Consider a fluid between two horizontal parallel flat plates 5 mm apart as shown in the figure. The top plate of dimensions 0.5 m $ \times $ 2 m is towed with an applied horizontal force $ F $ of $ 0.01 \, \text{N} $, while the infinitely long bottom plate is kept fixed. The horizontal velocity profile between the plates is assumed to be linear. If the dynamic viscosity ($ \mu $) of the fluid is $ 0.89 \times 10^{-3} \, \text{N}\text{-}\text{s}/\text{m}^2 $, then the towing velocity of the top plate is ………… $ \text{m/s} $ (rounded off to three decimal places).

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Step 1: Use Newton’s law of viscosity. The shear stress ($ \tau $) in a fluid is given by: $$ \tau = \mu \cdot \frac{du}{dy}, $$ where: - $ \tau $ is the shear stress, - $ \mu $ is the dynamic viscosity, - $ \frac{du}{dy} $ is the velocity gradient. For a linear velocity profile: $$ \frac{du}{dy} = \frac{u}{h}, $$ where: - $ u $ is the velocity of the top plate, - $ h $ is the distance between the plates. Step 2: Relate shear stress to force. The shear stress ($ \tau $) is related to the force ($ F $) acting on the top plate by: $$ \tau = \frac{F}{A}, $$ where: - $ A $ is the area of the top plate. Substitute $ \tau $ into the equation: $$ \frac{F}{A} = \mu \cdot \frac{u}{h}. $$ Step 3: Solve for the velocity of the top plate ($ u $). Rearranging the equation: $$ u = \frac{F \cdot h}{\mu \cdot A}. $$ Substitute the given values: - $ F = 0.01 \, \text{N} $, - $ h = 5 \, \text{mm} = 0.005 \, \text{m} $, - $ \mu = 0.89 \times 10^{-3} \, \text{N·s/m}^2 $, - $ A = 0.5 \times 2 = 1 \, \text{m}^2 $. $$ u = \frac{0.01 \cdot 0.005}{0.89 \times 10^{-3} \cdot 1}. $$ Simplify: $$ u = \frac{0.00005}{0.00089} = 0.053 \, \text{m/s}. $$ Conclusion: The towing velocity of the top plate is $ 0.053 \, \text{m/s} $ (rounded off to three decimal places).

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