Question 13: Properties of Matter

Question

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is $\_\_\_\_$ - ( $S$ is the surface tension of mercury).

Accepted Answer

When 8 small drops of radius $r$ combine to form one large drop of radius $R$, the total volume remains constant.
Volume of 8 small drops $=$ Volume of 1 large drop
$$ 8 	imes\left(\frac{4}{3} \pi r^3ight)=\frac{4}{3} \pi R^3 $$
$$\Rightarrow $$ $8 r^3=R^3$
$$ \Rightarrow R=2 r $$
The surface energy of a single drop is $E=S A$ where $S$ is the surface tension and $A$ is the surface area.
Total initial area for 8 drops $=8 	imes\left(4 \pi \mathrm{r}^2ight)=32 \pi \mathrm{r}^2$.
So, total initial surface energy is $\mathrm{U}_{\mathrm{i}}=32 \pi \mathrm{r}^2 \mathrm{~S}$
The final drop has radius $\mathrm{R}=2 \mathrm{r}$.
So, final surface area $=4 \pi R^2=4 \pi(2 r)^2=4 \pi\left(4 r^2ight)=16 \pi r^2$.
So, the final surface energy is $U_f=16 \pi r^2 S$
The energy released is the decrease in surface energy :
$$ \Delta U=U_i-U_f $$
$$\Rightarrow $$ $\Delta U=32 \pi r^2 S-16 \pi r^2 S$
$$\Rightarrow $$ $$ \Delta U=16 \pi r^2 S $$
Therefore, the surface energy released in this process is $16 \pi r^2 S$.
Hence, the correct option is (B).

Answer

$8 \pi r^2 \mathrm{~S}$

Answer

$16 \pi r^2 S$

Answer

$64 \pi r^2 S$

Answer

$4 \pi r^2 \mathrm{~S}$

Properties of Matter

Eight mercury drops, each of radius r, coalesce to form a bigger drop. The surface energy released in this process is \_\_\_\_ - ( S is the surface tension of mercury).

Choose the correct answer: