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### Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.13

Question 10.13:
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Solution:
Length of the horizontal tube, l = 1.5 m
Radius of the tube, r = 1 cm = 0.01 m
Diameter of the tube, d = 2r = 0.02 m
Glycerine is flowing at a rate of 4.0 × 10–3 kg s–1.
M = 4.0 × 10–3 kg s–1
Density of glycerine, ρ = 1.3 × 103 kg m–3
Viscosity of glycerine, η = 0.83 Pa s
Volume of glycerine flowing per sec:
V = M / ρ
= 4 X 10-3 / (1.3 X 103)  =  3.08 X 10-6 m3s-1
According to Poiseville’s formula, we have the relation for the rate of flow:
V = πpr4 / 8ηl
Where, p is the pressure difference between the two ends of the tube
∴ p = V8ηl / πr4
= 3.08 X 10-6 X 8 X 0.83 X 1.5 / [ π X (0.01)4 ]
= 9.8 × 102 Pa
Reynolds’ number is given by the relation:
R = 4pV / πdη
= 4 X 1.3 X 103 X 3.08 X 10-6 / ( π X 0.02 X 0.83)
= 0.3
Reynolds’ number is about 0.3. Hence, the flow is laminar.