MHT CET · Physics · Mechanical Properties of Fluids
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 \(\sqrt{\mathrm{n}}: 1\)
- B \(\sqrt[3]{\mathrm{n}}: 1\)
- C \(n: 1\)
- D \(\mathrm{n}^2: 1\)
Answer & Solution
Correct Answer
(B) \(\sqrt[3]{\mathrm{n}}: 1\)
Step-by-step Solution
Detailed explanation
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 \((E)=4 \pi R^2 T\)
\(\therefore \quad \frac{E_n}{E}=\frac{n r^2}{R^2}=\frac{n r^2}{\left(n^{\frac{1}{3}} r\right)^2}=\frac{n r^2}{n^{\frac{2}{3}} r^2}=n^{\frac{1}{3}}=\sqrt[3]{n}: 1\)
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 \((E)=4 \pi R^2 T\)
\(\therefore \quad \frac{E_n}{E}=\frac{n r^2}{R^2}=\frac{n r^2}{\left(n^{\frac{1}{3}} r\right)^2}=\frac{n r^2}{n^{\frac{2}{3}} r^2}=n^{\frac{1}{3}}=\sqrt[3]{n}: 1\)
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