Waves MCQs for NEET — Physics Questions with Answers

$ \nu = \sqrt T \Rightarrow T = \mu \nu^2 = { M \over L } \nu^2 = { 2.2 \over 11 } \times (340)^2 $ $ \therefore T = 2.31 \times 10^4 N $

$ f = { 1 \over 21 } \sqrt T \Rightarrow T = f^2 4 l^2 $ $ \therefore T = 4 f^2 ( M)^2 = 4 f^2 M^2 = 248 N$

$ In tube A , \lambda_A = 2 l $ $ In tube B , \lambda _B = 4 l $ $ \therefore \nu_A = { \nu \over \lambda_A } = { \nu \over 21} $ $ \therefore \nu_B = { \nu \over \lambda_B } = { \nu \over 41 } \Rightarrow { \nu_A \over \nu_B } ={ 2 \over 1 } $

The was decreases the frequency of unknown fork. The possible unknown frequencies are, (288+4) Hz and (288-4) Hz. Wax reduces 284 Hz and so beats should increases. It is not given in the question. This frequency is ruled out. Wax reduces 292 Hz and so beats should decrease. It is given that the beats decrease from 2 to 4. Hence the unknown fork has frequency 292 Hz. consider option (a)

Stationary wave : Y = a sin (wt-kx) + a sin(wt+kx) When x = 0, $Y \neq 0$. The option is not acceptable consider option (b) stationary wave : Y = a sin (wt-kx) - a sin(wt+kx) At x = 0, Y = 0. This option holds good. Option (c) gives Y = 2a sin(wt - kx) At x = 0, $Y \neq 0$ Option (d) gives Y = 0. Hence option (b) holds good

The possible frequency of piano are (256 + 5)Hz and (256 - 5)Hz. For a piano string $ \nu = { 1 \over 2l } \sqrt {T} $ When tension T increases v increases. (i) If 261 Hz increases, beats / second increase. This is not given. (ii) If 251 Hz increases due to tension, beats / second decrease. This is given.

$ By doppler's effect , { f_L \over f_S} = \left( { \nu + \nu_L \over \nu + \nu_S} \right) $ $ \therefore { f_L \over f_S} = { \nu + \nu_L \over \nu } = { \nu + \nu/5 \over \nu } = { 6 \over 5 } $ $ \therefore Fractional increase = { f_L - f_S \over f_S } = { f_L \over f_S } -1 = { 6 \over 5} -1 = {1 \over 5} $ $ \therefore Percentage increase = { 100 \over 5} = 20 \% $

$ \nu = \sqrt { \gamma P \over \rho } = \sqrt { \gamma RT \over M} $ $ \therefore { \nu_2 \over \nu_1 } = \sqrt { { \gamma_{He} \over 4 } \times { 32 \over \gamma_{02} } }$

In a longitudinal wave, pressure is maximum where displacement is minimum. Therefore pressure and displacement variations are $180 ^\circ $ out of phase

Frequency of tuning fork $f_1= 480 Hz$. Number of beats $s^{-1}$, n = 10 Frequency of string $f_2 = (480 + 10)Hz$. Aslight increase in tension increase $f_2$. $f_2 = 480 - 10 = 470 Hz.$