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Q.

Eight drops of mercury, each charged to potential V, coalesce to form a single drop. Then the potential of the drop is

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a

3.5 V

b

3 V

c

2 V

d

4 V

answer is D.

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Detailed Solution

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$V = \frac{1}{4 π \in_{0}} . \frac{q}{r} \Rightarrow q = 4 π \in_{0} r V$

Now, $\frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{3} \Rightarrow R = 2 r$

$∴ V^{1} = \frac{1}{4 π \in_{0}} \frac{Q}{R} = \frac{1}{4 π \in_{0}} \frac{8 \times 4 π \in_{0} r V}{2 r} = 4 V$

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