JEE Previous Year Questions With Solutions Of Newton's Laws Of Motion 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.

Previous Year's Questions And Solutions

1. A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (vertically). If the acceleration of the system is g/8, then the ratio of the masses is: (AIEEE 2002)

(A) 4 : 3

(B) 8 : 1

(C) 5 : 3

(D) 9 : 7

Sol: (D)

Given: a = g/8; tension in the string = T.

Equation of motion of m₁:

→ m₁a = T – m₁g

Or, m₁ (9/8) g = T

Or, m₁ = 8T/9g.

Equation of motion of m₂:

→ m₂a = m₂g – T

Or, m₂ (7/8) g = T

Or, m₂ = 8T/7g.

Therefore, m₂/m₁ = 9/7.

The ratio of the masses is 9 : 7.

2. When forces F₁, F₂, F₃ are acting on a particle of mass m such that F₂ and F₂ are mutually perpendicular, then the particle remains stationary. If the force F₁ is now removed then the acceleration of the particle is: (AIEEE 2002)

(A) F₁/m

(B) F₂ F₃/m F₁

(C) F₂/m

(D) (F₂ – F₃)/m

Sol: (A)

Resultant of forces F₂ and F₃ is equal to force F₁ [since the particle remains at rest].

So, if we remove F₁, then force acting on the particle is the resultant of forces F₂ and F₃ (= F₁).

Therefore, acceleration = F₁/m.

3. Three identical blocks of masses m = 2 kg are drawn by are drawn by a force F = 10.2 N with an acceleration of 0.6 m/s² on a frictionless surface, then what is the tension (in N) in the string between the blocks B and C? (AIEEE 2002)

(A) 7.8

(B) 9.2

(C) 4

(D) 9.8

Sol: (A)

Given: F = 10.2 N; a = 0.6 m/s²; m = 2 kg each.

Let T_AB = the tension in the string between the blocks A and B; T_Bc = the tension in the string between the blocks B and C.

We know, Force = mass * acceleration

From the FBD of block A:

→ F – T_AB = ma

Or, 10.2 – T_AB = 2 * 0.6

Or, T_AB = 9 N.

From the FBD of block B:

→ T_AB – T_BC = ma

Or, 9 – T_BC = 2 * 0.6

Or, T_BC = 7.8 N.

4. A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively: (AIEEE 2002)

(A) a, g

(B) g, g

(C) g – a, g

(D) g – a, g – a.

Sol: (C)

For the observer in the lift, acceleration = (g – a).

For the observer standing outside, acceleration = g.

5. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s², the reading of the spring balance will be: (AIEEE 2003)

(A) 15 N

(B) 24 N

(C) 74 N

(D) 49 N.

Sol: (B)

When the lift is stationary:

Weight of bag, W₁ = mg = 49 N.

When the lift is descents with a acceleration of 5 m/s²:

Weight of bag, W₂ = m (g – a).

Therefore, W₂/W₁ = m (g – a)/mg

Or, W₂/W₁ = 4.8/9.8

Or, W₂ = 24 N.

6. A body of mass 5 kg is moving with a velocity of 20 m/s. If a force of 100 N is applied on it for 10 s in the same direction as its velocity, what will now be the velocity of the body? (MP PMT 2000)

(A) 240 m/s

(B) 220 m/s

(C) 200 m/s

(D) 260 m/s.

Sol: (B)

Given: m = 5 kg; force, F = 100 N; time taken, t = 10 s; initial velocity, u = 20 m/s.

Acceleration, a = F/m = 100/5 = 20 m/s².

We have, v = u + at

Or, v = 20 + 20 * 10

Or, v = 220 m/s.

7. The mass of a lift is 500 kg. What will be the tension in its cable when it is going up with an acceleration of 2.0 m/s²? (g = 9.8 m/s²) (MP PMT 2000)

(A) 6200 N

(B) 5600 N

(C) 5900 N

(D) 5000 N.

Sol: (C)

Let the tension in the cable be T.

(i) Tension T acts upward.

(ii) Weight mg = 500 g downward.

Equation of motion of the lift (a = 2 m/s² upward):

→ Ma = T – mg

Or, 500 * 2 = T – 500 * 9.8

Or, T = 5900 N.

8. Three blocks of masses m₁, m₂, and m₃ are connected by massless strings as shown on a frictionless table. They are pulled with a force T₃ = 40 N. If m₁ = 10 kg, m₂ = 6 kg, and m₃ = 4 kg, the tension T₂ will be: (MP PMT 1998, Odisha JEE 2002)

(A) 10 N

(B) 40 N

(C) 20 N

(D) 32 N.

Sol: (D)

Taking m₁, m₂ and m₃ together as a system:

Force acting on this system is T₃ = 40 N and mass of the system is m = m₁ + m₂ + m₃ = 20 kg.

Therefore acceleration, a = T₃/m = 40/20 = 2 m/s².

Now the equation of motion of block m₃:

→ m₃a = T₃ – T₂

Or, 4 * 2 = 40 – T₂

Or, T₂ = 32 N.

9. Two blocks A (20 kg) and B (50 kg) lying on a frictionless table are connected by a light string. The system is pulled horizontally with an acceleration of 2 m/s² by a force F on B. The tension in the string is: (MP PMT 1993)

(A) 40 N

(B) 20 N

(C) 120 N

(D) 80 N.

Sol: (A)

Given: acceleration, a = 2 m/s²; mass of block A, m_A = 20 kg; mass of block B, m_B = 50 kg.

The equation of motion of block A:

→ T = m_Aa

Or, T = 20 * 2

Or, T₂ = 40 N.

10. A railway engine (mass 10⁴ kg) is moving with a speed of 72 km/h. The force which should be applied to bring it to rest over a distance of 20 m is: (SCRA 2000)

(A) 100000 N

(B) 3600 N

(C) 10000 N

(D) 7200 N.

Sol: (A)

Given: mass, m = 10⁴ kg; initial velocity, u = 72 km/h = 20 m/s; final velocity, v = 0; distance, s = 20 m.

We have, v² – u² = 2as

Or, 0² – 20² = 2 * a * 20

Or, a = – 10 m/s².

Required force, F = ma = 10⁴ * 10 = 10⁵ N. 

Given: mass, m = 5 kg; initial velocity, u = 65 cm/s = 0.65 m/s; final velocity, v = 15 cm/s = 0.15 m/s; time, t = 0.2 s. (EAMCET 1992, 2000)

We have, v – u = at

Or, 0.15 – 0.65 = a * 0.2

Or, a = – 2.5 m/s².

Required force, F = ma = 5 * 2.5 = 12.5 N. 

11. Three equal weights of mass 2 kg each are hanging on a string passing over a fixed pulley as shown in figure. What is the tension in the string connecting weights B and C? (SCRA 1996)

(A) 3.3 N

(B) 19.6 N

(C) 13 N

(D) Zero.

Sol: (C)

Given: mass of each blocks = m = 2 kg.

Let the acceleration of the blocks be a.

The equation of motion of block A:

→ T₂ – mg = ma --------- (1)

The equation of motion of block B:

→ T₁ + mg – T₂ = ma --------- (2)

The equation of motion of block C:

→ mg – T₁ = ma --------- (3)

From equation 1 and 2,

We get, T₁ = 2ma ----------- (4)

From equation 3 and 4,

We get, T₁ = 2mg/3 = 13.06 N ≈ 13 N.

12. A gun of mass 10 kg fires 4 bullets per second. The mass of each bullet is 20 g and the velocity of the bullet when it leaves the gun is 300 m/s. The force required to hold the gun while firing is: (EAMCET 2000)

(A) 6 N

(B) 24 N

(C) 240 N

(D) 8 N.

Sol: (B)

Given: mass of gun, m_g = 10 kg; mass of bullet, m_b = 20 g = 0.02 kg; no. of bullets fired per second, n = 4; final velocity of bullet, v = 300 m/s; initial velocity of bullet, u = 0 m/s.

Change in velocity, Δv = 300 m/s.

From Newton’s 3^rd Law:

Force required to hold the gun = force on the bullets

Or, Force required to hold the gun = Δp/Δt

Or, Force required to hold the gun = nm_bv

Or, Force required to hold the gun = 4 * 0.02 * 300 = 24 N.

13. A mass is suspended over a pulley on the inclined plane as shown in the figure. The acceleration is: (J & K CET 2000)

(A) 5 cm/s²

(B) Zero

(C) 5.33 cm/s²

(D) None of these.

Sol: (B) 

Let the acceleration of the blocks be a, tension in the string T.

The equation of motion of 10 kg block:

→ T – 10g sin 30 = 10a

Or, T – 50 = 10a --------- (1)

The equation of motion of 5 kg block:

→ 5g – T = 5a

Or, 50 – T = 5a --------- (2)

From equation 1 and 2,

We get, a = 0.

14. Two blocks m₁ = 5 g and m₂ = 10 g are hung vertically over a light frictionless pulley as shown in figure. What is the acceleration of the masses when left free? (CBSE 2000)

(A) g

(B) g/5

(C) g/2

(D) g/3.

Sol: (D)

Let the acceleration of the blocks be a, tension in the string T.

The equation of motion of m₁ block:

→ T – 5 g = 5a --------- (1)

The equation of motion of 5 kg block:

→ 10 g – T = 10a --------- (2)

From equation 1 and 2,

We get, a = g/3.

15. A mass 1 kg is suspended by a thread. It is (i) lifted up with an acceleration 4.9 m/s², (ii) lowered with an acceleration 4.9 m/s². The ratio of the tensions is: (CBSE 1998)

(A) 2: 1

(B) 3 : 1

(C) 1 : 3

(D) 1 : 2.

Sol: (B) 

(i) Lifted up with an acceleration 4.9 m/s²: Tension = T₁.

Equation of motion of the block:

→ T₁ – 1 g = 1a

Or, T₁ = 4.9 + 9.8 = 14.7 N.

(ii) lowered with an acceleration 4.9 m/s²: Tension = T₂.

Equation of motion of the block:

→ 1 g – T₂ = 1a

Or, T₂ = 9.8 – 4.9 = 4.9 N.

So, the ratio of the tensions is = T₁/T₂ = 14.7/4.9 = 3 : 1.

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