JEE Previous Year Questions With Solutions Of Kinematics Part - 2 | 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. The relation between time t and distance x is t = ax² + bx where a and b are constants. The acceleration is: (AIEEE 2005)

(A) – 2av²

(B) 2av³

(C) – 2av³

(D) 2av².

Sol: (C)

Given: t = ax² + bx --------- (1)

Differentiate the equation (1) w.r.t. ‘t’

We get, 1 = 2ax (dx/dt) + b(dx/dt)

Or, 1 = 2axv + bv [since v = (dx/dt)]

Or, v = 1/ (2ax +b) -------- (2)

Again, differentiate the equation (2) w.r.t. ‘t’

We get, (dv/dt) = – 2av³.

So, acceleration = – 2av³.

2. A projectile can have the same range R for two angle of projection. If t₁ and t₂ be the time of flights in the two cases, then the product of the two time of flights is proportional to: (AIEEE 2005, 2004, Karnataka CET 2003, EAMCET 2012)

(A) R

(B) 1/R

(C) 1/R²

(D) R².

Sol: (A)

Range is same for angle of projection θ and (90 – θ).

Range, R = (u² sin 2θ)/g

Time of flight for 1^st case:

→ t₁ = (2u sin θ)/g

And, Time of flight for 2^nd case:

→ t₂ = {2u sin (90 – θ)}/g

A.T.Q. t₁.t₂ = (4u² sin θ cos θ)/g²

Or, t₁.t₂ = (2/g) {(u² sin 2θ)/g} = 2R/g.

Therefore, t₁.t₂ is proportional to R.

3. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity v that varies as v = α√x. The displacement of the particle varies with time as: (AIEEE 2006)

(A) t          

(B) t^1/2

(C) t³

(D) t².         

Sol: (D)

Given: Velocity, v = α√x

Or, dx/dt = α√x

Or, dx/√x = α dt ------- (1)

Integrating equation (1)

We get, 2x^1/2 = αt + c

At t = 0, x = 0 → c = 0.

So, x = (αt)²/4.

Hence, the displacement of the particle varies with time as t².

4. The velocity of a particle is v = v₀ + gt + ft². If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is: (AIEEE 2007)

(A) v₀ + g/2 + f 

(B) v₀ + 2g + 3f

(C) v₀ + g + f

(D) v₀ + g/2 + f/3.     

Sol: (D)

Given: Velocity, v = v₀ + gt + ft²

Or, dx/dt = v₀ + gt + ft²

Or, dx = (v₀ + gt + ft²) dt ------- (1)

Integrating equation (1)

We get, x = v₀t + gt²/2 + ft³/3 + c

At t = 0, x = 0 → c = 0.

So, x = v₀t + gt²/2 + ft³/3.

Displacement (t = 1) = v₀ + g/2 + f/3.

5. A body starts from rest at time t = 0, the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be: (IITJEE 2004)

(A) 55 m/s

(B) 110 m/s

(C) 550 m/s

(D) 650 m/s.    

Sol: (A)

Area under the curve in acceleration-time graph gives the change in velocity.

Therefore, ΔV = ½ * 10 * 11 = 55 m/s.

Since initial velocity = 0.

Therefore maximum velocity attained = 55 m/s.

6. If the displacement of a particle changes with time as √x = t + 3, then the velocity of the particle will be proportional to: (WBJEE 2007)

(A) t^–1

(B) t

(C) √t

(D) t^–2.       

Sol: (B)

Given: √x = t + 3

Or, x = (t + 3)²

Or, dx/dt = 2 (t + 3)

Therefore, the velocity of the particle is proportional to t.

7. The distance travelled by an object along a straight line in time ‘t’ is given by S = 3 – 4t + 5t², the initial velocity of the object is: (WBJEE 2008)

(A) 3 unit

(B) – 3 unit

(C) 4 unit

(D) – 4 unit.     

Sol: (D)

Given: S = 3 – 4t + 5t²

Velocity, V = ds/dt = – 4 + 10t

Velocity, V (t = 0) = – 4 unit.

8. A body is fired vertically upwards. At half the maximum height, the velocity of the body is 10 m/s. The maximum height raised by the body is (g = 10 m/s^–2): (Odisha JEE 2008)

(A) 10 m

(B) 15 m

(C) 20 m

(D) Zero.   

Sol: (D)

Taking motion of the body from half the max height upto the highest point, we have u = 10 m/s, v = 0, a = – 10 m/s^–2, s = h/2.

We know, v² – u² = 2as

Or, 0² – 10² = 2 * (– 10) * (h/2)

Or, h = 10 m.

Hence, the maximum height raised by the body is 10 m.

9. 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: u_x = 5 m/s, u_y = 0, H = – 19.6 m, a = – 9.8 m/s².

Equation of motion along Y-axis:

We have, H = u_yt + ½ at²

Or, – 19.6 = 0 – ½ * 9.8 * t²

Or, t = 2s.

10. A bullet moving with a speed of 100 m/s can just penetrate two planks of equal thickness. Then, the number of such planks penetrated by the same bullet when the speed is doubled will be: (Karnataka CET 2004)

(A) 8

(B) 6

(C) 10

(D) 4

Sol: (A)

Given: initial speed, u = 100 m/s; final speed, v = 0.

Assume: thickness of plank be x.

We have, v² – u² = 2as

Or, 0² – 100² = 2a * (2x)

Or, a = – 2500/x.

Now, initial speed, u = 200 m/s; v = 0; s = nx.

Again, v² – u² = 2as

Or, 0² – 200² = 2 * (– 2500/x) * (nx)

Or, n = 8.

11. A metro train starts from rest and in five seconds achieves 108 km/h. After that is moves with constant velocity and goes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, find total time of travelling: (UPSEE 2006)

(A) 15.3 s

(B) 12.2 s

(C) 17.2 s

(D) 9 s

Sol: (C)

For 1^st 5 s.

Initial velocity, u = 0; final velocity, v = 108 km/h = 30 m/s; t₁ = 5s.

We have, v = u + at

Or, 30 = 0 + a * 5

Or, a = 6 m/s.

So, distance travelled, s = ut + ½ at²

Or, s = 75 m.

Let distance travelled with constant velocity be x.

→ 75 + x + 45 = 395

Or, x = 275 m.

And, time taken t₂ = 275/30 = 9.2 s.

For last 45 m:

Initial velocity, u = 30 m/s; final velocity, v = 0; s = 45 m.

We have, v² – u² = 2as

Or, 0² – 30² = 2 * a * 45

Or, a = – 10 m/s².

Time taken in travelling 45 m is

→ t₃ = (v – u)/a

Or, t₃ = 3 s.

So, total time taken in whole journey = t₁ + t₂ + t₃ = 5 + 9.2 + 3 = 17.2 s.

12. A projectile is thrown in the upward direction making an angle of 60⁰ with the horizontal direction with a velocity of 147 m/s. then the time after which its inclination with the horizontal is 45⁰, is? (UPSEE 2006)

(A) 2.745 s

(B) 15 s

(C) 5.49 s

(D) 10.98 s

Sol: (C)

During the projectile motion, horizontal component of velocities at different point of projectile motion remain same.

So, u Cos 60⁰ = v Cos 45⁰

Or, 147 * ½ = v * 1/√2

Or, v = 147/√2 m/s.

Vertical component of initial velocity is

→ u_y = u sin 60⁰ = 147√3/2 m/s

Vertical component of velocity @ 45⁰ is

→ v_y = v sin 45⁰ = (147/√2) (1/√2) = 147/2 m/s.

We have, v_y = u_y + at

Or, 147/2 = 147√3/2 – 9.8t

Or, t = 5.49 s.

13. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to: (UPSEE 2007)

(A) log e^x

(B) x²

(C) e^x

(D) x

Sol: (B)

Given: retardation, – a = kx

Where, k = proportionality constant and x = displacement.

Now, dv/dt = – kx

Or, (dv/dx) (dx/dt) = – kx

Or, v (dv/dx) = – kx

Or, v dv = – kx dx

Integrating both side we get,

→ v²/2 = – kx²/2 + c

Or, K.E. = mv²/2 = – mkx²/2 + mc

So, loss of kinetic energy for any displacement x is proportional to x².

14. A ball thrown by one player reaches the other in 2 s. The maximum height attained by the ball above the point of projection will be (g = 10 m/s²): (UPSEE 2008)

(A) 5 m

(B) 10 m

(C) 2.5 m

(D) 7.5 m

Sol: (A)

Given: time of flight, T = 2 s.

We have, T = (2u sin θ)/g

Or, 2 = (2u sin θ)/g

Or, u sin θ = g.

And, max height H = (u² sin² θ)/2g

Or, H = g²/2g = g/2 = 5 m.

15. A ball is dropped from a bridge at a height of 176.4 m over a river. After 2s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously? (UPSEE 2009)

(A) 14.5 m

(B) 49 m/s

(C) 2.45 m/s

(D) 24.5 m/s

Sol: (D)

For 1^st ball: H = – 176.4 m; u = 0; a = – g.

We have, H = ut + ½ at²

Or, – 176.4 = 0 – ½ gt²

Or, t = 5.9 s.

For 2^nd ball: t = 5.9 – 2 = 3.9 s; H = – 176.4 m; initial velocity, u = – u₀; a = – g.

Again, H = ut + ½ at²

Or, – 176.4 = – u₀ * 3.9 – ½ * g * 3.9²

Or, u₀ = 24.5 m/s.

16. A bullet is fired with a velocity u making an angle of 60⁰ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is: (WBJEE 2009)

(A) 0

(B) u

(C) (√3/2) u

(D) u/2

Sol: (D)

Horizontal component of the velocity in a projectile motion remains constant.

Horizontal component of the velocity = u cos θ = u cos 60⁰ = u/2.

17. A bullet emerges from a barrel of length 1.2 m with a speed of 640 m/s. Assuming constant acceleration, the approximate time that it spends in the barrel after the gun is fired is: (WBJEE 2008)

(A) 40 ms

(B) 4 ms

(C) 1ms

(D) 400 ms.

Sol: (B)

Given: u = 0; v = 640 m/s; s = 1.2 m.

Find: a =? t =?

We have, v² – u² = 2as

Or, 640² – 0² = 2 * a * 1.2

Or, a = 640²/2.4 m/s².

And, t = (v – u)/a

Or, t = (640 – 0)/ (640²/2.4)

Or, t = 0.00375 s ≈ 0.004 s = 4 ms.

18. A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is: (WBJEE 2009, WBJEE 2012)

(A) K

(B) Zero

(C) K/4

(D) K/2

Sol: (C)

Let mass of particle be m and velocity of projection v.

Given: kinetic energy = K = ½ mv²

Velocity at the highest point = v cos 60⁰ = v/2.

So, kinetic energy at the highest point = ½ m (v/2)² = (½ mv²)/4 = K/4.

19. A scooterist sees a bus 1 km ahead of him moving with a velocity of 10 m/s. With what speed the scooterist should move so as to overtake the bus in 100 s? (Odisha JEE 2008)

 (A) 10 m/s

(B) 50 m/s

(C) 20 m/s

(D) 30 m/s.

Sol: (C)

Distance travelled by bus in 100 s = 10 * 100 = 1000 m.

Let speed of scooter be u m/s.

Distance travelled by scooter in 100 s = u * 100 = 100u.

A.T.Q, 1000 + 1000 = 100u

Or, u = 20 m/s.

20. The maximum range of projectile fired with some initial velocity is found to be 1000 m, in the absence of wind and air resistance. The maximum height reached by the projectile is: (Odisha JEE 2009)

(A) 250 m

(B) 1000 m

(C) 500 m

(D) 2000 m.

Sol: (A)

We have, max range R_max = u²/g [for max range θ = 45⁰]

Or, 1000 = u²/10

Or, u = 100 m/s.

And max height H = (u² sin² θ)/2g

Or, H = (100² sin² 45⁰)/ (2*10)

Or, H = 250 m.

21. A car starting from rest, accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 15 s, then: (AIEEE 2005)

(A) s = ½ ft²

(B) s = 1/4 ft²

(C) s = 1/72 ft²

(D) s = 1/6 ft². 

Sol: (C)

For 1^st part of journey: u₁ = 0; a₁ = f; s₁ = s; t₁ = t₁.

Distance travelled, s₁ = u₁t₁ + a (t₁)²

Or, s = ½ f (t₁)² -------- (1)

And (v₁)² – (u₁)² = 2a₁s₁

Or, v₁ = √ (2fs) = ft₁.

For 2^nd part of journey: velocity is constant (v₂ = v₁); t₂ = t.

Distance travelled, s₂ = v₁t₂

Or, s₂ = f t₁ t

For 3^rd part of journey: v₃ = 0; a₃ = – f/2; u₃ = v₁ = ft₁.

Again, (v₃)² – (u₃)² = 2a₃s₃

Or, – (ft₁)² = 2 (– f/2) s₃

Or, s₃ = f (t₁)² = 2s.

A.T.Q, s₁ + s₂ + s₃ = 15s

Or, s + f t₁ t + 2s = 15s

Or, f t₁ t = 12s --------- (2)

From equation (1) & (2)

We get, t₁ = t/6.

So, s = ½ f(t₁)² = ½ f (t/6)² =  s = 1/72 ft².

22. A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball at T/3 second? (AIEEE 2004)

(A) 7h/9 metre from the ground

(B) 17h/18 metre from the ground

(C) 8h/9 metre from the ground

(D) h/9 metre from the ground.

Sol: (C)

We have, s = ut + ½ at²

Given: s = – h, t = T.

Therefore, – h = 0 – ½ gT²

Or, gT² = 2h -------- (1)

After T/3 sec,

→ – h’ = 0 – ½ g (T/3)²

Or, h’ = gT²/18 -------- (2)

From equation (1) and (2)

We get, h’ = h/9.

Therefore, height of ball from ground = h – h’ = 8h/9.

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