`n` small drops of same size are charged to `V` volts each .If they coalesce to from a single large drop, then its potential will be -

Options

A. V n
B. `V n^(2)`
C. `V//n`
D. `Vn^(2//3)`

Answer

D

Explanation

Let q be charge on each drop of radius r.
Thus, pot. Of each drop `V=q//r" "therefore q=Vr`
Now charge on n drops = nq = n V r = Q
Now volume of n drops = vol. of big drop
`therefore4/3pir^(3)n=4/3piR^(3)" "thereforeR=rn^(1//3)`
Thus,
pot. of big drop = `Q/R=(nVr)/(rn^(1//3))=Vn^(2//3)`