Eight drops of mercury of equal radii and possessing equal charges combine to form a single big drop. The ratio of the capacitance of the big drop to the capacity of a single drop is

Options

A. `1:1`
B. `2:1`
C. `3:1`
D. `4:1`

Answer

B

Explanation

`(4)/(3)piR^(3)=8xx(4)/(3)pir^(3)" "therefore R=2r`
`therefore ("Capacitance of the big drop")/("Capacitance of the small drop")=(4piepsi_(0)R)/(4piepsi_(0)r)=(2)/(1)`