Mercury drop of radius 1cm is sprayed into ${10^6}$ drops of equal size. The energy expended in joule is (surface tension of mercury is $460 \times {10^{ - 3}}N/m$):
(A) $0.057$
(B) $5.7$
(C) $5.7 \times {10^{ - 4}}$
(D) $5.7 \times {10^{ - 6}}$

Hint
We can find the energy expended by the product of the surface tension and the total difference in surface area of the bigger drop and all of the smaller droplets, where the radius of the smaller droplets can be found by equating the volume in the 2 cases.
In this solution we will be using the following formula,
$\Rightarrow V = \dfrac{4}{3}\pi {R^3}$ where $V$ is the volume and $R$ is the radius of a sphere.
$\Rightarrow A = 4\pi {R^2}$ where $A$ is the area of a sphere
and $E = T \times \Delta A$ where $E$ is the energy expended
$\Delta A$ is the difference in area and $T$ is the surface tension.

Complete step by step answer
In the question it is said that 1 big drop divides into a number of smaller droplets. So the volume of mercury into the bigger drop gets divided into $n$ numbers of smaller droplets. Therefore we can write, the volume of the drop is equal to $n$ times the volume of a single droplet. That is,
$\Rightarrow \dfrac{4}{3}\pi {R^3} = n \times \dfrac{4}{3}\pi {r^3}$ where $R$ and $r$ are the radius of the larger and the smaller drops respectively.
Therefore we can cancel $\dfrac{4}{3}\pi $ from both sides of the equation, and get,
$\Rightarrow {R^3} = n \times {r^3}$
On taking cube roots on both the sides we get
$\Rightarrow R = {n^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}r$
$\Rightarrow r = {n^{{{ - 1} \mathord{\left/
 {\vphantom {{ - 1} 3}} \right.} 3}}}R$
Now there is a difference in the total surface area of the larger drop and all of the smaller drops as the sum of the surface area of all the droplets is more than the surface area of the drop.
Let us denote this difference by $\Delta A$ and is given by,
$\Rightarrow \Delta A = n \times 4\pi {r^2} - 4\pi {R^2}$
Taking $4\pi $ common and substituting $r = {n^{{{ - 1} \mathord{\left/
 {\vphantom {{ - 1} 3}} \right.} 3}}}R$
$\Rightarrow \Delta A = 4\pi \left[ {n{{\left( {{n^{{{ - 1} \mathord{\left/
 {\vphantom {{ - 1} 3}} \right.} 3}}}R} \right)}^2} - {R^2}} \right]$
Taking ${R^2}$ common,
$\Rightarrow \Delta A = 4\pi {R^2}\left[ {\dfrac{n}{{{n^{{2 \mathord{\left/
 {\vphantom {2 3}} \right.} 3}}}}} - 1} \right]$
Therefore we get the difference in area as,
$\Rightarrow \Delta A = 4\pi {R^2}\left[ {{n^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}} - 1} \right]$
Substituting the values of the radius and the number of droplets given in the question as, $R = 1cm = {10^{ - 2}}m$ and $n = {10^6}$ we get
$\Rightarrow \Delta A = 4\pi {\left( {{{10}^{ - 2}}} \right)^2}\left[ {{{\left( {{{10}^6}} \right)}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}} - 1} \right]$
Which is equal to
$\Rightarrow \Delta A = 4\pi {10^{ - 4}}\left[ {{{10}^2} - 1} \right]$
On calculating we get
$\Rightarrow \Delta A = 4\pi {10^{ - 4}} \times 99$
Therefore the difference in area is
$\Rightarrow \Delta A = 0.1244{m^2}$
The formula of energy expended in terms of surface tension is given by,
$\Rightarrow E = T \times \Delta A$
We can substitute the previously obtained difference in area $\Delta A = 0.1244{m^2}$ and tension $T = 460 \times {10^{ - 3}}N/m$ and get,
$\Rightarrow E = 460 \times {10^{ - 3}} \times 0.1244$
This gives us the energy expended as,
$\Rightarrow E = 0.057J$
Therefore, the correct answer will be option (A); $0.057$

Note
We can also solve this problem alternatively by the direct formula of the energy expended given by,
$\Rightarrow W = 4\pi {R^2}\left( {{n^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}} - 1} \right)T$
So by substituting the values of the surface tension, radius and the number of smaller droplets we will directly get the answer.