A uniform rope of mass 0.1 kg and length 2.50 m hangs from a rigid support. The time taken by the transverse wave formed in the rope to travel through full length of the rope is ( $g=10m{s}^{-2}$ )

At the top of the rope, velocity is 

$\mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}} \& \mathrm{T}=\mathrm{mg}, \mathrm{\mu }=\frac{\mathrm{m}}{l},$

$\therefore \mathrm{V}=\sqrt{l\mathrm{g}},$ 

At the bottom of the rope, initial velocity is zero 

from $\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

$\Rightarrow l=0+\frac{1}{2}{\mathrm{at}}^2=\frac{\mathrm{a}}{2}{\mathrm{t}}^2$ 

$\mathrm{from} {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{aS}$

$l\mathrm{g}-0=2\mathrm{a}l\Rightarrow \mathrm{a}=\mathrm{g}/2$

$\mathrm{\therefore \hspace{0.17em}}\mathit{\hspace{0.17em}}l\mathrm{=}\frac{\mathrm{a}}{2}{\mathrm{t}}^2$

$\Rightarrow \hspace{0.17em}\mathit{\hspace{0.17em}}l=\frac{1}{2}\times \frac{\mathrm{g}}{2}\mathrm{\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}^2\Rightarrow \mathrm{t}=2\sqrt{l/\mathrm{g}}$

$t=2\sqrt{\frac{2.5}{10}}=1\text{\hspace{0.17em}\hspace{0.17em}second}$