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 pressure at point A and B respectively the pressure at point C.
`P_(C)=P_(A)-(2T)/(R_(1))` where `R_(1)=(r_(1))/(cos0^(0))=r_(1)`
The pressure at point `D,P_(D)=P_(B)-(2T)/(R_(2))`
Where, `R_(2)=(r_(2))/(cos0^(0))=r_(2)`
If `h` is the difference in levels of liquid in two limbs, then
`P_(D)-P_(C)=hrhogimplies(P_(B)-(2T)/(R_(2)))-(P_(A)-(2T)/(R_(1)))=hrhog`
As `P_(A)=P_(B)` and `R_(1)=r_(1)=1.5mm`
`R_(2)=r_(2)=3.0mm, so 2T((1)/(r_(1))-(1)/(r_(2)))=hrhog`
`0.2xx0.07((1)/(1.5xxx10^(-3))-(1)/(3xx10^(-3)))=hxx1000xx9.8`
After solving we get `h=4.76xx10^(-3)m`