Question

A capillary tube of inner diameter $0.4$mm is dipped in liquid of specific gravity $13.6$, surface tension $547\times {10}^{-3}$N/m and angle of contact ${135}^o$. Find the depression or elevation of the liquid in the tube.

Answer 1

Since, angle of contact is more than ${90}^o$, so there is a depression of the liquid in the tube.

Radius of the tube $r=\frac{d}{2}=\frac{0.4}{2}=0.2mm=2\times {10}^{-4}m$

Density of liquid $\rho =13.6\times 1000kg/m^3$

Depression in the tube $h=\frac{2T\cos \theta }{\rho gr}$

So, $h=\frac{2\times 547\times {10}^{-3}\times \cos {135}^o}{13.6\times 1000\times 10\times 2\times {10}^{-4}}=2.8\times {10}^{-2}m=2.8cm$