Fluid Mechanics

JEE Main level, numerical answer, medium

107 of 192

Question 107 of 192

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of energy of big drop is 10x\frac{10}{x}. The value of xx is __________

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Answer: 1


100043πr3=43πR310r=RR=10r\begin{aligned} 1000 \frac{4}{3} \pi r^3 & =\frac{4}{3} \pi R^3 \\ 10 r & =R \\ R & =10 r\end{aligned}
 S.E. of 1000 drops  S.E. of Big drop =1000(4πr2)T4πR2 T=1000×r2(10r)2=10=10x∴x=1\begin{aligned} \frac{\text { S.E. of } 1000 \text { drops }}{\text { S.E. of Big drop }} & =\frac{1000\left(4 \pi \mathrm{r}^2\right) \mathrm{T}}{4 \pi \mathrm{R}^2 \mathrm{~T}} \\ & =\frac{1000 \times \mathrm{r}^2}{(10 \mathrm{r})^2}=10=\frac{10}{\mathrm{x}} \\ \therefore \quad x & =1\end{aligned}

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