Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5µC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:
Text Solution
Verified by ExpertsThe correct answer is:
B
Let R = radius of combined drop
r = radius of smaller drop
Volume will remain same
R = 4r
Q=64q;
q : charge of smaller drop
Q : Charge of combined drop
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