Find the weight of water supported by surface tension in a capillary tube with a radius of 0.2 mm. Surface tension of water is `0.072Nm^(-1)` and angle of contact of water is `0^(0)`.

Explanation

Assume the weight of water to be .F.
Weight of water in capillary tube = upward force due to surface tension
i.e., `F=2pir(T cos theta)`
Surface tension of water `T-0.072Nm^(-1)`
Angle of contact `theta=0^(@)` , Radius of capillary tube
`(r)=(0.2)/(1000)m=0.2xx10^(-3)m`
`F=2pir(T cos theta)=2xx(22)/(7)xx0.2xx10^(-3)xx0.072xx1`
`=90.51xx10^(-6)NrArr F=90.51xx10^(-6)N`