Question

The amount of work done to break a big water drop of radius \( R \) into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be:

• A. 15 J
• B. 10 J
• C. 20 J
• D. 5 J

Hint:

When solving problems related to the work done to break a drop into smaller drops, remember that the work is proportional to the increase in surface area. The surface area is proportional to the square of the radius, and the volume is proportional to the cube of the radius.

Answer 1

The Correct Option is A

Step 1: Work for the first case (\( W_1 \))

The work to break the large drop into 27 smaller drops is given by: \[ W_1 = s \times 4\pi \left( \frac{R}{3} \right)^2 \times 27 - 4\pi R^2 \times s. \] Simplifying: \[ W_1 = 4\pi R^2 \times s \times (2) = 10 \, \text{Joule}. \]

Step 2: Work for the second case (\( W_2 \))

The work to break the large drop into 64 smaller drops is: \[ W_2 = s \times 4\pi R^2 \times (4 - 1) = 4\pi R^2 \times s \times 3. \] Using the value \( W_1 = 10 \) Joules, we substitute into the equation for \( W_2 \): \[ W_2 = 3 \times \frac{10}{2} = 15 \, \text{Joule}. \]

Final Answer:

The work required to break the large drop into 64 smaller drops is \( \boxed{15} \, \text{Joule} \).