Q 11-09-134JEE MainJEE Main 2020 (3 Sep, Shift 1)Medium
When a long glass capillary tube of radius $0.015\ \text{cm}$ is dipped in a liquid, the liquid rises to a height of $15\ \text{cm}$ within it. If the contact angle between the liquid and glass is close to $0^\circ$, the surface tension of the liquid, in milliNewton $\text{m}^{-1}$, is ______. $[\rho_{\text{liquid}} = 900\ \text{kg m}^{-3},\ g = 10\ \text{m s}^{-2}]$ (Give answer in closest integer)
Numerical value type. Enter your answer.
Answer: 101
$h = \dfrac{2S\cos\theta}{r\rho g}$ with $\cos\theta \approx 1$:
$$S = \frac{rh\rho g}{2} = \frac{1.5\times10^{-4}\times0.15\times900\times10}{2} \approx 0.101\ \text{N m}^{-1} = 101\ \text{mN m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics