Q106 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
A big water drop is formed by the combination of ' $n$ ' small water droplets of equal radii. The ratio of the surface energy of ' $n$ ' droplets to the surface energy of the big drop is
- A$\sqrt{\mathrm{n}}: 1$
- Correct answerB$\sqrt[3]{\mathrm{n}}: 1$
- C$n: 1$
- D$\mathrm{n}^2: 1$
Solution
Let R be the radius of bigger drop and r be the radius of single small water drop.
Volume of big drop $=n($ Volume of small drop $)$
$\begin{aligned}
\therefore \quad & \frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3 \\
& \Rightarrow R^3=n r^3 \\
& R=n^{\frac{1}{3}} r
\end{aligned}$
Surface energy of $n$ drops $\left(E_n\right)=n \times 4 \pi r^2 \times T$ Surface energy of big drop $(E)=4 \pi R^2 T$ $\therefore \quad \frac{E_n}{E}=\frac{n r^2}{R^2}=\frac{n r^2}{\left(n^{\frac{1}{3}} r\right)^2}=\frac{n r^2}{n^{\frac{2}{3}} r^2}=n^{\frac{1}{3}}=\sqrt[3]{n}: 1$
Surface energy of $n$ drops $\left(E_n\right)=n \times 4 \pi r^2 \times T$ Surface energy of big drop $(E)=4 \pi R^2 T$ $\therefore \quad \frac{E_n}{E}=\frac{n r^2}{R^2}=\frac{n r^2}{\left(n^{\frac{1}{3}} r\right)^2}=\frac{n r^2}{n^{\frac{2}{3}} r^2}=n^{\frac{1}{3}}=\sqrt[3]{n}: 1$
Hence, the correct answer is option: 'B'.