Questions
If n drops of mercury, each charged to a potential V, coalesce to form a single drop, the potential of the big drop will be
a
Vn2/3
b
Vn1/3
c
Vn1/3
d
Vn2/3
detailed solution
Correct option is D
If Q is the charge on each small drop, charge on the big drop is Q′=nQ. Now Q′=C′V′=Q=CV. Therefore V′V=Q′Q×CC′=nn1/3=n2/3Hence the correct choice is (4).Talk to our academic expert!
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