class 11 physics chapter 4 assignment

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Class 11 Physics Assignments

We have provided below free printable Class 11 Physics Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 11 Physics . These Assignments for Grade 11 Physics cover all important topics which can come in your standard 11 tests and examinations. Free printable Assignments for CBSE Class 11 Physics , school and class assignments, and practice test papers have been designed by our highly experienced class 11 faculty. You can free download CBSE NCERT printable Assignments for Physics Class 11 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Physics Class 11. Students can click on the links below and download all Pdf Assignments for Physics class 11 for free. All latest Kendriya Vidyalaya Class 11 Physics Assignments with Answers and test papers are given below.

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We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 11 Physics . Students and teachers can download and save all free Physics assignments in Pdf for grade 11th. Our expert faculty have covered Class 11 important questions and answers for Physics as per the latest syllabus for the current academic year. All test papers and question banks for Class 11 Physics and CBSE Assignments for Physics Class 11 will be really helpful for standard 11th students to prepare for the class tests and school examinations. Class 11th students can easily free download in Pdf all printable practice worksheets given below.

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NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane

NCERT Solutions Class 11 Physics Chapter 4 - CBSE Access Free PDF

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane: Welcome to the updated Class 11 Physics Chapter 4 exercise solutions on the Careers360 page. On this page, you will find comprehensive and easy-to-understand NCERT solutions for Chapter 4. The motion in a plane class 11 NCERT solutions cover a total of 32 questions, ranging from 4.1 to 4.25 in the exercise section and 4.26 to 4.32 in the additional exercise section. In addition to the solutions, we provide essential formulas and diagrams in PDF format for each chapter. These resources are designed to assist you during your revision, homework, and assignment work.

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NCERT Solutions for Class 11 Physics Chapter 4 Motion in A Plane

Significance of ncert solutions for class 11 physics chapter 4 motion in a plane, ncert solutions for class 11 physics chapter wise, subject wise ncert exemplar solutions.

Chapter 4 of Class 11 Physics is a part of the Class 11 Physics NCERT Solutions . The preceding chapter deals with motion along a straight line, and you should have already familiarized yourself with the concepts of position, velocity, and acceleration . These principles are applied throughout both Class 11 and Class 12 Physics. The solutions for NCERT Class 11 Physics Chapter 4, which pertains to "Motion in a Plane," build upon the foundation laid in chapters 2 and 3. This chapter begins by introducing the concept of vectors. To grasp the motion in a plane NCERT solutions, a clear understanding of vector addition, subtraction, and scalar multiplication is indispensable.

After introducing the concept of vectors, the NCERT describes motion in a plane, projectile motion, and uniform circular motion . The solutions for Motion in a Plane Class 11 are significant, as the concepts employed to solve these problems will prove valuable in forthcoming chapters. The NCERT solutions are indispensable for Class 11 exam preparation. The exercises in this chapter are divided into two parts. The questions in the additional exercises are more comprehensive compared to those in the main exercise.

Chapter 4 Physics Class 11: Important Formulas and Diagrams + eBook link

Motion in a Plane class 11 exercise solutions covers important formulas and diagrams, along with an eBook link. In this context, it's crucial to note that this chapter addresses significant concepts like projectile motion, relative motion, and vector notation. These foundational concepts pave the way for understanding the subsequent formulas and diagrams. Here, we provide the formula for projectile motion, relative motion, and vector notation, equipping you with essential tools to comprehend the content presented in this chapter.

Vector Notation

$\vec{a}=a_x \hat{\imath}+a_y \hat{\jmath}+a_z \hat{k}$

Vector terminology

Parallel vectors: Parallel vectors have the same or parallel direction but may have different magnitudes.

Unit vector: A unit vector is a vector with a magnitude of 1, often used to indicate a direction, denoted by adding a hat symbol (ˆ) to the vector symbol, e.g., ȳ.

Negative of a vector : The negative of a vector is a vector that has the same magnitude as the original vector but points in the opposite direction.

The magnitude of vector

$a=|\vec{a}|=\sqrt{a_x^2+a_y^2+a_z^2}$

Dot product of vector

$\vec{a} \cdot \vec{b}=a_x b_x+a_y b_y+a_z b_z=a b \cos \theta$

Cross product of vector

Projectile motion.

Time of flight:

Maximum range:

R= u 2 sin2Θ/g, R max =u 2 /g

Note: Range is maximum when sin2θ is maximum

Maximum height:

H= u 2 sin 2 Θ/2g

Relative velocity

V ab =v a -v b

Apart from the formulas covered in this chapter, students can also access a comprehensive collection of essential formulas for all Class 11 chapters. This valuable resource is available for download as a PDF by clicking the link provided below.

Download Ebook - Formula Sheet for Physics Class 11: Chapterwise Important Formulas With Examples, Graphs, And Points

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane - Important Topics

Scalars and Vectors: This section introduces the fundamental distinction between scalars and vectors. Vectors possess both magnitude and direction, while scalars are characterized solely by their magnitude.
Resolution of Vectors and Vector Addition: This topic delves into the technique of breaking down vectors into their individual components and subsequently explores how to combine vectors using addition.
Addition and Subtraction of Vectors – Graphical Method: Here, the chapter elaborates on how to graphically perform vector addition and subtraction, contributing to a deeper grasp of these operations.
Relative Velocity in Two Dimensions: This section introduces the concept of relative velocity, which pertains to the velocity of one object relative to another object. The complexities of relative motion within two dimensions are discussed.
Uniform Circular Motion: The chapter takes a closer look at the motion of an object within a circle, maintaining a constant speed. Concepts related to centripetal acceleration and angular velocity are explored.
Projectile Motion: This segment explores into the intricacies of projectile motion, wherein an object is propelled into the air under the influence of gravity. The chapter elaborates on the path, velocity, and other crucial aspects of such motion.

While these topics form the core of the chapter, it's important to note that there are additional significant concepts to consider. For instance, the equations of motion within two dimensions offer valuable insights into the mechanics of objects moving in a plane.

Here are some study recommendations to enhance your understanding of this motion in a plane ncert solutions chapter :

Grasp the distinction between vectors and scalars.
Develop proficiency in resolving vectors and performing vector addition and subtraction.
Cultivate the ability to interpret and construct vector diagrams.
Gain a solid comprehension of relative velocity principles.
Hone your problem-solving skills for scenarios involving uniform circular motion.
Practice tackling problems related to projectile motion.

By thoroughly engaging with these topics and adhering to the study tips provided, you can develop a strong foundation in comprehending motion within a plane.

Download physics chapter 4 class 11 exercise solutions for free.

** According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 4, "Motion in a Plane," has been renumbered as Chapter 3.

Download PDF

Access Physics Chapter 4 Class 11 Exercise Solutions – Motion in a Plane

Q. 4.1 State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity

Volume is a scalar quantity since it has only magnitude without any direction.

Mass is a scalar quantity because it is specified only by magnitude.

Speed is specified only by its magnitude not by its direction so it is a scalar quantity .

Acceleration is a vector quantity as it has both magnitude and direction associated.

Density is a scalar quantity as it is specified only by its magnitude.

The number of moles is a scalar quantity as it is specified only by its magnitude.

Velocity is a vector quantity as it has both magnitude and direction.

Angular frequency is a scalar quantity as it is specified only by its magnitude.

Displacement is a vector quantity since it has both magnitude and associated direction.

Angular velocity is a vector quantity as it has both magnitude and direction.

Q. 4.2 Pick out the two scalar quantities in the following list :

force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

The two scaler quantities are work and current , as these two don't follow laws of vector addition.

Q. 4.3 Pick out the only vector quantity in the following list :

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Among all, the impulse is the only vector quantity as it is the product of two vector quantities. Also, it has an associated direction.

Q. 4.4 (a) State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector

(a) Adding two scalars is meaningful if the two have the same unit or both represent the same physical quantity.

(b) Adding a scalar to a vector of the same dimensions is meaningless as vector quantity has associated direction.

(c) Multiplication of vector with scaler is meaningful as it just increases the magnitude of vector quantity and direction remains the same.

(d) Multiplication of scaler is valid and meaningful , unbounded of any condition. This is because, if we have two different physical quantity then their units will also get multiplied.

(e) Adding two vectors is meaningful if they represent the same physical quantity. This is because their magnitude will get added and direction will remain the same.

(f) Adding a component of a vector to the same vector is meaningful as this represents the same case of adding vectors with the same dimensions. In this, the magnitude of the resultant vector will increase and the direction will remain the same.

Q. 4.5 Read each statement below carefully and state with reasons, if it is true or false :

(a) The magnitude of a vector is always a scalar,

(b) each component of a vector is always a scalar,

(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same interval of time,

(e) Three vectors not lying in a plane can never add up to give a null vector.

Since the magnitude of a vector will not have any direction (also it is a number), so it will be scaler.

The component of a vector will always be a vector as it will also have a direction specified.

This is true only in case when the particle is moving in a straight line. This is because path length is a scalar quantity whereas displacement is vector.

From the above part (c) it is clear that total path length is either equal or greater than the displacement. As a result given statement is true.

Since they don't lie in the same plane so they cannot give null vector after addition.

Q. 4.6 (a) Establish the following vector inequalities geometrically or otherwise :

From (i) and (ii), we can conclude that,

Q. 4.6 (b) Establish the following vector inequalities geometrically or otherwise :

$|a+b|\geq |\; |a|-| b|\; |$

Consider the given image:

Sum of two sides of a triangle is greater than the length of another side.

$OB+BC > OC$

From (i) and (ii), we get :

Q. 4.6 (c) Establish the following vector inequalities geometrically or otherwise :

From (i) and (ii) we get :

Q. 4.6 (d) Establish the following vector inequalities geometrically or otherwise :

$\left | a-b \right |\geq \left | | a\right |-\left | b \right ||$

Consider the image given below :

Also, if the vectors are in a straight line then :

$\left | a-b \right |\ = \left | | a\right |-\left | b \right ||$

From (i) and (ii), we can conclude :

Q. 4.7 Given a + b + c + d = 0, which of the following statements are correct:

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a + c) equals the magnitude of ( b + d),

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

(a) Incorrect: - Sum of three vectors in a plane can be zero. So it is not a necessary condition that all of a,b,c,d should be null vector.

(b) Correct : We are given that a + b + c + d = 0

So, a + b = - (c + d)

Thus magnitude of a + c is equal to the c+d.

b + c + d = - a

So clearly magnitude of a cannot be greater than the sum of the other three vectors.

(d) Correct: - Sum of three vectors is zero if they are coplanar.

Thus, a + b + c + d = 0

or a + (b + c) + d = 0

Hence (b+c) must be coplaner with a and d

$200 m$

The displacement vector is defined as the shortest distance between two points which particle had covered.

In this case, the shortest distance between these points is the diameter of the circular ice ground.

Thus, Displacement = 400 m.

Girl B had travelled along the diameter so path travelled by her is equal to the displacement.

$O$

(a) net displacement,

The net displacement, in this case, will be zero because the initial and final position is the same.

Net displacement = Final position - Initial position .

$1\; km,$

(b) average velocity,

(b) Average velocity is defined as the net displacement per unit time. Since we have the net displacement to be zero so the avg. velocity will also be zero .

$Avg.\ Velocity\ =\ \frac{Net\ displacement}{Time\ taken}$

Total path = OP + arc PQ + OQ

$= 1+ \frac{1}{4}(2\Pi \times1)+ 1$

So the avg. speed is :

$= \frac{3.57}{\frac{1}{6}}\ =\ 21.42\ Km/h$

The track is shown in the figure given below:-

Let us assume that the trip starts at point A.

The third turn will be taken at D.

So displacement will be = Distance AD = 500 + 500 = 1000 m

Total path covered = AB + BC + CD = 500 + 500 + 500 = 1500 m

The sixth turn is at A.

So the displacement will be Zero

and total path covered will be = 6 (500) = 3000 m

The eighth turn will be at C.

So the displacement = AC

$=\ \sqrt{AB^2\ +\ BC ^2\ +\ 2(AB)(BC) \cos 60^{\circ}}$

And the total distance covered = 3000 + 1000 = 4000 m=4Km

$10 \; km$

(a) the average speed of the taxi,

(a) Avg. speed of taxi is given by:-

$=\ \frac{Total\ path\ travelled }{Total\ time\ taken}$

(b) the magnitude of average velocity? Are the two equal?

Total displacement = 10 Km

Total time taken in hours :

$=\ \frac{28}{60}\ hr$

Avg. velocity :

$=\ \frac{10}{\frac{28}{60}}\ hr$

It can be clearly seen that avg. speed and avg. velocity is not the same.

$30\; ms^{-1}$

The given situation is shown in the figure:-

Since both rain and woman are having some velocity so we need to find the relative velocity of rain with respect to woman.

$V\ =\ V_{rain}\ +\ (-V_{woman})$

And the angle is given by :

$\tan\Theta \ =\ \frac{V_{woman}}{V_{rain}}$

Hence woman needs to hold an umbrella at 18 degrees from vertical towards the south.

$4.0 \; km/h$

The speed of man is (swim speed ) = 4 Km/h.

Time taken to cross the river will be :

$=\ \frac{Distance}{Speed}$

Total distance covered due to the flow of the river:-

$=\ Speed\ of\ river \times Time\ taken$

According to the question the figure is shown below:-

The angle between velocity of wind and opposite of velocity of boat is (90 + 45) = 135 degree.

Using geometry,

$\tan \beta \ =\ \frac{51\sin (90+45)}{72\ +\ 51 \cos (90+45) }$

So the flag will be just 0.11 degree from the perfect east direction.

Q. 4.15 The ceiling of a long hall is 25 m 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?

It is known that the maximum height reached by a particle in projectile motion is given by :

$h\ =\ \frac{u^2sin^2\Theta }{2g}$

Putting the given values in the above equation :

$25\ =\ \frac{40^2sin^2\Theta }{2\times9.8}$

Now the horizontal range can be found from :

$R\ =\ \frac{U^2 \sin 2\Theta }{g}$

We are given the range of projectile motion.

$R\ =\ \frac{u^2\ \sin 2\Theta }{g}$

Substituting values :

$100\ =\ \frac{u^2\ \sin 90^{\circ} }{g}$

Now since deacceleration is also acting on the ball in the downward direction :

$v^2\ -\ u^2\ =\ -2gh$

Since final velocity is 0, so maximum height is given by :

$H\ =\ \frac{u^2}{2g}$

Frequency is given by :

$Frequency\ =\ \frac{No.\ of\ revolutions}{Total\ time\ taken}$

And, the angular frequency is given by :

$\omega \ =\ 2 \Pi f$

Hence the acceleration is given by :

$a\ =\ \omega ^2 r$

Convert all the physical quantities in SI units.

$Speed\ =\ 900\times \frac{5}{18}\ 250\ m/s$

So the acceleration is given by :

$a\ =\ \frac{v^2}{r}$

The ratio of centripetal acceleration with gravity gives :

$\frac{a}{g}=\ \frac{62.5}{9.8}\ =\ 6.38$

Q. 4.19 (a) Read each statement below carefully and state, with reasons, if it is true or false :

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

False :- Since the net acceleration is not directed only along the radius of the circle. It also has a tangential component.

Q. 4.19 (b) Read each statement below carefully and state, with reasons, if it is true or false :

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

True: - Because particle moves on the circumference of the circle, thus at any its direction should be tangential in order to move in a circular orbit.

Q. 4.19 (c) Read each statement below carefully and state, with reasons, if it is true or false :

True: - In a uniform circular motion, acceleration is radially outward all along the circular path. So in 1 complete revolution, all the vectors are cancelled and the null vector is obtained.

$r=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$

The velocity vector is given by:-

$v \ =\ \frac{dr}{dt}$

Now for acceleration :

$a \ =\ \frac{dv}{dt}$

Put the value of time t = 2 in the velocity vector as given below :

$v\ =\ 3\; \hat{i}-4(2)\; \hat{j}$

Thus the magnitude of velocity is :

$=\ \sqrt{3^2\ +\ (-8)^2}\ =\ 8.54\ m/s$

Direction :

$\Theta \ =\ \tan^{-1} \frac{8}{3}\ =\ -69.45^{\circ}$

And the acceleration is given as :

So, the velocity due to acceleration will be :

$a\ =\ \frac{dv}{dt}$

By integrating both sides,

$v\ =\ 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u$

Here u is the initial velocity (at t = 0 sec).

$v\ =\ \frac{dr}{dt}$

Integrating both sides, we get

$r\ =\ 8.0\times \frac{1}{2}t^2\ \widehat{i}\ +\ 2.0\times\frac{1}{2} t^2\ \widehat{j}\ +\ (10t\ ) \widehat{j}$

Comparing coefficients, we get :

$x\ =\ 4t^2$

In the question, we are given x = 16.

So t = 2 sec

and y = 10 (2) + 2 2 = 24 m.

$\left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}.$

(b) What is the speed of the particle at the time?

The velocity of the particle is given by :

Put t = 2 sec,

So velocity becomes :

$v\ =\ 8.0(2)\ \widehat{i}\ +\ 2.0(2)\ \widehat{j}\ +\ 10\widehat{j}$

Now, the magnitude of velocity gives :

$\left | v \right |\ =\ \sqrt{16^2\ +\ 14^2}$

Then we have:-

$\Theta \ =\ \tan^{-1}\left ( \frac{1}{1} \right )\ =\ 45^{\circ}$

Q. 4.23 For any arbitrary motion in space, which of the following relations are true :

$v_{average}=\left ( 1/2 \right )\left [ v\left ( t_{1} \right )+v\left ( t_{2} \right ) \right ]$

(a) False:- Since it is arbitrary motion so the following relation cannot hold all the arbitrary relations.

(b) True:- This is true as this relation relates displacement with time correctly.

(d) False:- The given equation is valid only in case of uniform acceleration motion. But this is arbitrary motion so acceleration can be no-uniform.

(e) True:- This is the universal relation between acceleration and velocity-time, as the definition of acceleration is given by this.

Q. 4.24 Read each statement below carefully and state, with reasons and examples, if it is true or false :

A scalar quantity is one that

(a) is conserved in a process

(b) can never take negative values

(d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axe

(a) False:- For e.g. energy is a scalar quantity but is not conserved in inelastic collisions.

(b) False:- For example temperature can take negative values in degree Celsius.

(d) False:- Gravitational potential varies in space from point to point.

(e) True:- Since it doesn't have direction.

$3400\; m$

For finding the speed of aircraft we just need to find the distance AC as we are given t = 10 sec.

$\tan 15^{\circ}\ =\ \frac{AB}{BD}$

Thus, the speed of aircraft :

$=\ \frac{1822.4}{10}\ =\ 182.24\ m/s$

Motion in plane class 11 NCERT solutions additional exercise :

$b$

No, a vector doesn't have a definite location as a vector can be shifted in a plane by maintaining its magnitude and direction.

Vector can change with time for e.g. displacement vector.

No, two equal vectors at a different location may not have identical physical effects. For e.g., two equal force vectors at a different location may have different torque but when they are applied together the net torque would be different.

Q. 4.27 A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? 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 that make any rotation a vector?

The main condition for a physical quantity to be a vector is that it should the law of vector addition. Also, the vector has both direction and, magnitude but these are not sufficient condition. For e.g. current has both magnitude and direction but is a scalar quantity as it doesn't follow the law of vector addition.

Rotation is not a vector on a large basis, as it is measured by an angle which follows the law of scaler addition.

Q. 4.28 (a) Can you associate vectors with (a) the length of a wire bent into a loop, Explain.

No, the length of a wire bent into a loop cannot be expressed in vector form as we have no direction associated with it.

Q. 4.28 (b) Can you associate vectors with (b) a plane area, Explain.

(b) The plane area can be expressed in vector form as direction can be associated as pointing outward or inward (normal to the plane) of the area.

Q. 4.28 (c) Can you associate vectors with (c) a sphere? Explain.

No, vector cannot be associated with a sphere as direction cannot be associated with sphere anyhow.

$30^{\circ}$

The range of bullet is given to be:- R = 3 Km.

$3\ =\ \frac{u^2\ \sin 60^{\circ} }{g}$

Now, we will find the maximum range (maximum range occurs when the angle of projection is 45 0 ).

$R_{max}\ =\ \frac{u^2\ \sin 2(45^{\circ}) }{g}$

Thus the bullet cannot travel up to 5 Km.

$1.5\; km$

According to the question the situation is shown below:-

Now, the horizontal distance traveled by the shell = Distance traveled by plane

$u \sin \Theta \ t\ =\ vt$

So, the required height will be:-

$H\ =\ \frac{u^2\ \sin^2(90\ -\ \Theta )}{2g}$

Speed of cycle = 27 Km/h = 7.5 m/s

The situation is shown in figure :-

The centripetal acceleration is given by :

$a_c\ =\ \frac{v^2}{r}$

So, the net acceleration becomes :

$a\ =\ \sqrt{a_c^2\ +\ a_T ^2}$

Now for direction,

$\tan \Theta \ =\ \frac{a_c}{a_T}$

where the symbols have their usual meaning.

Using the equation of motion in both horizontal and vertical direction.

$v_y\ =\ v_{oy}\ =\ gt$

(b) The maximum height is given by :

$h\ =\ \frac{u^2 \sin^2 \Theta }{2g}$

And, the horizontal range is given by :

$R\ =\ \frac{u^2 \sin 2\Theta }{g}$

Dividing both, we get :

$\frac{h}{R}\ =\ \frac{\tan \Theta }{4}$

The topic And Subtopic of Physics Chapter 4 Class 11 are listed below.

NCERT Chapter 4 Physics class 11 - Weightage in Exams

Class 11 physics chapter 4 exercise solutions in NCERT Class 11 Physics is extremely important and is weighted heavily in the exams. Vectors, relative velocity, uniform circular motion, and projectile motion are among the topics covered in this chapter.

The motion in a plane class 11 NCERT solutions is expected to carry roughly 10-12% weightage in the board exams, according to the CBSE marking plan for 2023-24. This means that in order to score well, students must be familiar with the topics and formulas taught in the chapter.

While studying this chapter try to understand the physical significance of the topic and relate to real-life examples which will be interesting.
As final exams for class 11 are considered the NCERT solutions for class 11 are important.
For exams like JEE Main and NEET one or two questions are expected from the chapter.
The CBSE NCERT solutions for class 11th physics chapter 4 motion will help to perform well in class and competitive exams.

Key Features of Motion in a Plane Class 11 NCERT Solutions

The C lass 11 physics chapter 4 exercise solutions provide easy-to-understand explanations for all the concepts discussed in the chapter. They break down complex solutions into simple terms.
Each problem is solved step by step, helping you follow the logic and understand the process of solving different types of questions.
The motion in a plane class 11 ncert solutions cover a wide range of problems related to motion in a plane, including vectors, circular motion, and projectile motion. This helps you practice and grasp different aspects of the chapter.
The solutions highlight important formulas related to motion in a plane. These formulas are essential for solving problems and understanding the underlying principles.
The motion in a plane class 11 exercise solutions act as a quick reference for revising the chapter before exams. You can use them to review the main concepts and formulas.
The language used in the physics chapter 4 class 11 exercise solutions is student-friendly and simple, making it easier for you to grasp the concepts without confusion.

Also Check NCERT Books and NCERT Syllabus here

NCERT Books Class 11 Physics
NCERT Syllabus Class 11 Physics
NCERT Books Class 11
NCERT Syllabus Class 11

NCERT Physics Exemplar Solutions Class 11 For All The Chapters:

Subject wise NCERT solutions

NCERT solutions for class 11 Mathematics
NCERT solutions for class 11 Chemistry
NCERT solutions for class 11 Physics
NCERT solutions for class 11 Biology
NCERT Exemplar Class 11th Solutions
NCERT Exemplar Class 11th Maths
NCERT Exemplar Class 11th Physics
NCERT Exemplar Class 11th Chemistry
NCERT Exemplar Class 11th Biology

Frequently Asked Question (FAQs)

One question from the chapter can be expected from motion in a plane for JEE Main. Some times the questions from Kinematics may be from both the NCERT class 11 chapter 3 and 4. In some papers only from one out the two chapters may be asked. But both the chapters are important because these chapters are pre-requirement for the coming chapters in mechanics.

Minimum one question can be expected from the chapter for NEET exam. All the topics discussed in the NCERT book are important for NEET. For practice question refer to previous year NEET papers and NCERT exemplar.

The type of questions present in class 11 physics chapter 4 exercise solutions are

Conceptual Questions

Numerical Questions

Multiple-Choice Questions

Diagram-Based Questions

Class 11 is a crucial stage for students as their performance in this grade can significantly impact their future career paths. To aid students in their preparation, the faculty at Careers360 has developed ncert solutions class 11 physics chapter 4 based on the latest CBSE Syllabus. These solutions are available in PDF format and are designed to be accessible to students of all intelligence levels. The solutions are created with the intention of aiding students in their exam preparation.

Motion in a Plane

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NCERT Solutions for Class 11 Physics chapter-4 Motion In A Plane

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NCERT Solutions for class-11 Physics Chapter 4 Motion In A Plane is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-11 in NCERT textbook, also do read theory of this Chapter 4 Motion In A Plane while going before solving the NCERT questions. You can download NCERT solution of all chapters from Physics Wallah in PDF.

Chapter 4 Motion In A Plane

Answer The Following Question Answer

Question 1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Solution : Scalar: Volume, mass, speed, density, number of moles, angular frequency Vector: Acceleration, velocity, displacement, angular velocity Question 2. Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Solution : Work and current are scalar quantities.

Question 3. Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Solution : Since, Impulse = change in momentum = force × time. As momentum and force are vector quantities, hence impulse is a vector quantity. Question 4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful: a. adding any two scalars, b. adding a scalar to a vector of the same dimensions, c. multiplying any vector by any scalar, d. multiplying any two scalars, e. adding any two vectors, f. adding acomponent of a vector to the same vector.

Solution : a. Yes, addition of two scalar quantities is meaningful only if they both represent the same physical quantity. b. No, addition of a vector quantity with a scalar quantity is not meaningful. c. Yes, scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse. d. Yes, scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions. e. Yes, addition of two vector quantities is meaningful only if they both represent the same physical quantity. f. Yes, component of a vector can be added to the same vector as they both have the same dimensions.

Question 5. Read each statement below carefully and state with reasons, if it is true or false:a. The magnitude of a vector is always a scalar, a. The magnitude of a vector is always a scalar, b. each component of a vector is always a scalar, c. the total path length is always equal to the magnitude of the displacement vector of a particle. d. the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, e. Three vectors not lying in a plane can never add up to give a null vector.

Solution : a. True; because magnitude is a pure number. b. False, each component of a vector is also a vector. c. True, only if the particle moves along a straight line in the same direction, if not then it is false. d. True, because the total path length is either greater than or equal to the magnitude of the displacement vector. e. True, as they can not be represented by the three sides of a triangle taken in the same order.

Question 6. Establish the following vector inequalities geometrically or otherwise: a. a + b| ≤ a| + b b. a + b| ≥ |a| − b| c. a − b| ≤ a| + b d. a − b| ≥ |a| − b| When does the equality sign above apply?

NCERT Solutions for Class 11 Physics chapter 4-Motion In A Plane

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOMN, we have:

Question 7. Given vectors a + b + c + d = 0, which of the following statements are correct: a. Vectors a, b, c, and d must each be a null vector, b. The magnitude of vectors (a + c) equals the magnitude of vectors(b+ d), c. The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, d. Vectors b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

Solution : a. In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero. b. Correct

a + b + c + d = 0 a + c = – (b + d) Taking modulus on both the sides, we get: | a + c | = | –(b + d)| = | b + d Hence, the magnitude of (a + c) is the same as the magnitude of (b + d). c. Correct a + b + c + d = 0 a = - (b + c + d) Taking modulus both sides, we get: | a | = | b + c + d | | a | ≤ | a | + | b | + | c | .... (i) Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d. Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d. d. Correct For a + b + c + d = 0 a + (b + c) + d = 0 The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle. If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

Question 8. Three girls skating on a circular ice ground of radius 200 m start from a point Pon the edge of the ground and reach a point Qdiametrically opposite to Pfollowing different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?

Solution : Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground. Radius of the ground = 200 m Diameter of the ground = 2 × 200 = 400 m Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B.

Question 9. A cyclist starts from the centre Oof a circular park of radius 1 km, reaches the edge Pof the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the

a. net displacement, b. average velocity, and c. average speed of the cyclist?

Solution : a. Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero. b. Average velocity is given by the relation: Average velocity = Net Displacement / Total time Since the net displacement of the cyclist is zero, his average velocity will also be zero. c. Average speed of the cyclist is given by the relation: Average speed = Total path length / Total time Total path length = OP + PQ + QO = 1 + (1 (2π × 1) / 4 ) + 1 = 2 + (π / 2) = 3.570 km Time taken = 10 min = 10 / 60 = 1 / 6 h ∴ Average speed = 3.570 / (1/6) = 21.42 km/h Question 10. On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Solution : The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure

The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure Let the motorist start from point A. The motorist takes the third turn at D. Magnitude of displacement =AD=BC+EF=500+500=1000m Total path length =AB+BC+CD=500+500+500=1500m The motorist takes the sixth turn at point A, which is the starting point. Magnitude of displacement =0 Total path length =6AB =6×500=3000m The motorist takes the eight turn at point C The magnitude of displacement =AC From triangle law, the magnitude of displacement is 866.03m at an angle of 30 with AC. It means AC makes an angle 30 with the initial direction. Total path length =8×500=4000m

Question 11. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

Solution : (a) Total distance travelled = 23 km Total time taken = 28 min = 28 / 60 h ∴ Average speed of the taxi = Total disttance travelled / Total time taken = 23 / (28/60) = 49.29 km/h (b) Distance between the hotel and the station = 10 km = Displacement of the car ∴ Average velocity = 10 / (28/60) = 21.43 km/h Therefore, the two physical quantities (averge speed and average velocity) are not equal.

Question 12. Rain is falling vertically with a speed of 30 m s-1. A woman rides a bicycle with a speed of 10 m s-1 in the north to south direction. What is the direction in which she should hold her umbrella ?

Solution : The described situation is shown in the given figure.

v c = Velocity of the cyclist v r = Velocity of falling rain In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman. v = v r + (-v c ) = 30 + (-10) = 20 m/s tan θ = v c / v r = 10 / 30 θ = tan -1 (1 / 3) = tan-1 (0.333) ≈ 180 Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical. Question 13. A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/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?

Solution : Speed of the man, v m = 4 km/h Width of the river = 1 km Time taken to cross the river = Width of the river / Speed of the river = 1/4 h = 1 × 60 / 4 = 15 min Speed of the river, v r = 3 km/h Distance covered with flow of the river = v r × t = 3 × 1 / 4 = 3 / 4 km = 3 × 1000 / 4 = 750 m Question 14. In a harbor, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbor flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?

Solution : The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (vwb) of the wind with respect to the boat.

Solution : Maximum height, h = 25 m In projectile motion, the maximum height reached by a body projected at an angle θ, is given by the relation: h = u 2 Sin 2 θ / 2g 25 = (40) 2 Sin 2 θ / 2 × 9.8 sin 2 θ = 0.30625 sin θ = 0.5534 ∴ θ = sin–1(0.5534) = 33.60° Horizontal range, R = u 2 Sin 2θ / g = (40) 2 × Sin (2 × 33.600) / 9.8 = 1600 × Sin (67.2) / 9.8 = 1600 × 0.922 / 9.8 = 150.53 m Question 16. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?

Solution : Maximum horizontal distance, R = 100 m The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°. The horizontal range for a projection velocity v, is given by the relation: R = u 2 Sin 2θ / g 100 = u 2 Sin 900 / g u 2 / g = 100 ....(i) The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H. Acceleration, a = –g Using the third equation of motion: v 2 - u 2 = -2gH H = u 2 / 2g = 100 / 2 = 50 m Question 17. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Solution : Length of the string, l = 80 cm = 0.8 m Number of revolutions = 14 Time taken = 25 s Frequency, v = Number of revolutions / Time taken = 14 / 25 Hz Angular frequency, ω = 2πν = 2 × (22/7) × (14/25) = 88 / 25 rad s -1 Centripetal acceleration, ac = ω 2 r = (88 / 25) 2 × 0.8 = 9.91 ms -2 The direction of centripetal acceleration is always directed along the string, toward the centre, at all points. Question 18. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Solution : Radius of the loop, r = 1 km = 1000 m Speed of the aircraft, v = 900 km/h = 900 × 5 / 18 = 250 m/s Centripetal acceleration, ac = v 2 / r = (250) 2 / 1000 = 62.5 ms -2 Acceleration due to gravity, g = 9.8 m/s 2 a c / g = 62.5 / 9.8 = 6.38 a c = 6.38 g Question 19. Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Solution : (a) False, the net acceleration of a particle is towards the centre only in case of a uniform circular motion. (b) True, because while leaving the circular path, the particle moves tangentially to the circular path. (c) True, The direction of acceleration vector in a uniform circular motion is directed towards the centre of circular path. It is constantly changing with time. The resultant of all these vectors will be zero vector. Question 20. The position of a particle is given by

b. What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

The negative sign indicates that the direction of velocity is below the x-axis.

Question 21. A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m/s and moves in the x-y plane with a constant acceleration of (8.0 î+ 2.0 ĵ) m s -2 .

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

Question 22. î and ĵ are unit vectors along x and y axis respectively. What is the magnitude and direction of the vectors î + ĵ, and î − ĵ ? What are the components of a vector A= 2î + 3ĵ along the directions of î + ĵ and î − ĵ? [You may use graphical method]

Question 23. For any arbitrary motion in space, which of the following relations are true : a. v average = (1/2) (v (t 1 ) + v (t 2 )) b. vaverage = [ r(t 2 ) - r(t 1 ) ] / (t 2 – t1) c. v (t) = v (0) + a t d. r (t) = r (0) + v (0) t + (1/2) a t 2 e. aaverage = [ v (t 2 ) - v (t 1 ) ] / ( t 2 – t 1 ) (The ‘average’ stands for average of the quantity over the time interval t 1 to t 2 )

Solution : a. It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation. b. The arbitrary motion of the particle can be represented by this equation. c.The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space. d. The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space. e. The arbitrary motion of the particle can be represented by this equation.

Question 24. Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that a. is conserved in a process b.can never take negative values c. must be dimensionless d. does not vary from one point to another in space e. has the same value for observers with different orientations of axes.

Solution : a.False Despite being a scalar quantity, energy is not conserved in inelastic collisions. b. False Despite being a scalar quantity, temperature can take negative values.

c. False Total path length is a scalar quantity. Yet it has the dimension of length.

d.False A scalar quantity such as gravitational potential can vary from one point to another in space.

e. True The value of a scalar does not vary for observers with different orientations of axes. Question 25. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ?

Solution : The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m Angle subtended between the positions, ∠POQ = 30° Time = 10 s In ΔPRO: tan 15 0 = PR / OR PR = OR tan 15 0 = 3400 X tan 15 0 ΔPRO is similar to ΔRQO. ∴PR = RQ PQ = PR + RQ = 2PR = 2 × 3400 tan 15° = 6800 × 0.268 = 1822.4 m ∴ Speed of the aircraft = 1822.4 / 10 = 182.24 m/s

Question 26. A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer.

Solution : (i) A vector in general has no definite location in space because a vector remains unaffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However, a position vector has a definite location in space. (ii) A vector can vary with time as example the velocity vector o an unaffected particle varies with time. (iii) Two equal vectors at different locations in space do not necessarily have same physical effects. For example, two equal forces acting at two different points on a body which can cause the rotation of a body about an axis will not produce equal turning effect.

Question 27. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? 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 that make any rotation a vector ?

Solution : No. A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition. Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

Question 28. Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain.

Solution : (a) No, one cannot associate a vector with the length of a wire bent into a loop.

(b) Yes, one can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area. (c) No, one cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.

Question 29. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.

Solution : Range, R = 3 km Angle of projection, θ = 30° Acceleration due to gravity, g = 9.8 m/s 2 Horizontal range for the projection velocity u 0 , is given by the relation: R = u 0 2 Sin 2θ / g 3 = u 0 2 Sin 60 0 / g u 0 2 / g = 2√3 .......(i) The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is, Rmax = u 0 2 / g ....(ii) On comparing equations (i) and (ii), we get: Rmax = 3√3 = 2 X 1.732 = 3.46 km Hence, the bullet will not hit a target 5 km away. Question 30. A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s -1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m s -2 ).

Solution : Height of the fighter plane = 1.5 km = 1500 m Speed of the fighter plane, v = 720 km/h = 200 m/s Let θ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s Time taken by the shell to hit the plane = t Horizontal distance travelled by the shell = u x t Distance travelled by the plane = vt The shell hits the plane. Hence, these two distances must be equal. uxt = vt u Sin θ = v Sin θ = v / u = 200 / 600 = 1/3 = 0.33 θ = Sin -1 (0.33) = 19.50 In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell. ∴ H = u 2 Sin 2 (90 - θ) / 2g = (600) 2 Cos 2 θ / 2g = 360000 X Cos 2 19.5 / 2 X 10 = 16006.482 m ≈ 16 km

Question 31. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Solution : 0.86 m/s 2 ; 54.46° with the direction of velocity Speed of the cyclist, v = 27 km/h = 7.5 m/s Radius of the circular turn, r = 80 m Centripetal acceleration is given as: ac = v 2 / r = (7.5) 2 / 80 = 0.7 ms -2 The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s 2 .

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist. Since the angle between ac and aT is 90 0 , the resultant acceleration a is given by: a = (ac 2 + aT 2 ) 1/2 = ( (0.7) 2 + (0.5) 2 ) 1/2 = (0.74) 1/2 = 0.86 ms -2 tan θ = ac / aT where θ is the angle of the resultant with the direction of velocity. tan θ = 0.7 / 0.5 = 1.4 θ = tan -1 (1.4) = 54.560

Question 32.a. Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by θ(t) = tan -1 ((v 0y - gt) / v 0x ) b. Shows that the projection angle θ 0 for a projectile launched from the origin is given by θ 0 = tan -1 (4hm / R) where the symbols have their usual meaning.

Solution : a. Let v 0x and v 0y respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions. Let vx and vy respectively be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P = t Applying the first equation of motion along the vertical and horizontal directions, we get: v y = v 0y = gt And v x = v 0x ∴ tan θ = v y / v x = (v 0y - gt) / v 0x θ = tan -1 (v 0y - gt) / v 0x b. Maximum vertical height, hm = u 0 2 Sin 2 θ / 2g ...(i) Horizontal range, R = u 0 2 Sin 2 2θ / g ... (ii) hm / R = Sin 2 θ / 2Sin 2 2θ = Sin θ X Sin θ / 2 X 2Sinθ Cosθ = Sin θ / 4 Cos θ = tan θ / 4 tan θ = (4hm / R) θ = tan-1 (4hm / R)

NCERT Solutions For Class-11 Physics Chapter Wise

Chapter 1 Physical World

Chapter 2 Units and Measurements

Chapter 3 Motion In A Straight Line

Chapter 5 Laws of Motion

Chapter 6 Work, Energy and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solid

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

Related Chapters

chapter-1 Physical World
chapter-2 Units And Measurements
chapter-3 Motion In A Straight Line
chapter-4 Motion In A Plane
chapter-5 Laws of Motion
chapter-6 Work, Energy And Power
chapter-7 System of Particles and Rotational Motion
chapter-8 Gravitation
chapter-9 Mechanical Properties Of Solids
chapter-10 Mechanical Properties of Fluids
chapter-11 Thermal Properties Of Matter
chapter-12 Thermodynamics
chapter-13 Kinetic Theory
chapter-14 Oscillations
chapter-15 Waves

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NCERT Books Free Download for Class 11 Physics Chapter 4 - Motion in a Plane

NCERT Books for Class 11 Physics Chapter 4 - Motion in a Plane - Free PDF Download

Free NCERT Books download for Class 11 Physics Chapter 4 - Motion in a Plane on Vedantu. Students can also download the NCERT Physics tuition to clear your doubts and score more in your exams.

Question: How to download Class 11 Physics Chapter 4 NCERT Book for CBSE?

Answer: Students can download it from Vedantu. On this page, Students can download Class 11 Physics Chapter-4 PDF Solutions of the Book.

Overview of NCERT Books for Class 11 Physics Chapter 4

As one of the top scoring Chapters in kinematics, NCERT Books for Class 11 Physics Chapter 4 – Motion in a plane is a crucial study material for Class 11 students. Students must strive to comprehend each topic in depth in order to write proper responses on their first-term assessment. To do well in the Class 11 term – I examination, students must answer the questions at the end of each Chapter in the NCERT book. Students will be able to formulate meaningful answers based on the marking scheme of the first term CBSE Syllabus for 2021-22 and its recommendations using the NCERT Solutions accessible at Vedantu.com

This study material includes detailed understanding of the principles and through revision of the important concepts. These solutions will assist you in preparation for the Class 11 term – I examinations as well as the entrance exams that will follow the Class 12 first and second term exams.

We will come into problems in physics about scalar and vector values, and we are quite well aware of the distinction between the two. On various scalar numbers, we shall do algebraic expressions. We'll come up with true or false questions, and students will have to explain why they believe these physical numbers are true or incorrect. Vector inequalities of various quantities will be drawn. For more Chapter-by-Chapter solutions, see NCERT Books for Class 11 Physics.

We'll also examine questions about vector magnitude and co-linearity, as well as finding the magnitude of the displacement vector of three females skating in a circular terrain. We'll be looking for a man's position after he cycles around a circle. We'll compare the many types of displacement that a driver can have. We'll show an exciting voyage of a passenger who has recently arrived in a town and how he was duped by a cab driver.

How do you feel when you're riding your bike in the rain and the rains fall at an odd angle on your face? We'll examine how long it takes a man to swim across a river and how far he needs to swim if the water isn't still.

Have you ever tossed a ball against the ceiling? You can now throw the ball considerable distances. We'll show how a cricketer throws a ball to the wicketkeeper for such a vast distance.

Do vector quantities in space still have location and position? Find out below. We will learn how to adjust our height when throwing a ball in order to attain a specific distance goal. How can you fire a helicopter in order to hit it, and how high can a pilot fly a helicopter from the ground? After reviewing the solutions offered here, you should be able to answer all of these questions. Do you know why a motorcycle slows down in the corners and what he'll do if he doesn't?

For the benefit of Class 11 Science students, Vedantu.com is dedicated to offering the greatest study material, notes, Class 11 Physics sample papers, previous year question papers, worksheets, MCQs, videos, and animations. Log on to Vedantu.com to gain access to all of Vedantu's resources.

Let's take a closer look at the principles covered by the Chapter.

Motion on a Plane is one of the most essential Chapters in Unit-II Kinematics in Term I of Class 11 Physics, and it carries a higher grade in the first term exam. Students are encouraged to use the NCERT Solutions from Vedantu’s , which are available in both online and offline formats, to get outstanding results. The following are some of the key concepts taught in NCERT books for Class 11 Physics Chapter 4:

SCALARS AND VECTORS

Position and Displacement Vectors

Equality of Vectors

MULTIPLICATION OF VECTORS BY REAL NUMBERS

ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD

RESOLUTION OF VECTORS

VECTOR ADDITION – ANALYTICAL METHOD

MOTION IN A PLANE

Position Vector and Displacement

MOTION IN A PLANE WITH CONSTANT ACCELERATION

RELATIVE VELOCITY IN TWO DIMENSIONS

PROJECTILE MOTION

UNIFORM CIRCULAR MOTION

NCERT Solutions for Class 11 Physics Physics Chapter 4 Motion on a Plane covers all of the major themes in detail to assist students better comprehend the concepts. NCERT Solutions for Class 11 Physics Physics Chapter 4 Motion on a Plane are required reading for students preparing for their Class 11 Physics exams. Going over the solutions on this page can assist you in understanding how to approach and solve the challenges.

Students can also find NCERT intext, activities, and problems at the end of each Chapter. I'm also working on Chapter 4 of Physics for Class 11: Motion in a Plane. Students would benefit greatly from NCERT Solutions in completing their homework and assignments on time. Students may also download the NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane PDF to access the solutions even when they are not connected to the internet.

FAQs on NCERT Books Free Download for Class 11 Physics Chapter 4 - Motion in a Plane

1. What types of questions may be found in NCERT Solutions for Class 11 Physics Chapter 4 Motion on a Plane?

The following are the kind of questions included in NCERT Solutions for Class 11 Physics Chapter 4 Motion on a Plane:

Reasoning questions – based on basic understanding of the concepts and using the principles to describe a certain work.

Short Answers – Mostly objective based questions which needs little description

Long Answers – Subjective based answers with a little description of why the phenomenon worked.

Very Long answers – Subjective questions with detailed explanation of the principle it is working with.

Correcting the statements that have been given

2. Where can I get the NCERT Books Free Download for Class 11 Physics Chapter 4 - Motion in a Plane online?

Class 11 is a pivotal year in the lives of every student. Students' professional paths will be determined by their performance in Class 11. The faculty at Vedantu.com has produced solutions for all of the Chapters based on the most recent term – I CBSE Syllabus for this purpose. Vedantu.com provides NCERT Solutions in PDF format for students to download. The solutions are written with first-term test preparation in mind for all students, regardless of cognitive level.

3. Why is it essential to use NCERT Books Free Download for Class 11 Physics Chapter 4 - Motion in a Plane?

NCERT Solutions Class II Physics Chapter 4: Motion on a Plane strengthens intellectual understanding. Because the textbook lacks a full description of the syllabus or answers, it gives a comprehensive perspective of the mandated curriculum. All of the main theorems and formulae in Class ll have been solved in their entirety, which aids in the formation of concepts and the discovery of new connections between them. Significantly, the Class Tl NCERT Solution Physics Chapter 4: Motion on a Plane is a useful resource for students preparing for competitive exams like NRRT, JEE Main/Advanced, BITSAT, and VITEEE, among others. The knowledge of ideas becomes straightforward and in-depth when the NCERT book is supported by answers. Furthermore, because it is meant to provide a step-by-step discussion, students can go right to the solutions.

4. Why should you choose NCERT Books Free Download for Class 11 Physics Chapter 4 - Motion in a Plane?

Here are several compelling reasons to use the NCERT Solutions Class Il Physics Chapter 4: Motion on a Plane.

Has 100 percent Reasoning Purely supervised responses

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Provides in-depth information with notes following the solutions that provide shortcuts, recommendations, alternate techniques, and important points to remember.

Provides a short review of the ideas covered in each Chapter, as well as crucial terminology and formulas, serving as a prepared reminder.

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5. How do you prepare for Class 11 Physics Chapter 4 'Motion in a Plane' using NCERT Books Free Download for Class 11 Physics Chapter 4 - Motion in a Plane?

Try to acquire the finest study guide for Chapter 4 of Class 11 Physics, 'Motion in a Plane.' Swinging around many novels may sometimes cause turmoil. Instead, start with one and finish it entirely before moving on to the others. You may also go to Vedantu, which has a definitive guide as well as examples of test problems to help you do well in Class 11 Physics Chapter 4 'Motion in a Plane.' The solutions supplied are completely free of charge. They may also be found on the Vedantu Mobile app.

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CBSE Worksheets for Class 11 Physics

CBSE Worksheets for Class 11 Physics: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 11 Physics Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. So in order to help you with that, we at WorksheetsBuddy have come up with Kendriya Vidyalaya Class 11 Physics Worksheets for the students of Class 11. All our CBSE NCERT Class 11 Physics practice worksheets are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance. These chapter wise test papers for Class 11 Physics will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 11 Physics Number of Worksheets: 103

CBSE Class 11 Physics Worksheets PDF

All the CBSE Worksheets for Class 11 Physics provided in this page are provided for free which can be downloaded by students, teachers as well as by parents. We have covered all the Class 11 Physics important questions and answers in the worksheets which are included in CBSE NCERT Syllabus. Just click on the following link and download the CBSE Class 11 Physics Worksheet. CBSE Worksheets for Class 11 Physics can also use like assignments for Class 11 Physics students.

CBSE Worksheets for Class 11 Physics Gravitation Assignment 1
CBSE Worksheets for Class 11 Physics Gravitation Assignment 2
CBSE Worksheets for Class 11 Physics Kinetic Theory Assignment 1
CBSE Worksheets for Class 11 Physics Kinetic Theory Assignment 2
CBSE Worksheets for Class 11 Physics Laws of Motion Assignment 1
CBSE Worksheets for Class 11 Physics Laws of Motion Assignment 2
CBSE Worksheets for Class 11 Physics Laws of Motion Assignment 3
CBSE Worksheets for Class 11 Physics Mechanical Properties of Fluids Assignment 1
CBSE Worksheets for Class 11 Physics Mechanical Properties of Fluids Assignment 2
CBSE Worksheets for Class 11 Physics Mechanical Properties of Fluids Assignment 3
CBSE Worksheets for Class 11 Physics Mechanical Properties of Fluids Assignment 4
CBSE Worksheets for Class 11 Physics Mechanical Properties of Fluids Assignment 5
CBSE Worksheets for Class 11 Physics Mechanical Properties of Solids Assignment 1
CBSE Worksheets for Class 11 Physics Mechanical Properties of Solids Assignment 2
CBSE Worksheets for Class 11 Physics Motion in a Plane Assignment 1
CBSE Worksheets for Class 11 Physics Motion in a Plane Assignment 2
CBSE Worksheets for Class 11 Physics Motion in a Plane Assignment 3
CBSE Worksheets for Class 11 Physics Motion in a Straight Line Assignment 1
CBSE Worksheets for Class 11 Physics Motion in a Straight Line Assignment 2
CBSE Worksheets for Class 11 Physics Oscillations Assignment 1
CBSE Worksheets for Class 11 Physics Oscillations Assignment 2
CBSE Worksheets for Class 11 Physics Oscillations Assignment 3
CBSE Worksheets for Class 11 Physics Physical World Assignment 1
CBSE Worksheets for Class 11 Physics Physical World Assignment 2
CBSE Worksheets for Class 11 Physics Physical World Assignment 3
CBSE Worksheets for Class 11 Physics Physical World Assignment 4
CBSE Worksheets for Class 11 Physics System of Particle and Rigid Body Assignment 1
CBSE Worksheets for Class 11 Physics System of Particle and Rigid Body Assignment 2
CBSE Worksheets for Class 11 Physics Systems of Particles and Rotational Motion Assignment 1
CBSE Worksheets for Class 11 Physics Systems of Particles and Rotational Motion Assignment 2
CBSE Worksheets for Class 11 Physics Systems of Particles and Rotational Motion Assignment 3
CBSE Worksheets for Class 11 Physics Thermal Properties of Fluids Assignment 1
CBSE Worksheets for Class 11 Physics Thermal Properties of Fluids Assignment 2
CBSE Worksheets for Class 11 Physics Thermal Properties of Fluids Assignment 3
CBSE Worksheets for Class 11 Physics Thermal Properties of Fluids Assignment 4
CBSE Worksheets for Class 11 Physics Thermodynamics Assignment 1
CBSE Worksheets for Class 11 Physics Thermodynamics Assignment 2
CBSE Worksheets for Class 11 Physics Units and Measurement Assignment 1
CBSE Worksheets for Class 11 Physics Units and Measurement Assignment 2
CBSE Worksheets for Class 11 Physics Units and Measurement Assignment 3
CBSE Worksheets for Class 11 Physics Waves Assignment 1
CBSE Worksheets for Class 11 Physics Waves Assignment 2
CBSE Worksheets for Class 11 Physics Waves Assignment 3
CBSE Worksheets for Class 11 Physics Waves Assignment 4
CBSE Worksheets for Class 11 Physics Waves Assignment 5
CBSE Worksheets for Class 11 Physics Work Energy and Power Assignment 1
CBSE Worksheets for Class 11 Physics Work Energy and Power Assignment 2
CBSE Worksheets for Class 11 Physics Work Energy and Power Assignment 3
CBSE Worksheets for Class 11 Physics Work Energy and Power Assignment 4
CBSE Worksheets for Class 11 Physics Work Energy and Power Assignment 5
CBSE Worksheets for Class 11 Physics Motion In A Straight Line Assignment 1
CBSE Worksheets for Class 11 Physics Motion In A Straight Line Assignment 2
CBSE Worksheets for Class 11 Physics Motion In A Straight Line Assignment 3
CBSE Worksheets for Class 11 Physics System of Particles Assignment 1
CBSE Worksheets for Class 11 Physics System of Particles Assignment 2
CBSE Worksheets for Class 11 Physics System Of Particles And Rotational Motion Assignment
CBSE Worksheets for Class 11 Physics Thermal Properties of Matter Assignment 1
CBSE Worksheets for Class 11 Physics Thermal Properties of Matter Assignment 2
CBSE Worksheets for Class 11 Physics Unit and Measurements Assignment 1
CBSE Worksheets for Class 11 Physics Unit and Measurements Assignment 2
CBSE Worksheets for Class 11 Physics Units And Measurement Assignment
CBSE Worksheets for Class 11 Physics Assignment 1
CBSE Worksheets for Class 11 Physics Assignment 2
CBSE Worksheets for Class 11 Physics Assignment 3
CBSE Worksheets for Class 11 Physics Assignment 4
CBSE Worksheets for Class 11 Physics Assignment 5
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CBSE Worksheets for Class 11 Physics Assignment 11
CBSE Worksheets for Class 11 Physics Assignment 12
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CBSE Worksheets for Class 11 Physics Assignment 14
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CBSE Worksheets for Class 11 Physics Assignment 16
CBSE Worksheets for Class 11 Physics Assignment 17
CBSE Worksheets for Class 11 Physics Assignment 18
CBSE Worksheets for Class 11 Physics Assignment 19
CBSE Worksheets for Class 11 Physics Assignment 20
CBSE Worksheets for Class 11 Physics Assignment 21
CBSE Worksheets for Class 11 Physics Assignment 22
CBSE Worksheets for Class 11 Physics Assignment 23

Advantages of CBSE Class 11 Physics Worksheets

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One of the helpful resources used in classroom revision.
CBSE Class 11 Physics Workbook Helps to improve subject-knowledge.
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Worksheets of CBSE Class 11 Physics are devised by experts of WorksheetsBuddy experts who have great experience and expertise in teaching Maths. So practising these worksheets will promote students problem-solving skills and subject knowledge in an interactive method. Students can also download CBSE Class 11 Physics Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 11 Physics Worksheets PDF .

Now that you are provided all the necessary information regarding CBSE Class 11 Physics Worksheet and we hope this detailed article is helpful. So Students who are preparing for the exams must need to have great solving skills. And in order to have these skills, one must practice enough of Class 11 Physics revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus. Working on CBSE Class 11 Physics Worksheets will be a great help to secure good marks in the examination. So start working on Class 11 Physics Worksheets to secure good score.

CBSE Worksheets For Class 11

Class XI Physics Notes

Class 11 physics notes, chapter 1 - measurements, numerical problems, important mcqs, chapter 2 - vectors and equilibrium, chapter 3 - motion and force, chapter 4 - work and energy, chapter 5 - rotational and circular motion, chapter 6 - fluid dynamics, chapter 7 - oscillations, chapter 8 - waves, chapter 9 - physical optics, chapter 10 - thermodynamics.

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Class 11 physics notes fbise & kpk board, 11th class physics chapter 1 measurement.

Chapter 1 Theory (complete)
Chapter 1 Assignments (complete)
Chapter 1 Conceptual Questions (complete)
Chapter 1 Numerical Questions (complete)

11th Class Physics Chapter 2 Vectors & Equilibrium

Chapter 2 Theory (complete)
Chapter 2 Assignments (complete)
Chapter 2 Conceptual Questions (complete)
Chapter 2 Numerical Questions (complete)

11th Class Physics Chapter 3 Forces & Motion

Chapter 3 Theory (complete)
Chapter 3 Assignments (complete)
Chapter 3 Conceptual Questions (complete)
Chapter 3 Numerical Questions (complete)

11th Class Physics Chapter 4 Work & Energy

Chapter 4 Theory (complete)
Chapter 4 Conceptual Questions (complete)

11th Class Physics Chapter 5 Rotational & Circular Motion

Chapter 5 Theory (complete)
Chapter 5 Conceptual Questions (complete)
Chapter 5 Assignments (selected)
Chapter 5 Numerical Questions (selected)

11th Class Physics Chapter 6 Fluid Dynamics

Chapter 6 Theory (complete)
Chapter 6 Conceptual Questions (complete)
Chapter 6 Assignments & Numericals (selected)

11th Class Physics Chapter 7 Oscillation

Chapter 7 Theory (complete)
Chapter 7 Conceptual Questions (complete)
Chapter 7 Assignments (selected)
Chapter 7 Numerical Question s (selected)

11th Class Physics Chapter 8 Waves

Chapter 8 Theory (complete)
Chapter 8 Conceptual Questions (complete)
Chapter 8 Assignments (selected)
Chapter 8 Numerical Questions (selected)

11th Class Physics Chapter 9 Physical Optics Notes

Chapter 9 Theory (complete)
Chapter 9 Conceptual Questions (complete)
Chapter 9 Assignments (selected)
Chapter 9 Numerical Questions (selected)

11th Class Physics Chapter 10 Thermodynamics

Chapter 10 Theory (complete)
Chapter 10 Conceptual Questions (complete)
Chapter 10 Assignments (selected)
Chapter 10 Numerical Questions (selected)

91 comments:

Very excillent information thanks brother

Assalamu alaikum

When will unit 10 full syllabus will be ready

How to get full note in hard form

sir can you tell me the password that you given to pdf files for locking the editing

So you not solve the MCQs

stop complaining

Its solved sister

Tusa saryan ni mao ch lan rakhan

hey are you mad ??????? STOP CURSING MAN

why you lots abusing bruvvv grow the hell up average 17 year olds cant even respect anyone these days.

Please upload HSSC 1 physics chapters full notes from chapter 4 to 10 ... Plz

Yes upload full sallybus

Yes😩😩we can't survive

Sir, Can u upload ch 4 to ch 10 full syllabus.

it is a good job from your side sir please upload the full sylabu.

please update the chapters

Numercels in short form pkz upload

please upload full syllabus

Please upload complete notes please

This notes are not show in pdf it want some password to your notes why?

Sir plz upload full syllabus conceptual as soon as possible

Sir plz upload full syllabus conceptual questions as soon as possible

Excellent notes Sir it's a humble request that our exams will held in june so plz fastly upload notes of all subjects (full syllabus)

Fastly isn't a word Einstein .....

Please upload full syllabus notes

Are these notes for smart syllabus or Full syllabus?

PLZ UPLOAD FULL NOTES PLZ PLZ PLZ PLZ PLZ

kindly upload full syllabus

Plzz upload important topics for 11 class physics

sir upload full syllabus notes

sir upload 11th physics important topics fbise

sir plzzzz upload chap 4 😭😭😭😭

chap 4 😭😭😭😭😭

Please upload full notes

Sir plzzz chapter 1and 2 ka notes full syllabus k according h

Complete 1st year course sir notes and lectures

Plz upload solutions of mcqs

Impressive Note sir plz plz upload the other chapter. Waiting 😶😶

sir plzz no response no result plzz sir give me a important topics and important question

khud nots bana lo bachy mery pass time nahi hy

beta sab yaad karo

Please upload full notes plz

sir plz upload full syllabus notes of class 11

Comprehensive question nai ha in notes ma

Sir u r notes r outstanding in how much time more notes will be prepare

sir I m a student of KPK board I am preparation for MDCAT TEST in these notes .are you make these notes from fadral board?

Bacha mera aap is english ky sath exam kesy pass karo gy , yes federal board is consistent with it

English doesn't matters

Plz upload full syllabus plzzzz

sir try to upload full syllabus notes soon extremely waiting for it

SIR NOTES OF COMPREHENSIVE QUESTIONS

Very good notes but not complete please complete the notes

Sir please upload full syllabus notes. Your notes are too good.

Lajawab ha sir

exams are coming sir , it would be alot more easy to prepare if u share the full syllabus. we all would be obliged. thank u

Kindly upload full syllabus notes from chapter 4 to 10

Sir these notes are great and easy to understand kindly upload full syllabus notes of remaining chapters it Wil be a lot easier for us to prepare for tests and exams

AOA sir your notes are exellent and usefull but kindly upload full syllabus

Chapter 10 Conceptual Questions And Chapter 9 Conceptual Questions Are Full Syllabus Not Reduced

please post full notes as soon as possible and DONT forget to include answers of quizes and and pont to ponder boxes ,, Waiting , JazakAllah

sir ap bilkul tension na lain, arram say notes bnayn, mai board ky papers ky baad inshaAllah tafseel say prhun ga

aoa sir please complete fast full syllabus becuase exams is near

Plz make complete notes

Board 2023 se pehle mil sakte hain sare?!!

sir can you please make full notes..it would be extremely helpful..

Pehla pta hota to parh leta ab akhri din muh marna Aya Hun🥷

where are numerical question of c.h.p 4

Notes are very good

Very Good Notes..............

Chapter 4 numerical

Chapter 4 numericals

aoa plz upload the comprehension questions answers of chapter no 1 measurement

numericals of chapter no 4 please

thanks for your notes

Awesome notes

Chapter 4 numericals please

Very informative notes Great

How to download ?

IT WAS A GREAT HELP

Very very very nice Sir MashaAllah .THANKU

Jazak Allah Sir

Where are the MCQs potions?

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AssignmentsBag.com

Assignments For Class 11 Physics

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You can get free PDF downloadable assignments for Grade 11 Physics from our website which has been developed by teachers after doing extensive research in each topic.

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IMAGES

NCERT Solutions for Class 11 Physics Chapter 4
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane
Ncert Solution for Class 11 Physics Chapter 4
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane
Motion in a Plane Class 11 Notes CBSE Physics Chapter 4 [PDF]

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Class 11 Physics समतल में गति
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Class 11 Physics Chapter 4 Motion in Plane
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Class-11 physics chapter 5 motion in plane notes 📚 #jee #neet #physics

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Class 11 Physics Assignments Download Pdf with Solutions
All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Physics Class 11. Students can click on the links below and download all Pdf Assignments for Physics class 11 for free. All latest Kendriya Vidyalaya Class 11 Physics Assignments with Answers and test papers are given below.
NCERT Solutions for Class 11 Physics Chapter 4 Motion in A Plane
NCERT Solutions Class 11 Physics Chapter 4 - Free PDF Download *According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 3. NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane is one of the important study materials for Class 11 students as it is one of the most scoring chapters in kinematics. Students must try to understand each and every topic in a ...
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane
Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane: Question 4. 1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane
Question 5. (a) The magnitude of a vector is always a scalar. (b) Each component of a vector is always a scalar. (e) Three vectors not lying in a plane can never add up to give a null vector.". (a) True; because magnitude of a vector is a pure number. (b) False; as each component of a given vector is always a vector.
NCERT Solutions for Class 11 Physics (Updated for 2023
NCERT Solutions for Class 11 Physics (Chapter-wise) Chapter 1: Units and Measurements. Chapter 2: Motion in a Straight Line. Chapter 3: Motion in a Plane. Chapter 4: Laws of Motion. Chapter 5: Work, Energy, and Power. Chapter 6: System of Particles and Rotational Motion. Chapter 7: Gravitation.
NCERT Solutions for Class 11 Physics Chapter 4 Motion In A Plane.
There are a variety of questions present in Chapter 4 'Motion in a Plane' in NCERT Solutions for Class 11 Physics. The questions will range from detailed to short answer-type. This will contain pick/choose the one out, true or false, reason stating questions, numerical or problem questions, long answer questions etc.
PDF NCERT Solutions for Class 11 Physics Chapter 4
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane Q-11: A passenger arriving in a new town wants to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. (a) What is the average speed of the taxi?
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane
The solutions for NCERT Class 11 Physics Chapter 4, which pertains to "Motion in a Plane," build upon the foundation laid in chapters 2 and 3. This chapter begins by introducing the concept of vectors. To grasp the motion in a plane NCERT solutions, a clear understanding of vector addition, subtraction, and scalar multiplication is indispensable.
NCERT Solutions for Class 11 Physics chapter-4 Motion In A Plane
NCERT Solutions For Class-11 Physics Chapter Wise. Chapter 1 Physical World. Chapter 2 Units and Measurements. Chapter 3 Motion In A Straight Line. Chapter 4 Motion In A Plane. Chapter 5 Laws of Motion. Chapter 6 Work, Energy and Power. Chapter 7 System of Particles and Rotational Motion. Chapter 8 Gravitation. Chapter 9 Mechanical Properties ...
Motion in a Plane Class 11 Notes Physics Chapter 4
Motion in a Plane Class 11 Notes Physics Chapter 4. • Motion in a plane is called as motion in two dimensions e.g., projectile motion, circular motion etc. For the analysis of such motion our reference will be made of an origin and two co-ordinate axes X and Y. • Scalar and Vector Quantities. Scalar Quantities.
NCERT Books Free Download for Class 11 Physics Chapter 4
Developed step-by-step to ensure a complete process to understand. Download PDF. Free NCERT Books download for Class 11 Physics Chapter 4 - Motion in a Plane on Vedantu.com. Students can also download the NCERT Textbooks Solutions in PDF for Class 6 to 12 all subjects. Register for Physics tuition to clear your doubts and score more in your exams.
Solved Assignments Chapter 4 Physics Class 11 Forces and ...
Assignment 4.1, 4.2 and 4.3 | Chapter 4 Work and Energy | Class 11 Physics | FBISE | Study With Me=====Chapter Wise All Assignments=====Chapter 1 https://you...
CBSE Worksheets for Class 11 Physics
These chapter wise test papers for Class 11 Physics will be useful to test your conceptual understanding. Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 11 Physics Number of Worksheets: 103. ... CBSE Worksheets for Class 11 Physics Waves Assignment 4;
NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane
Chapters Covered in Class 11 Physics. Chapter 1: Physical World. Chapter 2: Units and Measurements. Chapter 3: Motion in a Straight Line. Chapter 4: Motion in a Plane. Chapter 5: Laws of Motion. Chapter 6: Work, Energy, and Power. Chapter 7: System of Particles and Rotational Motion. Chapter 8: Gravitation.
Case Study Questions for Class 11 Physics Chapter 4 Motion In A Plane
Case Study Questions for Class 11 Physics Chapter 4 Motion in a Plane Case Study Questions: Question 1: Projectile motion is a form of motion in which an object or particle is thrown with some initial velocity near the earth's surface and it moves along a curved path under the action of gravity alone. The … Continue reading Case Study Questions for Class 11 Physics Chapter 4 Motion In A Plane
Assignment 4.1
Assignment 4.1 | Class 11 Physics Chapter 4 Work and Energy | Federal Board KP Board| Study with ME=====Chapter Wise All Assignments=====Chapter 1 https://yo...
Class 11 Physics Ch.4 Work and Energy
Download Class 11 Physics Ch.4 Work and Energy PDF Notes ⇒ Class 11 Physics Ch.4 Work and Energy Theory Notes ⇒ Class 11 Physics Ch.4 Work and Energy Conceptual and Numerical Questions
Class 11 Physics Notes for FBISE by ClassNotes
Class 11 Physics Notes are free and will always remain free. We will keep adding updated notes, past papers, guess papers and other materials with time. We will also introduce a mobile app for viewing all the notes on mobile. Make sure to comment down your experience regarding our website. Also tell us what other features and resources would ...
Class 11 Physics Notes FBISE & KPK Board
11th Class Physics Chapter 4 Work & Energy. Chapter 4 Theory (complete) Chapter 4 Conceptual Questions (complete) 11th Class Physics Chapter 5 Rotational & Circular Motion. Chapter 5 Theory (complete) Chapter 5 Conceptual Questions (complete) Chapter 5 Assignments (selected) Chapter 5 Numerical Questions (selected) 11th Class Physics Chapter 6 ...
Assignments For Class 11 Physics
Assignments Assignments for Class 11. Assignments for Class 11 Physics have been developed for Standard 11 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 11 Physics from our website as we have provided all topic ...