JEE Main 26 Feb 2021 Shift 2 Solved Paper

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)=‌

3kq12

5r

∴‌4∕3πr3‌=4∕3πr13×27
⇒‌r3‌=27r13
⇒‌r‌=3r1
where, k is Coulomb constant. Therefore, potential energy of small drop,
Us=‌

3

5

‌

kq12

r1

. . . (i)
and potential energy of bigger drop, UB=‌

3

5

‌

k(q12×27)2

r

UB=‌

3

5

‌

k(27)2q12

3r1

. . . (ii)
On dividing Eq. (ii) by Eq. (i), we get
‌

UB

Us

=‌

27×27

3

=243
‌UB=243Us
Hence, potential energy of big drop is 243 times of small drop.