NCERT Solutions Class 11 Physics

Physics is frequently rated as one of the most difficult subjects by students. However, if approached correctly, it may quickly become one of the most simple and fun subjects. Physics can not only improve your grades, but it can also spark a lifelong interest in science. As a result, when studying a conceptual subject such as physics, it is critical that the student take the proper approach. To assist you with this procedure, we have provided NCERT Solutions for Class 11 Physics.

In class 11, students are introduced to several fundamental physics ideas and laws. We will go through important topics like units of measurement, motion in a plane and straight line, gravitational force, kinetic energy, and so on.

NCERT Solutions Class 11 Physics for Download

Physics is a discipline that is filled with concepts and numerical issues. To solve the questions effortlessly, a full comprehension of the ideas is required. Extramarks looks to supply students with reliable information that is totally based on the CBSE board’s most recent syllabus. By following the links provided below, you may access NCERT Physics Class 11 for all chapters. Subject matter specialists with years of expertise explain each and every minute idea in the most straightforward manner possible, allowing students to approach the exam with confidence.

NCERT Class 11 Physics Chapter wise Solutions 

On this page, we will provide links to Class 11 Physics chapter-by-chapter solutions so that you can download them as needed. You can easily download them and refer to them as needed while studying. Continue reading to learn more about and get your Class 11 Physics Solutions.

NCERT Physics Class 11 Solutions

NCERT 11 Physics Solutions provide answers to every chapter end practise questions. When appropriate, diagrams are used to help students grasp the solutions. The solutions also offer some preparatory advice on how to use the solutions, which will help students pass the Class 11 exam with flying colours.

NCERT Solutions for Class 11 Physics Chapter 1 Physical World – Term I

Physical World is an introductory chapter that explains what Physics is and how it tries to explain various physical occurrences using a few concepts and laws.The chapter next discusses the main divisions of Physics – Classical Physics and Quantum Physics – and what they are all about. You’ll learn about the relationship between physics, technology, and society as well.

Topics Covered in Class 11 Physics Chapter 1 Physical World for First Term:

Physics-scope and excitement; nature of physical laws; Physics, technology and society. (To be discussed as a part of Introduction and integrated with other topics)

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement – Term I 

This chapter will teach you how to measure various physical quantities. A basic, internationally acknowledged reference standard has been adopted in order to quantify physical quantities. This is referred to as a unit. The physical quantity is compared to this unit during measurement. There are two types of units: fundamental and derived (which can be expressed as a combination of the fundamental units).

Topics Covered in Class 11 Physics Chapter 2 Units and Measurements for First Term:

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. Length, mass and time measurements; accuracy and precision of measuring instruments; errors in measurement; significant figures.

Dimensions of physical quantities, dimensional analysis and its applications.

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line – Term I

This chapter begins by describing motion, which is defined as the change in position of an object over time, before going on to discuss motion in a straight line. This is referred to as rectilinear motion. Graphs and equations are used to explain many terminology connected to Rectilinear Motion, such as location, path length, displacement, average and instantaneous velocity and speed, and acceleration. You’ll also learn about kinematic equations and relative velocity for uniformly accelerated motion.

Topics Covered in Class 11 Physics Chapter 3 Motion in a Straight Line  for First Term:

Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, average speed and instantaneous velocity, uniformly accelerated motion, velocity – time and position-time graphs.

Relations for uniformly accelerated motion (graphical treatment).

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

You will learn about the motion of an object in a plane, or in two dimensions, and how to quantify it in this chapter. To begin, you’ll learn about scalars (quantities with only one magnitude) and vectors (quantities that have both magnitude and directions). You’ll learn how to multiply vectors by real values, as well as how to add and subtract vectors using graphical approaches. The resolution of vectors and the analytical approach of vectors addition will come after that.

Topics Covered in Class 11 Physics Chapter 4 Motion in a Plane  for First Term:

Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, relative velocity, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors.

Motion in a plane, cases of uniform velocity and uniform acceleration-projectile motion, uniform circular motion.

NCERT Solutions for Class 11 Physics Chapter 5 Law of Motion – Term I

This chapter discusses how motion is caused. The law of inertia will be discussed, followed by Newton’s first, second, and third laws of motion. Students will gain a thorough understanding of how objects move, what force is, the direction, resultant motion, and the influence of gravitation. All of this will be done by using specific examples. The chapter also defines momentum and the conservation of momentum, before moving on to the idea of equilibrium, which occurs when the net force acting on a body is zero. Students will be able to succeed in their higher education levels if they have a thorough understanding of these ideas.

Topics Covered in Class 11 Physics Chapter 5 Laws of Motion  for First Term:

Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. (Recapitulation only)

Law of conservation of linear momentum and its applications.

Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.

Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power – Term I

From a scientific standpoint, you will study  work, power, and energy, three often used terminology, as well as the link between these three physical variables, in this chapter. You’ll receive a clear description of work using the Work-Energy Theorem, which is derived from equations studied in prior chapters. The two types of energy, kinetic and potential, will be discussed. Problems involving determining energy and power are solved in a step-by-step way based on the CBSE board’s mark weightage. Students will learn the chapter with ease if they practise these principles on a regular basis.

Topics Covered in Class 11 Physics Chapter 6 Work, Energy and Power  for First Term:

Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power.

Notion of potential energy, potential energy of a spring, conservative forces: conservation of mechanical energy (kinetic and potential energies); non-conservative forces: motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.

NCERT Solutions for Class 11 Physics Chapter 7 Systems of Particles and Rotational Motion – Term I

This chapter discusses the theoretical and mathematical elements of extended body motion. The linear momentum of a system of particles, the vector product of two vectors, angular velocity and its relation to linear velocity, torque and angular momentum, the equilibrium of a rigid body, moment of inertia, and theorems of perpendicular and parallel axes are all explained in detail with examples, diagrams, and graphs. Because this chapter covers so many subjects, students are strongly advised to use excellent study materials when solving the textbook’s exercise questions.

Topics Covered in Class 11 Physics Chapter 7 System of Particles and Rotational Motion  for First Term:

Centre of mass of a two-particle system, momentum conservation and centre of mass motion.

Centre of mass of a rigid body; centre of mass of a uniform rod.

Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications.

Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions.

Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation). 

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation – Term I

All items fall toward the earth due to gravity, as we all know. This chapter will provide you with a thorough understanding of the gravitational force and its different principles. You’ll begin with Kepler’s laws, then move on to the universal law of gravitation, the gravitational constant, the earth’s acceleration due to gravity, acceleration due to gravity below and above the earth’s surface, and gravitational potential energy.

Topics Covered in Class 11 Physics Chapter 8 Gravitation  for First Term:

Universal law of gravitation. Acceleration due to gravity (recapitulation only) and its variation with altitude and depth.

Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite, Geo-stationary satellites.

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids – Term II

You’ve learnt that a body’s motion is determined by how mass is distributed within it. However, your comprehension was limited to a simpler case of rigid bodies — you studied rigid body motion in which the hard solid body has a definite shape and size. In actual life, however, even a rod of iron can be bent, stressed, or compressed – it can be bent, stressed, or crushed. As a result, you will study the mechanical properties of solid things in this chapter. Problems based on these ideas arise in exams for higher marks, thus practising them on a daily basis is vital. Students will also learn about the stress-strain graph, which is critical for the exam.

Topics Covered in Class 11 Physics Chapter 9 Mechanical Properties of Solids  for Second Term:

Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids – Term II

Because they can flow, liquids and gases are termed fluids. Some basic physical features of fluids will be covered in this chapter. The chapter also discusses the differences between fluids and solids, as well as the parallels and differences between liquids and gases, among other topics. The chapter begins with a discussion of pressure before moving on to other topics such as streamline flow, Bernoulli’s principle, viscosity, and surface tension. Diagrams, equations, and numerical problems are used to teach all of these ideas in depth.

Topics Covered in Class 11 Physics Chapter 10 Mechanical Properties of Fluids for Second Term:

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure.

Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its applications.

Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter – Term II

In our everyday lives, we all use the phrases “hot” and “temperature.” This chapter explains these two words in a thorough and scientific manner. The chapter begins by explaining what heat and temperature are and how they are measured. The ideal gas equation and the concept of absolute temperature are next introduced. The thermal expansion of matter — how solids, liquids, and gases expand as the temperature rises – will be covered next.

Heat, temperature, (recapitulation only) thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv – calorimetry; change of state – latent heat capacity.

Heat transfer-conduction, convection and radiation (recapitulation only), thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law, Greenhouse effect.

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics – Term II

You will learn about the rules that control thermal energy in this chapter, as well as the mechanisms that convert work into heat and vice versa. The concept of thermal equilibrium is introduced first, followed by the Zeroth law of Thermodynamics. Thermodynamics will then teach you how heat and work aid in energy transmission to a system, resulting in a change in its internal energy.

Topics Covered in Class 11 Physics Chapter 12 Thermodynamics for Second Term:

Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. The first law of thermodynamics, isothermal and adiabatic processes.The second law of thermodynamics: reversible and irreversible processes.

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory – Term II

This chapter begins by addressing the molecular nature of matter and the behaviour of gases. The kinetic theory of an ideal gas, as well as the law of energy equipartition, will be studied after that. The specific heat capacities of monatomic gases, diatomic gases, polyatomic gases, solids, and water will be discussed after that. Finally, you’ll discover what the mean free path is.

Topics Covered in Class 11 Physics Chapter 13 Kinetic Theory for Second Term:

Equation of state of a perfect gas, work done in compressing a gas. Kinetic theory of gases – assumptions, the concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom; the law of equipartition of energy (statement only) and application to specific heat capacities of gases; the concept of mean free path, Avogadro’s number.

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations – Term II 

In this chapter, you’ll study about oscillations, also known as oscillatory motion, which is the to and fro motion of an object around a fixed point (Ex: the motion of a pendulum). This idea, sometimes known as vibrations, is crucial for comprehending a variety of physical processes. Students must use appropriate study resources to learn these ideas effectively. They will be better prepared to face more difficult questions in the exam if they memorise the chapter on a daily basis.

Topics Covered in Class 11 Physics Chapter 14 Oscillations for Second Term:

Periodic motion – time period, frequency, displacement as a function of time, periodic functions. Simple harmonic motion (S.H.M) and its equation; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period. Free, forced and damped oscillations (qualitative ideas only), resonance.

NCERT Solutions for Class 11 Physics Chapter 15 Waves – Term II

The oscillatory motion of a system or a collection of items, such as a material medium, will be studied here. You’ll learn about waves, which are patterns that travel without transferring stuff in its whole. You’ll learn about transverse and longitudinal waves, progressive wave displacement relationships, and travelling wave speed. The principles of wave superposition and reflection will be discussed as well. Important formulas and concepts are emphasised in the solutions to help students remember what they’ve learned.

Topics Covered in Class 11 Physics Chapter 15 Waves for Second Term:

Wave motion: Transverse and longitudinal waves, speed of travelling wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, Beats.

Important Theories and Topics in Physics Class 11

Students can make note of these important theories and topics for CBSE Class 11 Physics and prepare revision notes by referring to the NCERT Solutions.

  • Classical Mechanics studies particles and their motion. It investigates how fast they travel. Their speed is substantially slower than that of light.
  • Kinetic Theory is a theory that analyses and describes how gases behave. It is considered that gas is made up of swiftly moving atoms. The hypothesis explains the random movement of particles caused by the presence of significant intermolecular space.
  • The Relative Theory, proposed by Albert Einstein, describes how all moving objects are related to one another. It discusses the motion of particles travelling at a speed close to that of light.
  • Thermodynamics is the science of heat, temperature, and the conversion of heat into work.

Chapter-Wise Tips:

The first chapter is a simple one. Take your time with it and go over it thoroughly before the exam.

Measurement, error evaluation, and SI units are all covered in Chapter 2. From the standpoint of numerical and short-answer questions, all of these topics are incredibly significant. Make sure you’ve learned and understood all of the concepts and formulas, and that you’ve had plenty of practice with the questions and approaches. In the exam, you may also be asked about the dimensions of other quantities. Only a few questions from this chapter may appear in the exam, thus rigorous preparation is essential to achieving full marks.

Chapter 3 contains a graphical representation of motion, and questions about it can be asked. Make sure you’re familiar with all of the graphical representations.  The exam usually requires numericals from theories and topics like average velocity and speed, relative velocity, instantaneous velocity, and kinematic equations.Make sure you’ve had plenty of practice with these numerical questions and that you’ve gone over all of the formulas and equations.

Chapter 4 is crucial, and the principles covered here will be tested on the exam and will be applied in subsequent chapters. A strong understanding of vectors and all of the principles is necessary. For solving equations and numerical problems, the chapters on resolving vectors and projectile motion are also essential. This chapter requires you to practise vector addition, multiplication, differentiation, and integration. Almost all of the problems in this chapter employ these formulas. Make sure you’ve had plenty of practice and have a firm grasp on all of the formulas. Students can use the Physics 11 NCERT books and solutions to have a thorough understanding of how to solve numerical questions.

Chapter 5 is one of the most crucial chapters in the book. Students should be well-versed in Newton’s three laws. They should also practise as many numerical problems relating to momentum conversion as possible, as they will be questioned about them repeatedly in the exam. There are also questions about issues like friction and force. It is possible to ask students to solve and equalise them.

Chapter 6 focuses on the derivation of all laws and theorems. For full marks, students must have sufficient practice and know how to describe it step-by-step in a comprehensible manner. This chapter emphasises kinetic and potential energy. These topics ask for numerical answers. Derivation of Spring Potential Energy is particularly vital for students to understand because it may be questioned directly in the exam. It’s also a good idea to practise interpreting a potential energy curve. Students should attempt as many numerical and conversion questions as possible from these areas.

Chapter 7 covers all the vital topics such as the theorems, linear motion, kinematics and dynamics of rotational motion, rolling motion, inertia, torque and angular momentum, angular velocity and angular acceleration are all vital topics covered in Chapter 7. All students should concentrate on these topics and ensure that they understand them. Each of these topics’ numerical questions should be carefully practised as well.

Chapter 8 discusses Kepler’s Law in depth. You can directly question the law’s derivation. Students should figure out how to calculate gravity’s acceleration above and below the surface. Numericals  from Newton’s law of gravitation, as well as acceleration due to gravity, are topics where students should focus. Satellite escape speed and time period are also important topics to investigate numerically. Students should be familiar with all of the formulas and have sufficient numerical practice.

In Chapter 9, students should review all of the Mechanical Properties of Solids thoroughly.

Chapter 10 discusses Bernoulli’s Principle. In the exam, it can be used to ask both short answers and numerical questions. Venturi-metre and hydraulic lift can also be asked as application-based numerical questions.

All of the subjects covered in Chapter 11 are extremely significant, including the Ideal Gas Equation, Thermal Expansion, Specific Heat Capacity, Calorimetry, Latent Heat, Heat Transfer, and Newton’s Law of Cooling.Numerical and theoretical practise should be done and students should have a thorough understanding of all ideas. Calorimetry can feature numerical questions that require a lot of computations, so make sure you practise them enough.

The most significant chapter is Chapter 12, Thermodynamics. All of the laws and concepts can be directly challenged and quantified. While studying for the exam, students should answer as many questions as possible. All of the concepts and theories discussed in this chapter must be well understood. They should be thoroughly investigated by students. The interpretation of the Carnot Cycle should also be understood by students. It’s necessary when working with numbers and addressing challenges.

Focus on Kinetic Theory in Chapter 13 and make sure you grasp the material. It’s crucial to understand the Law of Equi-partition of Energy.

Questions about numerics and derivation can be found in Chapter 14 of SHM. Students should be well-versed in the topics and have enough practice with numbers. Simple Pendulum and Oscillations are equally important topics. All of the formulas should be memorised by the students. It’s possible that you’ll be asked for numbers.

In Chapter 15, students need to focus on Longitudinal Waves, their features, displacement relation, Principle of Superposition, Standing Waves, and Doppler’s effects in the field of waves. Derivation and numerical calculations for all these topics are important, so is the derivation of the nodes and antinodes from standing waves. In some cases, derivation of the Dooper’s effect is required to solve an issue. All of the concepts should be understood and practised by the pupils.

Features of Extramarks NCERT Solutions for Class 11 Physics

  • Senior faculty members prepare the subject matter, so students can get solutions and experience all on one plate
  • The language is simple and straightforward
  • Answers in the Class 11 Physics Solutions are simple and exact
  • A step-by-step procedure is followed in the solutions
  • Solutions are provided at the chapter and topic levels to ensure that students have a thorough understanding of the material
  • The syllabus covers the whole course, so students can access all the relevant material in one go. Students can consult the solutions offered here if they have any questions while practising. It will also aid in a quick review of the complete curriculum

CBSE Marking Scheme 2022-23

The CBSE Board has divided the academic syllabus equally into two terms. Students can find the split syllabus for Term I along with the marks assigned to each topic. 

Term I CBSE Class 11 Physics Syllabus Course Structure 2022-2023

Unit No Name of Unit Marks
Unit – I Physical World and Measurement 20
Chapter -1: Physical World
Chapter -2: Units and Measurements
Unit – II Kinematics
Chapter- 3: Motion in a Straight Line
Chapter- 4: Motion in a Plane
Unit – III Laws of Motion
Chapter- 5: Laws of Motion
Unit – IV Work, Energy and Power 15
Chapter- 6: Work, Energy and Power
Unit – V Motion of System of Particles and Rigid Body
Chapter- 7: System of Particles and Rotational Motion
Unit – VI Gravitation
Chapter- 8: Gravitation
TOTAL 35

At the beginning of Term II, students can refer to the marks distribution and begin planning their study schedule. 

Term II CBSE Class 11 Physics Syllabus Course Structure 2022-2023

Unit No Name of Unit Marks
Unit – VII Properties of Bulk Matter 23
Chapter -9: Mechanical Properties of Solids 
Chapter -10: Mechanical Properties of Fluids
Chapter -11: Thermal Properties of Matter
Unit – VIII Thermodynamics
Chapter- 12: Thermodynamics
Unit – IX Behaviour of Perfect Gases and Kinetic Theory of Gases
Chapter- 13: Kinetic Theory
Unit – X Oscillations and Waves 12
Chapter- 14: Oscillations
Chapter- 15: Waves
TOTAL 35

Why Extramarks?

Extramarks combines the benefits of The Learning App with Live Classes to provide you with a seamless and comprehensive learning experience. The NCERT Solutions Class 11 also includes questions from prior years’ sample papers and question papers. This allows students to get a sense of the types of questions they might expect on the exam. By attempting to answer these questions, they will gain a better understanding of how to write the responses. They can use the answers to these questions to evaluate and improve their own answers.

How to Use the NCERT Book While Preparing for Exams?

For students in Class 11 and 12, the NCERT textbook is the most important book. The majority of questions in board and school exams are taken from the NCERT textbook or are based on the same type of question pattern. Students should be thorough with all of the questions provided at the end of each chapter, as well as any additional questions or examples provided within the chapters. If you are thorough with these questions, it will be easy for you to attempt other questions of this type and variations on them. Practising NCERT questions will help you get good grades. You can refer to the NCERT Solutions and properly practise all of the diagrams,  numerical problems and examples from the book. 

Important Tips for the Exam:

  • Students should always have a schedule. The timeline will assist students in keeping track of their syllabus as well as determining how much time they will need for each course or topic. A timeline will also assist them in keeping track of the chapters they cover and ensure that no topic is missed
  • Make sure to take regular pauses and time to rejuvenate yourself. This can also help you study better when you return from your break. Always begin a new topic fresh, as this will help you retain more information. Have a restful sleep before the exam to be at your best
  • Make subject and chapter summaries using the Class 11 Physics Solutions– This will allow you to better revise for the exam. It will highlight key points from the chapters and present them in a succinct manner
  • Examine sample papers and papers from past years. This will help you understand the types of questions that may be asked in the exam and how you should respond to them. Always keep track of your progress while tackling the sample paper; this will help you manage your time effectively. After each attempt, evaluate yourself to determine how you might improve your performance and which topics require additional practise
  • Prepare your bag and any other exam-related items the day before the exam. Make sure to reach the centre at least 10-15 minutes before exam starts to keep your cool and focus
  • Make sure to answer all of the questions on your exam. Start with the questions you’re confident about. Spend no time on any questions about which you are unsure. Make a point of emphasising formulas, key words, and lines in your responses. A well-presented paper is usually preferable and offers you an advantage. Try all of the questions in the order they appear on the question paper and in the sections they belong to
  • After completing the paper, allow yourself adequate time to edit your answers. Make good use of your time and allow for some time for review. This is critical for numerical calculations . Revising your answers can assist you in identifying and correcting any errors that you may have made

Q.1 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Ans.

Given,lengthofthemetalsheet,l = 4.234m Breadthofthemetalsheet,b = 1.005m Thicknessofthemetalsheet,t = 2.01cm = 2 .01×10 -2 m Areaofthemetalsheet, A = 2 l×b + b×t + t×l A = 2 4.234 m×1.005 m + 1.005 m×2 .01×10 -2 m + 2 .01×10 -2 m×4.234 m A = 2 4 .3604739 m 2 = 8.7209478 m 2 Sinceareaandvolumecancontainamaximumofthreesignificantfigures, Onroundingoff,weobtain Area = 8.72 m 2 Volumeofmetalsheet,V = l×b×t V = 4.234 m×1.005 m×0.0201 m = 0.0855289 = 0.0855 m 3

Q.2 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 ×1030 kg, mass of the earth = 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

Ans.

Here, mass of sun, M S = 2×10 30 kg Mass of earth, M E = 6×10 24 kg Orbital radiusofEarth, r = 1 .5×10 11 m Letmass of the rocket be m If x is the distance from the centre of the Earth where the gravitational forceacting on satellite Q becomes equaltozero. Using Newton’s law of gravitation, we can equate gravitational forces acting onsatellite Q under the effect of the Sun and the Earth as: GmM S r-x 2 = G mM E x 2 r-x x 2 = M S M E r-x x = 2 × 10 30 kg 6 × 10 24 kg 1 2 = 577.35 1 .5×10 11 m – x = 577.35x 578.35x = 1 .5×10 11 m x = 2 .59×10 8 m

Q.3 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.

Ans.

Here,orbital radius of the Earth, r = 1 .5×10 11 m Periodofrevolutionof Earth around the Sun,T = 1year = 365.25 days = 365.25×24×60×60 s Universal gravitational constant,G = 6 .67×10 –11 Nm 2 kg –2 LetmassofSun be M andmassofEarth be m GravitationalforceactingonEarthdueto Sun, F g = GMm r 2 LetangularvelocityofEartharoundSun be ω Centripetalforceactingon Earth, F c = mrω 2 = mr T 2 = mr 2 T 2 SincegravitationalpullofSunonEarthprovidesthenecessarycentripetalforce, GMm r 2 = mr 2 T 2 M = 2 r 3 GT 2 M = 3.14 2 × 1 .5×10 11 m 3 6 .67×10 –11 Nm 2 kg –2 × 365.25×24×60×60 s 2 = 2 .0×10 30 kg Mass of the Sun = 2×10 30 kg

Q.4 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50×108 km away from the sun?

Ans.

Here, seperation of the earth from the sun, r E = 1 .5×10 8 km = 1 .5×10 11 m Let time period of the earth be T E Time period of saturn, T s = 29 .5 T E Letdistanceofsaturnfromthes un be r S UsingKepler’sthirdlawofplanetarymotion,wehave T = 2 r 3 GM 1/2 Incaseofsaturnandsun,wehave: r S 3 r E 3 = T S 2 T E 2 r S = r E T S T E 2/3 r S = 1 .5×10 11 m× 29 .5 T E T E 2/3 r S = 1 .5×10 11 m× 29.5 2/3 = 14 .32×10 11 m Distancebetweensaturnandsun = 14 .32×10 11 m

Q.5 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Ans.

Here,weight of body, W = 63 N Letradius of Earth be R E Acceleration due to gravity at height h from the Earth’s surface is given as: g’ = g 1+h R E 2 (i) Here,g = Acceleration due to gravity on thesurfaceof Earth Here,height,h = R E 2 (ii) Substitutingthevalueofhfromequation(ii)inequation(i),weobtain: g’ = g R E +h R E 2 = gR E 2 R E + R E 2 2 g’ = g 1+ 1 2 2 = 4 9 g Weight of a body of mass m at height h,W’ = mg’ W’ = m× 4 9 g = 4 9 W= 4 9 ×63N = 28N

Q.6 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Ans.

Letmassofthebody be m Given,weight of the bodyat the surfaceofEarth, W = mg = 250 N Here, R E = Radius of the Earth Acceleration due to gravity at depth d is given as: g’ = 1 – d R E g Thebody is located at depth,d = 1 2 R E g’ = 1 – R E 2×R E g = 1 2 g Weight of the body at depth disgivenas: W’ = mg’ = 1 2 mg = 1 2 W W’ = 1 2 ×250N = 125N

Q.7 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?

Ans.

Given,radiusofeachhydrogenatom,r = 0.5Å = 0 .5×10 -10 m Volumeofeachhydrogenatom,V = 4 3 πR 3 V = 4 3 ×3.14 0 .5×10 -10 3 = 5 .236×10 -31 m 3 NumberofHatomsin1gmoleofhydrogen = 6 .023×10 23 Atomicvolumeof1gmoleofHatom = 6 .023×10 23 ×V = 6 .023×10 23 ×5 .236×10 -31 m 3 = 3 .154×10 -7 m 3

Q.8 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Ans.

Given,massofgrocer’sbox,m = 2.3kg Massoffirstgoldpiece, m 1 = 20.15g = 0.02015kg Massofthesecondgoldpiece, m 2 = 20.17g = 0.02017kg (a)Totalmassofbox, m T = m + m 1 + m 2 m T = 2.3 kg + 0.02015 kg + 0.02017 kg = 2.34032kg Sincetheresultisaccurateonlyuptooneplaceofdecimal, Onroundingoff,weobtain: Totalmassofbox = 2.3kg (b)Differenceinmassesofthegold pieces = m 2 – m 1 = 20.17 g – 20.15 g = 0.02g

Q.9 A rocket is fired vertically with a speed of 5 kms–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G= 6.67 × 10–11 Nm2kg–2.

Ans.

Here, velocity of the rocket, v = 5 kms -1 = 5×10 3 ms -1 Mass of Earth, M E = 6 .0×10 24 kg Radius of Earth, R E = 6 .4×10 6 m Height attained by rocket mass, m be h At the Earth’surface, Total energy of rocket, E T = Kinetic energy + Potential energy E T = 1 2 mv 2 + -GM E m R E At highest point h,velocityofrocket,v = 0 At highest point h,Potentialenergyofrocket = – GM E m R E +h Total energy of rocket = 0 + -GM E m R E +h = – GM E m R E +h Accordingto the law of conservation of energy, we have Total energy of rocket at the Earth’s surface = Total energy at height h 1 2 mv 2 + -GM E m R E = – GM E m R E +h 1 2 v 2 = GM E 1 R E 1 R E +h 1 2 v 2 = gR E h R E +h Here,g = GM R E 2 =9.8 ms -2 v 2 R E + h = 2gR E h h = R E v 2 2gR E -v 2 h = 6 .4×10 6 5 × 10 3 ms -1 2 2×9 .8 ms -2 ×6 .4×10 6 m – 5 × 10 3 ms -1 2 = 1 .6×10 6 m Height attained by the rocket w.r.t. the centre of the Earth, H= R E +h H = 6 .4×10 6 m +1 .6×10 6 m = 8 .0×10 6 m

Q.10 A physical quantity P is related to four observables a, b, c and d as follows:

P = a 3 b 2 ( c d ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeqabeqabiqaceGabeqabeWabeqaeeaakeaajugGbiaabcfacaqGGaGaaeypaiaabccadaWcaaGcbaqcLbyacaqGHbWaaWbaaSqabeaajugGbiaabodaaaGaaeOyamaaCaaaleqabaqcLbyacaqGYaaaaaGcbaqcLbyadaqadaGcbaqcLbyadaGcaaGcbaqcLbyacaqGJbaaleqaaKqzagGaaeizaaGccaGLOaGaayzkaaaaaaaa@4525@

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error is the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Ans.

Given,P = a 3 b 2 c d Fractionalerrorinaquantityraisedtopower n isntimesthefractionalerror intheindividualquantity, MaximumfractionalerrorinP= ΔP P ΔP P = 3 Δa a + 2 Δb b + 1 2 Δc c + Δd d ΔP P = 3 1 100 + 2 3 100 + 1 2 4 100 + 2 100 = 13 100 = 0.13 PercentageerrorinP = ΔP P ×100 = 0.13×100 = 13% Sincetheresulthastwosignificantfigures, Theresult, P=3.763wouldberoundedoffto3.8.

Q.11 The escape speed of a projectile on the earth’s surface is 11.2 kms–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Ans.

Here,escape speed of the projectilefromEarth, v esc = 11.2 kms -1 Velocity of projection of the projectile, v P = 3v esc Letmass of the projectile be m Let velocity of the projectile far away from the Earth be v f Total energy of the projectile on the Earth = 1 2 mv P 2 1 2 mv esc 2 Gravitational potential energy of the projectile far away from the Earth = 0 Total energy of the projectile far away from the Earth = 1 2 mv f 2 Accordingtothelawofconservationofenergy,wehave 1 2 mv P 2 1 2 mv esc 2 = 1 2 mv f 2 v f = v P 2 – v esc 2 v f = 3v esc 2 v esc 2 = 8 v esc = 31.68 kms -1

Q.12 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4×106 m; G = 6.67×10–11 Nm2kg–2.

Ans.

Given, height of satellite, h = 400 km = 4×10 5 m = 0 .4×10 6 m Mass of Earth, M = 6 .0×10 24 kg Mass of satellite, m = 200 kg Radius of Earth, R = 6 .4×10 6 m Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 Total energy of the satellite at height h, E T = PE + KE E T = -GMm R+h + 1 2 mv 2 Sinceorbital velocity of the satellite, v = GM R+h Total energy oftheorbitingsatelliteat height, h = 1 2 m GM R+h GMm R+h = – 1 2 GMm R+h This is alsocalledasthebound energy of the satellite. Energy required to move the satellite out of its orbit = 1 2 GM e m R e +h = 1 2 × 6 .67 × 10 –11 Nm 2 kg –2 ×6 .0 × 10 24 kg×200 kg 6 .4 × 10 6 m+0 .4 ×10 6 m = 5 .9× 10 9 J

Q.13 A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion:

(i)y = a sin 2πt T (ii)y = a sin vt(iii)y = a T sin t a (iv)y = a 2 [ sin 2πt T +cos 2πt T ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@94E9@

(a = maximum displacement of particle, v = speed of particle, T = time-period of motion). Rule out the wrong formulae on dimensional grounds.

Ans.

Angleisadimensionlessphysicalquantity. (i) 2πt T = T T = 1 = M 0 L 0 T 0 Theaboveformulaisdimensionless. (ii)vt = LT -1 T = L = M 0 L 1 T 0 Theaboveformulaisnotdimensionless. (iii) t a = T L = L -1 T 1 (iv) 2πt T = T T = 1 = M 0 L 0 T 0 Theaboveformulaisnotdimensionless. Formulae(ii)and(iii)arewrong.

Q.14 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are at a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Ans.

Here, mass of each star, M = 2 × 10 30 kg Radius of each star, R = 10 4 km = 10 7 m Initial distance between the stars, r = 10 9 km = 10 12 m For negligible speeds, v = 0 Total initialenergy of two stars = -GMM r + 1 2 mv 2 = -GMM r + 0(i) Incase the stars are about to collide: Letvelocity of the stars be v Separation between the centers of the stars = 2R Final kinetic energy of both stars= 1 2 Mv 2 + 1 2 Mv 2 = Mv 2 Final potential energy of both stars= -GMM 2R Total final energy of the two stars = Mv 2 GMM 2R (ii) Accordingto the law of conservation of energy, we have Mv 2 GMM 2R = -GMM r v 2 = -GM r + GM 2R v 2 = GM 1 r + 1 2R v 2 = 6 .67×10 -11 Nm 2 kg -2 ×2×10 30 kg× 1 10 12 m + 1 2×10 7 m 6 .67×10 12 m 2 s -2 v = 6 .67×10 12 m 2 s -2 = 2 .58×10 6 ms -1

Q.15 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Ans.

The situation is shown in the given figure. Here,mass of each sphere, M = 100 kg Distance between the spheres, r = 1 m Here,weassumedS tobe the midpoint between the spheres. Sincegravitational force applied by each sphere atthemidpointwill beequaland opposite, Gravitationalforce at midpoint S=0 Gravitational potential atmidpoint S, V G = -GM r 2 GM r 2 = -4 GM r V G = 4×6 .67×10 -11 Nm 2 kg -2 ×100 kg 1 m = -2 .67×10 -8 Jkg -1 Since effective force acting on an object placed at the mid-point is zero, therefore, any object placed at point S will be in equilibrium. However, the equilibrium is unstable, because, any variation in the position of the object will change the effective force in that direction.

Q.16 As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 1024 kg, radius = 6400 km.

Ans.

Here,mass of the Earth, M = 6 .0×10 24 kg Radius of the Earth, R = 6400 km = 6 .4×10 6 m Height of thegeostationary satellite from the surface of Earth,h = 36000 km = 3 .6×10 7 m UniversalGravitationalconstant,G= 6 .67×10 -11 Nm 2 kg -2 Gravitational potential energy atheighthdue to Earth’s gravityisgivenas: V = -GM R+h V = -6 .67×10 -11 Nm 2 kg -2 ×6 .0 × 10 24 kg 3 .6×10 7 m + 0 .64×10 7 m = -9 .4× 10 6 Jkg -1

Q.17 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

m = m 0 1 – v 2 1/2 Guess where to put the missing c.

Ans.

As per the principle of Homogeneity of dimensions, dimensions of M, L, and T on one side of dimensional physical relation should be equal to their respective dimensions on the other side of the relation. On RHS, the denominator 1 – v 2 1/2 must be dimensionless. Inplaceof 1 – v 2 1/2 ,thereshouldbe 1 – v 2 c 2 1/2 Thecorrectrelationis:m = m 0 1 – v 2 c 2 1/2

Q.18 A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2×1030 kg)

Ans.

The body will stuck to the surface of a star if the inward gravitational force is morethan the outward centrifugalforce caused by the rotation of the star. Here,radius of the star,R = 12 km = 1 .2 ×10 4 m Mass of the star,M = 2 .5×2×10 30 = 5×10 30 kg Letmass of the body=m Gravitational force, f g = GMm R 2 Here,G = 6 .67×10 -11 Nm 2 kg -2 f g = 6 .67×10 -11 Nm 2 kg -2 ×5 × 10 30 kg×m 1 .2 ×10 4 m 2 = 2 .31×10 11 mN Centrifugal force, f c = mrω 2 Angular speed, ω = 2πν f c = mR 2πν 2 Here,angular frequency,ν = 1 .2 revs –1 f c = m×(1 .2 ×10 4 m)×4× 3.14 2 × 1 .2 revs –1 2 = 1 .7×10 5 mN As f g > f c , the body will remain stuck to the star.

Q.19 A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108 kg; G = 6.67 × 10–11 Nm2kg–2.

Ans.

Here, mass of spaceship, m S = 1000 kg Mass of Sun, M = 2×10 30 kg Mass of Mars, m M = 6 .4×10 23 kg Radiusoftheorbit of Mars, R = 2 .28×10 8 km = 2 .28 ×10 11 m Radius of Mars, r = 3395 km = 3 .395×10 6 m Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 P.E. of the spaceship becauseofthe gravitational attraction of the Sun, (PE) S = -GMm S R Potential energy of the spaceship becauseof the gravitational attraction of Mars, (PE) M = -Gm M m S r As the spaceship is stationed on Mars, its velocity =0 Kinetic energyof the spaceship = 0 Total energy of the spaceship, E T = (PE) S + (PE) M E T = -GMm S R -Gm M m S r = -Gm S M R + m M r The negative sign shows that the system is in bound state. Energy required for launching the spaceship from the solar system = – Total energy of the spaceship = – Gm S M R + m M r = Gm S M R + m S r = 6 .67×10 –11 Nm 2 kg –2 ×1000 kg× 2 × 10 30 kg 2 .28 ×10 11 m + 6 .4 × 10 23 kg 3 .395 × 10 6 m = 5 .98×10 11 J

Q.20 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Ans.

Given,sizeofhydrogenmolecule = 1Å Atomicradius,r = 0 .5×10 -10 m VolumeofeachHatom,V = 4 3 πR 3 V = 4 3 ×3.14× 0 .5×10 -10 m 3 = 5 .236×10 -31 m 3 Sincenumberofhydrogenatomsin1gmoleofhydrogen, Avogadro’sNumber, N 0 = 6 .023×10 23 Atomicvolumeof1gmoleofH atom = N 0 ×V = 5 .236×10 -31 m 3 ×6 .023×10 23 = 3 .154×10 -7 m 3 Molarvolume = 22.4L = 22 .4×10 -3 m 3 Molarvolume Atomicvolume = 22 .4×10 -3 m 3 3 .154×10 -7 m 3 = 7 .1×10 4 Thisratioislargebecauseoflargeintermolecularseparations.

Q.21 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 kms–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 Nm2kg2.

Ans.

Here,mass of theMars, M = 6 .4×10 23 kg Initial velocity of the rocket, v = 2 kms -1 = 2×10 3 ms -1 Radius ofthe Mars, R = 3395 km = 3 .395×10 6 m Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 Mass of the rocket be m Initial K.E. of the rocket = 1 2 mv 2 Initial P.E. of the rocket= -GMm R Total initial energyoftherocket= 1 2 mv 2 GMm R If 20 % of initial kinetic energy is lost becauseof Martian atmospheric resistance, thenonly 80 % of its kinetic energy helps in reaching a height. Totalamountof initial energy available = 80 100 × 1 2 mv 2 GMm R = 0.4 mv 2 GMm R Letmaximum height achieved by the rocket be h At maximum height, the velocity of the rocket = 0 At maximum height, the K.E. of the rocket = 0 At height h,total energy of the rocket = – GMm R+h Accordingto the law of conservation of energy for the rocket, we have: 0.4 v 2 = GM R GM R+h R+h h = GM 0.4 v 2 R h = 0.4 R 2 v 2 GM-0.4 v 2 R h = 0.4× 3 .395 × 10 6 m 2 × 2× 10 3 ms -1 2 6 .67×10 –11 Nm 2 kg –2 ×6 .4 × 10 23 kg -0.4× 2× 10 3 ms -1 2 × 3 .395 × 10 6 m = 495km

Q.22 Explain this common observation clearly. If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the moon, the stars etc.) seem to be stationary.

Ans.

The imaginary line joining an object to the eye is called the line of sight. When a train is moving rapidly, the line of sight of a nearby house changes its direction very rapidly. Therefore, the house seems to move rapidly in a direction opposite to the train’s motion.

Since distant objects such as the hill top, the moon, the stars, etc. are extremely large distances apart, therefore, the line of sight does not change its direction rapidly and they appear to be stationary.

Q.23 The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011 m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1’’ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1’’ (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Ans.

Given,dameterofEarth’s orbit = 3×10 11 m RadiusofEarth’sorbit,r = 1 .5×10 11 m Giventhat,parallaxangle,θ = 1″ θ = 1′ 60 = 1° 60×60 = π 180 × 1 60×60 rad = 4 .847×10 -6 rad Letthedistanceofstar be D Fromtherelation: θ = r D D = r θ D = 1 .5×10 11 4 .847×10 -6 = 3 .09×10 16 m 1parsec3 .09×10 16 m

Q.24 The nearest star to our solar system is 4.29 light years away. How much is the distance in terms of parsec? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Ans.

Distancebetweenthestarandthesolarsystem=4.29ly 1lightyear = speedoflight×1year 1light year = 3×10 8 ms -1 ×365×24×60×60 s = 94608×10 11 m 4.29ly = 4 .29×94608×10 11 m = 405868 .32×10 11 m 1parsec = 3 .08×10 16 m 4.29ly = 405868 .32×10 11 3 .08×10 16 = 1.32parsec Here,diameterofEarth’sorbit, d = 3×10 11 m DistanceofStarfromtheEarth,D = 405868.32× 10 11 m Fromtherelation: θ = d D θ = 3×10 11 405868 .32×10 11 = 7 .39×10 -6 rad 1sec = 4 .85×10 -6 rad 7 .39×10 -6 rad = 7 .39×10 -6 4 .85×10 -6 = 1.52″

Q.25 Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern Science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Ans.

It is true that precise measurements of physical quantities are essential for the development of laws of Physics or any other Science. For example:

(a) In various physical and chemical processes, ultra-shot LASER pulses are used to determine small time intervals.

(b) Mass spectrometer can be used to measure the mass of atoms precisely.

(c) The inter-atomic separation can be determined by using X-ray spectroscopy.

Q.26 Just as precise measurements are necessary in Science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.

Ans.

(a)Duringthemonsoonseason,ametrologistrecordsnearly 215cmofrainfallinIndia. Heightofwatercolumn,h = 215cm = 2.15m AreaoftheCountry,A = 3.3× 10 12 m 2 Volumeofrainwater,V = A×h = 7.09× 10 12 m 3 densityofrainwater,ρ = 1× 10 3 kgm -3 Massofrainwater = ρV = 7.09× 10 15 kg (b)LetusassumeaboatofknownbaseareaA. Letdepthofboatinsea be d 1 Volumeofwaterdisplacedbyboat, V 1 = Ad 1 Nowmovetheelephantontheboat Letdepthofboatwhenelephantismovedonthe boat be d 2 Volumeofwaterdisplacedbyboatandelephant, V 2 = Ad 2 Volumeofwaterdisplacedbyelephant, V = V 2 – V 1 = A d 2 – d 1 Letdensityofwater be ρ Massofelephant = Massofwaterdisplacedbyelephant = Vρ = A d 2 -d 1 ρ (c) With the help of an anemometer we can measure the wind speed. The wind speed is given by the rotation made by the anemometer in one second. (d)Letareaofheadcarryinghair be A Theradiusofahaircanbemeasuredwithhelpofascrewgauge. Letradiusofahair be r Areaofonehair = π r 2 Consideringthattheuniformdistributionofhairoverthehead, Numberofstrandsofhair = Totalsurfacearea Areaofonehair = A π r 2 (e)VolumeoccupiedbyonemoleofairatNTP = 22.4l = 22 .4×10 -3 m 3 Numberofmoleculesinonemoleofair = 6 .023×10 23 Letvolumeoftheroom be V NumberofairmoleculesinroomofvolumeV = 6 .023×10 23 22 .4×10 -3 m 3 ×V = V×1 .35×10 28 m -3

Q.27 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m.

Ans.

Given,MassoftheSun,M = 2 .0×10 30 kg RadiusoftheSun,R = 7 .0×10 8 m VolumeoftheSun,V = 4 3 πR 3 V = 4 3 × 22 7 × 7 .0×10 8 m = 1 .4377×10 27 m 3 DensityofSun = Mass Volume DensityofSun = 2 .0×10 30 kg 1 .4377×10 27 m 3 = 1 .3911×10 3 kgm -3 The density of sun is of order of the density of solids and liquids. The enormous gravitational attraction of the inner layers on the outerlayer of Sun is responsible for the high density of Sun.

Q.28 When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72” of arc. Calculate the diameter of Jupiter?

Ans.

Given,distancebetweenJupiterandEarth,r = 824 .7×10 6 km AngulardiameterofJupiter,θ = 35.72″ θ = 35.72 60×60 × π 180 rad LetdiameterofJupiter be l. Fromtherelation: θ = l r l = θr l = 35.72 60×60 × π 180 rad×824 .7×10 6 km = 1 .429×10 5 km

Q.29 A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit: as v → 0, θ → 0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man).

Do you think this relation can be correct? If not, guess the correct relation.

Ans.

The given relation is incorrect dimensionally . One way to make thegivenrelationcorrectisbydividing the R.H.Softhegivenrelationbythespeedofrainfallv’. Thegivenrelationbecomes: tanθ = v v’ Thisrelationisdimensionally correct.

Q.30 It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time interval of 1 s?

Ans.

Given,errorintimeofcesiumclocks= 0.02s Timetaken = 100years = 100×365 1 4 ×24×60×60s Errorintimeofcesiumclocksin100years = 0.02s Errorintimeofcesiumclocksin1s, Δt = 1 s× 0.02s 100×365 1 4 ×24×60×60 s Δt = 6 .337×10 -12 s 10 -11 s Accuracyofastandardcesiumclockformeasuringthe timeintervalof1s= 10 -11 s

Q.31 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kgm-3. Are the two densities of the same order of magnitude? If so, why?

Ans.

Given,sizeofsodiumatom = Diameterofsodiumatom = 2.5Å Radiusofsodiumatom,r = 1 2 ×2.5Å = 1.25Å = 1.25× 10 -10 m Volumeofsodiumatom,V = 4 3 πr 3 = 4 3 × 22 7 × 1 .25×10 -10 m 3 AsperAvogadrohypothesis, Numberofatomsinonemoleofsodium= 6 .023×10 23 Massofonemoleofsodium = 23 g = 23×10 -3 kg Massofoneatomofsodium = 23×10 -3 6 .023×10 23 kg Densityofsodiumatom = mass volume Densityofsodiumatom = 23×10 -3 6 .023×10 23 kg 4 3 × 22 7 × 1 .25×10 -10 m 3 = 4 .67×10 3 kgm -3 Duetotheinter-atomicseparationinthecrystallinephase,thetwodensitiesarenotofthesameorder.

Q.32 The unit of length convenient on the nuclear scale is a fermi: 1 f = 10-15 m. Nuclear sizes obey roughly the following empirical relation:
r = r0A 1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Ans.

Given,radiusofnucleus, r = r 0 A 1/3 Volumeofnucleus,V = 4 3 π r 3 = 4 3 π r 0 A 1/3 3 = 4 3 π r 0 3 A themassofthenucleus =Massnumber M = Aamu = A×1 .66×10 -27 kg Densityofsodiumnucleus,ρ = 3×1 .66×10 -27 kg 4×3.14× 1 .2×10 15 m 3 ρ = 4.98 21.71 ×10 18 kgm -3 = 2 .29×10 17 kgm -3 Densityofsodiumatom,ρ’ = 4 .67×10 3 kgm -3 ρ ρ’ = 2 .29×10 17 kgm -3 4 .67×10 3 kgm -3 = 4 .9×10 13

Q.33 A LASER is a source of very intense, monochromatic and unidirectional beam of light. These properties of a LASER light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a LASER as a source of light. A LASER light beamed at the moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?

Ans.

Given,timetakenbytheLASERbeamtoreturnearth,t = 2.56s Velocityoflightinvacuum, c = 3×10 8 ms -1 Radiusoflunarorbit = SeparationbetweenEarthandMoon Letradiusoflunarorbit be x Fromtherelation: x = c×t 2 x = 3×10 8 ms -1 ×2.56 s 2 = 3 .84×10 8 m

Q.34 A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine?
(Speed of sound in water = 1450 ms-1).

Ans.

Letdistancebetweentheshipandtheenemysubmarine = s Speedofsoundinwater,v = 1450 ms -1 TimedelaybetweentransmissionandreceptionofSONARwaves,t = 77s Fromtherelation: s = v×t 2 Distancebetweenshipandenemysubmarinesisgivenas: s = 1450 ms -1 ×77 s 2 = 55825m

Q.35 The farthest objects in our universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have yet not been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Ans.

Given,timetakenbyquasarlighttoreachEarth,t= 3billionyears t = 3×10 9 ×365×24×60×60s Velocityoflightinvacuum, c = 3×10 8 ms -1 As,distance = velocity×time DistancebetweenEarthandquasar, d = ct d = 3×10 8 ms -1 ×3×10 9 ×365×24×60×60s = 2 .84×10 22 km

Q.36 It is a well known fact that during a solar eclipse, the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Ans.

ThegivenfigureshowsthepositionsofSun,MoonandEarth atthetimeofalunareclipse. Here,distancebetweenMoonandEarth,UE = 3 .84×10 8 m DistancebetweenSunandEarth,SE = 1 .496×10 11 m DiameterofSun,AB = 1 .39×10 9 m ΔEMNandΔEABaresimilar AB MN = SE UE 1 .39×10 9 m MN = 1 .496×10 11 m 3 .84×10 8 m MN = 1.39×3.84 1.496 ×10 6 m = 3 .57×10 6 m DiameterofMoon = 3 .57×10 6 m

Q.37 A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic Physics(c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number,its magnitude being close to the present estimate on the age of the universe (~ 15 billion years).From the table of fundamental constants in this book, try to see if you can construct this number(or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Ans.

Onesuchquantityconsistingofsomefundamentalconstantsandhavingdimensionsoftimeisgivenas: t = e 2 4πε 0 2 × 1 m p m e 2 c 3 G Here, ε 0 representsabsolutepermittivity erepresentingchargeofelectrons = 1.6× 10 -19 C m e representingmassofelectrons = 9 .1×10 -31 kg m p representingmassofprotons = 1 .67×10 -27 kg 1 4πε 0 = 9×10 9 Nm 2 C -2 crepresentingspeedoflight = 3× 10 8 ms -1 andG = 6 .67×10 -11 Nm 2 kg -2 Puttingtheabovevaluesinthegivenrelation,weobtain: t = 1 .6×10 -19 4 × 9×10 9 2 9 .1×10 -31 2 ×1 .67×10 -27 × 10 8 3 ×6 .67×10 -11 = 2 .18×10 16 s

Q.38 In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.

Ans.

(a) The carriage moving without jerks is considered as a point object because its size is too small compared to the distance between the stations.

(b) The monkey is considered as a point object because the size of the monkey is too small compared to the size of a circular track.

(c) The spinning cricket ball is not considered as a point object as its size is comparable to the distance the ball might turn after hitting the ground.

(d) A beaker slipping off the edge of a table is not considered as a point object as its size is comparable to the height of the table in this case.

Q.39 The position-time (x-t) graphs for two children A and B returning from their school to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;
(a) (A/B) lives closer to school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).

Ans.

(a) In the given graph, distance OP<oq, therefore=””> </oq,>

(b) In the given graph, x = 0 and t = 0 for A, but t has a finite value for B. Therefore, A starts from the school earlier than B.

(c) The slope of x-t graph is equal to the speed. In the given x-t graph, the slope of B is greater than that of A. Therefore, B walks faster than A.

(d) From the given graph, it can be observed that A and B reach home at the same time.

(e) From the given graph we can deduce that A starts earlier than B, also, B has a greater speed than A. Thus, we can say that B overtakes A on the road once.

Q.40 A woman starts from her home at 9.00 am, walks with a speed of 5 kmh–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 kmh–1. Choose suitable scales and plot the x-t graph of her motion.

Ans.

Given,speedofthewoman = 5 kmh -1 Distancebetweenhomeandoffice = 2.5km Timetakentoreachoffice = Distance speed = 2.5km 5 kmh -1 = 0.5h Speedofauto = 25 kmh -1 Timetakentoreturnfromoffice = 2.5 km 25 kmh -1 = 0.1h = 6min The x-t graph of the motion of the woman is given below.

Q.41 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Ans.

Distance travelled in 1 step = 1 m

Time taken to walk 1 step = 1 s

Time taken to walk first 5 m forward = 5 s

Time taken to walk 3 m backward = 3 s

Total distance travelled = 5 m – 3 m = 2 m

Total time taken to travel 2 m = 8 s

Time taken to travel 2 m = 8 s

Time taken to travel 4 m = 16 s

Time taken to travel 6 m = 24 s

Time taken to travel 8 m = 32 s

Now, the drunkard will move 5 steps forward in the next 5 s and cover a distance of 5 m and a total distance of 13 m and fall into the pit.

Therefore, total time taken by the drunkard to cover 13 m = 32 s + 5 s = 37 s

Q.42 A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 kmh–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans.

Given, speed of the jet airplane, v A = 500 kmh -1 Relative speed of the combustion products ofjet airplane w.r.t . the jet airplane, v PA = -1500 kmh -1 Let relative speed of the combustion products w.r.t . the observer on the ground be v P Relative speed of the products of combustion w.r.t . the jet airplane, v PA = v P – v A v P = v A + v PA v P = 500 kmh -1 – 1500 kmh -1 = -1000 kmh -1 The negative sign shows that the direction of combustion products is opposite to that of the jet airplane.

Q.43 A car moving along a straight highway with speed of 126 kmh-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Ans.

Given, initial velocity of the car, u = 126 kmh -1 = 35 ms -1 Finalvelocity of the car,v = 0 Distancetraveledbythecar, S = 200 m Letaccelerationproducedinthecar be a Accordingtothethirdequationofmotion, v 2 – u 2 = 2aS 0 – 35 ms -1 2 = 2a 200 m a = – 35 ms -1 2 2×200 m = -3.06 ms -2 Here,thenegativesignshowsretardation. Accordingtothefirstequationofmotion,v = u + at 0 = 35 + 49 16 t t = 35×16 49 s = 11.43s

Q.44 Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 kmh-1 in the same direction with A ahead of B. The driver of B decides to overtake A and accelerate by 1 ms-2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Ans.

IncaseoftrainA, Given,initialvelocity, u = 72 kmh -1 = 20 ms -1 Time,t = 50s Asthetrainismovingwithuniformvelocity, Accelerationofthetrain,a = 0 Letdistancetravelledbytrain A be S A Accordingtothesecondequationofmotion,S = ut + 1 2 at 2 S A = 20 ms -1 ×50 s+ 1 2 ×0 ms -2 × 50 s 2 = 1000 m FortrainB, Given,initialvelocity, u = 72 kmh -1 = 20 ms -1 Time,t = 50s Accelerationofthetrain, a = 1 ms -2 Letdistancetravelledbytrain B be S B Accordingtothesecondequationofmotion,S = ut + 1 2 at 2 S B = 20 ms -1 ×50 s + 1 2 ×1 ms -2 × 50 s 2 = 2250m TheoriginaldistancebetweenthedriveroftrainAandguardoftrainB=2250 m – 1000 m = 1250m

Q.45 On a two lane road, car A is travelling with a speed of 36 kmh-1. Two cars B and C approach car A in opposite directions with a speed of 54 kmh-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Ans.

Given, velocity of car A, v A = 36 kmh -1 = 10 ms -1 VelocityofcarB, v B = 54 kmh -1 = 15 ms -1 VelocityofcarC, v C = 54 kmh -1 = 15 ms -1 RelativevelocityofcarBw.r.t.carAisgivenas: v BA = v B – v A v BA = 15 ms -1 – 10 ms -1 = 5 ms -1 RelativevelocityofcarCw.r.t.carAisgivenas: v CA = v C – (- v A ) = 15 ms -1 + 10 ms -1 = 25 ms -1 Atadefiniteinstantoftime,AB = AC = 1km = 1000m TimetakenbycarCtoovertakecarA = 1000 m 25 ms -1 = 40s Toavoidanaccident,carBmusttravelsamedistancewithin40s. Accordingtosecondequationofmotion,theminimumacceleration (a B )produced bycarBcanbecalculatedas: s = ut + 1 2 at 2 Putting, u = v BA = 5 ms -1 ;t = 40s;s = 1000 m; a = a B intheaboverelation,weobtain: 1000 = 5×40+ 1 2 ×a B × 40 2 a B = 1 ms -2

Q.46 Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 kmh–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

Ans.

LetthespeedofthebusrunningbetweentwotownsAandB be V Given,speedofthecyclist,v = 20 kmh -1 Relativevelocityofthebusmovinginthedirectionofthecyclist = V – v = V – 20 kmh -1 Relativevelocityofthebusmovingindirectionofthecyclist = V + v = V + 20 kmh -1 LetdistancetraveledbythebusintimeTminutes be d Accordingtothefirstconditionofthequestion: d v – 20 = 18 min d = 18v – 18×20(i) Accordingtothesecondconditionofthequestion: d v + 20 = 6min d = 6v + 20×6(ii) Fromequation(i)and(ii),weobtain: 18v – 18×20 = 6v + 20×6 12v = 20×6 + 18×20 = 480 v = 40 kmh -1 Substitutingthisvalueofvinequation(i),weobtain: 40T = 18×40 – 18×20 T = 18×20 40 = 9min

Q.47 A player throws a ball upwards with an initial speed of 29.4 ms-1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 and t = 0 be the location and time at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball returns to the player’s hands. (Take g = 9.8 ms-2 and neglect air resistance)

Ans.

(a)As the ball is moving upwards and gravitational forceis in downward direction. Thedirectionofaccelerationduetogravityisalwaysverticallydownwards. (b)Atthehighestpointofitsmotion,thevelocityoftheballbecomeszeroandthe accelerationduetogravityremainsconstantwithconstantvalue9.8 ms -2 . (c)Thesignofpositionwill bepositiveforboththeupwardanddownwardmotionoftheball. Thesignofvelocityisnegativeduringtheupwardmotionandpositiveduringthedownwardmotion. Thesignofacceleration ispositiveforboththeupwardanddownwardmotion. (d)Given,initialvelocityoftheball,u = 29.4 ms -1 Finalvelocityoftheball,v = 0 (Atmaximumheight,velocityofball = 0) Acceleration,a = -g = -9.8 ms -2 Lettbethetimetakenbytheballtoreachthehighestpoint,wheredistancefromthegroundisS. Accordingtothethirdequationofmotion, v 2 – u 2 = 2gS S = v 2 – u 2 2g = 0 ms -1 2 29 .4 ms -1 2 2 -9 .8 ms -2 = 44.1m Accordingtofirstequationofmotion,timeofascent(t)isgivenas: v = u + at t = v – u a t = 0 – 29 .4 ms -1 -9 .8 ms -2 = 3s As,timeofascent = timeofdescent Totaltimetakenbyballtocomebacktoplayer’shands = 3 s + 3 s = 6s

Q.48 Read each statement carefully and state with reasons and examples if it is true or false;
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity
(c) with constant speed may have zero acceleration
(d) With positive value of acceleration must be speeding up.

Ans.

(a) True, as when there is retardation, comes a point of inflection where the speed becomes zero.

(b) False, as the speed of the object at an instant will always be equal to the magnitude of velocity at that instant.

(c) True, since, constant speed can have a constant velocity while acceleration of an object is the rate of change of velocity with respect to time, therefore, acceleration of the object is equal to zero.

(d) The given statement is true if both velocity and acceleration are positive, at the instant of time taken as origin, example, when the body is falling vertically downwards.

Q.49 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Ans.

Given,Heightfromwhichtheballisdropped,s = 90m Initialvelocityoftheball,u = 0 Acceleration,a = 9.8 ms -2 Letfinalvelocityoftheball be v Accordingtosecondequationofmotion,weobtain: s = ut + 1 2 at 2 90 = 0 + 1 2 ×9 .8t 2 t = 18 .38 s 2 = 4.29s Accordingtofirstequationofmotion,finalvelocity(v)is givenas: v = u + at v = 0 + 9 .8 ms -2 ×4.29 s = 42 .04 ms -1 Reboundvelocityofballisgivenas:u’ = 9 10 v u’ = 9 10 ×42 .04 ms -1 = 37 .84ms -1 Time(t’)takenbytheballtoreachthehighestpointis givenby, t’ = u’ a t’ = 37 .84 ms -1 9 .8 ms -2 = 3.86 s Totaltime, T = t + t’ T = 4.29+3.86 = 8.15s As,timeofascent = timeofdescent Theballtakes3.86stofallbackonthefloor,where, Velocitybeforestrikingthefloor = 37 .84 ms -1 Velocityafterstrikingthefloor = 9 10 ×37 .84 ms -1 = 34 .05 ms -1 Totaltimetakenforsecondrebound = 8.15 s + 3.86 s = 12.01s Thespeed-timegraphofthemotionoftheballisshowninthegivenfigureas:

Q.50 Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality time true? [For simplicity, consider one dimensional motion only].

Ans.

(a) The magnitude of displacement will be the shortest distance between the initial and final positions of the particle over an interval of time.

The total length of the path covered is the actual total path covered by the particle in the given time interval.

For example, a particle moves from point P to point Q and then, comes back to a point R after the total time t, as shown in the given figure.

Then the displacement will be the length PR, that is, the distance between P and R while total length of path covered will be PQ + RQ (distance covered).

(b)Magnitudeofaveragevelocity = Magnitudeofdisplacement Time Incaseofthegivenparticle,averagevelocity = PR t Averagespeed = Totalpathlength Time = PR+RQ t As( PR + RQ )> PR Averagespeedisgreaterthanaveragevelocity. For the equality sign in a given time, the particle needs to move along a straight line. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@70AA@

Q.51 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh–1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time
(i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
[Note: You will appreciate from this exercise why itis better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

Ans.

Timetakenbythemantogotomarketfromhome, t 1 = 2.5 km 5 kmh -1 = 1 2 h = 0.5 h = 30min Timetakenbythemantogotohomefrommarket, t 2 = 2.5 km 7 .5 kmh -1 = 1 3 h = 20min Totaltimetakenbytheman = 30 min + 20 min = 50min (i)0to30min (a)Averagevelocity = displacement time = 2.5 km 0.5 h = 5 kmh -1 (b)Averagespeed = Total distance time = 2.5 km 0.5h = 5 kmh -1 (ii)0to50min Totaldistance = 2.5 km + 2.5 km = 5 km Totaldisplacement = 0 (a)Averagevelocity = displacement time = 0 (b)Averagespeed = total distance time = 5 km 5 6 h = 6 kmh -1 (iii)0to40min Distancemovedinfirst30min(fromhometomarket) = 2.5km Distancemovedinnext10min(frommarkettohome) = 7.5 km× 10 60 = 1.25km Totaldistancecovered = 2.5 km + 1.25 km = 3.75km Totaldisplacement = 2.5 km – 1.25 km = 1.25km (a)Averagevelocity = displacement time = 1.25 km 40 60 h = 1.875 kmh -1 (b)Averagespeed = distance time = 3.75 km 40 60 h = 5.625 kmh -1

Q.52 In the above questions 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Ans.

Instantaneous speed of a particle at any instant of timeis defined as the first derivative of distance w.r.t. timeat thatinstant of time i.e. v ins = lim Δt0 Δx Δt v ins = dx dt Since, in instantaneous speed, we take only a small interval of time dt during which the particle is not supposed to change its direction of motion. Therefore, there is no difference between the path length and magnitude of displacement of the particle during this interval of time. Therefore, thevalueofinstantaneous speed is always equal to the magnitude of instantaneous velocity.

Q.53 Figure 3.23 gives the x-t plot of a particle executing one dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t=0.3 s, 1.2 s, -1.2 s.

Ans.

(a)At time t = 0.3 s, x is negative. Therefore, the slope ofx-t graph is also negative. Thus, position and velocityare negative. Since, this graph represents a sinusoidal motion; for SHM of a particle, acceleration is given as: a = -ω 2 x Acceleration ispositive. (b) At time t = 1.2 s, x is positive, thus the slope ofx-t graph is also positive. Hence, position and velocityare positive. Since for SHM of a particle, acceleration is given as: a = -ω 2 x Acceleration isnegative. (c) At time t = -1.2 s, x is negative, therefore, the slope of x-t graph is also negative. But, as both x and t are negative, v becomes positive. A is also positive applying the SHM equation.

Q.54 Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one dimensional motion of a particle.

Ans.

(a) This graph cannot represent one dimensional motion of a particle, because in a one dimensional motion a particle cannot have two positions at the same instant of time.

(b) This graph cannot represent one dimensional motion of a particle, because a particle cannot have two values of velocity at the same instant of time in one dimensional motion.

(c) This graph cannot represent one dimensional motion of a particle, because according to this graph the particle can have negative speed, but speed being a scalar quantity cannot be negative.

(d) This graph cannot represent one dimensional motion of a particle, because the total path length of a particle cannot decrease with time.

Q.55 Figure 3.21 shows x-t plot of one dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Ans.

No, because the given x-t graph does not represent the trajectory of the path followed by a particle moving in a straight line as t = 0, x = 0. The given graph can represent the motion of a body with constant acceleration.

Q.56 A police van moving on a highway with a speed of 30 kmh-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 kmh-1. If the muzzle speed of the bullet is 150 ms-1, with what speed does the bullet hit the thief’s car? (Note: obtain that speed which is relevant for damaging the thief’s car).

Ans.

Given, speed of police van, vP = 30 kmh-1 = 8.33 ms-1

Muzzle speed of bullet, vB = 150 ms-1

Speed of the car of the thief, vT = 192 kmh-1 = 53.33 ms-1

Since the bullet is fired from the police van, its effective speed is, vB = vB + vP

vB = 150 ms-1 + 8.33 ms-1 = 158.33 ms-1

As, both the vehicles are traveling in the same direction, the speed of bullet w.r.t. thief’s car,

vBT = vB – vT

vBT = 158.33 ms-1 – 53.33 ms-1 = 105 ms-1

Q.57 Suggest a physical situation for each of the following graph.

Ans.

(a) The given x-t graph shows that initially x = 0, then it increases with time and attains a constant value for an instant and reduces to zero with time, then it increases in the opposite direction till it again attains a constant value i.e. comes to rest.

A similar physical situation arises when a football is kicked towards a player and he passes it back with a reduced speed but it misses the player and hits the goal finally coming to rest.

(b) In the given graph the velocity changes sign again and again with the passage of time and its magnitude decreases every time. A similar physical situation appears in the case of damping of a to and fro pendulum motion.

(c) The given graph shows that initially, the body moves with uniform velocity. Its acceleration increases for a small duration, which again reduces to zero. It shows that the body again starts moving with constant velocity. A similar situation appears when a hammer moving with a uniform velocity strikes a nail.

Q.58 Figure 3.24 gives the x-t plot of a particle in one dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Ans.

The average speed in a small interval of time is equal to the slope of x-t graph in that interval of time. From the graph, it is clear that the slope is maximum and minimum in intervals 3 and 2 respectively. Thus, the average speed of the particle is the maximum in interval 3 and the minimum in interval 2. The average speed is positive in both intervals 1 and 2 because the slope of x-t graph is positive there. But, the average speed is negative in interval 3 because the slope of x-t graph is negative in this interval.

Q.59 Figure 3.25 gives a speed time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of ‘u’ and ‘a’ in the three intervals. What are the accelerations at the points A, B, C and D?

Ans.

The slope of the speed-time graph gives acceleration. As the slope of the speed-time graph is the greatest in interval 2, hence the magnitude of the average acceleration is the maximum in this interval. The height of the curve from the time axis tells the average speed of the particle. From the given speed-time graph, it is clear that height is the greatest in interval 3. Therefore, during interval 3 the average speed of the particle is the greatest.

In interval 1, the slope of the given speed-time graph is positive. Therefore, acceleration is positive in this interval. In the same way, speed is also positive in this interval of time.

In interval 2, the slope of the given speed-time graph is negative. Therefore, the sign of acceleration is negative in this interval. But, speed is positive because it is a scalar physical quantity.

In interval 3, the slope of the given speed-time graph is zero. Therefore, acceleration is equal to zero in this interval. But, the particle gains some uniform speed and it is positive in this interval.

At points A, B, C and D, the speed-time graph is parallel to the time axis. Therefore, the acceleration of the particle is zero at points A, B, C and D.

Q.60 A three wheeler starts from rest, accelerates uniformly with 1 ms-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n= 1, 2, 3…..) versus n. What do you expect the plot to be during accelerated motion: a straight line or a parabola?

Ans.

In a straight line, the distance covered by a particle in n th secondisgivenas: D nth = D n – D n-1 , Dn = un + 1 2 an 2 Dn-1 = u(n – 1) + 1 2 a(n – 1) 2 D nth = u + a 2 2n – 1 Given,u = initialvelocity = 0 Acceleration,a = 1 ms -2 D n = 1 2 2n – 1 Substitutingn = 1,2,3,….,wecanfindthevalueof D n . Thedifferentvaluesofnandcorrespondingvaluesof D n areshownbelow:

Thegraphbetweennand D n willbeastraightlineasshownbelow. Asthethree-wheelergainstheuniformvelocityaftern = 10s,thelinewillbeparalleltothetimeaxisaftern = 10s.

Q.61 A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms–1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 ms-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

Ans.

Given,initialvelocityoftheball,u = 49 ms -1 Acceleration,a = -g = -9.8 ms -2 Case1:Whentheliftisstationary Theboythrowstheballupwards. Finalvelocityoftheballatthehighestpoint = 0 Accordingtothefirstequationofmotion,timeofascentisgivenas: v = u + at t = v – u a t = -49 ms -1 -9 .8 ms -2 = 5s But,timeofascent = timeofdescent Totaltimetakenbytheballtocomebacktotheboy’shand = 5 s + 5 s = 10s Case2:Whentheliftismovingupwardswithuniformvelocityof5 ms -1 Inthiscase,thereisnochangeintherelativevelocityoftheballw.r.t.theboyi.e., itremains49 ms -1 . Thus,inthiscasealso,theballwillcomebacktotheboy’shandafter10s.

Q.62 On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 kmh–1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 kmh–1. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time taken by the child in (a) and (b)? Which of the answers alter if motion is viewed by one of the parents?

Ans.

(a)Inthiscase, speedofthebelt, v B = 4 kmh -1 Speedofthechildw.r.t.thebelt, v C = 9 kmh -1 Themotionofthechildisinthedirectionofmotionofthemotionofthemovingbelt. Speedofthechildw.r.t.thestationary observer, v C’ = v C + v B v C’ = 9 kmh -1 + 4 kmh -1 = 13 kmh -1 (b)Inthiscase, speedofthebelt, v B = 4 kmh -1 Speedofthechildw.r.t.thebelt, v C = -9 kmh -1 Speedofthechildw.r.t.thestationary observer, v C’ = v C + v B v C’ = -9 kmh -1 + 4 kmh -1 = -5 kmh -1 Thenegativesignshowsthatthemotionofthechildisoppositetothedirectionofmotionofthebelt. (c)Given,distancebetweentheparents,s = 50m Asparentsandchildarelocatedonthesamebelt,thespeedofthechildineitherdirectionasobserved bytheparentswillremainthesamei.e.,9 kmh -1 = 2.5 ms -1 . Timetakenbythechildincase(a)and(b)is,t = 50 m 2 .5 ms -1 = 20s (d)Ifmotionisobservedbyanyoneoftheparents,answersobtainedin(a)and(b)willgetchanged. Thisisbecausethechildandparentsarelocatedonthesamebelt,therefore,thespeedofthechild w.r.t.eitherofthefatherormotherissamei.e.,9 kmh -1 . But,answer(c)remainsunchanged,asparentsandchildareonthesamebeltandallofthemareequally affectedbythemotionofthebelt.

Q.63 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms–1 and 30 ms–1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms–2. Give the equations for the linear and curved parts of the plot.

Ans.

Given,heightofthe cliff, x 0 = 200m Incaseoffirststone: Initialvelocity, u 1 = 15 ms -1 Acceleration,a = 10 ms -2 Accordingtotheequationofmotion,weobtain: x 1 = x 0 + u 1 t + 1 2 at 2 x 1 = 200 m+ 15t – 5t 2 (i) Whenthestonereachestheground, x 1 = 0 -5t 2 +15t + 200 = 0 t 2 – 3t – 40 = 0 t = 8sort =-5s Since,t=0istheinstant,whenthestonewasprojected,hencethenegativetimehasnomeaning. t = 8s Incaseofsecondstone, initialvelocity, u 2 = 30 ms -1 Acceleration,a = -10 ms -2 Accordingtotheequationofmotion: x 2 = x 0 + u 2 t + 1 2 at 2 x 2 = 200 + 30t – 5t 2 (ii) Whenthestonereachestheground; x 2 = 0 -5t 2 + 30t + 200 = 0 t 2 – 6t – 40 = 0 t = 10sort = -4s Since,negativesignismeaningless t = 10s Subtractingeqation(i)fromequation(ii),weget x 2 – x 1 = 200 + 30t – 5t 2 200 + 15t – 5t 2 x 2 – x 1 = 15t (iii) Equation(iii)showsthelinearpathofboththestones.Becauseofthislinearrelationbetweenx andt,thepathisastraightlinetillt = 8s. Formaximumseparationbetweenthetwostones, t = 8s Maximumseparation, x 2 – x 1 max = 15 ms -1 ×8 s = 120m After8s,onlysecondstoneisinmotionfor2s.Itsvariationwithtimeisgivenby thequadraticequation, x 2 – x 1 = 200 + 30 t – 5t 2 (Forintervaloftime8sto10s) Itrepresentsacurvedpath.

Q.64 The speed-time graph of a particle moving along a fixed direction is as shown in figure 3.28. Obtain the distance travelled by the particle between (a) T=0 s to 10 s, (b) T=2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

Ans.

(a)Distancecoveredbytheparticlebetween0to10s = Areaunderthegivengraph Areaunderthegivengraph = 1 2 ×10 ms -1 ×12 s = 60m Averagespeedofthe particle, v avg = Distance Time v avg = 60 10 ms -1 = 6 ms -1 (b)Let s 1 and s 2 bethedistancestraveledbytheparticleintime t 1 = 2sto5sand t 2 = 5sto6srespectively. Totaldistance(s)traveledbytheparticleintimeintervalt = 2sto6s, s = s 1 + s 2 (i) Tofinddistance s 1 : Letusconsider u 1 isthevelocityoftheparticleafter2sand a 1 istheaccelerationoftheparticleduringthetime interval0 to 5 s. Astheparticleisinuniformaccelerationintheintervalt = 0tot = 5 s, Accordingtothefirstequationofmotion,weobtain: v = u + at 12 ms -1 = 0 + a 1 ×5 s a 1 = 12 5 ms -2 = 2.4 ms -2 Again,usingfirstequationofmotion,weget u 1 = u + a 1 t u 1 = 0 + 2.4 ms -2 ×2 s = 4.8 ms -1 Distancecoveredbytheparticlebetweentime2sand5si.e.in3s s 1 = u 1 t + 1 2 a 1 t 2 s 1 = 4 .8 ms -1 ×3 s+ 1 2 ×2 .4 ms -2 × 3 s 2 = 25.2m (ii) Tofinddistance s 2 Let a 2 beaccelerationoftheparticleduringt = 5 stot = 10s, Accordingtothefirstequationofmotion, v = u + at 0 = 12 ms -1 + a 2 ×5 s a 2 = -12 5 ms -2 = -2.4 ms -2 Distancecoveredbytheparticleinduringt = 5stot = 6s s 2 = u 2 t + 1 2 a 2 t 2 s 2 = 12 ms -1 ×1 s+ 1 2 -2 .4 ms -2 × 1 s 2 s 2 = 12 m – 1.2 m = 10.8m (iii) Fromequation(i),(ii)and(iii),weobtain: Totaldistancetravelled,s = 25.2 m + 10.8 m = 36m Averagespeed = Totaldistance Totaltime = 36 m 4 s = 9 ms -1

Q.65 The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 :

(a) x(t 2 ) = x(t 1 ) + v(t 1 ) t 2 – t 1 + 1 2 a t 2 – t 1 2 (b) v(t 2 ) = v(t 1 ) + a t 2 – t 1 (c) v av = {r(t 2 ) – r(t 1 )} t 2 – t 1 (d) a av = {v(t 2 ) – v(t 1 )} t 2 – t 1 (e) x(t 2 ) = x(t 1 ) + v av (t 2 – t 1 ) + 1 2 a av t 2 – t 1 2 (f) x(t 2 ) – x(t 1 ) = areaunderthev-tcurveboundedbythetimeaxis anddottedlineshown.

Ans.

The given plot has a non-uniform slope, therefore the relations given in (a), (b) and (e) are not correct. However, relations given in (c), (d) and (f) are correct.

Q.66 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

Q.67 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.68 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.69 State with reasons whether the following algebraic operations with scalars and vectors are meaningful:
(a) adding any two scalars
(b) adding a scalar to a vector of the same dimension
(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

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.70 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.71

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 | | Whendoestheequalitysignaboveapply? MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3797

Ans.

(a)Considertwovectors, a and b representedbythesides OR and OT ofaparallelogramOTSRasshowninthe givenfigure(a). Here,wehave | OT |=| a |(i) | TS |=| OR | =| b |(ii) | OS |=| a + b |(iii) Incaseofatriangle,eachsideissmallerthanthesumof othertwosides. FromΔOTS,weobtain OS<( OT+TS ) | a + b |<| a |+| b |(iv) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D9F7@ Iftwovectors a and b areactingalongastraight lineinthesamedirection,thenwehave: | a + b |=| a |+| b |(v) Fromequation(iv)and(v),weobtain | a + b || a |+| b | (b)Considertwovectors, a and b representedby thesides OR and OT ofaparallelogramOTSRas showninthegivenfigure(a). Here, | OT |=| a |(i) | TS |=| OR | =| b |(ii) | OS |=| a + b |(iii) Incaseofatriangle,eachsideissmallerthanthesum ofothertwosides. FromΔOTS,weobtain OS+TS>OT OS+OT>TS | OS |>| OT OR |( OR=TS ) | a + b |<| a |+| b |(iv) Iftwovectors a and b areactingalongastraightlinein thesamedirection,thenwehave: | a + b |=| a |+| b |(v) Fromequation(iv)and(v),weobtain | a + b || a |+| b | MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1927@

(c)Considertwovectors, a and b representedby thesides OQ and OP ofaparallelogramOPQNas showninthegivenfigure(b). Here, | OP |=| a |(i) | PN |=| OQ | =| b |(ii) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@19C3@ Incaseofatriangle,eachsideissmallerthanthesumof othertwosides. FromΔOPN,weobtain ON<OP+PN | a b |<| a |+| b | | a b |<| a |+| b |(iii) Iftwovectorsareactingalongastraightlinebut inoppositedirections,then | a b |=| a |+| b |(iv) Fromequation(iii)and(iv),weobtain | a b || a |+| b | (d)Considertwovectors, a and b representedby thesides OQ and OP ofaparallelogramOPQNas showninthegivenfigure(b). Incaseofatriangle,eachsideissmallerthanthesum ofothertwosides. FromΔOPN,weobtain ON+PN>OP(i) ON>OPPN(ii) | a b |>| a || b |(iii) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0F9B@ TheLHSoftheaboveequationispositivebuttheRHS canbepositiveornegative.Inordertomakethe quantitiesonbothsidesoftheaboveequationpositive, wetakemodulusonbothsidesas: | | a b | |>| | a || b | | | a b |>| | a || b | |(iv) Ifthetwovectorsareactinginastraightlinebut inoppositedirections,then | a b |=| | a || b | |(v) Combiningequation(iv)and(v),weget | a b || | a || b | |

Q.72

Given a + b + c + d = 0,whichofthefollowing statementsarecorrect? (a) a , b , c and d musteachbeanullvector. (b)Themagnitudeof( a + c )equalsthemagnitudeof( b + d ). (c)Themagnitudeofacanneverbegreaterthan thesumofthemagnitudeof b , c and d . (d)b + cmustlieintheplaneof a and d ,if a and d are notcollinearandinthelineof a and d ,iftheyare collinear. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1765@

Ans.

( a ) Thegivenstatementisincorrect,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 isequaltozero. ( b ) Thegivenstatementiscorrect. As, a + b + c + d =0 a + c =( b + d ) Takingmodulusonbothsides, | a + c |=| ( b + d ) | =| ( b + d ) | Therefore,magnitudeof( a + c )isequaltothemagnitude of( b + d ). 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As a + b + c + d =0 a =( b + c + d ) Takingmodulusonbothsides | a |=| ( b + c + d ) | =| b + c + d | | a || b |+| c |+| d |(i) Equation(i)indicatesthatthemagnitudeof a canbelessthanorequaltothesumofthemagnitudesof b , c and d . Themagnitudeof a canneverbegreaterthanthesumofthemagnitudesof b , c and d . (d)Thegivenstatementiscorrect, As, a + b + c + d =0 a +( b + c )+ d =0 Theresultantsumofthreevectors a ,( b + c )and d canbezeroonlyif( b + c )lieintheplaneof a and d andthesethreevectorsarerepresentedbythethree sidesofatriangletakeninthesameorder. If a and d arecollinear,then( b + c )shouldbeinlineof a and d ,onlythenthevectorsumofallthevectorswillbezero. 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Q.73 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 ) Thegivenstatementisincorrect,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 isequaltozero. ( b ) Thegivenstatementiscorrect. As, a + b + c + d =0 a + c =( b + d ) Takingmodulusonbothsides, | a + c |=| ( b + d ) | =| ( b + d ) | Therefore,magnitudeof( a + c )isequaltothemagnitude of( b + d ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3B35@

Q.74 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)Inthegivencasethecyclistcomesbacktothestartingpointaftercyclingfor10minutes. Timetaken = 10min = 1 6 h Totaldisplacementofthecyclist=0 (b)Averagevelocityisgivenas: Averagevelocity = Totaldisplacement Totaltimetaken = 0 10 60 h =0 (c)Averagespeedisgivenas: Averagespeed = Totalpathlength Totaltimetaken Totalpathlength = OP + PQ + QO Given,OP = QO = 1km PQ = 1 4 2πr = 1 4 2π×1 km = 3.570km Averagespeed = 3.570 km 1 6 h = 21.42 kmh -1

Q.75 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.

Inthisproblem,thepathfollowedbythemotoristisaregularhexagonof lengthofeachside oftheregularhexagon = 500m LetAbethestartingpointofthemotorist. AtD,themotoristwilltakethethirdturn. MagnitudeofdisplacementatD = AG + GD = 500 m + 500 m = 1000 m TotalpathlengthfromAtoD = AB + BC + CD = 500 m + 500 m + 500 m = 1500 m AtA,themotoristtakessixthturn,whichisthestarting point Magnitudeofdisplacement = 0 TotalpathlengthfromAtoA = AB + BC + CD + DE + EF + FA = 6×500 m = 3000 m AtC,themotoristtakeseighthturn. Magnitudeofdisplacement = AC AC = AB 2 +BC 2 +2 AB . BC AC = 500 m 2 + 500 m 2 +2× 500 m × 500 m ×cos60 o = 866.03m tanβ = 500 m× sin60 o 500 m +500 m× cos60 o = 1 3 = tan30 o β = 30 o Therefore,themagnitudeofdisplacement(AC)is866.03matanangleof30°withAC. Totalpathlength = AB + BC + CD + DE + EF + FA + AB + BC = 8×500 m = 4000m

Q.76 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.

Maximumhorizontaldistance,R = 100m Theballwillcovermaximumhorizontaldistancewhen,angleofprojection = 45° As,horizontalrangeisgivenbytherelation, R = u 2 sin2θ g 100 m = u 2 g sin90° u 2 g = 100m Theballwillgainmaximumheightwhenheightwhenitisthrownverticallyupward. Forsuchmotion,thefinalvelocityiszeroatmaximumheight(H). Acceleration,a = -g Fromthirdequationofmotion, v 2 – u 2 = -2gH 0 – u 2 = -2gH H = 1 2 × u 2 g = 1 2 ×100m = 50m

Q.77 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,actualdistancetravelled = 23km Totaltimetaken = 28min = 28 60 h Displacementofthecar = Distancebetweenthehotelandthestation = 10km (a)Averagespeedofthetaxi = Actualdistancetravelled Totaltimetaken = 23 km 28 60 h = 49.29 kmh -1 (b)Magnitudeofaveragevelocity = Displacement Totaltimetaken = 10 km 28 60 h = 21.43 kmh -1 Averagespeedofthetaxiisnotequaltothemagnitudeofaveragevelocity.

Q.78 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.

Letvelocityof cyclist be v c Letvelocityof rain be v r Thewomanmustholdherumbrellain the opposite directionofrelativevelocity(v)oftherainw.r.t.herself, inorderto protectherselffromtherain, v = v r + -v c v = 30 ms -1 + -10 ms -1 = 20 ms -1 Fromthegivenfigure, tanθ = v c v r tanθ = 10 30 θ = tan -1 1 3 = tan -1 0.333 18° Thewomanshouldholdtheumbrellaatanangleofapproximately18°withtheverticaltowards southdirection.

Q.79 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,speedofman = 4 kmh -1 Widthofriver = 1km Timetakentocrosstheriver,t = Widthofriver Speedofman = 1 4 h t = 1 4 ×60 = 15min Given,speedofriver, v r =3 kmh 1 Distancetravelledalongwiththeflowofriverintimet= v r ×t =3× 1 4 km = 3 4 ×1000m =750m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7D88@

Q.80 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, velocityoftheboat, v b = 51 kmh -1 Velocityofthewind, v w = 72 kmh -1 As,theflagisflutteringinnorth-eastdirection, itimpliesthatthewindisblowingtowardsthe north-eastdirection. Whentheboatstartsmovingtowardsthenorthdirection, theflagwillflutter towardsthedirectionofrelativevelocityofwindw.r.t.theboat. Lettherelativevelocityofthewindw.r.t.theboat be v wb Letanglebetween v wb and v w be β Anglebetween 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° Angleformedw.r.t.eastdirection = 45.1° – 45° = 0.1° Theflagwillflutteralmosttowardseastdirection.

Q.81 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,speedofball,u = 40 ms -1 Maximumheight,h = 25m Letθbetheangleofprojectionwiththehorizontal. Forprojectilemotion, Maximumheight,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° Horizontalrange,R = u 2 sin2θ g R = 40 ms -1 2 ×sin2×33.6° 9 .8 ms -2 = 150.5m

Q.82 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,lengthofthestring, l = 80cm = 0.8m Numberofrevolutions = 14 Timetaken = 25s Frequencyofrevolution, ν = Numberofrevolutions Timetaken ν = 14 25 Hz Angularfrequency, ω = 2πν ω = 2× 22 7 × 14 25 Hz = 88 25 rads -1 Centripetalacceleration, a c = ω 2 r a c = 88 25 rads -1 2 ×0.8m = 9.91 ms -2 Thedirectionofcentripetalaccelerationisalwaysdirectedalongthestring,towardsthecentreofthecircularpath.

Q.83 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,radiusofhorizontalloop,r = 1km or 1000m Speedofaircraft,v = 900 kmh -1 = 250 ms -1 Centripetalaccelerationisgivenbythe relation, a c = v 2 r a c = 250 ms -1 2 1000m = 62.5 ms -2 Accelerationduetogravity, g = 9.8 ms -2 a c g = 62.5 ms -2 9.8 ms -2 = 6.38 Centripetalacceleration, a c = 6.38g

Q.84 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.85

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 oftheparticle? (b) What is the magnitude and direction of velocity of the particle at t = 2s? MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeqabeqabiqaceGabeqabeWabeqaeeaakqaabeqaaKqzagGaaeivaiaabIgacaqGLbGaaeiiaiaabchacaqGVbGaae4CaiaabMgacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaaeiCaiaabggacaqGYbGaaeiDaiaabMgacaqGJbGaaeiBaiaabwgacaqGGaGaaeyAaiaabohacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeOyaiaabMhaaOqaaKqbaoaaFiaakeaaieWajugGbiaa=jhacaaMe8oakiaawEniaKqzagGaaeypaiaaysW7caqGZaGaaeOlaiaabcdacaqG0bGaaGjbVlqa=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@3231@

Ans.

Acceleration a oftheparticleisgivenbytherelation: a = d v dt = d dt ( 3.0 i ^ 4.0t j ^ ) a =4.0 j ^ m s 2 Attime,t=2s v =3.0 i ^ 8.0tm s 1 Themagnitudeofvelocity,v= ( 3.0 ) 2 + ( 8 ) 2 v= 73 =8.54m s 1 Ifθistheanglebetween v andthex – 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 Thenegativesignshowsthatthevelocityisdirected belowthex – axis.

Q.86

Aparticlestartsfromoriginatt = 0withavelocityof10.0 j ms -1 andmovesinthex – yplanewitha constantaccelerationof( 8.0 i +2.0 j ) ms -2 .Atwhattimeisthex – coordinateoftheparticle16m? What isthey – coordinateoftheparticleatthattime?(b)Whatisthespeedoftheparticleatthattime? MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeqabeqabiqaceGabeqabeWabeqaeeaakqaabeqaaKqzagGaaeyqaiaaysW7caqGWbGaaeyyaiaabkhacaqG0bGaaeyAaiaabogacaqGSbGaaeyzaiaaysW7caqGZbGaaeiDaiaabggacaqGYbGaaeiDaiaabohacaaMe8UaaeOzaiaabkhacaqGVbGaaeyBaiaaysW7caqGVbGaaeOCaiaabMgacaqGNbGaaeyAaiaab6gacaaMe8UaaeyyaiaabshacaaMe8UaaeiDaiaabccacaqG9aGaaeiiaiaabcdacaaMe8Uaae4DaiaabMgacaqG0bGaaeiAaiaaysW7caqGHbGaaGjbVlaabAhacaqGLbGaaeiBaiaab+gacaqGJbGaaeyAaiaabshacaqG5bGaaGjbVlaab+gacaqGMbGaaGjbVlaabgdacaqGWaGaaeOlaiaabcdajuaGdaWfGaGcbaqcLbyacaWGQbaaleqabaqcLbyacqGHNis2aaGaaeyBaiaabohajuaGdaahaaWcbeqaaKqzagGaaeylaiaabgdaaaGaaGjbVlaabggacaqGUbGaaeizaiaaysW7caqGTbGaae4BaiaabAhacaqGLbGaae4CaiaaysW7caqGPbGaaeOBaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8UaaeiEaiaabccacaqGTaGaaeiiaiaabMhacaaMe8UaaeiCaiaabYgacaqGHbGaaeOBaiaabwgacaaMe8Uaae4DaiaabMgacaqG0bGaaeiAaiaaysW7caqGHbGaaGjbVdGcbaqcLbyacaqGJbGaae4Baiaab6gacaqGZbGaaeiDaiaabggacaqGUbGaaeiDaiaaysW7caqGHbGaae4yaiaabogacaqGLbGaaeiBaiaabwgacaqGYbGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaaMe8Uaae4BaiaabAgacaaMe8Ecfa4aaeWaaOqaaKqzagGaaGioaiaac6cacaaIWaqcfa4aaCbiaOqaaKqzagGaamyAaaWcbeqaaKqzagGaey4jIKnaaiabgUcaRiaaikdacaGGUaGaaGimaKqbaoaaxacakeaajugGbiaadQgaaSqabeaajugGbiabgEIizdaaaOGaayjkaiaawMcaaKqzagGaaGjbVlaab2gacaqGZbqcfa4aaWbaaSqabeaajugGbiaab2cacaqGYaaaaiaab6cacaaMe8UaaeyqaiaabshacaaMe8Uaae4DaiaabIgacaqGHbGaaeiDaiaaysW7caqG0bGaaeyAaiaab2gacaqGLbGaaGjbVlaabMgacaqGZbGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqG4bGaaeiiaiaab2cacaqGGaGaae4yaiaab+gacaqGVbGaaeOCaiaabsgacaqGPbGaaeOBaiaabggacaqG0bGaaeyzaiaaysW7caqGVbGaaeOzaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8UaaeiCaiaabggacaqGYbGaaeiDaiaabMgacaqGJbGaaeiBaiaabwgacaaMe8UaaeymaiaabAdacaaMe8UaaeyBaiaab+dacaaMe8oakeaajugGbiaabEfacaqGObGaaeyyaiaabshacaqGGaGaaeyAaiaabohacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaabMhacaqGGaGaaeylaiaabccacaqGJbGaae4Baiaab+gacaqGYbGaaeizaiaabMgacaqGUbGaaeyyaiaabshacaqGLbGaaGjbVlaab+gacaqGMbGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqGWbGaaeyyaiaabkhacaqG0bGaaeyAaiaabogacaqGSbGaaeyzaiaaysW7caqGHbGaaeiDaiaaysW7caqG0bGaaeiAaiaabggacaqG0bGaaGjbVlaabshacaqGPbGaaeyBaiaabwgacaqG=aGaaGjbVlaabIcacaqGIbGaaeykaiaaysW7caqGxbGaaeiAaiaabggacaqG0bGaaGjbVlaabMgacaqGZbGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqGZbGaaeiCaiaabwgacaqGLbGaaeizaiaaysW7caqGVbGaaeOzaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8UaaeiCaiaabggacaqGYbGaaeiDaiaabMgacaqGJbGaaeiBaiaabwgacaaMe8UaaeyyaiaabshacaaMe8UaaeiDaiaabIgacaqGHbGaaeiDaiaaysW7caqG0bGaaeyAaiaab2gacaqGLbGaae4paiaaysW7aaaa@8686@

Ans.

Given,initialvelocityofparticle, u =10.0 j m s 1 Accelerationofparticle, a = d v dt =( 8.0 i +2.0 j )m s 2 d v =( 8.0 i +2.0 j )dt Integratingitwithinthelimitsastimechangesfrom0totandvelocitychangesfromutov,weobtain: v ( t )=( 8.0t i +2.0t j + u )m s 1 Asvelocity, v = d r dt MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A171@ d r = v dt =( 8.0t i +2.0t j + u )dt Integratingitwithinthelimits:astimechangesfrom0tot,displacementchangesfrom0tor,wehave 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 orx i +y j =4.0 t 2 i +( 10t+ t 2 ) j Here,wehave,x=4.0 t 2 andy=10t+ t 2 t= ( x 4 ) 1 2 (a)whenx = 16m t= ( 16 4 ) 1 2 =2s y=10×2+ 2 2 = 24m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0E85@ (b)Velocityoftheparticleattimetisgivenas: v ( t )=( 8.0t i +2.0t j + u )m s 1 att = 2s v ( t )=8.0×2 i +2.0×2 j +10 j =( 16 i +14 j )m s 1 Speedoftheparticle= | v | = ( 16 ) 2 + ( 14 ) 2 = 256+196 = 452 =21.26m s 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@44CD@

Q.87

i and j areunitvectorsalongx –andy – axisrespectively. (a)Whatisthemagnitudeanddirectionofthevectors( i + j )and( i j )? (b)Whatarethecomponentsofavector B = 2 i +3 j alongthedirectionsof( i + j )and( i j )? MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5237@

Ans.

(a)Magnitudeof i + j = i + j = 1 2 + 1 2 = 2 Considerthatthevector i + j makesanangleθwith i , cosθ= i + j . i i + j i = 1 2 1 = 1 2 = cos45 o θ = 45 o Therefore,thevector i + j makesanangleof 45 o withthex-axis. (b)Given, B = 2 i +3 j Let a betheunitvectordirectedalongvector i + j . a = i + j i + j = i + j 1 2 + 1 2 = i + j 2 Magnitudeofthecomponentvectorof B alongvector i + j = B . a = 2 i +3 j . 1 2 i + j = 1 2 2+3 = 5 2 Componentof B alongthedirectionof i + j = B . a a = 5 2 i + j 2 = 5 2 i + j Let b betheunitvectordirectedalongvector i j . b = i j i j = i j 1 2 + 1 2 = 1 2 i j Magnitudeofthecomponentvectorof B alongthedirectionof i j = B . b = 2 i +3 j . 1 2 i j = 1 2 Componentof B alongthedirectionof i j = B . b b = 1 2 1 2 i j = 1 2 i j

Q.88

Foranarbitrarymotioninspace,whichofthefollowingrelationsaretrue: (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 [Theaveragestandsfortheaverageofthequantityoverthetimeinterval 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.89 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.90 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.

Thegivensituationisdescribedinthefigure. Given,heightoftheaircraftfromground,OS = 3400m Anglesubtendedbetweenthepositionsofaircraft,POS = 30° Timetakenbytheaircraft=10s InΔPSO, tan15° = PS OS PS = OStan15° PS = 3400m×tan15° = 3400m×0.2679 = 910.86m PQ = PS + SQ PQ = 2PS = 2×910.86m = 1821.72m Speedofaircraft,v = distancePQ time v = 1821.72m 10s = 182 .172 ms -1

Q.91 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.92 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.93 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.94 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,angleofprojectionofthebullet, θ = 30 o Rangeofthebullet,R= 3km Accelerationduetogravity,g = 9.8 ms -2 Horizontalrangeforprojectionvelocity u 0 ,R = u 0 2 sin2θ g 3km = u 0 2 sin60 o g = u 0 2 g 3 2 u 0 2 g = 2 3 (i) Thebulletachievesthemaximumrange (R max )whenfiredatanangleof 45 o withthehorizontal, R max = u 0 2 g (ii) Fromequation(i)and(ii),weobtain R max = 3 3 km = 2×1.732km = 3.46km Thebulletcannothitthetarget5kmaway.

Q.95 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,altitudeofplane = 1.5km = 1500m Speedofplane,v = 720 kmh -1 = 200 ms -1 Lettheshellhitstheplanewhenfiredatanangleθ. Thedescribedsituationisshowninthegivenfigure. Here,muzzlevelocityofgun,u = 600 ms -1 Lettimetakenbytheshelltohitthefighterplane = t Horizontaldistancecoveredbythe shell = u x t Distancecoveredbytheplane = vt Fortheshelltohittheplane,thesetwodistancesmustbeequal. 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 Inordertoavoidbeinghitbytheshell,thepilotmustflytheplaneataminimumheight whichisthemaximumheightattainedbytheshell. H = u 2 sin 2 90 o 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 = 16000m = 16km

Q.96 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,speedofthecyclist,v = 27 kmh -1 = 7.5 ms -1 Radiusofcircularturn,r = 80m Centripetalacceleration, a c = v 2 r = 7.5 2 80 ms -2 = 0.7 ms -2 SupposethecyclistappliesbrakesatpointPofthecircularturnthenthetangentialcomponentof acceleration a T willactoppositetovelocity. Here,tangentialcomponentofacceleration, a T = 0.5 ms -2 Anglebetween a c and a T = 90 o Magnitudeofresultantaccelerationaisgivenas: a = a c 2 +a T 2 a = 0 .7ms -2 2 + 0 .5ms -2 2 = 0.86 ms -2 Lettheresultantaccelerationmakesanangleθwiththedirectionofvelocity 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.97

(a)Showthatforaprojectiletheanglebetweenthevelocityandthex – axisasafunctionoftimeisgivenby θ (t) = tan 1 ( v 0y gt v 0x ) (b)Showthattheprojectionangle θ 0 foraprojectilelaunchedfromtheoriginisgivenby θ 0 = tan 1 ( 4 h m R ) Wherethesymbolshavetheirusualmeaning. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@682C@

Ans.

(a)Letinitialcomponentsofvelocityoftheprojectilealonghorizontal(OX)andvertical(OY)directionsbe v 0x and v 0y therespectively. Letthehorizontalandverticalcomponentsofvelocity atpointPbe v x and v y respectively. LettimetakenbyprojectiletoarriveatpointP = t Usingthefirstequationofmotionalongverticalandhorizontaldirections,wehave: 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 ) Inangularprojection Maximumverticalheight, h m = u 2 sin 2 θ 0 2g (i) Horizontalrangeisgivenas:R= u 2 sin2 θ 0 g (ii) Dividingequation(i)byequation(ii),weobtain: 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 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWabeqaeaaakqaabeqaaKqzagGaaeikaiaabggacaqGPaGaaGjbVlaabYeacaqGLbGaaeiDaiaaysW7caqGPbGaaeOBaiaabMgacaqG0bGaaeyAaiaabggacaqGSbGaaGjbVlaabogacaqGVbGaaeyBaiaabchacaqGVbGaaeOBaiaabwgacaqGUbGaaeiDaiaabohacaaMe8Uaae4BaiaabAgacaaMe8UaaeODaiaabwgacaqGSbGaae4BaiaabogacaqGPbGaaeiDaiaabMhacaaMe8Uaae4BaiaabAgacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaabchacaqGYbGaae4BaiaabQgacaqGLbGaae4yaiaabshacaqGPbGaaeiBaiaabwgacaaMe8UaaeyyaiaabYgacaqGVbGaaeOBaiaabEgacaaMe8UaaeiAaiaab+gacaqGYbGaaeyAaiaabQhacaqGVbGaaeOBaiaabshacaqGHbGaaeiBaiaaysW7caqGOaGaae4taiaabIfacaqGPaGaaGjbVlaabggacaqGUbGaaeizaiaaysW7caqG2bGaaeyzaiaabkhacaqG0bGaaeyAaiaabogacaqGHbGaaeiBaiaaysW7caqGOaGaae4taiaabMfacaqGPaGaaGjbVlaabsgacaqGPbGaaeOCaiaabwgacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohacaaMe8UaaeOyaiaabwgacaaMe8UaaeODaKqbaoaaBaaaleaajugGbiaaicdacaWG4baaleqaaKqzagGaaGjbVlaabggacaqGUbGaaeizaiaaysW7caqG2bqcfa4aaSbaaSqaaKqzagGaaGimaiaadMhaaSqabaqcLbyacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaabkhacaqGLbGaae4CaiaabchacaqGLbGaae4yaiaabshacaqGPbGaaeODaiaabwgacaqGSbGaaeyEaiaab6caaOqaaKqzagGaaeitaiaabwgacaqG0bGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqGObGaae4BaiaabkhacaqGPbGaaeOEaiaab+gacaqGUbGaaeiDaiaabggacaqGSbGaaGjbVlaabggacaqGUbGaaeizaiaaysW7caqG2bGaaeyzaiaabkhacaqG0bGaaeyAaiaabogacaqGHbGaaeiBaiaaysW7caqGJbGaae4Baiaab2gacaqGWbGaae4Baiaab6gacaqGLbGaaeOBaiaabshacaqGZbGaaGjbVlaab+gacaqGMbGaaGjbVlaabAhacaqGLbGaaeiBaiaab+gacaqGJbGaaeyAaiaabshacaqG5bGaaeiiaiaabggacaqG0bGaaGjbVlaabchacaqGVbGaaeyAaiaab6gacaqG0bGaaGjbVlaabcfacaaMe8UaaeOyaiaabwgacaaMe8UaaeODaKqbaoaaBaaaleaajugGbiaadIhaaSqabaqcLbyacaaMe8Uaaeyyaiaab6gacaqGKbGaaGjbVlaabAhajuaGdaWgaaWcbaqcLbyacaWG5baaleqaaKqzagGaaGjbVlaabkhacaqGLbGaae4CaiaabchacaqGLbGaae4yaiaabshacaqGPbGaaeODaiaabwgacaqGSbGaaeyEaiaab6caaOqaaKqzagGaaeitaiaabwgacaqG0bGaaGjbVlaabshacaqGPbGaaeyBaiaabwgacaaMe8UaaeiDaiaabggacaqGRbGaaeyzaiaab6gacaaMe8UaaeOyaiaabMhacaaMe8UaaeiCaiaabkhacaqGVbGaaeOAaiaabwgacaqGJbGaaeiDaiaabMgacaqGSbGaaeyzaiaaysW7caqG0bGaae4BaiaaysW7caqGHbGaaeOCaiaabkhacaqGPbGaaeODaiaabwgacaaMe8UaaeyyaiaabshacaaMe8UaaeiCaiaab+gacaqGPbGaaeOBaiaabshacaaMe8UaaeiuaiaabccacaqG9aGaaeiiaiaabshaaOqaaKqzagGaaeyvaiaabohacaqGPbGaaeOBaiaabEgacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaabAgacaqGPbGaaeOCaiaabohacaqG0bGaaGjbVlaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaGjbVlaab+gacaqGMbGaaGjbVlaab2gacaqGVbGaaeiDaiaabMgacaqGVbGaaeOBaiaaysW7caqGHbGaaeiBaiaab+gacaqGUbGaae4zaiaaysW7caqG2bGaaeyzaiaabkhacaqG0bGaaeyAaiaabogacaqGHbGaaeiBaiaaysW7caqGHbGaaeOBaiaabsgacaaMe8UaaeiAaiaab+gacaqGYbGaaeyAaiaabQhacaqGVbGaaeOBaiaabshacaqGHbGaaeiBaiaaysW7caqGKbGaaeyAaiaabkhacaqGLbGaae4yaiaabshacaqGPbGaae4Baiaab6gacaqGZbGaaeilaiaaysW7caqG3bGaaeyzaiaaysW7caqGObGaaeyyaiaabAhacaqGLbGaaeOoaaGcbaqcLbyacaWG2bqcfa4aaSbaaSqaaKqzagGaamyEaaWcbeaajugGbiabg2da9iaadAhajuaGdaWgaaWcbaqcLbyacaaIWaGaamyEaaWcbeaajugGbiabgkHiTiaadEgacaWG0baakeaajugGbiaadAhajuaGdaWgaaWcbaqcLbyacaWG4baaleqaaKqzagGaeyypa0JaamODaKqbaoaaBaaaleaajugGbiaaicdacaWG4baaleqaaaGcbaqcLbyacqGH0icxciGG0bGaaiyyaiaac6gacqaH4oqCcqGH9aqpjuaGdaWcaaGcbaqcLbyacaWG2bqcfa4aaSbaaSqaaKqzagGaamyEaaWcbeaaaOqaaKqzagGaamODaKqbaoaaBaaaleaajugGbiaadIhaaSqabaaaaaGcbaqcLbyacqGH9aqpjuaGdaWcaaGcbaqcLbyacaWG2bqcfa4aaSbaaSqaaKqzagGaaGimaiaadMhaaSqabaqcLbyacqGHsislcaWGNbGaamiDaaGcbaqcLbyacaWG2bqcfa4aaSbaaSqaaKqzagGaaGimaiaadIhaaSqabaaaaaGcbaqcLbyacqGH0icxcqaH4oqCcqGH9aqpciGG0bGaaiyyaiaac6gajuaGdaahaaWcbeqaaKqzagGaeyOeI0IaaGymaaaajuaGdaqadaGcbaqcfa4aaSaaaOqaaKqzagGaamODaKqbaoaaBaaaleaajugGbiaaicdacaWG5baaleqaaKqzagGaeyOeI0Iaam4zaiaadshaaOqaaKqzagGaamODaKqbaoaaBaaaleaajugGbiaaicdacaWG4baaleqaaaaaaOGaayjkaiaawMcaaaqaaKqzagGaaeysaiaab6gacaaMe8Uaaeyyaiaab6gacaqGNbGaaeyDaiaabYgacaqGHbGaaeOCaiaaysW7caqGWbGaaeOCaiaab+gacaqGQbGaaeyzaiaabogacaqG0bGaaeyAaiaab+gacaqGUbaakeaajugGbiaab2eacaqGHbGaaeiEaiaabMgacaqGTbGaaeyDaiaab2gacaaMe8UaaeODaiaabwgacaqGYbGaaeiDaiaabMgacaqGJbGaaeyyaiaabYgacaaMe8UaaeiAaiaabwgacaqGPbGaae4zaiaabIgacaqG0bGaaeilaiaaysW7caWGObqcfa4aaSbaaSqaaKqzagGaamyBaaWcbeaajugGbiabg2da9KqbaoaalaaakeaajugGbiaadwhajuaGdaahaaWcbeqaaKqzagGaaGOmaaaaciGGZbGaaiyAaiaac6gajuaGdaahaaWcbeqaaKqzagGaaGOmaaaacqaH4oqCjuaGdaWgaaWcbaqcLbyacaaIWaaaleqaaaGcbaqcLbyacaaIYaGaam4zaaaacqGHsgIRcaGGOaGaamyAaiaacMcaaOqaaKqzagGaaeisaiaab+gacaqGYbGaaeyAaiaabQhacaqGVbGaaeOBaiaabshacaqGHbGaaeiBaiaaysW7caqGYbGaaeyyaiaab6gacaqGNbGaaeyzaiaaysW7caqGPbGaae4CaiaaysW7caqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaaysW7caqGHbGaae4CaiaacQdacaaMe8UaamOuaiabg2da9KqbaoaalaaakeaajugGbiaadwhajuaGdaahaaWcbeqaaKqzagGaaGOmaaaaciGGZbGaaiyAaiaac6gacaaIYaGaeqiUdexcfa4aaSbaaSqaaKqzagGaaGimaaWcbeaaaOqaaKqzagGaam4zaaaacqGHsgIRcaGGOaGaamyAaiaadMgacaGGPaaakeaajugGbiaabseacaqGPbGaaeODaiaabMgacaqGKbGaaeyAaiaab6gacaqGNbGaaGjbVlaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaGjbVlaabIcacaqGPbGaaeykaiaaysW7caqGIbGaaeyEaiaaysW7caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaa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Q.98 Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Ans.

(a) As the rain drop is falling with a constant speed, its acceleration is zero. Thus, according to Newton’s second law of motion, the total force acting on the rain drop is zero.
(b) As the cork is floating on water, its weight is balanced by the buoyant force exerted by the water in the upward direction. Therefore, net force acting on the cork is zero.
(c) As the kite is stationary in the sky, therefore according to Newton’s first law of motion, no net force is acting on the kite.
(d) As the car is moving on a rough road with a constant velocity, its acceleration is zero. Therefore, according to Newton’s second law of motion, no net force is acting on the car.
(e) As the high speed electron is free from the effect of all fields, therefore, net force acting on the electron is zero.

Q.99 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest.
Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

Ans.

Acceleration due to gravity always acts in the downward direction, whatever may be the direction of motion of the object.
In all the three cases, the only force acting on the pebble is the gravitational force in vertically downward direction.

AccordingtoNewton’ssecondlawofmotion,itsmagnitudeisgivenby F=ma Here,F=Netforce m = Massofthepebble =0.05kg a=g =10 ms 2 F=0.05×10 =0.5N Inallthreecases,thenetforceactingonthepebbleis0.5inverticallydownwarddirection.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@30B0@

The answer remains unchanged even if pebble was thrown at an angle of 45Ëšwith the horizontal direction as the direction of force will not change.

Q.100 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) Just after it is dropped from the window of a stationary train,
(b) Just after it is dropped from the window of a train running at a constant velocity of 36 kmh-1,
(c) Just after it is dropped from the window of a train accelerating with 1 ms-2,
(d) Lying on the floor of a train which is acceleration with 1, the stone being at rest relative to the train. Neglect the resistance of air throughout.

Ans.

(a)Here,massofthestone,m=0.1kg Accelerationofthestone,a=g =10m s 2 AccordingtoNewton’ssecondlawofmotion,Netforceactingonthestone, F = ma =mg =0.1×10 =1N F =1N,verticallydownwards (b)Asthetrainismovingwithconstantvelocity, Accelerationofthetrain = 0 Forceonthestone,F = weightofstone = mg = 0.1×10 = 1N;verticallydownwards (c)Asthetrainisacceleratingwiththerateof1 ms -2 ,anadditionalforce F=ma =0.1×10 =0.1N actsonthestoneinthehorizontaldirection.But,whenthestoneisdropped fromthetrain,F’=0andthenetforceactingonthestoneisduetoitsweightonly. Itisgivenas: F = mg = 0.1×10 = 1N;verticallydownwards (d)Asthestoneislyingonthetrain,itsweightis balancedbythenormal reactionofthetrain. Accelerationofthestone = Accelerationofthetrain = 1 ms -2 Forceactingonthestone,F = ma = 0.1×1 = 0.1N Thisforceactsinthedirectionofmotionofthetrain. 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Q.101

One end of a string of length r is connected to a particle of mass m and the other end to a small pegonasmoothhorizontalsurface. Iftheparticlemovesinacirclewithspeedv,thenetforceonthe particle(directedtowardscentre)is (i)T(ii)T – mv 2 r (ii)T + mv 2 r (iv)0 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaac H8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFf ea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeqa beqabiqaceGabeqabeWabeqaeeaakqaabeqaaKqzagGaae4taiaab6 gacaqGLbGaaeiiaiaabwgacaqGUbGaaeizaiaabccacaqGVbGaaeOz aiaabccacaqGHbGaaeiiaiaabohacaqG0bGaaeOCaiaabMgacaqGUb Gaae4zaiaabccacaqGVbGaaeOzaiaabccacaqGSbGaaeyzaiaab6ga caqGNbGaaeiDaiaabIgacaqGGaGaaeOCaiaabccacaqGPbGaae4Cai aabccacaqGJbGaae4Baiaab6gacaqGUbGaaeyzaiaabogacaqG0bGa aeyzaiaabsgacaqGGaGaaeiDaiaab+gacaqGGaGaaeyyaiaabccaca qGWbGaaeyyaiaabkhacaqG0bGaaeyAaiaabogacaqGSbGaaeyzaiaa bccacaqGVbGaaeOzaiaabccacaqGTbGaaeyyaiaabohacaqGZbGaae iiaiaab2gaaOqaaKqzagGaaeyyaiaab6gacaqGKbGaaeiiaiaabsha caqGObGaaeyzaiaabccacaqGVbGaaeiDaiaabIgacaqGLbGaaeOCai aabccacaqGLbGaaeOBaiaabsgacaqGGaGaaeiDaiaab+gacaqGGaGa aeyyaiaabccacaqGZbGaaeyBaiaabggacaqGSbGaaeiBaiaabccaca qGWbGaaeyzaiaabEgacaaMe8Uaae4Baiaab6gacaaMe8Uaaeyyaiaa ysW7caqGZbGaaeyBaiaab+gacaqGVbGaaeiDaiaabIgacaaMe8Uaae iAaiaab+gacaqGYbGaaeyAaiaabQhacaqGVbGaaeOBaiaabshacaqG HbGaaeiBaiaaysW7caqGZbGaaeyDaiaabkhacaqGMbGaaeyyaiaabo gacaqGLbGaaeOlaiaabccaaOqaaKqzagGaaeysaiaabAgacaaMe8Ua aeiDaiaabIgacaqGLbGaaGjbVlaabchacaqGHbGaaeOCaiaabshaca qGPbGaae4yaiaabYgacaqGLbGaaGjbVlaab2gacaqGVbGaaeODaiaa bwgacaqGZbGaaGjbVlaabMgacaqGUbGaaGjbVlaabggacaaMe8Uaae 4yaiaabMgacaqGYbGaae4yaiaabYgacaqGLbGaaGjbVlaabEhacaqG PbGaaeiDaiaabIgacaaMe8Uaae4CaiaabchacaqGLbGaaeyzaiaabs gacaaMe8UaaeODaiaabYcacaaMe8UaaeiDaiaabIgacaqGLbGaaGjb Vlaab6gacaqGLbGaaeiDaiaaysW7caqGMbGaae4BaiaabkhacaqGJb GaaeyzaiaaysW7caqGVbGaaeOBaiaaysW7caqG0bGaaeiAaiaabwga aOqaaKqzagGaaeiCaiaabggacaqGYbGaaeiDaiaabMgacaqGJbGaae iBaiaabwgacaaMe8UaaeikaiaabsgacaqGPbGaaeOCaiaabwgacaqG JbGaaeiDaiaabwgacaqGKbGaaGjbVlaabshacaqGVbGaae4Daiaabg gacaqGYbGaaeizaiaabohacaaMe8Uaae4yaiaabwgacaqGUbGaaeiD aiaabkhacaqGLbGaaeykaiaaysW7caqGPbGaae4CaiaaysW7aOqaaK qzagGaaeikaiaabMgacaqGPaGaaGjbVlaabsfacaaMe8Uaaeikaiaa bMgacaqGPbGaaeykaiaaysW7caqGubGaaeiiaiaab2cajuaGdaWcaa GcbaqcLbyacaqGTbGaaeODaKqbaoaaCaaaleqabaqcLbyacaqGYaaa aaGcbaqcLbyacaqGYbaaaiaaysW7caqGOaGaaeyAaiaabMgacaqGPa GaaGjbVlaabsfacaqGGaGaae4kaKqbaoaalaaakeaajugGbiaab2ga caqG2bqcfa4aaWbaaSqabeaajugGbiaabkdaaaaakeaajugGbiaabk haaaGaaGjbVlaabIcacaqGPbGaaeODaiaabMcacaaMe8Uaaeimaaaa aa@49ED@

Ans.

Whenaparticleconnectedtoastringrevolvesinacircularpath,tensionin thestringprovidesthenecessarycentripetalforce.Inthiscase,thenetforce ontheparticledirectedtowardsthecentreistheforceoftension,T. F = T Here,Fisthetotalforceactingontheparticle. (i)Tisthecorrectoption. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaac H8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFf ea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaa baqaaiaacaGaaeqabaWabeqaeaaakqaabeqaaKqzagGaae4vaiaabI gacaqGLbGaaeOBaiaaysW7caqGHbGaaGjbVlaabchacaqGHbGaaeOC aiaabshacaqGPbGaae4yaiaabYgacaqGLbGaaGjbVlaabogacaqGVb GaaeOBaiaab6gacaqGLbGaae4yaiaabshacaqGLbGaaeizaiaaysW7 caqG0bGaae4BaiaaysW7caqGHbGaaGjbVlaabohacaqG0bGaaeOCai aabMgacaqGUbGaae4zaiaaysW7caqGYbGaaeyzaiaabAhacaqGVbGa aeiBaiaabAhacaqGLbGaae4CaiaaysW7caqGPbGaaeOBaiaaysW7ca qGHbGaaGjbVlaabogacaqGPbGaaeOCaiaabogacaqG1bGaaeiBaiaa bggacaqGYbGaaGjbVlaabchacaqGHbGaaeiDaiaabIgacaaMe8Uaae ilaiaabshacaqGLbGaaeOBaiaabohacaqGPbGaae4Baiaab6gacaaM e8UaaeyAaiaab6gaaOqaaKqzagGaaeiDaiaabIgacaqGLbGaaGjbVl aabohacaqG0bGaaeOCaiaabMgacaqGUbGaae4zaiaaysW7caqGWbGa aeOCaiaab+gacaqG2bGaaeyAaiaabsgacaqGLbGaae4CaiaaysW7ca qG0bGaaeiAaiaabwgacaaMe8UaaeOBaiaabwgacaqGJbGaaeyzaiaa bohacaqGZbGaaeyyaiaabkhacaqG5bGaaGjbVlaabogacaqGLbGaae OBaiaabshacaqGYbGaaeyAaiaabchacaqGLbGaaeiDaiaabggacaqG SbGaaGjbVlaabAgacaqGVbGaaeOCaiaabogacaqGLbGaaeOlaiaabM eacaqGUbGaaGjbVlaabshacaqGObGaaeyAaiaabohacaaMe8Uaae4y aiaabggacaqGZbGaaeyzaiaabYcacaaMe8UaaeiDaiaabIgacaqGLb GaaGjbVlaab6gacaqGLbGaaeiDaiaaysW7caqGMbGaae4Baiaabkha caqGJbGaaeyzaaGcbaqcLbyacaqGVbGaaeOBaiaaysW7caqG0bGaae iAaiaabwgacaaMe8UaaeiCaiaabggacaqGYbGaaeiDaiaabMgacaqG JbGaaeiBaiaabwgacaaMe8UaaeizaiaabMgacaqGYbGaaeyzaiaabo gacaqG0bGaaeyzaiaabsgacaaMe8UaaeiDaiaab+gacaqG3bGaaeyy aiaabkhacaqGKbGaae4CaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8 Uaae4yaiaabwgacaqGUbGaaeiDaiaabkhacaqGLbGaaGjbVlaabMga caqGZbGaaGjbVlaabshacaqGObGaaeyzaiaaysW7caqGMbGaae4Bai aabkhacaqGJbGaaeyzaiaaysW7caqGVbGaaeOzaiaaysW7caqG0bGa aeyzaiaab6gacaqGZbGaaeyAaiaab+gacaqGUbGaaeilaiaabsfaca qGUaaakeaajugGbiaabAeacaqGGaGaaeypaiaabccacaqGubaakeaa jugGbiaabIeacaqGLbGaaeOCaiaabwgacaqGSaGaaGjbVlaabAeaca aMe8UaaeyAaiaabohacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaa bshacaqGVbGaaeiDaiaabggacaqGSbGaaGjbVlaabAgacaqGVbGaae OCaiaabogacaqGLbGaaGjbVlaabggacaqGJbGaaeiDaiaabMgacaqG UbGaae4zaiaaysW7caqGVbGaaeOBaiaaysW7caqG0bGaaeiAaiaabw gacaaMe8UaaeiCaiaabggacaqGYbGaaeiDaiaabMgacaqGJbGaaeiB aiaabwgacaqGUaGaaGjbVdGcbaqcLbyacqGH0icxcaqGOaGaaeyAai aabMcacaaMe8UaaeivaiaaysW7caqGPbGaae4CaiaaysW7caqG0bGa aeiAaiaabwgacaaMe8Uaae4yaiaab+gacaqGYbGaaeOCaiaabwgaca qGJbGaaeiDaiaaysW7caqGVbGaaeiCaiaabshacaqGPbGaae4Baiaa b6gacaqGUaaaaaa@7639@

Q.102 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?

Ans.

Here,retardingforceappliedtothebody,F = -50N Massofthebody,m = 20kg Initialspeedofthebody,u = 15 ms -1 Finalspeedofthebody,v = 0 AccordingtoNewton’ssecondlawofmotion, F = ma 50=20×a a= 50 20 =2.5m s 2 Accordingtothefirstequationofmotion, v=u+at t= u a = 15 2.5 =6s MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@76F0@

Q.103 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?

Ans.

Here,massofbody,m = 3.0kg initialspeedofbody,u = 2.0 ms -1 Finalspeedofbody,v = 0 Accordingtothefirstequationofmotion: v=u+at a= vu t = 3.52 25 = 1.5 25 =0.06m s 2 AccordingtoNewton’ssecondlawofmotion: F = ma =3×0.06 =0.18N Theforceisalongthedirectionofmotionofthebody. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWabeqaeaaakqaabeqaaKqzagGaaeisaiaabwgacaqGYbGaaeyzaiaabYcacaaMe8UaaeyBaiaabggacaqGZbGaae4CaiaaysW7caqGVbGaaeOzaiaaysW7caqGIbGaae4BaiaabsgacaqG5bGaaeilaiaaysW7caqGTbGaaeiiaiaab2dacaqGGaGaae4maiaab6cacaqGWaGaaGjbVlaabUgacaqGNbGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7aOqaaKqzagGaaeyAaiaab6gacaqGPbGaaeiDaiaabMgacaqGHbGaaeiBaiaaysW7caqGZbGaaeiCaiaabwgacaqGLbGaaeizaiaaysW7caqGVbGaaeOzaiaaysW7caqGIbGaae4BaiaabsgacaqG5bGaaeilaiaaysW7caqG1bGaaeiiaiaab2dacaqGGaGaaeOmaiaab6cacaqGWaGaaGjbVlaab2gacaqGZbqcfa4aaWbaaSqabeaajugGbiaab2cacaqGXaaaaaGcbaqcLbyacaqGgbGaaeyAaiaab6gacaqGHbGaaeiBaiaaysW7caqGZbGaaeiCaiaabwgacaqGLbGaaeizaiaaysW7caqGVbGaaeOzaiaaysW7caqGIbGaae4BaiaabsgacaqG5bGaaeilaiaaysW7caqG2bGaaeiiaiaab2dacaqGGaGaaeimaaGcbaqcLbyacaqGbbGaae4yaiaabogacaqGVbGaaeOCaiaabsgacaqGPbGaaeOBaiaabEgacaaMe8UaaeiDaiaab+gacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaabAgacaqGPbGaaeOCaiaabohacaqG0bGaaGjbVlaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaGjbVlaab+gacaqGMbGaaGjbVlaab2gacaqGVbGaaeiDaiaabMgacaqGVbGaaeOBaiaabQdaaOqaaKqzagGaamODaiabg2da9iaadwhacqGHRaWkcaWGHbGaamiDaaGcbaqcLbyacqGH0icxcaWGHbGaeyypa0tcfa4aaSaaaOqaaKqzagGaamODaiabgkHiTiaadwhaaOqaaKqzagGaamiDaaaaaOqaaKqzagGaeyypa0tcfa4aaSaaaOqaaKqzagGaaG4maiaac6cacaaI1aGaeyOeI0IaaGOmaaGcbaqcLbyacaaIYaGaaGynaaaaaOqaaKqzagGaeyypa0tcfa4aaSaaaOqaaKqzagGaaGymaiaac6cacaaI1aaakeaajugGbiaaikdacaaI1aaaaaGcbaqcLbyacqGH9aqpcaaIWaGaaiOlaiaaicdacaaI2aGaaGjbVlaad2gacaWGZbqcfa4aaWbaaSqabeaajugGbiabgkHiTiaaikdaaaaakeaajugGbiaabgeacaqGJbGaae4yaiaab+gacaqGYbGaaeizaiaabMgacaqGUbGaae4zaiaaysW7caqG0bGaae4BaiaaysW7caqGobGaaeyzaiaabEhacaqG0bGaae4Baiaab6gacaqGNaGaae4CaiaaysW7caqGZbGaaeyzaiaabogacaqGVbGaaeOBaiaabsgacaaMe8UaaeiBaiaabggacaqG3bGaaGjbVlaab+gacaqGMbGaaGjbVlaab2gacaqGVbGaaeiDaiaabMgacaqGVbGaaeOBaiaabQdaaOqaaKqzagGaaeOraiaabccacaqG9aGaaeiiaiaab2gacaqGHbaakeaajugGbiaabccacqGH9aqpcaaIZaGaey41aqRaaGimaiaac6cacaaIWaGaaGOnaaGcbaqcLbyacaqGGaGaeyypa0JaaGimaiaac6cacaaIXaGaaGioaiaaysW7caqGobaakeaajugGbiaabsfacaqGObGaaeyzaiaaysW7caqGMbGaae4BaiaabkhacaqGJbGaaeyzaiaaysW7caqGPbGaae4CaiaaysW7caqGHbGaaeiBaiaab+gacaqGUbGaae4zaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8UaaeizaiaabMgacaqGYbGaaeyzaiaabogacaqG0bGaaeyAaiaab+gacaqGUbGaaGjbVlaab+gacaqGMbGaaGjbVlaab2gacaqGVbGaaeiDaiaabMgacaqGVbGaaeOBaiaaysW7caqGVbGaaeOzaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8UaaeOyaiaab+gacaqGKbGaaeyEaiaab6caaaaa@9816@

Q.104 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Ans.

Here,massofthebody,m = 5kg Thegivensituationcanberepresentedbythegivenfigure Let, F 1 = OS =8N and F 2 = OT =6N Theresultantoftwoperpendicularforcesisgivenas: R= F 1 2 + F 2 2 Resultantforce,R= ( 8 ) 2 + ( 6 ) 2 = 64+36 =10N LetθbeanglemadebyRwith F 1 tanθ= SC OS = OT OS = 6 8 θ= tan 1 ( 6 8 ) = 36.87 o Thisgivesthedirectionoftheresultantforceandthereforethedirectionoftheaccelerationofthebody. AccordingtoNewton’ssecondlawofmotion,wehave: F=ma Acceleration,a= F m = 10 5 =2m s 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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