Question

Eight drops of mercury, each of the same radius and same charge, combine to form a bigger drop. The ratio of the capacitance of the bigger drop to that of each smaller drop is:

• A. \( 8:1 \)
• B. \( 4:1 \)
• C. \( 2:1 \)
• D. \( 1:1 \)

Hint:

When drops combine, their volume is conserved, and the radius of the bigger drop is the cube root of the sum of the volumes. The capacitance is directly proportional to the radius.

Answer 1

The Correct Option is C

The capacitance of a spherical drop is proportional to its radius: \[ C \propto r \] When eight smaller drops combine to form a bigger drop, the radius of the bigger drop is \( r_{\text{big}} = 2r_{\text{small}} \). Thus, the capacitance of the bigger drop is: \[ C_{\text{big}} = 2C_{\text{small}} \] The ratio of the capacitance of the bigger drop to that of the smaller drop is: \[ \frac{C_{\text{big}}}{C_{\text{small}}} = 2 \]