A capillary tube made of glass of radius $0.15\ \text{mm}$ is dipped vertically in a beaker filled with methylene iodide (surface tension $= 0.05\ \text{N m}^{-1}$, density $= 667\ \text{kg m}^{-3}$) which rises to height $h$ in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opposite sides of the capillary) make an angle of $60^\circ$ with one another. Then $h$ is close to $(g = 10\ \text{m s}^{-2})$
Answer: (B) $0.087\ \text{m}$
The two tangents are symmetric about the vertical axis and meet at $60^\circ$, so each makes $30^\circ$ with the wall: the angle of contact is $\theta = 30^\circ$.
$$h = \frac{2S\cos\theta}{r\rho g} = \frac{2\times0.05\times\cos30^\circ}{0.15\times10^{-3}\times667\times10} \approx 0.087\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics