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

To solve this problem, we first need to understand the relationship between the volume of a drop and its capacitance. The capacitance of a spherical conductor is given by:

\(C = 4 \pi \varepsilon R\) 

where \(R\) is the radius of the sphere, and \(\varepsilon\) is the permittivity of the medium.

Let's proceed step-by-step:

1. The volume of a single small drop of mercury can be written as: \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius of each small drop.
2. For eight such identical drops, the total volume becomes: \(V_{\text{total}} = 8 \left( \frac{4}{3} \pi r^3 \right)\).
3. These eight drops combine to form a larger drop whose volume is equal to the total volume of the eight smaller drops. Thus, the volume of the large drop is: \(V_{\text{large}} = \frac{4}{3} \pi R^3\), where \(R\) is the radius of the large drop.
4. Equating the total volume of the small drops to the volume of the large drop: \(8 \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi R^3\).

 

1. Cancel out the common terms and solve for \(R\): \(R^3 = 8r^3\), which gives \(R = 2r\).
2. Recalculate the capacitance. The capacitance of the larger drop is: \(C_{\text{large}} = 4 \pi \varepsilon R = 4 \pi \varepsilon (2r)\) \(= 2 \times 4 \pi \varepsilon r = 2C_{\text{small}}\).

Hence, the capacitance of the bigger drop is 2 times that of each smaller drop. The correct answer is "2 times".