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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)
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Solution
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.
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