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)andP_(B)` are the pressure at points A and B respectively. The pressure at point C.
`P_(C)=P_(A)-(2S)/(R_(1))`
Where `R_(1)=(r_(1))/(cos0^(@))=r`
The pressure at point D.
`P_(D)=P_(B)-(2S)/(R_(2))`
where, `R_(2)=(r_(2))/(cos0^(@))=r_(2)`
If .h. is the difference between heights rise in two limbs, then `P_(D)-P_(C )=hrhog` or
`(P_(B)-(2S)/(R_(2)))-(P_(A)-(2S)/(R_(1)))=hrhog`
As `P_(A)=P_(B)andR_(1)=r_(1)=1.5mm`,
`R_(2)=r_(2)=3.0mm`, so `2S((1)/(r_(1))-(1)/(r_(2)))=hrhog`
`0.2xx0.07((1)/(1.5xx10^(-3))-(1)/(3xx10^(-3)))`
`=hxx1000xx9.8`
After solving, we get `h=4.76xx10^(-3)m`