1
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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
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B
8 times
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C
4 times
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D
2 times
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Solution

The correct option is D 2 timesLet the radius of the big drop is R and the radius of the small drop is r. The volume is given as, 43πR3=8×43πr3 R=2r The capacitance of small drop is given as, Cs=4πεr The capacitance of big drop is given as, Cb=4πεR Cb=4πε(2r) Cb=2Cs Thus, the capacitance of the big drop relative to small drop is 2 times.

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