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Mechanical Properties of Fluids - Viscosity and Stokes' Law

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Viscosity: It is the property of a fluid by virtue of which an internal frictional force comes into play when the fluid is in motion and opposes the relative motion between its different layers.

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Velocity Gradient: The rate of change of velocity with distance measured in the direction perpendicular to the flow, represented as dvdx\frac{dv}{dx}.

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Coefficient of Viscosity (η\eta): Defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel layers of a liquid. Its SI unit is Pa⋅sPa \cdot s or N⋅s⋅m−2N \cdot s \cdot m^{-2} (Poiseuille).

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Stokes' Law: States that when a small spherical body moves through a viscous medium, the backward dragging force (FF) acting on it is directly proportional to the radius of the sphere (rr), its velocity (vv), and the coefficient of viscosity (η\eta) of the medium.

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Terminal Velocity: The maximum constant velocity acquired by a body while falling through a viscous medium. This occurs when the viscous force and buoyant force together balance the weight of the body.

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Effect of Temperature: The viscosity of liquids decreases with an increase in temperature, whereas the viscosity of gases increases with an increase in temperature.

📐Formulae

F=−ηAdvdxF = -\eta A \frac{dv}{dx} (Newton's Law of Viscosity)

F=6πηrvF = 6\pi \eta r v (Stokes' Law)

vt=2r2(ρ−σ)g9ηv_t = \frac{2r^2 (\rho - \sigma) g}{9\eta} (Terminal Velocity formula, where ρ\rho is density of the body and σ\sigma is density of the fluid)

η=2r2(ρ−σ)g9vt\eta = \frac{2r^2 (\rho - \sigma) g}{9v_t} (Coefficient of Viscosity from terminal velocity)

💡Examples

Problem 1:

A steel ball of radius 2×10−3 m2 \times 10^{-3} \text{ m} falls through a column of glycerin with a terminal velocity of 0.04 m/s0.04 \text{ m/s}. If the density of steel is 8000 kg/m38000 \text{ kg/m}^3 and the density of glycerin is 1200 kg/m31200 \text{ kg/m}^3, find the coefficient of viscosity of glycerin. (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

Given: r=2×10−3 mr = 2 \times 10^{-3} \text{ m}, vt=0.04 m/sv_t = 0.04 \text{ m/s}, ρ=8000 kg/m3\rho = 8000 \text{ kg/m}^3, σ=1200 kg/m3\sigma = 1200 \text{ kg/m}^3. Using the formula: η=2r2(ρ−σ)g9vt\eta = \frac{2r^2(\rho - \sigma)g}{9v_t}. Substituting values: η=2×(2×10−3)2×(8000−1200)×9.89×0.04\eta = \frac{2 \times (2 \times 10^{-3})^2 \times (8000 - 1200) \times 9.8}{9 \times 0.04}. η=2×4×10−6×6800×9.80.36\eta = \frac{2 \times 4 \times 10^{-6} \times 6800 \times 9.8}{0.36}. η≈1.48 Pa⋅s\eta \approx 1.48 \text{ Pa}\cdot\text{s}.

Explanation:

The terminal velocity is reached when the net force on the ball is zero. By rearranging the terminal velocity formula, we can solve for the unknown coefficient of viscosity η\eta.

Problem 2:

What is the viscous force acting on a rain drop of radius 1 mm1 \text{ mm} falling through air with a velocity of 2 m/s2 \text{ m/s}? Given ηair=1.8×10−5 N s/m2\eta_{\text{air}} = 1.8 \times 10^{-5} \text{ N s/m}^2.

Solution:

Using Stokes' Law: F=6πηrvF = 6\pi \eta r v. Here, r=1×10−3 mr = 1 \times 10^{-3} \text{ m}, v=2 m/sv = 2 \text{ m/s}, η=1.8×10−5 Pa⋅s\eta = 1.8 \times 10^{-5} \text{ Pa}\cdot\text{s}. F=6×3.14×1.8×10−5×10−3×2F = 6 \times 3.14 \times 1.8 \times 10^{-5} \times 10^{-3} \times 2. F=6.78×10−7 NF = 6.78 \times 10^{-7} \text{ N}.

Explanation:

Stokes' Law allows us to calculate the resistive force experienced by a spherical object moving through a fluid based on its dimensions and the fluid's properties.

Viscosity and Stokes' Law - Revision Notes & Key Formulas | CBSE Class 11 Physics