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← JEE Main 2022 · 30 Jul · Shift 2
PhysicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 30 Jul · Shift 2, Physics question 27

A water drop of radius 1 cm is broken into eight equal droplets. Surface tension of water is 0.075 N m−1^{-1}. The gain in surface energy is _______ ×10−7\times10^{-7} J. (Take π = 3.14)

Answer and worked solution

Official answer: 942

Areas. Each droplet has radius 0.5 cm: new area 8×4π(0.5)2=8π8\times4\pi(0.5)^2=8\pi cm² against 4π4\pi cm², a gain of 4π×10−44\pi\times10^{-4} m².

Energy 0.075×4×3.14×10−4=942×10−70.075\times4\times3.14\times10^{-4}=942\times10^{-7} J.

Answer: 942

Topics:Surface tension
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