JEE Physics Waves MCQs Set 24

Answer: (a) \( 2\times10^{-6}\sin(0.1\pi)\approx6.2\times10^{-7} \) m, 12500 Hz

Explanation:
1. \( v=\sqrt{Y/\rho}=5000 m/s \). Nodes at 0, \( \pm0.2 \), \( \pm0.4 m \), so \( \lambda/2=0.2 m \), \( \lambda=0.4 m \), \( f=v/\lambda=12500 \) Hz.
2. \( k=5\pi m ^{-1} \): amplitude at \( x=0.02 m \) is \( 2\times10^{-6}\sin(5\pi\times0.02)=2\times10^{-6}\sin(0.1\pi)\approx6.2\times10^{-7} m \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) \( 2\sqrt3 \) cm; x = 0, 15, 30, 45, 60 cm

Explanation:
1. At \( x = 5 \) cm: amplitude \( =4\sin\dfrac\pi3=2\sqrt3 \) cm.
2. Nodes where \( \sin\dfrac{\pi x}{15}=0\Rightarrow x=15n \): 0, 15, 30, 45, 60 cm.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (c) zero

Explanation:
1. \( \dfrac{\partial y}{\partial t}=-4\cdot96\pi\sin\dfrac{\pi x}{15}\sin(96\pi t) \).
2. At \( t = 0.25 s \), \( 96\pi t=24\pi \), \( \sin24\pi=0 \), so the velocity is zero.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (d) \( y_1=2\sin\left(\dfrac{\pi x}{15}-96\pi t\right),\ y_2=2\sin\left(\dfrac{\pi x}{15}+96\pi t\right) \)

Explanation:
1. Use \( \sin(A-B)+\sin(A+B)=2\sin A\cos B \) with \( A=\pi x/15 \), \( B=96\pi t \).
2. Hence \( 4\sin A\cos B \) is the sum of two travelling waves of amplitude 2 moving in opposite directions.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (a) 4 °C

Explanation:
1. At \( 0 ^{\circ}C \): \( v=\dfrac{1000}3=333.3 m/s \). \( v\propto\sqrt T \).
2. \( T=273\left(\dfrac{336}{333.3}\right)^2\approx277.4 K \approx4 ^{\circ}C \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) The wave is propagating along the negative x-axis with speed 25 m/s

Explanation:
1. The \( + \) sign (kx + \( \omega t \)) means propagation along negative \( x \).
2. \( v=\omega/k=50/2=25 m/s \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) 17.3 GHz

Explanation:
1. Relativistic Doppler shift for approach: \( f'=f\sqrt{\dfrac{1+\beta}{1-\beta}} \) with \( \beta=0.5 \).
2. \( f'=10\sqrt3=17.3 \) GHz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) 600 Hz

Explanation:
1. Frequency is determined by the source and does not change when the wave crosses a boundary.
2. Only the speed and wavelength change, so the observer hears 600 Hz.

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (a) \( \sqrt{\dfrac{m_2}{m_1}} \)

Explanation:
1. \( v=\sqrt{\gamma RT/M} \), with the same \( \gamma=5/3 \) and \( T \).
2. \( \dfrac{v_1}{v_2}=\sqrt{\dfrac{m_2}{m_1}} \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) \( \dfrac{\sqrt3}5 \)

Explanation:
1. \( v\propto\sqrt{\gamma/M} \): \( \gamma_{N_2}=\dfrac75,M=28 \); \( \gamma_{He}=\dfrac53,M=4 \).
2. \( \dfrac{v_N}{v_{He}}=\sqrt{\dfrac{7/5}{5/3}\cdot\dfrac4{28}}=\sqrt{\dfrac{21}{25}\cdot\dfrac17}=\dfrac{\sqrt3}5 \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) \( \dfrac{\sqrt{42}}5 \)

Explanation:
1. \( \dfrac{v_H}{v_{He}}=\sqrt{\dfrac{7/5}{5/3}\cdot\dfrac42}=\sqrt{\dfrac{21}{25}\cdot2} \).
2. \( =\dfrac{\sqrt{42}}5\approx1.30 \), not \( \sqrt{21/5} \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (d) 336 m/s

Explanation:
1. \( l+0.3d=\dfrac\lambda4\Rightarrow16+1.5=17.5 \) cm, so \( \lambda=70 \) cm.
2. \( v=480\times0.70=336 m/s \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (a) 0.14 s

Explanation:
1. \( \mu_1=0.0125 \) kg/m \( \Rightarrow v_1=\sqrt{80/0.0125}=80 m/s \); \( \mu_2=0.078125 \) kg/m \( \Rightarrow v_2=32 m/s \).
2. \( t=\dfrac{4.8}{80}+\dfrac{2.56}{32}=0.06+0.08=0.14 s \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) 1.5 cm and 2.0 cm

Explanation:
1. \( A_r=\dfrac{v_2-v_1}{v_2+v_1}A=\dfrac{-48}{112}(3.5)=-1.5 \) cm (magnitude 1.5 cm).
2. \( A_t=\dfrac{2v_2}{v_1+v_2}A=\dfrac{64}{112}(3.5)=2.0 \) cm.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (c) 11 Hz

Explanation:
1. Thermal tension \( T=YA\alpha\Delta T=2\times10^{11}\times10^{-6}\times1.21\times10^{-5}\times20=48.4 N \).
2. \( \mu=0.1 \) kg/m, \( v=\sqrt{484}=22 m/s \), \( f=\dfrac v{2L}=11 \) Hz.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (d) 0.12 m

Explanation:
1. Tension at the bottom \( =2g \) and at the top \( =8g \). \( \lambda\propto v\propto\sqrt T \) (frequency constant).
2. \( \lambda_{top}=0.06\sqrt{8/2}=0.12 m \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (a) 70.1 m/s

Explanation:
1. Strain \( =\alpha\Delta T=1.7\times10^{-5}\times20=3.4\times10^{-4} \).
2. \( v=\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{Y\varepsilon}\rho}=\sqrt{\dfrac{1.3\times10^{11}\times3.4\times10^{-4}}{9\times10^3}}\approx70.1 m/s \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) 2.5 m/s

Explanation:
1. Moving away from \( S_1 \): \( f_1=f\dfrac{v-u}{v} \); towards \( S_2 \): \( f_2=f\dfrac{v+u}v \).
2. Beat frequency \( =\dfrac{2uf}v=10\Rightarrow u=\dfrac{10\times330}{2\times660}=2.5 m/s \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) 4 maxima and 20 oscillations

Explanation:
1. Beat frequency \( =11-9=2 \) Hz, so there are 4 intensity maxima in \( 2 s \).
2. The resultant oscillates at the mean frequency 10 Hz, giving 20 oscillations in \( 2 s \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) 750 Hz

Explanation:
1. Approach: \( 1000=f\dfrac{350}{300}\Rightarrow f=\dfrac{6000}{7}\approx857 \) Hz.
2. Recede: \( f'=857\times\dfrac{350}{400}=750 \) Hz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) 12 and 18

Explanation:
1. Receding observer: \( 480=500\dfrac{300-u_1}{300}\Rightarrow u_1=12 m/s \).
2. Approaching observer: \( 530=500\dfrac{300+u_2}{300}\Rightarrow u_2=18 m/s \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) 2250 Hz

Explanation:
1. Source and observer recede: \( f'=f\dfrac{340-20}{340+20} \).
2. \( f=2000\times\dfrac{360}{320}=2250 \) Hz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) \( \dfrac{19}{18} \)

Explanation:
1. \( f_1=f\dfrac{340}{340-34} \), \( f_2=f\dfrac{340}{340-17} \).
2. \( \dfrac{f_1}{f_2}=\dfrac{323}{306}=\dfrac{19}{18} \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) 666 Hz

Explanation:
1. Second harmonic of an open pipe: \( f=\dfrac vL=\dfrac{330}{0.5}=660 \) Hz.
2. \( u=10\ \text{km/h}=2.78 m/s \): \( f'=660\times\dfrac{332.78}{330}\approx666 \) Hz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) 12%

Explanation:
1. Approach: \( 1000\times\dfrac{320}{300}=1066.7 \) Hz; recede: \( 1000\times\dfrac{320}{340}=941.2 \) Hz.
2. Change \( =\dfrac{125.5}{1000}\approx12.5\%\approx12\% \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) 8.50 kHz

Explanation:
1. Building receives \( f_1=8\times\dfrac{320}{320-10}=8.258 \) kHz.
2. Driver (moving towards the wall at \( 10 m/s \)) hears \( f_2=f_1\dfrac{330}{320}=8.50 \) kHz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) 344 Hz

Explanation:
1. The string frequency equals the third harmonic of the closed pipe: \( \dfrac{3\times340}{4\times0.75}=340 \) Hz.
2. Beats of 4 mean \( n=336 \) or 344 Hz. Raising the tension raises the string frequency, and beats reduce only if \( n>340 \), so \( n=344 \) Hz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) 22 m/s

Explanation:
1. Horn heard by the motorcyclist (moving away from the car, car chasing): \( 176\dfrac{330-v}{330-22} \). Siren heard: \( 165\dfrac{330+v}{330} \).
2. No beats means equality: \( 0.5714(330-v)=0.5(330+v)\Rightarrow v=22 m/s \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) 2

Explanation:
1. \( f'=f\dfrac{v+u}{v} \): \( 5.5=5\dfrac{v+u_A}v\Rightarrow u_A=0.1v \); \( 6.0=5\dfrac{v+u_B}v\Rightarrow u_B=0.2v \).
2. \( u_B/u_A=2 \).

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) 500

Explanation:
1. \( f'=450\times\dfrac{330}{330-33} \).
2. \( f'=450\times\dfrac{330}{297}=500 \) Hz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.