Question

A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :

• A. $8\pi R^2 T$
• B. $3\pi R^2 T$
• C. $\frac{1}{8} \pi R^2 T$
• D. $4\pi R^2 T$

Answer 1

Answer: (A)

$8\pi R^2 T$

Answer 2

Step 1. Volume Conservation:
Since the total volume of the liquid remains constant, we equate the volume of the original drop to the combined volume of the 27 smaller drops.

For the original drop:

$\frac{4}{3}\pi R^3$
For the 27 smaller drops (each of radius $r$):

$27 \times \frac{4}{3}\pi r^3$
Equating the volumes:

$\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$

$R^3 = 27r^3 \implies r = \frac{R}{3}$

Step 2. Calculate the Surface Areas:
- Surface area of the original drop:
$A_{\text{initial}} = 4\pi R^2$
- Surface area of the 27 smaller drops:
$A_{\text{final}} = 27 \times 4\pi r^2 = 27 \times 4\pi \left(\frac{R}{3}\right)^2 = 27 \times 4\pi \frac{R^2}{9} = 12\pi R^2$

Step 3. Calculate the Work Done :
The work done in increasing the surface area is given by: $\text{Work done} = T\Delta A = T(A_{\text{final}} - A_{\text{initial}})$

$= T(12\pi R^2 - 4\pi R^2) = T \times 8\pi R^2$

Thus, the work done in the process is $8\pi R^2 T$.

The Correct Answer is:$8\pi R^2 T$