Question

A spherical liquid drop of radius R is divided into 64 equal droplets. If surface tension is \(T\) then the work done in this process will be

• A. \(8\pi R^{2}T\)
• B. \(12\pi R^{2}T\)
• C. \(4\pi R^{2}T\)
• D. \(16\pi R^{2}T\)

Hint:

Work done equals surface tension times the increase in surface area.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Work done \(W=T\times\Delta A\), where \(\Delta A\) is the increase in surface area.

Step 2: Radius of small drops
Volume is conserved: \(\dfrac{4}{3}\pi R^{3}=64\times\dfrac{4}{3}\pi r^{3}\Rightarrow r=\dfrac{R}{4}\).

Step 3: Area change
\(\Delta A=64\times4\pi\dfrac{R^{2}}{16}-4\pi R^{2}=16\pi R^{2}-4\pi R^{2}=12\pi R^{2}\).

Step 4: Work done
\(W=12\pi R^{2}T\).

Final Answer:
The work done is \(12\pi R^{2}T\). \[ \boxed{\text{(B) }12\pi R^{2}T} \]