Twenty-seven drops of mercury are charged simultaneously to the same potential of 10 volts. what will be potential if all the charged drops are made to combine to form one large drop ?

 Let us assume the potential of each small drop = Kq/r = V

The volume of n (here 27 small drops) drops = Volume of one big drop.

which implies:

n * 4/3 pi r³ = 4/3 pi R³

here r is the radius of a small drop 

R is the radius of the big drop.

Therefore , R = n⅓ * r 

here charge remains conserved.

charge on 1 small drop = q and charge on big drop containing n drops = nq .

The potential on big drop = K (nq)/ (n⅓)r

= (n⅔) Kq/r

= (n⅔) V

Now substitute the n with the number of drops you wish to combine and do the math.

Potential of each small drop = 10V.

No. of drops: 27.

Potential on big drop = 10*K(27q)/(27⅓)r.

=(10 * 27)/3 Kq/r

=(10 * 27)/3 V

= 270/3

= 90 V