Answer

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**Hint:**Here we have to use the concept that the total volume and total charge of single drop would be the same as combined volume and charges of 27 drops. Then we can find the potential of the bigger drop using the formula. $V = \dfrac{{kq}}{r}$

**Complete step by step answer:**

Let us assume the charge on one small drop be q. Let the radius be r. Then, the potential of smaller drop, V is

$V = \dfrac{{kq}}{r}$

Now, we are given that the potential of each small drop is 10 volts.

$

10 = \dfrac{{kq}}{r} \\

\Rightarrow q = \dfrac{{10r}}{k} \\

$

Let the radius of the bigger drop be R. Now, because all the smaller drops are combined , the volume of the bigger drop would be the same as the total volume of smaller drops.

$

\dfrac{{4\pi }}{3}{R^3} = 27 \times \dfrac{{4\pi }}{3}{r^3} \\

\Rightarrow R = 3r \\

$

Let the total charge on bigger drop be $Q = 27q = \dfrac{{270r}}{K}$

Now, the potential of bigger drop, $V'$ can be written as,

$

V' = \dfrac{{kQ}}{R} \\

\therefore V' = \dfrac{{k \times \dfrac{{270r}}{k}}}{{3r}} = 90V \\

$

**Therefore, the correct answer is 90 volts.**

**Additional information:**

The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as the amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration. When the body is charged, either electric electrons are supplied to it, or they are removed from it. In both the cases, the work is done. This work is stored in the body in the form of electric potential. Thus, the body can do the work by exerting a force of attraction or repulsion on the other charged particles.

**Note:**We can assume the spheres to be point charges while calculating their potential as the potential of a sphere is much more complicated. That is the key point in solving this question, also we assume that there is no loss of liquid while combining the drops.

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