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.

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

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HCVERMA SOLUTION (CHAPTERWISE), JEE Main and advance questions and solutions.

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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.

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HCVERMA SOLUTION (CHAPTERWISE), JEE Main and advance questions and solutions.

Read More

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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