Question

A mercury drop of radius \( 1 \) cm is divided into \( 10^6 \) droplets of equal size. If the surface tension of mercury is \( 35 \times 10^{-3} \) N/m, then the change in surface energy in the process is:

• A. \( 8712 \mu \)J
• B. \( 8712 \) erg
• C. \( 4356 \mu \)J
• D. \( 4356 \) erg

Hint:

For problems involving surface energy, compute the change in surface area using volume conservation principles before applying \( \Delta U = S \Delta A \).

Answer 1

The Correct Option is C

Step 1: Surface Energy Formula Surface energy is given by: \[ \Delta U = S \Delta A \] where: - \( S = 35 \times 10^{-3} \) N/m (surface tension), - \( \Delta A \) = Change in surface area. Step 2: Surface Area Calculations Initial surface area: \[ A_{\text{initial}} = 4\pi R^2 \] where \( R = 1 \) cm, \[ A_{\text{initial}} = 4\pi (1)^2 = 4\pi \text{ cm}^2 \] Final surface area for \( n = 10^6 \) droplets of radius \( r \): \[ A_{\text{final}} = n \times 4\pi r^2 \] Since volume is conserved: \[ \frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3 \] Solving for \( r \): \[ r = R n^{-1/3} = 1 \times (10^6)^{-1/3} = 10^{-2} \text{ cm} \] Final surface area: \[ A_{\text{final}} = 10^6 \times 4\pi (10^{-2})^2 = 4\pi \times 10^4 \text{ cm}^2 \] Change in surface area: \[ \Delta A = A_{\text{final}} - A_{\text{initial}} = (4\pi \times 10^4 - 4\pi) \text{ cm}^2 \] \[ = 4\pi(10^4 - 1) \text{ cm}^2 \] Step 3: Compute \( \Delta U \) \[ \Delta U = S \Delta A = (35 \times 10^{-3}) \times (4\pi \times 9999 \times 10^{-4}) \text{ J} \] \[ = 4356 \mu \text{J} \] Conclusion Thus, the correct answer is: \[ 4356 \mu \text{J} \]