Q138 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics

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

  • A
    $\sqrt{2}: 1$
  • B
    $2^{2 / 3}: 1$
  • Correct answer
    C
    $2^{1 / 3}: 1$
  • D
    $2: 1$

Solution

Total surface energy before coalesce $$ E_1=2\left(4 \pi R^2\right) T $$ But $\frac{4}{3} \pi \mathrm{R}^3 \times 2=\frac{4}{3} \pi \mathrm{R}^{13}$ $$ R^{\prime}=(2)^{1 / 3} R $$ Total surface energy after coalescing $$ \begin{aligned} & \mathrm{E}_2=4 \pi R^{\prime 2} \mathrm{~T}=4 \pi(2)^{2 / 3} R^2 \mathrm{~T} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{2\left(4 \pi R^2\right) \mathrm{T}}{4 \pi(2)^{2 / 3} \mathrm{R}^2 \mathrm{~T}}=2^{1-\frac{2}{3}}=2^{1 / 3} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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