:
1. Open pipe: fn = nv/2L = n × 330/6 = 55n Hz, which gives 110 Hz for n = 2, so (b) resonates.
2. Closed pipe: f = (2n - 1)v/4L = 27.5(2n - 1) Hz, and 110 Hz would need 2n - 1 = 4, which is not odd.
3. String: μ = 0.015 kg m-1, v = \( \sqrt{90/0.015} \) = 77.5 m s-1, f1 = 19.4 Hz, and 110/19.4 is not an integer.

Teacher's Note:
1. Open pipes support all harmonics; closed pipes only odd ones.

Question 15: At a distance of 5 m from a source the intensity of a sound wave is I. At what distance is the intensity 4I?
(a) 1.25 m
(b) 10 m
(c) 20 m
(d) 2.5 m
Show Answer & Explanation

Answer: (d) 2.5 m

Explanation:
1. I ∝ 1/r2, so 4I needs r to be halved.
2. r = 5/2 = 2.5 m.

Teacher's Note:
1. Intensity is inversely proportional to the square of distance.

Question 16: The velocity of sound in a gas is 1360 m s-1, and in O2 under the same conditions it is 340 m s-1. Which gas is it? (molar masses: H2 = 2, N2 = 28, CO2 = 44, Cl2 = 71)
(a) CO2
(b) H2
(c) N2
(d) Cl2
Show Answer & Explanation

Answer: (b) H2

Explanation:
1. v ∝ 1/√M, so vgas/vO2 = 4 = √(32/M).
2. M = 32/16 = 2, which is hydrogen.

Teacher's Note:
1. Lighter gases carry sound faster.

Question 17: A car moves at 108 km h-1 away from a wall. When it is 100 m from the wall the driver blows the horn and hears the echo after 2/3 s. What is the speed of sound in air?
(a) 310 m s-1
(b) 320 m s-1
(c) 330 m s-1
(d) 340 m s-1
Show Answer & Explanation

Answer: (c) 330 m s-1

Explanation:
1. 108 km h-1 = 30 m s-1.
2. In 2/3 s the car moves 20 m, so the sound travels 100 + (100 + 20) = 220 m in 2/3 s.
3. v = 220/(2/3) = 330 m s-1.

Teacher's Note:
1. The echo path includes the extra distance the car has moved.

Question 18: The velocity of a wave through a string is 15 m s-1. The string is held under a tension of 200 N and has a mass of 50 g. What is its length?
(a) 62.5 cm
(b) 5.6 cm
(c) 16.3 cm
(d) 22.5 cm
Show Answer & Explanation

Answer: (b) 5.6 cm

Explanation:
1. μ = T/v2 = 200/225 = 0.889 kg m-1.
2. Length = m/μ = 0.05/0.889 = 0.0563 m = 5.6 cm.

Teacher's Note:
1. Use v = √(T/μ) and μ = m/L.

Question 19: Choose the incorrect statement.
(a) The velocity of sound is least in solids.
(b) The velocity of sound in air increases with temperature.
(c) When sound travels through air, the air molecules vibrate longitudinally.
(d) Under similar conditions of temperature and pressure, the velocity of sound in hydrogen is greater than in oxygen.
Show Answer & Explanation

Answer: (a) The velocity of sound is least in solids.

Explanation:
1. Sound travels fastest in solids and slowest in gases, so (a) is wrong.
2. The other three statements are correct.

Teacher's Note:
1. Speed order: solids > liquids > gases.

Question 20: The 3rd and 5th resonances in a resonating air column (closed at one end) occur at air-column lengths of 82 cm and 122 cm. If the speed of sound is 320 m s-1, what is the frequency (in Hz) of vibration?
(a) 800
(b) 400
(c) 200
(d) 350
Show Answer & Explanation

Answer: (a) 800

Explanation:
1. The resonant lengths are ℓn = (2n - 1)λ/4, so ℓ5 - ℓ3 = (9λ - 5λ)/4 = λ.
2. λ = 122 - 82 = 40 cm = 0.4 m.
3. f = v/λ = 320/0.4 = 800 Hz.

Teacher's Note:
1. Two resonances apart means a difference of a full wavelength.

Question 21: A string vibrates in its first harmonic with a tuning fork of frequency 10 Hz. If the speed of the wave on the string is 10 m s-1, what is the length of the string?
(a) 0.2 m
(b) 0.4 m
(c) 0.5 m
(d) 0.6 m
Show Answer & Explanation

Answer: (c) 0.5 m

Explanation:
1. λ = v/f = 1 m.
2. In the first harmonic L = λ/2 = 0.5 m.

Teacher's Note:
1. Fundamental: L = λ/2.

Question 22: Arrange the steps to determine the velocity of a wave on a string.
(A) Use v = √(T/m) to find the velocity.
(B) Determine the linear density of the string.
(C) Fix a string to a rigid support and hang a known weight from the other end.
(D) Set the string into vibration.
(a) A C B D
(b) A D B C
(c) C B D A
(d) C A D B
Show Answer & Explanation

Answer: (c) C B D A

Explanation:
1. Set up the string with a known weight (C), find its linear density (B), set it vibrating (D) and then calculate v (A).
2. The order is C, B, D, A.

Teacher's Note:
1. Here m denotes the mass per unit length.

Question 23: Match Column A with Column B.
(P) Reverberation (Q) Doppler effect (R) Velocity of sound in air
(p) apparent change in frequency (q) changes with temperature (r) reflection of sound
(a) P → q; Q → p; R → r
(b) P → r; Q → p; R → q
(c) P → q; Q → r; R → q
(d) P → q; Q → r; R → p
Show Answer & Explanation

Answer: (b) P → r; Q → p; R → q

Explanation:
1. Reverberation is the persistence of sound due to repeated reflections (r).
2. The Doppler effect is the apparent change of frequency (p).
3. The speed of sound in air changes with temperature (q).

Teacher's Note:
1. Reverberation time matters in hall design.

Question 24: Assertion (A): The velocity of sound in solid A is less than in solid B if the Young's modulus of A is less than that of B and their densities are equal.
Reason (R): The velocity of sound in solids is directly proportional to the square root of Young's modulus and inversely proportional to the square root of density.
(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) Both A and R are false.
Show Answer & Explanation

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

Explanation:
1. v = √(Y/d), so for equal densities the solid with the smaller Y has the smaller v; A is true.
2. R states the formula that proves A, so R is the correct explanation.

Teacher's Note:
1. v ∝ √Y and v ∝ 1/√d.

Question 25: Which of the following is the unit of frequency?
(a) cm
(b) m
(c) Hz
(d) s
Show Answer & Explanation

Answer: (c) Hz

Explanation:
1. Frequency is the number of oscillations per second, and its unit is the hertz (Hz).

Teacher's Note:
1. 1 Hz = 1 s-1.

Question 26: An observer moves towards a stationary source and hears a frequency of 200 Hz. If he changes his speed from 5 m s-1 to 10 m s-1, then (speed of sound = 330 m s-1)
(a) the change in apparent frequency is 3 Hz.
(b) the apparent frequency at 10 m s-1 is 203 Hz.
(c) the frequency of the source is 197 Hz.
(d) all of the above
Show Answer & Explanation

Answer: (d) all of the above

Explanation:
1. f' = f(v + vo)/v: 200 = f × 335/330, so f = 197 Hz (c is true).
2. At 10 m s-1: f' = 197 × 340/330 = 203 Hz (b is true).
3. The change is 203 - 200 = 3 Hz (a is true), so all are correct.

Teacher's Note:
1. An approaching observer meets more wavefronts per second.

Question 27: A string resonates with a tuning fork of frequency 500 Hz in its 6th harmonic. If the distance between the 1st node and the 3rd antinode is 2.5 m, what is the length of the string? (the string is fixed at both ends, and the first node is at one end)
(a) 6 m
(b) 2.5 m
(c) 3 m
(d) 7 m
Show Answer & Explanation

Answer: (a) 6 m

Explanation:
1. Counting from the fixed end, the 3rd antinode is at 5λ/4 from the 1st node, so 5λ/4 = 2.5 m and λ = 2 m.
2. In the 6th harmonic L = 6 × λ/2 = 6 m.

Teacher's Note:
1. Antinodes are at λ/4, 3λ/4, 5λ/4 ... from a fixed end.

Question 28: An open pipe of length 2 m and a closed pipe of length 2.5 m are both in resonance with a tuning fork of frequency 150 Hz. What are the harmonic numbers in the open pipe and the closed pipe? (speed of sound = 300 m s-1)
(a) 5th, 2nd
(b) 2nd, 5th
(c) 3rd, 9th
(d) 9th, 3rd
Show Answer & Explanation

Answer: (b) 2nd, 5th

Explanation:
1. Open pipe: f = nv/2L = n × 75 Hz, so 150 Hz is the 2nd harmonic.
2. Closed pipe: f = (2n - 1)v/4L = 30(2n - 1) Hz, so 150 Hz needs 2n - 1 = 5, the 5th harmonic.

Teacher's Note:
1. Closed pipes have only odd harmonics.

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