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NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11 PDF Download

The Laws of Motion is an important chapter in NEET Physics. It is a fundamental concept that forms the basis for understanding many other topics in Physics. Let's have a look at Previous Year Questions of the chapter:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 1. A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is           [2023]
A: Along eastward
B: Along northward
C: Along north-east
D: Along south-west
Ans:
C
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Direction of change of momentum and direction of force acting on the player will be same, so correct answer is North east direction

Q 2. Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 ( g = 10 m s–2).           [2023]
A: 1.2 m s–2 
B: 150 m s–2 
C: 1.5 m s–2 
D: 50 m s–2
Ans:
C
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 3. An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms–1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m s–2 )           [2022]
A: 20000 
B: 34500 
C: 23500 
D: 23000 
Ans: B
Solution:

The power delivered by the motor to the lift can be calculated using the formula:

Power = Force x Velocity

In this case, the force is the sum of the weight of the lift and the passengers and the frictional force opposing the motion:

Force = (Weight of lift + passengers) + Frictional force

Weight of lift + passengers = Mass x gravity

= (2000 kg / 1000) x 10 m/s2

= 20,000 N

Therefore,

Force = (20,000 N) + (3000 N)

= 23,000 N

Now, using the formula for power, we get:

Power = Force x Velocity

= (23,000 N) x (1.5 m/s)

= 34,500 W

Therefore, the minimum power delivered by the motor to the lift in watts is 34,500 W.


Q 4. A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly :     [2021]
A: 2.1 kg m/s
B: 1.4 kg m/s
C: 0 kg m/s
D: 4.2 kg m/s
Ans: 
D
Solution:
The impulse imparted to an object is given by the change in momentum of the object. In this case, when the ball strikes the ground, its velocity changes from a downward velocity to an upward velocity. Therefore, the change in momentum of the ball is given by:
Δp = 2mv

where m is the mass of the ball and v is the velocity of the ball just after it rebounds.

The velocity of the ball just before it hits the ground can be calculated using the equation of motion:

v= u+ 2as
where u is the initial velocity (which is zero since the ball is dropped from rest), a is the acceleration due to gravity (which is -g since the ball is moving downward), and s is the distance traveled (which is equal to the height of the drop, 10 m).
Substituting the values, we get:

v2 = 0 + 2(-10 m/s)(10 m)

v = 200

v = √200 ≈ 14.14 m/s

The velocity of the ball just after it rebounds is also equal to v, since it rebounds to the same height. Therefore, the change in momentum of the ball is:

Δp = 2mv = 2(0.15 kg)(14.14 m/s) ≈ 4.24 kg m/s

The magnitude of the impulse imparted to the ball is therefore approximately 4.24 kg m/s.

So, the answer is closest to option (D) 4.2 kg m/s.


Q 5. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :     [2020]
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

A: g/5
B: g/10
C: g
D: g/2

Ans: A
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11


Q 6.  A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be : (g 10 m/s2)     [2019]
A: √10 rad/s
B: 10/2π rad/s
C: 10 rad/s
D: 10π rad/s
Ans: 
C
Solution:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

For equilibrium of the block limiting friction
fL≥ mg
⇒ μN ≥ mg
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11


Q 7. A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:    [2019]
A: The mass is at the highest point
B: The wire is horizontal
C: The mass is at the lowest point
D: Inclined at an angle of 60° from vertical
Ans: 
C:
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
The tension is maximum at the lowest position of mass, so the chance of breaking is
maximum.

Q 8. When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance.
Then x1 : x2 will be:    [2019]
A: 1: √2
B: √2: 1
C: 1: √3
D: 1: 2√3
Ans:
C
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 9. Which one of the following statements is incorrect?
A: Rolling friction is smaller than sliding friction
B: Limiting value of static friction is directly proportional to normal reactions
C: Frictional force opposes the relative motion
D: Coefficient of sliding friction has dimensions of length    [2018]
Ans:
D
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 10. A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is :-    [2018]
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
B: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
C: a = g cos θ
D: a = g tan θ
Ans:
D
Solution:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 11. Two blocks A and B of masses 3 m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively :-    [2017]
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
B: g, g
C: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
D: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Ans:
A
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 12. What is the minimum velocity with which a body of mass m must enter a vertcal loop of radius R so that it can complete the loop ? [2016]
A: √5gR
B: √gR
C: √2gR
D: √3gR
Ans:
A
Solution: 
The question is illustrated in the figure below,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Let, the tension at point A be TA.
Using Newton's second law, we have
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Energy at point C is,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
At point C, using Newton's second law,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
In order to complete a loop, T≥ 0
so,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
From equation (i) and (ii)
Using the principle of conservation of energy,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 13. A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10-4J by the end of the second revolution after the beginning of the motion ?    [2016]
A: 0.2 m/s2

B: 0.1 m/s2
C: 0.15 m/s2
D: 0.18 m/s2
Ans: B
Solution:

Given, mass of particle. m = 0.01 kg
Radius of circle along which particle is moving , r = 6.4 cm
Kinetic energy of particle, K.E. = 8 x 10-4 J
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Given that, KE of particle is equal to 8 x 10-4 J by the end of second revolution after the beginning of the motion of particle.
It means, initial velocity (u) is 0 m/s at this moment.
Now, using the Newton's 3rd equation of motion,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 14. A car is negotiating a curved road of radius R. The road is banked at an angle .. The coefficient of friction between the tyres of the care and the road is 1s. The maximum safe velocity on this road is:    [2016]
A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
B: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
C: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
D: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Ans:
C
Solution: A car is negotiating a curved road of radius R. The road is banked at angle straight theta and the coefficient of friction between the tyres of car and the road is straight mu subscript straight s.
The given situation is illustrated as:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
In the case of vertical equilibrium,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Dividing Eqns. (i) and (ii), we get
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 15. Three blocks A, B, and C, of masses 4 kg, 2 kg, and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is :    [2015]

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
A: 18 N
B: 2 N
C: 6 N
D: 8 N
Ans:
C
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
a =14/7 = 2m/s2
∴ 14 - N1 = 4 x 2
N1 = 6N

Q 16. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is :    [2015]
A: W(D-x)/d
B: Wx/d
C: Wd/x
D: W(d-x)/x
Ans:
A
Solution:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 17. A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is :    [2015]
A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
B: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
C: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
D: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
Ans:
C
Solution:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 18. A block A of mass m1 rests on a horizontal table. A lights string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is μk. When the block A is sliding on the table, the tension in the string is:   [2015]
A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
B: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
C: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
D: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11 
Ans:
D
Solution:
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

See figure alongside
Let T be the tension in the string.
Let a be the acceleration of the combination

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
From Equation (2) and (3) we get,
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 19. A balloon with mass ‘m’ is descending down with an acceleration ‘a’(where a<g). How much mass should be removed from it so that it starts moving up with an acceleration ‘a’ ?    [2014]
A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
B: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
C:NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
D: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Ans: C
Solution:
Let the up thrust on balloon be U.
mq - U = ma     ...(i)
If Δm is removed
U = (m - Δm)g = (m - Δm)a    ..(ii)
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 20. A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is :
(Assume m1 = m2 = m3 = m)    [2014]
NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11    


A: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11 
B: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
C: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
D: NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11 
Ans:
D
Solution:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

Q 21. The force F acting on a particle of mass m is indicated by the force-time graph as shown below. The climate change in momentum of the particle over the time interval from zero to 8s is,    [2014]

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11
A: 12 Ns
B: 6 Ns
C: 24 Ns
D: 20 Ns
Ans:
A
Solution:

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11

The document NEET Previous Year Questions (2014-2023): Laws of Motion | Physics Class 11 is a part of the NEET Course Physics Class 11.
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FAQs on NEET Previous Year Questions (2014-2023): Laws of Motion - Physics Class 11

1. What are the laws of motion?
Ans. The laws of motion, also known as Newton's laws of motion, are three fundamental principles that describe the relationship between the motion of an object and the forces acting upon it. These laws were formulated by Sir Isaac Newton in the 17th century. The laws are as follows: 1. First Law (Law of Inertia): An object at rest will remain at rest, and an object in motion will continue to move in a straight line at a constant velocity, unless acted upon by an external force. 2. Second Law (Law of Acceleration): The acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. It can be expressed as F = ma, where F is the net force, m is the mass, and a is the acceleration. 3. Third Law (Law of Action and Reaction): For every action, there is an equal and opposite reaction. This means that whenever an object exerts a force on another object, the second object exerts a force of the same magnitude but in the opposite direction on the first object.
2. What is the difference between mass and weight?
Ans. Mass and weight are two different concepts in physics: - Mass: Mass is a measure of the amount of matter in an object. It is a scalar quantity and is usually measured in kilograms (kg). Mass remains constant regardless of the location of the object and is an inherent property of the object. - Weight: Weight, on the other hand, is a measure of the force of gravity acting on an object. It is a vector quantity and is usually measured in newtons (N). Weight can vary depending on the gravitational field strength of the location. The weight of an object can be calculated using the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. In summary, mass is a measure of the amount of matter an object contains, while weight is the force exerted on an object due to gravity.
3. How does Newton's third law of motion apply to everyday life?
Ans. Newton's third law of motion states that for every action, there is an equal and opposite reaction. This law can be observed in various everyday life situations: - Walking: When we walk, we push the ground backward with our feet. As a result, the ground exerts an equal and opposite force on us, propelling us forward. - Swimming: In swimming, we push the water backward with our arms and legs. The water exerts an equal and opposite force on our bodies, allowing us to move forward. - Balloon Rocket: When air is released from a balloon, it exerts a force in one direction, causing the balloon to move in the opposite direction. - Jumping: When we jump, we exert a downward force on the ground. The ground exerts an equal and opposite force, propelling us upward. These examples demonstrate how Newton's third law applies to everyday life by showing that every action has a reaction of equal magnitude but in the opposite direction.
4. What is the significance of Newton's laws of motion in the field of engineering?
Ans. Newton's laws of motion are of great significance in the field of engineering. They provide the foundation for understanding and analyzing the behavior of objects in motion, enabling engineers to design and build structures, machines, and systems that function effectively and safely. Some key applications of Newton's laws in engineering include: - Structural Analysis: Engineers use Newton's second law to calculate the forces acting on structures, such as bridges and buildings, to ensure their stability and structural integrity. - Vehicle Design: Newton's laws are crucial in designing vehicles, including cars, airplanes, and spacecraft. They help engineers determine the forces involved in propulsion, braking, and steering, ensuring optimal performance and safety. - Robotics: Engineers apply Newton's laws to design and control the motion of robots, allowing them to perform tasks efficiently and accurately. - Aerospace Engineering: Newton's laws are fundamental in designing rockets, satellites, and other spacecraft, guiding their trajectory, propulsion, and navigation. In summary, Newton's laws of motion play a vital role in engineering by providing the principles and tools necessary for analyzing and designing various structures, machines, and systems.
5. What is the relationship between force, mass, and acceleration according to Newton's second law?
Ans. Newton's second law states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. Mathematically, it can be expressed as F = ma, where F is the net force, m is the mass of the object, and a is the resulting acceleration. This equation shows that the force applied to an object is directly proportional to the acceleration it experiences. If the force acting on an object increases, its acceleration will also increase, assuming its mass remains constant. Conversely, if the mass of an object increases, its acceleration will decrease for a given force. In other words, the greater the force applied to an object, the greater the acceleration it will experience, and the greater the mass of an object, the smaller the acceleration it will have for a given force. This relationship between force, mass, and acceleration is crucial in understanding and predicting the motion of objects under the influence of external forces.
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