Question

A capillary tube of radius 1 mm is immersed in a beaker of mercury. The mercury level inside the tube is found to be 0.536 cm below the level in the reservoir. Determine the angle of contact between the mercury and glass. (Given: T = 0.485 N/m, p = 13.6 x 10³ kg/m³)

Answer 1

Given: r = 1 mm = 10^-3 m,

h = 0.536 cm= 0 .536 x 10^-2 m

T = 0.485 N/m, p = 13.6 x 10³ kg/m³

`T = -(hpgr)/(2costheta)`

`-costheta = (hpgr)/(2T)`

`= (0.536xx10^-2xx13.6xx10^3xx9.8xx10^-3)/(2xx0.485)`

`=(71.38xx10^-2)/0.97`

`-costheta = 0.7359`

`cos(pi-theta)=0.7359        [because cos(pi-theta) = -costheta]`

`pi - theta = cos^-1(0.7359)`

`pi - theta = 42°34`

`theta = pi -42°34`

`theta = 180° - 42°34`

`theta = 179°60^'-42°34`

`theta = 137°26^'`