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Fluid Mechanics

JEE Main level, numerical answer, medium

112 of 192

Question 112 of 192

A big drop is formed by coalescing $1000$ small identical drops of water. If ${E}_{1}$ be the total surface energy of $1000$ small drops of water and ${E}_{2}$ be the surface energy of single big drop of water, the ${E}_{1}$ : ${E}_{2}$ is $x:1$ , where $x=$ ________.

Show answer and solution

Answer: 10

Let radius of the small drop be

r

and the radius of the big drop be

R

. Then, total volume will remain constant, therefore

1000\times \left[\rho \left(\dfrac{4}{3}\pi {r}^{3}\right)\right]=\dfrac{4}{3}\pi {R}^{3}\rho

\Rightarrow R=10r

Initial total surface energy,

{E}_{1}=1000\times 4\pi {r}^{2}\times S

and

{E}_{2}=4\pi (10r{)}^{2}S

\Rightarrow \dfrac{{E}_{1}}{{E}_{2}}=\dfrac{10}{1} \text{Hence},x=10.

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