A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m^{-1}, then the excess force required to take it away from the surface is (1) 19.8 mN (2) 198 N (3) 1.98 mN (4) 99 N

Correct Answer: A — 19.8 mNStep-by-step Solution:When a thin disc is placed on water, the surface tension acts along the circumference. The force required to lift the disc is the surface tension force = T × (circumference) = T × 2πr. Here, r = 4.5 cm = 0.045 m, T = 0.07 N/m. So F = 0.07 × 2 × 3.14 × 0.045 = 0.07 × 0.2826 = 0.01978 N = 19.8 mN.Key Concepts: Surface tension, force due to surface tension along a line of contact.

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A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m^{-1}, then the excess force required to take it away from the surface is (1) 19.8 mN (2) 198 N (3) 1.98 mN (4) 99 N

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A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m^{-1}, then the excess force required to take it away from the surface is (1) 19.8 mN (2) 198 N (3) 1.98 mN (4) 99 N

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