Two small drops of mercury each of radius ' $R$ ' coalesce to from a large single drop. The ratio of the…
Two small drops of mercury each of radius ' $R$ ' coalesce to from a large single drop. The ratio of the total surface energies before and after the change is
$\sqrt{2}: 1$
$2^{2 / 3}: 1$
$2^{1 / 3}: 1$
$2: 1$
Solution
Total surface energy before coalesce
$E_1=2\left(4 \pi R^2\right) T$
But $\frac{4}{3} \pi \mathrm{R}^3 \times 2=\frac{4}{3} \pi \mathrm{R}^{\prime 3}$
$R^{\prime}=(2)^{1 / 3} R$
Total surface energy after coalescing
$\begin{aligned}
& \mathrm{E}_2=4 \pi R^{\prime 2} \mathrm{~T}=4 \pi(2)^{2 / 3} R^2 \mathrm{~T} \\
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{2\left(4 \pi R^2\right) \mathrm{T}}{4 \pi(2)^{2 / 3} \mathrm{R}^2 \mathrm{~T}}=2^{1-\frac{2}{3}}=2^{1 / 3}
\end{aligned}$