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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate is
  • a)
    0.651 × 10–3
  • b)
    0.651
  • c)
    6.51
  • d)
    0.651 × 103
Correct answer is option 'B'. Can you explain this answer?
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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2...

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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2...
Given data:

Kinematic viscosity (ν) = 7.4 × 10^-7 m^2/s

Specific gravity (S.G) = 8.88

Velocity of top plate (u) = 0.5 m/s

Gap between the plates (h) = 0.5 mm = 0.0005 m

We can find the shear stress (τ) on the surface of the top plate using the formula:

τ = μ × du/dy

where μ is the dynamic viscosity of the fluid and du/dy is the velocity gradient.

Velocity gradient (du/dy) can be found using the formula:

du/dy = u/h

Substituting the given values, we get:

du/dy = 0.5/0.0005 = 1000 m/s^2

Dynamic viscosity (μ) can be found using the formula:

μ = ν × ρ

where ρ is the density of the fluid.

Density (ρ) can be found using the formula:

ρ = S.G × 1000 kg/m^3

Substituting the given values, we get:

ρ = 8.88 × 1000 = 8880 kg/m^3

Substituting the values of μ and du/dy in the formula for shear stress, we get:

τ = μ × du/dy

τ = (7.4 × 10^-7) × 8880 × 1000 = 6.5512 Pa

Rounding off to three significant figures, we get:

τ = 0.651 Pa

Therefore, the shear stress on the surface of the top plate is 0.651 Pa (option B).
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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2...
Answer is C
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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 × 10–3b)0.651c)6.51d)0.651 × 103Correct answer is option 'B'. Can you explain this answer?
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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 × 10–3b)0.651c)6.51d)0.651 × 103Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 × 10–3b)0.651c)6.51d)0.651 × 103Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 × 10–3b)0.651c)6.51d)0.651 × 103Correct answer is option 'B'. Can you explain this answer?.
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