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An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m²/s, specific gravity, 0.88) 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
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- 0.651 × 10–3
- 0.651
- 6.51
- 0.651 ×103
Correct Option: B
Given:
υ = 7.4 × 10–7 m2 /s
ρ = 0.88 × 1000 = 880
v = 0.5 m/s
Shear stress, τ = μ⋅ | = (υ × ρ) × | ||
dy | dy |
= 7.4 × 10–7 × 800 × | = 0.6512. | |
(0.0005) |