Physics - Fluid Question with Solution | TestHub

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

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

When all the drops forms the bigger drop the radius of the new drop $\left(r\text{'}\right)$ can be calculated as follows:

$\begin{array}{rcl}\mathrm{Volume} \mathrm{of} \mathrm{big} \mathrm{drop}& =& 64\times \mathrm{volume} \mathrm{of} \mathrm{each} \mathrm{tiny} \mathrm{drop}\\ & \Rightarrow & \frac{4}{3}\pi r{\text{'}}^3=64\times \frac{4}{3}\pi r^3\\ & \Rightarrow & r\text{'}=4r\end{array}$

So, the potential of the new big drop $\left(V\text{'}\right)$ can be written as

$\begin{array}{rcl}V\text{'}& =& k\frac{64q}{r\text{'}}\\ & =& k\frac{64q}{4r}\\ & =& 16k\frac{q}{r} ...\left(2\right)\end{array}$

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

$\begin{array}{rcl}\frac{V\text{'}}{V}& =& \frac{16k{\displaystyle \frac{q}{r}}}{k{\displaystyle \frac{q}{r}}}\\ & \Rightarrow & V\text{'}=16V ...\left(3\right)\end{array}$

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

$\begin{array}{rcl}V\text{'}& =& 16\times 10 \mathrm{mV}\\ & =& 160 \mathrm{mV}\end{array}$