NEET Practice Questions (MCQs) with Answers & Solutions

$ { f_L \over f_s } = { \nu + \nu_L \over \nu + \nu_s } or { f_L \over f_s } = { \nu - nu_L \over \nu - \nu_s } $ $ but , \nu_L = \nu_s $ $ \therefore { f_L \over f_s } = 1 $

Since the rope is heavy, the tension at the lower end & top end of the rope will be different. Mass of rope $ m_2 $ = 3kg Mass of block $ m_1 $ = 1 kg $ \therefore tension at the lower end T_1 = m_1 g = 1 g N and at the upper end in T_2 = (m_1 + m_2 ) g = 4 g N $ Now speed of wave in rope $ \nu = \sqrt T \Rightarrow f \lambda = \sqrt T $ $ \therefore \lambda = \sqrt T ( \therefore f , \mu are constants ) $ $ \therefore Wave length at lower end and \lambda_1 = \sqrt T_1 and at the upper end \lambda_2 = \sqrt T_ 2$ $ \therefore { \lambda_2 \over \lambda_1 } = \sqrt { T_2 \over T_1 } \Rightarrow \lambda_2 = \sqrt { T_2 \over T_1 } = \lambda_1 = \lambda_1 = 0.1 m $

$ Speed of sound = \sqrt { \gamma P \over \rho } $ $ \rho = { mass of 1 mole air \over volume of 1 mole air } = { 29 \times 10^{-3} kg \over 22.4 \times 10^{-3} m^3 } = 1.3 $ $ \therefore speed = \sqrt { { 7 \over 5} \times { 1.01 \times 10^5 \over 1.3 } } = 330 ms^{-1} $

$ From the phase angle ( 40 -2t ) , we get k =40 \, OR\, { 2 \pi \over \lambda } = 40 \Rightarrow \lambda = { \pi \over 20 } $ $ and \omega = 2 \, OR \, 2 \pi f = 2 \Rightarrow f = \pi^{-1} Hz $

Increase in tension of string increases its frequency. If the original frequency of $B(f_B)$ were greater than that of $A(f_A)$, further the increase in $f_B$ should have resulted in increase in the beat frequency. But the beat frequency is found to decrease. This shows that $f_A-f_B = 5 Hz$ and $f_A=427 Hz$, we get $f_B = 422 Hz$

$ f_L = \left( { \nu - \nu_L \over \nu - \nu_s } \right) f_s = \left[ { 330 - 0 \over 330 -130 } \right] \times 800 = 1320 Hz $

$ \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)