Explanation

(d) When source approaches the observer, the apparent frequency heard by observer is
$v^{'} = v\left( \frac{v}{v - v_{s}} \right)\ \ \ \ldots(i)$
$v_{s} =$speed of source of sound
During its recession, apparent frequency
$v^{''} = v\left( \frac{v}{v + v_{s}} \right)\ \ \ \ \ldots(ii)$
Accordingly
$v^{'} = {v^{'}}^{'} = \frac{2}{100}v\ \ \ \ (given)$
$\therefore v\left( \frac{v}{v - v_{s}} \right) - v\left( \frac{v}{v + v_{s}} \right) = \frac{2}{100}v$
Or
$v\left\lbrack \frac{v + v_{s} - v + v_{s}}{\left( v - v_{s} \right)v + v_{s}} \right\rbrack = \frac{2}{100}$
Or
$\frac{2vv_{s}}{\left( v - v_{s} \right)\left( v + v_{s} \right)} = \frac{2}{100}$
Or
$100vv_{s} = v^{2} - v_{s}^{2}$
But speed of sound in air v=$300\ ms^{- 1}$
$\therefore 3000v_{s} = (300)^{2} - v_{s}^{2}$
$\Longrightarrow v_{s}^{2} + 30000v_{s} - 90000$
$v = \frac{- 30000 \pm \sqrt{(30000)^{2} + 4 \times 90000}}{2}$
$= - \frac{30000 \pm 30006}{2} = \frac{6}{2} = 3ms^{- 1}$
(Taking +ve sign only)