Question

Water is flowing on a horizontal fixed surface such that its flow velocity varies with \(y\) (vertical direction) as \[ v = k\!\left(\frac{2y^2}{a^2}-\frac{y^3}{a^3}\right). \] If coefficient of viscosity for water is \(\eta\), what will be the shear stress between layers of water at \(y=a\)?

• A. \(\dfrac{\eta k}{a}\)
• B. \(\eta ka\)
• C. \(\dfrac{\eta a}{k}\)
• D. None of these

Hint:

Shear stress depends on velocity gradient: \[ \tau=\eta\frac{dv}{dy} \] Not on velocity itself.

Answer 1

The Correct Option is A

Step 1: Shear stress: \[ \tau = \eta \frac{dv}{dy} \]
Step 2: Differentiate velocity: \[ \frac{dv}{dy} = k\left(\frac{4y}{a^2}-\frac{3y^2}{a^3}\right) \]
Step 3: At \(y=a\): \[ \frac{dv}{dy} = k\left(\frac{4}{a}-\frac{3}{a}\right)=\frac{k}{a} \]
Step 4: Hence: \[ \tau = \frac{\eta k}{a} \]