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^(@)`.

Answer

C, D

Explanation

Suppose pressure at the points, `A, B, C`, and `D` be `P_(A), P_(B), P_(C)`and `P_(D)`respectively.
The pressure on the concave side of the liqiud surface is greater than on the other side by `(2T)/(R)`.
An angle of contact `theta` is given to be `0^(@)`, hence
`R cos 0^(@)` or` R=r`
`:. P_(A)=P_(B)+(2T)/(r_(1))` and `P_(C)=p_(D)+(2T)/(r_2)`

where, `r_(1)` and `r_(2)` are the radii of the two limbs
But `p_(A)=P_(C)`
`:. P_(B)+(2T)/(r_1)=P_(D)+(2T)/(r_2)`
or, `P_(D)-p_(B)-2T((1)/(r_1)-(1)/(r_2))`
where, h is the difference in water levels in the two limbs
Now, `h=(2T)/(rho g)((1)/(r_1)-(1)/(r_2))`
Given, that `T=0.07Nm^(-1),rho=1000kgm^(-3)`
`r_(1)=(3)/(2) mm=(3)/(20)cm=(3)/(20xx100)m`
`=1.5xx10^(-3)m.r_(2)=3xx10^(-3)m`
`:. h=(2xx0.07)/(1000xx9.8)((1)/(1.5xx10^(-3))-(1)/(3xx10^(-3)))m`
`=4.76xx10^(-3)m=4.76mm`