MHT CET · Physics · Mechanical Properties of Fluids
Two small drops of mercury each of radius ' \(R\) ' coalesce to from a large single drop. The ratio of the total surface energies before and after the change is
- A \(\sqrt{2}: 1\)
- B \(2^{2 / 3}: 1\)
- C \(2^{1 / 3}: 1\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(C) \(2^{1 / 3}: 1\)
Step-by-step Solution
Detailed explanation
Total surface energy before coalesce
\(
E_1=2\left(4 \pi R^2\right) T
\)
But \(\frac{4}{3} \pi \mathrm{R}^3 \times 2=\frac{4}{3} \pi \mathrm{R}^{13}\)
\(
R^{\prime}=(2)^{1 / 3} R
\)
Total surface energy after coalescing
\(
\begin{aligned}
& \mathrm{E}_2=4 \pi R^{\prime 2} \mathrm{~T}=4 \pi(2)^{2 / 3} R^2 \mathrm{~T} \\
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{2\left(4 \pi R^2\right) \mathrm{T}}{4 \pi(2)^{2 / 3} \mathrm{R}^2 \mathrm{~T}}=2^{1-\frac{2}{3}}=2^{1 / 3}
\end{aligned}
\)
\(
E_1=2\left(4 \pi R^2\right) T
\)
But \(\frac{4}{3} \pi \mathrm{R}^3 \times 2=\frac{4}{3} \pi \mathrm{R}^{13}\)
\(
R^{\prime}=(2)^{1 / 3} R
\)
Total surface energy after coalescing
\(
\begin{aligned}
& \mathrm{E}_2=4 \pi R^{\prime 2} \mathrm{~T}=4 \pi(2)^{2 / 3} R^2 \mathrm{~T} \\
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{2\left(4 \pi R^2\right) \mathrm{T}}{4 \pi(2)^{2 / 3} \mathrm{R}^2 \mathrm{~T}}=2^{1-\frac{2}{3}}=2^{1 / 3}
\end{aligned}
\)
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