N identical drops of mercury are charged simultaneously to 10 V. When combined to form one large drop, the potential is found to be 40 V, the value of N is:

Options

A. 4
B. 6
C. 8
D. 10

Answer

D

Explanation

Volume of bigger drop = N `xx` volume of smaller drop
or `(4)/(3)pi R^(3) = N xx (4)/(3) pi r^(3)`
or `N = ((R )/(r ))^(3)`
Charge is conserved. Hence Q = Nq
Capacity of bigger drop `= 4pi epsilon_(0)R`
Capacity of smaller drop `= 4pi epsilon_(0)r`
`:. (4pi epsilon_(0)R) xx 40 - N(4pi epsilon_(0)r) xx 10`
or `4R = Nr` or `(R )/(r ) = (N)/(4)`
from (i) and (ii), we get `N = ((N)/(4))^(3) = (N^(3))/(64)`
or, `N^(2) = 64` or N = 8