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NCERT Solutions for Class 8 Science Chapter 9 Friction

NCERT Solutions for Class 8 Science Chapter 9 Friction

Edited By Vishal kumar | Updated on Jun 26, 2025 09:30 AM IST

Ever wondered why your shoes grip the ground or why it is hard to push a heavy box(or heavy object)? That is all because of friction. Class 8 Science Chapter 9 Friction question answer helps you understand what friction is, how it works and why it is both helpful and sometimes a problem. The NCERT Solutions for Class 8 Science Chapter 9 are prepared by subject experts based on the latest NCERT syllabus.

This Story also Contains
  1. NCERT Solutions for Class 8 Science Chapter 9 Friction: Exercise Questions
  2. Friction Class 8 NCERT Science Chapter: Topics
  3. Class 8 Science Chapter 9 NCERT Solutions: Formulas
  4. Approach to solve the NCERT Class 8 Science Chapter 9 Questions
  5. NCERT Chapter-Wise Solutions For Class 8th Science
NCERT Solutions for Class 8 Science Chapter 9 Friction
NCERT Solutions for Class 8 Science Chapter 9 Friction

These NCERT Solutions for class 8 science Chapter 9 Friction provide clear answers to all exercise questions along with important topics like types of friction, factors affecting it, and methods to reduce or increase it and important formulas. With step-by-step explanations and easy language, these NCERT solutions make problem solving simple and help you prepare better for exams. We have also included the approach to solve the chapter 9 questions so you can understand how to tackle different types of problems confidently.

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NCERT Solutions for Class 8 Science Chapter 9 Friction: Exercise Questions


Below are the solved exercise questions from NCERT Class 8 Chapter 9 Friction. Go through each solution carefully to improve your concept clarity and perform better in exams.

Q1. Fill in the blanks.

(a) Friction opposes the _____________ between the surfaces in contact with each other.
(b) Friction depends on the _____________ of surfaces.
(c) Friction produces __________.
(d) Sprinkling of powder on the carrom board ________ friction.
(e) Sliding friction is ___________ than the static friction.

Answer:

(a) Friction opposes the motion between the surfaces in contact with each other.
(b) Friction depends on the nature of surfaces.
(c) Friction produces heat.
(d) Sprinkling of powder on the carrom board reduces friction.
(e) Sliding friction is less than the static friction.

Q2. Four children were asked to arrange forces due to rolling, static and sliding frictions in a decreasing order. Their arrangements are given below. Choose the correct arrangement.

(a) rolling, static, sliding
(b) rolling, sliding, static
(c) static, sliding, rolling
(d) sliding, static, rolling

Answer:

(c) is the correct arrangement.

Static friction > sliding friction > rolling friction.

Q3. Alida runs her toy car on dry marble floor, wet marble floor, newspaper and towel spread on the floor. The force of friction acting on the car on different surfaces in increasing order will be

(a) wet marble floor, dry marble floor, newspaper and towel.
(b) newspaper, towel, dry marble floor, wet marble floor.
(c) towel, newspaper, dry marble floor, wet marble floor.
(d) wet marble floor, dry marble floor, towel, newspaper

Answer:

The force of friction acting on the car on different surfaces in increasing order will be wet marble floor, dry marble floor, newspaper and towel. Therefore (a) is the correct answer.

Q5. You spill a bucket of soapy water on a marble floor accidentally. Would it make it easier or more difficult for you to walk on the floor? Why?

Answer:

It would make it more difficult for us to walk on the marble floor since the frictional force would further decrease and thus our grip on the floor while walking would become even lesser.

Q6. Explain why sportsmen use shoes with spikes.

Answer:

Sportsmen use shoes with spikes to increase the grip between their feet and the ground by increasing the friction with the help of spikes. This aids them while running on the ground.

Q7. Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why?

Answer:

Seema would have to apply a larger force as she would have to overcome a greater frictional force. The frictional force between two surfaces exists due to the interlocking of the irregularities on the surface. The magnitude of this frictional force depends on how tight this interlocking is which in this case depends on the weight of the boxes which is more in the case of Seema.

Q8. Explain why sliding friction is less than static friction.

Answer:

Sliding friction is less than static friction as the frictional force between two surfaces depends on the interlocking between the irregularities on the surfaces. This interlocking in case of sliding decreases as the contact points does not get enough time to interlock properly thus reducing the frictional force.

Q9. Give examples to show that friction is both a friend and a foe.

Answer:

Friction acting as a friend:

(a) It aids us to walk and cars to move on the roads.

(b)Fire can be ignited using friction.

Friction acting as a foe:

(a) It causes the efficiency of engines to decrease and also damages machinery.

(b) Friction causes the soles of shoes and tires to wear off.

Q10. Explain why objects moving in fluids must have special shapes.

Answer:

When an object is moving through a liquid there is an opposing force offered by the liquid against the motion of the object. To reduce this opposing force objects must have special shapes.

Friction Class 8 NCERT Science Chapter: Topics

Below are the important topics covered in Class 8 NCERT Science Chapter 9 Friction. These topics explain what friction is, how it affects motion, and ways to increase or reduce it. Go through each section thoroughly to understand the role of friction in everyday life and do well in your exams.

9.1 Force of Friction
9.2 Factors Affecting Friction
9.3 Friction: A Necessary Evil
9.4 Increasing and Reducing Friction
9.5 Wheels Reduce Friction
9.6 Fluid Friction

Class 8 Science Chapter 9 NCERT Solutions: Formulas

Below are the simple formulas from Chapter 9, Friction. These will help you understand the nature of friction, its effects, and how it can be managed.

Limiting Friction (fl) (maximum static friction before motion starts):

fl=μs×N

Where:
fl= Limiting friction
μs= coefficient of static friction
N= normal force

Kinetic Friction (fk) (friction when the object is moving):

fk=μk×N
Where:
μk= coefficient of kinetic friction

Approach to solve the NCERT Class 8 Science Chapter 9 Questions

To solve questions from Class 8 Science Chapter 9: Friction start by identifying what the question is focusing on whether it is about types of friction , factors affecting friction or methods to increase or reduce it. Carefully read the problem to understand the situation and If the question involves formulas, apply the relevant one like
Frictional force =μ× Normal force,
where μ is the coefficient of friction. Use proper SI units and make sure the values given match the formula requirements.

NCERT Chapter-Wise Solutions For Class 8th Science

The chapter-wise NCERT solutions for all the science chapters are listed here along with their links:

NCERT Solutions For Class 8: Subject-Wise

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. Is friction an important topic for higher studies?

Yes, friction, force, motion and pressure are important topics that will be studied in Class 9 NCERT Science Syllabus  and also in Class 11 NCERT Physics book

2. What is the role of friction?

Friction opposes the motion between the surfaces in contact with each other. If there is no frictional force we wont be able to walk. 

3. What are the topic covered in friction class 8

The following topics are covered in the chapter friction:

 9.1 Force of Friction
9.2 Factors Affecting Friction
9.3 Friction: A Necessary Evil
9.4 Increasing and Reducing Friction
9.5 Wheels Reduce Friction
9.6 Fluid Friction

4. What is μ in friction?

 μ  represents the coefficient of friction between two surfaces.

5. What is the value of μ?

The value of the coefficient of friction (μ) varies depending on the materials of the surfaces in contact and other factors such as temperature and pressure. It can range from 0 to 1 or even higher.

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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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