Question:
Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is :
Correct Option: 1
Solution:
$\frac{4}{3} \pi \mathrm{R}^{3}+\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \mathrm{R}^{13}$
$R^{\prime}=2^{\frac{1}{3}} R$ ............(i)
$\mathrm{A}_{\mathrm{i}}=2\left[4 \pi \mathrm{R}^{2}\right]$
$\mathrm{A}_{\mathrm{f}}=4 \pi \mathrm{R}^{\prime 2}$
$\frac{U_{i}}{U_{f}}=\frac{A_{i}}{A_{f}}=\frac{2 R^{2}}{2^{2 / 3} R^{2}}=2^{1 / 3}$
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