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NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

Edited By Vishal kumar | Updated on Sep 08, 2023 03:19 PM IST

NCERT Solutions Class 9 Science Chapter 9 – Access and Download Free PDF

NCERT Solutions for Class 9 Science Chapter 9 Force And Laws Of Motion: Solution for force and laws of motion is part of NCERT Solutions for Class 9 Science. This Solution for class 9 chapter 9 science force and laws of motion are designed by the highly experienced faculty of Careeas360 in a very easy and systematic manner. By utilizing these NCERT solutions, students can prepare for exams even if they have not previously studied the material covered in the chapter. Students can download force and laws of motion class 9 pdf and use them according to their convenience which are free of cost.

This chapter will help you to understand the concepts of force and laws of motion. Have you ever thought about what causes a body to move? How can we stop a moving body? How can we change the direction of a body? The answer to all these three questions is by applying a force, With the help of chapter 9 science class 9, you will be able to understand the concepts of the chapter. Along with this chapter's solutions, you can also check NCERT solutions for other chapters and subjects.

CBSE class 9 chapter 9 science question answer Force and Laws of Motion, will help you to understand Newton's 3 laws of motion and another main topic like inertia. The class 9 science chapter 9 question answer given here as per the latest syllabus & exam pattern for the questions mentioned at the end of the chapter Force And Laws Of Motion will help you to revise the chapter.

NCERT Solutions for Class 9 Science: Important Formulas and Diagrams + eBook link

Formulas for chapter 9 class 9 science are essential for understanding and solving problems involving the laws of motion and forces. They provide an easy-to-understand explanation of the mathematical relationships behind these topics. Here are some important formulas from the chapter on the laws of motion:

  • Force (F) = mass (m) Ă— acceleration (a) [SI unit of force = N]

  • Momentum (p) = mass (m) Ă— velocity (v) [SI unit of momentum=kg.m/s]

  • Law of Conservation of Momentum

Total initial momentum = Total final momentum

Mathematically,

(m1*u1) + (m2*u2) = (m1*v1) + (m2*v2)

where:

m1 and m2 are the masses of the objects involved,u1 and u2 are the initial velocities of the objects, v1 and v2 are the final velocities of the objects.

The above-mentioned formulae are some of the more essential ones from Chapter 9 class 9 Science Force and Laws of Motion. Careers360 experts have compiled all of the crucial science formulas chapter by chapter. These formulas are extremely beneficial for exam preparation, revision, and answering difficult questions. Students can download the Force and Laws of Motion Class 9 pdf of chapter by chapter formula by clicking on the given below link and use them online/ offline according to their comfort for free of cost.

Download Ebook - NCERT Class 9 Science: Chapterwise Important Formulas, Diagrams, And Points

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws Of Motion - Important Topics

  • Force: Understanding the concept of force, including its definition and effects on objects.
  • Inertia: Studying the concept of inertia and its function in the resistance of objects to changes in their state of motion.

Newton's Laws of Motion: Studying Newton's three laws of motion:

a. First Law: Law of Inertia, which states that an object at rest or in motion will continue in that state unless acted upon by an external force.

b. Second Law: Relationship between force, mass, and acceleration, as described by the equation F = ma.

c. Third Law: Law of Action and Reaction, which states that every action has an equal and opposite reaction.

  • Momentum: understanding momentum as the product of an object's mass and velocity, as well as its importance in terms of motion.
  • Conservation of Momentum: Exploring the concept that the total momentum of a system remains constant unless external forces act upon it.
  • Friction: Examining the force of friction and its different types, such as static friction and kinetic friction, and their effects on motion.
  • Applications of Force and Laws of Motion: Understanding how the principles of force and laws of motion apply to real-life situations, such as the working of vehicles, sports activities, and safety measures.

This chapter has been renumbered as Chapter 8 in accordance with the CBSE Syllabus 2023–24.

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws Of Motion

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NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion: Solved In-text Questions

Class 9 Science Chapter 9 Solutions: Topic 9.3(Inertia and Mass)

Q1. Which of the following has more inertia:

(a) a rubber ball and a stone of the same size?

(b) a bicycle and a train?

(c) a five rupees coin and a one-rupee coin?

Answer:

We know that inertia is defined by the mass of the body. So the body has more mass/weigth will have more inertia.

(a) In the case of rubber ball and stone of the same size, the stone will have greater mass so its inertia will be more as compared to a rubber ball.

(b) The train has much more mass as compared to the bicycle so the inertia of the train is larger.

(c) A five rupees coin has more weight than one rupee coin so the inertia of five rupees coin is more.

Q2. In the following example, try to identify the number of times the velocity of the ball changes:

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.

Also, identify the agent supplying the force in each case.

Answer:

Velocity has both magnitude and direction. Thus change in any of magnitude or direction results in a change in velocity.

In the given case, when the player kicks the ball to another player the direction of the ball is changed thus velocity is changed.

The second player then pushes the ball in a forward direction, so velocity changes. Then the goalkeeper brings ball at rest hence velocity changes (magnitude is changed). After that he kicks the ball thus increase the speed and direction is opposite, hence velocity changes again.

The agent of supplying force is 1st player, 2nd player, goalkeeper, and goalkeeper respectively.

Q3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer:

This can be explained with the help of the concept of inertia. When the tree is at rest, its leaves are also at rest. But when we bring the tree in motion by means of shaking it, due to the inertia of leaves they still tend to be in rest. Thus force is acting on leaves vigorously with changing direction rapidly. This results in detaching of leaves from the tree.

Q4 . Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer:

In the situation of a moving bus, the passengers are also in motion along with the bus. Now when brakes are applied the bus comes to rest but due to the inertia of passengers they still tend to move in the forward direction, Thus they move in the forward direction after applying brakes.

Now when the bus is rest, the passengers are also at rest. The inertia tend them to be at rest. That's why when bus accelerates they move in a backward direction.


Force And Laws of Motion Class 9 Topic 9.5

Q1. If action is always equal to the reaction, explain how a horse can pull a cart.

Answer:

For motion, horse applies force on the ground and thus the frictional force of the ground pushes the horse forward (action-reaction pair). Thus horse applies force on the cart and cart applies backward force on the horse but due to unbalanced force, the horse moves in the forward direction.

Q2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer:

This can be explained with the help of Newton's third law of motion (action-reaction pair). When the velocity of water is more i.e., water is coming out with higher force thus the reaction force on the hose is also large. This makes it difficult for a fireman to hold the hose.

Q3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s -1 . Calculate the initial recoil velocity of the rifle.

Answer:

Here we will use conservation of momentum :

Initial momentum of system ( rifle + bullet ) = Final momentum of system

m_1u_1\ +\ m_2u_2\ =\ m_1v_1\ +\ m_2v_2

or 4\times 0\ +\ 0.05\times 0\ =\ 4\times v_1\ +\ 0.05\times 35

or -4 \times v_1\ =\ 1.75

or v_1 \ =\ -\ 0.4375\ m/s

The negative sign shows that velocity is opposite to that of the bullet. This is because the force on the rifle will be opposite by bullet (action-reaction pair).

Force and Laws of Motion: Solved Exercise Questions-

Q 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:

When a net zero external unbalanced force is applied and the object is in motion then there will be a condition in which its motion is unopposed by any external force. Then as a result object will continue to remain in motion. The velocity, in this case, will be constant and in a particular direction.

Q 2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

During beating the carpet we bring it in the motion. But the dust particles have a tendency to be at rest so they resist the motion. This is why they come out of the carpet. The factor responsible for this is inertia .

Q 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

This is because the luggage has some mass so the inertia. When the bus is in motion, the luggage is also in motion. But when the bus stops suddenly, the luggage still tends to move with the same velocity due to inertia. Thus it may fell down. Similarly, when the bus starts from the rest, the luggage tends to be in rest thus it is pushed in a backward direction.

Q 4. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball would want to come to rest

Answer:

The frictional force present between the ground and the ball acts in the opposite direction of the motion of the ball. Thus balls stop after travelling a few distances.

Hence option (c) is correct.

Q 6. A stone of 1 kg is thrown with a velocity of 20 m s -1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer:

By the equation of motion, we know that,

v^2\ =\ u^2\ +\ 2as

or 0^2\ =\ 20^2\ +\ 2a\times 50 (Since ball comes to rest thus final velocity is zero.)

or a\ =\ \frac{-400}{100}\ =\ -4\ m/s^2

The acceleration is negative. This implies that the force is opposing the motion.

The force is given by :

F\ =\ ma

or F\ =\ 1\times \ (-4)\ =\ -4\ N

Hence the frictional force between stone and ice is 4 N.

Q 7. (a) A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:

  • the net accelerating force and

Answer:

The net accelerating force is given by :

Net accelerating force = Force by engine - Frictional force

=\ 40000\ -\ 5000\ =\ 35000\ N

Q 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s -2 ?

Answer:

By Newton's law of motion, we have :

F\ =\ ma

and =\ 1500\times 1.7

or F\ =\ -2550\ N

Thus the retarding force required is 2550 N in the direction opposite to the motion of the vehicle

Q 9. What is the momentum of an object of mass m, moving with a velocity v?

(a) (mv)^{2} (b) mv^{2} (c) \frac{1}{2}mv^{2} (d) mv

Answer:

We know that the momentum of a body is given by:-

Momentum = Mass \times Velocity

Thus we have : p\ =\ mv

Hence option (D) is correct.

Q 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer:

By Newton's third law it is clear that the frictional force acting on the cabinet will be 200 N. (Note that the limit of frictional force depends upon the mass of the body and the coefficient of friction between the bodies. So it might be possible that if frictional force acting is less than 200 N).

Q 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer :

The student is correct. The force applied by him is less than the static friction present between tyres of the truck and the road (as the mass of truck is large thus the frictional force is also high). Thus the force applied by the student is balanced by the static frictional force which is the action-reaction pair.

Q 14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s -1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer:

By the equation of motion we know that:-

v\ =\ u\ +\ at

Putting values of v, u and t in the equation :

0\ =\ 150\ +\ a\times 0.03

or a\ =\ \frac{-150}{0.03}

or a\ =\ -5000\ m/s^2

Now we have :

s\ =\ ut\ +\ \frac{1}{2}at^2

or =\ 150\times 0.03\ +\ \frac{1}{2}\times (-5000)\times 0.03^2

or =\ 2.25\ m

Hence the distance of penetration of the bullet in the block is 2.25 m.

Now, for the force we have :

F\ =\ ma

or =\ 0.01\times (-5000)\ =\ -50\ N

Thus the retarding force acting on the bullet is 50 N.

Q 16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s -1 to 8 m s -1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

The momentum is given by :

p\ =\ mv

Initial momentum is : p_i\ =\ mu\ =\ 100\times 5\ =\ 500\ Kg\ m/s

And the final momentum is :

p_f\ =\ mv\ =\ 100\times 8 \ =\ 800\ Kg\ m/s

Also, the force is defined as rate of change of momentum.

Thus F\ =\ \frac{p_f\ -\ p_i}{t}

or =\ \frac{800\ -\ 500}{6}\ =\ 50\ N

Hence the force exerted on the body is 50 N.

Q 17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions

Answer:

Kiran’s observation:- The insect experiences a greater change in its velocity as the mass of insects is much less as compared to the motorcar so it experiences a greater change in its momentum. Thus Kiran's observation is correct.

Akhtar’s observation:- We know that motorcar has a larger velocity and has a bigger mass as compared to the insect. Also, the motorcar continues to move in the same direction even after the collision. This suggests that the change in momentum of a motorcar is very less, whereas the insect experiences a great change in its momentum.

Rahul’s observation:- The momentum gained by the insect is equal to the momentum lost by the motorcar (law of conservation of momentum). Thus his observation is also correct.

Q 18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s -2 .

Answer :

For calculating momentum, we need the final velocity of the bell.

By equation of motion we can write :

v^2\ =\ u^2\ +\ 2as

or v^2\ =\ 0^2\ +\ 2\times 10\times 0.8

or v\ =\ 4\ m/s

Thus the momentum is : p\ =\ mv

or =\ 10\times 4\ =\ 40\ Kg\ m/s


NCERT Class 9 Science Chapter 9 Exercise Solutions: Solved Additional Exercises

Q A1. (a) The following is the distance-time table of an object in motion:

  • What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
Time in seconds Distance in metre
0 0
1 1
2 8
3 27
4 64
5 125
6 216
7 343

Answer:

From the table, the relation between time and distance can be seen.

s\ =\ t^3

Thus the velocity of the particle is increasing with time.

v\ =\ \frac{ds}{dt}\ =\ 3t^2

and a\ =\ \frac{dv}{dt}\ =\ 6t

Hence acceleration increases linearly with time.

Q A1. (b) The following is the distance-time table of an object in motion:

  • What do you infer about the forces acting on the object?
Time in seconds Distance in metre
0 0
1 1
2 8
3 27
4 64
5 125
6 216
7 343

Answer:

i)Velocity is the change in distance with time. So the velocity is increasing non uniformly, so the acceleration increases with the time.

F\ =\ ma

ii) Thus the net force acting on the body is also increasing.

Q A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s -2 . With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Answer:

From the question, it is clear that two-person pushes the car with constant velocity. But when the third person pushes it has some acceleration. Thus the third person is responsible for the acceleration generated.

Force by the third person is given by :

F\ =\ ma

=\ 1200\times 0.2\ =\ 240\ N

Hence the force from each man is 240 N.

Q A3. A hammer of mass 500 g, moving at 50 m s -1 , strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer:

Thus the force is given by :

\\F = m\frac{v-u}{t}\\=500\times10^{-3}(\frac{0-50}{0.01})=-2500N

or =\ 0.5\times (-5000)\ =\ -2500\ N

The negative sign implies that the force by the nail on the hammer is in the opposite direction.


Force And Laws of Motion Class 9 Topics:

topics for ncert solutions for class 9 science chapter 9 exercise questions are listed below:

9.1 Balanced and Unbalanced Forces

9.2 First Law of Motion

9.3 Inertia and Mass

9.4 Second Law of Motion

9.5 Third Law of Motion

9.6 Conservation of Momentum

Key feathers Of NCERT solutions For Class 9 Science Chapter 9 Force And Laws of Motion

  • Expertly Created Solutions: The NCERT solutions for Class 9 Science Chapter 9 are written by subject matter experts who have a thorough understanding of the subjects covered in the chapter. The solutions are correct, dependable, and adhere to the NCERT curriculum.
  • Comprehensive Coverage: The class 9 science chapter 9 solutions cover all the important topics and questions mentioned in the chapter. They provide step-by-step explanations and solutions for better understanding and clarity.
  • Easy to Understand: The class 9 science chapter 9 question answer are designed in a simple and concise manner, making them easier for students to understand. The language used is student-friendly, ensuring that you can effectively grasp the topics.
  • Doubt Clarification: If you have any doubts or queries about the force and laws of motion class 9 numericals, you can seek assistance from subject matter experts via chat support. They are available 24 hours a day, seven days a week to provide clarifications and support, ensuring that your questions are immediately handled.
  • Enhances Learning Experience: By utilizing the NCERT class 9 science chapter 9 exercise solutions, you can enhance your learning experience and develop a strong foundation in the concepts of force and laws of motion. The solutions provide additional practice and reinforcement of the chapter's content.
  • Accessible Anytime: The ncert solutions for class 9 science chapter 9 exercise questions are easily accessible online, allowing you to study and revise at your own convenience. You can access them on various devices like smartphones, tablets, or computers.

Also Check -

NCERT Books and NCERT Syllabus here:

NCERT Solutions For Class 9 Science- Chapter Wise

Chapter No. Chapter Name
Chapter 1 Matter in Our Surroundings
Chapter 2 Is Matter Around Us Pure
Chapter 3 Atoms and Molecules
Chapter 4 Structure of The Atom
Chapter 5 The Fundamental Unit of Life
Chapter 6 Tissues
Chapter 7 Diversity in Living Organisms
Chapter 8 Motion
Chapter 9 Force and Laws of Motion
Chapter 10 Gravitation
Chapter 11 Work and Energy
Chapter 12 Sound
Chapter 13 Why Do We Fall ill?
Chapter 14 Natural Resources
Chapter 15 Improvement in Food Resources

NCERT Solutions for Class 9 - Subject Wise

NCERT Science Exemplar Solutions Class 9 - Chapter Wise


Frequently Asked Question (FAQs)

1. What are the important topics of the chapter Force And Laws Of Motion

All the NCERT book topics are important from Force And Laws Of Motion Class 9 Science chapter . The topics are

  • Balanced and Unbalanced Forces   
  • First Law of Motion   
  • Inertia and Mass   
  • Second Law of Motion   
  • Third Law of Motion   
  • Conservation of Momentum 
2. What is the acceleration of a body with constant velocity along a straight line?

Constant velocity means the change in velocity is zero. There for the acceleration is zero.

3. What is the name of Chapter 9 in Class 9 science?

The name of chapter 9 in class 9 science is "Force And Laws Of Motion".

4. What is the SI unit of Force?

The SI unit of force is Newton (N).

5. What are the 3 laws of motion Class 9?

The three laws of motion, as stated by Sir Isaac Newton, are:

  • Newton's First Law: An object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity (that is, it will continue moving at the same speed and in the same direction) unless acted upon by an external force.

  • Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In other words, F = ma, where F is the net force, m is the mass of the object, and a is the acceleration produced.

  • Newton's Third Law: For every action, there is an equal and opposite reaction. This means that whenever one object exerts a force on another object, the second object exerts an equal and opposite force on the first object.

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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