JEE Physics Waves MCQs Set 04

Answer: (c) a trough of one wave is superimposed on a crest of the other

Explanation:
1. Destructive interference occurs when the waves are in opposite phase (phase difference \( \pi \)), so crest meets trough and the displacements cancel.

Teacher's Note:
1. Key fact to remember: Destructive interference occurs when the waves are in opposite phase (phase difference \( \pi \)), so crest meets trough and the displacements cancel.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 7/4 Hz

Explanation:
1. \( v = 7 \) m/s, \( f_1 = v/2L = 7/4 \) Hz.

Teacher's Note:
1. Key fact to remember: \( v = 7 \) m/s, \( f_1 = v/2L = 7/4 \) Hz.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) increases

Explanation:
1. Speed of sound increases with temperature (\( v \propto \sqrt{T} \)), while the pipe length is nearly constant.
2. Frequency \( f = v/\lambda \) therefore increases.

Teacher's Note:
1. Key fact to remember: Speed of sound increases with temperature (\( v \propto \sqrt{T} \)), while the pipe length is nearly constant.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) 1/2 m

Explanation:
1. \( k = 2\pi \), so \( \lambda = 1 \) m.
2. The minimum length with both ends fixed is \( \lambda/2 = 1/2 \) m.

Teacher's Note:
1. Key fact to remember: \( k = 2\pi \), so \( \lambda = 1 \) m.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 2 cm

Explanation:
1. \( k = \pi/2 \) cm\(^{-1}\), so \( \lambda = 4 \) cm.
2. Consecutive nodes are \( \lambda/2 = 2 \) cm apart.

Teacher's Note:
1. Key fact to remember: \( k = \pi/2 \) cm\(^{-1}\), so \( \lambda = 4 \) cm.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 47 cm

Explanation:
1. \( \lambda/4 = 15 + 1 = 16 \) cm, so \( \lambda = 64 \) cm.
2. Next resonance: \( 3\lambda/4 - 1 = 48 - 1 = 47 \) cm.

Teacher's Note:
1. Key fact to remember: \( \lambda/4 = 15 + 1 = 16 \) cm, so \( \lambda = 64 \) cm.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) 1024 Hz

Explanation:
1. Closed: \( f = v/4L = 512 \). Open both ends: \( f' = v/2L = 2\times512 = 1024 \) Hz.

Teacher's Note:
1. Key fact to remember: Closed: \( f = v/4L = 512 \). Open both ends: \( f' = v/2L = 2\times512 = 1024 \) Hz.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 31.25 cm

Explanation:
1. \( L = v/4f = 330/(4\times264) = 0.3125 \) m = 31.25 cm.

Teacher's Note:
1. Key fact to remember: \( L = v/4f = 330/(4\times264) = 0.3125 \) m = 31.25 cm.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) a, b and d only

Explanation:
1. Closed pipe: \( f = (2n-1)v/4L = (2n-1)\times80 \) Hz: 80, 240, 400, 560 Hz...
2. So 80, 240 and 400 Hz are possible and 500 Hz is not.

Teacher's Note:
1. Key fact to remember: Closed pipe: \( f = (2n-1)v/4L = (2n-1)\times80 \) Hz: 80, 240, 400, 560 Hz...
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 550 Hz

Explanation:
1. \( f = v/2L = 330/0.6 = 550 \) Hz.

Teacher's Note:
1. Key fact to remember: \( f = v/2L = 330/0.6 = 550 \) Hz.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 20 m/s

Explanation:
1. The fixed end is a node, and the next node is \( \lambda/2 \) away: \( \lambda/2 = 0.1 \) m, so \( \lambda = 0.2 \) m.
2. \( v = f\lambda = 100\times0.2 = 20 \) m/s.

Teacher's Note:
1. Key fact to remember: The fixed end is a node, and the next node is \( \lambda/2 \) away: \( \lambda/2 = 0.1 \) m, so \( \lambda = 0.2 \) m.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 2

Explanation:
1. \( f \propto \sqrt{T} \), so \( f'/f = \sqrt4 = 2 \).

Teacher's Note:
1. Key fact to remember: \( f \propto \sqrt{T} \), so \( f'/f = \sqrt4 = 2 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 6 kg

Explanation:
1. Tension \( T = Mg \) must stay the same for the same frequency and length.
2. \( M\,g/6 = 1\times g \), so \( M = 6 \) kg.

Teacher's Note:
1. Key fact to remember: Tension \( T = Mg \) must stay the same for the same frequency and length.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) 5 Hz

Explanation:
1. 5 loops means \( 5\lambda/2 = 10 \) m, so \( \lambda = 4 \) m.
2. \( f = v/\lambda = 20/4 = 5 \) Hz.

Teacher's Note:
1. Key fact to remember: 5 loops means \( 5\lambda/2 = 10 \) m, so \( \lambda = 4 \) m.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 1.21 Å

Explanation:
1. Three nodes and two antinodes mean two loops, so the length equals \( 2\times\lambda/2 = \lambda \).
2. \( \lambda = 1.21 \) Å.

Teacher's Note:
1. Key fact to remember: Three nodes and two antinodes mean two loops, so the length equals \( 2\times\lambda/2 = \lambda \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 100 Hz

Explanation:
1. \( f \propto \sqrt T \): \( f_2/f_1 = \sqrt{160/129.6} = 10/9 \).
2. \( f_2 - f_1 = 10 \) gives \( f_1 = 90 \) Hz and \( f_2 = 100 \) Hz, the fork frequency.

Teacher's Note:
1. Key fact to remember: \( f \propto \sqrt T \): \( f_2/f_1 = \sqrt{160/129.6} = 10/9 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) four times

Explanation:
1. \( f \propto \sqrt T \), so \( T \propto f^2 \) and doubling f needs 4T.

Teacher's Note:
1. Key fact to remember: \( f \propto \sqrt T \), so \( T \propto f^2 \) and doubling f needs 4T.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 150 cm

Explanation:
1. Resonances occur at \( \lambda/4, 3\lambda/4, 5\lambda/4 \ldots \) with \( \lambda/4 = 50 \) cm.
2. Next length \( = 3\times50 = 150 \) cm.

Teacher's Note:
1. Key fact to remember: Resonances occur at \( \lambda/4, 3\lambda/4, 5\lambda/4 \ldots \) with \( \lambda/4 = 50 \) cm.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 4.2 m

Explanation:
1. \( L = v/4f = 336/(4\times20) = 4.2 \) m.

Teacher's Note:
1. Key fact to remember: \( L = v/4f = 336/(4\times20) = 4.2 \) m.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) decrease

Explanation:
1. The buoyant force reduces the effective tension \( T = Mg - F_b \).
2. \( f \propto \sqrt T \), so the frequency decreases.

Teacher's Note:
1. Key fact to remember: The buoyant force reduces the effective tension \( T = Mg - F_b \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) pluck at \( \ell/4 \) and touch at \( \ell/2 \)

Explanation:
1. The second harmonic has a node at the centre and antinodes at \( \ell/4 \) and \( 3\ell/4 \).
2. So touch the centre (a node) and pluck at an antinode, \( \ell/4 \).

Teacher's Note:
1. Key fact to remember: The second harmonic has a node at the centre and antinodes at \( \ell/4 \) and \( 3\ell/4 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) \( y = a\sin(kx + \omega t) \)

Explanation:
1. \( \sin(kx-\omega t) + \sin(kx+\omega t) = 2\sin kx\cos\omega t \), which vanishes at x = 0 for all t, so x = 0 is a node.
2. The other options give a non-zero value at x = 0 at some instants.

Teacher's Note:
1. Key fact to remember: \( \sin(kx-\omega t) + \sin(kx+\omega t) = 2\sin kx\cos\omega t \), which vanishes at x = 0 for all t, so x = 0 is a node.
2. Revise this point with the matching definition or example in the chapter notes.