A capilaary tube of radius 0.35cm is dipped in water. To what height will water inside the capilary raise? (Given S.T. of water `=0.072 Nm^(-1), theta` between water-glass contact `theta=0^(@))`

Explanation

We know that, for water-glass, `theta=0^(@), cos 0^(@)=1`
and `T=1/2 (pgr(h+r/3))/(cos theta)=1/2 xx 10^(3) xx 9.8 xx 0.035 xx 10^(-2)(3h+0.035 xx 10^(-2))/(3)`
`=0.072=0.517(3h+0.035 xx 10^(-2))`
i.e, `3h+0.035 xx 10^(-2)=0.126`
3h=(0.126)-(0.0035)
`h=(0.2157)/(3)=h=0.0419m`