Question

In identical mercury droplets charged to the same potential V coalesce to form a single bigger drop. The potential of the new drop will be:
A. $\dfrac{V}{n}$
B. $nV$
C. $N{V^2}$
D. ${n^{\dfrac{2}{3}}}V$

Answer 1

In this question, we have to find the potential of the new drop. First, we have to find the volume of the one drop and then the volume of n drops. Now, compare the two equations that is the charge on n drops and charge on one drop. Equate both equations to get the further result. Using the formula of potential, we get the potential of new drop.

Complete step by step answer:
First, we have to find the volume of the small drop is given as,
${V_1} = \dfrac{4}{3}\pi {r^3}$
Here, $r$ is the radius of the small drops.
Now, we have to find the volume of the n drop can is given as,
${V_2} = n\dfrac{4}{3}\pi {R^3}$
Here, $R$ is the radius of the big drops.

Now, equate the volume of small drop with the volume of n drop we get,
$\begin{array}{l} \dfrac{4}{3}\pi {r^3} = n\dfrac{4}{3}\pi {R^3}\\\ R = {n^{\dfrac{1}{3}}}r \end{array}$
At the time of coalesce the charge remains conserved because there is only volume increases not charge.
Therefore, the charge on one drop is $q$.
Similarly, the charge on $n$ bigger drop is $nq$
We know that, the potential on one drop is given as,
$V = \dfrac{{Kq}}{r}$
Therefore, the potential on $n$ big drop is given as,
${V_n} = k\dfrac{{nq}}{R}$
Substitute the value of $R$ in the above equation we get,
${V_n} = k\dfrac{{nq}}{{{n^{\dfrac{1}{3}}}r}}\\\ \implies {V_n} = {n^{\dfrac{2}{3}}}V$
Thus, the potential on $n$ big drop is ${n^{\dfrac{2}{3}}}V$.

So, the correct answer is “Option D.

Note:
In this question, students must have knowledge of the expression of potential. We have to know that the charge remains conserved and cannot change during the coalesce of a bigger drop only volume of the drop changes.