Physics MCQs for NEET — Practice Questions with Answers

Speed of sound in an ideal gas $ \nu = sqrt { \gamma RT \over m } $ $ \therefore { \nu_1 \over \nu_2 } = \sqrt { m_2 \over m_1 } \left ( \therefore v { 1 \over \sqrt m } \right) $

$ E = { 1 \over 2 } m \omega^2 A^2 = {1 \over 2 } m 4 \pi^2 f^2 A^2 $ $ \therefore E \alpha f^2 \Rightarrow { E_1 \over E_2 } = \left( { f_1 \over f_2 } \right)^2 = \left( { f \over 2 f } \right)^2 = { 1 \over 4 } \therefore E_2 = 4 E_1 $

Let hte freq. of 1st fork be $f_1$ $ \therefore frequency of 2nd fork = f_1 + 6 = f_1 + 6 (2 -1 ) $ $ \therefore freq. of th 24th fork = f_1 + 6(24 -1 )=f_1 + 138$ Now, freq. of 24th fork = 2 x freq. of 1st fork (given) $ \therefore f_1 + 138 = 2 f_1 \therefore f_1 = 138 Hz $

$ wave number = { 1 \over \lambda } = { 1 \over 0.005 } = 200 m^{-1} $

Frequency heard by the listener $ f_L = \left ( { \nu + \nu_L \over \nu } \right) f_s $ $ ( \therefore \nu_s = 0 ) $ $ \therefore { f_L \over f_s } = { \nu + \nu_2 \over \nu } = { \nu + { \nu \over 4 } \over \nu } = { 5 \over 4 } $ $ \therefore \% increase = { f_L - f_s \over f_s } \times 100 = \left( { 5 - 4 \over 4 } \right) \times 100 = 25 \% $

$ From \nu = \sqrt { \gamma RT \over M } , \nu \alpha \sqrt T $ In summer, velocity increases & hence decreases and so L increases. The length of 2nd halmonics $ x = 3L_1 = 3 \times 16 = 48 cm $ In summer, velocity being more, $ x \gt 3L_1$ $ \therefore x \gt 48 $

$ In f_L = \left( { \nu + \nu_L \over \nu + \nu_s } \right) f_s $ $ putting \nu_L = 0 , f_L = 2 f_s , \nu = \nu, \nu_s = - \nu_s $ $ 2 f_s = \left( { \nu \over \nu - \nu_s } \right) f_s \Rightarrow 2 \nu_s = \nu \therefore \nu_s = { \nu \over 2 } $

$ { 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} $