Fluid Mechanics

JEE Main level, single correct, medium

111 of 192

Question 111 of 192

A big drop is formed by coalescing 10001000 small droplets of water. The surface energy will become :

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Answer: (D) 110th\dfrac{1}{10}th

Lets say radius of small droplets is rr and that of big drop is RR. Total volume of the drops will remains unchanged. Therefore,
43πR3=1000(43πr3)\dfrac{4}{3}\pi {R}^{3}=1000\left(\dfrac{4}{3}\pi {r}^{3}\right)
⇒R=10r\Rightarrow R=10r
Now, initial surface energy is, Ui=1000(4πr2S){U}_{i}=1000\left(4\pi {r}^{2}S\right) and the final surface energy is,
Uf=4πR2S{U}_{f}=4\pi {R}^{2}S
=100(4πr2S)=100\left(4\pi {r}^{2}S\right)
Clearly, Uf=110Ui{U}_{f}=\dfrac{1}{10}{U}_{i}.

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