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
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a
8 times
b
4 times
c
2 times
d
32 times
answer is C.
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Detailed Solution
Volume of 8 small drops = Volume of big drop8×43πr3=43πR3⇒R=2rAs capacity is proportional to r, hence capacity becomes 2 times.