JEE Physics Waves MCQs Set 03

Answer: (b) frequency

Explanation:
1. Frequency depends only on the source; velocity and wavelength change with the medium.

Teacher's Note:
1. Key fact to remember: Frequency depends only on the source; velocity and wavelength change with the medium.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) \( v = \sqrt{p/\rho} \)

Explanation:
1. Newton assumed isothermal elasticity \( E = p \), so \( v = \sqrt{E/\rho} = \sqrt{p/\rho} \).
2. Laplace later corrected it with the factor \( \gamma \).

Teacher's Note:
1. Key fact to remember: Newton assumed isothermal elasticity \( E = p \), so \( v = \sqrt{E/\rho} = \sqrt{p/\rho} \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 1000

Explanation:
1. \( \beta = 10\log_{10}(I/I_0) = 30 \), so \( \log_{10}(I/I_0) = 3 \).
2. \( I/I_0 = 10^3 = 1000 \).

Teacher's Note:
1. Key fact to remember: \( \beta = 10\log_{10}(I/I_0) = 30 \), so \( \log_{10}(I/I_0) = 3 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) \( v_m > v_d \)

Explanation:
1. Water vapour is lighter than the nitrogen and oxygen it displaces, so moist air has lower density.
2. \( v \propto 1/\sqrt{\rho} \), so sound is faster in moist air.

Teacher's Note:
1. Key fact to remember: Water vapour is lighter than the nitrogen and oxygen it displaces, so moist air has lower density.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 8a

Explanation:
1. Energy per unit time \( \propto A^2f^2 \) (equal here).
2. \( A_Bf_B = A_Af_A \), so \( A_B = a\times8 = 8a \).

Teacher's Note:
1. Key fact to remember: Energy per unit time \( \propto A^2f^2 \) (equal here).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 1/2

Explanation:
1. For the same pressure and the same \( \gamma \), \( v \propto 1/\sqrt{\rho} \).
2. \( v_1/v_2 = \sqrt{\rho_2/\rho_1} = 1/2 \).

Teacher's Note:
1. Key fact to remember: For the same pressure and the same \( \gamma \), \( v \propto 1/\sqrt{\rho} \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) \( \pi/6 \)

Explanation:
1. \( \omega = 2\pi/12 = \pi/6 \) rad/s.
2. \( \Delta\phi = \omega\,\Delta t = \pi/6 \).

Teacher's Note:
1. Key fact to remember: \( \omega = 2\pi/12 = \pi/6 \) rad/s.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) \( V/\sqrt{2} \)

Explanation:
1. \( v = \sqrt{E/\rho} \propto 1/\sqrt{\rho} \).
2. Doubling \( \rho \) divides v by \( \sqrt2 \).

Teacher's Note:
1. Key fact to remember: \( v = \sqrt{E/\rho} \propto 1/\sqrt{\rho} \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 170 m

Explanation:
1. The sound travels to the mountain and back in 1 s.
2. Distance \( = vt/2 = 340\times1/2 = 170 \) m.

Teacher's Note:
1. Key fact to remember: The sound travels to the mountain and back in 1 s.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) \( \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos\phi} \)

Explanation:
1. Adding the two phasors gives resultant amplitude by the cosine rule: \( A^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi \).

Teacher's Note:
1. Key fact to remember: Adding the two phasors gives resultant amplitude by the cosine rule: \( A^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 2 : 1

Explanation:
1. \( I = I_0 + I_0 + 2I_0\cos 90° = 2I_0 \).
2. So \( I/I_0 = 2 \).

Teacher's Note:
1. Key fact to remember: \( I = I_0 + I_0 + 2I_0\cos 90° = 2I_0 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 49 : 1

Explanation:
1. \( I_{max}/I_{min} = \left(\dfrac{a_1+a_2}{a_1-a_2}\right)^2 = \left(\dfrac{7}{1}\right)^2 = 49 \).

Teacher's Note:
1. Key fact to remember: \( I_{max}/I_{min} = \left(\dfrac{a_1+a_2}{a_1-a_2}\right)^2 = \left(\dfrac{7}{1}\right)^2 = 49 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) 4 : 1

Explanation:
1. Amplitude ratio = 3 : 1.
2. \( I_{max}/I_{min} = ((3+1)/(3-1))^2 = 4 \).

Teacher's Note:
1. Key fact to remember: Amplitude ratio = 3 : 1.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) \( n + x \)

Explanation:
1. Loading B with wax lowers its frequency.
2. The beats decrease, so B's frequency was higher than that of A: \( f_B = n + x \).

Teacher's Note:
1. Key fact to remember: Loading B with wax lowers its frequency.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 258 Hz

Explanation:
1. Possible values are 242 or 258 Hz (from 250 ± 8).
2. With 270 Hz: 270 - 258 = 12 beats matches, whereas 270 - 242 = 28 does not. So X is 258 Hz.

Teacher's Note:
1. Key fact to remember: Possible values are 242 or 258 Hz (from 250 ± 8).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 404 m/s

Explanation:
1. \( f_1 - f_2 = v\left(\dfrac{1}{2} - \dfrac{1}{2.02}\right) = 2 \).
2. \( v = 2/0.0049505 = 404 \) m/s.

Teacher's Note:
1. Key fact to remember: \( f_1 - f_2 = v\left(\dfrac{1}{2} - \dfrac{1}{2.02}\right) = 2 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (b) 292 Hz

Explanation:
1. Before loading, \( f_A = 288 \pm 4 \): 292 or 284 Hz.
2. Loading lowers \( f_A \), and the beats decrease, so \( f_A \) must have been above 288 Hz: 292 Hz.

Teacher's Note:
1. Key fact to remember: Before loading, \( f_A = 288 \pm 4 \): 292 or 284 Hz.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 4

Explanation:
1. \( I_{max} \propto (a + a)^2 = 4a^2 \) and one wave has \( I \propto a^2 \).
2. So \( n = 4 \).

Teacher's Note:
1. Key fact to remember: \( I_{max} \propto (a + a)^2 = 4a^2 \) and one wave has \( I \propto a^2 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 346.5 m/s

Explanation:
1. Closed pipe: \( f = v/4L \). \( \dfrac{v}{4}\left(\dfrac{1}{0.75} - \dfrac{1}{0.77}\right) = 3 \).
2. \( v = 12/0.034632 = 346.5 \) m/s.

Teacher's Note:
1. Key fact to remember: Closed pipe: \( f = v/4L \). \( \dfrac{v}{4}\left(\dfrac{1}{0.75} - \dfrac{1}{0.77}\right) = 3 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (c) \( 2\pi/3 \)

Explanation:
1. Resultant amplitude \( A = 2a\cos(\Delta\phi/2) = a \), so \( \cos(\Delta\phi/2) = 1/2 \).
2. \( \Delta\phi/2 = \pi/3 \), giving \( \Delta\phi = 2\pi/3 \).

Teacher's Note:
1. Key fact to remember: Resultant amplitude \( A = 2a\cos(\Delta\phi/2) = a \), so \( \cos(\Delta\phi/2) = 1/2 \).
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (d) 0.5 s

Explanation:
1. Beat frequency = 258 - 256 = 2 Hz.
2. Interval between maxima = 1/2 = 0.5 s.

Teacher's Note:
1. Key fact to remember: Beat frequency = 258 - 256 = 2 Hz.
2. Revise this point with the matching definition or example in the chapter notes.

Answer: (a) 180

Explanation:
1. \( f_1 = 250 \) Hz and \( f_2 = 253 \) Hz, so 3 beats per second.
2. Per minute: 3 × 60 = 180.

Teacher's Note:
1. Key fact to remember: \( f_1 = 250 \) Hz and \( f_2 = 253 \) Hz, so 3 beats per second.
2. Revise this point with the matching definition or example in the chapter notes.