A large number of identical droplets of a liquid, each of radius $r$, coalesce to form a bigger drop of radius $R$. It $\sigma$ is the surface tension of the liquid and $\rho$ its density and if $60\%$ of the energy released is used up in increasing the kinetic energy of the big drop, the velocity of drop will be 

If n droplets coalesce, the energy released is 

         $E=(n\times 4{\mathrm{\pi r}}^2-4{\mathrm{\pi R}}^2)\mathrm{\sigma }$ 

 Since there is no change in volume in this process, 

                n x $\frac{4\mathrm{\pi }}{3}{r}^3$ = $\frac{4\mathrm{\pi }}{3}{R}^3$

$$\begin{gathered} \Rightarrow n=\frac{R^3}{r^3} \\ \therefore E=\left(\frac{R^3}{r^3}\times 4{\mathrm{\pi r}}^2-4{\mathrm{\pi R}}^2\right)\sigma \\ =4{\mathrm{\pi R}}^2\left(\frac{R}{r}-1\right)\sigma \end{gathered}$$

$60\%$ of $E$$=0.6\times 4{\mathrm{\pi R}}^2\left(\frac{R}{r}-1\right)\sigma$. 

Mass of big drop is $M=\frac{4\pi }{3}\rho R^3$

If v is the velocity of the drop. 

           $$\begin{gathered} \frac{1}{2}M{v}^2=0.6\times 4{\mathrm{\pi R}}^2\left(\frac{R}{r}-1\right)\sigma \\ \Rightarrow \frac{1}{2}\times \frac{4\mathrm{\pi }}{3}\rho R^3{v}^2=0.6\times 4{\mathrm{\pi R}}^2\left(\frac{R}{r}-1\right)\sigma \\ \Rightarrow v={\left[\frac{3.6\sigma }{\rho rR}\left(R-r\right)\right]}^{1/2} \end{gathered}$$

So the correct choice is (a).