JEE Advanced Previous Year Questions with Solutions on Vectors

All physical quantities can be classified into two categories – scalars and vectors. The quantities which have both magnitude and direction, and obey the rules of vectors, are known as vector quantities. While writing, the vectors are represented by bold-faced letters like A and by putting an arrow over the letter. Since the vectors have both magnitude and direction, they cannot be added by ordinary algebra. If the two vectors are represented in magnitude and direction by the two sides of the triangle in the same order, then the third side drawn in the opposite direction represents the vector sum. Vectors can also be added by the parallelogram law of vector addition. If two vectors are represented both in magnitude and direction by the adjacent sides of the parallelogram drawn from the point, then the resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the point.

Download JEE Advanced Previous Year Questions with Solutions on Vectors PDF

Question 1) A train of 150 m length is going towards the North direction at a speed of 10 ms^-1. A parrot flies at a speed of 5 ms^-1 towards the South direction parallel to the railway track. The time taken by the parrot to cross the train is equal to

1) 12 s

2) 8 s

3) 15 s

4) 10 s

5) 5 s

Answer: 4) 10 s

Solution:

The relative velocity of the parrot w.r.t the train = 10 – (-5) = 15 m/s

Time taken to cross the train = 150/15 = 10 s

Question 2) A passenger train is moving at 5 ms^-1 . An express train is travelling at 30 ms^-1, on the same track and rear side of the passenger train at some distance. The driver in the express train applied brakes to avoid collision. If the retardation due to brakes is 4 ms^-2, the time in which the accident is avoided after the application of brakes is

1) 4.25 s

2) 5.25 s

3) 6.25 s

4) 7.25 s

Answer: 3) 6.25 s

Solution:

Speed of passenger train, U_p= 5 ms^-1

Speed of express train, U_e= 30 ms^-1

The initial relative velocity of the express train w.r.t the passenger train, U = U_e−U_p=30−5=25ms⁻¹

Final relative velocity of the express train w.r.t the passenger train, V

(because the express train comes to rest relative to the passenger train )

From the first equation of motion,

V= U−at

⇒0=25−4t,

4t=25

⇒t=6.25s

Question 3) A boat crosses a river with a velocity of 8 km/hr. If the resulting velocity of boat is 10 km/hr, then the velocity of river water is

1) 4 km/hr

2) 6 km/hr

3) 8 km/hr

4) 10 km/hr

Answer: 2) 6 km/hr

Solution:

Boat crosses a river with a velocity, V_B = 8 km/hr

Resultant velocity, V_R= 10 km/h

Velocity of river,

V_r² = 10² – 8²

= 36

V_r = 6 km/hr

Question 4) A river is flowing from East to West at a speed of 5m/min. A man on the south bank of river, capable of swimming 10 m/min in still water, wants to swim across to the river in the shortest time. He should swim

1) Due North

2) Due North- East

3) Due North-East with double the speed of river

4) None of the above

Answer: 1) Due North

Solution:

To cross the river in the shortest time the man has to swim perpendicular to the river current. Therefore in this case it is due North.

Question 5) A thief is running away on a straight road on a jeep moving with a speed of 9 m/s. A policeman chases him on a motorcycle moving at a speed of 10 m/s. If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take for the policemen to catch the thief?

1) 1 sec

2) 19 sec

3) 90 sec

4) 100 sec

Answer: 4) 100 sec

Solution:

The relative velocity of the policeman w.r.t the thief is 10 – 9 = 1 m/s .

Instantaneous separation of jeep from the motorcycle is 100 m

The time taken by the policeman to catch the thief = 100/1 = 100 sec

Question 6) A motorboat covers a given distance in 6 hr moving downstream on a river. It covers the same distance in 10 hours moving upstream. The time it takes to cover the same distance in still water is

1) 9 hr

2) 7.5 hr

3) 6.5 hr

4) 8 hr

Answer: 2) 7.5 hr

Solution:

Velocity of water = v_w

Velocity of the motorboat in still water. = v_b

The distance covered by motorboat in moving downstream in 6h

x=(v_b+v_w)×6…(1)

The same distance covered by motorboat in moving upstream in 10h is

x=(v_b−v_w)×10…(2)

From Eqs. (1) and (2), we have

(v_b+v_w)×6=(v_b−v_w)×10

v_w=v_b/4

x=(v_b+v_w)×6=7.5vb

Time taken by the motorboat to cover the same distance in still water is

t=x/v_b

=7.5v_b/v_b=7.5h

Question 7) 100 coplanar forces each equal to 10 N act on a body. Each force makes an angle π/50 with the preceding force. What is the resultant of the forces?

1) 1000 N

2) 500 N

3) 250 N

4) Zero

Answer: 4) Zero

Solution:

Total angle=100x (π/50)=2π

All the forces will pass through one point and all the forces will be balanced.

Resultant Force will be Zero.

Question 8) The sum of the magnitudes of two forces acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at 90^o with a force of smaller magnitude, what are the magnitudes of forces?

1) 2, 5

2) 14, 4

3) 5, 13

4) 10, 8

Answer: 3) 5, 13

Solution:

Let P be the smaller force and Q be the greater force. Then

P+Q=18 ……..(1)

Therefore, P + Q cos θ = 0 ———(3)

By solving (i),(ii),and (iii),we will get P = 5 and Q = 13.

Question 9) The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is

1) 120°

2) 150°

3) 135°

4) None of these

Answer: 2) 150°

Solution:

The magnitude of resultant of A and B is given by,

R = √(A² + B² + 2ABcosФ)

According to question,

Magnitude of resultant = half of Magnitude of B

e.g., √(A² + B² + 2A.BcosФ) = B/2

Taking square both sides,

A² + B² + 2A.BcosФ = B²/4

A² + 2ABcosФ + 3B²/4 = 0 ——–(1)

tan 90⁰ = (BsinФ)/(A + BcosФ)

⇒ A + BcosФ = 0

cosФ = – A/B , put it in equation (1)

A² + 2A.B(-A/B) + 3B²/4 = 0

A² – 2A² + 3B²/4 = 0

A = √3B/2

Now, cosФ = -A/B = -√3B/2B = -√3/2

cosФ = cos150°

⇒Ф = 150°

Hence angle between A and B = 150°

Question 10) A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical component will be

1) 10 N

2) 3 N

3) 4 N

4) 2.5 N

Answer: 4) 2.5 N

Solution:

The component of force is vertical direction = F cosθ

= F cos 60⁰

= 5 x (1/2) = 2.5 N

Polygon Law Of Vector Addition