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A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodide (surface tension = 0.05 $N m^{- 1}$ , density = 667 kg $m^{- 3}$ ) which rises to height h in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. Sides of the capillary) make an angle of $60^{0}$ with one another. Then h is close to (g = 10 $m s^{- 2}$ ).

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0.049 m

0.172 m

0.087 m

0.137 m

answer is C.

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Detailed Solution

$\begin{array}{l} R i s e = h = \frac{2 S \cos θ}{ρ r g} \\ C o n t a c t a n g l e = θ = 30^{0} \\ \Rightarrow h = \frac{2 \times 0.05 \cos 30^{0}}{667 \times 0.15 \times 10^{- 3} \times 10} = 0.087 m \end{array}$

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