NEET Previous Year Questions (2014-2024): Motion in a Plane | Physics Class 11 PDF Download

2024

Q2: A particle moving with uniform speed in a circular path maintains:           [2024]
(a) Constant velocity
(b) Constant acceleration
(c) Constant velocity but varying acceleration
(d) Varying velocity and varying acceleration
Ans: (d)
When a particle moves along a circular path with a uniform speed, it is important to understand the motion characteristics in terms of both velocity and acceleration. Velocity is a vector quantity which means it has both magnitude and direction, while acceleration is the rate of change of velocity with respect to time.
First, let's analyze the velocity:
Although the speed (magnitude of the velocity vector) remains constant in uniform circular motion, the direction of the velocity vector continuously changes as the particle progresses along the circle. Since velocity includes both the magnitude and the direction, any change in either results in a change in velocity. Consequently, in uniform circular motion, the velocity of the particle is not constant but varies due to the continuous change in direction.
Next, consider the acceleration:
In circular motion, there is always an acceleration directed towards the center of the circle, known as centripetal acceleration. This acceleration is responsible for changing the direction of the velocity vector, thereby keeping the particle moving in a circle, despite the speed being constant. The formula for centripetal acceleration is:
a_c = v²/r
where v is the speed of the particle and r is the radius of the circle. This acceleration is always directed towards the center of the circle and varies with the square of the speed and inversely with the radius of the circle.
Therefore, given that both the velocity and acceleration change, the correct choice is:
Option D - Varying velocity and varying acceleration
It states that in uniform circular motion, both the velocity and acceleration of the particle vary – velocity due to continuous changes in direction and acceleration due to its consistent inward (centripetal) direction towards the center of the circle.

2023

Q1: A ball is projected from point A with velocity   20 ms⁻¹ at an angle 60∘ to the horizontal direction. At the highest point B of the path (as shown in figure), the velocity vms⁻¹ of the ball will be:     [2023]

(a) 20
(b) 10√3
(c) Zero
(d) 10
Ans: (d)

At the top most point of its trajectory particle will have only horizontal component of velocity

Q2: A particle is executing uniform circular motion with velocity  and acceleration .     [2023]
Which of the following is true?
(a)  is a constant;  is not a constant

(b)  is not a constant;  is not a constant

(c)  is a constant;  is a constant

(d)  is not a constant;  is a constant
Ans: (b)
Direction of velocity and centripetal acceleration changes continuously so  is not a constant;  is not a constant. 

Q3: A bullet is fired from a gun at the speed of 280 m s^–1 in the direction 30° above the horizontal. The maximum height attained by the bullet is (g = 9.8 m s^–2, sin30° = 0.5)           [2023]
(a) 2800 m
(b) 2000 m
(c) 1000 m
(d) 3000 m
Ans: (c)
Solutions:


Q4: The angular acceleration of a body, moving along the circumference of a circle, is           [2023]
(a) Along the radius, away from centre
(b)  Along the radius towards the centre
(c) Along the tangent to its position
(d) Along the axis of rotation
Ans: (d) 
Solutions: Angular acceleration of a body, moving along the circumference of a circle is along the axis of rotation.

Q5: A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s⁻¹. The ball strikes the water surface after  4 s. The height of bridge above water surface is (Take   g = 10 m s⁻² )       [2023]
(a) 60 m
(b) 64 m
(c) 68 m
(d) 56 m
Ans: (b)

2022

Q1: A cricket ball is thrown by a player at a speed of 20 m/s in a direction 30∘ above the horizontal. The maximum height attained by the ball during its motion is (g = 10 m/s²)             [2023]
(a) 25 m
(b) 5 m
(c) 10 m
(d) 20 m
Ans: (b)

Maximum height reached by the ball

Q2: A ball is projected with a velocity, 10 ms^–1 , at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be           [2022]
(a) 5√3 ms^–1
(b) 5 ms–1 
(c) 10 ms–1 
(d) Zero
Ans: (a)
Solutions: At highest point vertical component of velocity become zero.  

At highest point speed of object = 10cos30°
= 5√3 ms^–1

Q3: The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s² is             [2022]

(a) 4𝝅
(b) 12𝝅
(c) 104𝝅
(d) 2𝝅
Ans: (a)
Solutions:

= 4𝝅

2021

Q1: A car starts from rest and accelerates at 5 m/s2. At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s ?    [2021]
(a) 20√2m/s, 0 m/s²
(b) 20√2m/s, 10 m/s²
(c) 20 m/s,5 m/s²
(d) 20 m/s, 0
Ans: (b)
Solutions:
velocity of car at t = 4 sec is
v = u + at  v = 0 + 5(4)
= 20 m/s
At t = 6 sec
acceleration is due to gravity
∴ a = g = 10 m/s
vx = 20 m/s (due to car)
vy = u + at
= 0 + g(2) (downward)
= 20 m/s (downward)

= 20√2 m/s.

Q2: A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 'θ'to the horizontal, the maximum height attained by it equals 4R. The angle of projection,θ, is then given by:    [2021]
(a)
(b)
(c)
(d)
Ans: (b)
Solutions:

2019

Q1: When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60^o with horizontal, it can travel a distance x₁ along the plane. But when the inclination is decreased to 30^o 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)
Assume initial velocity = u

Q2: The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by:    [2019]
(a) 30° west
(b) 0° west
(c) 60° west
(d) 45° west
Ans: (a)
Solution:
V_SR = 20m/s
V_RG =  10m/s

sin θ = 10/20
sin θ = 1/2
θ = 30⁰ west

Q3: A particle moving with velocity  is acted by three forces shown by the vector triangle PQR. The velocity of the particle will:    [2019]

(a) Increase
(b) Decrease
(c) Remain constant
(d) Change according to the smallest force
Ans: (c)
Solution

As forces are forming closed loop in same order

2017

Q1: The x and y coordinates of the particle at any time are x = 5t – 2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is:-    [2017]
(a) 5 m/s² 
(b) – 4 m/s² 
(c) – 8 m/s² 
(d) 0
Ans: (b)
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) 
Solution: Net force on the particle in uniform circular motion is centripetal force, which is provided by the tension in string.

2016

Q1: In the given figure, a = 15 m s^$-$2 represents the total acceleration of particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is                   [2016]

(a) 4.5 m s^$-$1 
(b) 5.0 m s^$-$1 
(c) 5.7 m s^$-$1 
(d) 6.2 m s^$-$1 
Ans: (c)

Q2: If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is:    [2016]
(a) 180 deg
(b) 0 deg
(c) 90 deg
(d) 45 deg
Ans: (c)
Solution:

2015

Q1: If vectors   are functions of time, then the value of t at which they are orthogonal to each other is     [2015]
(a) t = π/ω
(b) t = 0
(c) t = π/4ω
(d) t = π/2ω
Ans: (a)

Q2: The positions vector of a particle  as a function of time is given by + . Where R is in meters, t is in seconds and  denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle? 
(a) Magnitude of the velocity of particle is 8 meter/second.
(b) Path of the particle is a circle of radius 4 meter. 
(c) Acceleration vector is along  -
(d) Magnitude of acceleration vector is v² / R, where v is the velocity of particle.           [2015]
Ans: (a)

Its magnitude is

Q3: A ship A is moving Westwards with a speed of 10 km h^-1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h^-1. The time after which the distance between them becomes shortest, is:    [2015]
(a) 10√2h
(b) 0 h
(c) 5h
(d) 5√2h
Ans: (c) 
Solution:

2014

Q1: A particle is moving such that its position coordinates (x, y) are (2m, 3m) at time t = 0, (6m, 7m) at time t = 2s and (13m, 14m) at time t= 5s. Average velocity vector  from t = 0 to t = 5s is:    [2014]
(a)
(b)
(c)
(d)
Ans: (b)
Solution:
Average velocity Vector,

Q2: A projectile is fired from the surface of the earth with a velocity of 5 ms^$-$1 and angle $�$ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms^$-$1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms^$-$2) is
(Given g = 9.8 m s^$-$2) 
(a) 3.5
(b) 5.9
(c) 16.3
(d) 110.8
Ans: (a)
The equation of trajectory is

where θ is the angle of projection and u is the velocity with which projectile is projected. For equal trajectories and for same angles of projection,