Question

The work done in splitting a water drop of radius 'R' into 64 droplets is (T is the surface tension of water)

• A. \(8πR^2T\)
• B. \(12πR^2T\)
• C. \(4πR^2T\)
• D. \(16πR^2T\)

Hint:

Volume is conserved, so the radius of each droplet is R divided by the cube root of 64. Work equals tension times the increase in area.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Splitting a drop creates new surface area. The work done equals surface tension times the increase in area.

Step 2: Find the droplet radius.
\[ \frac{4}{3}\pi R^3 = 64\cdot\frac{4}{3}\pi r^3 \Rightarrow r = \frac{R}{4} \]

Step 3: Find the increase in area.
Final area: \(64\times 4\pi\left(\dfrac{R}{4}\right)^2 = 64\times 4\pi\times\dfrac{R^2}{16} = 16\pi R^2\). Initial area: \(4\pi R^2\).
\[ \Delta A = 16\pi R^2 - 4\pi R^2 = 12\pi R^2 \]

Step 4: Work done.
\[ W = T\,\Delta A = 12\pi R^2 T \]

Step 5: Check the options.
Option (A) is \(8\pi R^2T\), and (C) is only the initial area term. Option (D) is the final area term without subtracting the initial area.

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