Question

64 small drops of mercury each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface charge density of each small drop with that of big drop is
(A) 4:1
(B) 1:4
(C) 1:64
(D) 64:1

Answer 1

Hint : The one large drop will have a charge 64 times that of the individual small drops. The volume of the large drop will be equal to the sum of the volume of the individual small drops.

Formula used: In this solution we will be using the following formula;
$\Rightarrow \sigma = \dfrac{q}{A}$ where $\sigma$ is the surface charge density, $q$ is the total charge on the surface, and $A$ is the surface area of the surface.
$\Rightarrow A = 4\pi {r^2}$ where $A$ is the surface area of the sphere and $r$ is its radius.
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$ where $V$ is the volume of a sphere, $r$ is the radius of the sphere.

Complete step by step answer
One of the 64 small drops has a surface area of $A = 4\pi {r^2}$ $r$ is its radius. Now these small drops coalesce into a large drop, the total volume hence must be the same as in
$V = 64 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$ , where $R$ must be the radius of the larger sphere. $\dfrac{4}{3}\pi {r^3}$ is the volume of one small drop. Hence,
$64{r^3} = {R^3}$
$\Rightarrow R = \sqrt[3]{{64{r^3}}} = 4r$
Also, the total charge on the larger drop is the sum of the charge on the individual drop, hence
$\Rightarrow Q = 64q$ where $Q$ is the charge on the larger drop and $q$ is the charge on the smaller drop.
Hence, the surface densities of the two surfaces are
$\Rightarrow {\sigma _s} = \dfrac{q}{{4\pi {r^2}}}$ ( for small surface) and ${\sigma _l} = \dfrac{{64q}}{{4\pi {R^2}}}$ ( for large surface)
Hence, the ratio of small drop to large drop is
$\Rightarrow \dfrac{{{\sigma _s}}}{{{\sigma _l}}} = \dfrac{q}{{4\pi {r^2}}} \div \dfrac{{64q}}{{4\pi {R^2}}} = \dfrac{q}{{4\pi {r^2}}} \times \dfrac{{4\pi {{\left( {4r} \right)}^2}}}{{64q}}$
By elimination and reduction, we have
$\Rightarrow \dfrac{{{\sigma _s}}}{{{\sigma _l}}} = \dfrac{1}{4}$ hence,
$\Rightarrow {\sigma _s}:{\sigma _l} = 1:4$
Thus, the correct option is B.

Note
In actuality, the surface of the large drop will become a little larger than as predicted hence reducing the surface charge density than as predicted. This is because as the many charges surface density of the charge increases, the forces on each other will increase repelling each other away and thus increasing the surface area of the drop. This stops when the surface tension of the drop becomes equal to the repulsive force.