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Properties of Bulk Matter - Viscosity (Stokes' Law, Terminal Velocity)

Grade 11ICSEPhysics

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

🔑Concepts

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Viscosity is the property of a fluid by virtue of which an internal frictional force (viscous drag) comes into play when the fluid is in motion, opposing the relative motion between its layers.

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Newton's Law of Viscosity states that the viscous force FF between two layers of area AA and velocity gradient dvdx\frac{dv}{dx} is given by F=−ηAdvdxF = -\eta A \frac{dv}{dx}, where η\eta is the coefficient of viscosity.

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The S.I. unit of the coefficient of viscosity is Pascal-second (Pa s)\text{Pascal-second (Pa s)} or kg m−1s−1\text{kg m}^{-1}\text{s}^{-1}. The CGS unit is the Poise (1 Pa s=10 Poise1 \text{ Pa s} = 10 \text{ Poise}).

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Stokes' Law states that for a small spherical body of radius rr moving through a fluid of viscosity η\eta with velocity vv, the viscous drag force is F=6πηrvF = 6\pi \eta r v.

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Terminal Velocity is the maximum constant velocity acquired by a body while falling through a viscous medium. This occurs when the weight of the body is balanced by the sum of the buoyant force and the viscous drag.

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Factors affecting Terminal Velocity: It is directly proportional to the square of the radius of the body (vt∝r2v_t \propto r^2) and the difference in densities between the body and the fluid.

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

📐Formulae

F=ηAdvdxF = \eta A \frac{dv}{dx}

F=6πηrvF = 6\pi \eta r v

vt=2r2(ρ−σ)g9ηv_t = \frac{2r^2(\rho - \sigma)g}{9\eta}

η=2r2(ρ−σ)g9vt\eta = \frac{2r^2(\rho - \sigma)g}{9v_t}

Dimensions of η:[ML−1T−1]\text{Dimensions of } \eta: [M L^{-1} T^{-1}]

💡Examples

Problem 1:

A steel ball of radius 0.3 mm0.3 \text{ mm} falls through a column of oil of density 0.94×103 kg/m30.94 \times 10^3 \text{ kg/m}^3. If the density of steel is 7.8×103 kg/m37.8 \times 10^3 \text{ kg/m}^3 and the terminal velocity is 0.05 m/s0.05 \text{ m/s}, calculate the coefficient of viscosity of the oil. (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

Given: r=0.3×10−3 mr = 0.3 \times 10^{-3} \text{ m}, ρ=7.8×103 kg/m3\rho = 7.8 \times 10^3 \text{ kg/m}^3, σ=0.94×103 kg/m3\sigma = 0.94 \times 10^3 \text{ kg/m}^3, vt=0.05 m/sv_t = 0.05 \text{ m/s}. Using the formula: η=2r2(ρ−σ)g9vt\eta = \frac{2r^2(\rho - \sigma)g}{9v_t} η=2×(0.3×10−3)2×(7.8×103−0.94×103)×9.89×0.05\eta = \frac{2 \times (0.3 \times 10^{-3})^2 \times (7.8 \times 10^3 - 0.94 \times 10^3) \times 9.8}{9 \times 0.05} η=2×9×10−8×6.86×103×9.80.45\eta = \frac{2 \times 9 \times 10^{-8} \times 6.86 \times 10^3 \times 9.8}{0.45} η≈0.269 Pa s\eta \approx 0.269 \text{ Pa s}

Explanation:

The coefficient of viscosity is determined by substituting the terminal velocity, radius, and the difference in densities of the sphere and the fluid into the rearranged terminal velocity equation.

Problem 2:

Eight rain drops of radius 1 mm1 \text{ mm} each falling down with terminal velocity 5 cm/s5 \text{ cm/s} coalesce to form a single bigger drop. Find the terminal velocity of the bigger drop.

Solution:

Let the radius of the small drop be rr and the big drop be RR. Since volume is conserved: 43πR3=8×43πr3  ⟹  R=2r\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \implies R = 2r. Terminal velocity vtv_t is proportional to r2r^2. Thus, vRvr=R2r2=(2r)2r2=4\frac{v_R}{v_r} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4. Therefore, vR=4×vr=4×5 cm/s=20 cm/sv_R = 4 \times v_r = 4 \times 5 \text{ cm/s} = 20 \text{ cm/s}.

Explanation:

When drops coalesce, the new radius is calculated via volume conservation. Since terminal velocity is directly proportional to the square of the radius, the new velocity is the old velocity multiplied by the square of the radius ratio.

Viscosity (Stokes' Law, Terminal Velocity) Revision - Class 11 Physics ICSE