Question

A mercury drop of water has potential 'V' on its surface. $1000$ such drops combine to form a new drop. Find the potential on the surface of the new drop.

• A. V
• B. $10$V
• C. $100$V
• D. $1000$V

Answer 1

Answer: (C)

$100$V

Potential of single smaller Hg drop of water =$\frac{Kq}{r}$=V-----(1)

and we know potential of n smaller Hg drop of water $=\frac{Knq}{R}$------(2)

Volume of 1000 small drops = volume of larger drop

$1000\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

Thus radius of bigger droplet $R=10r$

Now putting the value of R in equation 2 we get

Potential of larger drop$=\frac{1000Kq}{10r}$=$100V$