JEE Previous Year Questions With Solutions Of Kinematics Part - 1 | IIT JEE | AIEEE | NEET | JEE Main & Advance | AIIMS | WBJEE | KCET | PMT | OTHER JE EXAMS

It is very important to have the idea of any examination to get success in that. The best way to have that idea is to have a look at previous year Questions of that examination. That's why we have collected some previous year Questions with Solutions of JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams so that you can get clear pattern and level of question that will be asked in forthcoming JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams.

JEE Previous Year's Questions With Solutions

1. If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? (AIEEE 2002, WBJEE 2012)

(A) 1 cm

(B) 2 cm

(C) 3 cm

(D) 4 cm.

Sol: (A)

Let initial velocity be u; Velocity of body after penetration of 3 cm is v = u/2; Acceleration = a; s = 3 cm.

We have, v² – u² = 2as

Or, (u/2)² – u² = 2 * a * 3

Or, a = – u²/8 cm/s².

Now, initial velocity = u/2; final velocity = 0; s =?.

Again, v² – u² = 2as

Or, 0 – (u/2)² = 2 * (– u²/8) * s

Or, s = 1 cm.

2. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are: (AIEEE 2002)

(A) 12 N, 6 N

(B) 13 N, 5 N

(C) 10 N, 8 N

(D) 16 N, 2 N.

Sol: (B)

Given: Resultant R = 12 N is perpendicular to smaller force B and (A + B) = 18 N ------ (1).

From right angle triangle (acd)

We have, A² = B² + R²

Or, A² – B² = R²

Or, (A + B) (A – B) = R²

Or, 18 (A – B) = 12²

Or, A – B = 8 ---------- (2)

Solving equation 1 and 2 we get

A = 13 N and B = 5 N.

3. Speeds of two identical cars are u and 4u at a specific instant. If the same deceleration is applied on both the cars, the ratio of the respective distances in which the two cars are stopped from that instant is: (AIEEE 2002)

(A) 1 : 1

(B) 1 : 4

(C) 1 : 8

(D) 1 : 16.

Sol: (D)

For car A: initial speed = u; final speed = 0; acceleration = – a; distance travelled = S_A.

We have, v² – u² = 2as

Or, (0)² – u² = 2 * (– a) * S_A

Or, S_A = u²/2a.

For car B: initial speed = 4u; final speed = 0; acceleration = – a; distance travelled = S_B.

Again, v² – u² = 2as

Or, (0)² – (4u)² = 2 * (– a) * S_B

Or, S_B = 8u²/a.

Therefore, the ratio of the respective distances is 1 : 16.

4. From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If v_A and v_B are their respective velocities on reaching the ground, then: (AIEEE 2002)

(A) v_A = v_B

(B) v_A > v_B

(C) v_A < v_B

(D) Their velocities depend on their masses.

Sol: (A)

Let height of the building be h.

Ball A projected upwards with velocity u falls back to building top with velocity u downwards. It completes its journey to ground under gravity.

Therefore, (v_A)² = u² + 2gh ------- (1)

Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h.

Therefore, (v_B)² = u² + 2gh ------- (2)

From equation 1 and 2

We get, v_A = v_B.

5. A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30⁰ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? [g = 10 m/s; sin 30⁰ = ½; cos 30⁰ = (√3)/2]: (AIEEE 2003)

(A) 5.20 m

(B) 4.33 m

(C) 2.60 m

(D) 8.66 m.

Sol: (A)

Given: initial velocity, u = 10 m/s; angle of projection, θ = 30⁰.

Initial and final points are in same horizontal plane.

We have,

Range, R = (u² Sin 2θ)/g

Or, R = (10² sin 60⁰)/10

Or, R = 8.66 m.

6. A particle moves in a circle of radius R. In half the period of revolution its displacement is ____ and distance covered is ___. (IITJEE 1983)

Sol:

Displacement = shortest path = AB = 2R.

Distance = Total path = ACB = πR.

7. Two balls of different masses are thrown vertically upwards with the same the speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance). (IITJEE 1983)

Sol: (True)

Since the initial speed and acceleration are same for both the balls. So, they pass through the point of projection in their downward motion with the same speed.

8. A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest path in 15 minutes. The velocity of the river in km/h is: (IITJEE 1988, CBSE 2000, 1998)

(A) 1

(B) 4

(C) √41

(D) 3.

Sol: (D) 

V_{b, r} = velocity of boat w.r.t river = 5 km/h.

V_{r, g} = velocity of river w.r.t ground.

V_{b, g} = velocity of boat w.r.t ground = 1 km/15 min = 4 km/h.

We have, V_{b, g} = V_{b, r} + V_{r, g} -------- (1)

Taking vertical component of the equation (1)

→ V_{b, g} = V_{b, r} sin θ + V_{r, g} cos 90⁰

Or, 4 = (5 sin θ)

Or, sin θ = 4/5

And, cos θ = 3/5.

Taking horizontal component of the equation (1)

→ V_{b, g} cos 90⁰ = – V_{b, r} cos θ + V_{r, g}

Or, 0 = – (5 * 3/5) + V_{r, g}

Or, V_{r, g} = 3 km/h.

9. A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is: (IITJEE 1992)

(A) 4.0 m/s

(B) 5.5 m/s

(C) 5.0 m/s

(D) 4.8 m/s.

Sol: (D)

Let the total distance be x.

Time taken for 1^st half, t₁ = (x/2)/3 = x/6.

Time taken for 2^nd half, t₂ = (x/2)/ (4.5 + 7.5) = x/24.

Total time, t = 5x/24.

So, average speed = x/t = 24/5 = 4.5 m/s.

10. The area under the acceleration time graph represents: (MP PMT 1993)

(A) Velocity

(B) The displacement

(C) Distance travelled

(D) Change in velocity.

Sol: (D)

We have,

Acceleration, a = ΔV/Δt

Or, ΔV = a * Δt.

So, the area under the acceleration time graph represents the change in velocity.

11. A particle moves along the x-axis such that its coordinates (x) varies with time (t), according to the expression: x = 2 – 5t + 6t² where x is in metres and t is in seconds. The initial velocity of the particle is: (MP PET 1996)

(A) 40 m/s

(B) Zero

(C) 8 m/s

(D) 20 m/s.

Sol: (B)

Distance travelled, x = 2 – 5t + 6t²

Velocity, v = dx/dt = – 5 + 12t

Initial velocity, u = velocity at t = 0

Or, u = – 5 m/s.

12. For the velocity time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds?(MP PMT 1998)

(A) 1/2

(B) 2/3

(C) 1/4

(D) 1/3.

Sol: (C)

Distance covered = area under velocity time graph.

Total distance covered = ½ * 2 * 10 + 10 * 2 + ½ * 2 * 10 = 40 m.

Distance covered in last 2s = ½ * 2 * 10 = 10 m.

Fraction = 10/40 = 1/4.

13. The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 m/s. If the change in velocity of the body is 0.18 m/s during this time, its uniform acceleration is: (EAMCET 2000)

(A) 0.04 m/s²

(B) 0.03 m/s²

(C) 0.02 m/s²

(D) 0.01 m/s².

Sol: (C)

Given: distance travelled, S = 3.06 m; average velocity, V_avg = 0.34 m/s; change in velocity, ΔV = 0.18 m/s.

Time taken, t = S / V_avg = 9 s.

Acceleration, a = ΔV/t = 0.18/9 = 0.02 m/s².

14. A constant force acts on a body of mass 0.9 kg at rest for 10 s. If the body moves a distance of 250 m, the magnitude of the force is: (EAMCET 2000)

(A) 4.5 N

(B) 5 N

(C) 3 N

(D) 4 N.

Sol: (A)

Given: initial velocity, u = 0; time taken, t = 10 s; distance travelled, S = 250 m; mass, m = 0.9 kg.

We have, S = ut + ½ at²

Or, 250 = 0 + ½ * a * 10²

Or, a = 5 m/s².

And, Force = mass * acceleration

Or, Force = 0.9 * 5 = 4.5 N.

15. The equation of motion of a projectile are given by x = 36t m and 2y = 96t – 9.8t² m. The angle of projection is: (EAMCET 2003)

(A) Sin^{– 1} (4/3)

(B) Sin^{– 1} (4/5)

(C) Sin^{– 1} (3/5)

(D) Sin^{– 1} (3/4).

Sol: (B)

Horizontal velocity, V_h = dx/dt = 36 m/s.

Vertical velocity, V_v = dy/dt = 48 – 9.8t.

Vertical velocity at time of projection (t = 0) = 48 m/s.

Let the angle of projection be θ.

So, tan θ = (V_v/ V_h)_{t = 0} = 48/36 = 4/3

Or, sin θ = 4/5

Or, θ = sin^{– 1} (4/5).

16. The displacement of a particle is represented by the following equation: s = 3t³ + 7t² + 5t + 8 where s is in metre and t in second. The acceleration of the particle at t = 1s is: (CBSE 2000)

(A) Zero

(B) 18 m/s²

(C) 32 m/s²

(D) 14 m/s².

Sol: (C)

Given: s = 3t³ + 7t² + 5t + 8

Or, velocity, v = ds/dt = 9t² + 14t + 5

Or, acceleration, a = d²s/dt² = 18t +14.

Therefore, acceleration, a (t = 1 s) = 32 m/s².

17. Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is π/3 the maximum height reached by it is 102 m. Then the maximum height reached by the other in metre is: (EAMCET 2003)

(A) 34

(B) 224

(C) 336

(D) 56.

Sol: (A)

Horizontal range will be same if angle of projection is θ or (90 – θ).

Angle of projection for 1^st stone, θ₁ = 60⁰.

Angle of projection for 2^nd stone, θ₂ = (90 – 60⁰) = 30⁰.

For 1^st stone, Max height = 102 = (u² sin² 60⁰)/2g.

For 2^nd stone, Max height = h = (u² sin² 30⁰)/2g.

Therefore, h/102 = sin² 30⁰/ sin² 60⁰

Or, h = 102/3 = 34 m.

18. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first? (Odisha JEE 2003)

(A) Both will reach simultaneously

(B) Depends on their mass

(C) The slower one

(D) The faster one.

Sol: (A)

Both the bullet have same vertical displacement and acceleration. Therefore, both will have the same time and t = √ (2h/g).

19. A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is: (AIEEE 2003)

(A) 18 m

(B) 6 m

(C) 24 m

(D) 12 m.

Sol: (C)

For 1^st case: u = 50 km/hr = 180 m/s; v = 0; s = 6 m; a =?

We have, v² – u² = 2as

Or, 180² – 0² = 2 * a * 6

Or, a = 180²/12 = 2700 m/s².

For 2^nd case: u = 100 km/hr = 360 m/s; v = 0; a = 2700 m/s²; s =?

Again, v² – u² = 2as

Or, 360² – 0² = 2 * 2700 * s

Or, s = 360²/5400 = 24 m.

20. The co-ordinates of a moving particle at any time t are given by x = αt³ and y = βt³. The speed of the particle at time t is given by: (AIEEE 2003)

(A) √ (α² + β²)

(B) 3t² √ (α² + β²)

(C) 3t √(α² + β²)

(D) t² √ (α² + β²).

Sol: (B)

Velocity along x-axis, v_x = dx/dt = 3 αt².

Velocity along y-axis, v_y = dy/dt = 3 βt².

So, resultant velocity = √ [(v_x)² + (v_y)²]

Or, resultant velocity = √ [(3αt²)² + (3βt²)²]

Or, resultant velocity = 3t² √ (α² + β²).

21. An automobile travelling with a speed of 60 km/h can brake to stop with a distance of 20 m. If the car is going twice as fast i.e. 120 km/h, the stopping distance will be: (AIEEE 2004)

(A) 20 m

(B) 60 m

(C) 40 m

(D) 80 m.

Sol: (D)

In the 1^st case: initial velocity, u = 60 km/h = 100/6 m/s; final velocity, v = 0; distance travelled, s = 20 m.

We have, v² – u² = 2as

Or, 0² – (100/6)² = 2 * a * 20

Or, a = – 6.94 m/s².

In the 2^nd case: initial velocity, u = 120 km/h = 100/3 m/s; final velocity, v = 0; distance travelled, s =? Acceleration, a = – 6.94 m/s².

Again, v² – u² = 2as

Or, 0² – (100/3)² = 2 * (– 6.94) * s

Or, s = 80 m.

22. Which of the following statements is false for a particle moving in a circle with a constant angular speed? (AIEEE 2004)

(A) The velocity vector is tangent to the circle

(B) The acceleration vector is tangent to the circle

(C) The acceleration vector points to the centre of the circle

(D) The velocity and acceleration vectors are perpendicular to each other.

Sol: (B)

The acceleration vector is tangent to the circle is a false statements.

Discussion - If you have any Query or Feedback comment below.

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