Question

Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger drop as compared to each smaller drop is

• A. 2 times
• B. 8 times
• C. 4 times
• D. 16 times

Answer 1

The Correct Option is A

The capacitance of a sphere is given by $C = 4\pi\epsilon_0 r$, where $r$ is the radius.

Let $r$ be the radius of each smaller drop and $R$ be the radius of the big drop. Since the volume of the combined drop is equal to the sum of the volumes of the smaller drops:

$\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$

$R^3 = 8r^3$

$R = 2r$

The capacitance of a smaller drop is $C_\text{small} = 4\pi\epsilon_0 r$.

The capacitance of the bigger drop is $C_\text{big} = 4\pi\epsilon_0 R = 4\pi\epsilon_0 (2r) = 2(4\pi\epsilon_0 r) = 2C_\text{small}$.

The correct answer is (A) 2 times.