Question

$27$ identical drops of mercury are charged simultaneously with the same potential of $10$ Volt. Assuming the drop to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be _____ Volt

• A. $90$
• B. $40$
• C. $160$
• D. $10$

Answer 1

The correct option is A $90$

Let the charge on one smaller drop be $q$.
Potential at each small drop $V=\frac{Kq}{r}$
where $r$ is the radius of each small drop and $V$ is $10$ volts.
$\therefore$ $10=\frac{Kq}{r}$
We get $q=\frac{10r}{K}$
Let radius of bigger drop be $R$.
Volume of 27 small drops is equal to volume of one bigger drop.
$\therefore$ $\frac{4\pi }{3}{R}^3=27\times \frac{4\pi }{3}{r}^3$
$⟹R=3r$
Total charge on bigger drop $Q=27q=\frac{270r}{K}$
Potential of bigger drop $V_b=\frac{KQ}{R}$
$⟹V_b=\frac{K\times \frac{270r}{K}}{3r}=90$ Volts