JEE Main 2026 · 5 Apr · Shift 2 · Physics Q33

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PhysicsSingle correct+4 / -1Question 33 of 75

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).

8\pi r^2S

16\pi r^2S

64\pi r^2S

4\pi r^2S

Answer and worked solution

Official answer: B

Radius. The volume is conserved: $8\cdot\tfrac43\pi r^3=\tfrac43\pi R^3$ , so $R=2r$ .

Surface area. Before: $8\cdot4\pi r^2=32\pi r^2$ . After: $4\pi(2r)^2=16\pi r^2$ .

Energy released: $16\pi r^2S$ .

(B)

Topics:Surface tension

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JEE Main 2026 · 5 Apr · Shift 2, Physics question 33