**Viscosity**

- Viscosity is the property of a fluid that resists the force tending to cause the fluid to flow.
- It is analogous to friction in solids.
- Example:-
- Consider 2 glasses one filled with water and the other filled with honey.
- Water will flow down the glass very rapidly whereas honey won’t.This is because honey is more viscous than water.
- Therefore in order to make honey flow we need to apply greater amount of force. Because honey has the property to resist the motion.

- Viscosity comes into play when there is relative motion between the layers of the fluid.The different layers are not moving at the same pace.

__Coefficient of Viscosity__

- Coefficient of viscosity is the measure of degree to which a fluid resists flow under an applied force.
- This means how much resistance does a fluid have to its motion.
- Ratio of shearing stress to the strain rate.
- It is denoted by ‘η’.
- Mathematically
- Δt=time , displacement =Δx
- Therefore,
- shearing stress =Δx/l where l= length
- Strain rate=Δx/lΔt

- η=shearing stress/strain rate
- (F/A)/(Δx/lΔt) = (Fl)/vA where Δx/t=v

- Therefore
**η=(Fl)/ vA** - I. Unit:- Poiseiulle (PI)/Pa/Nsm
^{-2} - Dimensional Formula: [ML
^{-1}T^{-1}]

__Answer:-__

The metal block moves to the rightbecause of the tension in the string. The tension

T is equal in magnitude to the weight of thesuspended mass m. Thus the shear force F is

F = T = mg = 0.010 kg × 9.8 ms^{–2} = 9.8 × 10^{-2} N

Shear stress on the fluid = F/A =9.8 x10^{–2}/0.10

Strain rate =v/l =0.085/0.030

η= stress/strain rate

= (9.8×10^{-2}x 0.30 x10^{-3}m)/ (0.085ms^{-1}x0.10m^{2})

= 3.45 × 10^{-3} Pa s

**Stokes Law**

- The force that retards a sphere moving through a viscous fluid is directly ∝to the velocity and the radius of the sphere, and the viscosity of the fluid.
- Mathematically:-
**F =6πηrv**where- Let retarding force F∝v where v =velocity of the sphere
- F ∝ r where r=radius of the sphere
- F∝η where η=coefficient of viscosity
- 6π=constant

- Stokes law is applicable only to laminar flow of liquids.It is not applicable to turbulent law.
- Example:-Falling raindrops
- Consider a single rain drop, when rain drop is falling it is passing through air.
- The air has some viscosity; there will be some force which will try to stop the motion of the rain drop.
- Initially the rain drop accelerates but after some time it falls with constant velocity.
- As the velocity increases the retarding force also increases.
- There will be viscous force F
_{v}and bind force F_{b}acting in the upward direction.There will also be F_{g}gravitational force acting downwards. - After some time F
_{g}= F_{r}(F_{v}+F_{b}) - Net Force is 0. If force is 0 as a result acceleration also becomes 0.

- Let retarding force F∝v where v =velocity of the sphere

**Terminal Velocity**

- Terminal velocity is the maximum velocity of a body moving through a viscous fluid.
- It is attained when force of resistance of the medium is equal and opposite to the force of gravity.
- As the velocity is increasing the retarding force will also increase and a stage will come when the force of gravity becomes equal to resistance force.
- After that point velocity won’t increase and this velocity is known as terminal velocity.
- It is denoted by ‘v
_{t}’.Where_{t}=terminal. - Mathematically:-
- Terminal velocity is attained when Force of resistance = force due to gravitational attraction.
- 6πηrv =mg
- 6πηrv = densityxVg (Because density=m/V), density=ρ – σ where ρ and σ are the densities of the sphere and the viscous medium resp.
- 6πηrv = (ρ – σ)x4/3πr
^{3}g where Volume of the sphere(V) =4/3πr^{3} - By simplifying
- =(ρ – σ)gx4/3r
^{2}x1/(6η)

- =(ρ – σ)gx4/3r
**v**Where(v=v_{t}=2r^{2}(ρ – σ)g/9 η .This is the terminal velocity._{t})

** Problem: **The terminal velocity of acopper ball of radius 2.0 mm fallingthrough a tank of oil at 20

^{o}C is 6.5 cm s

^{-1}.Compute the viscosity of the oil at 20

^{o}C.Density of oil is 1.5 × 10

^{3}kg m

^{-3}, density of

copper is 8.9 × 10^{3} kg m^{-3}.

__Answer:-__

Given: – v_{t} = 6.5 × 10^{-2} ms^{-1}, a = 2 × 10^{-3} m,g = 9.8 ms^{-2}, ρ = 8.9 × 10^{3} kg m^{-3},

σ =1.5 × 10^{3} kg m^{-3}. From Equation: -v_{t} =2r^{2 }(ρ – σ) g/9 η

=2/9(2 x10^{-3} m^{2} x 9.8ms^{-2}/ 6.5×10^{-2}ms^{-1}) x 7.4 10^{3} kgm^{-3}

=9.9 x 10^{-1} kgm^{-1}s^{-1}

** Problem: **Calculate the terminal velocity in air of an oil drop of radius 2×10

^{5}m from the following data g=9.8m/s

^{2}; coefficient of viscosity of air =1.8×10

^{-5}Pas; density of oil=900kg/m

^{3}.The up thrust of air may be neglected?

__Answer:__

Radius r =2×10^{-5} m

g=9.8m/s^{2}

η=1.8×10^{-5}Pas

ρ =900kg/m^{3}

v=v_{t}when 6πηrv = mg

6πηrv =ρx4/3r^{3}g

Simplifying: – v_{t} =2r^{2}gρ/9π = (2 x (2×10^{-5})^{2}x9.8×900)/9×1.8×10^{-5}

=4.36cm/s