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

Step 1. Volume Conservation:  
  Since the total volume of the liquid remains constant, we equate the volume of the original drop to the combined volume of the 27 smaller drops.  

  For the original drop:  

  \(\frac{4}{3}\pi R^3\)
  For the 27 smaller drops (each of radius \( r \)):  

  \(27 \times \frac{4}{3}\pi r^3\)
  Equating the volumes:  

  \(\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3\)

  \(R^3 = 27r^3 \implies r = \frac{R}{3}\) 

Step 2. Calculate the Surface Areas:  
  - Surface area of the original drop:  
    \(A_{\text{initial}} = 4\pi R^2\)
  - Surface area of the 27 smaller drops:  
    \(A_{\text{final}} = 27 \times 4\pi r^2 = 27 \times 4\pi \left(\frac{R}{3}\right)^2 = 27 \times 4\pi \frac{R^2}{9} = 12\pi R^2\)

Step 3. Calculate the Work Done :  
  The work done in increasing the surface area is given by:  \(\text{Work done} = T\Delta A = T(A_{\text{final}} - A_{\text{initial}})\)
 
  \(= T(12\pi R^2 - 4\pi R^2) = T \times 8\pi R^2\)

Thus, the work done in the process is \( 8\pi R^2 T \).

The Correct Answer is:\( 8\pi R^2 T \)