MHT CET · Physics · Mechanical Properties of Fluids
Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is
- A \(2\frac{2}{3}:1\)
- B \(\sqrt{2}: 1\)
- C \(2\frac{1}{3}: 1\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(C) \(2\frac{1}{3}: 1\)
Step-by-step Solution
Detailed explanation
The volume of mercury remains same.
\(
\begin{array}{l}
\therefore \frac{4 \pi}{3} R^{3}=2\left(\frac{4 \pi}{3} r^{3}\right) \\
\therefore R^{3}=2 r^{3} \quad \text { or } \quad r^{3}=\frac{R^{3}}{2} \\
\therefore r=\frac{R}{2^{\frac{1}{3}}}
\end{array}
\)
Initial surface energy \(\mathrm{E}_{1}=4 \pi \mathrm{R}^{2} \cdot \mathrm{T}\)
Final surface energy
\(\mathrm{E}_{2}=4 \pi \mathrm{r}^{2} \mathrm{~T}=4 \pi\left(\frac{\mathrm{R}}{2^{\frac{1}{3}}}\right)^{2} \mathrm{~T}=4 \pi \frac{\mathrm{R}^{2}}{2^{\frac{2}{3}} \mathrm{~T}}\)
\(\therefore \frac{E_{1}}{E_{2}}=2^{\frac{2}{3}}\)
\(
\begin{array}{l}
\therefore \frac{4 \pi}{3} R^{3}=2\left(\frac{4 \pi}{3} r^{3}\right) \\
\therefore R^{3}=2 r^{3} \quad \text { or } \quad r^{3}=\frac{R^{3}}{2} \\
\therefore r=\frac{R}{2^{\frac{1}{3}}}
\end{array}
\)
Initial surface energy \(\mathrm{E}_{1}=4 \pi \mathrm{R}^{2} \cdot \mathrm{T}\)
Final surface energy
\(\mathrm{E}_{2}=4 \pi \mathrm{r}^{2} \mathrm{~T}=4 \pi\left(\frac{\mathrm{R}}{2^{\frac{1}{3}}}\right)^{2} \mathrm{~T}=4 \pi \frac{\mathrm{R}^{2}}{2^{\frac{2}{3}} \mathrm{~T}}\)
\(\therefore \frac{E_{1}}{E_{2}}=2^{\frac{2}{3}}\)
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