JEE Class main Answered

If q is charge on the surface of mercury spherical drops of radius r , then potential on the surface of drop is

V = K × (q/r) = 10 Volt ( given )   ................................(1)

where K = 1/(4πε_o) is Coulomb's constant .

If 27 drops are combined together to form a large drop radius R , then new potential V_new is

V_new = K × [ (27q) / R ] ...............................(2)

Volume of large drop is given as

Hence we get,  R = 3r

If we substitute R = 3r in eqn.(2) , then we get

V_new = (27/3) × K × ( q/R ) = 9 × K × ( q/R ) .........................(3)

From eqn.(1) and (3) , we conclude that potential of large drop is 9 times the potential of small drop