Question

Eight drops of mercury of same radius and having same charge collide to form a big drop. Capacitance of big drop relative to that of small drop will be :

• A. $16$ times
• B. $8$ times
• C. $4$ times
• D. $2$ times

Answer 1

The correct option is D $2$ times

Let the radius of the big drop is $R$ and the radius of the small drop is $r$.

The volume is given as,

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

$R=2r$

The capacitance of small drop is given as,

$C_s=4\pi \epsilon r$

The capacitance of big drop is given as,

$C_b=4\pi \epsilon R$

$C_b=4\pi \epsilon \left(2r\right)$

$C_b=2C_s$

Thus, the capacitance of the big drop relative to small drop is $2$ times.