(i) Determine the height h to which water rises when capillary tube of radius 0.2 mm is dipped vertically in a beaker containing water (Surface tension of water 7.0xx10^(-3)Nm^-1) and angle of contact =0^(@)) (ii) Discuss what will happen if the capillary tube is now pushed down till its height above thesurface of water is 4 cm. Assume that the density of water is 1000 kgm^(-3) and g is 10ms^(-2)

(i) Determine the height h to which water rises when capillary tube of

1.	(i) Determine the height h to which water rises when capillary tube of radius 0.2 mm is dipped vertically in a beaker containing water (Surface tension of water 7.0xx10^(-3)Nm^-1) and angle of contact =0^(@)) (ii) Discuss what will happen if the capillary tube is now pushed down till its height above thesurface of water is 4 cm. Assume that the density of water is 1000 kgm^(-3) and g is 10ms^(-2)
Answer» Solution :(i) Formula `h=(2Scos THETA)/(r rho g)` Angle of contact `theta=0^(@):.cos 0^(@)=1` Surface tension `S=7xx10^(-2)Nm^(-1)` Density of water `d=1000kgm^(-3)` Acceleration due to gravity `g=10ms^(-2)` Radius of the tube `r=(0.2//1000)m` `:.` Height `h=(2xx(7xx10^(-2))xx1)/((0.2/1000)xx(1000)xx(10))=7xx10^(-2)m ` or 7 cm (i) However when the capillary tube is pushed down till its height above the surface of water is 4 cm, the water RISES up to 4 cm only and it will not over flow. In this sepcial CASE the angle of contact will CHANGE `(h_(2))/(h_(1))=(cos theta_(2))/(cos theta_(1))implies cos theta_(2)=(h_(2))/(h_(1))cos theta_(1)` here `h_(1)=7cm,h_(2)=4cm, theta_(1)=0^(@)` `cos theta_(2)=cos^(-1)(0.5714)=55^(@)`