JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Fluid Mechanics

A capillary pulls a liquid up until the upward pull of surface tension around the rim balances the weight of the raised column, giving the rise $h=\frac{2T\cos \theta }{r\rho g}$. With the contact angle close to $0^{\circ }$ we take $\cos \theta \approx 1$, so $T=\frac{hr\rho g}{2}$. Putting $h=0.15$ m, $r=0.015$ cm $=1.5\times 10^{-4}$ m, $\rho =900$ kg m${}^{-3}$ and $g=10$ m s${}^{-2}$ gives $T=\frac{0.15\times 1.5\times 10^{-4}\times 900\times 10}{2}=0.10125$ N m${}^{-1}$, which is $101.25$ mN m${}^{-1}$. Rounded to the closest integer as the question asks, the surface tension is $101$ mN m${}^{-1}$.