# Important Questions Class 11 Physics Chapter 4

## Class 11 Physics Chapter 4 Important Questions

### Important Questions for CBSE Class 11 Physics Chapter 4 – Motion in a Plane

For exam preparation, students can refer to Important Questions for Class 11 Physics Chapter 4 because these questions are organised by important topics and marks distribution. These questions are constructed by subject matter experts in accordance with the CBSE Syllabus. Students can revise the main topics of the chapter with the step-by-step solutions provided with these questions.

Chapter 4 Class 11 Physics Important Questions are a set of questions based on the concepts, terminologies, formulae, and numericals related to “Motion in a Plane”. On the Extramarks website, these questions are easily available, which can help students save time during exam preparations.

CBSE Class 11 Physics Chapter 4 Important Questions

Study Important Questions for Class 11 Physics Chapter 4 – Motion in a Plane

Students can preview the set of Class 11 Physics Chapter 4 Important Questions given below. They can also click on the link provided to access CBSE Extra Questions for the main topics of Chapter 4 Motion in a Plane of Class 11 Physics.

[1 – 2 Marks]

Q1. A stone is thrown vertically upwards, and then it returns to the thrower. Is it a projectile? Justify.

Ans: In this scenario, the stone has only one direction of velocity when rising or falling. Furthermore, a projectile shall comprise two perpendicular components of velocity. Therefore, a stone thrown vertically upwards is not considered to be a projectile.

Q2. At the highest point of the trajectory, what makes the direction of motion of a projectile go horizontal?

Ans: The projectile’s direction of motion changes to horizontal at the highest point of the trajectory as the velocity vertical component turns zero.

Q3. When the initial velocity is increased twofold and the angle of projection is constant, how will the horizontal range of a projectile be affected?

Ans: The horizontal range will increase by four times in comparison to the initial horizontal range.

Q4. Select the vector quantity in the list mentioned below.

Time, temperature, energy, gravitational potential, pressure, power, impulse, charge, coefficient of friction, and total path length.

Ans:  Impulse is the vector quantity in the preceding lists. The force is a vector quantity, as its product with time (a scalar quantity) results in a vector quantity. Hence, the impulse is the product of force and time.

Q5. A stone tied at the end of the string is whirled in a circle. State the reason why the stone flies away tangentially when the string breaks.

Ans: When a stone is tied at the end of a string that moves in a circular path, its velocity is tangent to the circle. The centripetal force does not work when the string breaks. The stone moves along the tangent to the circular path and flies away tangentially due to inertia.

Q6. What does “uniform circular motion” mean? What do you mean by the terms ‘time period’, ‘frequency’, and ‘angular velocity’? Explain the relationship between them.

Ans: The motion is considered to be a uniform circular motion when an object rotates in a circular path with the same speed.

The total time taken by the object to complete one revolution is considered a time period.

The total number of revolutions completed per second is termed “frequency.”

The angular velocity is defined as the time rate of transformation of angular displacement.

=2T=2

1T=, which is the required relation.

Q7. A man can swim with a speed of 4.0 km/h in still water. How long does it take to cross a river 1km/h wide if the river flows steadily at 3.0km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Ans. The speed of the man is given as m=4.0km/h

Width of the river is given as 1km/h

Total time needed to cross the river =width of the riverspeed of the river

Time needed to cross the river =14h

=1460

=15 min

Speed of the river, r=3 km/h

Distance covered with the flow of the river rt

314=34

341000=750m, which is the required distance.

Q8. a) A vector has both magnitude and direction. From this statement, can we conclude that anything that has magnitude and direction is a vector?

Ans: No, a physical quantity having both magnitude and direction is not necessarily a vector.

For example, regardless of its magnitude and direction, the current is a scalar quantity. Any physical quantity must follow the law of vector addition to be considered a vector.

Q8. b) The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does it lead to any rotation of a vector?

Ans: No.

As the law of vector addition is not followed, the rotation of a body across an axis is not considered to be a vector quantity.

[3 – 4 Marks]

Q1. a)Define time of flight and horizontal range?

Ans: The time of flight is the time required by the projectile to finish its trajectory.

The  horizontal range is the maximum horizontal distance travelled by the projectile from the bottom of the tower to the point the projectile touches the ground.

Q1. b) From a certain height above the ground, stone A is gently dropped. Simultaneously, another stone, B, is fired horizontally. Which of the two stones will touch the ground first?

Ans: Because the initial vertical velocity in both scenarios is zero, and the stones fall with a similar acceleration equal to gravity’s acceleration. Therefore, both stones will simultaneously touch the ground.

Q2. The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40m/s can go without hitting the ceiling of the hall?

Ans. Given that,

Speed of the ball is given as, =40m/s

Maximum height is given as, h=25m

In projectile motion, the maximum height attained by a body projected at an angle , is given by the formula:

h=u2sin22g

25=40229.8

sin2=0.30625

sin=0.5534

=sin-1(0.5534)=33.60°

Horizontal Range,

R=u2sin2g

R=402 sin2(33.60)9.8

R=1600 sin(67.2)9.8

R=1600 0.9229.8=150.53m

Q3. Derive expressions for velocity and acceleration for uniform circular motion.

Ans. We can write,

If PQ=l,

V=lt

And angular velocity can be given as

=l

Using, =lr,

=lr                                                                                                                ………….(1)

l=Vt and =t

Substituting in (1)

t=V tr

V=r, which is the required expression for velocity.

Also,

a=dVdt=rddt=drdt=V=VrV=V2r

a=V2r, which is the required expression for acceleration.

[5 – 6 Marks]

Q1. A particle starts from the origin at t=0s with a velocity of 10 ĵm/s and moves in the x-y plane with a constant acceleration of 8î+2ĵm/s2.

1. a) At what time is the x-coordinate of the particle 16m? What is the y-coordinate of the particle at that time?

Ans. The velocity of the particle is given as =10 ĵm/s

Acceleration of the particle is given as a=ddt=8î+2ĵ

Also, a=8î+2ĵ

But,

a=ddt=8î+2ĵ

d=8î+2ĵdt

Integrating both sides:

(t)=8tî+2tĵ+u

Where,

= Velocity vector of the particle at t = 0

u= Velocity vector of the particle at time t

However,

=drdt

=dt=(8tî+2tĵ+u)dt

Integrating the equations with the conditions: at t=0;  r=0 and at t=t; and

r=r

r=ut+128t2î+122t2ĵ

r=ut+4t2î+t2ĵ

=(10ĵ)t+4t2î+t2ĵ

xî+yĵ=4t2î+(10t+t2

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of î and ĵ we get,

x=4t2

t=x412

And y=10t+t2

When x=16m:

t=16412=2s, which is the required time.

y=102+22=24m, which is the required y coordinate.

1. b) What is the speed of the particle at the time?

Ans: Velocity of the particle can be given by:

(t)=8tî+2tĵ+u

At t=2s

(2)=8(2)î+2(2)ĵ+10ĵ

=16î+14ĵ

=162+142

=256+196

=452

=21.26m/s, which is the required speed.

A set of questionnaires is constructed for students to get a better understanding of the Chapter 4 of Class 11 Physics. These questions are easily accessible on the Extramarks website.

Class 11 Physics Chapter 4 Important Questions

Motion in a Plane Class 11 Important Questions

Class 11 Physics Chapter 4 Important Questions cover various topics such as the following.

• The basic concepts of scalars and vectors
• The displacement and acceleration
• Multiplication of vectors
• Uniform circular motion
• Concepts of projectile motion
• Three-dimensional motion in a plane
• Addition and subtraction of vector

Why go for Extramarks?

Extramarks provides comprehensive study material which includes numericals, concepts, reason-based questions, etc. Students can directly access these reliable materials from the website while preparing for the exams. This set of questionnaires are made by the subject matter experts at Extramarks.

Conclusion

Students can refer to Class 11 Physics Chapter 4 Important Questions as it will help them cover all the important topics while revising the chapter.

#### Both assertion and reason are true but reason is not the correct explanation of the assertion.

Marks:1

Ans

For getting a zero-resultant vector, the three vectors can be arranged in a polygon. If three vectors are coplanar then their resultant vector can be zero, that is when the resultant of two of them is equal in magnitude and opposite to the direction of the third vector.

Q.2 If an angle of projection of a projectile is 90°, then the horizontal range, R is

zero.

infinite.

one-fourth of its maximum height.

four times its maximum height.

Marks:1

Ans

zero.

### Marks:1

Ans

Q.4

<– [if gtevml 1]> <[endif]–><– [ifgtemso 9]><[endif]–>

Marks:2

Ans

Marks:3

Ans

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### 1. Explain linear motion.

A one-dimensional motion across a straight line that uses only one spatial dimension is considered to be a linear motion. This motion is also called a rectilinear motion. Uniform linear motion with a constant velocity and zero acceleration and non-uniform linear motion with a variable velocity and non-zero acceleration are the two most evident types of linear motion. It is the most basic  of all the motions. Linear motion is considered to be nearly similar to general motion. The rolling of the ball on the ground is an example of linear motion.

### 2. A passenger arriving in a new town wants to go from the station to a hotel located 10km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23km long and reaches the hotel in 28 min. (a) What is the average speed of the taxi? (b) What is the magnitude of average velocity? Are the two equal?

(a) Total distance travelled =23km

Total time taken = 28 min=2860h

Therefore,

Average speed =Total distance travelledTotal time taken

=23286049.29km/h

(b) Distance between the hotel and the station =10km= Displacement of the car

Therefore,

Average velocity =102860=21.34km/h

The two physical quantities are not equal.

### 3. After carefully reading each statement below and providing appropriate reasons, state whether it is true or false. (a) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector. (b) The net acceleration of a particle in a circular motion is always along the radius of the circle towards the centre. (c) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.

(a) True

The acceleration vector direction faces toward the centre of the circle in a uniform circular motion. Later, it frequently changes with time, and over one cycle, the average of these vectors is a null vector.

(b) False

In a uniform circular motion, a particle’s net acceleration in circular motion is always guided across the radius of the circle, pointing to the centre.

(c) True

A particle visibly moves tangentially at a point on a circular path.