64 identical drops each charged upto potential of 10mV are combined...

The formula to calculate the potential $\left(V\right)$ of each tiny drop is given by

$V = k\dfrac{q}{r} . . . \left(1\right)$

When all the drops forms the bigger drop the radius of the new drop \left(r '\right)$can be calculated as follows:</p><p>$\begin{cases}Volumeofbigdrop & = & 64 \times volumeofeachtinydrop \\ ⇒ & \dfrac{4}{3}\pir{'}^{3} = 64 \times \dfrac{4}{3}\pi{r}^{3} \\ ⇒ & r ' = 4r\end{cases}$</p><p>So, the potential of the new big drop \left(V '\right)$ can be written as

$\begin{cases}V ' & = & k\dfrac{64q}{r '} \\ = & k\dfrac{64q}{4r} \\ = & 16k\dfrac{q}{r} . . . \left(2\right)\end{cases}$

Divide equation (2) by equation (1) and simplify to obtain the potential of the new drop.

$\begin{cases}\dfrac{V '}{V} & = & \dfrac{16k\dfrac{q}{r}}{k\dfrac{q}{r}} \\ ⇒ & V ' = 16V . . . \left(3\right)\end{cases}$

Substitute the value of the known parameter into equation (3) to calculate the required value of the potential for the new drop.

$\begin{cases}V ' & = & 16 \times 10mV \\ = & 160mV\end{cases}$