A tube of 1mm bore is dipped into a vessel containing a liquid of density `rho = 800 kg m^(-3)` and of surface tension , `S=49 xx 10^(-3)Nm^(-1)` and angle of contact , `theta = 0^(@)`. The tube is held inclined to the verticle at an angle of 60^(@), find the height to which the liquid can rise and the length which the liquid will oc cupy in the tube.

Explanation

Here, `r=1.2 mm = 1/2 xx 10^(-3)m, rho = 800 kg//m^(3), S=49 xx 10^(-3) N//m, theta = 0^(@) , h=?`
Angle with the horizontal of inclined tube `alpha= 90^(@)-60^(@)=30^(@)`
Alsom `h=(2S cos theta)/(r rho g) = 2 xx 49 xx 10^(-30 xx cos 0^(@))/((1//2 xx 10^(-))xx 800 xx 9.8) = 0.025 m = 2.5 cm`
if l is the length oc cupied by the liquid in the tube sin `alpha = h//l`
or `l=(h)/(sin alpha) = (2.5) /(sin 30^(@) ) = 2.5 xx 2 = 5 cm`.