Class 11 Physics Chapter 14: Waves MCQs with answers
10 multiple-choice questions on Waves, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (b) longitudinal, since air cannot sustain a shearing stress
Transverse waves need a medium that can sustain shearing stress, which a fluid such as air cannot. Fluids sustain compressive strain, so only longitudinal mechanical waves travel through air.
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Answer: (c) energy, while each layer of air only oscillates about its mean position
In a wave, energy and not matter is transferred. Sound travels as compressions and rarefactions; each layer of air oscillates about its mean position, unlike a wind, which is a motion of air as a whole.
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Answer: (d) 4.0 m s⁻¹
A crest and the nearest trough are half a wavelength apart, so λ = 0.80 m. v = λν = 0.80 × 5 = 4.0 m s⁻¹. 2.0 m s⁻¹ takes 0.40 m as the wavelength; 0.16 m s⁻¹ divides λ by ν; 8.0 m s⁻¹ takes the wavelength as 1.60 m.
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Answer: (a) is halved
The speed v = √(T/μ) depends only on the string, so it does not change. The source fixes the frequency, and λ = v/ν: doubling ν halves λ.
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Answer: (a) 30 m s⁻¹
v = √(T/μ) with μ unchanged, so v₂ = v₁√(T₂/T₁) = 20 × √(90/40) = 20 × 1.5 = 30 m s⁻¹. 45 m s⁻¹ takes v ∝ T; 13 m s⁻¹ inverts the ratio; 20 m s⁻¹ ignores the tension.
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Answer: (c) 40 g
From v = √(T/μ), μ = T/v² = 50/2500 = 0.020 kg m⁻¹. Mass = μL = 0.020 × 2.0 = 0.040 kg = 40 g. 20 g forgets to multiply by the length; 80 g doubles the length again; 2.0 kg uses μ = T/v.
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Answer: (c) y = 0.030 sin(4πx − 80πt) m
k = 2π/λ = 2π/0.50 = 4π m⁻¹. ν = v/λ = 20/0.50 = 40 Hz, so ω = 2πν = 80π s⁻¹. A wave along +x has the form a sin(kx − ωt). The + sign gives a wave along −x; 40πt drops the 2 in ω = 2πν; 0.50x − 20t puts λ and v where k and ω belong.
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Answer: (a) 15 cm
Successive nodes are λ/2 apart and an antinode lies midway between two nodes, so a node and the adjacent antinode are λ/4 = 60/4 = 15 cm apart. 30 cm is the node-to-node distance.
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Answer: (d) the period and the amplitude
A displacement–time graph for one fixed element shows its oscillation in time, with amplitude a and period T. The wavelength is a property of the shape in space at one instant, so it, and hence the speed, cannot be read from this graph.
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Answer: (b) 36 m s⁻¹
A phase difference of 2π corresponds to one wavelength, so π/3 corresponds to λ/6. λ = 6 × 0.15 = 0.90 m and v = λν = 0.90 × 40 = 36 m s⁻¹. 18 m s⁻¹ takes π/3 as a third of a wavelength; 6.0 m s⁻¹ multiplies the separation by the frequency.