A glass U-tube is such that the diameter of one limb is 3.0 mm and that of the other is 6.00mm. The tube is inverted with the open ends below the surface of water in a beaker. What is the difference between the heightto which water rises in the two limbs? Surface tension of water is 0.07N/m. Assume that the angle of contact between water and glass is 0^(@).

A glass U-tube is such that the diameter of one limb is 3.0 mm and tha

1.	A glass U-tube is such that the diameter of one limb is 3.0 mm and that of the other is 6.00mm. The tube is inverted with the open ends below the surface of water in a beaker. What is the difference between the heightto which water rises in the two limbs? Surface tension of water is 0.07N/m. Assume that the angle of contact between water and glass is 0^(@).
Answer» Solution :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`