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The space between two parallel plates kept 4 mm apart is filled with an oil of dynamic viscosity 0.2 Pa-s. What is the shear stress on the lower fixed plate, if the upper one is moved with a velocity of 1.6 m/s? .
  • a)
    100 Pa
  • b)
    90 Pa
  • c)
    80 Pa
  • d)
    60 Pa
Correct answer is option 'C'. Can you explain this answer?
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Given:
Distance between two parallel plates, d = 4 mm = 0.004 m
Dynamic viscosity of oil, μ = 0.2 Pa-s
Velocity of the upper plate, v = 1.6 m/s

Shear stress can be calculated using the formula:
Shear stress = μ x (velocity gradient)

Calculation:
The velocity gradient can be calculated as:
Velocity gradient = (velocity of upper plate - velocity of lower plate) / distance between plates
As the lower plate is fixed, its velocity is 0. Thus, the velocity gradient can be calculated as:
Velocity gradient = (1.6 m/s - 0 m/s) / 0.004 m = 400 s^-1

Substituting the given values in the formula, we get:
Shear stress = 0.2 Pa-s x 400 s^-1 = 80 Pa

Therefore, the shear stress on the lower fixed plate is 80 Pa.
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