First slide

Pontial due to

Question

A spherical drop of mercury of diameter 8 cm is charged to a potential of 12 volt. The drop is broken into 8 equal drops, then potential of each drop is

Moderate

A

4 volt
B

6 volt
C

1.5 volt
D

3 volt

Solution

$8 \times \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times {(\frac{8}{2})}^{3} \Rightarrow r = 2 cm$
If Q be the charge on the larger drop, then charge on each smaller drop is Q/8.
Now,
$\begin{array}{l} \frac{1}{4 π \in_{o}} . \frac{Q}{4} = 12 \Rightarrow Q = 48 \times 4 π \in_{o} coul . \\ ∴ V' = \frac{1}{4 π \in_{o}} \times \frac{Q / 8}{2} = \frac{1}{16} \times \frac{48 \times 4 π \in_{o}}{4 π \in_{o}} = 3 volt \end{array}$

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