JEE Physics Waves MCQs Set 05

Answer: (d) If the maximum velocity of a particle of the wire is \( 80\pi \) m/s, the displacement amplitude of the particle is 10 cm

Explanation:
1. Two consecutive particles with the same displacement and direction are one wavelength apart, so \( \lambda = 1 \) m, \( v = f\lambda = 400 \) m/s and \( T = \mu v^2 = 0.05\times 160000 = 8000 \) N. So A, B and C are false.
2. Maximum particle velocity \( A\omega = 80\pi \) with \( \omega = 800\pi \) gives \( A = 0.1 \) m = 10 cm, so D is correct.

Teacher's Note:
1. Consecutive same-phase particles: one wavelength.
2. v = sqrt(T/mu).

Answer: (d) \( \dfrac{9}{4} \)

Explanation:
1. Fixed frequency: \( f = \tfrac{n}{2L}\sqrt{T/\mu} \). Increasing T raises the wave speed, so the next mode has fewer loops: n = 2.
2. \( n\sqrt{T} \) is constant, so \( T_f/T_i = (3/2)^2 = 9/4 \).

Teacher's Note:
1. Fewer loops at higher tension.
2. n squared times T is constant.

Answer: (d) \( \dfrac{5\pi}{3k} \)

Explanation:
1. Amplitude is \( A

Teacher's Note:
1. \sin kx| = A/2 \), so \( kx = \pi/6, 5\pi/6, 7\pi/6, 11\pi/6, \ldots \).
2. Differences of these values include \( 11\pi/6 - \pi/6 = 5\pi/3 \), so the distance \( 5\pi/(3k) \) is possible. The other options do not appear.|List all positions.
3. Take differences.

Answer: (a) \( \sqrt{45} \) and \( \tan^{-1}(1/2) \)

Explanation:
1. Write all in terms of cos and sin: \( y_2 = -4\sin\omega t \), \( y_3 = -2\cos\omega t \), \( y_4 = \sin\omega t \).
2. Sum \( = 6\cos\omega t - 3\sin\omega t = A\cos(\omega t + \phi) \) with \( A\cos\phi = 6 \), \( A\sin\phi = 3 \). So \( A = \sqrt{45} \) and \( \tan\phi = 1/2 \).

Teacher's Note:
1. Add components.
2. Compare with A cos(wt + phi).

Answer: (d) \( \pi/2 \)

Explanation:
1. \( \lambda = v/f = 10/100 = 0.1 \) m.
2. \( \Delta\phi = 2\pi\Delta x/\lambda = 2\pi(0.025)/0.1 = \pi/2 \).

Teacher's Note:
1. Phase difference from path difference.
2. 2 pi times x over lambda.

Answer: (c) all particles between consecutive nodes vibrate in the same phase

Explanation:
1. In a stationary wave all particles between two consecutive nodes vibrate in phase, and particles on opposite sides of a node are in opposite phase.
2. Antinodes of adjacent loops are therefore not in phase.

Teacher's Note:
1. Loop moves as a whole.
2. Adjacent loops are opposite.

Answer: (a) 96 m/s

Explanation:
1. The frequencies differ by 60 Hz, which equals \( v/2L \) (both for the string fixed at both ends).
2. \( v = 60\times 2\times 0.8 = 96 \) m/s.

Teacher's Note:
1. Successive harmonics differ by the fundamental.
2. Check: 90 = 3 x 30 is not the fundamental, so fundamental is 30 Hz.

Answer: (d) 4 m

Explanation:
1. Three loops mean \( L = 3\lambda/2 \).
2. \( \lambda = 2L/3 = 12/3 = 4 \) m.

Teacher's Note:
1. Each loop is half a wavelength.
2. Amplitude is not needed.

Answer: (d) 0.7 m

Explanation:
1. \( v = \sqrt{T/\mu} = \sqrt{490/0.1} = 70 \) m/s.
2. Successive resonant frequencies differ by \( v/2L = 50 \) Hz, so \( L = 70/100 = 0.7 \) m.

Teacher's Note:
1. Speed from tension.
2. Frequency gap is the fundamental.

Answer: (d) 3/4

Explanation:
1. \( f \propto \sqrt{T} \). Halving f requires \( T' = T/4 \).
2. Fractional change is \( (T - T/4)/T = 3/4 \) (a decrease).

Teacher's Note:
1. Frequency square gives tension.
2. Magnitude of the fractional change.

Answer: (a) \( 4I_0 \)

Explanation:
1. Path difference at M is \( QM - PM = 2 - 1 = 1 \) m, which is one wavelength, so the waves arrive in phase.
2. Amplitudes add: \( 2A \), so intensity is \( 4I_0 \).

Teacher's Note:
1. Long diagonal of a regular hexagon is twice the side.
2. Constructive interference.

Answer: (a) 200 Hz

Explanation:
1. Open pipe fundamental: \( f_0 = v/2L \). Closed pipe third harmonic: \( 3v/4L = 1.5f_0 \).
2. \( 1.5f_0 - f_0 = 100 \), so \( f_0 = 200 \) Hz.

Teacher's Note:
1. Closed pipe has odd harmonics only.
2. 0.5 f0 = 100.

Answer: (a) 75 Hz

Explanation:
1. Open pipe: \( f_n = n\,v/2L = n\times 330/2.2 = 150n \) Hz.
2. Resonances are 150, 300, 450, 600, 750 Hz; 75 Hz is not a multiple of 150, so no resonance.

Teacher's Note:
1. Fundamental is 150 Hz.
2. Check multiples.

Answer: (c) 10

Explanation:
1. Path difference for maxima is \( n\lambda \) with \(

Teacher's Note:
1. n| \le 2.7 \), so n = 0, ±1, ±2.
2. Each of these five curves cuts the circle at 2 points, giving 10 points.|The path difference cannot exceed the source separation.
3. Count the intersections.

Answer: (a) 30 dB

Explanation:
1. Intensity is inversely proportional to the square of the distance. At 10 km, intensity is 1/100 of that at 1 km.
2. The level falls by \( 10\log 100 = 20 \) dB, so Y hears 30 dB.

Teacher's Note:
1. Inverse square law.
2. Decibel difference.

Answer: (c) 927 °C

Explanation:
1. \( v \propto \sqrt{T} \). For 2V, T must be four times: \( 4\times 300 = 1200 \) K.
2. 1200 K is 927 degrees Celsius.

Teacher's Note:
1. Use kelvin.
2. Speed doubles when T quadruples.

Answer: (b) 404 m/s

Explanation:
1. \( v/\lambda_1 - v/\lambda_2 = 4 \), so \( v(1 - 1/1.01) = 4 \).
2. \( v = 4\times 1.01/0.01 = 404 \) m/s.

Teacher's Note:
1. Frequencies are v over wavelength.
2. Beat frequency is the difference.

Answer: (b) \( 8n_0\,\Delta\ell/\ell \)

Explanation:
1. Segment lengths are \( \ell/2 \pm \Delta\ell \), so frequencies are \( v/(\ell \pm 2\Delta\ell) \) with \( n_0 = v/2\ell \).
2. Their difference is \( \approx 2n_0\times 4\Delta\ell/\ell = 8n_0\Delta\ell/\ell \).

Teacher's Note:
1. Fundamental is v over 2L.
2. Binomial expansion for small shift.

Answer: (a) 6

Explanation:
1. Closed pipe has odd harmonics: 1400, 4200, 7000, ... that is \( 1400(2k+1) \).
2. Audible up to 20 000 Hz: \( 2k + 1 \le 14.3 \), so odd harmonics up to the 13th: 7 harmonics. Overtones exclude the fundamental, giving 6.

Teacher's Note:
1. Fundamental is not an overtone.
2. Odd multiples only.

Answer: (d) 1/12 s

Explanation:
1. Beat frequency is 6 Hz, so beat period is 1/6 s.
2. Interval from a maximum to the next minimum is half of that, 1/12 s.

Teacher's Note:
1. Half the beat period.
2. Maximum and minimum alternate.

Answer: (d) 25 cm

Explanation:
1. Air column resonates with \( L = 3\lambda/4 \) where \( \lambda = 340/680 = 0.5 \) m, which gives 3/8 m.
2. The next resonance is \( 5\lambda/4 = 0.625 \) m, so the pipe is raised by \( 0.625 - 0.375 = 0.25 \) m = 25 cm.

Teacher's Note:
1. The water surface acts as the closed end.
2. Raising the pipe lengthens the air column.

Answer: (b) \( n_1 > n_2 \)

Explanation:
1. Approaching: \( n_1 - n = nv/(c - v) \). Receding: \( n - n_2 = nv/(c + v) \).
2. The first change is larger, so \( n_1 > n_2 \).

Teacher's Note:
1. Smaller denominator, larger change.
2. True for any v.

Answer: (d) 10 Hz

Explanation:
1. Frequency heard directly: \( 350\times 350/355 \). Frequency of the reflected sound: \( 350\times 350/345 \).
2. Beats = \( 122500(1/345 - 1/355) \approx 10 \) Hz.

Teacher's Note:
1. Source moves away from the person but towards the wall.
2. Subtract the two.

Answer: (c) zero

Explanation:
1. All move with the same velocity, so there is no relative motion between the sources and the observer.
2. There is no Doppler shift, both sources are heard at 100 Hz and the beat frequency is zero.

Teacher's Note:
1. Relative velocity is zero.
2. Same frequency, no beats.