Question

$27$ similar drops of mercury are maintained at $10\text{V}$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is _______ times that of a smaller drop.

Answer 1

Let $q$ and $r$ be the charge and radius of a smaller drop, respectively.

Also, $R$ be the radius of the bigger drop.
Charge on the bigger drop will be $27q$.

Volume of $27$ smaller drops will be equal to volume of bigger drop.

$\therefore \frac{4}{3}\pi r^3\times 27=\frac{4}{3}\pi R^3$
$\Rightarrow R={\left(27\right)}^{\frac{1}{3}}r=3r$

We know that, potential energy,
$U\propto \frac{q^2}{r}$
So,
$\frac{U_f}{U_i}=\frac{q_f^2{r}_i}{r_f{q}_i^2}$

$\Rightarrow \frac{U_f}{U_i}=\frac{(27q{)}^2\times r}{3r\times q^2}=243$

$\Rightarrow U_f=243U_i$