If n drops, each of capacitance C and charged to a potential V, coalesce to form a big drop, the ratio of the energy stored in the big drop to that in each small drop will be

Options

A. `n^(5//3):1`
B. `n^(4//3):1`
C. n:1
D. `n^(3) : 1`

Answer

A

Explanation

Let r be radius of each small drop and R be radius of bigger drop
As the volume remains constant
`:. (4)/(3) pi R^(3) = n xx (4)/(3) pi r^(3)`
`R = n^(1//3) r`
Capacitance of each small drop, `C = 4pi epsilon_(0)r`
Capacitance of bigger drop, `C. = 4pi epsilon_(0)R = 4pi epsilon_(0)n^(1//3)r = n^(1//3)C`
Charge on each small drop, Q = CV
Charge on bigger drop, Q. = nQ
Potential of bigger drop, `V. = (Q.)/(C.) = (nQ)/(n^(1//3)C) = n^(2//3)V`
Energy stored in each small drop, `U = (1)/(2)CV^(2)`
Energy stored in bigger drop, `U. = (1)/(2) C.V.^(2)`
`= (1)/(2) n^(1//3) C(n^(2//3)V)^(2) = (1)/(2)n^(5//3)CV^(2)`
`:. (U.)/(U) = (1)/(2)n^(5//3) CV^(2) xx (2)/(CV^(2)) = (n^(5//3))/(1)`