NCERT Solutions For Class 9 Science Chapter 9 Force And Laws Of Motion
| 1. Exercise Questions |
| 2. Intext Questions |
NCERT Solutions For Class 9 Science Chapter 9 Force And Laws Of Motion on this step-by-step Force And Laws Of Motion answer guide . In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those NCERT Solutions. It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science Force And Laws Of Motion so you can be searching for assist from them. Students have to solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Force And Laws Of Motion below and prepare for your tests easily.
NCERT Solution for class 9
Chapter 9: Force and Laws Of Motion
Exercise Questions
Q. 1. An object experiences a net zero external unbalanced force . Is it possible for the object to be travelling 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 .
Ans . Yes , it is possible , but only when the object has been moving with a constant velocity in a particular direction .
Then , there is no net unbalanced force applied on the body . The object will keep moving with a non – zero velocity . To change the state of motion , a net non – zero external unbalanced force must be applied on the object .
Q. 2 . When a carpet is beaten with a stick , dust comes out of it . Explain .
Ans . According to the law of inertia , when a carpet is beaten with a stick , then the carpet comes to motion .
But the dust particles try to resist any change to their inertia . Hence the dust particles come out of the carpet , as they do not move with the carpet .
Q. 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope ?
Ans . When the bus accelerates and moves forward , it acquires a state of motion . However , the luggage kept on the roof , owing to its inertia , tends to remain in its state of rest .
So , with the forward movement of the bus , the luggage tends to remain at its original position and ultimately falls from the roof of the bus . To avoid this , it is advised to tie any luggage kept on the roof of a bus with a rope .
Q. 4. A batsman hits a cricket ball which then rolls on a 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 .
Ans . Correct option : ( c )
Explanation : A batsman hits a cricket ball , which then rolls on a level ground . After covering a short distance , the ball comes to rest because there is frictional force on the ball opposing its motion .
Frictional force always acts in the direction opposite to the direction of motion . Hence , this force is responsible for stopping the cricket ball .
Q. 5. A truck starts from rest and rolls down a hill with a constant acceleration . It travels a distance of 400 m in 20 seconds . Find its acceleration . Find the force acting on it if its mass is 7 metric tonnes ( Hint : 1 metric tonne = 1,000 kg . )
Ans . Given that ,
Initial velocity of the truck , u = 0 m / s
Time taken , t = 20 s
Distance covered by the stone , s = 400 m
According to the second equation of motion :
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 ?
Ans . Given that ,
Initial velocity of the stone , u = 20 m / s
Final velocity of the stone , v = 0 m / s
Distance covered by the stone , s = 50 m
According to the third equation of motion :
v² = u² + 2as
( 0 ) ² = ( 20 ) ² + 2 × a × 50
or a = -4 m / s²
The negative sign indicates that acceleration is acting against the motion of the stone .
Mass of the stone , m = 1 kg
From Newton’s second law of motion :
Force ,
F = Mass x Acceleration
F = ma
F = 1 × ( -4 ) = – 4 N
Hence , the force of friction between the stone and the ice is -4 N.
Q. 7. An 8,000 kg engine pulls a train of five wagons , each of 2,000 kg , along a horizontal track . If the engine exerts a force of 40,000 N and the track offers a friction force of 5,000 N , then calculate :
( a ) the net accelerating force ;
( b ) the acceleration of the train ; and
( c ) the force of wagon 1 on wagon 2 .
Ans . ( a ) Given that ,
Force exerted by the engine , F = 40,000 N
Frictional force offered by the track , Ff = 5,000 N
Net accelerating force , Fa = F – Ff
= 40,000 – 5,000 = 35,000 N
Hence , the net accelerating force is 35,000 N.
( b ) Given that ,
Acceleration of the train = a
The engine exerts a force of 40,000 N on all the five wagons .
Net accelerating force on the wagons , Fa = 35,000 N
Mass of the wagons , m = Mass of a wagon x Number of wagons
Mass of one wagon = 2,000 kg
Number of wagons = 5
∴ m = 2,000 × 5 = 10,000 kg
Total mass ( including the mass of engine ) ,
M = m + 8,000 = 18,000 kg
( c ) Given that ,
Mass of four wagons ( excluding wagon 1 ) ,
m = 2,000 × 4 = 8,000 kg
Acceleration of the train = 1.944 m / s²
∴ The force of wagon 1 on , wagon 2
ma = 8,000 x 1.944 = 15,552 N
Q. 8. An automobile vehicle has a mass of 1,500 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 ms – ² ?
Ans . Given that ,
Mass of the automobile vehicle , m = 1,500 kg
Final velocity , v = 0 m / s
Acceleration of the automobile , a = -1.7 m s – ²
From Newton’s second law of motion :
Force Mass × Acceleration
= 1,500 × ( -1.7 ) = -2,550 N
So , the force between the automobile and the road is -2,550 N , in the direction opposite to the motion of the automobile .
Q. 9. What is the momentum of an object of mass m , moving with a velocity v ?
( a ) ( mv ) ² . ( b ) mv² .
( c ) ½ mv² .
( d ) mv .
Ans . Correct option : ( d )
Explanation :
If ,
Mass of the object = m
Velocity = v
Momentum = Mass x Velocity
Momentum = mv
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 ?
Ans . A force of 200 N is applied in the forward direction . An equal amount of force will act in the opposite direction , so that the net force will be zero .
This opposite force is the fictional force exerted on the cabinet . So , a frictional force of 200 N is exerted on the cabinet .
Q. 11. Two objects , each of mass 1.5 kg are moving in the same straight line but in opposite directions . The velocity of each object is 2.5 m s– ¹ before the collision during which they stick together . What will be the velocity of the combined object after collision ?
Ans . Given that ,
Mass of one of the objects , m₁ = 1.5 kg
Mass of the other object , m₂ = 1.5 kg
Velocity of m1 before collision , u₁ = 2.5 m / s
Velocity of m₂ , moving in opposite direction before collision , u₂ = -2.5 m / s
( Negative sign arises because mass m₂ is moving in an opposite direction )
After collision , the two objects stick together .
Total mass of the combined object = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum :
Total momentum before collision = Total momentum after collision
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) v
1.5 ( 2.5 ) + 1.5 ( -2.5 ) = ( 1.5 +1.5 ) v
3.75 – 3.75 = 3 v
v = 0
Hence , the velocity of the combined object after collision is 0 m / s .
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 .
Ans . The truck has a large mass . So , the static friction between the truck and the road is also very high . To move the car , one has to apply a force more than the static friction .
When someone pushes the truck and the truck does not move , then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction .
Therefore , the student is right in justifying that the two opposite and equal cancel each other .
Q. 13. A hockey ball of mass 200 g travelling at 10 m s – ¹ is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s– ¹ . Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick .
Ans . Given that ,
Mass of the hockey ball , m = 200 g = 0.2 kg
Hockey ball travels with velocity , v₁ = 10 m / s
Initial momentum = mv₁
Hockey ball travels in the opposite direction with velocity , v₂ = -5 m / s
Final momentum = mv₂
Change in momentum = mv₂ – mv₁
= 0.2 [ -5 – 10 ] = 0.2 ( -15 )
= -3 kg m s – 1
So , the change in momentum of the hockey ball is 3 kg m s – ¹ .
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 seconds . 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 .
Ans . Given that ,
The bullet is travelling with a velocity of 150 m / s .
Thus , when the bullet enters the block , its velocity Initial velocity , u = 150 m / s
Final velocity , v = 0 ( since the bullet finally comes to rest )
Time taken to come to rest , t = 0.03 second
According to the first equation of motion ,
v = u + at
( Negative sign indicates that the velocity of the bullet is decreasing . )
According to the third equation of motion :
Hence , the distance of penetration of the bullet into the block is 2.25 m .
From Newton’s second law of motion :
Force , F = Mass x Acceleration
Mass of the bullet , m = 10 , ( g ) = 0.01 kg
Acceleration of the bullet , a = -5,000 m / s²
F = ma = -0.01 x 5,000 = -50 N
Hence , the force exerted by the wooden block on the bullet is 50 N , opposing bullet’s motion .
Q. 15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s– ¹ collides with , and sticks to , a stationary wooden block of mass 5 kg . Then they both move off together in the same straight line . Calculate the total momentum just before the impact and just after the impact . Also , calculate the velocity of the combined object .
Ans . Given that ,
Mass of the object , m₁ = 1 kg
Velocity of the object before collision ,
v₁ = 10 m / s
Mass of the stationary wooden block , m₂ = 5 kg
Velocity of the wooden block before collision , v₂ = 0 m / s
∴Total momentum before collision
= m₁v₁ + m₂v₂
= 1 ( 10 ) + 5 ( 0 ) = 10 kg m s – ¹
It is given that after collision , the object and the wooden block stick together .
Total mass of the combined system = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum :
Total momentum before collision = Total momentum after collision
The total momentum after collision is 10 kg m s – ¹ and velocity of combined object is 5/3 m / s .
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 .
Ans . Given that ,
Initial velocity of the object , u = 5 m / s
Final velocity of the object , v = 8 m / s
Mass of the object , m = 100 kg
Time take by the object to accelerate , t = 6 s
Initial momentum , mu = 100 x 5
= 500 kg m s – 1
Final momentum , mv = 100 x 8
= 800 kg ms – 1
Force exerted on the object ,
Initial momentum of the object is 500 kg ms – 1 . Final momentum of the object is 800 kg ms – 1 . Force exerted on the object 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 a 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 .
Ans . According to the law of conservation of momentum :
Momentum of the car and insect system before collision = Momentum of the car and insect system after collision
Hence , the change in momentum of the car and insect system is zero . The insect gets stuck on the wind screen . This means that the direction . of the insect is reversed . As a result , the velocity of the insect changes to a great amount . On the other hand , the car continues moving with a constant velocity .
Hence , Kiran’s suggestion that the insect suffers from a great change in momentum as compared to the car is correct . The momentum of the insect after collision becomes very high because the car is moving at a high speed .
Therefore , the momentum gained by the insect is equal to the momentum lost by the car . Akhtar made a correct conclusion because the mass of the car is very large as compared to the mass of the insect .
Rahul gave a correct explanation as both the car and the insect experienced equal forces caused by the Newton’s action – reaction law . But , he made an incorrect statement as the system suffers a change in momentum because the momentum before . the collision is equal to the momentum after the collision .
Q. 18. How much momentum will a dumbbell 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 .
Ans . Given that ,
Mass of the dumbbell , m = 10 kg
Distance covered by the dumbbell ,
s = 80 cm = 0.8 m
Acceleration in the downward direction , a = 10 m / s²
Initial velocity of the dumbbell , u = 0
Final velocity of the dumbbell ( when it was about to hit the floor ) = v
According to the third equation of motion :
v² = u² + 2as
v² = 0 + 2 ( 10 ) 0.8
v = 4 m / s
Hence , the momentum with which the dumbbell hits the floor is
= mv
= 10 x 4 kg m s – ¹
= 40 kg m s – ¹
So , the momentum with which the dumbbell hits the floor is 40 kg m s¯¹ .
Additional Exercise
A. 1. The following is the distance – time table of an object in motion :
( a ) What conclusion can you draw about the acceleration ? Is it constant , increasing , decreasing , or zero ?
( b ) What do you infer about the forces acting on the object ?
Ans . ( a ) There is an unequal change of distance in an equal interval of time .
Thus , the given object is having a non – uniform motion . The equation of motion with constant decleration is
( b ) According to Newton’s second law of motion , the force acting on an object is directly proportional to the acceleration produced in the object .
In the given case , the increasing acceleration of the given object indicates that the force acting on the object is also increasing .
A. 2. Two persons manage to push a motorcar of mass 1,200 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 ms – 2 . With what force does each person push the motorcar ? ( Assume that all persons push the motorcar with the same muscular effort )
Ans . Given that ,
Mass of the motor car = 1,200 kg
Only two persons manage to push the car . So , the acceleration acquired by the car is given by the third person alone .
Acceleration produced by the car , when it is pushed by the third person will be , a = 0.2 m / s²
Let the force applied by the third person be
From Newton’s second law of motion :
Force Mass × Acceleration
F = 1,200 x 0.2 = 240 N
Thus , the third person applies a force of magnitude 240 N.
So , each person applies a force of 240 N to push the motor car . 480 N was used to overcome friction .
A. 3. A hammer of mass 500 g , moving at 50 ms¯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 ?
Ans . Given that ,
Mass of the hammer , m = 500 g = 0.5 kg
Initial velocity of the hammer , u = 50 m / s
Time taken by nail to stop the hammer , t = 0.01 s
Final velocity of the hammer , v = 0 ( because the hammer finally comes to rest )
From Newton’s second law of motion :
The hammer stop due to this force acting on it , which means that the nail exerts a force of 2500 N on the hammer , in a direction opposite to the motion of the hammer .
A. 4. A motorcar of mass 1,200 kg is moving along a straight line with a uniform velocity of 90 km / h . Its velocity is slowed down to 18 km / h in 4 seconds by an unbalanced external force . Calculate the acceleration and change in momentum . Also calculate the magnitude of the force required .
Ans . Given that ,
Mass of the motor car , m = 1,200 kg
Initial velocity of the motor car ,
Negative sign indicates that it’s a retarding motion . So , the velocity is decreasing .
Change in momentum = mv – mu = m ( v – u )
= 1,200 ( 5 – 25 ) = -24,000 kg m s – 1
Newton’s second law :
Force = Mass X Acceleration
= 1200 × ( -5 ) = -6000 N
Acceleration of the motor car = -5 m / s²
Change in momentum of the motor car = -24,000 kg m s – ¹
Hence , the force required to decrease the velocity is 6,000 N.
( Negative sign indicates retardation , decrease in momentum and retarding force )
Intext Question
Page 118
Q. 1. 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 . [ NCERT Q. 1 , Page 118 ] [ KVS , Patna Region , SA – I , 2015-16 ]
Ans . Inertia is the measure of the mass of the body . The greater is the mass of the body , the greater is its inertia and vice – versa .
( a ) Mass of a stone is more than the mass of a rubber ball for the same size . So , inertia of the stone is greater than that of a rubber ball .
( b ) Mass of a train is more than the mass of a bicycle . So , inertia of the train is greater than that of the bicycle .
( c ) Mass of a five – rupee coin is more than that of a one – rupee coin . So , inertia of the five – rupee coin is greater than that of the one – rupee coin .
Q. 2. 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 . [ NCERT Q. 2 , Page 118 ]
Ans . The velocity of football changes four times in following ways :
( i ) First , when a football player kicks the football to another player .
( ii ) Second , when that player kicks the football to the goalkeeper .
( iii ) Third , when the goalkeeper stops the football .
( iv ) Fourth , when the goalkeeper kicks the football towards a player of his own team .
As a football player kicks the football , its speed changes from zero to a certain value . So that , the velocity of the ball gets changed . In this case , the player applied a force to change the velocity of the ball . Another player kicks the ball towards the goal post . As a result , the direction of the ball gets changed .
Therefore , its velocity also changes . In this case , the player applied a force to change the velocity of the ball . The goalkeeper collects the ball . In other words , the ball comes to rest . So , its speed comes to zero from a certain value . The velocity of the ball has changed . In this case , the goalkeeper applied an opposite force to stop / change the velocity of the ball .
The goalkeeper kicks the ball towards his team players . Hence , the speed of the ball increases from zero to a certain value . Hence , its velocity changes once again . In this case , the goalkeeper applied a force to change the velocity of the ball .
Agent supplying the force are –
( i ) First case – First player
( ii ) Second case – Second player
( iii ) Third case – Goalkeeper
( iv ) Fourth case – Goalkeeper
Q. 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch . [ NCERT Q. 3 , Page 118 ]
Ans . Some leaves of a tree get detached when we shake its branches vigorously . This is due to the Newton’s first law of motion .
Initially , leaves and tree both are in rest . So , when the branches of a tree are shaken , it moves to and fro , but its leaves tend to remain at rest .
This is because the inertia of the leaves tends to resist the ‘ to and fro ‘ motion . Due to this reason , some leaves fall down from the tree when shaken some vigorously .
Q. 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest ? [ NCERT Q. 4 , Page 118 ]
Ans . According to the Law of inertia- ” If a body is at rest , then it tries to remain at rest . If a body is moving , then it tries to remain in motion ” .
In a moving bus , passengers are in motion along with bus . When brakes are applied to stop a moving bus , it comes to rest . So the lower of a person in contact with the bus also stops . But the upper body remains in motion due to inertia , caused the person fall forwards .
Similarly , when a bus is accelerated from rest , the lower body is contact with the bus gets dragged forward , while the upper body remains in a state rest due to inertia . So the person falls backwards .
Page 126
Q. 5. If action is always equal to the reaction , explain how a horse can pull a cart . [ NCERT Q. 1 , Page 126 ]
Ans . Horse pushes the ground in backward direction . In reaction to this , the ground pushes the horse moves forward and cart moves along with horse in forward direction .
According to Newton’s third law of motion , a reaction force is exerted by the ground on the horse in the forward direction .
Q. 6. Explain , why is it difficult for a fireman to hold a hose , which ejects large amounts of water at a high velocity . [ NCERT Q. 2 , Page 126 ]
Ans . According to the Newton’s third law of motion , when large amount of water is ejected from a hose at a high velocity , water pushes the hose in backward direction with the same force .
As a result of the backward force , the stability of the fireman decreases . Hence , it is difficult for him to remain stable while holding the hose .
Q. 7. 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 . [ NCERT Q. 3 , Page 126 ] [ KVS , Patna Region , SA – I , 2015-16 ]
Ans . Given that ,
Mass of the rifle , m₁ = 4 kg
Mass of the bullet , m₂ = 50 g = 0.05 kg
Recoil ( final ) velocity of the rifle = v₁
Bullet is fired with an initial velocity , v₂ = 35 m / s
Initially , the rifle and bullet are at
rest , so u₁ = u₂ = 0
Total initial momentum of the rifle and bullet system
= m₁u₁ + m₂u₂
= 0
Total momentum of the rifle and bullet system after firing : =
m₁v₂ + m₂v₂
= 4 ( v₁ ) + 0.05 x 35
= 4v₁ +1.75
According to the law of conservation of momentum :
Total momentum after the firing = Total momentum before the firing
m₁u₁ + m₂u₂ = m₁v₂ + m₂v₂
4v₁ + 1.75 = 0
v₁ = -1.75 / 4
= -0.4375 m / s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m / s .
Page 127
Q. 8. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s – ¹ and 1 m s – ¹ , respectively . They collide and after the collision , the first object moves at a velocity of 1.67 m s – ¹ . Determine the velocity of the second object . [ NCERT Q. 4 , Page 127 ]
Ans . Given that ,
Mass of one of the objects , m₁ = 100 g = 0.1 kg
Mass of the other object , m₂ = 200 g = 0.2 kg
Velocity of m₁ before collision , u₁ = 2 m / s
Velocity of m₂ before collision , u₂ = 1 m / s
Velocity of m₁ after collision , v₁ = 1.67 m / s
Velocity of m₂ after collision = v₂
According to the law of conservation of momentum :
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
⇒ 0.1 x 2 + 0.2 x 1 = 0.1 x 1.67 + 0.2 × v₂
⇒ 0.4 = 0.67 + 0.2 × v₂
⇒ v₂ = 1.165 m / s
Hence , the velocity of the second object becomes 1.165 m / s after the collision .
Benefits of 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 contains extremely important points, and for each chapter, each concept has been simplified to make it easier to remember and increase your chances of achieving excellent exam results. Exam Preparation References Here are some tips on how these solutions can help you prepare for the exam.
- This helps students solve many of the problems in each chapter and encourages them to make their concepts more meaningful.
- NCERT Solutions for Class 9 Science Chapter 9 Force And Laws Of Motion encourage you to update your knowledge and refine your concepts so that you can get good results in the exam.
- These NCERT Solutions For Class 9 Science Chapter 9 Force And Laws Of Motion are the best exam materials, allowing you to learn more about your week and your strengths. To get good results in the exam, it is important to overcome your weaknesses.
- Most of the questions in the exam are formulated in a similar way to NCERT textbooks. Therefore, students should review the solutions in each chapter in order to better understand the topic.
- It is free of cost.
Tips & Strategies for Class 9 Exam Preparation
- Plan your course and syllabus and make time for revision
- Please refer to the NCERT solution available on the cbsestudyguru website to clarify your concepts every time you prepare for the exam.
- Use the cbsestudyguru learning app to start learning to successfully pass the exam. Provide complete teaching materials, including resolved and unresolved tasks.
- It is important to clear all your doubts before the exam with your teachers or Alex (an Al study Bot).
- When you read or study a chapter, write down algorithm formulas, theorems, etc., and review them quickly before the exam.
- Practice an ample number of question papers to make your concepts stronger.
- Take rest and a proper meal. Don’t stress too much.
Why opt for cbsestudyguru NCERT Solutions for Class 9 ?
- cbsestudyguru provide NCERT Solutions for all subjects at your fingertips.
- These solutions are designed by subject matter experts and provide solutions to every NCERT textbook questions.
- cbsestudyguru especially focuses on making learning interactive, effective and for all classes.
- We provide free NCERT Solutions for class 9 and all other classes.
