Mixing Droplets

Assuming drops to be a spherical in shape.
Let charge on smaller drop be $\color{#3D99F6}{q_1}$ and charge on bigger drop be $\color{#D61F06}{q_2}$ .
On mixing all the small drops, $\color{#D61F06}{q_2} = 1000\color{#3D99F6}{q_1}$ .
Volume of 1000 drops (radius $\color{#3D99F6}{r_1}$ ) = Volume of bigger drop (radius $\color{#D61F06}{r_2}$ )
$\Rightarrow 1000 \times \dfrac{4}{3}\pi \color{#3D99F6}{r_1}^3 = \dfrac{4}{3}\pi \color{#D61F06}{r_2}^3$ $\Rightarrow 10 \color{#3D99F6}{r_1} = \color{#D61F06}{r_2}$

Therefore, radius of bigger drop is 10 times the radius of smaller drops.

As, $\Rightarrow V_1 = \dfrac{k \color{#3D99F6}{q_1}}{\color{#3D99F6}{r_1}} \quad \quad \quad \quad \quad \quad \quad (V_1 = \text{potential of smaller drop})$

$\Rightarrow V_2 = \dfrac{k \color{#D61F06}{q_2}}{\color{#D61F06}{r_2}} = \dfrac{1000k\color{#3D99F6}{q_1}}{10\color{#3D99F6}{r_1}} = 100V_1 \quad \quad (V_2 = \text{potential of bigger drop})$

Its a popular question and answer is= n^(2/3)* v. Putting n and v we get the answer.