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A capillary tube of radius $0.1 m m$ is dipper in water. The water rises to a height of $2 c m$ in the tube. If the surface tension of water is $0.072 N / m$ , The contact angle between water and wall of the tube is $(g = 10 m / s^{2})$

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An Intiative by Sri Chaitanya

$θ = \cos^{- 1} (\frac{1}{1.8})$

$θ = \cos^{- 1} (\frac{1}{3.6})$

$θ = \cos^{- 1} (\frac{1}{6.2})$

$θ = \cos^{- 1} (\frac{1}{7.2})$

answer is B.

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Detailed Solution

$h = \frac{2 T \cos θ}{r ρ g}$

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