Sixty four equal charged drops are combined to form a big drop. If the potential on each drop is 10 volt, then potential on big drop will be...

Dear student,

64 4/3 pie r3=4/3 pie R3

R=4r

Similarly we can find the charge on bigger sphere.

Hence potential on bid srop will be 16* 10=160 V

let the radius of each small drop is r ...

let radius of drop formed after combination of all drops is R..

in this process volume remains same so,

       initial  volume=64(4/3xpi(r)³)............1

       final volume=4/3pi(R)³ ...................2

equating 1 and 2 we get

            R=4r                     ..............3

now potential of small drop is given 10 volts

potential of small drop =V_s=kq/r      (charge of each drop is q)

now potential of big drop=V_b=kQ/R

Q=total charge=64q and R=4r

now V_b=64kq/4r=16kq/r

                     V_b =16V_s

                         =16x10=160volts

THE POTENTIAL OF THE BIGGER DROP IS 160 VOLT................  USE THE FORMULA N^2/3V

APPROVE MY ANSWER...............

For one drop V=(4/3)*(3.14)*r^3

For big one V=64*(4/3)*(3.14)*r^3=(4/3)*(3.14)*(r')^3

                       =r'/r=4

        therefore r'=4*r

Potential due to one drop=V=K*q/r

For big drop= V'=K*64*q/r'

                        =K*64*q/(4*r)

                        =16*V

                        =16*10

                        =160 V

This is what I think.