Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total…
Two small drops of mercury each of radius 'R' coalesce to form a large single drop.
The ratio of the total surface energies before and after the change is
$\sqrt{2}: 1$
$2^{1 / 3}: 1$
$2: 1$
$2^{2 / 3}: 1$
Solution
$\frac{4}{3} \pi R^{3} \times 2=\frac{4}{3} \pi R^{\prime 3}$
$R^{\prime}=2^{1 / 3} R$
Total surface energy before $=2 \times 4 \pi R^{2} T \quad(T=$ surface tension $)$
Total surface energy after $=4 \pi R^{\prime 2} T$
$\therefore \quad$ Ratio $=\frac{2 \times R^{2}}{R^{\prime 2}}=\frac{2 \times R^{2}}{2^{2 / 3} R^{2}}=\frac{2}{2^{2 / 3}}=2^{1-\frac{2}{3}}=2^{1 / 3}$
$\therefore$ Ratio $=2^{1 / 3}: 1$