### Q.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}).

**Answer:-**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 × 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 × 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} = 2*s *Cosθ / r_{1}ρg …..**(i)**

*h*_{2} = 2*s *Cosθ / r_{2}ρg …..**(ii)**

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

^{-3}m

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