VITEEE 2012 Paper

Volume of 8 small drops = Volume of big drop
∴ (

4

3

πr3) × 8 =

4

3

πR3
⇒ 2r = R ...(i)
According to charge conservation
8q = Q ... (ii)
Potential of one small drop (V') =

q

4πε0r

Similarly, potential of big drop (V) =

Q

4πε0R

Now,

V′

V

=

q

Q

×

R

r

⇒

V′

20

=

9

8q

×

2r

r

∴ V' = 5V