27 small drops each having charge q and radius r coalesce to form big drop. How many times charge and capacitance respectively will become?

Options

A. 3 and 27
B. 27 and 3
C. 27 and 27
D. 3 and 3

Answer

B

Explanation

Let R and r be the radii of bigger and each smaller drop.
Charge remains conserved.
Hence, charge on bigger drop
`27xx` charge on smaller drop
i.e. `q'=27q`
Now, before and after coalescing, volume remains same. That is,
`(4)/(3) pi R^(3) =27 xx (4)/(3) pi r^(3)`
`therefore R=3r`
Hence, capacitance of bigger drop
`C'=4pi epsilon_(0)R=4pi epsilon_(0)(3r)`
`=3(4pi epsilon_(0)r)=3C`