Question

If T is the surface tension of a fluid, then the energy needed to break a liquid drop of radius R into 64 equal drops is:

(A) $6\pi {R^2}T$
(B) $\pi {R^2}T$
(C) $12\pi {R^2}T$
(D) $8\pi {R^2}T$

Answer 1

Using the formula of the energy in terms of the surface tension we get the value of the energy in terms of the surface tension for the big drop and the small individual drops and then we can calculate the value of the energy required in terms of surface tension.

Complete Step-by-step solution

We know that the surface tension of a fluid is defined as the work required to extend a surface under isothermal conditions.
Given T is the surface tension of a fluid.
Also, the radius of liquid drops is R.
After breaking down let the radius of liquid is r.
Since the liquid drop is breaking down into small drops of 64 equal drops, the volume remains constant.
That is the volume of a liquid drop of radius R is equal to the volume of 64 drops of radius r
Then $\dfrac{4}{3}\pi {R^3} = 64 \times \dfrac{4}{3}\pi {r^3}$
From this we get $R = 4r$
Then the energy of a big drop $= T.4\pi {R^2}$
The energy of 64 drops $= 64\left( {T \times 4\pi {r^2}} \right) = 64\left[ {T \times 4\pi {{\left( {\dfrac{R}{4}} \right)}^2}} \right]$
Therefore the energy of 64 drops $= 16\pi T{R^2}$

The energy needed to break the drop into 64 droplets is $\Delta W = {E_2} - {E_1} = 12\pi {R^2}T$

Note: We need to find the energy of the big drop before breaking and then find the energy of the total individual drops after breaking and then we need to find the difference in the area between them to get the value of the energy difference. We need to take care of the point that we should not only consider only the energy of the single drop after breaking.