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 points A and B respectively. The pressure at poin 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 in levels of liquid in two limbs, then
`P_(D)-P_(C)=hrho grArr (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)=0.3mm," so "2T((1)/(r_(1))-(1)/(r_(2)))=hrhog`
`0.2xx0.07((1)/(1.5xx10^(-3))-(1)/(3xx10^(-3)))=hxx1000xx9.8`
After solving, we get `h=4.7xx10^(-3),m`