Question

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

• A. 8 times
• B. 4 times
• C. 2 times
• D. 32 times

Answer 1

Answer: (C)

2 times

Answer 2

Volume of 8 small drops = Volume of big drop

$8 \times \frac { 4 } { 3 } \pi r ^ { 3 } = \frac { 4 } { 3 } \pi R ^ { 3 }$ ⇒ R = 2r

As capacity is r, hence capacity becomes 2 times.