NEET Previous Year Questions (2014-2024): Laws of Motion

# NEET Previous Year Questions (2014-2024): 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:

## 2024

Q1: A horizontal force  10 N is applied to a block A as shown in figure. The mass of blocks A and B are  2kg and 3kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is :               [2024]
(a) Zero
(b) 4 N
(c) 6 N
(d) 10 N
Ans:
(c)

## 2023

Q1: 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:

Direction of change of momentum and direction of force acting on the player will be same, so correct answer is North east direction

Q2: 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:

Q3: A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels  $24 cm$ through the block along its length horizontally, velocity of bullet becomes u/2. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is :      [2023]
(a) 24 cm
(b) 28 cm
(c) 30 cm
(d) 27 cm
Ans:
(d)

Similarly from starting

## 2022

Q1: If  , then the scalar and vector products of  and  have the magnitudes respectively as                   [2022]
(a) 10, 2
(b) 5, √3
(c) 4, √5
(d) 10, √2
Ans:
(d)

Q2: In the diagram shown, the normal reaction force between 2 kg and 1 kg is (Consider the surface, to be smooth) : (Given g = 10 ms−2)

(a) 10 N
(b) 25
N
(c) 39 N

(d) 6 N                          [2022]
Ans: (b)
Assuming all three blocks as system

Q3: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.

## 2021

Q1: 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.

## 2020

Q1: 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]

(a) g/5
(b) g/10
(c) g
(d) g/2

Ans: (a)
Solution:

## 2019

Q1: 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]
Ans:
(c)
Solution:

For equilibrium of the block limiting friction
fL≥ mg
⇒ μN ≥ mg

Q2: 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:

The tension is maximum at the lowest position of mass, so the chance of breaking is
maximum.

Q3: 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:

## 2018

Q1: 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:

Q2: 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]

(a)
(b)
(c) a = g cos θ
(d) a = g tan θ
Ans:
(d)
Solution:

## 2017

Q1: 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]

(a)
(b) g, g
(c)
(d)
Ans:
(a)
Solution:

Q2: One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v', the net force on the particle (directed towards centre) will be (T represents the tension in the string)                    [2017]
(a)
(b)
(c) zero
(d) T
Ans:
(d)
Centripetal force  is provided by tension so net force on the particle will be equal to tension T.

## 2016

Q1: 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)
(b)
(c)
(d)
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:

In the case of vertical equilibrium,

Dividing Eqns. (i) and (ii), we get

Q2: 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,

Let, the tension at point A be TA.
Using Newton's second law, we have

Energy at point C is,

At point C, using Newton's second law,

In order to complete a loop, T≥ 0
so,

From equation (i) and (ii)
Using the principle of conservation of energy,

Q3: 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

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,

## 2015

Q1: Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius  r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is      [NEET / AIPMT 2015]
(a) 4
(b) 1
(c) 2
(d) 3
Ans:
(c)
According to question, two stones experience same centripetal force  i.e. FC1 = FC2

or
So, V1 = 2V2 i.e., n = 2

Q2: A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30o, the box starts to slip and slides 4.0 m down the plank in 4.0 s.      [NEET / AIPMT 2015]

The coefficients of static and kinetic friction between the box and the plank will be, respectively
(a) 0.5 and 0.6
(b) 0.4 and 0.3
(c) 0.6 and 0.6
(d) 0.6 and 0.5
Ans:
(d)
Coefficient of static friction,

Q3: 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]

(a) 18 N
(b) 2 N
(c) 6 N
(d) 8 N
Ans:
(c)
Solution:

a =14/7 = 2m/s2
∴ 14 - N1 = 4 x 2
N1 = 6N

Q4: 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:

Q5: 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)
(b)
(c)
(d)
Ans:
(c)
Solution:

Q6: 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)
(b)
(c)
(d)
Ans:
(d)
Solution:

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

From Equation (2) and (3) we get,

## 2014

Q1: 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)
(b)
(c)
(d)

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)

Q2: 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]

(a)
(b)
(c)
(d)
Ans:
(d)
Solution:

Q3: 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]

(a) 12 Ns
(b) 6 Ns
(c) 24 Ns
(d) 20 Ns
Ans:
(a)
Solution:

The document NEET Previous Year Questions (2014-2024): Laws of Motion | Physics Class 11 is a part of the NEET Course Physics Class 11.
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## Physics Class 11

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## FAQs on NEET Previous Year Questions (2014-2024): Laws of Motion - Physics Class 11

 1. What are the three laws of motion?
Ans. The three laws of motion, formulated by Sir Isaac Newton, are: 1. The first law, also known as the law of inertia, states that an object at rest will remain at rest, and an object in motion will remain in motion unless acted upon by an external force. 2. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. 3. The third law states that for every action, there is an equal and opposite reaction.
 2. How are forces classified in the context of laws of motion?
Ans. Forces are classified into different categories based on their characteristics. Some common classifications include contact forces (such as friction and normal force) and non-contact forces (such as gravitational force and electromagnetic force). Additionally, forces can be categorized as balanced forces (resulting in no acceleration) or unbalanced forces (causing acceleration).
 3. What is the principle of conservation of momentum?
Ans. The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces are acting on it. This means that in a closed system, the total momentum before an event must be equal to the total momentum after the event, regardless of any internal interactions.
 4. How does Newton's third law of motion apply to everyday situations?
Ans. Newton's third law of motion can be observed in various everyday situations. For example, when a person jumps off a boat onto the shore, the person exerts a force on the boat (action), causing the boat to move in the opposite direction (reaction). Similarly, when a rocket propels itself forward by expelling gas backward, the force of the expelled gas propels the rocket forward.
 5. How can the laws of motion be used to explain the motion of objects in free fall?
Ans. The laws of motion can be applied to explain the motion of objects in free fall, such as a ball dropped from a height. The force of gravity acting on the ball causes it to accelerate towards the ground (second law), while the ball remains in motion until it hits the ground unless acted upon by another force (first law). Additionally, the force of the ball hitting the ground and rebounding is an example of Newton's third law in action.

## Physics Class 11

102 videos|411 docs|121 tests

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