JEE Previous Year Questions With Solutions Of Kinematics 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.

JEE Previous Year's Questions With Solutions

1. If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? (AIEEE 2002, WBJEE 2012)

(A) 1 cm

(B) 2 cm

(C) 3 cm

(D) 4 cm.

Sol: (A)

Let initial velocity be u; Velocity of body after penetration of 3 cm is v = u/2; Acceleration = a; s = 3 cm.

We have, v² – u² = 2as

Or, (u/2)² – u² = 2 * a * 3

Or, a = – u²/8 cm/s².

Now, initial velocity = u/2; final velocity = 0; s =?.

Again, v² – u² = 2as

Or, 0 – (u/2)² = 2 * (– u²/8) * s

Or, s = 1 cm.

2. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are: (AIEEE 2002)

(A) 12 N, 6 N

(B) 13 N, 5 N

(C) 10 N, 8 N

(D) 16 N, 2 N.

Sol: (B)

Given: Resultant R = 12 N is perpendicular to smaller force B and (A + B) = 18 N ------ (1).

From right angle triangle (acd)

We have, A² = B² + R²

Or, A² – B² = R²

Or, (A + B) (A – B) = R²

Or, 18 (A – B) = 12²

Or, A – B = 8 ---------- (2)

Solving equation 1 and 2 we get

A = 13 N and B = 5 N.

3. Speeds of two identical cars are u and 4u at a specific instant. If the same deceleration is applied on both the cars, the ratio of the respective distances in which the two cars are stopped from that instant is: (AIEEE 2002)

(A) 1 : 1

(B) 1 : 4

(C) 1 : 8

(D) 1 : 16.

Sol: (D)

For car A: initial speed = u; final speed = 0; acceleration = – a; distance travelled = S_A.

We have, v² – u² = 2as

Or, (0)² – u² = 2 * (– a) * S_A

Or, S_A = u²/2a.

For car B: initial speed = 4u; final speed = 0; acceleration = – a; distance travelled = S_B.

Again, v² – u² = 2as

Or, (0)² – (4u)² = 2 * (– a) * S_B

Or, S_B = 8u²/a.

Therefore, the ratio of the respective distances is 1 : 16.

4. From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If v_A and v_B are their respective velocities on reaching the ground, then: (AIEEE 2002)

(A) v_A = v_B

(B) v_A > v_B

(C) v_A < v_B

(D) Their velocities depend on their masses.

Sol: (A)

Let height of the building be h.

Ball A projected upwards with velocity u falls back to building top with velocity u downwards. It completes its journey to ground under gravity.

Therefore, (v_A)² = u² + 2gh ------- (1)

Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h.

Therefore, (v_B)² = u² + 2gh ------- (2)

From equation 1 and 2

We get, v_A = v_B.

5. A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30⁰ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? [g = 10 m/s; sin 30⁰ = ½; cos 30⁰ = (√3)/2]: (AIEEE 2003)

(A) 5.20 m

(B) 4.33 m

(C) 2.60 m

(D) 8.66 m.

Sol: (A)

Given: initial velocity, u = 10 m/s; angle of projection, θ = 30⁰.

Initial and final points are in same horizontal plane.

We have,

Range, R = (u² Sin 2θ)/g

Or, R = (10² sin 60⁰)/10

Or, R = 8.66 m.

6. A particle moves in a circle of radius R. In half the period of revolution its displacement is ____ and distance covered is ___. (IITJEE 1983)

Sol:

Displacement = shortest path = AB = 2R.

Distance = Total path = ACB = πR.

7. Two balls of different masses are thrown vertically upwards with the same the speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance). (IITJEE 1983)

Sol: (True)

Since the initial speed and acceleration are same for both the balls. So, they pass through the point of projection in their downward motion with the same speed.

8. A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest path in 15 minutes. The velocity of the river in km/h is: (IITJEE 1988, CBSE 2000, 1998)

(A) 1

(B) 4

(C) √41

(D) 3.

Sol: (D) 

V_{b, r} = velocity of boat w.r.t river = 5 km/h.

V_{r, g} = velocity of river w.r.t ground.

V_{b, g} = velocity of boat w.r.t ground = 1 km/15 min = 4 km/h.

We have, V_{b, g} = V_{b, r} + V_{r, g} -------- (1)

Taking vertical component of the equation (1)

→ V_{b, g} = V_{b, r} sin θ + V_{r, g} cos 90⁰

Or, 4 = (5 sin θ)

Or, sin θ = 4/5

And, cos θ = 3/5.

Taking horizontal component of the equation (1)

→ V_{b, g} cos 90⁰ = – V_{b, r} cos θ + V_{r, g}

Or, 0 = – (5 * 3/5) + V_{r, g}

Or, V_{r, g} = 3 km/h.

9. A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is: (IITJEE 1992)

(A) 4.0 m/s

(B) 5.5 m/s

(C) 5.0 m/s

(D) 4.8 m/s.

Sol: (D)

Let the total distance be x.

Time taken for 1^st half, t₁ = (x/2)/3 = x/6.

Time taken for 2^nd half, t₂ = (x/2)/ (4.5 + 7.5) = x/24.

Total time, t = 5x/24.

So, average speed = x/t = 24/5 = 4.5 m/s.

10. The area under the acceleration time graph represents: (MP PMT 1993)

(A) Velocity

(B) The displacement

(C) Distance travelled

(D) Change in velocity.

Sol: (D)

We have,

Acceleration, a = ΔV/Δt

Or, ΔV = a * Δt.

So, the area under the acceleration time graph represents the change in velocity.

11. A particle moves along the x-axis such that its coordinates (x) varies with time (t), according to the expression: x = 2 – 5t + 6t² where x is in metres and t is in seconds. The initial velocity of the particle is: (MP PET 1996)

(A) 40 m/s

(B) Zero

(C) 8 m/s

(D) 20 m/s.

Sol: (B)

Distance travelled, x = 2 – 5t + 6t²

Velocity, v = dx/dt = – 5 + 12t

Initial velocity, u = velocity at t = 0

Or, u = – 5 m/s.

12. For the velocity time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds?(MP PMT 1998)

(A) 1/2

(B) 2/3

(C) 1/4

(D) 1/3.

Sol: (C)

Distance covered = area under velocity time graph.

Total distance covered = ½ * 2 * 10 + 10 * 2 + ½ * 2 * 10 = 40 m.

Distance covered in last 2s = ½ * 2 * 10 = 10 m.

Fraction = 10/40 = 1/4.

13. The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 m/s. If the change in velocity of the body is 0.18 m/s during this time, its uniform acceleration is: (EAMCET 2000)

(A) 0.04 m/s²

(B) 0.03 m/s²

(C) 0.02 m/s²

(D) 0.01 m/s².

Sol: (C)

Given: distance travelled, S = 3.06 m; average velocity, V_avg = 0.34 m/s; change in velocity, ΔV = 0.18 m/s.

Time taken, t = S / V_avg = 9 s.

Acceleration, a = ΔV/t = 0.18/9 = 0.02 m/s².

14. A constant force acts on a body of mass 0.9 kg at rest for 10 s. If the body moves a distance of 250 m, the magnitude of the force is: (EAMCET 2000)

(A) 4.5 N

(B) 5 N

(C) 3 N

(D) 4 N.

Sol: (A)

Given: initial velocity, u = 0; time taken, t = 10 s; distance travelled, S = 250 m; mass, m = 0.9 kg.

We have, S = ut + ½ at²

Or, 250 = 0 + ½ * a * 10²

Or, a = 5 m/s².

And, Force = mass * acceleration

Or, Force = 0.9 * 5 = 4.5 N.

15. The equation of motion of a projectile are given by x = 36t m and 2y = 96t – 9.8t² m. The angle of projection is: (EAMCET 2003)

(A) Sin^{– 1} (4/3)

(B) Sin^{– 1} (4/5)

(C) Sin^{– 1} (3/5)

(D) Sin^{– 1} (3/4).

Sol: (B)

Horizontal velocity, V_h = dx/dt = 36 m/s.

Vertical velocity, V_v = dy/dt = 48 – 9.8t.

Vertical velocity at time of projection (t = 0) = 48 m/s.

Let the angle of projection be θ.

So, tan θ = (V_v/ V_h)_{t = 0} = 48/36 = 4/3

Or, sin θ = 4/5

Or, θ = sin^{– 1} (4/5).

16. The displacement of a particle is represented by the following equation: s = 3t³ + 7t² + 5t + 8 where s is in metre and t in second. The acceleration of the particle at t = 1s is: (CBSE 2000)

(A) Zero

(B) 18 m/s²

(C) 32 m/s²

(D) 14 m/s².

Sol: (C)

Given: s = 3t³ + 7t² + 5t + 8

Or, velocity, v = ds/dt = 9t² + 14t + 5

Or, acceleration, a = d²s/dt² = 18t +14.

Therefore, acceleration, a (t = 1 s) = 32 m/s².

17. Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is π/3 the maximum height reached by it is 102 m. Then the maximum height reached by the other in metre is: (EAMCET 2003)

(A) 34

(B) 224

(C) 336

(D) 56.

Sol: (A)

Horizontal range will be same if angle of projection is θ or (90 – θ).

Angle of projection for 1^st stone, θ₁ = 60⁰.

Angle of projection for 2^nd stone, θ₂ = (90 – 60⁰) = 30⁰.

For 1^st stone, Max height = 102 = (u² sin² 60⁰)/2g.

For 2^nd stone, Max height = h = (u² sin² 30⁰)/2g.

Therefore, h/102 = sin² 30⁰/ sin² 60⁰

Or, h = 102/3 = 34 m.

18. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first? (Odisha JEE 2003)

(A) Both will reach simultaneously

(B) Depends on their mass

(C) The slower one

(D) The faster one.

Sol: (A)

Both the bullet have same vertical displacement and acceleration. Therefore, both will have the same time and t = √ (2h/g).

19. A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is: (AIEEE 2003)

(A) 18 m

(B) 6 m

(C) 24 m

(D) 12 m.

Sol: (C)

For 1^st case: u = 50 km/hr = 180 m/s; v = 0; s = 6 m; a =?

We have, v² – u² = 2as

Or, 180² – 0² = 2 * a * 6

Or, a = 180²/12 = 2700 m/s².

For 2^nd case: u = 100 km/hr = 360 m/s; v = 0; a = 2700 m/s²; s =?

Again, v² – u² = 2as

Or, 360² – 0² = 2 * 2700 * s

Or, s = 360²/5400 = 24 m.

20. The co-ordinates of a moving particle at any time t are given by x = αt³ and y = βt³. The speed of the particle at time t is given by: (AIEEE 2003)

(A) √ (α² + β²)

(B) 3t² √ (α² + β²)

(C) 3t √(α² + β²)

(D) t² √ (α² + β²).

Sol: (B)

Velocity along x-axis, v_x = dx/dt = 3 αt².

Velocity along y-axis, v_y = dy/dt = 3 βt².

So, resultant velocity = √ [(v_x)² + (v_y)²]

Or, resultant velocity = √ [(3αt²)² + (3βt²)²]

Or, resultant velocity = 3t² √ (α² + β²).

21. An automobile travelling with a speed of 60 km/h can brake to stop with a distance of 20 m. If the car is going twice as fast i.e. 120 km/h, the stopping distance will be: (AIEEE 2004)

(A) 20 m

(B) 60 m

(C) 40 m

(D) 80 m.

Sol: (D)

In the 1^st case: initial velocity, u = 60 km/h = 100/6 m/s; final velocity, v = 0; distance travelled, s = 20 m.

We have, v² – u² = 2as

Or, 0² – (100/6)² = 2 * a * 20

Or, a = – 6.94 m/s².

In the 2^nd case: initial velocity, u = 120 km/h = 100/3 m/s; final velocity, v = 0; distance travelled, s =? Acceleration, a = – 6.94 m/s².

Again, v² – u² = 2as

Or, 0² – (100/3)² = 2 * (– 6.94) * s

Or, s = 80 m.

22. Which of the following statements is false for a particle moving in a circle with a constant angular speed? (AIEEE 2004)

(A) The velocity vector is tangent to the circle

(B) The acceleration vector is tangent to the circle

(C) The acceleration vector points to the centre of the circle

(D) The velocity and acceleration vectors are perpendicular to each other.

Sol: (B)

The acceleration vector is tangent to the circle is a false statements.

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 Vector Analysis 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. If A x B = B x A, then the angle between A and B is: (AIEEE 2004)

(A) π/2

(B) π/3

(C) π

(D) π/4. 

Sol: (C)

Given: A x B = B x A

Or, (A x B) – (B x A) = 0

Or, (A x B) + (A x B) = 0 [same magnitude & opposite direction]

Or, 2 (A x B) = 0

Or, 2AB Sin θ n = 0

| A| and |B| are non-zero value.

Therefore it is possible if Sin θ = 0 → θ = 0 or π.

2. A force F = 5i + 3j + 2k N is applied over a particle which displaces it from its origin to the point r = 2i – j m. The work done on the particle in joules is: (AIEEE 2004)

(A) + 7

(B) – 7

(C) + 10

(D) + 13. 

Sol: (A)

Displacement, s = (2i – j) – (0i + 0j) = 2i – j m.

Force, F = 5i + 3j + 2k N.

So, work done, W = F.s

Or, W = (5i + 3j + 2k). (2i – j)

Or, W = + 7 J.

3. A particle has an initial velocity of 3i + 4j and acceleration of 0.4i + 0.3j. Its speed after 10s is: (AIEEE 2009)

(A) 7√2 units

(B) 7 units

(C) 8.5 units

(D) 10 units.   

Sol: (D)

Initial velocity, u = 3i + 4j

Magnitude of initial velocity, u = 5 units.

Acceleration, a = 0.4i + 0.3j.

Magnitude of acceleration, a = 0.5 units.

Time, t = 10 s.

We have, v = u + at

Or, v = 5 + 0.5 * 10 = 10 units.

4. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then (IIT JEE 2009)

(A) V_C – V_A = 2 (V_A – V_C)

(B) V_C – V_B = V_B – V_A

(C) |V_C – V_A| = 2 |V_B – V_C|

(D) |V_C – V_A| = 4 |V_B|. 

Sol: (C)

Let ω be the angular velocity of rotation of sphere V_A, V_B, V_C be the velocities at A, B and C respectively.

Then V_A = 0; V_B = ωr; V_C = 2ωr.

|V_C – V_A| = 2ωr and |V_B – V_C| = ωr.

Therefore, |V_C – V_A| = 2 |V_B – V_C|.

5. Which of the following is a vector quantity? (WBJEE 2007)

(A) Temperature

(B) Flux density

(C) Magnetic field intensity

(D) Time.  

Sol: (C)

Magnetic field intensity is a vector quantity.

6. The angle subtended by the vector A = 4i + 3j + 12k with the x-axis is: (WBJEE 2008)

(A) Sin^-1 (3/13)

(B) Sin^-1 (4/13)

(C) Cos^-1 (3/13)

(D) Cos^-1 (4/13).  

Sol: (D)

A = 4i + 3j + 12k and B = i (x- axis)

→ A.B = (4i + 3j + 12k). (i)

Or, A.B = 4

Magnitude of A = 13

Magnitude of B = 1

We have, Cos θ = (A.B)/AB

Or, Cos θ = 4/13

Or, θ = Cos^-1 (4/13).   

7. A and B are two vectors given by A = 2i + 3j and B = i + j. The magnitude of the components of A along B is: (WBJEE 2009, 2011)

(A) 5/√2

(B) 3/√2

(C) 7/√2

(D) 1/√2.  

Sol: (D)

The magnitude of the components of A along B is

= (A.B)/|B|

= 5/√2.

8. Given, C = A x B and D = B x A. what is the angle between C and D? (WBJEE 2009)

(A) 30⁰

(B) 60⁰

(C) 90⁰

(D) 180⁰.  

Sol: (D)

Since A x B = – B x A [same magnitude & opposite direction]

Or, C = – D 

Therefore, the angle between C and D is 180⁰.

9. The value of ‘λ’ for which the two vectors a = 5i + λj + k and b = i − 2j + k are perpendicular to each other is: (WBJEE 2010)

(A) 2

(B) – 2

(C) 3

(D) – 3

Sol: (C)

a.b = 0 (since two vectors perpendicular to each other)

→ (5i + λj + k).(i − 2j + k) = 0

Or, 5 − 2λ + 1 = 0

Or, λ = 3. 

10. If a + b = c and a + b = c, then the angle included between a and b is: (WBJEE 2010)

(A) 90⁰

(B) 180⁰

(C) 120⁰

(D) Zero

Sol: (D)

Given: a + b = c and a + b = c

Now, c = √ (a² + b² + 2ab cos θ)

Or, (a + b) = √ (a² + b² + 2ab cos θ)

Or, a² + b² + 2ab = a² + b² + 2ab cos θ

Or, cos θ = 1

Or, θ = 0⁰.

11. Three vectors satisfy the relation A.B and A.C = 0, then A is parallel to: (Karnataka CET 2003)

(A) B x C

(B) B

(C) C

(D) B.C

Sol: (A)

Given: A.B = 0, A.C = 0.

→ A is perpendicular to B and C.

And, B x C is perpendicular to the plane containing B and C.

Hence, A is parallel to B x C.

12. Two equal forces (F each) act a point inclined to each other at an angle of 120⁰. The magnitude of their resultant is: (Karnataka CET 2004)

(A) F

(B) F/2

(C) F/4

(D) 2F

Sol: (A)

Resultant = √ (F² + F² + 2.F.F Cos 120)

Resultant = √ (F²) = F.

13. If a and b are two vectors then the value of (a + b) x (a – b) is: (BHU 2002)

(A) (a x b)

(B) – 2 (b x a)

(C) 2 (b x a)

(D) (b x a)

Sol: (C)

Given: (a + b) x (a – b)

= a x a – a x b + b x a – b x b 

= 2 (b x a) [since, a x a = b x b = 0; – a x b = b x a].

14. If |A x B| = √3 (A.B), then the value of |A + B| is: (CBSE 2004)

(A) (A² + B² + AB/√3)^1/2

(B) (A + B)

(C) (A² + B² + √3AB)^1/2

(D) (A² + B² + AB)^1/2

Sol: (C)

Given: |A x B| = √3 (A.B)

Or, AB Sin θ = √3AB Cos θ

Or, Tan θ = √3 → θ = 60⁰.

Now, |A + B| = (A² + B² + 2AB Cos θ)^1/2

Or, |A + B| = (A² + B² + 2AB Cos 60)^1/2

Or, |A + B| = (A² + B² + AB/√3)^1/2.

15. If the scalar and vector products of two vectors A and B are equal in magnitude, then the angle between the two vectors is: (J & K CET 2009)

(A) 180⁰

(B) 90⁰

(C) 360⁰

(D) 45⁰.

Sol: (D)

Given: |A x B| = (A.B)

Or, AB Sin θ = AB Cos θ

Or, Tan θ = 1 → θ = 45⁰.

16. The component of vector A = a_xi + a_yj + a_zk along the direction of (i – j) is: (EAMCET 2008)

(A) (a_x + a_y)

(B) (a_x – a_y)/√2

(C) (a_x – a_y + a_z)
(D) (a_x + a_y + a_z).

Sol: (B)

Unit vector of (i – j) = (i – j)/|(i – j)| = 1/√2 (i – j).

Component of A along (i – j) = A. ((i – j)/|(i – j)|)

Component of A along (i – j) = (a_x – a_y)/√2.

17. Given A = 2i + 3j and B = i + j. The component of vector A along vector B is: (WBJEE 2011)

(A) 1/√2

(B) 3/√2

(C) 5/√2

(D) 7/√2

Sol: (C)

Component of A along B = A. (B/|B|)

Component of A along B = 5/√2.

18. If A = B + C and A, B, C have scalar magnitudes of 5, 4, 3 units respectively then the angle between A and C is: (WBJEE 2012)

(A) cos^-1 (3/5)

(B) cos^-1 (4/5)

(C) π/2

(D) sin^-1 (3/5)

Sol: (A)

Given: A = B + C

Or, 5² = 4² + 3² + 2 * 4 * 3 cos θ

Or, angle between B and C is = 90⁰.

And A is the resultant of B + C.

So, the angle between A and C is = cos^-1 (3/5).

19. Two vectors are given by A = i + 2 j + 2k and B = 3i + 6j + 2k. Another vector C has the same magnitude as B but has the same direction as A. Then which of the following vectors represents C? (WBJEE 2013)

(A) (7/3) (i + 2j + 2k)

(B) (7/3) (i – 2j + 2k)

(C) (7/9) (i – 2j + 2k)

(D) (9/7) (i + 2j + 2k)

Sol: (A)

Magnitude of B = √ (3² + 6² + 2²) = 7.

Unit vectors of A = A/| A| = (i + 2 j + 2k)/3.

So, vector C = (7/3) (i + 2 j + 2k).

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 Vector Analysis 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.

JEE Previous Year's Questions With Solutions

1. The magnitudes of vectors A, B and C are 3, 4 and 5 units respectively. If A + B = C, the angle between A and B is: (CBSE 1990, CPMT 1997, UPSEAT 1999)

(A) π/2

(B) π/4

(C) Cos^{– 1} 0.6

(D) Tan^{– 1} (7/5) 

Sol: (A)

Magnitude and direction of A + B same as the magnitude and direction of C.

Therefore, √ (A² + B² + 2AB Cos θ) = C

Or, 3² + 4² + 24 Cos θ = 5²

Or, Cos θ = 0/24 = 0

Or, θ = π/2.

2. The angle between A and B is θ. The value of triple product A. (B x A) is: (CBSE 1991)

(A) 0

(B) A²B cos θ

(C) A²B

(D) A²B sin θ

Sol: (A)

Since (B x A) is perpendicular to vector A.

Therefore, A. (B x A) = 0.

3. What is the torque due to a force F = 2i – 3j + 4k N acting at a point r = 3i + 2j + 3k about the origin? (CBSE 1995, 1997, Karnataka CET 1998)

(A) 6i – 6j + 12k

(B) – 6i + 6j + 12k

(C) – 17i – 6j + 13k

(D) 17i – 6j – 13k

Sol: (D)        

We have, torque = r x F

Or, torque = (3i + 2j + 3k) x (2i – 3j + 4k)

Or, torque = 17i – 6j – 13k. 

4. A force vector applied on a mass is represented as F = 6i – 8j + 10k N and accelerates with 1 m/s². What will be the mass of the body in kg? (CBSE 1996)

(A) 20

(B) 2√10

(C) 10

(D) 10√2

Sol: (D)

Magnitude of force, F = √ (6² + (– 8)² + 10²) = 10√2 N.

We have, F = ma

Or, m = F/a = 10√2/1 = 10√2 kg.

5. What is the value of linear velocity, if ω = 3i – 4j + k and r = 5i – 6j + 6k? (CBSE 1999)

(A) 4i – 13j + 6k

(B) 6i + 2j – 3k

(C) – 18i – 13j + 2k

(D) 6i – 2j + 8k

Sol: (C)

We have, v = r x ω

Or, v = (5i – 6j + 6k) x (3i – 4j + k)

Or, v = – 18i – 13j + 2k. 

6. The angle between two vectors – 2i + 3j + k and i + 2j – 4k is: (EAMCET 1990, EAMCET 1995, CPMT 1997)

(A) 90⁰

(B) 180⁰

(C) 0⁰

(D) None of the above.

Sol: (A)

Let, A = – 2i + 3j + k and B = i + 2j – 4k.

→ A.B = (– 2i + 3j + k).( i + 2j – 4k)

Or, A.B = (– 2 + 6 – 4) = 0

Magnitude of A = √14

Magnitude of B = √21

We have, Cos θ = (A.B)/AB

Or, Cos θ = 0

Or, θ = 90⁰.

7. If A and B are perpendicular vectors and vector A = 5i + 7j – 3k and B = 2i + 2j – ak. The value of ‘a’ is: (EAMCET 1991)

(A) 2

(B) – 7

(C) – 8

(D) 8.

Sol: (C)

A.B = 0 (since two vectors perpendicular to each other)

→ (5i + 7j – 3k).( 2i + 2j – ak) = 0

Or, 10 + 14 + 3a = 0

Or, a = – 8.

8. If 0.5i + 0.8j + ck is a unit vector, then c is: (EAMCET 1994)

(A) √0.89

(B) 0.2

(C) 0.3

(D) √0.11.

Sol: (D)

Magnitude of a unit vector is one.

Therefore, √ (0.5² + 0.8² + c²) = 1

Or, 0.5² + 0.8² + c² = 1

Or, c = √0.11. 

9. A vector is represented by 3i + j + 2k. Its length in X-Y plane is: (EAMCET 1994)

(A) 2

(B) √14

(C) √10

(D) √5.

Sol: (C)

The projection of vector in X-Y plane is 3i + j.

So the length of vector in X-Y is √ (3² + 1²) = √10.

10. Two forces F₁ = 5i + 10j – 20k and F₂ = 10i – 5j – 15k act on a single point. The angle between F₁ and F₂ is nearly: (AMU 1995)

(A) 90⁰

(B) 60⁰

(C) 45⁰

(D) 30⁰.

Sol: (C)

→ F₁. F₂ = (5i + 10j – 20k).(10i – 5j – 15k)

Or, F₁. F₂ = (50 – 50 + 300) = 300

Magnitude of F₁ = √525

Magnitude of F₂ = √350

We have, F₁. F₂ = F₁ F₂ Cos θ

Or, 300 = 428 Cos θ

Or, Cos θ = 0.7 → θ = 45⁰.

11. Two adjacent sides of a parallelogram are represented by the two vectors i + 2j + 3k and 3i – 2j + k. What is the area of the parallelogram? (AMU 1997)

(A) 8√3

(B) 8

(C) 3√8

(D) 192.

Sol: (A)

We have,

Area of parallelogram = |a x b|

So, A = |( i + 2j + 3k) x (3i – 2j + k)|

Or, A = |8 i + 8j – 8k |

Or, A = 8√3.

12. Magnitude of vector which comes on addition of two vectors, 6 i + 7j and 3 i + 4j is: (BHU 2000)

(A) √13.2

(B) √136

(C) √160

(D) √202.

Sol: (D)

Vector, V = (6 i + 7j) + (3 i + 4j)

Or, V = 9 i + 11j.

Magnitude of V = √ (9² + 11²) = √202.

13. The position vectors of radius are 2 i + j + k and 2 i – 3j + k while those of linear momentum are 2i + 3j – k. Then the angular momentum is: (BHU 1997)

(A) 4 i – 8j

(B) – 4 i – 8k

(C) 2 i – 4k

(D) 2 i – 4j + 2k.

Sol: (B)

Radius vector, r = (2 i + j + k) – (2 i – 3j + k) = 4j.

Linear momentum, p = 2 i + 3j – k.

We have, angular momentum, L = r x p

Or, L = (4j) x (2 i + 3j – k) = – 4 i – 8k.

14. A particle moves from position r₁ = 3 i + 2j – 6k to position r₂ = 14 i + 13j + 9k under the action of force 4i + j + 3k N. The work done will be: (Punjab PMT 2002, 2003)

(A) 50 J

(B) 75 J

(C) 100 J

(D) 200 J.

Sol: (C)

Displacement, s = (14 i + 13j + 9k) – (3 i + 2j – 6k) = 11 i + 11j + 15k.

Force, F = 4i + j + 3k N.

So, work done, W = F.s 

Or, W = (4i + j + 3k). (11 i + 11j + 15k) 

Or, W = 100 J.

15. If the vectors P = ai + aj + 3k and Q = ai – 2j – k are perpendicular to each other, then the positive value of a is: (AIIMS 2002)

(A) 0

(B) 3

(C) 2

(D) 1.

Sol: (B)

Since P and Q are perpendicular to each other.

→ P.Q = 0

Or, (ai + aj + 3k).( ai – 2j – k) = 0

Or, a² – 2a – 3 = 0

Or, a = – 1, 3.

So, the positive value of a is 3.

16. If |A x B| = √3 A.B, then the value of |A + B| is: (CBSE 2004)

(A) A + B

(B) (A² + B² + AB/√3)^1/2

(C) (A² + B² + AB)^1/2  

(D) (A² + B² + √3 AB)^1/2.

Sol: (C)

Given: |A x B| = √3 A.B

Or, AB sin θ = √3 AB cos θ

Or, tan θ = √3

Or, θ = 60⁰.

Therefore, |A + B| = (A² + B² + 2AB cos 60⁰)^1/2  

Or, |A + B| = (A² + B² + AB)^1/2.

17. When A.B = – |A|.|B|, then: (Odisha JEE 2003)

(A) A and B act in the same direction

(B) A and B can act in any direction

(C) A and B act in the opposite direction

(D) A and B are perpendicular to each other.

Sol: (C)

Given: A.B = – |A|.|B|

Or, |A|.|B| cos θ = – |A|.|B|

Or, cos θ = – 1

Or, θ = 180⁰.

Therefore, A and B act in the opposite direction.

18. With respect to a rectangular Cartesian coordinate system, three vectors are expressed as a = 4i – j; b = – 3i + 2j; c = – k where i, j, k are unit vectors, along the x, y, z-axis respectively. The unit vector r along the direction of sum of these vector is: (Kerala CET 2003)

(A) r = 1/√3 (i + j – k)

(B) r = 1/3 (i – j + k)

(C) r = 1/√2 (i + j – k)

(D) r = 1/2 (i – j + k)

(E) r = 1/3 (i – j + k).

Sol: (A)

Sum of three vectors = a + b + c = i + j – k.

Magnitude of sum of three vectors = √ [1² + 1² + (– 1)²] = √3.

So, the unit vector r = 1/√3 (i + j – k).

19. If |A + B| = |A| + |B|, then the angle between A and B will be: (CBSE 2001)

(A) 60⁰

(B) 90⁰

(C) 120⁰

(D) 0⁰.

Sol: (D)

Given: |A + B| = |A| + |B|

Or, √ (|A|² + |B|² + 2|A||B| cos θ) = |A| + |B|

Or, |A|² + |B|² + 2|A||B| cos θ = |A|² + |B|² + 2|A||B|

Or, cos θ = 1 → θ = 0⁰.

20. The unit vector parallel to the resultant of the vectors A = 4i + 3j + 6k and B = – i + 3j – 8k is: (EAMCET 2000)

(A) 1/49 (3i + 6j – 2k)

(B) 1/49 (3i – 6j + 2k)

(C) 1/7 (3i + 6j + 2k)

(D) 1/7 (3i + 6j – 2k).

Sol: (D)

Resultant of A and B = A + B

Or, A + B = 3i + 6j – 2k

And, | A + B | = √ [3² + 6² + (– 2)²] = √49 = 7.

The unit vector parallel to the resultant = (A + B)/ (| A + B |).

Therefore, unit vector = 1/7 (3i + 6j – 2k).

21. A body of mass 5 kg starts from the origin with an initial velocity u = 30i + 40j m/s. If a constant force – 6i – 5j N acts on the body, the time in which the y-component of the velocity becomes zero is: (EAMCET 2000)

(A) 20 s

(B) 40 s

(C) 80 s

(D) 5 s.

Sol: (B)

Given: mass, m = 5 kg, y-component of initial velocity, u_y = 40 m/s, y-component of force = – 5 N, y-component of acceleration, a_y = – 5/5 = – 1 m/s², v_y = 0.

We have, v_y – u_y = a_yt

Or, 0 – 40 = – 1 * t

Or, t = 40 s.

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

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