Question

The velocity of upper layer of water in a river is \(36\ kmh^{–1}\). Shearing stress between horizontal layers of water is \(10^{–3} Nm^{–2}\). Depth of the river is ______ m. (Co-efficient of viscosity of water is \(10^{–2} Pa.s\))

Answer 1

Correct Answer: 100

To find the depth of the river, we need to understand the relationship between the velocity gradient, shearing stress, and viscosity. The velocity gradient is given by \(\frac{dv}{dy}\), where \(v\) is the velocity and \(y\) is the distance or depth in this context. The formula relating shearing stress \(\tau\), velocity gradient, and viscosity \(\eta\) is:

\(\tau = \eta \cdot \frac{dv}{dy}\)

Given:

• \(\tau = 10^{-3} \ Nm^{-2}\)
• \(\eta = 10^{-2} \ Pa \cdot s\)
• Velocity of the upper layer, \(v = 36 \ km/h = 10 \ m/s\) (converted from km/h to m/s)

Assuming the bottom layer of the river is stationary, the velocity gradient is given by:

\(\frac{dv}{dy} = \frac{v}{d}\)

Here, \(d\) is the depth we need to find. Substitute the known values into the shearing stress equation:

\(10^{-3} = 10^{-2} \cdot \frac{10}{d}\)

Solving for \(d\):

\(d = \frac{10^{-2} \cdot 10}{10^{-3}} = 100 \ m\)

The calculated depth of the river, \(100 \ m\), fits within the expected range of 100 to 100 meters, indicating a correct solution.