Step 1: Understanding the Question:
Work is done to increase the surface area of the water. We must find the change in area. Step 2: Key Formula or Approach:
1. Conservation of volume: \( V = n v \).
2. Work done \( W = T \Delta A = T (n \cdot 4\pi r^2 - 4\pi R^2) \). Step 3: Detailed Explanation:
Volume of large drop \( = 64 \times \) Volume of small drop.
\[ \frac{4}{3}\pi R^3 = 64 \left( \frac{4}{3}\pi r^3 \right) \implies R^3 = 64 r^3 \implies r = \frac{R}{4} \]
Change in area \( \Delta A = 64(4\pi r^2) - 4\pi R^2 \).
Substitute \( r = R/4 \):
\[ \Delta A = 64 \left[ 4\pi \left( \frac{R}{4} \right)^2 \right] - 4\pi R^2 \]
\[ \Delta A = 64 \left[ \frac{4\pi R^2}{16} \right] - 4\pi R^2 \]
\[ \Delta A = 16\pi R^2 - 4\pi R^2 = 12\pi R^2 \]
Work done \( W = T \Delta A = 12\pi T R^2 \). Step 4: Final Answer:
The work done is \( 12\pi T R^2 \).