64 small drops of mercury, each of radius r and charge q coalesce to form a big drop the ratio of the surface density of charge of each small drop with that of the big drop is

Options

A. `64:1`
B. `1:64`
C. `1:4`
D. `4:1`

Answer

C

Explanation

Radius (R) of the bigger drop is given by
`(4)/(3)piR^(3)=64xx(4)/(3)pir^(3)impliesR=4r`.
charge on bigger drop `=64q`.
Surface density of charge of a small drop,
`sigma_(1)=(q)/(4pir^(2))`
Surface density of charge of the bigger drop,
`sigma_(2)=(64q)/(4piR^(2))=(64q)/(4pixx(4r)^(2))`
`therefore(sigma_(1))/(sigma_(2))=(q)/(4pir^(2))x(4pixx(4r)^(2))/(64q)=(1)/(4)`
`thereforesigma_(1):sigma_(2)=1:4`.