Question

A liquid drop of diameter D breaks up into 27 drops. Find the resultant change in energy

• A. $2\pi TD^2$
• B. $\pi TD^2$
• C. $\frac{\pi TD^2}{2}$
• D. $4\pi TD^2$

Answer 1

Answer: (A)

$2\pi TD^2$

Answer 2

Radius Calculation:
The radius of the larger drop $= \frac{D}{2}$.
The radius of each smaller drop $= r = \frac{D}{6}$.

Initial Surface Area of the Large Drop:
$A_{\text{initial}} = 4 \pi \left( \frac{D}{2} \right)^2 = \pi D^2$
This is the surface area of the original large drop.

Final Surface Area of the 27 Small Drops:
$A_{\text{final}} = 27 \times 4 \pi r^2 = 27 \times 4 \pi \left( \frac{D}{6} \right)^2 = 27 \times 4 \pi \cdot \frac{D^2}{36} = 3 \pi D^2$
This is the total surface area of the 27 smaller drops.

Change in Surface Area (and Energy):
$\Delta A = A_{\text{final}} - A_{\text{initial}} = 3 \pi D^2 - \pi D^2 = 2 \pi D^2$
Therefore, the change in energy $\Delta E$ is equal to the increase in surface area $\Delta A$:
$\Delta E = 2 \pi T D^2$

So, the correct option is (A): $2\pi TD^2$.