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

Edited By Vishal kumar | Updated on Aug 24, 2023 11:52 AM IST

## 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.

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 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.

• #### Projectile motion

Time of flight:

T=2usinΘ/g

Maximum range:

R= u2sin2Θ/g, Rmax =u2/g

Note: Range is maximum when sin2θ is maximum

Maximum height:

H= u2sin2Θ/2g

• #### Relative velocity

Vab=va-vb

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.

### 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.

**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.

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

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

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 :

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 :

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

(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.

(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 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.

(a) True.

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

(b) False.

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

(c) False.

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.

(d) True

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.

(e) True

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

( a) $\left | a+b \right |\leq \left | a \right |+\left | b \right |$

Consider the image given below :-

In $\Delta$ OCB,

$OB < OC+BC$

or $\left | a+b \right | < \left | a \right |+\left | b \right |$ ................................................(i)

But if $a\ and\ b$ are in a straight line then $|a+b|\ =\ |a|\ +\ |b|$ ...............................................(ii)

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

$\left | a+b \right |\leq \left | a \right |+\left | b \right |$

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

Consider the given image:

In $\Delta$ OCB, we have:

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

or $OB+BC > OC$

or $OB > \left | OC - BC \right |$

or $|a+b|\ >\ |\; |a|-| b|\; |$ ............................................................(i)

Also, if $a\ and\ b$ are in a straight line but in the opposite direction then

$|a+b|\ =\ |\; |a|-| b|\; |$ ...............................................................(ii)

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

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

(c) $\left | a-b \right |\leq \left | a \right |+\left | b \right |$

Consider the image given below:-

In $\Delta$ OAB, we have

$OB < OA+AB$

or $\left | a-b \right |\ < \left | a \right |+\left | b \right |$ ...........................................................(i)

For vectors in a straight line, $\left | a-b \right |\ =\ \left | a \right |+\left | b \right |$ ...........................................................(ii)

From (i) and (ii) we get :

$\left | a-b \right |\leq \left | a \right |+\left | b \right |$

(d) $\left | a-b \right |\geq \left | | a\right |-\left | b \right ||$

Consider the image given below :

In $\Delta$ OAB, we have :

$OB +AB > OA$

or $OB > \left | OA - AB \right |$

or $\left | a-b \right |\ > \left | | a\right |-\left | b \right ||$ ...................................................(i)

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

$\left | a-b \right |\ = \left | | a\right |-\left | b \right ||$ .........................................................(ii)

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

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

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

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

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and 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.

(c) Correct :- We have a + b + c + d = 0

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

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.

(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 .

(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 .

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

(c) the average speed of the cyclist?

(c) For finding average speed we need to calculate the total path travelled.

Total path = OP + arc PQ + OQ

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

$= 3.57\ Km$

$Time\ taken\ in\ hour\ = \frac{1}{6}$

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

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

or $=\ \sqrt{(500)^2\ +\ (500) ^2\ +\ 2(500)(500) \cos 60^{\circ}}$

$=\ 866.03\ m$

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

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

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

$=\ \frac{23\ Km}{\frac{28}{60}\ h}$

$=\ 49.29\ Km/h$

Total displacement = 10 Km

Total time taken in hours :

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

Avg. velocity :

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

$=\ 21.43\ Km/h$

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

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.

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

$=\ 30\ +\ (-10)$

$=\ 20\ m/s$

And the angle is given by :

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

$\tan\Theta \ =\ \frac{10}{30}$

$\Theta \ \approx \ 18^{\circ}$

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

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

Time taken to cross the river will be :

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

$=\ \frac{1}{4}\ =\ 15\ min.$

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

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

$=\ 3\times\frac{1}{4}\ =\ 0.75\ Km$

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,

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

$\tan \beta \ =\ \frac{51}{50.8 }$

Thus $\beta \ =\ \tan^{-1} \frac{51}{50.8 }$

$=\ 45.11^{\circ}$

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

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 :

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

So, we get

$\sin \Theta \ =\ 0.5534$ and $\Theta \ =\ 33.60^{\circ}$

Now the horizontal range can be found from :

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

or $=\ \frac{40^2 \sin (2\times 33.6 )}{9.8}$

$=\ 150.53\ m$

We are given the range of projectile motion.

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

Substituting values :

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

So, $\frac{u^2 }{g}\ =\ 100$

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 :

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

or $H\ =\ 50\ m$

Frequency is given by :

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

$=\ \frac{14}{25}\ Hz$

And, the angular frequency is given by :

$\omega \ =\ 2 \Pi f$

Thus, $=\ 2 \times\ \frac{22}{7}\times\frac{14}{25}$

$=\ \frac{88}{25}\ rad/s$

Hence the acceleration is given by :

$a\ =\ \omega ^2 r$

or $=\ \left ( \frac{88}{25} \right ) ^2 \times 0.8$

or $a\ =\ 9.91\ m/s$

Convert all the physical quantities in SI units.

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

So the acceleration is given by :

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

$=\ \frac{(250)^2}{1000}$

$=\ 62.5\ m/s^2$

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 :

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 :

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.

(a) Find the $v$ and $a$ of the particle?

(a) We are given the position vector $r=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$

The velocity vector is given by:-

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

$v\ =\ \frac{d\left ( 3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k} \right )}{dt}$

or $v\ =\ 3\; \hat{i}-4t\; \hat{j}$

Now for acceleration :

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

$=\ d(\frac{3\; \hat{i}-4t\; \hat{j})}{dt}$

$=\ -4\; \hat{j}$

(b) What is the magnitude and direction of velocity of the particle at $t=2.0\; s ?$

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

$v\ =\ 3\; \hat{i}-4t\; \hat{j}$

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

or $=\ 3\; \hat{i}-8\; \hat{j}$

Thus the magnitude of velocity is :

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

Direction :

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

(a) At what time is the x- coordinate of the particle $\inline 16 \; m?$ What is the y-coordinate of the particle at that time

We are given the velocity of the particle as $10.0\; \hat{j}\; m/s$ .

And the acceleration is given as :

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

So, the velocity due to acceleration will be :

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

So, $dv\ =\ \left ( 8.0\ \widehat{i}\ +\ 2.0\ \widehat{j} \right )dt$

By integrating both sides,

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

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

Now,

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

or $dr\ =\ \left ( 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u \right )dt$

Integrating both sides, we get

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

or $r\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}$

or $x\widehat{i}\ +\ y\widehat{j}\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}$

Comparing coefficients, we get :

$x\ =\ 4t^2$ and $y\ =\ 10t\ +\ t^2$

In the question, we are given x = 16.

So t = 2 sec

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

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

The velocity of the particle is given by :

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

Put t = 2 sec,

So velocity becomes :

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

or $v\ =\ 16\ \widehat{i}\ +\ 14\ \widehat{j}$

Now, the magnitude of velocity gives :

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

$=\ \sqrt{256+196}$

$=\ 21.26\ m/s$

Let A be a vector such that:- $\overrightarrow{A}\ =\ \widehat{i}\ +\ \widehat{j}$

Then the magnitude of vector A is given by : $\left | A \right |\ =\ \sqrt{1^2\ +\ 1^2}\ =\ \sqrt{2}$

Now let us assume that the angle made between vector A and x-axis is $\Theta$ .

Then we have:-

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

Similarly, let B be a vector such that:- $\overrightarrow{B}\ =\ \widehat{i}\ -\ \widehat{j}$

The magnitude of vector B is : $\left | B\right |\ =\ \sqrt{1^2\ +\ (-1)^2}\ =\ \sqrt{2}$

Let $\alpha$ be the angle between vector B and x-axis :

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

Now consider $\overrightarrow{C}\ =\ 2\widehat{i}\ +\ 3\widehat{j}$ :-
Then the required components of a vector C along the directions of is:- $=\ \frac{2+3}{\sqrt{2}}\ =\ \frac{5}{\sqrt{2}}$
and the required components of a vector C along the directions of is:-
$\frac{2-3}{\sqrt{2}}\ =\ \frac{-1}{\sqrt{2}}$

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

(b) $v_{average}=\left [ r\left ( t_{2} \right ) -r\left ( t_{1} \right )\right ]/\left ( t_{2}-t_{1} \right )$

(c) $v(t)=v\left ( 0 \right )+a\; t$

(d) $v(t)=r\left ( 0 \right )+v\left ( 0 \right )t+\left ( 1/2 \right )a\; t^{2}$

(e) $a_{average}=\left [ v(t_{2})-v(t_{1}) \right ]/\left ( t_{2}-t_{1} \right )$

(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.

(c) False: - The given equation is valid only in case of uniform acceleration motion.

(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.

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 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.

(c) False:- Since speed is a scalar quantity but has dimensions.

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

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

The given situation is shown in the figure:-

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

Consider $\Delta$ ABD,

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

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

or $AC\ =\ 2AB\ =\ \ 2BD\ \times \tan 15^{\circ}$

or $=\ \ 2\times 3400\times \tan 15^{\circ}$

or $=\ 1822.4\ m$

Thus, the speed of aircraft :

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

### Motion in plane class 11 NCERT solutions additional exercise :

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.

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.

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

(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.

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

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

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

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

or $\frac{u^2}{g}\ =\ 2\sqrt{3}$

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

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

or $=\ 3.46\ Km$

Thus the bullet cannot travel up to 5 Km.

According to the question the situation is shown below:-

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

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

or $\sin \Theta \ =\ \frac{v}{u}$

or $=\ \frac{200}{600}$

So, $\Theta \ =\ 19.5^{\circ}$

So, the required height will be:-

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

or $=\ \frac{600^2\ \cos^2 \Theta }{2g}$

or $=\ 16006.48\ m$

or $=\ 16\ Km$

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

The situation is shown in figure :-

The centripetal acceleration is given by :

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

$=\ \frac{(7.5)^2}{80}$

$=\ 0.7\ m/s^2$

And the tangential acceleration is given as $0.5\ m/s^2$ .

So, the net acceleration becomes :

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

or $=\ \sqrt{(0.7)^2\ +\ (0.5) ^2}$

or $=\ 0.86\ m/s^2$

Now for direction,

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

or $=\ \frac{0.7}{0.5}$

Thus, $\Theta \ =\ 54.46^{\circ}$

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

$v_y\ =\ v_{oy}\ =\ gt$ and $v_x\ =\ v_{ox}$

Now,

$\tan \Theta \ =\ \frac{v_y}{v_x}$

or $=\ \frac{v_{oy}\ -\ gt}{v_{ox}}$

Thus, $\Theta \ =\ \tan^{-1} \left ( \frac{v_{oy}\ -\ gt }{v_{ox}} \right )$

(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 :

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

Hence $\Theta \ =\ \tan^{-1} \left ( \frac{4h }{R} \right )$

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

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

 Section Name Topic Name 4 Motion in a plane 4.1 Introduction 4.2 Scalars and vectors 4.3 Multiplication of vectors by real numbers 4.4 Addition and subtraction of vectors – graphical method 4.5 Resolution of vectors 4.6 Vector addition – analytical method 4.7 Motion in a plane 4.8 Motion in a plane with constant acceleration 4.9 Relative velocity in two dimensions 4.10 Projectile motion 4.11 Uniform circular motion

### 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.

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

• 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 Class 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.

## NCERT Solutions for Class 11 Physics Chapter Wise

 Chapter 1 Physical world Chapter 2 Chapter 3 Motion in a straight line Chapter 4 Motion in a Plane Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15

NCERT Physics Exemplar Solutions Class 11 For All The 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

## Subject wise NCERT Exemplar solutions

1. What is the weightage of the chapter for JEE Main

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.

2. How important is the chapter for NEET

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.

3. Which type of questions is present in chapter 4 physics class 11 ncert solutions ?

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

1. Conceptual Questions

2. Numerical Questions

3. Multiple-Choice Questions

4. Diagram-Based Questions

4. Where can I get the class 11 physics chapter 4 exercise solutions?

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.

5. What is the name of chapter 4 physics class 11?

Motion in a Plane

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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9
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3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### .NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### DevOps Architect

A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties.

2 Jobs Available
##### Cloud Solution Architect

Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

2 Jobs Available