Answer:
[tex]d=0.414\times 10^{-4}\ m[/tex]
Explanation:
Given that
P = 4 KPa
Contact angle = 6°
Surface tension = 1 N/m
Lets assume that atmospheric pressure = 100 KPa
Lets take that density of water =[tex]1000\ kg/m^3[/tex]
So the capillarity rise h
[tex]h=\dfrac{\Delta P}{\rho g}[/tex]
[tex]h=\dfrac{100\times 1000-4\times 1000}{1000\times 10}[/tex]
h= 9.61 m
We know that for capillarity rise h
[tex]h=\dfrac{2\sigma cos\theta }{r\rho g}[/tex]
[tex]r=\dfrac{2\sigma cos\theta }{h\rho g}[/tex]
[tex]r=\dfrac{2\times 1 cos4^{\circ} }{9.61\times 1000\times 10}[/tex]
[tex]r=0.207\times 10^{-4}\ m[/tex]
[tex]d=0.414\times 10^{-4}\ m[/tex]
