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**Hint:**First we need to know the formula for surface energy of liquid drop. We need to represent the final surface energy and the initial surface energy. After that, we need to find a relationship between the radius of bigger drop and radius of small droplets, then we need to put that relationship into the equation for final surface energy. After equating we will get our required result.

**Formula used:**

$S{{E}_{i}}=T\times A$

$\dfrac{4}{3}\pi {{R}^{3}}=1000\times \dfrac{4}{3}\pi {{r}^{3}}$

**Complete step by step answer:**

The initial surface energy of the liquid drop is E

$S{{E}_{i}}=T\times A$ ,

Now let us consider surface tension T, as $\sigma $

So,

$S{{E}_{i}}=\sigma \times 4\pi {{R}^{2}}$, R represents the radius of the bigger drop.

$E=\sigma \times 4\pi {{R}^{2}}$.

Now, it is sprayed into 1000 equal droplets, by this statement it means that the droplets now have equal radius

Therefore the final surface energy for 1 droplet must be,

$S{{E}_{f}}=\sigma \times 4\pi {{r}^{2}}$, r is the radius of the small drop.

Now, for 1000 droplets,

$S{{E}_{f}}=\sigma \times 4\pi {{r}^{2}}\times 1000$,

Now we have to find a relation between ‘r’ and ‘R’, we generally do it by equating the volume of both the bigger with the smaller droplets, so

Volume of bigger drop = volume of smaller droplets

$\dfrac{4}{3}\pi {{R}^{3}}=1000\times \dfrac{4}{3}\pi {{r}^{3}}$ ,

On equating the above solution, we get the relation as,

$r=\dfrac{R}{10}$ ,

Now, final surface energy is

$S{{E}_{f}}=\sigma \times 4\pi {{r}^{2}}\times 1000$,

$S{{E}_{f}}=\sigma \times 4\pi {{\left( \dfrac{R}{10} \right)}^{2}}\times 1000$, on substituting the value $r=\dfrac{R}{10}$.

Now, we can write

$S{{E}_{f}}=\sigma \times 4\pi {{R}^{2}}\times \dfrac{1000}{100}$

$S{{E}_{f}}=E\times 10$, as ($E=\sigma \times 4\pi {{R}^{2}}$)

So, when a single droplet with surface energy E is sprayed into 1000 droplets, the final surface energy becomes 10E.

**Hence the correct option is option B.**

**Note:**

In the formula for surface energy, $S{{E}_{i}}=T\times A$, here ‘T’ is the surface tension and ‘A’ is the area of the drop. We are equating the volume for the relation between ‘r’ and ‘R’ because when the drop is sprayed into 1000 droplets everything changes the radius, area and all the required elements but the volume always remains constant hence volume is equated.

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