Question

A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :

• A. \( 8\pi R^2 T \)
• B. \( 3\pi R^2 T \)
• C. \( \frac{1}{8} \pi R^2 T \)
• D. \( 4\pi R^2 T \)

Answer 1

The Correct Option is A

To determine the work required to divide a liquid drop into smaller ones, analyze the alteration in surface energy resulting from the change in total surface area.

1. Determine the initial surface area of the original drop:
   • The original drop is a sphere with radius \(R\).
   • The surface area \(A_1\) of a sphere is computed using the formula \(A_1 = 4\pi R^2\).
2. Determine the aggregate surface area of the smaller drops:
   • The original drop fragments into 27 identical smaller drops.
   • Let the radius of each small drop be \(r\). Volume conservation dictates:
   • \(\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3\)
   • This simplifies to: \(R^3 = 27r^3 \rightarrow r = \frac{R}{3}\)
   • The surface area of each individual small drop is \(A_r = 4\pi r^2 = 4\pi \left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}\).
   • The cumulative surface area of 27 smaller drops is \(A_2 = 27 \times \frac{4\pi R^2}{9} = 12\pi R^2\).
3. Calculate the variation in surface area and the associated work:
   • The change in surface area is \(\Delta A = A_2 - A_1 = 12\pi R^2 - 4\pi R^2 = 8\pi R^2\).
   • Work done (equivalent to the change in surface energy) is calculated as \(W = T \times \Delta A = T \times 8\pi R^2\).

Consequently, the work expended in this process totals \( 8\pi R^2 T \), aligning with the provided correct answer.