# 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. $4\pi T{D^2}$

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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 $= 4\pi {r^2}$.

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.