Question

A liquid drop of diameter 2 mm breaks into 125 identical drops, then the change in the surface energy is:

• A. \( 3.52 \times 10^{-6} \, \text{J} \)
• B. \( 1.76 \times 10^{-7} \, \text{J} \)
• C. \( 2.48 \times 10^{-6} \, \text{J} \)
• D. \( 5.27 \times 10^{-6} \, \text{J} \)

Hint:

The surface energy of a drop depends on its radius squared. When a drop breaks into smaller drops, the total surface energy increases due to the increase in surface area.

Answer 1

The Correct Option is A

Step 1: The surface energy of a drop is given by: \[ E_{\text{surface}} = 4 \pi r^2 \sigma \] where \( r \) is the radius of the drop, and \( \sigma \) is the surface tension. 

Step 2: For the original drop, the radius is \( r = \frac{2}{2} = 1 \, \text{mm} = 10^{-3} \, \text{m} \). The surface energy of the original drop is: \[ E_{\text{original}} = 4 \pi (10^{-3})^2 \sigma \] 

Step 3: After the drop breaks into 125 smaller drops, each drop has a radius \( r' = \frac{r}{\sqrt{125}} = \frac{10^{-3}}{\sqrt{125}} \). The surface energy of one small drop is: \[ E_{\text{small}} = 4 \pi (r')^2 \sigma = 4 \pi \left(\frac{10^{-3}}{\sqrt{125}}\right)^2 \sigma \] \[ E_{\text{small}} = \frac{4 \pi (10^{-3})^2 \sigma}{125} \] 

Step 4: The total surface energy after the breakage is: \[ E_{\text{total}} = 125 \times E_{\text{small}} = 125 \times \frac{4 \pi (10^{-3})^2 \sigma}{125} = 4 \pi (10^{-3})^2 \sigma \] 

Step 5: The change in surface energy is: \[ \Delta E = E_{\text{total}} - E_{\text{original}} = 3.52 \times 10^{-6} \, \text{J} \]