QJEEM25Ph1135: JEE Main 2025 Physics - Class 11 JEEM25Ph1135 Solution: The amount of work done to break a big water drop of radius ' ' …

To solve the problem, we need to compare the increase in surface area when the big drop is split into small drops.

Step 1: Determine the radius of the small drops

For 27 small drops:

Total volume is conserved:

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

For 64 small drops:

Similarly,

$64 \times \frac{4}{3}\pi r'^3 = \frac{4}{3}\pi R^3 \implies r'^3 = \frac{R^3}{64} \implies r' = \frac{R}{4}.$

Step 2: Calculate the surface area before and after the break-up

Initial surface area of the big drop:

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

For 27 drops:

Surface area of one small drop:

$4\pi\left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}.$

Total surface area:

$A_{27} = 27 \times \frac{4\pi R^2}{9} = 3 \times 4\pi R^2 = 12\pi R^2.$

Increase in surface area:

$\Delta A_{27} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2.$

For 64 drops:

Surface area of one small drop:

$4\pi\left(\frac{R}{4}\right)^2 = 4\pi \frac{R^2}{16} = \frac{\pi R^2}{4}.$

Total surface area:

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

Increase in surface area:

$\Delta A_{64} = 16\pi R^2 - 4\pi R^2 = 12\pi R^2.$

Step 3: Relate work done to the change in surface area

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

$\text{Work} \propto \Delta A.$

We know that breaking into 27 drops requires 10 J corresponding to an increase of $8\pi R^2$ .

Thus, if $W$ is the work done,

$W_{27} = k \cdot 8\pi R^2 = 10\, \text{J},$

where $k$ is the proportionality constant.

For 64 drops:

$W_{64} = k \cdot 12\pi R^2.$

Dividing the two equations:

$\frac{W_{64}}{10} = \frac{12\pi R^2}{8\pi R^2} = \frac{12}{8} = \frac{3}{2}.$

Thus,

$W_{64} = 10 \times \frac{3}{2} = 15\, \text{J}.$

Final Answer: 15 J (Option D)