NEET Practice Questions (MCQs) with Answers & Solutions

Let the number of loops obtained for 315Hz and 420Hz n and (n+1) respectively. $ \therefore f_n = nf_1 = 315 $ $ \therefore f_{n+1} = (n+1) f_1 = 420 $ $ \therefore f_{n+1} - f_n = f_1 =105 Hz $

When, sound waves travel from one medium to another, its frequency does not change. $ \therefore f = { \nu \over \lambda } = constant $ $ \therefore { \nu_a \over \lambda_a} = { \nu_b \over \lambda_b } $ $ \lambda_b = { \nu_b \over \nu_a } \lambda_a = 10 \lambda_a $

Wave velocity = max. velo.of particle $ { \omega \over k } = A \omega $ $ \therefore A = { 1 \over k } = { \lambda \over 2 \pi } $ $ \therefore \lambda = 2 \pi A $

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 } $