JEE Main 2026 Apr 5 Shift 2, Physics Q8: Properties of Fluids

JEE Main2026Apr 5, Shift 2Physics

Eight mercury drops, each of radius $r$ , coalesce to form a bigger drop. The surface energy released in this process is _____. ( $S$ is the surface tension of mercury).

A
$8 π r^{2} S$
B
$16 π r^{2} S$
C
$64 π r^{2} S$
D
$4 π r^{2} S$

Solution

Volume conservation: $8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2 r$ .

Initial surface energy: $U_{i} = 8 \times 4 π r^{2} S = 32 π r^{2} S$ .

Final surface energy: $U_{f} = 4 π R^{2} S = 4 π (2 r)^{2} S = 16 π r^{2} S$ .

Energy released $= U_{i} - U_{f} = 16 π r^{2} S$ .

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JEE Main 2026 Apr 5 Shift 2, Physics Q8: Properties of Fluids

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