Physics - Fluid Question with Solution | TestHub

Radius of one drop of mercury is $R$. $\therefore$ The volume of one drop $=\frac{4}{3}\pi R^3$ $\therefore$ Total volume of the two drops, $$V=2\times \frac{4}{3}\pi R^3=\frac{8}{3}\pi R^3$$ Let the radius of the large drop formed be $R^{\prime }$. The volume of the large drop is also $V$. $$\therefore \frac{4}{3}\pi R^{\prime 3}=\frac{8}{3}\pi R^3\Rightarrow R^{\prime 3}=2R^3\Rightarrow R^{\prime }=2^{1/3}R\text{. }$$ Now the surface area of the two drops is $$S_1=2\times 4\pi R^2=8\pi R^2$$ and the surface area of the resultant drop is $$S_2=4\pi R^{\prime 2}=4\pi 2^{2/3}{R}^2$$ Let $T$ be the surface tension of mercury. Therefore the surface energy of the two drops before coalescing is $$U_1=S_{1}T=8\pi R^{2}T$$ and the surface energy after coalescing, $$\begin{array}{c}{U}_2=S_{2}T=2^{2/3}\times 4\pi R^{2}T\\ \therefore \frac{U_1}{U_2}=\frac{8\pi R^{2}T}{2^{2/3}\times 4\pi R^{2}T}=\frac{2}{2^{2/3}}=2^{1/3}.\end{array}$$