A glass rod of radius 0.1 mm is inserted symmetrically into a vertical capillar tube of radius 0.2 mm such that their lower ends are at same level. If this arrangement is dipped in water then what will be the height of water rised in the tube? (Take surface tension of water `=0.07N//m,g=10m//s^(2))`

Answer

`00.14`

Explanation

Let radius of glass rob be `r_(1)` and of tube be `r_(2)`.
Total upward force due to surface tension
`F=T(2pir_(1)+2pir_(2))`
Let water rise into tube upto height h weight of water column
`W=h(pir_(2)^(2)-pir_(1)^(2))rho g`
At equilibrium `F=W`
`:.T2pi(r_(1)+r_(2))`
`=hpi(r_(1)+r_(2))(r_(2)-r_(1))rhog`
`:.h=(2T)/((r_(2)-r_(1))rhog)=(2xx0.07)/((0.2-0.1)xx10^(-3)xx10^(3)xx10)`
`:.h=0.14m`