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MHT CET · Physics · Mechanical Properties of Fluids

When a mercury drop of radius ' \(R\) ' splits up into 1000 droplets of radius ' \(r\) ', the change in surface energy is ( \(T=\) surface tension of mercury)

  1. A \(8 \pi \mathrm{R}^2 \mathrm{~T}\)
  2. B \(16 \pi \mathrm{R}^2 \mathrm{~T}\)
  3. C \(\quad 34 \pi \mathrm{R}^2 \mathrm{~T}\)
  4. D \(\quad 36 \pi R^2 T\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\quad 36 \pi R^2 T\)

Step-by-step Solution

Detailed explanation

As volume remains constant,
\(\mathrm{R}^3=1000 \mathrm{r}^3\)
\(\therefore \quad \mathrm{n}^{\prime}=10 \mathrm{r}\) or \(\mathrm{r}=\frac{\mathrm{R}}{10}\)
Change in surface area
\(\begin{aligned}
& =\left(1000 \times 4 \pi r^2\right)-4 \pi R^2 \\
& =4 \pi\left(1000 \times \frac{R^2}{100}-R^2\right)=36 \pi R^2
\end{aligned}\)
Surface energy, \(T \Delta A=36 \pi R^2 T\)