JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Fluid Mechanics

Two water drops each of radius 'r' coalesce to form a bigger drop. If 'T' is the surface tension, the surface energy released in this process is :

Answer: (A)

Solution

Let each small drop have radius

r

. Conserving volume when two of them merge into one drop of radius

R

2 \cdot \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}

, so

R = 2^{1/3} r

. The initial surface energy of the two drops is

U_{i} = 2 (4 π r^{2} T) = 8 π r^{2} T

, and the final surface energy of the single drop is

U_{f} = 4 π R^{2} T = 4 π r^{2} T \cdot 2^{2/3}

. The energy released is

U_{i} - U_{f} = 8 π r^{2} T - 4 π r^{2} T \cdot 2^{2/3} = 4 π r^{2} T (2 - 2^{2/3})

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option