If a `5 cm` long capillary tube with `0.1 mm` internal diameter, open at both ends, is slightly dipped in water having surface tension `75 dyn//cm`, state whether:
(i) water will rise halfway in the capillary,
(ii) water will rise up to the upper end of capillary,
(iii) water will overflow out of the upper end of capillary.
Explain your answer.

Answer

A

Explanation

Given that surface tension of water , `T=75 dyne//cm`
Radius `r=(0.1)/(2)nm=0.05mm=0.005cm`
Density `rho=1 gm//cm^(3)`, angle of contact, `theta=0^(@)`
Let h be the height to which water water rises in the capillary tube, Then
`h=(2T costheta)/(r rho g)=(2xx75xxcos 0^(@))/(0.005xx1xx981) cm`
`=30.58 cm`
But lenght of capilliary tube is `h'=5cm`
(a) Because `hgt(h')/(2)` therefore the first possibility does not exist.
(b) Because the tube is of insufficient length therefore the water will rise upto the upper end of the tube.
(c ) The water will not overflow of the upper end of the capilliary. It will rise only up to the upper end of the capillary,
The liquid meniscus will adjust its radius of curvature `R'` in such a way that
`R' h'=Rh [ :. hR=(2T)/(rho g)=constant]`
Where, `R` is the radius of curvature that the liquid meniscus would possess if the capillary tube were of sufficient lenght.
`:. R'=(Rh)/(h')=(rh)/(h')=(0.005xx30.58)/(5) =0.0306cm [ :. R=(r)/(costheta)=(r)/(cos 0^(@))=r]`.