Given,
n=27 V1=10V Let
q1 be the charge of one drop,
r1 be its radius,
r be the radius of bigger drop and
q be its charge.
As, volume remains constant.
‌ Electric potential ‌(V)=‌ ∴‌4∕3πr3‌=4∕3πr13×27 ⇒‌r3‌=27r13 ⇒‌r‌=3r1 where,
k is Coulomb constant. Therefore, potential energy of small drop,
Us=‌‌ . . . (i)
and potential energy of bigger drop,
UB=‌‌ UB=‌‌. . . (ii)
On dividing Eq. (ii) by Eq. (i), we get
‌=‌=243 ‌UB=243Us Hence, potential energy of big drop is 243 times of small drop.