A water drop of 0.01 cm³ is squeezed between two glass plates and spreads in to area of 10 cm². If surface tension of water is 70 dyne /cm then the normal force required to separate glass plates from each other will be (a) 12 N (b) 14 N (c) 16 N (d) 28 N

Answer: (b). Thickness t = V/A = (0.01)/(10) = 0.001 cm. F = 2AT/t = (2 × 10 × 70)/(0.001) = 1.4×10⁶ dyne. F = 14 N. So the answer is (b).

A water drop of 0.01 cm³ is squeezed between two glass plates and… | MHT CET 2025 (20 Apr, Shift 1)

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Force due to Surface Tension

A water drop of $0.01\text{ cm}^{3}$ is squeezed between two glass plates and spreads in to area of $10\text{ cm}^{2}$ . If surface tension of water is 70 dyne $/\text{cm}$ then the normal force required to separate glass plates from each other will be

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Answer: (b)

Thickness $t = \frac{V}{A} = \frac{0.01}{10} = 0.001\text{ cm}$ .
$F = \dfrac{2AT}{t} = \dfrac{2 \times 10 \times 70}{0.001} = 1.4\times10^{6}\text{ dyne}$ .
$F = 14\text{ N}$ . So the answer is (b).

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