Question

The shear stress at a point in a liquid is found to be 0.03 N/m². The velocity gradient at the point is 0.15 s⁻¹. What will be its viscosity (in Poise)?

• A. 2
• B. 0.5
• C. 0.2
• D. 20

Answer 1

Answer: (A)

The viscosity is 2 Poise.

Answer 2

The relationship between shear stress ($\tau$), dynamic viscosity ($\mu$), and velocity gradient ($\frac{du}{dy}$) is given by Newton's Law of Viscosity:

$\tau = \mu \frac{du}{dy}$

Given values:

• Shear stress ($\tau$) = 0.03 N/m²
• Velocity gradient ($\frac{du}{dy}$) = 0.15 s⁻¹

Calculate dynamic viscosity ($\mu$) in SI units:

Rearrange the formula: $\mu = \frac{\tau}{\frac{du}{dy}}$

Substitute the values:

$$\mu = \frac{0.03 \, \text{N/m}^2}{0.15 \, \text{s}^{-1}} = 0.2 \, \text{N s/m}^2$$

Convert viscosity to Poise:

1 Poise = 0.1 N s/m²

$$\mu (\text{in Poise}) = 0.2 \, \text{N s/m}^2 \times \frac{1 \, \text{Poise}}{0.1 \, \text{N s/m}^2} = 2 \, \text{Poise}$$