Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop

The correct option is 1980 V.

Let the radius of each small drop be $r$ and the charge on each small drop be $q$.
The electrostatic potential ($V$) of a single small drop is given by the formula:
$V=\frac{kq}{r}$
We are given that the potential of each small drop is $V=220\text{V}$.

When 27 identical small drops combine to form a single larger drop, the total volume remains conserved.
Let $R$ be the radius of the larger drop.
Volume of the larger drop = 27 × Volume of a small drop
$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$
Simplifying the expression by dividing both sides by $\frac{4}{3}\pi$:
$R^3=27r^3$
Taking the cube root on both sides:
$R=3r$

By the principle of conservation of charge, the total charge on the larger drop ($Q$) is the sum of the charges on all 27 small drops:
$Q=27q$

The potential of the larger drop ($V^{\prime }$) is given by:
$V^{\prime }=\frac{kQ}{R}$
Substituting $Q=27q$ and $R=3r$ into the potential equation:
$V^{\prime }=\frac{k\left(27q\right)}{3r}$
$V^{\prime }=9\times \left(\frac{kq}{r}\right)$
Since $\frac{kq}{r}=V=220\text{V}$:
$V^{\prime }=9\times 220\text{V}$
$V^{\prime }=1980\text{V}$