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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 × 10^{3} kg m^{–3} (g = 9.8 m s^{–2})

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#### Solution

Diameter of the first bore, *d*_{1} = 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 × 10^{3} kg/m^{–3}

Acceleration due to gravity, g = 9.8 m/s^{2}

Let *h*_{1} and *h*_{2 }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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