What is the ratio of the surface energy of one small drop and one large drop, if $1000$ small drops combine to form one large drop?
$1)100:1$
$2)1000:1$
$3)10:1$
$4)1:100$

Hint: We need to find the ratio of the surface energy of the one small drop to the surface energy of the combined large drop. We know that the volume of the small drops combined will be equal to the volume of the large drop and we need to use that concept to calculate the required ratio.

Complete step by step solution:
Let the radius of the small drop be $r$.
Now the radius of the big drop formed by the combination of the $1000$ small drops will be equal to $1000$ times the radius of the small drops.
Let the radius of the large drop be $R$
Therefore we have $R = 1000r$
Now the volume of the small drop will be $\dfrac{4}{3}\pi {r^3}$
And the volume of the large drop will be $\dfrac{4}{3}\pi {R^3}$
Now equating the values of volume of large drop and the volume of $1000$ small drops we get
$1000\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$
$ \Rightarrow 1000{r^3} = {R^3}$
$ \Rightarrow R = 10r$
$ \therefore \dfrac{r}{R} = \dfrac{1}{{10}}$
This gives us the ratio of the radius of the small drop to the radius of the large drop.
Now the ratio of the surface energy of one small drop to the surface energy of one large drop gives us
$\dfrac{{4\pi {r^2}T}}{{4\pi {R^2}T}} = \dfrac{1}{{100}}$
Where $T$ is the surface tension working on the small drop.
This gives us the required ratio.
Hence option 4) is the correct option.

Note: Note that the surface energy is the net energy of the bonds that are between the molecules at the surface and the surface tension is the net intermolecular force present on the molecules at the surface. Both are two different things and should not be used interchangeably.