NEET Previous Year Questions (2015-2024): Laws of Motion | Physics Class 11 PDF Download

Ans: (c)
m_total= m_A + m_B = 2 kg + 3 kg = 5kg
 Since both blocks move together on a frictionless surface, they will have the same acceleration. Using Newton's second law:
F_total = m_total ⋅ a
10N = 5kg ⋅ a
Block B is being accelerated by the force exerted by block A. The force on block B can be calculated using Newton’s second law again, but this time for block B alone:
F on B  = m_B ⋅a
F on B  = 3kg ⋅ 2m/s² = 6 N
Thus, the force exerted by block A on block B is 6 N.

Ans: (c)
Concept:

• Centripetal Force: When an object changes direction, there is a force directed towards the center of the circular path, which is called centripetal force.
• The player is changing direction, meaning their velocity vector is also changing direction. In this case, the player is turning from south to east, which implies a change in velocity towards the east.

Force Direction:

• The force responsible for this change in direction would act towards the north-east because the player is transitioning between the southward and eastward directions. The force is responsible for changing both the magnitude and direction of the velocity as the player turns.

Thus, the force acts along the north-east direction.

Ans: (c)
For the body to remain stationary relative to the car, the maximum static friction force must equal the force required to accelerate the body along with the car. The friction force is what keeps the body stationary relative to the moving car.
F_friction = μ⋅N
Where N is the normal force, and for a horizontal surface, N = m ⋅ g.
F_friction = μ ⋅ m ⋅ g
The force required to accelerate the body is:
F = m ⋅a
For the maximum acceleration, the static friction force must equal the force required for acceleration:
μ ⋅ m ⋅g = m⋅a
 ⇒ a = μ ⋅ g
⇒ a = 0.15 ⋅ 10 m/s² 
⇒ a = 1.5 m/s²
Thus, the maximum acceleration of the car is 1.5 m/s².

Ans:(d)

The bullet is fired with an initial velocity u and travels through the block. After penetrating 24 cm, its velocity reduces to u/3. It then continues to penetrate until it comes to rest.

We can apply the equation of motion to the first part of the bullet's journey through the block. The equation :
v² = u²+2as
Where:- v = final velocity after penetrating 24 cm = u/3
u = initial velocity = u
a = acceleration (deceleration in this case, so it will be negative)
s = distance travelled = 24 cm = 0.24 m
Substituting the known values into the equation:
This simplifies to:
Rearranging gives:
Thus,
a = - 8 u² / 9 × 0.48 = - 8 u² / 4.32 = - 1.85 u²

Now, the bullet continues to penetrate until it comes to rest. The initial velocity for this part is u/3 and the final velocity is 0. We will again use the equation of motion:
Where s′ is the additional distance travelled before coming to rest. Substituting the values:

0 = u² / 9 + 2 ( -1.85 u²) s'

Ans: (d)
Given:
The dot product of two vectors and  is given by:
Substitute the components of

So, the scalar product is 10.
Now, The cross product of two vectors  is given by the determinant of the following matrix:
Substitute the components of
Now, calculate the determinant:





The vector product is
Therefore, the magnitude of

Thus, the correct answer is: (d) 10, √2

Ans: (b)
Given:

Maximum load (lift + passengers) = 2000 kg

Constant speed v=1.5 m/s

Frictional force f = 3000 N

Gravitational acceleration g = 10 m/s²

 The force required to lift the load at a constant speed is equal to the gravitational force acting on the lift:

The total force the motor must overcome includes both the gravitational force and the frictional force:

Power is given by the product of force and velocity:
Substituting the values:
P = 23000 N × 1.5 m/s = 34500 W
Thus, the minimum power delivered by the motor to the lift is 34500 W.

Ans:(d)
Given: 

Mass of the ball, m = 0.15kg 

Initial height, h = 10m 

Gravitational acceleration, g = 10m/s²  
The ball falls from a height of 10 meters, and we can use the kinematic equation to calculate the velocity just before it hits the ground:

This is the velocity just before hitting the ground (downward).

Since the ball rebounds to the same height, the velocity after rebounding is also 14.14 m/s, but in the opposite direction (upward).
The change in velocity Δv is the difference between the final velocity and the initial velocity,

Impulse J is the change in momentum, which is given by:
J = m ⋅ Δv
Substituting the values:
J = 0.15 kg × 28.28 m/s = 4.2 kg⋅m/s

Ans: (a)

For the 4 kg mass (m₁): The force acting on it is F₁ = m₁g.

For the 6 kg mass (m₂): The force acting on it is F₂ = m₂g.

The net force acting on the system is due to the difference in the weights of the two masses. The heavier mass (6 kg) will pull the lighter one (4 kg) upwards while it descends. 
F_net = m₂ g − m₁ g = (6g − 4g) = 2g
The total mass of the system is the sum of the masses of both bodies:
M_total = m₁ + m₂ = 4 kg + 6 kg = 10 kg
 Using Newton’s second law, the net force is related to the total mass and the acceleration of the system:
F_net = M_total ⋅ a
Substituting the values:
2g = 10a
Solving for a:

Ans:(c)
The gravitational force acting on the block is:
F_gravity = mg = 10 kg × 10 m/s² = 100N
The frictional force must balance the gravitational force:
F_friction = μ N
Where N is the normal force, which is the centrifugal force in this case.
The centrifugal force acting on the block due to the rotation is:
N = mω²r
Where ω is the angular velocity, and r is the radius of the cylinder.
For the block to remain stationary, the frictional force must balance the gravitational force:
F_friction = F_gravity 
Substituting the expressions:
μN = mg 
μ(mω²r)=mg
Simplifying:
μω²r=g
Solving for ω:

Substitute the Values: Substituting g = 10m/s², μ=0.1, and r = 1m:

Ans: (c)
Concept: The tension in the wire is responsible for keeping the mass moving in a circular path. At different points in the circle, the forces on the mass vary due to the direction of gravity. The wire will break when the tension in the wire is at its maximum.
 At the highest point:

The forces acting are the tension in the wire T_top  and the weight of the mass mg, both acting downward.

The net force provides the centripetal force to keep the mass moving in a circle.

The tension is given by: 

Comparison:

• At the highest point, the tension is lower because gravity acts in the same direction as the centripetal force, reducing the required tension.
• At the lowest point, the tension is greater because the wire must support the mass’s weight and provide the centripetal force. This is where the tension is maximum.

Since the wire is more likely to break at the point where the tension is highest, the wire is most likely to break when the mass is at the lowest point of the vertical circle.

Ans: (c)

The distance traveled by an object on an inclined plane is influenced by the acceleration due to gravity acting along the plane. The distance can be expressed using the kinematic equation for motion along an inclined plane:
where:

v₀  is the initial velocity, 

θ is the angle of inclination, 

g is the acceleration due to gravity.

The distance x₁ along the plane when the angle is 60° is given by:

Simplifying the trigonometric values:

Thus:
The distance x₂ along the plane when the angle is 30° is given by:
Simplifying the trigonometric values:
Thus:
Now, divide x₁  by  x₂ :
Therefore:
x₁ : x₂ = 1: √3

Ans: (d)
(a) Rolling friction is smaller than sliding friction:

This is a correct statement. Rolling friction is generally much smaller than sliding friction because the contact area in rolling is reduced, and there is less interlocking between surfaces compared to sliding.

(b) Limiting value of static friction is directly proportional to normal reaction:
This is also correct. The limiting value of static friction f_s is given by: f_s = μ_sN
where μ s  is the coefficient of static friction and N is the normal reaction force. So, static friction is directly proportional to the normal force.
(c) Frictional force opposes the relative motion:
This is a correct statement. The frictional force always acts in the direction opposite to the relative motion or the attempted relative motion between two surfaces.
(d) Coefficient of sliding friction has dimensions of length:
This is incorrect. The coefficient of friction, whether static or sliding, is a dimensionless quantity. It is simply a ratio of the frictional force to the normal force and does not have any physical dimensions.

Ans: (d)

Forces acting on the block:

Gravitational force: The gravitational force acting on the block is mg, which acts vertically downward.

Normal force: The normal force N acts perpendicular to the inclined plane. 

Pseudo force due to acceleration: Since the wedge is accelerating to the right with acceleration a, a pseudo force ma acts on the block towards the left (relative to the wedge), parallel to the incline.

Resolving forces:

Parallel to the inclined plane: mg sinθ

Perpendicular to the inclined plane: mg cosθ
The pseudo force ma acts parallel to the wedge, providing a component in the opposite direction along the incline.

The net force along the incline must be zero. This means the force component along the incline due to the pseudo force must balance the force component along the incline due to gravity:

ma cosθ = mg sinθ
a = g tanθ
Thus, the relationship between the acceleration $a$a of the wedge and the angle of inclination $\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}\backslash theta$θ is: a = g tanθ

Ans: (d)

When an object moves in a circular path, the net force directed towards the center (centripetal force) is responsible for keeping the object in circular motion.

The centripetal force  F_c required to keep a particle of mass 𝑚 m moving in a circle of radius l at a speed v is given by:
This force is directed towards the center of the circle.

Tension in the String: The tension in the string, T, provides the necessary centripetal force to keep the particle moving in a circular path.

Net Force Towards the Center: The net force on the particle, which is directed towards the center, must account for both the tension in the string and the centripetal force. The tension in the string must balance the centripetal force entirely. Therefore, the tension in the string T equals the centripetal force mv²/l.

Thus, the net force acting on the particle towards the center is: T = mv²/l

Ans: (a)
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_c ≥ 0
so,

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

Ans: (b)
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,

Ans: (c)

The centripetal force F_c acting on a body moving in a circular path is given by:
where:

• m is the mass of the stone,
• v is the tangential speed,
• r is the radius of the circular path.

For the lighter stone of mass m and radius r, the centripetal force is:

For the heavier stone of mass 2m and radius r/2, the centripetal force is:
Since both stones experience the same centripetal force:
F₁ = F₂
Substituting the expressions for F₁  and F₂:
Simplifying:
Taking the square root of both sides:
v₁ = 2v₂