If n identical drops of mercury are combined to form a bigger drop then find the capacity of bigger drop, if capacity of each drop of mercury is C.

Options

A. `n^(1//3) C`
B. `n^(2//3)C`
C. `n^(1//4)C`
D. nC

Answer

C

Explanation

Let a radius of small drop = r
Volume of n drops `= n(4)/(3)pi r^(3)`
or `R = n^(1//3) r :. C = 4pi epsilon_(0)r`
`C. = 4pi epsilon_(0)R`
Hence, `C. = Cn^(1//3)`