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

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.

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NCERT Solutions Class 11 Physics Chapter 4 - CBSE Access Free PDF
NCERT Solutions for Class 11 Physics Chapter 4 Motion in A Plane
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.

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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.

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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.

1692853404976

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

$\begin{gathered} \vec{a} \times \vec{b}=\left(a_y b_z-a_z b_y\right) \hat{\imath}+\left(a_z b_x-a_x b_z\right) \hat{\jmath}+\left(a_x b_y-a_y b_x\right) \hat{k} \\ |\vec{a} \times \vec{b}|=a b \sin \theta \end{gathered}$

Projectile motion

Time of flight:

T=2usinΘ/g

Maximum range:

R= u²sin2Θ/g, R_max =u²/g

Note: Range is maximum when sin2θ is maximum

Maximum height:

H= u²sin²Θ/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.

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

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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

Answer:

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.

Answer:

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.

Answer:

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

Answer:

(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.

Answer:

(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.

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.

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

Answer:

Consider the image given below :- 1650435211918

In $\Delta$ OCB,

$OB < OC+BC$

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

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

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

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

Answer:

Consider the given image:

1650435245232

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 :

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

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

Answer:

Consider the image given below:- 1650435294332

In $\Delta$ OAB, we have

$OB < OA+AB$

From (i) and (ii) we get :

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

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

Answer:

Consider the image given below :

1650435343919

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 :

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

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?

Answer:

(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

Q. 4.8 Three girls skating on a circular ice ground of radius $200 m$ start from a point $P$ on the edge of the ground and reach a point $Q$ diametrically opposite to $P$ following 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 path skate?

1650435369363

Answer:

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.

Q. 4.9 (a) A cyclist starts from the centre $O$ of a circular park of radius $1\; km$ , reaches the edge $P$ of 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,

1650435405375

Answer:

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

Net displacement = Final position - Initial position .

Q. 4.9 (b) A cyclist starts from the centre $O$ of a circular park of radius $1\; km,$ reaches the edge $P$ of 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

(b) average velocity,

1650435442444

Answer:

(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 .

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

Q. 4.9 (c) A cyclist starts from the centre $O$ of a circular park of radius $1\; km,$ reaches the edge $P$ of 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

1650435479236

Answer:

(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 :

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

Q. 4.10 On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ 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.

Answer:

The track is shown in the figure given below:-

1650435677641

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

Q. 4.11 (a) 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,

Answer:

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

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

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

$=\ 49.29\ Km/h$

Q. 4.11 (b) 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

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

Answer:

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.

Q. 4.12 Rain is falling vertically with a speed of $30\; ms^{-1}$ A woman rides a bicycle with a speed of $10\; ms^{-1}$ . in the north to south direction. What is the direction in which she should hold her umbrella?

Answer:

The given situation is shown in the figure:-

1650435706988

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.

Q. 4.13 A man can swim with a speed of $4.0 \; km/h$ in still water. How long does he take to cross a river $1.0 \; km/h$ 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?

Answer:

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:-

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

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

Q. 4.14 In a harbour, wind is blowing at the speed of $72\; km/h$ and the flag on the mast of a boat anchored in the harbour 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?

Answer:

According to the question the figure is shown below:-

1650435734447

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.

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?

Answer:

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

$=\ 150.53\ m$

Q. 4.16 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?

Answer:

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 $H\ =\ 50\ m$

Q. 4.17 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?

Answer:

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 $a\ =\ 9.91\ m/s$

Q. 4.18 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.

Answer:

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 :

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

Answer:

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

Answer:

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 :

Answer:

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.

Q. 4.20 (a) The position of a particle is given by $r=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$ where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres.

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

Answer:

(a) We are given the position vector

The velocity vector is given by:-

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

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

Now for acceleration :

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

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

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

Q. 4.20 (b) The position of a particle is given by $r=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$ where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres.

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

Answer:

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 :

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

Q. 4.21 (a) A particle starts from the origin at $t=0\; s$ with a velocity of $10.0\; \hat{j}\; m/s$ and moves in the x-y plane with a constant acceleration of $\left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}$ .

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

Answer:

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

And the acceleration is given as :

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

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.

Q. 4.21 (b) A particle starts from the origin at $t=0\; s$ with a velocity of $10.0\; \hat{j}\; m/s$ and moves in the x-y plane with a constant acceleration of $\left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}.$

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

Answer:

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 :

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

$=\ \sqrt{256+196}$

$=\ 21.26\ m/s$

Q 4. 22 $\hat i$ and $\hat j$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat i + \hat j$ , and $\hat i - \hat j$ ? What are the components of a vector $A=2\hat i +3 \hat j$ along the directions of $\hat i + \hat j$ and $\hat i - \hat j$ ? [You may use graphical method]

Answer:

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 :-
Then the required components of a vector C along the directions of is:-
and the required components of a vector C along the directions of is:-

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

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

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

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

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

Answer:

(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

Answer:

(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.

Q. 4.25 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 $30^{\circ},$ what is the speed of the aircraft?

Answer:

The given situation is shown in the figure:-

1650436373264

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 :

Q. 4.26 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

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

(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.

Answer:

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

Q. 4.29 A bullet fired at an angle of $30^{\circ}$ 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 be fixed, and neglect air resistance.

Answer:

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.

Q. 4.30 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? $(Take\; g=10\; m\; s ^{-2}).$

Answer:

According to the question the situation is shown below:-

1650436404814

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 $=\ 16\ Km$

Q. 4.31 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?

Answer:

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

The situation is shown in figure :-

1650436449943

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 .

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

Q. 4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by $\theta (t)=tan^{-1}\left [ \frac{v_{0y-gt}}{v_{0x}} \right ]$

where the symbols have their usual meaning.

Answer:

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

Q. 4.32 (b) Shows that the projection angle $\theta _{0}$ for a projectile launched from the origin is given by

$\theta _{0}=tan^{-1}\left [ \frac{4h_{m}}{R} \right ]$

where the symbols have their usual meaning.

Answer:

(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.

Also Check NCERT Books and NCERT Syllabus here

NCERT Solutions for Class 11 Physics Chapter Wise

Chapter 1	Physical world
Chapter 2	Units and Measurement
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

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 solutions

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

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.

Also Read

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

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NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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

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

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

Vector Notation

Vector terminology

The magnitude of vector

Dot product of vector

Cross product of vector

Projectile motion

Relative velocity

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

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

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

Motion in plane class 11 NCERT solutions additional exercise :

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

NCERT Chapter 4 Physics class 11 - Weightage in Exams

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

Key Features of Motion in a Plane Class 11 NCERT Solutions

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NCERT Solutions for Class 11 Physics Chapter Wise

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