A U-tube is supported with its limbs vertical and is partly filled with water. If internal diameters of the limbs are `1xx10^(-2)mand1xx10^(-4)m` respectively, what will be the difference in height of water in the two limbs ? Surface tension of water is `0.07Nm^(-1)` ?

Explanation

Surface tension, `sigma=0.07Nm^(-1)`, Density, `rho=1000kgm^(-3),g=9.8ms^(-2)`
Angle of contact `theta=0^(@)`, radius, `r_(1)=0.5xx10^(-2)m`, Radius, `r_(2)=0.5xx10^(-4)m`,
Let `h_(1)` be the height of water in the limb of radius `r_(1)`.
Then, `h_(1)=(2sigmacostheta)/(r_(1)rhog)=(2xx0.07xxcos0^(@))/(0.5xx10^(-2)xx1000xx9.8)m=2.86xx10^(-3)m`
Let `h_(2)` be the height of water in the limb of radius `r_(2)`.
Then, `h_(2)=(2sigmacostheta)/(r_(2)rhog)=(2xx0.07xxcos0^(@))/(0.5xx10^(-4)xx1000xx9.8)m=2.86xx10^(-1)m`
Difference in heights = `h_(2)-h_(1)=2.68xx10^(-1)m-2.86xx10^(-3)m=(0.286-0.00286)m=0.283m`