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 :-

Step 1: Understanding Work Done in Breaking a Drop

The work done to break a drop into smaller drops is given by the formula:

$$W \propto \text{Increase in Surface Area}$$

Since the surface energy is proportional to the surface area, we calculate the total surface area before and after breaking.

Step 2: Surface Area Before Breaking

The surface area of the original drop (before breaking):

$$A_{\text{initial}} = 4\pi R^2$$

Step 3: Surface Area After Breaking

Let the radius of each small drop be $r$. Since volume is conserved, we equate the total volume before and after:

$$\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi r^3$$

 $$R^3 = 27 r^3$$ $$r = \frac{R}{3}$$

Now, the total surface area after breaking into 27 drops:

$$A_{\text{final}} = 27 \times 4\pi r^2 = 27 \times 4\pi \left(\frac{R}{3}\right)^2$$

 $$A_{\text{final}} = 27 \times 4\pi \frac{R^2}{9} = 12\pi R^2$$

Step 4: Work Done Ratio

The increase in surface area:

$$\Delta A_1 = A_{\text{final}} - A_{\text{initial}} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$$

For the second case (64 small drops):

$$R^3 = 64 r^3$$

 $$r = \frac{R}{4}$$

Total surface area after breaking into 64 drops:

$$A_{\text{final}} = 64 \times 4\pi r^2 = 64 \times 4\pi \left(\frac{R}{4}\right)^2$$

 $$A_{\text{final}} = 64 \times 4\pi \frac{R^2}{16} = 16\pi R^2$$

Increase in surface area:

$$\Delta A_2 = 16\pi R^2 - 4\pi R^2 = 12\pi R^2$$

Since work done is proportional to the increase in surface area:

$$\frac{W_2}{W_1} = \frac{\Delta A_2}{\Delta A_1} = \frac{12\pi R^2}{8\pi R^2} = \frac{12}{8} = 1.5$$

 $$W_2 = 1.5 \times 10 = 15 \text{ J}$$

Final Answer:

The work done required to break the drop into 64 smaller drops is 15 J.