8 drops of Hg are combined to form a single big drop. What is the ratio of the capacitance of a single small drop and that of the single big drop?

Options

A. `1:2`
B. `1:8`
C. `8:1`
D. None of these

Answer

A

Explanation

`(4)/(3)piR^(3)=8((4)/(3)pir^(3))`
`therefore R^(3)=8r^(3)" "therefore R=2r`
Capacitance of a small drop`=4piepsi_(0)(r)=C_(1)`
Capacitance of a single big drop `=4piepsi_0(2r)=C_(2)`
`therefore (C_(1))/(C_(2))=(r)/(2r)=(1)/(2)`