# NCERT Solutions Class 11 Physics Chapter 4

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

Chapter 4 Motion in a Plane is one of the most important chapters in the Class 11 NCERT textbook. The major topics covered in the chapter are – Scalars and Vectors, Vector Multiplication by Real Numbers, Vector Resolution, Vector Addition – Analytical Method, Relative Velocity in Two Dimensions, Projectile Motion, and Uniform Circular Motion.

We learned about the concepts of displacement, position, acceleration, and velocity in the previous chapter, which are important in explaining the motion of an object in a straight line. The ‘+’ and ‘-‘ signs can be used to take care of the direction aspect. Vectors, on the other hand, can be used to explain the motion of an object in two or three dimensions. Students can refer to NCERT Solutions to understand the basic concepts from this chapter.

### Motion in a Plane Class 11 NCERT Solutions

An understanding of problems in Chapter Motion in a Plane in class 11 helps students to answer theoretical questions as well as solve numerical problems.

 Section Number Section Title 4.1 Introduction 4.2 Scalars and Vectors 4.2.1 Position and Displacement Vectors 4.2.2 Equality of 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 4.7.1 Position Vector and Displacement 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

### 4.1 Introduction

Students will learn about oscillatory motion and its equations and properties in this section.

### 4.2 Scalars and Vectors

Quantities in physics are classified as scalars or vectors. A vector quantity has magnitude as well as direction.

### 4.2.1: Vectors of Position and Displacement:

The vector quantities of position and displacement are the same.

### 4.2.2: Vector Equality:

It’s possible that two vectors are equal. When their magnitude and direction are equal, it is possible.

### 4.3: Multiplication of Vectors by Real Numbers

In this section, students will learn that whenever a vector is multiplied with a real number, there will only be an increase in the magnitude, while the direction remains constant.

### 4.4 Addition and Subtraction of Vectors – Graphical Method

Using the graphical method, students will learn how to add and subtract two vectors. The law of parallelogram will apply to vectors.

### 4.5: Vector Resolution

Students will learn how to divide a given vector into two component vectors and a set of two vectors all in the same plane.

### 4.6 Vector Addition – Analytical Method

Analytical methods can also be used to add vectors. Combining the components of vectors makes it easier to add them.

### 4.7: Motion in a Plane

Students will learn how to use vectors to describe motion in two dimensions.

### 4.7.1: Position Vector and Displacement

A particle’s position vector in a plane with respect to the origin of an x-y reference frame.

### 4.8: Motion in a Plane with Constant Acceleration

With constant acceleration, if an object moves in the x-y plane with constant acceleration, the equation of motion is derived by dividing the average acceleration over a period of time.

### 4.9: Relative Velocity in Two Dimensions

Students will learn about relative velocity in two dimensions for motion along a straight line. They will also learn to extend motion in a plane or three dimensions.

### 4.10 Projectile Motion

A projectile is an object that flies after being thrown. The projectile motion and its equation will be studied by the students.

### 4.11 Uniform Circular Motion

Uniform circular motion occurs when an object follows a circular path at a constant speed.

Students are introduced to the concepts of scalars and vectors in Chapter 4. To gauge their understanding of the concepts, students should practise the questions and exercises given at the end of the chapter.

Extramarks also helps students in finding out the right answers to the questions. They can download NCERT Solutions Class 11 Physics Chapter 4 from the website and check if they derived the right answer or to get answers to the question.

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

While the graphical method of adding vectors aids in visualising the vectors and the resultant vector, it can be time-consuming and inaccurate. Adding vectors is much simpler when their components are combined. Class 11 Physics Chapter 4 NCERT Solutions will teach you that in one dimension, an object’s velocity and acceleration always follow the same straight line. However, for two or three-dimensional motion, velocity and acceleration vectors can have any angle between them between 0 and 180 degrees.

Motion in a plane is a crucial topic in CBSE class 11 Physics. Students must thoroughly prepare for this chapter in order to perform well in their board examination. The answers to the textbook questions can be found in NCERT Solutions for Class 11 Physics Chapter 4. It will also provide answers to key questions from previous year’s exam papers. The various types of problems and examples will assist the student in developing strong concepts of motion.

### CBSE Class 11 Physics Chapter 4 NCERT Solutions

We come across a variety of objects with oscillatory motions in our daily lives. NCERT Solutions for Class 11 Physics Chapter 4 assist students in comprehending this fundamental chapter of Physics. Students will learn about various terms and equations for planar motion in a two-dimensional plane.

### Benefits of Motion in a Plane Class 11 Solutions

NCERT Solutions Class 11 Physics Chapter 4 is a fantastic resource for finishing the syllabus and improving your understanding. It provides a step-by-step solution to each exercise question and assists students in completing their homework quickly.

Class 11 Physics Chapter 4 NCERT Solutions will improve students’ understanding of this chapter of Physics. It will assist students in applying their conceptual knowledge to solve the problems presented in the chapter in a simple and straightforward manner. Other benefits students will receive with these solutions include:

• The detailed Class 11 Physics Solutions will assist you in comprehending the topics in a better way.
• Students can refer to the solutions provided here if they get stuck on any question  while practising. It will also facilitate a thorough review of the entire syllabus.
• Students can improve their problem-solving skills by using the detailed and step-by-step solutions provided here.
• By using these solutions, you will be able to gain a better understanding of the entire subject.
• These solutions will also assist you in preparing for competitive examinations such as NEET and JEE.

## Tips to Prepare for Class 11 Physics

### Learn about the exam pattern and syllabus

Understanding the exam pattern and syllabus for Class 11 Physics is the most important thing for students. Students can prepare for their exams more effectively if they are familiar with the syllabus for class 11 Physics. The CBSE has changed the exam pattern for the current school year. As a result, students must be aware of the new exam pattern and syllabus for class 11 Physics.

### Use the NCERT physics textbooks for Class 11

NCERT books cover the entire class 11 Physics syllabus. Students can easily understand the concepts of Physics after reading all of the chapters in the books. The information provided in NCERT books is accurate and complete.

### Focus on important topics

To score good marks, students should focus more on the topics that are important from the exam point of view. Work, Energy, and Power, Thermal Properties of Matter, Rotational Motion, Thermodynamics, Kinetic Theory, Gravitation, Laws of Motion, and Waves are the most important topics.

### Make a study schedule

Students must create a study plan in order to prepare for the CBSE class 11 Physics syllabus. They should divide the syllabus appropriately so that they can cover all topics. Students should try to complete the entire syllabus one or two months before the final exams.

### Make a list of key points

Students should jot down all important formulas, theorems, definitions, and principles from the class 11 Physics book. This will allow them to properly revise before the exams. Revision notes will also help you understand the concepts better.

### Diagrams should be practised

Circuit diagrams and ray diagrams are important in physics, so students need to practise drawing them before they take the exam. To get good grades, students should memorise the labelling of these diagrams.

### Revision as needed

Students must thoroughly review the entire CBSE class 11 Physics Syllabus. Regular review of important topics will assist students in memorising information for final exams. This will also help to reduce the students’ stress and boost their confidence. It can also help them perform better in exams.

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

Q.1 (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) Show that the projection angle  θ 0  for a projectile launched from the origin is given by θ 0 = tan −1 ( 4 h m R ) Where the symbols have their usual meaning.

Ans.

(a) Let initial components of velocity of the projectile along horizontal (OX) and vertical (OY) directions be  v 0x  and  v 0y  the respectively. Let the horizontal and vertical components of velocity at point P be  v x  and  v y  respectively. Let time taken by projectile to arrive at point P = t Using the first equation of motion along vertical and horizontal directions, we have: v y = v 0y −gt v x = v 0x ∴tanθ= v y v x = v 0y −gt v 0x ∴θ= tan −1 ( v 0y −gt v 0x ) In angular projection Maximum vertical height,  h m = u 2 sin 2 θ 0 2g →(i) Horizontal range is given as: R= u 2 sin2 θ 0 g →(ii) Dividing equation (i) by equation (ii), we obtain: h m R = sin 2 θ 0 2sin2 θ 0 = sin 2 θ 0 4sin θ 0 cos θ 0 = sin θ 0 4cos θ 0 ∴ h m R = tan θ 0 4   ∴ θ 0 = tan −1 ( 4 h m R )

Q.2 A cyclist is riding with a speed of 27 kmh-1. 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 ms-1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Ans.

Given, speed of the cyclist, v = 27  kmh -1 = 7.5  ms -1 Radius of circular turn, r = 80 m ∴Centripetal acceleration,  a c = v 2 r = 7.5 2 80 ms -2 = 0.7  ms -2 Suppose the cyclist applies brakes at point P of the circular turn then the tangential component of  acceleration  a T  will act opposite to velocity. Here, tangential component of acceleration,   a T = 0.5  ms -2 Angle between  a c  and  a T   = 90 o

∴Magnitude of resultant acceleration a is given as: a = a c 2 +a T 2 a = 0 .7ms -2 2 + 0 .5ms -2 2 = 0.86  ms -2 Let the resultant acceleration makes an angle θ with the direction of velocity ∴tanθ = a c a T tanθ = 0 .7 ms -2 0 .5 ms -2 = 1.4 ∴ θ = tan -1 1.4 = 54 .46 o

Q.3 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 kmh-1 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 ms-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 ms-2)

Ans.

Given, altitude of plane = 1.5 km = 1500 m Speed of plane, v = 720  kmh -1 = 200  ms -1 Let the shell hits the plane when fired at an angle θ. The described situation is shown in the given figure. Here, muzzle velocity of gun, u = 600  ms -1 Let time taken by the shell to hit the fighter plane = t Horizontal distance covered by the  shell = u x t Distance covered by the plane = vt For the shell to hit the plane, these two distances must be equal. u x t = vt ∴usinθ = v ∴sinθ = v u = 200  ms -1 600  ms -1 = 0.33 θ = sin -1 0.33 = 19 .3 o In order to avoid being hit by the shell, the pilot must fly the plane at a minimum height  which is the maximum height attained by the shell. ∴H = u 2 sin 2 90 o -θ 2g = 600  ms -1 2 cos 2 θ 2g H = 600  ms -1 2 cos 2 19 .3 o 2g = 18000× 8 3 2  m H = 16000 m = 16 km

Q.4 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 be fixed, and neglect air resistance.

Ans.

Given, angle of projection of the bullet,  θ = 30 o

Range of the bullet, R = 3 km Acceleration due to gravity, g = 9.8  ms -2 Horizontal range for projection velocity  u 0 , R = u 0 2 sin2θ g ∴3 km = u 0 2 sin60 o g = u 0 2 g 3 2∴ u 0 2 g = 2 3

(i) The bullet achieves the maximum range  (R max ) when fired at an angle of  45 o  with the horizontal, R max = u 0 2 g

(ii) From equation (i) and (ii), we obtain R max = 3 3  km = 2×1.732 km = 3.46 km ∴The bullet cannot hit the target 5 km away.

Q.5 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere? Explain.

Ans.

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

(b) Yes, we can associate a vector with a plane area. Such a vector is called an area vector and its direction is normal, inward or outward to the plane area.

(c) No, we cannot associate a vector with the volume of a sphere. But, we can associate an area vector with the area of a sphere.

Q. 6 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?

Ans.

A physical quantity having both magnitude and direction is not necessarily a vector. There are many physical quantities that have both magnitude and direction, but they are not vectors. The essential condition for a physical quantity to be a vector is that it should follow the laws of vector addition. For example, the rotation of a body about an axis is not a vector quantity, because it does not follow the laws of vector addition. However, the rotation of a body by a definite small angle follows the laws of vector addition and is therefore considered a vector.

Q.7 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

Ans.

A vector has no fixed location in space because it remains invariant when displaced in such a way that its magnitude and direction remain the same.

A vector can change with time. For example, the displacement vector of an accelerated particle changes with time.

Two equal vectors at different locations in space do not necessarily produce the same physical effect. For example, two equal forces acting at two different points on a body can cause the rotation of the body, but their combination cannot produce the same turning effect.

Q.8 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?

Ans.

The given situation is described in the figure. Given, height of the aircraft from ground, OS = 3400 m     Angle subtended between the positions of aircraft, ∠POS = 30°  Time taken by the aircraft =10 s In ΔPSO, tan 15° = PS OS ∴PS = OStan15° ∴PS = 3400 m×tan15° = 3400 m×0.2679 = 910.86 m PQ = PS + SQ PQ = 2PS = 2×910.86 m = 1821.72 m ∴Speed of aircraft, v = distance PQ time v = 1821.72 m 10 s = 182 .172 ms -1

Q.9 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.

Ans.

(a) The given statement is false because, temperature being a scalar quantity is not conserved during inelastic collisions.

(b) The given statement is false because, temperature being a scalar quantity can take negative values.

(c) The given statement is false because, distance is a scalar quantity, yet it has the dimension of length.

(d) The given statement is false because, intensity of light being a scalar quantity change from one point to another in space.

(e) The given statement is true because, the value of a scalar does not change for observers with different orientations of axes.

Q.10 For an arbitrary motion in space, which of the following relations are true :

(a)  v → average = 1 2 v → t 1 + v → t 2

(b)  v → average = r → t 2 + r → t 1 t 2 – t 1

(c)  v → t = v → 0 + a → t

(d)  r → t = r → 0 + v → 0 t + 1 2 a → t 2

(e)  a → average = v → t 2 + v → t 1 t 2 – t 1 [The average stands for the average of the quantity over the time interval  t 1  and  t 2 ]

Ans.

Out of the given relations, the relations (b) and (e) are true. The relations (a), (c) and (d) are false as they hold for uniform acceleration only.

Q.11 i ∧  and  j ∧  are unit vectors along x - and y – axis respectively.

(a) What is the magnitude and direction of the vectors ( i ∧ + j ∧ ) and ( i ∧ − j ∧ ) ?

(b) What are the components of a vector B → = 2 i ∧ +3 j ∧  along the directions of ( i ∧ + j ∧ ) and ( i ∧ − j ∧ ) ?

Ans.

(a) Magnitude of  i ∧ + j ∧ = i ∧ + j ∧ = 1 2 + 1 2 = 2 Consider that the vector  i ∧ + j ∧  makes an angle θ with  i ∧ , ∴cosθ= i ∧ + j ∧ . i ∧ i ∧ + j ∧ i ∧ = 1 2 1 = 1 2 = cos45 o ∴ θ = 45 o Therefore, the vector  i ∧ + j ∧  makes an angle of  45 o  with the x-axis.

(b) Given, B → = 2 i ∧ +3 j ∧ Let  a ∧  be the unit vector directed along vector  i ∧ + j ∧ . ∴ a ∧ = i ∧ + j ∧ i ∧ + j ∧ = i ∧ + j ∧ 1 2 + 1 2 = i ∧ + j ∧ 2 Magnitude of the component vector of  B →  along vector  i ∧ + j ∧ = B →  . a ∧   = 2 i ∧ +3 j ∧ .  1 2 i ∧ + j ∧ =  1 2 2+3 = 5 2 ∴Component of  B →  along the direction of  i ∧ + j ∧ = B →  . a ∧ a ∧     = 5 2 i ∧ + j ∧ 2 = 5 2 i ∧ + j ∧ Let  b ∧  be the unit vector directed along  vector  i ∧ − j ∧ . ∴ b ∧ = i ∧ − j ∧ i ∧ − j ∧ = i ∧ − j ∧ 1 2 + −1 2 = 1 2 i ∧ − j ∧ Magnitude of the component vector of  B →  along the direction of  i ∧ − j ∧ = B → . b ∧ = 2 i ∧ +3 j ∧ . 1 2 i ∧ − j ∧ =− 1 2 ∴Component of  B →  along the direction of  i ∧ − j ∧ = B →  . b ∧ b ∧ = − 1 2 1 2 i ∧ − j ∧ =− 1 2 i ∧ − j ∧

Q.12 A particle starts from origin at t = 0 with a velocity of 10.0 j ∧ ms -1  and moves in the x – y plane with a  constant acceleration of ( 8.0 i ∧ +2.0 j ∧ )  ms -2 . At what time is the x – coordinate of the particle 16 m?

(a) What is the y – coordinate of the particle at that time?

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

Ans.

Given, initial velocity of particle,  u → =10.0 j ∧  m s −1 Acceleration of particle,  a → = d v → dt =( 8.0 i ∧ +2.0 j ∧ ) m s −2 ∴d v → =( 8.0 i ∧ +2.0 j ∧ )dt Integrating it within the limits as time changes from 0 to t and velocity changes from u to v, we obtain: v → ( t )=( 8.0t i ∧ +2.0t j ∧ + u → ) m s −1 As velocity,  v → = d r → dt

∴d r → = v → dt =( 8.0t i ∧ +2.0t j ∧ + u → )dt Integrating it within the limits: as time changes from 0 to t,displacement changes from 0 to r, we have r → = u → t+ 1 2 ×8.0 t 2 i ∧ + 1 2 ×2.0 t 2 j ∧ = u → t+4.0 t 2 i ∧ + t 2 j ∧   =( 10.0 j ∧ )t+4.0 t 2 i ∧ + t 2 j ∧ or x i ∧ +y j ∧ =4.0 t 2 i ∧ +( 10t+ t 2 ) j ∧ Here, we have, x=4.0 t 2  and y=10t+ t 2 ∴t= ( x 4 ) 1 2 (a) when x = 16 m t= ( 16 4 ) 1 2 =2 s ∴y=10×2+ 2 2 = 24 m

(b) Velocity of the particle at time t is given as: v → ( t )=( 8.0t i ∧ +2.0t j ∧ + u → ) m s −1 at t = 2 s v → ( t )=8.0×2 i ∧ +2.0×2 j ∧ +10 j ∧ =( 16 i ∧ +14 j ∧ ) m s −1 ∴Speed of the particle = | v → | = ( 16 ) 2 + ( 14 ) 2 = 256+196 = 452 =21.26 m s −1

Q.13 The position of a particle is given by r  → = 3.0t  i ^  - 2 .0t 2 j ^  + 4.0 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? (b) What is the magnitude and direction of velocity of the particle at t = 2s?

Ans.

Acceleration  a →  of the particle is given by the relation:   a →  =  d v → dt = d dt ( 3.0  i ^ −4.0t  j ^ ) ∴ a → =−4.0 j ^  m s −2 At time, t=2 s v → =3.0  i ^ −8.0t m s −1 The magnitude of velocity, v= ( 3.0 ) 2 + ( −8 ) 2 ∴ v= 73 =8.54 m s −1 If θ is the angle between  v →  and the x – axis, thentanθ= v y v x = −8 3 =−2.667 ∴θ= tan −1 ( v y v x ) = tan −1 ( −8 3 ) =− tan −1 ( 2.667 ) ∴θ=− 69.45 o The negative sign shows that the velocity is directed below the x – axis.

Q.14 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 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.

Ans.

(a) The given statement is false because the net acceleration of a particle is directed towards the centre only in case of uniform circular motion.

(b) The given statement is true because, when a particle leaves the circular path, it moves tangentially to the circular path. Therefore, the velocity vector of a particle at a point is always directed along the tangent at that point.

(c) The given statement is true because, the direction of the acceleration vector in a uniform circular motion is always directed towards the centre of the circle, but it constantly changes with time. The resultant of these vectors over one cycle is a null vector.

Q.15 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.

Ans.

Given, radius of horizontal loop, r = 1 km  or  1000 m Speed of aircraft, v = 900  kmh -1 = 250  ms -1 Centripetal acceleration is given by the  relation, a c = v 2 r a c = 250  ms -1 2 1000 m = 62.5  ms -2 Acceleration due to gravity, g = 9.8  ms -2 a c g = 62.5  ms -2 9.8  ms -2 = 6.38 ∴Centripetal acceleration,  a c = 6.38 g

Q.16 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?

Ans.

Given, length of the string, l = 80 cm = 0.8 m Number of revolutions = 14 Time taken = 25 s Frequency of revolution, ν = Number of revolutions Time taken ν = 14 25  Hz Angular frequency, ω = 2πν ω = 2× 22 7 × 14 25  Hz = 88 25   rads -1 Centripetal acceleration,  a c = ω 2 r a c = 88 25   rads -1 2 ×0.8 m = 9.91  ms -2 The direction of centripetal acceleration is always directed along the string,towards the centre of the circular path.

Q.17 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms-1 can go without hitting the ceiling of the hall?

Ans.

Given, speed of ball, u = 40  ms -1Maximum height, h = 25 m Let θ be the angle of projection with the horizontal. For projectile motion, Maximum height, H = u 2 sin 2 θ 2g 25 m = 40 ms -1 2 sin 2 θ 2×9 .8 ms -2 ∴ sin 2 θ = 0.30625 sinθ = 0.5534 ∴ θ = sin -1 0.5534 = 33.60° Horizontal range, R = u 2 sin2θ g R = 40 ms -1 2 ×sin2×33.6° 9 .8 ms -2 = 150.5 m

Q.18 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?

Ans.

Given, velocity of the boat, v → b = 51  kmh -1 Velocity of the wind,  v → w = 72  kmh -1 As, the flag is fluttering in north-east direction, it implies that the wind is blowing towards the north-east direction. When the boat starts moving towards the north direction, the flag will flutter  towards the direction of relative velocity of wind w.r.t. the boat. Let the relative velocity of the wind w.r.t. the boat be v → wb Let angle between  v → wb  and   v → w be β Angle between  v w  and  -v b = 90° + 45° ∴tanβ = 51 kmh -1 ×sin 90° + 45° 72 kmh -1   +51 kmh -1 ×cos 90° + 45° = 1.0038 ∴ β = tan -1 1.0038 = 45.1° Angle formed w.r.t. east direction = 45.1° – 45° = 0.1° ∴The flag will flutter almost towards east direction.

Q.19 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?

Ans.

Given, speed of man = 4  kmh -1 Width of river = 1 km Time taken to cross the river, t = Width of river Speed of man = 1 4 h ∴t = 1 4 ×60 = 15 min Given, speed of river,  v r =3  kmh −1 Distance travelled along with the flow of river in time t= v r ×t =3× 1 4  km = 3 4 ×1000 m =750 m

Q.20 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?

Ans.

Let velocity of  cyclist be v c Let velocity of  rain be v r The woman must hold her umbrella in the opposite direction of relative velocity (v) of the rain w.r.t. herself,  in order to protect herself from the rain, v = v r + -v c v = 30 ms -1 + -10 ms -1 = 20  ms -1 From the given figure, tanθ = v c v r tanθ = 10 30 ∴ θ = tan -1 1 3 = tan -1 0.333 ≈18° ∴The woman should hold the umbrella at an angle of approximately 18° with the vertical towards south direction.

Q.21 A passenger arriving in a new town wishes to go from the station to a hotel located10 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?

Ans.

Given, actual distance travelled = 23 km Total time taken = 28 min = 28 60 h Displacement of the car = Distance between the hotel and the station = 10 km (a) Average speed of the taxi = Actual distance travelled Total time taken = 23 km 28 60 h = 49.29  kmh -1 (b) Magnitude of average velocity = Displacement Total time taken = 10 km 28 60 h = 21.43  kmh -1 ∴Average speed of the taxi is not equal to the magnitudeof average velocity.

Q.22 A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same speed how high above the ground can the cricketer throw the same ball?

Ans.

Maximum horizontal distance,R = 100 m The ball will cover maximum horizontal distance when, angle of projection = 45° As, horizontal range is given by the relation, R = u 2 sin2θ g 100 m = u 2 g sin90° ∴ u 2 g = 100 m The ball will gain maximum height when height when it is thrown vertically upward. For such motion, the final velocity is zero at maximum height (H). Acceleration, a = -g From third equation of motion, v 2 – u 2 = -2gH 0 – u 2 = -2gH ∴H = 1 2 × u 2 g = 1 2 ×100  m = 50 m

Q.23 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.

Ans.

In this problem, the path followed by the motorist is a regular hexagon of length of each side of the regular hexagon = 500 m Let A be the starting point of the motorist. At D, the motorist will take the third turn .  ∴Magnitude of displacement at D = AG + GD = 500 m + 500 m = 1000 m Total path length from A to D = AB + BC + CD = 500 m + 500 m + 500 m = 1500 m At A, the motorist takes sixth turn, which is the starting point ∴Magnitude of displacement = 0 Total path length from A to A = AB + BC + CD + DE + EF + FA = 6×500 m = 3000 m At C, the motorist takes eighth turn. ∴Magnitude of displacement = AC AC = AB 2 +BC 2 +2 AB . BC ∴AC = 500 m 2 + 500 m 2 +2× 500 m × 500 m ×cos60 o = 866.03 m tanβ = 500 m× sin60 o 500 m +500 m× cos60 o = 1 3 = tan30 o ∴ β = 30 o Therefore, the magnitude of displacement (AC) is 866.03 m at an angle of 30° with AC. Total path length = AB + BC + CD + DE + EF + FA + AB + BC = 8×500 m = 4000 m

Q.24 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge Pof the park, then cycles along the circumference, and returns to the centre along QOas 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?

Ans.

(a) In the given case the cyclist comes back to the starting point after cycling for 10 minutes. Time taken = 10 min = 1 6  h ∴Total displacement of the cyclist = 0 (b) Average velocity is given as: Average velocity = Total displacement Total time taken = 0 10 60 h =0 (c) Average speed is given as: Average speed = Total path length Total time taken Total path length = OP + PQ + QO Given, OP = QO = 1 km PQ = 1 4 2πr = 1 4 2π×1 km = 3.570 km ∴Average speed = 3.570 km 1 6 h = 21.42  kmh -1

Q.25 Three girls skating on a circular ice ground of radius200 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?

Ans.

( a ) The given statement is incorrect, because in order to make a → + b → + c → + d → = 0, it is not necessary to have all the four vectors to be null vectors. There are many other combinations whose sum is equal to zero. ( b ) The given statement is correct. As, a → + b → + c → + d → =0 ∴ a → + c → =−( b → + d → ) Taking modulus on both sides, | a → + c → |=| −( b → + d → ) | =| ( b → + d → ) | Therefore, magnitude of ( a → + c → ) is equal to the magnitude of ( b → + d → ). Given  a → + b → + c → + d → = 0,

Q.26 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 → ).

(c) The magnitude of a can never be greater than the sum of the magnitude 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.

Ans.

( a ) The given statement is incorrect, because in order to make a → + b → + c → + d → = 0, it is not necessary to have all the four vectors to be null vectors. There are many other combinations whose sum is equal to zero.

( b ) The given statement is correct. As, a → + b → + c → + d → =0 ∴ a → + c → =−( b → + d → ) Taking modulus on both sides, | a → + c → |=| −( b → + d → ) | =| ( b → + d → ) | Therefore, magnitude of ( a → + c → ) is equal to the magnitude of ( b → + d → ).

(c) The given statement is correct. As  a →  +  b →  +  c →  +  d →  =0 ∴ a →  = -( b →  +  c →  +  d → ) Taking modulus on both sides  | a → |=| -( b →  +  c →  +  d → ) | =| b →  +  c →  +  d → | ∴| a → |≤| b → |+| c → |+| d → |→(i) Equation (i) indicates that the magnitude of  a →  can be less than or equal to the sum of the magnitudes of b → ,  c →  and  d → . ∴The magnitude of  a →  can never be greater than the sum of the magnitudes of  b → ,  c →  and  d → .

(d) The given statement is correct, As,  a →  +  b →  +  c →  +  d →  =0  ∴ a → +( b →  +  c → )+ d → =0 The resultant sum of three vectors  a → , ( b →  +  c → ) and  d →  can be zero only if ( b →  +  c → ) lie in the plane of  a →  and  d →  and these three vectors are represented by the three sides of a triangle taken in the same order.

If  a →  and  d →  are collinear, then ( b →  +  c → ) should be in line of  a →  and  d → , only then the vector sum of all the vectors will be zero.

Q.27 Establish the following 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?

Ans.

(a) Consider two vectors,  a →  and  b →  represented by the sides OR →  and  OT →  of a parallelogram OTSR as shown in the  given figure (a). Here, we have  | OT → |=|   a → |→(i) | TS →   |=| OR → | =|   b → |→(ii) | OS → |=|   a → +  b → |→(iii) In case of a triangle, each side is smaller than the sum of  other two sides. ∴From ΔOTS, we obtain OS<( OT+TS ) |   a → +  b → |<|   a → |+|   b → |→(iv)

If two vectors   a →  and  b →  are acting along a straight line in the same direction, then we have: |   a → +  b → |=|   a → |+|   b → |→(v) From equation (iv) and (v), we obtain |   a → +  b → |≤|   a → |+|   b → | (b) Consider two vectors,  a →  and  b →  represented by  the sides  OR →  and  OT →  of a parallelogram OTSR as  shown in the given figure (a). Here,   | OT → |=|   a → |→(i) | TS →   |=| OR → | =|   b → |→(ii) | OS → |=|   a → +  b → |→(iii) In case of a triangle, each side is smaller than the sum  of other two sides. ∴From ΔOTS, we obtain OS+TS>OT OS+OT>TS | OS → |>| OT → − OR → |( ∵OR=TS ) |   a → +  b → |<|   a → |+|   b → |→(iv)

If two vectors   a →  and  b →  are acting along a straight line in  the same direction, then we have: |   a → +  b → |=|   a → |+|   b → |→(v) From equation (iv) and (v), we obtain |   a → +  b → |≤|   a → |+|   b →

(c) Consider two vectors,  a →  and  b →  represented by the sides  OQ →  and  OP →  of a parallelogram OPQN as  shown in the given figure (b). Here,  | OP → |=|   a → |→(i) | PN →   |=| OQ → | =|   b → |→(ii)

In case of a triangle, each side is smaller than the sum of  other two sides. ∴From ΔOPN, we obtain ON<OP+PN |   a →  − b → |<|   a → |+| −  b → | |   a →  − b → |<|   a → |+|   b → |→(iii) If two vectors are acting along a straight line but  in opposite directions, then  |   a →  − b → |=|   a → |+|   b → |→(iv) From equation (iii) and (iv), we obtain |   a →  − b → |≤|   a → |+|   b → | (d) Consider two vectors,  a →  and  b →  represented by  the sides  OQ →  and  OP →  of a parallelogram OPQN as  shown in the given figure (b). In case of a triangle, each side is smaller than the sum  of other two sides. ∴From ΔOPN, we obtain ON+PN>OP→(i) ON>OP−PN→(ii) |   a →  − b → |>|   a → |−|   b → |→(iii)

The LHS of the above equation is positive but the RHS     can be positive or negative. In order to make the  quantities on both sides of the above equation positive,  we take modulus on both sides as: | |   a →  − b → | |>| |   a → |−|   b → | | |   a →  − b → |>| |   a → |−|   b → | |→(iv) If the two vectors are acting in a straight line but  in opposite directions, then  |   a →  − b → |=| |   a → |−|   b → | |→(v) Combining equation (iv) and (v), we get |   a →  − b → |≥| |   a → |−|   b → | |

Q.28 Read each statement 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.

(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.

Ans.

(a) It is true because the magnitude of a vector is a pure number.

(b) It is false because each component of a vector is also a vector.

(c) It is false because the total path length is either greater than or equal to the magnitude of displacement. It is true only if the particle is moving in a straight line.

(d) It is true because the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) It is true because three vectors that do not lie in a plane cannot be represented by the three sides of a triangle taken in the same order. Thus, it can never follow the triangle method of vector addition to give a null vector.

Q.29 State with reasons whether the following algebraic operations with scalars and vectors are meaningful:

(b) adding a scalar to a vector of the same dimension

(c) multiplying any vector by any scalar

(d) multiplying any two scalars

(f) adding a component of a vector to the same vector

Ans.

(a) No, the addition of any two scalars is meaningful only if they represent the same physical quantity.

(b) No, the addition of a vector physical quantity with a scalar physical quantity is not meaningful.

(c) Yes, a scalar quantity can be multiplied by a vector quantity. Velocity multiplied by time to give displacement is one of the examples of it.

(d) Yes, multiplication of two scalars is meaningful. For example, power is multiplied by time to give work, which is a useful operation.

(e) No, addition of two vectors is meaningful only if both of them represent the same physical quantity.

(f) Yes, addition of a component of a vector to the same vector is meaningful as both of them have the same dimensions.

Q.30 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.

Ans.

Impulse = change in momentum = force × time

As force is a vector quantity, its product with time will also be a vector quantity. Thus, the impulse is a vector quantity.

Q.31 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.

Ans.

Work and current are the scalar quantities in the given list.

Work is defined as the dot product of force and displacement. As the dot product of two physical quantities is always a scalar physical quantity, so work is a scalar physical quantity.

Current is explained by its magnitude only. So current is a scalar physical quantity.

Q.32 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.

Ans.

Out of the given physical quantities,

Scalar physical quantities: Volume, mass, speed, density, number of moles and angular frequency

Vector physical quantities: Acceleration, velocity, displacement and angular velocity

## 1. What can we expect to learn in Chapter 4 of the NCERT Solutions for Class 11 Physics?

In Chapter 4 Physics Class 11, you will learn the following topics or concepts. Let’s take a look at a few of them right now;

1. Displacement, Scalar and Vector
2. Using real numbers to multiply vectors
3. Using a graphical method – vector addition and subtraction
4. Vector addition using a triangle and an analytical method
5. Motion in a plane, velocity, acceleration, and finding averages

## 2. What types of questions and answers can be found in Chapter 4 of the NCERT Solutions for Class 11 Physics, "Motion in a Plane"?

The NCERT Solutions for Class 11 Physics Chapter 4 ‘Motion in a Plane’ contains a variety of questions and answers. The questions will range in complexity from detailed to short-answer. Pick/choose the one out, true or false, reason stating questions, numerical or problem questions, long answer questions, and so on will all be included. The answers to these questions will also be given in the solution. As a result, you will find everything in one place and will not need to go elsewhere to study thoroughly.

## 3. How Do Scalar and Vector Quantities Differ?

A vector quantity has both magnitude and direction, while a scalar quantity has magnitude only. A vector quantity is always the cross product of two vector quantities, whereas a scalar quantity is always the dot product of two vector quantities. Mass, density and current are examples of scalar quantities. Vector quantities include acceleration, velocity, and displacement.

## 4. Where can I find online NCERT Solutions for Class 11 Physics Chapter 4?

Class 11 is a pivotal point in every student’s life. Students’ career paths will be determined by their performance in Class 11. To that end, the faculty at Extramarks has developed solutions for all chapters based on the updated term – I CBSE Syllabus. Extramarks has NCERT Solutions Class 11 available for students. The solutions are designed with students’ first-term exam preparation in mind, regardless of intelligence level.

## 5. How should I prepare for Class 11 Physics Chapter 4 'Motion in a plane'?

Try to get the best study material for Class 11 Physics Chapter 4 ‘Motion in a Plane.’ Studying multiple books at the same time can be a disaster. Instead, pick one and finish it before moving on to the others. You can also refer to Class 11 Physics Chapter 4 NCERT Solutions by Extramarks.