Question

The work done in splitting a spherical liquid drop of radius \( a \) into eight liquid droplets of the same size (surface tension of the liquid = \( S \)) is

• A. \( 8\pi Sa^2 \)
• B. \( \pi Sa^2 \)
• C. \( 2\pi Sa^2 \)
• D. \( 4\pi Sa^2 \)
• E. \( 16\pi Sa^2 \)

Hint:

Always conserve volume when a drop splits.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Work done against surface tension is equal to the surface tension (S) multiplied by the increase in the surface area (\(\Delta A\)). When a large drop splits into smaller droplets, the total surface area increases, and work must be done to create this new surface area.
Step 2: Key Formula or Approach:
1. Work Done, \(W = S \times \Delta A = S \times (A_{final} - A_{initial})\). 2. Initial surface area of the large drop of radius 'a': \(A_{initial} = 4\pi a^2\). 3. The volume of the liquid is conserved. The initial volume must equal the total final volume of the 8 droplets. Let 'r' be the radius of a small droplet. \[ V_{initial} = 8 \times V_{final\_droplet} \] \[ \frac{4}{3}\pi a^3 = 8 \times \frac{4}{3}\pi r^3 \] 4. Solve for `r` in terms of `a`. 5. Calculate the final total surface area, \(A_{final} = 8 \times (4\pi r^2)\). 6. Calculate \(W\).
Step 3: Detailed Explanation:
1. Find the radius of the small droplets. From the conservation of volume: \[ \frac{4}{3}\pi a^3 = 8 \times \frac{4}{3}\pi r^3 \] \[ a^3 = 8r^3 \implies a = 2r \implies r = \frac{a}{2} \] 2. Calculate the initial and final surface areas. Initial surface area: \[ A_{initial} = 4\pi a^2 \] Final surface area of one small droplet is \(4\pi r^2\). Since there are 8 droplets: \[ A_{final} = 8 \times (4\pi r^2) = 32\pi r^2 \] Substitute \(r = a/2\): \[ A_{final} = 32\pi \left(\frac{a}{2}\right)^2 = 32\pi \frac{a^2}{4} = 8\pi a^2 \] 3. Calculate the increase in surface area. \[ \Delta A = A_{final} - A_{initial} = 8\pi a^2 - 4\pi a^2 = 4\pi a^2 \] 4. Calculate the work done. \[ W = S \times \Delta A = S \times 4\pi a^2 = 4\pi S a^2 \] Step 4: Final Answer:
The work done is \(4\pi S a^2\).