__Question 10.30__:
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}).

__Solution__:*d*

_{1}= 3.0 mm = 3 × 10

^{–3}m

Hence, the radius of the first bore, r

_{1}= d

_{1}/ 2 = 1.5 X 10

^{-3}m

Diameter of the first bore,

*d*

_{2}= 6.0 mm = 6 × 10

^{–3}mm

Hence, the radius of the first bore, r

_{2}= d

_{2}/ 2 = 3 X 10

^{-3}m

Surface tension of water,

*s*= 7.3 × 10

^{–2}N m

^{–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}= 2sCosθ / r

_{1}ρg .....(i)

h

_{2}= 2sCosθ / r

_{2}ρg .....(ii)

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

^{-3}m

= 4.97 mm

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

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