# 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

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)

#### 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.

Is there an error in this question or solution?

#### APPEARS IN

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