Question

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10^–2 N m^–1. Take the angle of contact to be zero and density of water to be 1.0 × 10³ kg m^–3 (g = 9.8 m s^–2)

Answer 1

Diameter of the first bore, d₁ = 3.0 mm = 3 × 10^–3 m

Hence the radius of the first bore, `r_1 = d_1/2 = 1.5 xx 10^(-3) m`

Diameter of the second bore, `d_2`= 6.0 mm

Hence the radius of the second bore, `r_2 = d_2/2 = 3xx 10^(-3) m`

Surface tension of water `s = 7.3 xx 10^(-2) Nm^(-1)`

Angle of contact between the bore surface and water, θ= 0

Density of water, ρ =1.0 × 10³ kg/m^–3

Acceleration due to gravity, g = 9.8 m/s²

Let h₁ and h₂ be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:

`h_1 = (2s cos theta)/(r_1rhog)`   ...(i)

`h_2 = (2scos theta)/(r_2rhog)` ...(ii)

The difference between the levels of water in the two limbs of the tube can be calculated as:

`= (2 s cos theta)/(r_1rhog) - (2 s cos theta)/(r_2rhog)`

`= (2 s cos theta)/(rhog)[1/r_1 - 1/r_2]`

`= (2xx 7.3 xx 10^(-2) xx 1)/(1xx10^3xx9.8) [1/(1.5xx10^(-3)) - 1/(3xx10^(-3))]`

`= 4.966 xx 10^(-3) m`

= 4.97 mm

Hence, the difference between levels of water in the two bores is 4.97 mm.