JEE Previous Year Questions With Solutions Of Newton's Laws Of Motion 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. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is: (AIEEE 2003)

(A) Pm/ (M – m)

(B) P

(C) PM/ (M + m)

(D) Pm/ (M + m)

Sol: (C)

Taking (Block + Rope) as a system:

Acceleration of the system, a = P/ (M + m)

F_B,R = the force exerted by the rope on the block.

From the F.B.D. of the block:

→ Ma = F_B,R

Or, F_B,R = M * [P/ (M + m)]

Or, F_B,R = PM/ (M + m).

2. A rocket with a lift- off mass 3.5 * 10⁴ kg is blasted upwards with an initial acceleration of 10 m/s². Then the initial thrust of the blast is: (AIEEE 2003)

(A) 7.0 * 10⁵ N

(B) 3.5 * 10⁵ N

(C) 1.75 * 10⁵ N

(D) 14.0 * 10⁴ N.

Sol: (B)

Given: mass, m = 3.5 * 10⁴ kg; acceleration, a = 10 m/s².

→ Initial thrust, F = ma

Or, F = 3.5 * 10⁴ * 10 = 3.5 * 10⁵ N.

3. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statements about the scale reads is: (AIEEE 2003)

(A) Both the scales read M/2 kg

(B) Both the scales read M kg each

(C) The scale of the lower one reads M kg and of the upper one zero

(D) The reading of the two scales can be anything but the sum of the reading will be M kg.

Sol: (B)

Both the scales read M kg each.

4. A machine gun fires a bullet of mass 40 g with a velocity 1200 m/s. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most? (AIEEE 2004)

(A) Four

(B) Two

(C) Three

(D) One.

Sol: (C)

Given: mass of bullet, m = 40 g = 0.04 kg; final velocity, v = 1200 m/; initial velocity, u = 0.

Let he can fire n bullets per second.

We have, Force = change in momentum per second

Or, 144 = n * 0.04 * (1200 – 0)

Or, n = 3.

5. Two masses m₁ = 5 kg and m₂ = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move? (AIEEE 2004, Odisha JEE 2002, UPSEAT 2002)

(A) 5 m/s²

(B) 0.2 m/s²

(C) 4.8 m/s²

(D) 9.8 m/s².

Sol: (C) 

The equation of motion of m₁ = 5 kg:

→ m₁g – T= m₁a

Or, 5g – T = 5a ------- (1)

The equation of motion of m₁ = 4.8 kg:

→ T – m₂g = m₂a

Or, T – 4.8g = 4.8a ------- (2)

Solving equation (1) and (2)

We get, a = 0.2 m/s².

6. A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (take g = 10 m/s²): (Kerala CET 2001)

(A) Zero

(B) 3 m/s²

(C) 30 m/s²

(D) 10 m/s²

(E) 100 m/s².

Sol: (B) 

Let T be the tension in the string and a be the acceleration.

The equation of motion of A:

→ T = Ma

Or, T = 7a ------- (1)

The equation of motion of B:

→ Mg – T = Ma

Or, 3g – T = 3a ------- (2)

Solving equation (1) and (2)

We get, a = 3 m/s².

7. A cricketer catches a ball of mass 150 g in 0.1 s moving with speed 20 m/s, then he experiences force of: (CBSE 2001)

(A) 0.3 N

(B) 3 N

(C) 30 N

(D) 300 N.

Sol: (C)

Given: mass of ball, m = 150 g = 0.15 kg; change of velocity, ΔV = 200 – 0 = 200 m/s; time, t = 0.1 s.

We know, Force, F = ma

Or, F = 0.15 * 200/0.1 = 30 N

8. A body of mass 1 kg is falling with an acceleration of 10 m/s². Its apparent weight will be (g = 10 m/s²): (MP PMT 2002)

(A) Zero

(B) 0.5 kg wt

(C) 2.0 kg wt

(D) 1.0 kg wt.

Sol: (A)

Given: mass, m = 1 kg; acceleration, a = 10 m/s²; g = 10 m/s².

Apparent weight, W’ = m (g – a)

Or, W’ = 1 * (10 – 10)

Or, W’ = 0.

9. A body whose mass is 50 kg stands on a spring balance inside a lift. The lift starts to ascend with an acceleration of 2 m/s². The reading of the machine or balance (g = 10 m/s²): (Kerala CET 2002)

(A) 60 kg

(B) 49 kg

(C) 50 kg

(D) 45 kg

(E) Zero.

Sol: (A)

Given: mass, m = 50 kg; acceleration, a = 2 m/s²; g = 10 m/s².

Apparent weight, W’ = m (g + a)

Or, W’ = 50 * (10 + 2)

Or, W’ = 600 N.

Now, W’ = m’g

Or, m’ = 60 kg.

10. A ball of mass 0.5 kg moving with a velocity of 2 m/s strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and wall is 10^{– 3} s, the average force exerted by the wall on the ball is: (Kerala CET 2002, Haryana CEET 2002)

(A) 5000 N

(B) 1000 N

(C) 1125 N

(D) 2000 N

(E) 500 N.

Sol: (D)

Given: mass, m = 0.5 kg; time, t = 10^{– 3} s; change in velocity, ΔV = 2 – (–2) = 4 m/s.

Acceleration, a = ΔV/t = 4/10^{– 3} = 4 * 10³ m/s².

Force, F = ma = 0.5 * 4 * 10³ = 2000 N.

11. You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface? (J&K CET 2002)

(A) By running on the plane

(B) By spitting or sneezing

(C) By jumping

(D) By rolling your body on the surface.

Sol: (B)

By spitting or sneezing.

12. A 1 kg particle strikes a wall with velocity 1 m/s at an angle 30⁰ and reflects at the same angle. If it remains in contact with wall for 0.1 s, then the force is: (CBSE 2000, CPMT 2001)

(A) 10√3 N

(B) 30√3 N

(C) 40√3 N

(D) Zero.

Sol: (A) 

Horizontal motion:

Change in velocity, ΔV = [(V cos 30⁰) – (– V cos 30⁰)] = 2 * 1 * √3/2 = √3 m/s.

Acceleration, a = (√3)/ 0.1 = 10√3 m/s².

Therefore, Force F = ma = 1 * 10√3 = 10√3 N.

13. A lift of mass 1000 kg which is moving with acceleration of 1 m/s² in upward direction, then the tension developed in string which is connected to lift is: (CBSE 2002)

(A) 10800 N

(B) 11000 N

(C) 10000 N

(D) 9800 N.

Sol: (B)

Given: mass, m = 1000 kg; acceleration, a = 1 m/s².

The equation of motion of lift:

→ ma = T – mg

Or, 1000 * 1 = T – 1000 * 10

Or, T = 11000 N.

14. A ball of mass 150 g moving with acceleration of 20 m/s² is hit by a force which acts on it for 0.1 s. The impulsive force is: (Punjab PMT 2003)

(A) 0.1 Ns

(B) 1.2 Ns

(C) 0.5 Ns

(D) 0.3 Ns.

Sol: (D)

Given: mass, m = 150 g = 0.15 kg; acceleration, a = 20 m/s²; time, t = 0.1 s.

We know, Force = ma = 0.15 * 20 = 3 N.

Therefore, Impulse = Ft = 3 * 0.1 = 0.3 Ns.

15. Two masses M and M/2 are joined together by means of light inextensible string passed over a frictionless pulley as shown in figure. When the bigger mass is released, the small one will ascend with an acceleration of: (Kerala CET 2005)

(A) g/3

(B) g/2

(C) 3g/2

(D) g.

Sol: (A) 

The equation of motion of M kg block:

→ Ma = Mg – T ------- (1)

The equation of motion of m₁ = 4.8 kg:

→ Ma/2 = T – Mg/2 ------- (2)

Solving equation (1) and (2)

We get, a = g/3.

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