Q135 / 160 · AP EAMCET 2023 (19 May Shift 1) · Physics
Two mercury drops of radii $\mathrm{r}$ and $2 \mathrm{r}$ merge to form a bigger drop. The surface energy released in the process is nearly
(Surface tension mercury is $\mathrm{S}$ and take $9^{2 / 3}=4.326$ )
- A$1.6 \pi \mathrm{r}^2 \mathrm{~S}$
- B$3.2 \pi \mathrm{r}^2 \mathrm{~S}$
- Correct answerC$17.1 \pi r^2 S$
- D$2.7 \pi \mathrm{r}^2 \mathrm{~S}$
Solution
Radii of drop, $r_1=r$ and $r_2=2 r$
Sum of volume of two drops = Volume of the bigger drop
$$
\begin{aligned}
& \frac{4}{3} \pi r^3+\frac{4}{3} \pi(2 r)^3=\frac{4}{3} \pi R^3 \\
& r^3+8 r^3=R^3 \\
& R^3=9 r^3=(9)^{1 / 3} r
\end{aligned}
$$
Surface energy, $E=4 \pi R^2 S$
$$
\begin{aligned}
& =4 \pi \times(9)^{2 / 3} \mathrm{r}^2 \mathrm{~S} \\
& =4 \pi \times 4.326 \mathrm{r}^2 \mathrm{~S} \\
& =17.1 \pi \mathrm{r}^2 \mathrm{~S}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.