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

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

1.	(i) Determine the height 'h' to which water rises when capillary tube of radius 0.2 mm is dipped vertically in a beaker containg water (Surface tension of water = 7.0xx10^(-2)Nm^(-1) and angle of contact 0^(@)) (ii) Discuss what will happen if the capillary tube is now pushed down till its height above the surface of water is 4cm. Assume that the density of water is 1000kgm^(-3) and g is 10ms^(-2).
Answer» Solution :i) FORMULA : `h=(2Scostheta)/(rrhog)` Angle of contact `theta=0^(@) therefore cos0^(@)=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` `therefore` Height `h=(2XX(7xx10^(-2))xx1)/(((0.2)/(1000))xx(1000)xx(10))=7xx10^(-2)mor7cm`. II) 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 SPECIAL case the angle of contact will change. `(h_(2))/(h_(1))=(costheta_(2))/(costheta_(1))impliescostheta_(2)=(h_(2))/(h_(1))costheta_(1)` here `h_(1)=7cm, h_(2)=4cm, theta_(1)=0^(@)` `costheta_(2)=cos^(-1)(0.5714)=55^(@)`