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

1. Volume Conservation:

When eight small drops of mercury combine to form a big drop, the total volume of mercury remains conserved. Let $r$ be the radius of each small drop and $R$ be the radius of the big drop. The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.

Volume of each small drop, $V_s = \frac{4}{3}\pi r^3$.

Total volume of 8 small drops, $V_{total} = 8 \times V_s = 8 \times \frac{4}{3}\pi r^3 = \frac{32}{3}\pi r^3$.

Volume of the big drop, $V_b = \frac{4}{3}\pi R^3$.

Since the volume is conserved, $V_b = V_{total}$:

$\frac{4}{3}\pi R^3 = \frac{32}{3}\pi r^3$

2. Relation between Radii:

Divide both sides of the volume equation by $\frac{4}{3}\pi$:

$R^3 = \frac{32}{4} r^3 = 8 r^3$

Take the cube root of both sides to find the relation between $R$ and $r$:

$R = \sqrt[3]{8 r^3} = \sqrt[3]{8} \times \sqrt[3]{r^3} = 2r$

So, the radius of the big drop is twice the radius of each small drop.

3. Capacitance of a Spherical Conductor:

The capacitance of a spherical conductor of radius $r$ is given by the formula $C = 4\pi\epsilon_0 r$, where $\epsilon_0$ is the permittivity of free space.

Capacitance of each small drop, $C_s = 4\pi\epsilon_0 r$.

Capacitance of the big drop, $C_b = 4\pi\epsilon_0 R$.

4. Comparing Capacitances:

Substitute $R = 2r$ into the expression for $C_b$:

$C_b = 4\pi\epsilon_0 (2r) = 2 \times (4\pi\epsilon_0 r)$

Since $C_s = 4\pi\epsilon_0 r$, we can write:

$C_b = 2 \times C_s = 2 C_s$

Final Answer: Thus, the capacitance of the bigger drop is 2 times the capacitance of each smaller drop.

Answer 2

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.