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The velocity distribution near the solid wall at a section in a laminar flow is given by u = 5 sin(5πy) for y < 0.1 m. The shear stress at a section y = 0.05m (if the dynamic viscosity of the fluid is 5 poise) will be
  • a)
    27.66 N/m2
  • b)
    39.27 N/m2
  • c)
    7.85 N/m2
  • d)
    zero
Correct answer is option 'A'. Can you explain this answer?
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Πy) where u is the velocity in m/s and y is the distance from the wall in m.

To find the maximum velocity and its location, we need to take the derivative of the velocity distribution with respect to y and set it equal to zero:

du/dy = 25π cos(5πy) = 0

cos(5πy) = 0

y = 0, 0.2, 0.4, 0.6, 0.8 m

These are the locations where the velocity is maximum.

To find the maximum velocity, we can substitute these values of y into the velocity distribution equation:

u(0) = 5 sin(0) = 0 m/s

u(0.2) = 5 sin(π) = 0 m/s

u(0.4) = 5 sin(2π) = 0 m/s

u(0.6) = 5 sin(3π) = -5 m/s

u(0.8) = 5 sin(4π) = 0 m/s

Therefore, the maximum velocity is -5 m/s and it occurs at y = 0.6 m.
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The velocity distribution near the solid wall at a section in a laminar flow is given by u = 5 sin(5πy) for y < 0.1 m. The shear stress at a section y = 0.05m (if the dynamic viscosity of the fluid is 5 poise) will bea)27.66 N/m2b)39.27 N/m2c)7.85 N/m2d)zeroCorrect answer is option 'A'. Can you explain this answer?
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