JEE Previous Year Questions With Solutions Of Kinematics Part - 3 | 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. A particle has an initial velocity 3i + 4j and an acceleration of 0.4i + 0.3j. Its speed after 10 s is: (AIEEE 2009)

(A) 7√2 units

(B) 7 units

(C) 10 units

(D) 8.5 units.   

Sol: (A)

Given: initial velocity, u = 3i + 4j; acceleration, a = 0.4i + 0.3j; t = 10 s.

We have, v = u + at

Or, v = 3i + 4j + (0.4i + 0.3j) * 10

Or, v = 7i + 7j

Magnitude of v is = √ (7² + 7²) = 7√2 units.

2. A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v₀ in the x-y plane as shown in the figure. At a time t < (v₀ sin θ/g), the angular momentum of the particle is: (AIEEE 2010)

(A) – mgv₀t² cos θ j

(B) ½ mgv₀t² cos θ i

(C) – ½ mgv₀t² cos θ k

(D) mgv₀t cos θ k.    

Where i, j, and k are unit vectors along x, y and z-axis respectively.

Sol: (C)

The position vector of the particle from the origin at any time t is r = v₀t cos θ i + (v₀t sin θ – ½ gt²) j.

And velocity v = v₀ sin θ i + (v₀ cos θ – gt) j.

We have,

Angular momentum, L = r x mv

Or, L = m(r x v)

Or, L = m [(v₀t cos θ i + (v₀t sin θ – ½ gt²) j) x (v₀ sin θ i + (v₀ cos θ – gt) j)]

Or, L = – ½ mgv₀t² cos θ k.

3. An object moving with a speed of 6.25 m/s, is decelerated at a rate given by dv/dt = – 2.5√v, where v is the instantaneous speed. The time taken by the object, to come to rest, would be: (AIEEE 2011)

(A) 8 s

(B) 1 s

(C) 2 s

(D) 4 s.       

Sol: (C)

Given: dv/dt = – 2.5√v

Or, dv/√v = – 2.5 dt

Integrating both side and putting the limit (at t₁ = 0, v₁ = 6.25 m/s and at t₂ = t, v₂ = 0)

We get, t = 2 s.

4. A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be: (AIEEE 2012, WBJEE 2011)

(A) 10 m

(B) 20 m

(C) 10√2 m

(D) 20√2 m.

Sol: (B)

Let u be the velocity of projection.

We have, max height, H_max = u²/2g = 10 m.

And max range, R_max = u²/g = 2 * H_max = 20 m.

5. A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for: (WBJEE 2010)

(A) 6 s

(B) 5 s

(C) 7 s

(D) 4 s

Sol: (B)

Given: u = 0; t = 3 s; a = – g = – 10 m/s².

We know, s = ut + ½ at²

Or, s = 0 – ½ * 10 * 3²

Or, s = – 45 m (–ve sign shows downward displacement).

Distance travel in n^th second is given by

→ S_n = u + (2n – 1) g/2

Or, 45 = 0 + (2n – 1) * 5

Or, 2n – 1 = 9

Or, n = 5 (last second).

So the stone remains in the air for 5 seconds.

6. A body is projected with a speed u m/s at an angle β with the horizontal. The kinetic energy at the highest point is 3/4th of the initial kinetic energy. The value of β is: (WBJEE 2010)

(A) 30⁰

(B) 45⁰

(C) 60⁰

(D) 120⁰

Sol: (A)

Let mass of the body be m.

Initial kinetic energy = ½ mu².

Velocity at the highest point = u cos β.

Therefore, kinetic energy at the highest point = ½ m (u cos β)².

A.T.Q., Kinetic energy at the highest point = (3/4) * Initial kinetic energy

Or, ½ m (u cos β)² = (3/4) * ½ mu²

Or, cos² β = ¾

Or, β = cos^–1 (√3/2) = 30⁰.

7. A ball is projected horizontally with a velocity of 5 m/s from the top of a building 19.6 m high. How long will the ball take of hit the ground? (WBJEE 2010)

(A) √2 s

(B) 2 s

(C) 3 s

(D) √3 s

Sol: (B)

Given: horizontal velocity, u_h = 5 m/s; vertical velocity, u_v = 0; H = – 19.6 m.

Vertical motion:

We have, s = u_vt + ½ at²

Or, – 19.6 = 0 – ½ * 9.8 * t²

Or, t = 2 s.

8. From the top of a tower, 80 m high from the ground, a stone is thrown in the horizontal direction with a velocity of 8 ms^–1. The stone reaches the ground after a time‘t’ and falls at a distance of ‘d’ from the foot of the tower. Assuming g = 10 m/s², the time t and distance d are given respectively by: (WBJEE 2012)

(A) 4s, 16 m

(B) 6 s, 64 m

(C) 4 s, 32 m

(D) 6 s, 48 m.

Sol: (C)

Given: horizontal velocity, u_h = 8 m/s; vertical velocity, u_v = 0; H = – 80 m.

Vertical motion:

We have, s = u_vt + ½ at²

Or, – 80 = 0 – ½ * 10 * t²

Or, t = 4 s.

And horizontal distance, d = u_h * t = 8 * 4 = 32 m.

9. A particle is travelling along a straight line OX. The distance x (in metres) of the particle from O at a time t is given by x = 37 + 27t – t³ where t is time in seconds. The distance of the particle from O when it comes to rest is: (WBJEE 2012)

(A) 81 m

(B) 91 m

(C) 101 m

(D) 111 m.

Sol: (B)

Given: x = 37 + 27t – t³

Velocity, v = dx/dt = 27 – 3t²

When particle comes to rest, then v = 0.

So, 27 – 3t² = 0 → t = 3 s.

And distance, x (t = 3 s) = 37 + 27 * 3 – 3³ = 91 m.

10. A particle moves along X-axis and its displacement at any time is given by x(t) = 2t³ – 3t² + 4t in SI units. The velocity of the particle when its acceleration is zero, is: (WBJEE 2013)

(A) 8.5 m/s

(B) 4.5 m/s

(C) 3.5 m/s

(D) 2.5 m/s

Sol: (D)

Given: x (t) = (2t³ – 3t² + 4t)

Velocity, v = dx/dt = 6t² – 6t + 4

Acceleration, a = dv/dt = 12t – 6

Now, a = dv/dt = 12t – 6 = 0

Or, t = 0.5 s.

So, v (at a = 0) = (6 * 0.5² – 6 * 0.5 + 4) = 2.5 m/s.

11. A body is projected from the ground with a velocity v = (3i + 10j) m/s. The maximum height attained and the range of the body respectively are (given g = 10 ms^–2): (WBJEE 2013)

(A) 3 m and 5 m

(B) 6 m and 5 m

(C) 3 m and 10 m

(D) 5 m and 6 m

Sol: (D)

Given: V = 3i + 10j

Vertical component velocity, V_v = 10 m/s.

Horizontal component velocity, V_h = 3 m/s.

Max height, H_max = (V_v)²/2g = 5 m.

Time of flight, T = 2V_v/g = 2 s.

Therefore, range = V_h * T = 6 m.

12. Two persons A and B start from the same location and walked around a square in opposite directions with constant speeds. The square has side 60m. Speeds of A and B are 4m/s and 2m/s respectively. When will they meet first time? (UPSEE 2016)

(A) 10 sec

(B) 20 sec

(C) 30 sec

(D) 40 sec

Sol: (D)

Given: side of square, L = 60 m; Speeds of A = 4 m/s and Speeds of B = 2 m/s.

Total length = perimeter of square = 4L = 240 m.

Let A travel a distance x.

Then distance travel by B = 240 – x.

Time taken by A = x/4 and time taken by B = (240 – x)/2.

A.T.Q. they will take same time.

Therefore, x/4 = (240 – x)/2

Or, x = 160 m.

Time = 160/4 = 40 seconds.

13. A projectile is projected with an initial velocity (4i + 5j) m/s. Here j is the unit vector directed vertically upwards and unit vector i is in the horizontal direction. Velocity of the projectile (in m/s) just before it hits the ground is: (UPSEE 2016)

(A) 4i + 5j

(B) – 4i + 5j

(C) 4i – 5j

(D) – 4i – 5j.

Sol: (C)

Horizontal component remains constant, whereas vertical component changes its sign. 

14. Suppose you drive to Delhi (200 km away) at 400 km/hr and return at 200 km/hr. What is yours average speed for the entire trip? (UPSEE 2016)

(A) Zero

(B) 300 Km/hr

(C) Less than 300 km/hr

(D) More than 300 km/hr.

Sol: (C)

Total distance, S = 200 + 200 = 400 km

Time taken for 1^st 200 km = 200/400 = 0.5 hr.

Time taken for 2^nd 200 km = 200/200 = 1 hr.

Total time taken, T = 0.5 + 1 = 1.5 hrs.

Therefore, average speed = S/T = 400/1.5 = 266.6 km/hr.

Which is Less than 300 km/hr.

15. For a freely falling body, the vertical velocity at the fifth second is: (Kerala CET 2003)

(A) 49 m/s

(B) 39.2 m/s

(C) 94.9 m/s

(D) 245 m/s

(E) 19.6 m/s.

Sol: (A)

Given: initial velocity, u = 0; time, t = 5 s; a = – 9.8 m/s².

We have, v – u = at

Or, v – 0 = – 9.8 * 5

Or, v = – 49 m/s (– ve sign shows downward motion).

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

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