Question 129 of 192

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

A${2}^{\dfrac{1}{3}}:1$B$1:{2}^{\dfrac{1}{3}}$C$2:1$D$1:2$

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Answer: (A) ${2}^{\dfrac{1}{3}}:1$

$\dfrac{4}{3}\pi {R}^{3}+\dfrac{4}{3}\pi {R}^{3}=\dfrac{4}{3}\pi {R}^{'3}$
${R}^{'}={2}^{\dfrac{1}{3}}R\ldots \left(i\right)$
${A}_{i}=2\left[4\pi {R}^{2}\right]$
${A}_{f}=4\pi {R}^{'2}$
$\dfrac{{U}_{i}}{{U}_{f}}=\dfrac{{A}_{i}}{{A}_{f}}=\dfrac{2{R}^{2}}{{2}^{\dfrac{2}{3}}{R}^{2}}={2}^{\dfrac{1}{3}}$

JEE Main Physics practice question on Fluid Mechanics (Surface tension), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.