Question

Find the kinematic viscosity of an oil having density 981 kg/m3. The shear stress at
a point in oil is 0.2452 N/m2 and velocity gradient at t 0.2 per second?​

Answer 1

Answer:

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Answer 2

Answer:

The kinematic viscosity of oil, γ, measured is 1.24 × 10⁻³ m²/s.

Step-by-step explanation:

Given,

The density of oil, ρ = 981 Kg/m³

The shear stress at a point in oil, σ = 0.2452 N/m²

The velocity gradient, $\frac{dv}{dy}$ = 0.2 s⁻¹

 The kinematic viscosity of oil, γ =?

As we know,

•  The kinematic viscosity (γ) is defined as the ratio of dynamic viscosity (μ) and the density of the liquid (ρ).
• $\nu = \frac{\mu}{\rho}$       ------- equation (1)

Now,

• The dynamic viscosity (μ) can be calculated by using the formula for shear stress given below:
• $\sigma = \mu \frac{dv}{dy}$    ------- equation (2)

Here,

• σ = The shear stress
• μ = The dynamic viscosity
• $\frac{dv}{dy}$ = The velocity gradient

After putting all the given values in equation (2), we get:

• $0.2452 = \mu\times 0.2$
• μ = 1.226 N-s/m²

Now, after putting the value of μ in equation (1), we get:

• $\nu = \frac{1.226}{981}$
• γ = 0.00124 m²/s = 1.24 × 10⁻³ m²/s

Therefore, the kinematic viscosity of oil, γ = 1.24 × 10⁻³ m²/s.