Explanation

(c) Frequency of sonometer wire is given by
$v = \frac{1}{2l}\sqrt{\frac{T}{m}} = \frac{1}{2l}\sqrt{\frac{T}{\pi r^{2}p}}$
$v_{1} = \frac{1}{2l_{1}}\sqrt{\frac{T_{1}}{\pi r_{1}^{2}\rho_{1}}}$
$v_{2} = \frac{1}{2l_{2}}\sqrt{\frac{T_{2}}{\pi r_{2}^{2}\rho_{2}}}$
$\therefore\ \ \frac{v_{1}}{v_{2}} = \frac{l_{2}}{l_{1}}\sqrt{\frac{T_{1}}{T_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}} \times \frac{\rho_{2}}{\rho_{1}}}$
$\frac{v_{1}}{v_{2}} = \frac{35}{36}\sqrt{\frac{8}{1} \times \frac{1}{16} \times \frac{2}{1}}$
$\because v_{1} < v_{2}\ and\ v_{2} = 360Hz$
Therefore,
$v = 360 \times \frac{35}{36}$
$v_{1}$=350 Hz
So, number of beats produced =$v_{1} - v_{2}$
=360-350=10