Question

Twenty seven drops of the same size are charged to a potential of $220\text{V}$ each. They combine to form a bigger drop. The potential of the bigger drop is-

• A. $660\text{V}$
• B. $1320\text{V}$
• C. $1520\text{V}$
• D. $1980\text{V}$

Answer 1

The correct option is D $1980\text{V}$

Given, potential of each smaller drop is, $220\text{V}$

The potential of each small drop is,

$V_s=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

$\Rightarrow 220=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}.....\left(1\right)$

In this process volume remains the same,

$27V_{small}=V_{big}$

$27\times \left(\frac{4}{3}\pi r^3\right)=\left(\frac{4}{3}\pi R^3\right)$

$\Rightarrow R=3r$

Potential of the bigger drop is,

$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}=\frac{nq}{4\pi {\epsilon }_0\left(3r\right)}$

On dividing (2) $÷$ (1)

$\frac{V}{220}=\frac{27}{3}$

$\Rightarrow V=9\times 220=1980\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.