Important Questions Class 11 Physics Chapter 5

Important Questions Class 11 Physics Chapter 5

Important Questions for CBSE Class 11 Physics Chapter 5 – Law of Motion

You can find the best revision questions for Laws of Motion Class 11 Physics Chapter 5 Important Questions here at Extramarks. The questions contain step-by-step solutions that are chosen by subject matter experts. The following Class 11 Physics Chapter 5 Important Questions will help students score well in your final exams. 

All of the important topics of the Chapter Law of Motion are covered by the Important Questions and Answers. This set of questions is prepared in such a way that they will aid students in studying the chapter while keeping the important questions and answers in mind. The study material has been created with the types of questions asked in the CBSE Class 11 examination in mind.

This article also includes many important concepts and formulas for Class 11 Physics Chapter 5 that are explained in detail for better understanding.

CBSE Class 11 Physics Chapter 5 Important Questions

These are examples of Chapter 5 Class 11 Physics Important Questions, click here to get the full set of Important Questions for Class 11 Physics Chapter 5.

1 Mark Answers and Questions

Q1. Name the factor on which the coefficient of friction depends.

Ans: The coefficient of friction will mainly depend on two factors, which are as follows:

1. The materials of the surfaces in contact.
2. The characteristics of the surfaces.

Q2. What provides the centripetal force to a car taking a turn on a level road?

Ans: Centripetal force is provided by the frictional contact between the tyres and the road.

Q3. Why does a swimmer push the water backwards?

Ans: From Newton’s 3rd Law of Motion, we know that “when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted”. As a result, the swimmer pushes water backward with his hands in order to swim ahead.

Q4. Action and reaction forces do not balance each other. Why?

Ans: Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

Q5. The two ends of a spring-balance are pulled by a force of 10 kg each. What will be the reading of the balance?

Ans: As the spring balancing is based on the tension in the spring, it gauges weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg, and the reading will be 10kg.

Q6. A lift is an acceleration upward. Will the apparent weight of a person inside the lift increase, decrease, or remain the same relative to its real weight? What happens if the lift is going at a uniform speed?

Ans: There will be an increase in perceived weight. The apparent weight will stay the same as the true weight if the lift moves at a constant pace.

Q7.  Why is it desirable to hold a gun tight to one’s shoulder when it is being fired?

Ans. Because the gun recoils after being fired, it must be held softly on the shoulder. Because the gun and the shoulder are combined into one mass system, the back kick is reduced. When shooting, a gunman must keep his weapon securely against his shoulder.

Q8. Justify that friction is a self-adjusting force. 

Ans: Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit friction.

2 Marks Answers and Questions

Q1. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with constant speed.

(b) a kite skillfully held stationary in the sky.

Ans: (a) As the raindrop is falling at a constant speed, its acceleration will be 0. The net force acting on the raindrop will be 00 because the force acting on a particle is given by.

(b) As the kite is held stationary, by Newton’s first law of motion, the algebraic sum of forces acting on the kite is zero.

Q2. Write two consequences of Newton’s second law of motion.

Ans: The two consequences of Newton’s Second Law of Motion are as follows.

1.It demonstrates that the motion is only accelerated when force is applied to it.

2. It introduces the notion of a body’s inertial mass.

Q3. A bird is sitting on the floor of a wire cage, and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?

Ans: When the bird begins to fly within the cage, the weight of the bird is no longer felt since the air inside is in direct touch with ambient air, making the cage look lighter.

Q4. Why does a cyclist lean to one side, while going along a curve? In what direction does he lean?

Ans: A cyclist leans while riding along a curve because a component of the ground’s natural response supplies him with the centripetal force he needs to turn.

He must lean inward from his vertical posture, towards the circular path’s centre.

Q5. A soda water bottle is falling freely. Will the gas bubbles rise to the surface of the water in the bottle?

Ans: As the water in a freely falling bottle is in a state of weightlessness, As a result, there is no upthrust force on the bubbles, and the bubbles do not ascend in the water.

Q6. Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly.

Ans: When a fast bus comes to a complete stop, the bottom half of the body in touch with the seat comes to a complete halt, while the upper section of the passengers’ bodies prefer to retain their uniform motion. As a result, the passengers are pushed forward.

Q7. How does road banking reduce tyre wear and tear?

Ans: When a curving road is not banked, friction between the tyres and the road provides centripetal force.

Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground’s natural response supplies the necessary centripetal force, reducing tyre wear and tear.

Q8. A force is being applied to a body, but it causes no acceleration. What possibilities might be considered to explain the observation?

Ans: (1) If the force is a deforming force, no acceleration is produced.

(2) Internal force is incapable of causing acceleration.

3 Marks Answers and Questions

Q1. In which of the following cases is centripetal force provided?

(i) Motion of planet around the sun

(ii) Motion of moon around the Earth

(iii) Motion of an electron around the nucleus in an atom

Ans: (i) The centripetal force is provided by the gravitational force acting on the Earth and the sun.

(ii) The centripetal force is provided by the Earth’s gravitational attraction to the moon.

(iii) The centripetal force is provided by the electrostatic attraction between the electron and the proton.

Q2. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g10 g floating on water,

(c) a kite skillfully held stationary in the sky,

(d) a car moving with a constant velocity of 30 km/h−130 km/h−1 on a rough road, and

(e) a high-speed electron in space, far from all material objects, and free of electric and magnetic fields.

Ans: (a) Zero net force

The raindrops are falling at a steady rate. As a result, the acceleration is zero. The net force acting on the raindrop is zero, according to Newton’s second law of motion.

(b) Zero net force

The cork’s weight is acting downward. The buoyant force exerted by the water in an upward direction balances it. As a result, there is no net force acting on the floating cork.

(c) Zero net force

In the sky, the kite is stationary, i.e. it is not moving at all. As a result, according to Newton’s first law of motion, there is no net force acting on the kite.

(d) Zero net force

The car is going at a constant speed down a bumpy route. As a result, it has no acceleration. There is no net force operating on the car, according to Newton’s second law of motion.

(e) Zero net force

All fields have no effect on the high-speed electron. As a result, there is no net force acting on the electron.

Q3. A train runs along an unbanked circular bend of radius 30 m at a speed of 54 km/hr. The mass of the train is 106 kg. What provides the necessary centripetal force required for this purpose, the engine or the rails? What is the angle of banking required to prevent the rail from wearing out?Ans: From the question, we have the radius of a circular bend given as, r=30 m.

Speed of train = v = 54 km h−1 = 54×518 = 15 ms−1

Mass of train given as, m = 106 kg

Then we need to find the angle of banking θ.

(1) The centripetal force is generated by the lateral force exerted by rails on the train’s wheels.

(2) The centripetal force is provided by the lateral thrust of the outer rail.

(3) According to Newton’s third law of motion, the train exerts (i.e., causes) an equal and opposite thrust on the outer rail, causing its wear and tear.

Therefore, the angle of baking:

tanθ = v2rg= 15213×19.8

Θ = 37.4

4 Marks Answers and Questions

Q1. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg

(a) just after it is dropped from the window of a stationary train,

(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,

(c) just after it is dropped from the window of a train accelerating with 1 m s−2,

(d) lying on the floor of a train which is accelerating with 1 m s−2, the stone being at rest relative to the train. Neglect air resistance throughout.

Ans: (a) 1 N; vertically downward

From the question, we have the mass of the stone given as, m=0.1 kg

The acceleration of the stone is given as, a = g = 10 m/s2

The net force exerted on the stone, according to Newton’s second law of motion, is

F = ma = m g

= 0.1×10 = 1 N

Gravitational acceleration always works in the downward direction.

(b) 1 N; vertically downward

The train is travelling at a constant speed. As a result, its acceleration in the horizontal direction, where it is moving, is zero. As a result, there is no horizontal force acting on the stone.

The net force acting on the stone is due to gravity’s acceleration, and it is always vertically downward. This force has a magnitude of 1 N.

(c) 1 N; vertically downward

It is given that the train is accelerating at the rate of 1 m/s2.

Hence, the net force acting on the stone will be equal to, F′ = ma = 0.1×1 = 0.1 N

This force has a horizontal component to it. The horizontal force F′, no longer acts on the stone when it is dropped. This is due to the fact that the force acting on a body at any one time is determined by the current circumstances rather than previous ones.

As a result, the net force acting on the stone is determined only by gravity’s acceleration.

F = mg = 1 N

This force acts vertically downward.

(d) 0.1 N; in the direction of motion of the train

The typical reaction of the floor balances the weight of the stone. The train’s horizontal motion is the only source of acceleration.

Acceleration of the train, a = 0.1 m/s2

The net force acting on the stone will be directed in the train’s direction of travel. Its magnitude is given by:

F = ma

= 0.1×1 = 0.1 N

5 Marks Answers and Questions

Q1. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms−2. The crew and the passengers each weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

Ans: (a) Mass of the helicopter is given as, mh= 1000 kg

Mass of the crew and passengers is given as, mp= 300 kg

Therefore, the total mass of the system, m = 1300 kg

As the acceleration of the helicopter is given as, a = 15 m/s2

The reaction force R exerted on the system by the floor may be computed using Newton’s second equation of motion.

R−mpg = ma

= mp(g+a)

= 300(10+15) = 300×25

= 7500 N

The response force will likewise be directed upward because the helicopter is accelerating vertically.As a result, the force exerted on the floor by the crew and passengers is 7500 N, directed downward, according to Newton’s third law of motion.

(b) The reaction force R experienced by the helicopter may be computed using Newton’s second equation of motion as follows = 32500 N

The helicopter is being pushed higher by the reaction force of the surrounding air. As a result, the rotor’s action on the surrounding air will be 32500N, directed downward, according to Newton’s third law of motion.

(c) The surrounding air exerts a force of 32500N on the helicopter, which is directed upward.