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 - Physics

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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 hbe 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.

Concept: Surface Tension
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Chapter 10: Mechanical Properties of Fluids - Exercises [Page 271]

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NCERT Physics Class 11
Chapter 10 Mechanical Properties of Fluids
Exercises | Q 30 | Page 271

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