Question

The work done to split a liquid drop of radius R into N identical drops is (take σ as the surface tension of the liquid:-
\[\begin{align}
  & A)4\pi {{R}^{2}}({{N}^{^{\dfrac{1}{3}}}}-1)\sigma \\
 & B)4\pi {{R}^{2}}N\sigma \\
 & C)4\pi {{R}^{2}}({{N}^{^{\dfrac{1}{2}}}}-1) \\
 & D)None \\
\end{align}\]

Answer 1

Hint: In this type of question whenever you see that large liquid drop gets split into N number of small liquid drops then first we to equalize the initial volume of drop and final volume of drop so that we can get relation in radius between them then we have apply the relation of work done with surface tension.

Complete answer:
Let us assume the radius of each small drop is r.
When a liquid drop of radius R is split into N identical small drops of radius r then volume remains the same before and after the splitting of liquid drop.
So we can say that,
Volume of a liquid drop of radius R is equal to Volume of N identical small drops of radius r.
Since all drops are spherical on the surface. So formula for volume of sphere is used.
\[\dfrac{4}{3}\pi {{R}^{3}}=N\times \dfrac{4}{3}\pi {{r}^{3}}\]
\[\Rightarrow {{R}^{3}}=N{{r}^{3}}\]
Cube root on both sides, we get
\[\therefore R={{N}^{\dfrac{1}{3}}}r\]
In terms of radius r this formula can be written as,
\[\therefore r={{N}^{\dfrac{-1}{3}}}R\](Equation 1)
so we get the relation between radius R of liquid drop and radius r of liquid drop.
Since, Work done in splitting the liquid drop = Change in area of liquid drop x Surface Tension.
Let us assume work done in splitting the liquid drop is represented by W.
Let us assume change in area is represented by\[\Delta A\].
Surface tension is represented by σ as it is given in the question.
So work done can be represented by following relation: -
\[W=\Delta A\times \sigma \] (Equation 2).
First we calculate change in area-
\[\Delta A\]= Final area of N small drops of radius r – Initial area of big drop of radius R
\[\Delta A=N\times 4\pi {{r}^{2}}-4\pi {{R}^{2}}\]
Put value of r from equation 1 in this equation, we get
\[\Rightarrow \Delta A=N\times 4\pi {{({{N}^{\dfrac{-1}{3}}}R)}^{2}}-4\pi {{R}^{2}}\]
\[\begin{align}
  & \Rightarrow \Delta A=N\times 4\pi {{N}^{\dfrac{-2}{3}}}{{R}^{2}}-4\pi {{R}^{2}} \\
 & \therefore \Delta A=4\pi {{R}^{2}}({{N}^{\dfrac{1}{3}}}-1) \\
\end{align}\]
This is the required value of change in area of liquid drop.
Put this value of change in area in equation 2, we get
\[\therefore \]\[W=4\pi {{R}^{2}}({{N}^{\dfrac{1}{3}}}-1)\sigma \].

So Correct Option is A.

Note:
The cohesive forces acting between same molecules i.e. between water molecules or between any liquid molecules exerting the phenomenon of surface tension. While adhesive forces acting between different molecules. The molecules at the surface try to minimize their area so they can become stable.