Question

If 216 drops of the same size are charged at 200 V each and they combine to form a bigger drop, the potential of the bigger drop will be

• A. 2400 V
• B. 7200 V
• C. 1200 V
• D. 8200 V

Hint:

When smaller drops combine to form a larger drop, the potential increases in proportion to the size of the drop. The potential depends on the radius of the drop, and the relationship follows from the conservation of volume.

Answer 1

The Correct Option is B

When smaller drops combine to form a bigger drop, the volume remains constant, but the radius increases. The potential \( V \) of a drop is related to its radius \( r \) by the formula: \[ V \propto \frac{1}{r} \] The volume of a spherical drop is given by: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \] When 216 drops combine, their total volume is conserved. Therefore, the volume of the bigger drop will be 216 times the volume of one small drop. Since volume is proportional to \( r^3 \), the radius of the bigger drop will be: \[ r_{\text{big}} = 216^{1/3} r_{\text{small}} \]
Thus, the potential of the bigger drop will be: \[ V_{\text{big}} = 216^{1/3} V_{\text{small}} = 6 \times 200 = 1200 \, \text{V} \] The potential of the bigger drop is \( 7200 \, \text{V} \).