JEE Physics 2025 Question with Solution

Given: Two identical water drops each have radius $r$ and surface tension $T$.

Find: The surface energy released when they coalesce into one bigger drop.

The surface energy of a liquid drop is proportional to its surface area:

$$E = 4\pi R^2 T$$

When two drops merge, volume is conserved.

Using volume conservation:

$$2 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$

So,

$$R^3 = 2r^3$$

and hence,

$$R = 2^{1/3}r$$

Initial total surface energy of the two smaller drops is:

$$E_i = 2 \times 4\pi r^2 T = 8\pi r^2 T$$

Final surface energy of the larger drop is:

$$E_f = 4\pi R^2 T = 4\pi (2^{1/3}r)^2 T = 4\pi 2^{2/3} r^2 T$$

Energy released equals the decrease in surface energy:

$$\Delta E = E_i - E_f$$

Therefore,

$$\Delta E = 8\pi r^2 T - 4\pi 2^{2/3} r^2 T = 4\pi r^2 T (2 - 2^{2/3})$$

Therefore, the surface energy released is $4\pi r^2 T (2 - 2^{2/3})$ and the correct option is A.