INPhO 2000 if i amnt mistaken......dekho..when i first saw the Q. this is all that i cud figure out.....

assume that there is a sphere and we are depositing charge over it uniformly on its surfare.....

assume there to be a charge q already and we are adding a charge of dq......work done to do so will be

dW = Kqdq/R..........integrating from 0 to Q

and W = KQ²/2R

this is stored as potential energy of the system

so when u splitt into 2 equal spheres.i assume that the total volume is unaltered......so

4/3pi R³ = 2 * 4/3pi r³

or r= R/2^1/3......

and the energy of  first sphere = E = KQ²/R + TA₁........(A1 and A2 are the surface areas of the first and the second 2 spheres)(TA= energy due to surface tension)

where T= surface tension of water...(given in the value list and not used in any other problem)

and energy of the smaller sphere = U = KQ²/4(R/2^1/3) + TA₂

and for this change to be feasible we need E = 2U

equating which and solving we get Q and for 3 bubbles i guess u shud do it using r= R/3^1/3

and E = 3U....

third part main mere palle nahi padh rahe(dielectric break down)

but i have given a very brief method for the first 2 parts.....

correct if wrong....

rate if useful......