Question

If 1000 droplets of water of surface tension \( 0.07 \). having same radius 1mm each, combine to from a single drop. In the process the released surface energy is-
(Take \( \pi = \frac{22}{7} \)):

• A. \( 7.92 \times 10^{-6} \) J
• B. \( 7.92 \times 10^{-4} \) J
• C. \( 9.68 \times 10^{-4} \) J
• D. \( 8.8 \times 10^{-5} \) J

Hint:

For problems involving merging of droplets, remember: - Total volume remains constant. - Radius of new drop: \( R = n^{1/3} r \). - Surface energy change: \( \Delta U = {Surface Tension} \times \Delta A \).

Answer 1

The Correct Option is B

Step 1: Understanding the formula for surface energy. The surface energy is given by: \[ U = {Surface Tension} \times {Surface Area} \] For a sphere, surface area is: \[ A = 4\pi R^2 \] Step 2: Calculate initial total surface area. - Given 1000 small droplets, each of radius \( r = 1 \) mm. - Total surface area before merging: \[ A_{{initial}} = 1000 \times 4\pi r^2 \] \[ A_{{initial}} = 1000 \times 4 \times \frac{22}{7} \times (1 \times 10^{-3})^2 \] \[ A_{{initial}} = 1000 \times 4 \times \frac{22}{7} \times 10^{-6} \] \[ A_{{initial}} = \frac{88000}{7} \times 10^{-6} \] \[ A_{{initial}} \approx 12.57 \times 10^{-3} { m}^2 \] Step 3: Calculate final surface area. The final drop has a total volume equal to the sum of all small drops: \[ \frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3 \] Cancelling \( \frac{4}{3} \pi \) from both sides: \[ R^3 = 1000 r^3 \] \[ R = 10r = 10 \times 1 { mm} = 10 { mm} = 10^{-2} { m} \] Final surface area: \[ A_{{final}} = 4\pi R^2 \] \[ A_{{final}} = 4 \times \frac{22}{7} \times (10^{-2})^2 \] \[ A_{{final}} = 4 \times \frac{22}{7} \times 10^{-4} \] \[ A_{{final}} = \frac{880}{7} \times 10^{-6} \] \[ A_{{final}} \approx 1.257 \times 10^{-3} { m}^2 \] Step 4: Compute the released surface energy. Change in surface area: \[ \Delta A = A_{{initial}} - A_{{final}} \] \[ \Delta A = (12.57 - 1.257) \times 10^{-3} \] \[ \Delta A = 11.313 \times 10^{-3} { m}^2 \] Energy released: \[ \Delta U = {Surface Tension} \times \Delta A \] \[ \Delta U = (0.07) \times (11.313 \times 10^{-3}) \] \[ \Delta U = 7.92 \times 10^{-4} { J} \] Final Answer: \[ \boxed{7.92 \times 10^{-4} { J}} \]