Flow of Liquids Numerical on Stokes’ Law and Terminal Velocity Class-11 Nootan ISC Physics

Flow of Liquids Numerical on Stokes’ Law and Terminal Velocity 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 Stokes’ Law and Terminal Velocity 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 Stokes’ Law and Terminal Velocity Academic Session 2024-2025

Numericals on Stokes’ Law and Terminal Velocity

Flow of Liquids Class-11 Nootan ISC Physics Solutions Ch-15 Solutions of Kumar and Mittal Physics, Nageen Prakashan

Que-11: What will be the terminal velocity of a steel ball of radius 2 mm in glycerin ?

Ans- v = 2/9 [r²(ρ – σ) g] / η

=> [2 x (2 x 10^-3)² (8.0 – 1.2) x 10³] / (9 x 0.85)

=> (2 x 4 x 10^-6 x 6.8 x c) / (9 x 0.85)

=> 0.07 m/s = 7 cm/s

Que-12: An iron ball of radius 0.3 cm falling through a column of oil of density 0.94 g/cm³ attains a terminal velocity of 0.5 cm/s. What is the coefficient of viscosity of the oil? The density of iron is 7.8 g/cm³.

Ans- v = 2/9 [r²(ρ – σ) g] / η

=> η = [2 r² g (ρ – σ)] / 9 v

=> [2 (7.8 x 0.94) x 10³ x (3 x 10^-2)² x 9.8] / 9 x (0.5 x 10^-2)

=> 269 x 10^-1 (SI unit)

=> 269 poise

Que-13: A steel shot of diameter 2 mm is dropped in a viscous liquid filled in a drum. Find the terminal speed of the shot. Density of the material of the shot = 8.0 × 103 kg/m³, density of liquid=1.0 x 10³ kg/m³. Coefficient of viscosity of liquid = 1.0 kg/m-s, g=10 m/s².

Ans- v = 2/9 [r²(ρ – σ) g] / η

=> [2 x (8 – 1) x 10³ x (2 x 10^-3)² x 10] / (9 x 1)

=> 1.55 x 10^-2 m/s = 1.55 cm/s

Que-14: If an oil drop of density 0.95 x 10³ kg/m³ and radius 10^-4 cm is falling in air whose density is 1.3 kg/m³ and coefficient of viscosity is 18 x 10^-6 kg/m-s. Calculate the terminal speed of the drop.

Ans- v = 2/9 [r²(ρ – σ) g] / η

=> [2 x 0.95 x 10³ x (10^-6)² x 9.8] / (9 x 18 x 10^-6)

=> 0.0115 x 10^-2 m/s = 0.0115 cm/s

density of air is negligible as compared to liquid.

Que-15: The terminal velocity of a copper ball of radius 2.0 mm in falling through a tank of oil is 6.5 cm/s. Find the viscosity of the oil. The densities of copper and oil are 8.9 x 10³ kg/m³ 3 and 1.5 x 10³ kg/m³ respectively.

Ans- v = 2/9 [r²(ρ – σ) g] / η

=> η = [2 r² g (ρ – σ)] / 9 v

=> [2 (8.9 – 1.5) x 10³ x (2 x 10^-3)² x 9.8] / 9 x (6.5 x 10^-2)

=> 0.99 N-s/m²

Que-16: Two drops of equal radii are falling through air with a terminal velocity of 5 cm/s. If they coalesce into one drop, what will be the terminal velocity of the new drop?

Ans- V ∝ r² => V ∝ v^(2/3)    {v = volume}

=> (V1 / V2) = (v1 / 2v1)

=> (5 / V2) = (1 / 2)^(2/3)

=> (5 / V2) = (1 / 1.59)

=> V2 = 5 x 2^(2/3) cm/s

Que-17: An air bubble (radius 0.4 mm) rises up in water. Determine the terminal speed of the bubble. Density of air is negligible.

Ans- v = 2/9 [r²(ρ – σ) g] / η

=> [2 x (0 – 1000) x (0.4 x 10^-3)² x 9.8] / 0.001

=> – 0.348 m/s

negative sign shows that bubble is rising up.

Que-18:  With what terminal velocity will an air bubble of diameter 0.8 mm rise in a liquid of density 900 kg/m³ 13 and viscosity 0.15 N s/m² ? What will be the terminal velocity of the same bubble in water? Ignore density of air.

Ans- v = 2/9 [r²(0 – σ) g] / η

(air density is negligible so ρ = 0)

=> [-2 x 900 x (0.4 x 10^-3)² x 9.8] / (9 x 0.15)

=> -2.1 x 10^-3 m/s

in water,

v = [-2 x 1000 x (0.4 x 10^-3)² x 9.8] / (9 x 0.001)

=> -0.348 m/s

—:  end of Flow of Liquids Numerical on Stokes’ Law and Terminal Velocity Class-11 Nootan ISC Solutions :—

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