A liquid drop of diameter \[D\] breaks up into \[27\] drops. Find the resultant change in energy. Surface tension of the liquid is \[T\].
a. \[ 2\pi T{D^2}\]
b. \[ \pi T{D^2}\]
c. \[ \dfrac{{\pi T{D^2}}}{2}\]
d. \[ 4\pi T{D^2}\]

Hint: First we have to find the total volume of large drop, so we have to construct the volume of small drop and volume of \[27\] drops is multiplied. Then we have to equate that and by doing some simplification we get the required answer.

Formula Used:
Change in energy \[ = \]\[(\]Change in Area \[ \times \] Surface Tension \[)\]
Volume of a sphere is \[ = \dfrac{4}{3}\pi {r^3}\], here \[r\] is the radius of the given sphere.
Area of a sphere is \[ = 4\pi {r^2}\].

Complete step by step answer:
A liquid drop is like a sphere.
So, the volume of the drop will be equal to the volume of the sphere.
Now we have to use the formula for volume of a sphere is \[ = \dfrac{4}{3}\pi {r^3}\], here \[r\] is the radius of the given sphere.
Let's say, the radius of each small drop is\[d\]unit.
The diameter of the large drop is \[ = D\] .
Radius of the large drop is \[ = \dfrac{D}{2}\] .
Volume of the large drop is \[ = \dfrac{4}{3}\pi {\left( {\dfrac{D}{2}} \right)^3}\] .
Volume of each small drop is \[ = \dfrac{4}{3}\pi {d^3}\].
Total volume of \[27\] drops will be\[ = \dfrac{4}{3}\pi {d^3} \times 27 = 9 \times 4\pi {d^3} = 36\pi {d^3}\] .
So, the total volume of \[27\]drops will be equal to the total volume of the large drop.

So, we derive the following equation:
\[ \Rightarrow 36\pi {d^3} = \dfrac{4}{3}\pi {\left( {\dfrac{D}{2}} \right)^3}\]
By simplifying the cubic form we get:
\[ \Rightarrow 36\pi {d^3} = \dfrac{4}{3}\pi \left( {\dfrac{{{D^3}}}{8}} \right)\]
Calculate the denominator parts in R.H.S:
\[ \Rightarrow 36\pi {d^3} = \pi \left( {\dfrac{{{D^3}}}{6}} \right)\]
Cancel out the \[\pi \]from both sides:
 \[ \Rightarrow 36{d^3} = \left( {\dfrac{{{D^3}}}{6}} \right)\]
Divide R.H.S by \[36\]we get:
\[ \Rightarrow {d^3} = \left( {\dfrac{{{D^3}}}{6}} \right) \times \dfrac{1}{{36}}\]

Simplifying the above form:
\[ \Rightarrow {d^3} = \left( {\dfrac{{{D^3}}}{{{6^3}}}} \right)\]
Applying the rule of indices:
\[ \Rightarrow d = \left( {\dfrac{D}{6}} \right)\]
Initial area of the large liquid was \[ = 4\pi {\left( {\dfrac{D}{2}} \right)^2} = \pi {D^2}\] .
After division in \[27\] drops, the total area of all these drops becomes \[ = 27 \times 4\pi {d^2}\] .
But \[d = \left( {\dfrac{D}{6}} \right)\].
The changed area will be \[ = 27 \times 4\pi {\left( {\dfrac{D}{6}} \right)^2} = 3\pi {D^2}\] .
Change in total area will be \[ = (3\pi {D^2} - \pi {D^2}) = 2\pi {D^2}\] .
Resultant change in energy will be \[ = \]( Change in area \[ \times \] Surface Tension of the liquid) .
So, change in energy \[ = (2\pi {D^2} \times T) = 2\pi T{D^2}\].

Hence, the correct answer is option (A).

Note: A surface tension is the tendency of minimums possible and the liquid surfaces to shrink into it. So, change in energy is calculated alongside the change in area of liquid.