NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | PHYSICS IMPORTANT QUESTIONS | CHAPTER – 4 | MOTION IN A PLANE | EDUGROWN |

In This Post we are  providing Chapter-4 MOTION IN A PLANE NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON MOTION IN A PLANE

Question 1.
A body is acted upon by the following, velocities:
(i) 7 ms^-1 due to E,
(ii) 10 ms^-1 due S,
(iii) 52–√ ms^-1 due N.E.
Find the magnitude and direction of the resultant velocity.

Answer:
Let OA, OB and OC represent the velocities given in the statement i.e.
OA = 7 ms^-1
OB = 10 ms^-1
and OC = 52–√ ms^-1
To find their resultant velocity, resolve OC into two rectangular components along east and north.

Hence resultant velocity along east = 7 + 5 = 12 ms^-1 and resultant velocity along south = OB – OF = 10 – 5 = 5 ms^-1.

If R be the resultant velocity, then the magnitude of R is obtained by applying the parallelogram law of vector addition as

When OG = 12ms^-1 and OH 5ms^-1.

The direction of R: Let θ be the angle made by R with the east.
∴ in rt. ∠d ΔOGI,

Question 2.
A projectile is fired horizontally with a velocity of 98 ms^-1 from the top of a hill 490 m high. Find:
(i) the velocity with which it strikes the ground.
(ii) the time is taken to reach the ground.
(iii) the distance of the target from the hill.
Answer:

(i) h = 490 m, a = g = 9.8 ms²
U_y = initial velocity along the y-axis at the top of the tower = 0

(ii) Let v be the velocity along the y-axis with which the projectile hits the ground.

If V be the resultant velocity of hitting the ground

Let θ be the angle made by V with the horizontal

(iii) Let x, be the distance of the target from the hill.
∴ x = horizontal distance covered with u in a time t.
ut = 98 × 10 = 980 m.

Question 3.
A boy stands at 78.4 m from a building and throws a ball which just enters a window 39.2 m above the ground. Calculate the velocity of the projection of the ball.

Answer:
Let the boy standing at A throw a ball with initial velocity u.
θ = angle of the projection made with the horizontal.

As the boy is at 78.4 m from the building and the ball just enters above the ground.

Question 4.
Two particles located at a point begin to move with velocities 4 ms^-1 and 1 ms^-1 horizontally in opposite directions. Determine the time when their velocity vectors become perpendicular. Assuming that the motion takes place in a uniform gravitational field of strength g.

Answer:
Let v₁ and v₂ be the velocities of first and 2nd particles respectively after a time t.
∴ v₁ = 4î – gt ĵ
v₂ = – î – gt ĵ
For v₁ and v₂ to be ⊥ to each other, then their dot product must be zero.

Question 5.
A body is projected with a velocity of 40 ms^-1. After two seconds, it crosses a verticle pole of 20.4 m. Find the angle of projection and the horizontal range.

Answer:
Here, u = 40 ms^-1
height of vertical pole, h = 20.4 m
t = 2 seconds

Let us take vertical motion

∴ The horizontal range is given by the relation,

Question 6.
The greatest and the least resultant of two forces acting at a point are 29 N and 5 N respectively. If each force is increased by 3 N, find the resultant of two new forces when acting at a point at an angle of 90° with each other.

Answer:
Let A and B be the two forces.
∴ Greatest Resultant = A + B = 29 N ….(1)
least Resultant = A – B = 5 N ….(2)

Let A and B be the new forces such that
A’ = A + 3 = 17 + 3 = 20N and
B’ = B + 3 = 12 + 3 = 15 N

Here, θ = angle between A’ and B’ = 90°
Let R be the resultant of A’ and B’.
∴ according to parallelogram law of vector addition

The direction of R:
Let β be the angle made by R with A’

Question 7.
An aircraft is trying to fly due north at a speed of 100 ms^-1 but is subjected to a crosswind blowing from west to east at 50 ms^-1. What is the actual velocity of the aircraft relative to the surface of the earth?

Answer:
Let V_a and V_w be the velocities of aircraft and wind respectively.
∴ V_a = 100 ms^-1 along N direction
V_w = 50 ms^-1 along E direction

If V be the resultant velocity of the aircraft, then these may be represented as in the figure given below. So the magnitude of V is given by,


Let ∠AOB = θ be the angle which the resultant makes with the north direction.

Question 8.
Calculate the total linear acceleration of a particle moving in a circle of radius 0.4 m at the instant when its angular velocity is 2 rad s^-1 and angular acceleration is 5 rad s^-2.

Answer:
Since the particle possesses angular acceleration, so its total linear acceleration (a) is the vector sum of the tangential acceleration (a,) and the centripetal acceleration (ac). a₁ and a_c, are at right angles to each other.
a = \sqrt{a_{t}^{2}+a_{c}^{2}} …. (1)

Question 9.
An airplane flies 400 km west from city A to city B then 300 km north-east to city C and finally 100 km north to city D. How far is it from city A to city D? In what direction must the airplane go to return directly to the city A from city D?

Answer:
Given, AB = 400 km
BC = 300 km
CD = 100 km
AD =?

Let N₁, N₂ represent north directions.
∠ABC = 45°
Draw CC’ ⊥ AB, And CB’ ⊥ BN₂
Now in ΔBC’ C

From AAC’D, AD is given by

Question 10.
Which of the following quantities are independent of the choice of the orientation of the coordinates axes:
a + b, 3a_x + 2b_y, [a + b – c], angle between b and c, a?

Answer:
a + b, |a + b – c|, angle between b and c, a are the quantities that are independent of the choice of the orientation of the coordinate axes.

But the value of 3a_x + 2b_y depends on the orientation of the axes.