N identical dropsof mercury are charged sumultaneously 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. A)4
B. B)6
C. C)8
D. D)10

Answer

C

Explanation

After combining, the volume remains same i.e.,
volume of bigger drop = `Nxx` Volume of smaller drop
or `(4)/(3)piR^(3)=Nxx(4)/(3)pir^(3)`
or `N=((R)/(r))^(3)...(i)`
As charge is conserved, hence
`Q=Nq…(ii)`
Capacity of bigger drop = `4piepsilon_(0)R`
Capacity of smaller drop = `4piepsilon_(o)r`
From Eq. (ii), we have
`(4piepsilon_(0)R)V_("big")=N(4piepsilon_(0)r)V_("small")`
or `(4piepsilon_(0)R)xx40=N(4piepsilon_(0)r)xx10`
or `4R=Nr`
`(R)/(r)=(N)/(4)...(iii)`
From Eqs. (i) and (iii),
`N=((N)/(4))^(3)`
or `N=(N^(3))/(64)`
or `N^(2)=64`
or `N=8`