Explanation

(c) Velocity of longitudinal waves,
$v_{1} = \sqrt{\frac{Y}{\rho}}$
Velocity of transverse waves
$v_{2} = \sqrt{\frac{T}{m}}$
If$a$ is area of cross-section of string, then
$m = \frac{mass}{length} = \frac{mass}{volume} \times area = \rho a$
$v_{2} = \sqrt{\frac{T}{\rho a}}$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{Y}{\rho}.\frac{\rho a}{T}} = \sqrt{\frac{Ya}{T}}$
As $Y = \frac{F}{\frac{a\mathrm{\Delta}l}{l}} = \frac{T}{a(\mathrm{\Delta}l/l)}$
$\therefore\frac{v_{1}}{v_{2}} = \sqrt{\frac{T}{a\left( \frac{\mathrm{\Delta}l}{l} \right)}\frac{a}{T} =}\left( \frac{\mathrm{\Delta}l}{l} \right)^{- 1/2}$
We are given, $\frac{\mathrm{\Delta}l}{l} = \frac{1}{n}$
$\therefore\frac{v_{1}}{v_{2}} = \left( \frac{1}{n} \right)^{- 1/2} = \sqrt{n}$
If $f_{1},f_{2}$ are the corresponding fundamental frequencies of longitudinal and transverse vibration, then
$v_{1} = f_{1}\lambda\ and\ v_{2} = f_{2}\lambda$
$\therefore\frac{v_{1}}{v_{2}} = \frac{f_{1}}{f_{2}} = \sqrt{n}$