A glass U-tube is such that the diameter of one limb is `3.0mm` and that of the other is 6.0mm. The tube is inverted vertically with the open ends below the surface of water in a beaker. What is the difference between the height to which water rises in the two limbs? Surface tension of water is `0.07Nm^(-1)`. Assume that the angle of contact between water and glass is `0^(@)`.

Explanation

Let `P_(A)` and `P_(B)` are the pressures at points A and B respectively. The pressure at point C
`P_(C)=P_(A)-(2T)/(R_(1))`
where `R_(1)=(r_(1))/(cos 0^(@))=r_(1)`
The pressure at point `D, P_(D)=P_(B)-(2T)/(R_(2))`

where `R_(2)=(r_(2))/(cos 0^(@))=r_(2)`
If h is the difference between heights rise in two limbs, then
`P_(D)-P_(C)=h rho g`
or `(P_(B)-(2T)/(R_(2)))-(P_(A)-(2T)/(R_(1)))=h rho g`
As `P_(A)=P_(B)` and `R_(1)=r_(1)=1.5mm , R_(2)=r_(2)=3.0 mm, ` so
`2T(1/(r_(1))-1/(r_(2)))=h rho g`
`0.3xx0.07(1/(1.5xx10^(-3))-1/(3xx10^(-3)))9.8=hxx1000xx9.8`
After solving we get
`h=4.76xx10^(-3)m`