91Ó°ÊÓ

First, we will find the radius of the drop then after merging two drops, we will find the radius of new drop.

The new drop radius can be calculated by comparing the volume of new drop with the two mercury drops, and at last we can find the potential at the surface of the new drop.

Given:

Charge of each drop, $q=0.10\mathrm{nC}=0.10×{10}^{-9}\mathrm{C}$

Potential at the surface, $V=300\mathrm{V}$

Formula used:

The potential is calculated by the formula

$V=K\frac{q}{r}$

Or, radius of the new drop is calculated as

$r=K\frac{4}{V}$

Where

" V " is the potential at the surface

${}^{\mathrm{Î\pm }}K=9×{10}^9\mathrm{N}-{\mathrm{m}}^2/{\mathrm{C}}^{2"}$ is the electric constant

" q " is the charge of the drop

" r " is the radius of the drop

Volume of a drop is calculated by the formula

$vol=\frac{4}{3}\mathrm{Ï€}{r}^3$

Where "$r^n$is the radius of the drop

The radius of the drop

$r=K\frac{q}{v}$

Plugging the values in the above equation

$$\begin{gathered} r=9×{10}^9\frac{0.10×{10}^{-7}}{300} \\ =0.003\mathrm{m} \end{gathered}$$

Volume of new drop =$2×volumeofthedrop$(before merging)

$$\begin{gathered} \frac{4}{3}\mathrm{π}{\left(r_{\text{aeN}}\right)}^3=2×\frac{4}{3}\mathrm{π}{r}^3 \\ {r}_{\text{new}}=\sqrt[3]{2}×r \end{gathered}$$

By putting the value of " r " from equation (1)

$$\begin{gathered} r_{\text{new}}=\sqrt[3]{2}×3×{10}^{-3} \\ =0.0037\mathrm{m} \end{gathered}$$

Now, we can calculate the potential of the new drop, but in this for two drops the charge will be double so instead of$"q""2q^{\text{'}\text{'}}$will be there

$V=K\frac{2\mathrm{q}}{r}$

Plugging the values in the above equation