Explanation

(b) As v=nλ
$\therefore\ \ \ \ \lambda = \frac{v}{n} = \frac{300}{500} = \frac{3}{5}m$
Now, phase difference
$= \frac{2\pi}{\lambda} \times path\ difference$
$\therefore\ \ \ 60{^\circ} = \frac{2\pi}{\lambda} \times path\ difference\ $
$or\ \ \ \frac{60{^\circ} \times \pi}{180{^\circ}} = \frac{2\pi \times 5}{3} \times path\ difference$
$path\ difference = \frac{3 \times 60 \times \pi}{2\pi \times 5 \times 180} = 0.1$