Question

The velocity of water in a river is 18 km/hr near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water is \( 10^{-2} \) poise:

• A. \( 10^{-1} \) poise
• B. \( 10^{-2} \) N/m\(^2\)
• C. \( 10^{-3} \) N/m\(^2\)
• D. \( 10^{-4} \) N/m\(^2\)

Hint:

Shearing stress in a fluid is directly related to the coefficient of viscosity and the velocity gradient.

Answer 1

The Correct Option is C

Step 1: Shearing stress is given by the formula: \[ \tau = \eta \dfrac{du}{dy}, \] where \( \eta \) is the coefficient of viscosity, \( du \) is the change in velocity, and \( dy \) is the distance between the layers.
Step 2: The velocity gradient is \( \dfrac{du}{dy} = \dfrac{18 \times 10^3}{5} = 3.6 \times 10^3 \, \text{m/s/m} \).
Step 3: The shearing stress is: \[ \tau = 10^{-2} \times 3.6 \times 10^3 = 3.6 \times 10^{-2} \, \text{N/m}^2 = 10^{-3} \, \text{N/m}^2. \]
Final Answer: \[ \boxed{10^{-3} \, \text{N/m}^2} \]