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9-1.Fluid Mechanics
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Two small drops of mercury, each of radius $R$ coalesce to form a single large drop. The radio of the total surface energies before and after the change is
A
$1 : 2^{1/3}$
B
$2^{1/3} : 1$
C
$2 : 1$
D
$1 : 2$
Solution
Let $\mathrm{r}$ be radius of common drop
$\frac{4}{3} \pi r^{3}=2 \times \frac{4}{3} \pi R^{3}$
$\mathrm{r}=(2)^{\frac{1}{3}} \mathrm{R}$
Surface energy before the coalesce
$=2 \times 4 \pi \mathrm{R}^{2} \mathrm{T}$
Surface energy after the coalesce $=4 \pi r^{2} T$
Ratio $=\frac{2 \times 4 \pi \mathrm{R}^{2} \mathrm{T}}{4 \pi \mathrm{r}^{2} \mathrm{T}}=\frac{2 \mathrm{R}^{2}}{2^{2 / 3} \mathrm{R}^{2}}$
$=\frac{2^{\frac{1}{3}} \cdot 2^{\frac{2}{3}}}{2^{\frac{2}{3}}}=\frac{2^{\frac{1}{3}}}{1} $
Standard 11
Physics
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