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 find the work done in dividing a liquid drop into smaller drops, we need to consider the change in surface energy due to the change in total surface area.

1. Calculate 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 given by the formula \(A_1 = 4\pi R^2\).
2. Calculate the total surface area of the smaller drops:
   • The original drop is divided into 27 identical smaller drops.
   • If the radius of each small drop is \(r\), then using the volume conservation:
   • \(\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3\)
   • This implies: \(R^3 = 27r^3 \rightarrow r = \frac{R}{3}\)
   • The surface area of each small drop is \(A_r = 4\pi r^2 = 4\pi \left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}\).
   • The total surface area of 27 smaller drops is \(A_2 = 27 \times \frac{4\pi R^2}{9} = 12\pi R^2\).
3. Calculate the change in surface area and work done:
   • 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 (which is the change in surface energy) is given by \(W = T \times \Delta A = T \times 8\pi R^2\).

Therefore, the work done in the process is \( 8\pi R^2 T \), which matches the correct answer given in the options.

Answer 2

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:\( 8\pi R^2 T \)