Question

A liquid drop having surface energy E is sprayed into \(512\) droplets of the same size. Then the final surface energy is

• A. \(4E\)
• B. \(8E\)
• C. \(2E\)
• D. \(E\)

Hint:

Volume is conserved, so r = R/8; surface area (and energy) increases by a factor of 8.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Surface energy equals surface tension times surface area. When the drop splits, the volume stays the same but the area changes.

Step 2: Find the small drop radius:
\(512\times\dfrac43\pi r^3=\dfrac43\pi R^3\), so \(r^3=\dfrac{R^3}{512}\) and \(r=\dfrac R8\).

Step 3: Compare areas:
Final area \(=512\times4\pi r^2=512\times4\pi\dfrac{R^2}{64}=8\times4\pi R^2\). So the area is 8 times the initial area.

Step 4: Find the energy:
Final surface energy \(=8E\). Option B.

Step 5: Why the other options are wrong.
4E, 2E and E are less than the correct factor. Energy rises because many small drops have more total surface than one big drop.

Final Answer:
The final surface energy is 8E. \[ \boxed{\text{(B) }8E} \]