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

The work done to break a drop into smaller drops is related to the surface area of the drops. The formula for the work done in breaking a drop into smaller drops is: \[ W \propto \Delta A = 4\pi R^2 \left( \text{final surface area} - \text{initial surface area} \right), \] where the surface area of a sphere is given by \( A = 4\pi r^2 \), and \( r \) is the radius of the drop.
Step 1: We are given that the work done to break a big drop of radius \( R \) into 27 smaller drops is 10 J. The radius of each smaller drop will be \( \frac{R}{3} \) because the volume of the drop is conserved (volume is proportional to \( r^3 \)).
Thus, for 27 smaller drops, the radius of each smaller drop is \( \frac{R}{3} \).
Step 2: The surface area of the big drop is \( 4\pi R^2 \), and the total surface area of the 27 smaller drops is \( 27 \times 4\pi \left( \frac{R}{3} \right)^2 = 27 \times 4\pi \times \frac{R^2}{9} = 12\pi R^2 \).
The work done to break the big drop into 27 smaller drops is proportional to the increase in surface area:
\[ W_{27} \propto 12\pi R^2 - 4\pi R^2 = 8\pi R^2. \] Since \( W_{27} = 10 \, \text{J} \), we have: \[ W_{27} = k \times 8\pi R^2 = 10, \] where \( k \) is the proportionality constant.
Step 3: Now, let's consider the case where the big drop is broken into 64 smaller drops. The radius of each smaller drop will be \( \frac{R}{4} \), since the volume is conserved (the volume is proportional to \( r^3 \)).
The surface area of the 64 smaller drops is \( 64 \times 4\pi \left( \frac{R}{4} \right)^2 = 64 \times 4\pi \times \frac{R^2}{16} = 16\pi R^2 \).
The work done to break the big drop into 64 smaller drops is proportional to the increase in surface area: \[ W_{64} \propto 16\pi R^2 - 4\pi R^2 = 12\pi R^2. \] Thus, the work done is: \[ W_{64} = k \times 12\pi R^2. \] From Step 2, we know that \( k \times 8\pi R^2 = 10 \), so \( k = \frac{10}{8\pi R^2} \). Now, substitute \( k \) into the equation for \( W_{64} \): \[ W_{64} = \frac{10}{8\pi R^2} \times 12\pi R^2 = 15 \, \text{J}. \] Thus, the work done to break the big drop into 64 smaller drops is \( \boxed{15 \, \text{J}} \).