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

Equating pressure at point `A` and `B` which are in same horizontal level
`P - (2T)/(r_(1)) + rhogh = P - (2T)/(r_(2)) rArr h = (2T)/(rhog) ((1)/(r_(1)) - (1)/(r_(2)))`
Given that `T = 0.07 Nm^(-1), rho = 1000 kgm^(-3)`
`r_(1) = (3)/(2) mm = (3)/(20) cm = (3)/(20 xx 100) m = 1.5 xx 10^(-3)m, r_(2) = 3 xx 10^(-3)m`
`:. h = (2 xx 0.07)/(1000 xx 9.8)((1)/(1.5 xx 10^(-3)) - (1)/(3 xx 10^(-3)))m = 4.76 xx 10^(-3)m = 4.76 xx 10^(-3)m = 4.76 mm`