JEE Main 2025 Apr 2 Shift 2, Physics Q4: Properties of Fluids

JEE Main2025Apr 2, Shift 2Physics

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

A
$4 π r^{2} T [2 - 2^{2/3}]$
B
$4 π r^{2} T [2 - 2^{1/3}]$
C
$4 π r^{2} T [1 + 2]$
D
$4 π r^{2} T [2 - 1]$

Solution

Volume is conserved when the two small drops merge into one larger drop of radius $R$ :

$2 \cdot \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2^{1/3} r$ .

Initial surface energy (two drops): $U_{i} = 2 \times 4 π r^{2} T = 8 π r^{2} T$ .

Final surface energy (one drop): $U_{f} = 4 π R^{2} T = 4 π r^{2} \cdot 2^{2/3} T$ .

Energy released $= U_{i} - U_{f} = 4 π r^{2} T (2 - 2^{2/3})$ .

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JEE Main 2025 Apr 2 Shift 2, Physics Q4: Properties of Fluids

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