Notes : Define Stokes' Law and Terminal Velocity | Derivation, Formula, MCQ, JEE, NEET
Stokes' Law and Terminal Velocity Class 11 Physics Notes | Derivation, Formula, MCQ, Questions Answers for JEE, NEET chapter 9 Mechanical Properties of Fluids - Physicskund
Introduction
When a body moves through a viscous fluid, it drags the fluid layers in contact with it. The fluid layers farther away remain almost undisturbed. As a result, relative motion is produced between different layers of the fluid.
Due to this relative motion, a backward dragging force acts on the body and opposes its motion. This force is called viscous force or drag force.
The magnitude of this force increases with the velocity of the body.
Stokes' Law
Statement
According to Stokes,
The viscous drag force acting on a small spherical body moving through a viscous fluid is directly proportional to the coefficient of viscosity of the fluid, radius of the sphere and velocity of the sphere.
Mathematical Form
$$F \propto \eta$$
$$F \propto r$$
$$F \propto v$$
Combining,
$$F \propto \eta rv$$
or
$$F=k\eta rv \qquad ...(1)$$
where k is a constant.
Experimentally,
$$k=6\pi$$
Therefore,
$$\boxed{F=6\pi\eta rv} \qquad ...(2)$$
Where
- F = Viscous force
- η = Coefficient of viscosity
- r = Radius of sphere
- v = Velocity of sphere
Proof of Stokes' Law by Dimensional Analysis
Assume,
$$F=k\eta^a r^b v^c \qquad ...(3)$$
Dimensions
Force:
$$[F]=[MLT^{-2}]$$
Coefficient of viscosity:
$$[\eta]=[ML^{-1}T^{-1}]$$
Radius:
$$[r]=[L]$$
Velocity:
$$[v]=[LT^{-1}]$$
Substituting dimensions in equation (3),
$$ [MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c $$
$$ = M^aL^{-a+b+c}T^{-a-c} $$
Comparing Powers
For M,
$$a=1 \qquad ...(4)$$
For T,
$$-a-c=-2$$
Using equation (4),
$$-1-c=-2$$
$$c=1 \qquad ...(5)$$
For L,
$$-a+b+c=1$$
Using equations (4) and (5),
$$-1+b+1=1$$
$$b=1 \qquad ...(6)$$
From equations (4), (5) and (6),
$$a=b=c=1$$
Hence,
$$F=k\eta rv$$
Experimentally,
$$k=6\pi$$
Therefore,
$$\boxed{F=6\pi\eta rv}$$
Important Features of Stokes' Law
- Applicable only to small spherical bodies.
- Viscous force always acts opposite to the direction of motion.
- Viscous force increases with velocity.
- Viscous force depends on viscosity of the fluid.
- Viscous force depends on radius of the sphere.
Terminal Velocity
When a body is dropped in a viscous fluid, it is initially accelerated due to gravity.
As the velocity increases, the viscous force also increases.
After some time, the acceleration becomes zero and the body starts moving with a constant velocity.
Definition
The constant velocity attained by a body while moving through a viscous fluid when its acceleration becomes zero is called Terminal Velocity.
Expression for Terminal Velocity
Consider a spherical body of radius r and density ρ falling through a viscous fluid of density σ and coefficient of viscosity η.
Forces Acting on the Sphere
(i) Weight of Sphere (W)
The weight acts vertically downward.
$$W=mg \qquad ...(7)$$
Mass of sphere,
m=(Volume of sphere) ×(Density of sphere) ...(8)
$$m=\frac{4}{3}\pi r^3\rho \qquad ...(9)$$
Substituting equation (9) into equation (7),
$$W=\left(\frac{4}{3}\pi r^3\rho\right)g$$
Therefore,
$$W=\frac{4}{3}\pi r^3\rho g \qquad ...(10)$$
(ii) Upward Thrust (T)
According to Archimedes' Principle, a body immersed in a fluid experiences an upward force equal to the weight of the fluid displaced by it.
$$T=\text{Weight of fluid displaced} \qquad ...(11)$$
Mass of fluid displaced,
m=(Volume of sphere) ×(Density of fluid) .......(12)
$$m=\frac{4}{3}\pi r^3\sigma \qquad ...(13)$$
Since,
$$T=mg \qquad ...(14)$$
Substituting equation (13) into equation (14),
$$T=\left(\frac{4}{3}\pi r^3\sigma\right)g$$
Therefore,
$$T=\frac{4}{3}\pi r^3\sigma g \qquad ...(15)$$
(iii) Viscous Force (F)
According to Stokes' Law,
$$F=6\pi\eta rv \qquad ...(16)$$
where v is the terminal velocity.
Derivation of Terminal Velocity
Net downward force on the sphere is
$$W-T-F$$
At terminal velocity, acceleration becomes zero.
Hence,
$$W-T-F=0$$
or
$$F=W-T \qquad ...(17)$$
Substituting equations (10), (15) and (16) into equation (17),
$$ 6\pi\eta rv = \frac{4}{3}\pi r^3\rho g - \frac{4}{3}\pi r^3\sigma g \qquad ...(18) $$
$$ 6\pi\eta rv = \frac{4}{3}\pi r^3(\rho-\sigma)g \qquad ...(19) $$
$$ v = \frac{\frac{4}{3}\pi r^3(\rho-\sigma)g} {6\pi\eta r} \qquad ...(20) $$
$$ v = \frac{2r^2(\rho-\sigma)g} {9\eta} \qquad ...(21) $$
$$ \boxed{ v=\frac{2r^2(\rho-\sigma)g}{9\eta} } \qquad ...(22) $$
Conclusions from Terminal Velocity Formula
Dependence on Radius
$$v\propto r^2$$
Terminal velocity increases with square of radius.
Dependence on Density Difference
$$v\propto(\rho-\sigma)$$
Terminal velocity increases with density difference.
Dependence on Gravity
$$v\propto g$$
Terminal velocity increases with acceleration due to gravity.
Dependence on Viscosity
$$v\propto\frac{1}{\eta}$$
Terminal velocity decreases with increase in viscosity.
SI Unit and Dimensional Formula
Coefficient of Viscosity
SI Unit:
$$\boxed{\text{N s m}^{-2}}$$
or
$$\boxed{\text{Pa s}}$$
Dimensional Formula:
$$\boxed{[ML^{-1}T^{-1}]}$$
Important Formulae
Stokes' Law
$$F=6\pi\eta rv$$
Weight of Sphere
$$W=\frac{4}{3}\pi r^3\rho g$$
Upward Thrust
$$T=\frac{4}{3}\pi r^3\sigma g$$
Terminal Velocity
$$v=\frac{2r^2(\rho-\sigma)g}{9\eta}$$
Coefficient of Viscosity
$$\eta=\frac{2r^2(\rho-\sigma)g}{9v}$$
Frequently Asked Questions (FAQs)
1. What is Stokes' Law?
Stokes' Law states that the viscous drag force acting on a small spherical body moving through a viscous fluid is directly proportional to the coefficient of viscosity, radius of the sphere and velocity of the sphere.
2. What is terminal velocity?
The constant velocity attained by a body moving through a viscous fluid when its acceleration becomes zero is called terminal velocity.
3. Why does a body attain terminal velocity?
A body attains terminal velocity when the upward forces (viscous force and buoyant force) become equal to its weight.
4. On which factors does terminal velocity depend?
Terminal velocity depends on radius of sphere, density difference between sphere and fluid, acceleration due to gravity and coefficient of viscosity.
5. What is the SI unit of viscosity?
The SI unit of viscosity is N s m-2 or Pa s.
Multiple Choice Questions (MCQs)
1. According to Stokes' law, viscous force is
A) \(F \propto \eta r^2v\)
B) \(F \propto \eta rv\)
C) \(F \propto \eta^2rv\)
D) \(F \propto \eta rv^2\)
Answer: B) \(F \propto \eta rv\)
2. The SI unit of viscosity is
A) N m
B) N s m-2
C) N s
D) Pa
Answer: B) N s m-2
3. Terminal velocity is attained when
A) Weight = Buoyant force
B) Weight = Viscous force
C) Weight = Buoyant force + Viscous force
D) Viscous force = Buoyant force
Answer: C) Weight = Buoyant force + Viscous force
4. Terminal velocity is proportional to
A) r
B) r2
C) r3
D) 1/r
Answer: B) r2
5. The dimensional formula of viscosity is
A) [MLT-2]
B) [ML-1T-2]
C) [ML-1T-1]
D) [M0LT-1]
Answer: C) [ML-1T-1]
Fill in the Blanks
1. The force opposing the motion of a body in a fluid is called __________ force.
Answer: Viscous
2. According to Stokes' law, F=__________
Answer: \(6\pi\eta rv\)
3. The constant velocity attained by a body in a viscous fluid is called __________ velocity.
Answer: Terminal
4. The SI unit of viscosity is __________.
Answer: N s m-2
5. Terminal velocity is inversely proportional to __________.
Answer: Viscosity
True / False
1. Viscous force acts in the direction of motion.
Answer: False
2. Stokes' law is applicable to spherical bodies.
Answer: True
3. Terminal velocity increases with viscosity.
Answer: False
4. Terminal velocity is directly proportional to r2.
Answer: True
5. The dimensional formula of viscosity is [ML-1T-1].
Answer: True
Quick Revision Points
- Stokes' Law: \(F=6\pi\eta rv\)
- Viscous force always opposes motion.
- Terminal velocity is attained when net force becomes zero.
- At terminal velocity, \(W=T+F\).
- Terminal velocity is directly proportional to \(r^2\).
- Terminal velocity is inversely proportional to viscosity.
- SI unit of viscosity = N s m-2 or Pa s.
- Dimensional formula of viscosity = [ML-1T-1].
Very Short Answer Questions
1. Define Stokes' Law.
Answer:
Stokes' Law states that the viscous drag force acting on a small spherical body moving through a viscous fluid is directly proportional to the coefficient of viscosity, radius of the sphere and velocity of the sphere.
2. What is terminal velocity?
Answer:
The constant velocity attained by a body moving through a viscous fluid when its acceleration becomes zero is called terminal velocity.
3. Write the SI unit of viscosity.
Answer:
N s m-2 or Pa s.
4. Write Stokes' Law.
Answer:
$$F=6\pi\eta rv$$
5. Write the dimensional formula of viscosity.
Answer:
$$[ML^{-1}T^{-1}]$$
6. What is viscous force?
Answer:
The force that opposes the motion of a body through a fluid is called viscous force.
7. What is the SI unit of terminal velocity?
Answer:
m s-1
8. Name the forces acting on a sphere falling through a viscous fluid.
Answer:
Weight, buoyant force (upthrust) and viscous force.
Short Answer Questions
1. State Stokes' Law.
Answer:
According to Stokes, the viscous drag force acting on a small spherical body moving through a viscous fluid is directly proportional to the coefficient of viscosity of the fluid, radius of the sphere and velocity of the sphere.
Mathematically,
$$F=6\pi\eta rv$$
2. What are the forces acting on a sphere falling through a viscous fluid?
Answer:
Three forces act on the sphere:
- Weight of sphere (W)
- Upward thrust or buoyant force (T)
- Viscous force (F)
Weight acts downward while buoyant force and viscous force act upward.
3. Why does a body attain terminal velocity?
Answer:
As the velocity of a body increases, the viscous force acting on it also increases. Eventually, the upward forces become equal to the downward weight.
$$W=T+F$$
At this stage, the net force becomes zero and the body moves with a constant velocity called terminal velocity.
4. Write the expression for terminal velocity.
Answer:
The expression for terminal velocity is
$$ v=\frac{2r^2(\rho-\sigma)g}{9\eta} $$
where:
- r = radius of sphere
- ρ = density of sphere
- σ = density of fluid
- η = coefficient of viscosity
5. State Archimedes' Principle.
Answer:
A body wholly or partially immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by it.
6. Why does terminal velocity decrease with viscosity?
Answer:
Terminal velocity is inversely proportional to viscosity.
$$ v\propto \frac{1}{\eta} $$
Therefore, when viscosity increases, the viscous resistance increases and terminal velocity decreases.
Long Answer Questions
1. Derive Stokes' Law by dimensional analysis.
Answer:
Assume that viscous force depends upon coefficient of viscosity, radius and velocity.
$$ F=k\eta^a r^b v^c $$
Dimensions:
$$ [F]=[MLT^{-2}] $$
$$ [\eta]=[ML^{-1}T^{-1}] $$
$$ [r]=[L] $$
$$ [v]=[LT^{-1}] $$
Substituting dimensions and comparing powers of M, L and T,
$$ a=b=c=1 $$
Therefore,
$$ F=k\eta rv $$
Experimentally,
$$ k=6\pi $$
Hence,
$$ \boxed{F=6\pi\eta rv} $$
2. Derive the expression for terminal velocity.
Answer:
Consider a sphere of radius r and density ρ falling through a viscous fluid of density σ.
Weight of sphere,
$$ W=\frac{4}{3}\pi r^3\rho g $$
Upward thrust,
$$ T=\frac{4}{3}\pi r^3\sigma g $$
Viscous force,
$$ F=6\pi\eta rv $$
At terminal velocity,
$$ W=T+F $$
Substituting values,
$$ 6\pi\eta rv = \frac{4}{3}\pi r^3(\rho-\sigma)g $$
Therefore,
$$ v= \frac{2r^2(\rho-\sigma)g}{9\eta} $$
Hence,
$$ \boxed{ v= \frac{2r^2(\rho-\sigma)g}{9\eta} } $$
3. Explain terminal velocity and discuss the factors affecting it.
Answer:
Terminal velocity is the constant velocity attained by a body moving through a viscous fluid when its acceleration becomes zero.
The expression for terminal velocity is
$$ v= \frac{2r^2(\rho-\sigma)g}{9\eta} $$
From this relation:
- $$v\propto r^2$$ (depends on square of radius)
- $$v\propto (\rho-\sigma)$$ (depends on density difference)
- $$v\propto g$$ (depends on gravity)
- $$v\propto \frac{1}{\eta}$$ (inversely proportional to viscosity)
Thus, terminal velocity increases with radius, density difference and gravity, but decreases with viscosity.
Important Exam Questions
- State and explain Stokes' Law.
- Derive Stokes' Law using dimensional analysis.
- Define terminal velocity.
- Derive an expression for terminal velocity.
- Explain the forces acting on a sphere falling through a viscous fluid.
- State Archimedes' Principle.
- Write the SI unit and dimensional formula of viscosity.
- Discuss the factors affecting terminal velocity.
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