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 co-efficient of viscosity of water $= 10^{-2}$ poise.

Question

A metal block of base area $0.20 \,m^2$ is placed on a table, as shown in figure. A liquid film of thickness $0.25\, mm$ is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N and moves with a constant speed. If the viscosity of the liquid is $5.0×10^{−3}Pl$, the speed of block is ________ $×10^{−3}m/s$

Answer 1

To find the shearing stress between the horizontal layers of water, we use the formula for shearing stress in a fluid, which is given by:

\[ \tau = \eta \frac{dv}{dy} \]

Where:

$\tau$ is the shearing stress.
$\eta$ is the coefficient of viscosity.
$\frac{dv}{dy}$ is the velocity gradient perpendicular to the flow direction.

Given:

The velocity of water near the surface, $v = 18 \, \text{km/hr} = 18000 \, \text{m/hr} = 5 \, \text{m/s}$ (using conversion $1 \, \text{km/hr} = \frac{5}{18} \, \text{m/s}$).
The depth of the river, $d = 5 \, \text{m}$.
The coefficient of viscosity, $\eta = 10^{-2} \, \text{poise} = 10^{-3} \, \text{Ns/m}^2$ (using conversion $1 \, \text{poise} = 0.1 \, \text{Ns/m}^2$).

The velocity at the bottom layer is considered to be zero since the bottom layer is in contact with the riverbed, hence not moving. Therefore, the velocity gradient is given by:

\[ \frac{dv}{dy} = \frac{v}{d} = \frac{5 \, \text{m/s}}{5 \, \text{m}} = 1 \, \text{s}^{-1} \]

Substituting the values into the shearing stress formula:

\[ \tau = \eta \frac{dv}{dy} = 10^{-3} \, \text{Ns/m}^2 \times 1 \, \text{s}^{-1} = 10^{-3} \, \text{N/m}^2 \]

Therefore, the shearing stress between the horizontal layers of water is 10^{-3} \, \text{N/m}^2.

The correct answer is: 10^{-3} \, \text{N/m}^2

Answer 2