Flow of Liquids Numerical on Viscosity Poiseuille’s Formula and Reynold’s Number Class-11 Nootan ISC Physics

Flow of Liquids Numerical on Viscosity Poiseuille’s Formula and Reynold’s Number Class-11 Nootan ISC Physics Solutions Ch-15. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Flow of Liquids Numerical on Viscosity Poiseuille's Formula and Reynold's Number Class-11 Nootan ISC Physics

Flow of Liquids Numerical on Viscosity Poiseuille’s Formula and Reynold’s Number Class-11 Nootan ISC Physics

Board ISC
Class 11
Subject Physics
Writer Kumar and Mittal
Publication  Nageen Prakashan
Chapter-15 Flow of Liquids
Topics Numericals on Viscosity, Poiseuille’s Formula and Reynold’s Number
Academic Session 2024-2025

Numericals on Viscosity, Poiseuille’s Formula and Reynold’s Number

Que-19: There is a 1 mm thick layer of oil between a flat plate of area 10^-2 m² and a big plate. How much force is required to move the plate with a velocity of 1.5 cm/s. The coefficient of viscosity of oil is 1 poise.

Ans- F = η A v / l

=> (1 x 10^-1 x 10^-2 x 1.5 x 10^-2) / (1 x 10^-3)

=> 1.5 x 10^-2 N

Que-20: A square plate of 10 cm side moves parallel to another plate with a relative velocity of 10 cm/s, both plates immersed in water. If the viscous force is 2 x 10³ N, calculate the perpendicular distance between the plates.

Ans- F = η A v / l

=> l = η A v / F

=> [0.001 x (0.1 x 0.1) x 10 x 10^-2] / (2 x 10^-3)

=> 0.05 x 10^-2 m = 0.05 cm 

Que-21: Water is flowing through a horizontal tube 4 km in length and 4 cm in radius at a rate of 20 liter/sec. Calculate the pressure required to maintain the flow in terms of the height of mercury column.

Ans- According to Poiseuille’s Formula

Q = π p r^4 / 8 η l

p = Q . 8 η l / π r^4

=> [20 x 10^-3 x 8 x 0.001 x 4000] / [3.14 x (4 x 10^-2)^4]

=> (20 x 8 x 4 x 10^-3) / (3.14 x 256 x 10^-8)

=> 0.8 x 10^5 pa

if height of mercury column is h

then h ρ g = 0.8 x 10^5

=> h = (0.8 x 10^5) / (13.6 x 10³ x 9.8)

=> 0.006 x 10²

=> 0.6 m = 60 cm

Que-22: The level of blood in a bottle used to give blood to a patient is 1.3 m high above the needle which is 3 cm in length and has an internal diameter of 0.36 mm. If 4.5 cm³ of blood is passing through the needle per minute, calculate the viscosity of blood. The density of blood is 1020 kg/m³. Ignore fall in the level of blood in the bottle.

Ans- Q = π p r^4 / 8 η l

η =  π p r^4  / 8 Q l

=> [3.14 x 1.3 x 1020 x 9.8 x (0.18 x 10^-3)^4] / (8 x 4.5 x 10^-6 x 3 x 10^-2)

=> 2.38 x 10^-3 N-s/m²

—:  end of Flow of Liquids Numerical on Viscosity Poiseuille’s Formula and Reynold’s Number Class-11 :—

Return to :  Nootan Solutions for ISC Physics Class-11 Nageen Prakashan

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