assignment 1.5 physics class 11

NCERT Solutions for Class 11 Physics

NCERT Solutions for Class 11 Physics in Hindi and English Medium PDF format are available to download updated for new academic session 2024-25. CBSE Class 11 Physics NCERT Solutions are a set of comprehensive and detailed answers, explanations, and solutions for the exercises and problems presented in the Class 11 Physics textbook published by the National Council of Educational Research and Training (NCERT). These solutions are designed to help students understand and master the concepts covered in the Physics curriculum for Class 11 under the Central Board of Secondary Education (CBSE) in India. Get here Solutions of Exercises, Additional Exercises, Supplementary material and NCERT books. Class 11 Physics Chapter Wise NCERT Solutions Chapter 1. Units and Measurements Chapter 2. Motion in a Straight Line Chapter 3. Motion in a Plane Chapter 4. Laws of Motion Chapter 5. Work, Energy and Power Chapter 6. System of Particles and Rotational Motion Chapter 7. Gravitation Chapter 8. Mechanical Properties of Solids Chapter 9. Mechanical Properties of Fluids Chapter 10. Thermal Properties of Matter Chapter 11. Thermodynamics Chapter 12. Kinetic Theory Chapter 13. Oscillations Chapter 14. Waves The CBSE Class 11 Physics NCERT Solutions offer step-by-step explanations for the various types of questions and problems found in the textbook. These solutions cover a wide range of topics in physics, including mechanics, thermodynamics, waves, optics, electromagnetism, and more. By providing clear explanations, diagrams, graphs, and examples, these solutions aim to make complex physics concepts accessible and understandable to students.

Offline Apps based on latest NCERT Solutions for (+1) are available to download along with the answers given at the end of the book. Key features and benefits of CBSE Class 11 Physics NCERT Solutions are given here. The solutions help students develop a strong understanding of fundamental physics concepts, laying the foundation for advanced topics in Class 12 and beyond. Revision books, based on Latest CBSE Syllabus, for each chapter as well as notes of each chapter are available to download in PDF format. Chapter tests and assignments will be uploaded time to time. Class 11 Physics NCERT Solutions align with the CBSE curriculum, making them valuable study materials for exams, including school exams and board exams. If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback.

The solutions include practice exercises that allow students to apply their knowledge and practice problem-solving skills. These follow a logical sequence, guiding students through the thought process required to solve different types of physics problems. Class 11 Physics  NCERT text book solutions  are given here to download in PDF. Extra study material covering the whole syllabus divided into 10 units will help the students in revising the complete course.  Unit 1  contains the material related to dimension and measurement,  Unit 2  is covering Kinematics,  Unit 3  covers the newton’s law of motion, similarly the  Unit 4 ,  Unit 5 ,   Unit 6 ,  Unit 7 ,  Unit 8 ,  Unit 9  and  Unit 10  covers the entire curriculum. Many physics concepts are explained using diagrams, graphs, and illustrations to enhance understanding. The solutions often include real-world examples that demonstrate the practical applications of physics concepts.

NCERT Book solutions are prepared by experienced educators and experts in the field. This ensures that the content is accurate, reliable, and follows the best pedagogical practices. Strong comprehension of Class 11 physics concepts facilitated by these solutions can help students prepare for various competitive entrance exams. CBSE Class 11 Physics NCERT Solutions are designed to be self-explanatory, making them suitable for self-study and independent learning. In essence, Class 11 Physics NCERT solutions play a crucial role in helping students lay a solid foundation in physics, providing them with the necessary skills and knowledge to excel in their studies and future pursuits.

Chapter Wise Study material for Class 11 Physics Unit 1 to 10

• Class 11 Physics Study Material Unit 1
• Class 11 Physics Study Material Unit 2
• Class 11 Physics Study Material Unit 3
• Class 11 Physics Study Material Unit 4
• Class 11 Physics Study Material Unit 5
• Class 11 Physics Study Material Unit 6
• Class 11 Physics Study Material Unit 7
• Class 11 Physics Study Material Unit 8
• Class 11 Physics Study Material Unit 9
• Class 11 Physics Study Material Unit 10

11th Physics solutions are prepared by subject matter experts, ensuring accuracy and quality in the explanations and solutions provided. Overall, CBSE Class 11 Physics NCERT Solutions serve as a valuable resource to help students master the subject, develop problem-solving skills, and prepare for exams and future academic pursuits in the field of physics or related disciplines.

Chapter Wise Revision Books for Class 11 Physics

• Revision Book Chapter 1: Units and Measurements
• Revision Book Chapter 2: Motion in a Straight Line
• Revision Book Chapter 3: Motion in a Plane
• Revision Book Chapter 4: Laws of Motion
• Revision Book Chapter 5: Work, Energy and Power
• Revision Book Chapter 6: System of Particles and Rotational Motion
• Revision Book Chapter 7: Gravitation
• Revision Book Chapter 8: Mechanical Properties of Solids
• Revision Book Chapter 9: Mechanical Properties of Fluids
• Revision Book Chapter 10: Thermal Properties of Matter
• Revision Book Chapter 11: Thermodynamics
• Revision Book Chapter 12: Kinetic Theory
• Revision Book Chapter 13: Oscillations
• Revision Book Chpater 14: Waves
• Revision Books Solutions and Answers Part 1
• Revision Books Solutions and Answers Part 2

The NCERT solutions for Class 11 Physics hold significant importance for students. Physics is a fundamental subject that forms the basis for many scientific and technical fields. Here’s why Class 11 Physics Tiwari Academy NCERT solutions are important. Class 11 Physics introduces students to essential concepts that lay the groundwork for more advanced topics in higher classes.

The work ‘Physics’ has originated from Greek work fusis, meaning nature. So, in physics we deal with nature and natural phenomena. We understand nature only through the study matter, energy and their interactions. The solutions provided in NCERT textbooks and their solutions align closely with the CBSE Class 11 Physics syllabus. As a result, studying these solutions thoroughly prepares students for their exams, including the board exams and competitive entrance exams.

Practice Test based on all Chapters of 11 Physics NCERT

• 11th Physics Chapter 1 Test 1
• 11th Physics Chapter 2 Test 1
• 11th Physics Chapter 2 Test 2
• 11th Physics Chapter 3 Test 1
• 11th Physics Chapter 4 Test 1
• 11th Physics Chapter 5 Test 1
• 11th Physics Chapter 5 Test 2
• 11th Physics Chapter 6 Test 1
• 11th Physics Chapter 6 Test 2
• 11th Physics Chapter 7 Test 1
• 11th Physics Chapter 7 Test 2
• 11th Physics Chapter 8 Test 1
• 11th Physics Chapter 8 Test 2
• 11th Physics Chapter 9 Test 1
• 11th Physics Chapter 10 Test 1
• 11th Physics Chapter 10 Test 2
• 11th Physics Chapter 11 Test 1
• 11th Physics Chapter 12 Test 1
• 11th Physics Chapter 13 Test 1
• 11th Physics Chapter 14 Test 1
• 11th Physics Chapter 15 Test 1
• 11th Physics Chapter 15 Test 2

Physics involves problem-solving and critical thinking. NCERT solutions include a variety of exercises that require students to apply concepts to solve problems. Practicing these problems hones their analytical and problem-solving skills. It often involves visual concepts, such as diagrams, graphs, and illustrations. NCERT solutions use these visual aids to explain concepts, making it easier for students to grasp abstract ideas.

Mechanics and properties of matter, Heat and thermodynamics, Sound or acoustics, Electricity and magnetism, Modern physics, Biophysics, Astrophysics, Geophysics, Nuclear physics , etc. Physics is based on mathematical equations and formulas. NCERT solutions break down these equations step by step, making it easier for students to understand how they are derived and applied.

1. Classical Mechanics: It explains the motion of particles which travel with velocities much less than that of light. 2. Theory of Relativity: It explains the in-variance in nature and also the motion of particles which travel with velocities close to that of light. 3. Thermodynamics: It is the theory of heat, temperature and conversion of heat into work and vice-versa. It also explains the behaviours of systems containing very large number of particles. NCERT solutions often include real-life examples that demonstrate the practical applications of physics concepts. This helps students relate theoretical concepts to the world around them.

4. Electromagnetism: It is theory of electricity, magnetism and the electromagnetic radiation including optics. 5. Quantum Mechanics: It explains the behaviours of atomic and subatomic systems of particles. According to Albert Einstein, Science is not just a collection of laws, a catalogue of unrelated facts. It is a creation of human mind with its freely invented ideas and concepts. About modern physics, he said, the reality created by modern physics is indeed, far removed from the reality of the early days. Many competitive entrance exams for engineering and other science-related fields include physics questions. Strong foundational knowledge gained from NCERT solutions can give students an advantage in such exams.

Class 1 Physics Chaperwise Revision Notes

• Class 11 Physics Chapter 1 Revision Questions
• Class 11 Physics Chapter 1 Revision Notes
• Class 11 Physics Chapter 2 Revision Notes
• Class 11 Physics Chapter 3 Revision Notes
• Class 11 Physics Chapter 4 Revision Notes
• Class 11 Physics Chapter 5 Revision Notes
• Class 11 Physics Chapter 6 Revision Notes
• Class 11 Physics Chapter 7 and 8 Revision Notes
• Class 11 Physics Chapter 9 Revision Notes
• Class 11 Physics Chapter 10 Revision Notes

The animate and ananimate – all objects are made up of matter and energy. Therefore, any type of study of an object is basically the study of some manifestation of matter and energy. The concepts learned in Class 11 Physics are built upon in Class 12 and higher education. Understanding Class 11 concepts thoroughly using NCERT solutions ensures a smooth transition to more advanced topics. The matter and energy is the subject of study of physics, hence the laws of physics are at work at the root of the all sciences. Therefore, it need not be surprising if some direct modification of the laws of physics appear to be underlying the human behaviour also. Due to these reasons physics has intimate relationship with all the sciences and hence it is called a fundamental science. Physics requires logical reasoning and critical analysis. By working through NCERT solutions, students develop logical thinking skills that are valuable in various academic and professional pursuits.

Class 11 Physics Chapterwise Assignments

• Physics Mathematical Tools
• 11 Physics Chapter 4 Assignment 1
• 11 Physics Chapter 7 Assignment 1
• 11 Physics Chapter 7 Assignment 2
• 11 Physics Chapter 7 Assignment 3
• 11 Physics Chapter 7 Assignment 4
• 11 Physics Chapter 7 Assignment 5
• 11 Physics Chapter 8 Assignment 1
• 11 Physics Chapter 8 Assignment 2
• 11 Physics Chapter 8 Assignment 3
• 11 Physics Chapter 8 Assignment 4
• 11 Physics Chapter 8 Assignment 5
• 11 Physics Chapter 9 Assignment 1
• 11 Physics Chapter 9 Assignment 2
• 11 Physics Chapter 10 Assignment 1
• 11 Physics Chapter 10 Assignment 2
• 11 Physics Chapter 10 Assignment 3
• 11 Physics Chapter 10 Assignment 4
• 11 Physics Chapter 10 Assignment 5
• 11 Physics Chapter 12 Assignment 1
• 11 Physics Chapter 12 Assignment 2
• 11 Physics Chapter 13 Assignment 1
• 11 Physics Chapter 13 Assignment 2
• 11 Physics Chapter 13 Assignment 3
• 11 Physics Chapter 14 and 15 Assignment 1
• 11 Physics Chapter 14 and 15 Assignment 2
• 11 Physics Chapter 14 and 15 Assignment 3
• 11 Physics Chapter 14 and 15 Assignment 4
• 11 Physics Chapter 14 and 15 Assignment 5

NCERT solutions are designed to be self-explanatory. This makes them a great resource for self-study, enabling students to learn and revise concepts independently. Ask your doubts related to NIOS or CBSE Board and share your knowledge with your friends and other users through Discussion Forum . Download NCERT Solutions and CBSE Offline Apps based on latest NCERT Books. Class 11 Physics covers a wide range of topics, including mechanics, thermodynamics, waves, and more. NCERT textbook solutions provide a comprehensive understanding of these topics, helping students build a holistic view of the subject.

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NCERT Solutions for Class 11 Physics

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Unlock the magic of NCERT Solutions for Class 11 Physics ! Crafted by experienced teachers, these solutions dive into every idea, getting you ready for exams. Physics might seem tough, but with our Class 11 Physics NCERT Solutions , complex ideas become easy. Our content matches the latest CBSE syllabus, giving you real insights.

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From crucial exams to getting ahead, Class 11 Physics is super important. Our NCERT 11 Physics solutions are online or PDFs you can download, letting you learn your way. Topics cover a lot, so you pick what to study. Whether it’s motion, thermodynamics, or laws of motion, our NCERT 11 Physics solution has your back.

Physics, all about matter, is key for science. Class 11 shapes your future. Our NCERT Physics Class 11 solutions make it simple, reducing stress and boosting understanding. Revise easily with PDF solutions. Be great at concepts for higher classes and tough exams.

Our curated NCERT Physics Class 11 solutions , made by experts, guide you to understand. Access chapters now, download the whole thing, or just a chapter. Our teachers help with problem-solving, showing steps. Key questions and tips help you learn smoothly.

Know Physics, from basics to formulas. Our NCERT Physics Class 11 PDF solutions help you stride into your final school year and beyond with confidence. Boost your skills, tackle the trickiest stuff, and learn more with Infinity Learn’s NCERT Solutions for Class 11 Physics today!

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

NCERT Solutions Class 11 Physics Chapter Details and Exercises

Ncert solutions for class 11 physics chapter 1: physical world.

We’ll look at how things work and why they work the way they do in this Chapter. You’ll learn more about the world around y and the beginnings and history of science. You’ll also find various aphorisms and their reasons in this Chapter, all of which use Physics concepts in real-life situations. The strong and weak nuclear and gravitational, and electromagnetic forces are described. Listed below are some of the numerous scientists’ theories, observations, and conclusions. You will have a more profound knowledge of the factors influencing our country’s scientific and technological progress. In order to equip students with basic conceptual knowledge, this Chapter expands on the existence of electrons.

The following topics were covered:

Physics-scope and excitement; nature of physical laws; Physics, technology, and society.

Also, access the following resources for Class 11 Physics Chapter 1 The Solid State at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 1 – Introduction
• Class 11 Notes Physics Chapter 1 Physical World

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

Scientists use their senses, such as their ears and eyes, to collect data and make observations. Some are straightforward, such as determining color and texture, whereas others are more difficult and require measurement. It is a basic scientific concept without which scientists would be unable to conduct research. Students will be taught the units of physical quantities as well as how to evaluate them. It will also give you a good idea of the types of errors that can occur when measuring things and large amounts.

Units of measurement; systems of units, SI units, and fundamental and derived units are all required for measurement. Measurements of length, mass, and time; measuring device accuracy and precision; measurement errors; important figures

Dimensions of physical quantities, dimensional analysis, and its applications.

Also, access the following resources for Class 11 Physics Chapter 2 Units and Measurements at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 2 – Units and Measurement
• Units and Measurement Class 11 Notes Physics Chapter 2

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

This Chapter will go over crucial topics like comparing items as point objects and how to draw graphs and calculate values from them. You may easily Learn how to plot an x-t motion graph using the NCERT Solutions from Infinity Learn. By consulting these answers, you can quickly determine how long it takes a bike to travel, how fast a car goes, and how long it takes a bus to travel. Students will Learn the distinction between the magnitude of displacement and the total length of the path traveled. According to CBSE requirements, they will also study average speed, average velocity, instantaneous speed, and velocity.

Motion along a straight line, frame of reference: Speed and velocity, as well as a position-time graph.

Differentiation and integration are covered for characterizing uniform and non-uniform motion, average speed, instantaneous velocity, evenly accelerated motion, velocity-time, and position-time graphs. For uniformly accelerated motion, the following equations apply (graphical treatment).

Also, access the following resources for Class 11 Physics Chapter 3 Motion in a Straight Line at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 3 – Motion in a Straight Line
• Motion in a Straight Line Class 11 Notes Physics Chapter 3
• Motion In A Straight Line Questions for CBSE Class 11th

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

This Chapter will go over crucial topics like comparing items as point objects and how to draw graphs and calculate values from them. You may easily Learn how to plot an x-t motion graph using NCERT Solutions from Infinity LearnBy consulting these answers; you can quickly determine how long it takes a bike to travel, how fast a car goes, and how long it takes a bus to travel. Students will Learn the distinction between the magnitude of displacement and the total length of the path traveled according to CBSE requirements; they will also study average speed, average velocity, instantaneous speed, and velocity.

Position and displacement vectors; general vectors and associated notations; equality of vectors; multiplication of vectors by an actual number; Resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors; addition and subtraction of vectors, relative velocity, Unit vector;

Cases of uniform velocity and uniform acceleration in-plane motion homogeneous circular motion, projectile motion

Also, access the following resources for Class 11 Physics Chapter 4 Motion in a Plane at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 4 – Motion in a Plane
• Motion in a Plane Class 11 Notes Physics Chapter 4
• Motion In A Plane Questions for CBSE Class 12th

NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

This Chapter is critical in mechanics. Usually, problems based on conservation of momentum are asked on the exam. After reading this Chapter, students will understand the laws of motion and how they apply to our daily lives. To assist students in preparing for exams, this Chapter solves various numerical problems based on mechanics. Students with a good comprehension of these concepts will be able to excel in their higher education levels.

Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion.

The law of linear momentum conservation and its applications.

Concurrent forces equilibrium, static and kinetic friction, friction laws, rolling friction, and lubrication

Uniform circular motion dynamics: centripetal force, circular motion examples (vehicle on a level circular road, vehicle on a banked road).

Also, access the following resources for Class 11 Physics Chapter 5 Laws of Motion at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 5 – Law of Motion

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

The Work-Energy Theorem is an important concept in this Chapter. Because this Chapter will be studied again in future sessions, it is critical that students memorize it completely. To ensure that students comprehend the ideas, the labour that a person performs, and the energy required to execute that work is quickly explained with appropriate examples. The CBSE board’s mark weightage is used to answer problems involving determining energy and power in a step-by-step manner. They will have no trouble learning the Chapter if they apply these ideas regularly.

The following subjects were discussed:

Kinetic energy, work-energy theorem, and power are terms used to describe the work done by a constant force and a variable force.

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

Also, access the following resources for Class 11 Physics Chapter 6 Work, Energy, and Power at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 6 – Work, Energy and Power
• Work, Energy and Power Class 11 Notes Physics Chapter 6

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

This Chapter will help you understand how extended bodies, such as a particle system, move. The central mass of a system of particles is the main topic of this Chapter. The concept of motion and its utility are briefly defined in the CBSE board’s syllabus. Students are strongly urged to use excellent study materials when answering the textbook’s practice questions because this Chapter covers many topics.

Momentum conservation, centre of mass motion, and the centre of mass of a two-particle system A rigid body’s centre of mass; a uniform rod’s centre of mass. Force moment, torque, angular momentum, the law of conservation of angular momentum, and its applications

Rigid body equilibrium, rigid body rotation, rotational motion equations, and comparison of linear and rotational motions

Inertia moment, gyration radius, and moment of inertia values for simple geometrical objects (no derivation). Theorems of parallel and perpendicular axes and their applications are stated.

Also, access the following resources for Class 11 Physics Chapter 7 System of Particles and Rotational Motion at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 7 – Systems of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

All material objects are attracted to the earth, as we already know. Things we throw up end up on the ground. It is more difficult to climb a hill than it is to descend one. Gravitation is a fairly common idea and anunderpins most of the concepts we’ll be studying in the future. In order to understand the more advanced issues in this Chapter, students must first understand the difference between gravity and gravitation. Other concepts explored in this Chapter include the potential energy differential between two points, acceleration due to gravity, and interplanetary motion.

The universal law of gravitation, Kepler’s rules of planetary motion. Gravitational acceleration and how it varies with altitude and depth.

Geo-stationary satellites, gravitational potential energy, gravitational potential escape velocity, orbital velocity of a spacecraft

Also, access the following resources for Class 11 Physics Chapter 8 Gravitation at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 8 – Gravitation
• Gravitation Class 11 Notes Physics Chapter 8

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

All material objects are attracted to the earth, as we already know. Things we throw up end up on the ground. It is more difficult to climb a hill than it is to descend one. Gravitation is a fairly common idea and underpins most of the concepts we’ll be studying in the future. To understand the more advanced issues in this Chapter, students must first understand the difference between gravity and gravitation. Other concepts explored in this Chapter include the potential energy differential between two points, acceleration due to gravity, and interplanetary motion.

Geostationary satellites, gravitational potential energy, gravitational potential escape velocity, orbital velocity of a spacecraft

Also, access the following resources for Class 11 Physics Chapter 9 Mechanical Properties of Solids at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 9 – Mechanical Properties of Solids
• Mechanical Properties of Solids Class 11 Notes Physics Chapter 9
• Mechanical Properties Of Solids Questions for CBSE Class 11th

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

In this part, we’ll look at the physical properties of liquids and gases. Fluids are defined as substances that can flow. This is an important distinction between solids and liquids, and gases. Other important concepts covered are Bernoulli’s principle, Reynold’s number, streamline flow, viscosity, and surface tension. The ability to flow is the most basic characteristic of a fluid. It cannot change its shape. Students will have a good understanding of fluid mechanical properties, which will aid their performance on the final exam.

Pascal’s l, its applications (hydraulic lift and hydraulic brakes), and the influence of gravity on fluid pressure are all discussed.

Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem, and applications, Surface energy, and surface tension, angle of contact, excess pressure across a curved surface, application of surface tension ideas to drops, bubbles, and capillary rise.

Also, access the following resources for Class 11 Physics Chapter 10 Mechanical Properties of Fluids at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 10 – Mechanical Properties of Fluids
• Mechanical Properties Of Fluids Questions for CBSE Class 11th
• Mechanical Properties of Fluid | Class 11 Physics

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

We’re all familiar with the concepts of heat and temperature. The procedure for determining the temperature is covered in this Chapter. Students will also obtain a solid comprehension of Newton’s Law of Cooling, which is an important concept for many competitive assessments. Temperature is the hotness of a body that can be measured with a thermometer. This Chapter walks students through the various thermal properties of matter in a step-by-step format to help them grasp the concepts.

Heat, temperature, and thermal expansion of solids, liquids, and gases, as well as anomalous expansion of water; specific heat capacity. Change of state – latent heat capacity; Cp, Cv – calorimetry

Heat transfer-conduction, convection radiation, thermal conductivity, qualitative conceptions of Blackbody radiation, Wein’s displacement law, Stefan’s law, and the Greenhouse effect are alf heat transfer-conduction.

Also, access the following resources for Class 11 Physics Chapter 11 Thermal Properties of Matter at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 11 – Thermal Properties of Matter
• Thermal Properties Of Matter Questions for CBSE Class 11th

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

The study of heat and temperature, as well as their conversion to various forms of energy, is known as thermodynamics. Most exam questions are located in this Chapter, one of the most popular themes among question paper writers. In order to get great grades, students must understand all of these principles. This Chapter discusses, among other things, thermodynamic rules, specific heat capacity, various thermodynamic processes, and the Carnot engine.

The zeroth law of thermodynamics, thermal equilibrium, heat, work, and internal energy are discussed—isothermal and adiabatic processes, as well as the first law of thermodynamics.

Reversible and irreversible processes, heat engines, and refrigerators are all examples of the second law of thermodynamics.

Also, access the following resources for Class 11 Physics Chapter 12 Thermodynamics at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 12 – Thermodynamics
• Thermodynamics Questions for CBSE Class 11th

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

This Chapter is more important when it comes to exam preparation. Unlike solids and liquids, the properties of gases are easy to comprehend. The molecular nature of matter, the behaviour of gases, the kinetic theory of an ideal gas, the law of energy equipartition, specific heat capacity, and the mean free path are all covered in this Chapter. The concerns are also addressed systematically and CBSE-compliantly, making it easy for Class 11 students to earn decent scores.

Work done in compressing a gas, equation of state of a perfect gas.

Assumptions and the concept of pressure in the kinetic theory of gases. 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; 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

Also, access the following resources for Class 11 Physics Chapter 13 Kinetic Theory at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 13 – Kinetic Theory
• Kinetic Theory Class 11 Notes Physics Chapter 13

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Oscillations are a basic physics concept. Because the principles provided in this Chapter will be revisited in the following lectures, it is vital that students comprehend them. This Chapter covers periodic and oscillatory motions, basic harmonic motion, uniform circular motion, and other concepts. It is critical for students to use adequate study materials in order to master these topics. If students study the Chapter daily, they will be better prepared to face the more challenging questions on the exam.

Period, frequency, displacement as a function of time, and periodic functions are all examples of periodic motion.

Phase; oscillations of a loaded spring-restoring force and force constant; energy in S.H.M. Simple pendulum derivation of expression for its period; kinetic and potential energies Resonance, free, forced, and damped oscillations (qualitative notions only).

Also, access the following resources for Class 11 Physics Chapter 14 Oscillations at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 14 – Oscillations
• Oscillations Class 11 Notes Physics Chapter 14
• Important Topic of Physics – Oscillation

NCERT Solutions for Class 11 Physics Chapter 15 Waves

This Chapter includes in-depth questions on key subjects, including wave dynamics and other relevant topics. This Chapter’s questions will walk you through concepts like wave types, sound speed in the air, string tension, and sound speed in air dependence. Important formulas and concepts are highlighted in the solutions to make it easier for students to revise them. Numerous wave characteristics are discussed in this Chapter, including amplitude, frequency, phase, wavelength, wave displacement, and resonance.

Transverse and longitudinal waves, traveling wave speed, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect

Also, access the following resources for Class 11 Physics Chapter 15 Waves at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 15 – Waves

Because Infinity Learn is a learning organization, we understand the significance of study materials and how they influence academic performance. As a result, we developed Class 11 NCERT Solutions, a learning resource aimed at helping students study and achieve their objectives. Because each Chapter covers a huge number of topics, students must study them frequently to ensure that they understand them. It will help students understand how to approach difficult questions on the final test in the right way. NCERT Class 11 Books might help students better understand the syllabus.

CBSE Marking Scheme 2023-24

The CBSE board has thoughtfully organized the academic sessions into well-defined segments, strategically covering the entire syllabus. This meticulous approach is designed to empower students with effective learning experiences. The subject matter experts have ingeniously connected various topics and concepts, establishing a strong foundation for the learners. This interconnectedness not only aids in comprehension but also highlights the beautiful harmony within the subject.

The culmination of these sessions is the eagerly awaited board examinations, masterminded by the CBSE board itself. These examinations, held at the culmination of the academic year, serve as the ultimate assessment of the student’s grasp of the subjects. These exams are structured in accordance with the thoughtfully divided syllabus, ensuring that students are evaluated on the complete spectrum of what they have learned. This approach adds an extra layer of clarity and coherence to the entire academic process.

The significance of this approach lies in its ability to enhance the overall quality of education. By conducting these comprehensive assessments, students are encouraged to engage more profoundly with the subjects throughout the year. The anticipation of the year-end examinations acts as a guiding light, propelling students toward consistent learning and preparation.

Why Should One Opt for Infinity Learn NCERT Solutions?

Opting for Infinity Learn NCERT Solutions offers a multitude of benefits that can significantly enhance your learning experience and academic performance. Here are some compelling reasons why you should consider choosing Infinity Learn’s NCERT Solutions:

• Comprehensive Coverage: Infinity Learn’s NCERT Solutions cover the entire NCERT curriculum for various subjects and classes. This ensures that you have access to accurate and detailed explanations for every topic, chapter, and concept, leaving no gaps in your understanding.
• Clarity and Simplicity: The solutions provided by Infinity Learn are designed to be clear, concise, and easy to understand. Complex concepts are broken down into simple explanations, making it easier for you to grasp and apply the knowledge effectively.
• Expertly Crafted: Infinity Learn’s NCERT Solutions are crafted by subject-matter experts who have a deep understanding of the curriculum and the intricacies of each topic. This ensures that you receive accurate and reliable information that aligns with the NCERT syllabus.
• Step-by-Step Approach: The solutions follow a step-by-step approach, guiding you through each problem or concept methodically. This approach not only helps you solve problems but also teaches you the thought process behind it, enhancing your problem-solving skills.
• Variety of Questions: Infinity Learn’s solutions cover a wide range of questions, including examples from the NCERT textbook as well as additional practice questions. This variety prepares you for different types of questions that might appear in exams.
• Time Management: By providing efficient solutions, Infinity Learn helps you save time in understanding concepts and solving problems. This gives you more time to focus on revision, practice, and exploring additional study materials.
• Visual Aids: Some solutions may include diagrams, graphs, and illustrations to enhance your understanding of certain concepts. These visual aids make learning more engaging and help in retaining information better.
• Concept Reinforcement: Infinity Learn’s solutions aim to reinforce the core concepts of each topic. This helps in building a strong foundation, which is crucial for understanding more complex concepts in the future.
• Exam Preparation: The solutions are structured in a way that prepares you for exams. They not only provide answers but also offer insights into the right approach to answering questions and highlighting key points that examiners might look for.
• Self-Assessment: Most solutions include exercises and practice questions. By attempting these questions and comparing your answers with the provided solutions, you can assess your own understanding and identify areas that need more attention.
• Convenient Accessibility: Infinity Learn’s NCERT Solutions are often available online, making them accessible anytime, anywhere. This convenience allows you to study and practice at your own pace and convenience.
• Supplement to Classroom Learning: These solutions can serve as a supplement to your classroom learning, helping you reinforce what you’ve learned in school and clarifying any doubts you might have.

In conclusion, Infinity Learn’s NCERT Solutions offer a wealth of advantages, from comprehensive coverage to expert guidance and efficient learning. They can be a valuable tool to complement your studies, enhance your understanding, and boost your academic success.

Vijayi Bhava Course | Sample Video for Class 11 Physics

Frequently Asked Questions on NCERT Solutions for Class 11 Physics

Why should i consider using ncert solutions for class 11 physics.

NCERT Solutions for Class 11 Physics serve as a valuable resource to tackle textbook problems effectively. The challenges posed by Class 11 Physics chapters can be daunting initially, given the vast syllabus and introduction to new concepts. To navigate this, students can turn to NCERT Solutions available on e-learning platforms like Infinity Learn for comprehensive clarity. These solutions play a crucial role in resolving doubts related to any question, thereby aiding students in better exam preparation. Since Class 11 Physics content holds significant weightage in competitive exams like NEET and JEE, it's vital for students to focus on mastering the subject. By opting for NCERT Solutions, students can enrich their understanding and gain the proficiency needed to solve problems accurately.

Which chapter should I begin with if I'm finding Class 11 Physics challenging?

Within the NCERT Solutions for Class 11 Physics, you'll discover 15 diverse chapters, each with its level of complexity tailored to individual students. If you're seeking a starting point, consider Chapter 1: Physical World, which offers a straightforward introduction. Additionally, Chapter 2: Units and Measurements is both manageable and essential. Infinity Learn's solutions provide clear and accessible explanations for all chapters, offering substantial support to students. Initiating your journey with these chapters can boost your confidence and provide a solid foundation for progressing through the rest of the material with greater ease.

Which chapter is important in Class 11 Physics?

In the realm of Physics, the significance of each chapter is uniform and noteworthy. Despite varying levels of complexity, these chapters collectively contribute to your academic achievement and lay the groundwork for future endeavors. Adhering to the NCERT-prescribed syllabus is essential, as it aligns with examination content. Infinity Learn’s NCERT solutions for Class 11 Physics echo this approach, furnishing students with confidence in their studies. This assurance stems from the alignment with curriculum, ensuring a holistic understanding of the subject matter.

Is Class 11 Physics Challenging?

No subject becomes challenging when approached with proper guidance and consistent practice. Within the Class 11 curriculum, Physics might present a slightly more intricate landscape compared to other subjects. It demands a firm grasp of concepts and a sturdy foundation to conquer its intricacies. Infinity Learn's Class 11 Physics solutions serve as a valuable resource for demystifying these complex topics. These solutions present the material in a clear, engaging manner, ensuring that students can readily comprehend and retain these fundamental principles for a lifetime.

Which chapter in the NCERT Solutions for Class 11 Physics is considered less challenging?

When students dedicate consistent effort to understand all concepts, no chapter becomes overly difficult. The extensive syllabus might initially seem overwhelming, but as you progress, you'll find comfort in the subject. Should any uncertainties arise, you can seek guidance from experienced faculty members at Infinity Learn. They have years of expertise in this domain. Furthermore, you can freely access the solutions PDF on the Infinity Learn website

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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements - FREE PDF Download

NCERT for Chapter 1 Units and Measurements Class 11 Solutions by Vedantu, forms the basis for all experimental and theoretical studies in physics. Understanding units and measurements is important as it ensures consistency and accuracy in scientific observations and calculations. This chapter emphasizes the importance of standardized units and accurate measurements in ensuring consistency and precision in scientific observations and calculations. It provides students with the essential tools and techniques to measure physical quantities correctly, which is critical for validating scientific theories and conducting experiments. By practising with Class 11 Physics NCERT Solutions , students can build confidence in their understanding and excel in their studies.

Note: ➤ Predict your NEET rank effortlessly with our NEET Rank Predictor 2024 !

Glance on Physics Chapter 1 Class 11 - Units and Measurements

The chapter introduces the concept of base units for fundamental physical quantities and derived units for quantities that are combinations of base quantities. 

Various techniques and instruments for measuring physical quantities accurately are discussed. This includes the use of vernier calipers, micrometers, and other precise tools.

The chapter also covers different types of errors—systematic and random—and methods to minimize and account for these errors in measurements.

The chapter discusses the applications of dimensional analysis in checking the correctness of equations, converting units from one system to another, and deriving relationships between physical quantities. Also, it includes significant figures and rules for determining the number of significant figures in a given measurement.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 1 - Units And Measurements, which you can download as PDFs.

There are 17 fully solved questions in class 11th physics chapter 1 Units And Measurements Exercise.

System of Units:

Access NCERT Solutions for Class 11 Physics Chapter 1 – Units and Measurements

a) The Volume of a Cube of 1cm is Equal To ……………… ${{m}^{3}}$. 

We know that, 

$1cm=\frac{1}{100}m$

Volume of a cube of side 1cm would be, 

$V=1cm\times 1cm\times 1cm=1c{{m}^{3}}$

On converting it into unit of ${{m}^{3}}$, we get, 

$1c{{m}^{3}}={{\left( \frac{1}{100}m \right)}^{3}}={{\left( {{10}^{-2}}m \right)}^{3}}$

$\therefore 1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$

Therefore, the volume of a cube of side 1cm is equal to  ${{10}^{-6}}{{m}^{3}}$.

b) The Surface Area of a Solid Cylinder of Radius 2.0cm and Height 10.0cm is Equal To ……………….${{\left( mm \right)}^{2}}$

We know the formula for the total surface area of cylinder of radius r and height h to be,

$S=2\pi r\left( r+h \right)$

We are given:

$r=2cm=20mm$

$h=10cm=100mm$

On substituting the given values into the above expression, we get, 

$S=2\pi \times 20\left( 20+100 \right)=15072m{{m}^{2}}=1.5\times {{10}^{4}}m{{m}^{{{2}^{{}}}}}$

Therefore, the surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to $1.5\times {{10}^{4}}{{\left( mm \right)}^{2}}$. 

c) A Vehicle Moving With a Speed of $18km{{h}^{-1}}$Covers…………………. m in 1s. 

We know the following conversion:

$1km/h=\frac{5}{18}m/s$

$\Rightarrow 18km/h=18\times \frac{5}{18}=5m/s$

Now we have the relation:

$\text{Distance = speed }\times \text{ time }$

Substituting the given values, $\text{Distance = 5}\times \text{1 =5m}$

Therefore, a vehicle moving with a speed of $18km{{h}^{-1}}$covers 5m in 1s.

d) The Relative Density of Lead is 11.3. Its Density Is ………………. $gc{{m}^{-3}}$or………………… $kg{{m}^{-3}}$. 

We know that the relative density of substance could be given by, 

$\text{Relative density = }\frac{density\text{ of substance}}{density\text{ of water}}$

$density\text{ of water = 1kg/}{{\text{m}}^{3}}$

$\text{density of lead = Relative density of lead }\times \text{ density of water = 11}\text{.3}\times \text{1= 11}\text{.3g/c}{{\text{m}}^{3}}$

But we know, 

$1g={{10}^{-3}}kg$

$1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$

$\Rightarrow 1g/c{{m}^{3}}=\frac{{{10}^{-3}}}{{{10}^{-6}}}kg/{{m}^{3}}={{10}^{3}}kg/{{m}^{3}}$

$\therefore 11.3g/c{{m}^{3}}=11.3\times {{10}^{3}}kg/{{m}^{3}}$

Therefore, the relative density of lead is 11.3. Its density is $11.3gc{{m}^{-3}}$or$11.3\times {{10}^{3}}kg{{m}^{-3}}$.

2. Fill ups.

a) $1kg{{m}^{2}}{{s}^{-2}}=..................gc{{m}^{2}}{{s}^{-2}}$

We know that:

$1kg={{10}^{3}}g$

$1{{m}^{2}}={{10}^{4}}c{{m}^{2}}$

$1kg{{m}^{2}}{{s}^{-2}}={{10}^{3}}g\times {{10}^{4}}c{{m}^{2}}\times 1{{s}^{-2}}={{10}^{7}}gc{{m}^{2}}{{s}^{-2}}$

Therefore, $1kg{{m}^{2}}{{s}^{-2}}={{10}^{7}}gc{{m}^{2}}{{s}^{-2}}$

b) $1m=.................ly$

We know that light year is the total distance covered by light in one year. 

$1ly=\text{Speed of light }\times \text{ one year}$

$\Rightarrow 1ly=\left( 3\times {{10}^{8}}m/s \right)\times \left( 365\times 24\times 60\times 60s \right)=9.46\times {{10}^{15}}m$

$\therefore 1m=\frac{1}{9.46\times {{10}^{15}}}=1.057\times {{10}^{-16}}ly$

Therefore, $1m=1.057\times {{10}^{-16}}ly$

c) $3.0m/{{s}^{2}}=.................km/h{{r}^{2}}$ 

$3.0m/{{s}^{2}}=$ ………….$km/h{{r}^{2}}$

We have, $1m={{10}^{-3}}km$

$1hr=3600s$

$\Rightarrow 1{{s}^{2}}={{\left( \frac{1}{3600} \right)}^{2}}h{{r}^{2}}$

Then, 

$3.0m/{{s}^{2}}=\frac{3\times {{10}^{-3}}}{{{\left( \frac{1}{3600}h \right)}^{2}}}km/h{{r}^{2}}$

$\therefore 3.0m/{{s}^{2}}=3.9\times {{10}^{4}}km/h{{r}^{2}}$ 

d)\[6.67\times{{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=.............{{g}^{-1}}c{{m}^{3}}{{s}^{-2}}\] 

$1N=1kgm{{s}^{-2}}$

$1kg={{10}^{-3}}g$

$1{{m}^{3}}={{10}^{6}}c{{m}^{3}}$

\[\Rightarrow 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}=6.67\times {{10}^{-11}}\times \left( 1kgm{{s}^{-2}} \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)\]

\[=6.67\times {{10}^{-11}}\times \left( 1kg\times 1{{m}^{3}}\times 1{{s}^{-2}} \right)\]

\[=6.67\times {{10}^{-11}}\times \left( {{10}^{-3}}{{g}^{-1}} \right)\left( {{10}^{6}}c{{m}^{3}} \right)\left( 1{{s}^{-2}} \right)\]

$\therefore 6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=6.67\times {{10}^{-8}}c{{m}^{3}}{{s}^{-2}}{{g}^{-1}}$

3. A Calorie is a Unit of Heat or Energy and Is Equivalent to 4.2 J Where $1J=1kg{{m}^{2}}{{s}^{-2}}$. Suppose We Employ a System of Units in Which the Unit of Mass Equals $\alpha \text{ kg}$, the Unit of Length Equals $\beta $ m, the Unit of Time is $\gamma \text{ s}$ . Show That a Calorie Has a Magnitude $4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}$ In Terms of the New Unit.

We are given that,

\[1calorie=4.2\left( 1kg \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)\] 

Let the new unit of mass $=\alpha \text{ kg}$. 

So, one kilogram in terms of the new unit, $1\text{ kg}=\frac{1}{\alpha }={{\alpha }^{-1}}$.

One meter in terms of the new unit of length can be written as, $\text{1m}=\frac{1}{\beta }={{\beta }^{-1}}$  or $\text{1}{{\text{m}}^{2}}={{\beta }^{-2}}$.

And, one second in terms of the new unit of time,

$1\text{ s}=\frac{1}{\gamma }={{\gamma }^{-1}}$

$1\text{ }{{\text{s}}^{2}}={{\gamma }^{-2}}$

$1\text{ }{{\text{s}}^{-2}}={{\gamma }^{2}}$

\[\therefore 1calorie=4.2\left( 1{{\alpha }^{-1}} \right)\left( 1{{\beta }^{-2}} \right)\left( 1{{\gamma }^{2}} \right)=4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}\] 

Therefore, the value equivalent to one calorie in the mentioned new unit system is \[4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}\].

4. Explain This Statement Clearly:

“To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:

The given statement is true because a dimensionless quantity may be large or small, but there should be some standard reference to compare that. 

For example, the coefficient of friction is dimensionless but we could say that the coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

a) Atoms Are Very Small Objects.

Ans: An atom is very small compared to a soccer ball.

b) A Jet Plane Moves With Great Speed.

Ans: A jet plane moves with a speed greater than that of a bicycle.

c)The Mass of Jupiter is Very Large.

Ans: Mass of Jupiter is very large compared to the mass of a cricket ball.

d) The Air Inside This Room Contains a Large Number of Molecules.

Ans: The air inside this room contains a large number of molecules as compared to that contained by a geometry box.

e) A Proton is Much More Massive than an Electron.

Ans: A proton is more massive than an electron.

f) The Speed of Sound is Much Smaller than the Speed of Light.          

Ans:  Speed of sound is less than the speed of light. 

5. A New Unit of Length Is Chosen Such That the Speed of Light in Vacuum is Unity. What is the Distance Between the Sun and the Earth in Terms of the New Unit If Light Takes 8 Min and 20 S to Cover This Distance?

Ans:  

Distance between the Sun and the Earth:

\[\text{x= Speed of light}\times \text{Time taken by light to cover the distance}\] 

It is given that in the new system of units, the speed of light \[c=1\text{ }unit\]. 

Time taken, \[t=8\text{ }min\text{ }20\text{ }s=500\text{ }s\]

Thus, the distance between the Sun and the Earth in this system of units is given by\[x'=c\times t=1\times 500=500\text{ }units\] 

6. Which of the Following is the Most Precise Device for Measuring Length?

Ans: A device which has the minimum least count is considered to be the most precise device to measure length.

a) A Vernier Caliper With 20 Divisions on the Sliding Scale.

Least count of vernier calipers is given by

\[LC=1\text{ }standard\text{ }division\left( SD \right)-1\text{ }vernier\text{ }division\left( VD \right)\] 

$\Rightarrow LC=1-\frac{19}{20}=\frac{1}{20}=0.05cm$

b) A Screw Gauge of Pitch 1 Mm and 100 Divisions on the Circular Scale.

Least count of screw gauge $=\frac{\text{Pitch}}{\text{No of divisions}}$

$\Rightarrow LC=\frac{1 mm}{100}=\frac{0.1 cm}{100}$

$\Rightarrow LC=\frac{1}{1000}=0.001cm$

c) An Optical Instrument that Can Measure Length to Within a Wavelength of Light.

Least count of an optical device $=\text{Wavelength of light}\sim \text{1}{{\text{0}}^{-5}}cm$ 

$\Rightarrow LC=0.00001cm$ 

Hence, it can be inferred that an optical instrument with the minimum least count among the given three options is the most suitable device to measure length.

7. A Student Measures the Thickness of a Human Hair Using a Microscope of Magnification 100. He Makes 20 Observations and Finds that the Average Width of the Hair in the Field of View of the Microscope is 3.5 Mm. Estimate the Thickness of Hair.

We are given that: 

Magnification of the microscope \[=100\] 

Average width of the hair in the field of view of the microscope \[=3.5\text{ }mm\] 

$\therefore $ Actual thickness of the hair would be, $\frac{3.5}{100}=0.035\text{ }mm.$ 

8. Answer the Following:

a) YOu Are Given a Thread and a Meter Scale. How Will You Estimate the Diameter of the Thread?

Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. 

Measure the length that is wounded by the thread using a metre scale. 

The diameter of the thread is given by the relation,

Diameter $=\frac{\text{Length of thread}}{\text{Number of turns}}$ 

B) A Screw Gauge Has a Pitch of 1.0 Mm and 200 Divisions on the Circular Scale. Do You Think it Is Possible to Increase the Accuracy of the Screw Gauge Arbitrarily by Increasing the Number of Divisions on the Circular Scale?

Increasing the number divisions of the circular scale will increase its accuracy to a negligible extent only.

C) The Mean Diameter of a Thin Brass Rod Is to Be Measured by Vernier Calipers. Why Is a Set of 100 Measurements of the Diameter Expected to Yield a More Reliable Estimate Than a Set of 5 Measurements Only?

A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved will be reduced on increasing the number of measurements.

9. The Photograph of a House Occupies an Area of $1.75c{{m}^{2}}$ On a 35 Mm Slide. the Slide Is Projected Onto a Screen, and the Area of the House on the Screen is $1.55{{m}^{2}}$. What is the Linear Magnification of the Projector-Screen Arrangement?

We are given, 

The area of the house on the $35mm$ slide (area of the object) is given by, 

${{A}_{O}}=1.75c{{m}^{2}}$.

The area of the image of the house that is formed on the screen is given by, ${{A}_{I}}=1.55{{m}^{2}}=1.55\times {{10}^{4}}c{{m}^{2}}$

We know that areal magnification is given by, 

${{m}_{a}}=\frac{{{A}_{I}}}{{{A}_{O}}}$

Substituting the given values, 

${{m}_{a}}=\frac{1.55\times {{10}^{4}}}{1.75}$

Now, we have the expression for Linear magnification as, ${{m}_{l}}=\sqrt{{{m}_{a}}}$ 

$\Rightarrow {{m}_{l}}=\sqrt{\frac{1.55}{1.75}\times {{10}^{4}}}$ 

$\therefore {{m}_{l}}=94.11$

Thus, we found the linear magnification in the given case to be, ${{m}_{l}}=94.11$.

10. State the Number of Significant Figures in the Following:

a) $0.007{{m}^{2}}$

Ans: We know that when the given number is less than one, all zeroes on the right of the decimal point are insignificant and hence for the given value, only 7 is the significant figure. So, the number of significant figures in this case is 1. 

b) $2.64\times {{10}^{24}}kg$

Ans: We know that the power of 10 is considered insignificant and hence, 2, 6 and 4 are the significant figures in the given case. So, the number of significant figures here is 3. 

c) $0.2370gc{{m}^{-3}}$

Ans: For decimal numbers, the trailing zeroes are taken significantly. 2, 3, 7 and 0 are the significant figures. So, the number of significant figures here is 4. 

d) $6.320J$

Ans: All figures present in the given case are significant. So, the number of significant figures here is 4. 

e) $6.032N{{m}^{-2}}$

Ans: Since all the zeros between two non-zero digits are significant, the number of significant figures here is 4. 

f) $0.0006032{{m}^{2}}$

Ans: For a decimal number less than 1, all the zeroes lying to the left of a non-zero number are insignificant. Hence, the number of significant digits here is 4. 

11. The Length, Breadth and Thickness of a Rectangular Sheet of Metal Are 4.234m, 1.005m and 2.01cm Respectively. Give the Area and Volume of the Sheet to Correct Significant Figure. 

Length of sheet, $l=4.234m$; number of significant figures: 4

Breadth of sheet, $b=1.005m$; number of significant figures: 4

Thickness of sheet, $h=2.01cm=0.0201m$; number of significant figures: 3

So, we found that area and volume should have the least significant figure among the given dimensions, i.e., 3. 

Surface area, $A=2\left( l\times b+b\times h+h\times l \right)$

$\Rightarrow A=2\left( 4.234\times 1.005+1.005\times 0.0201+0.0201\times 4.234 \right)=2\left( 4.25517+0.02620+0.08510 \right)$

$\therefore A=8.72{{m}^{2}}$

Volume, $V=l\times b\times h$

$\Rightarrow V=4.234\times 1.005\times 0.0201$

$\therefore V=0.0855{{m}^{3}}$

Therefore, we found the area and volume with 3 significant figures to be $A=8.72{{m}^{2}}$

and $V=0.0855{{m}^{3}}$respectively. 

12. The Mass of a Box Measured by a Grocer's Balance is 2.300 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?

Mass of grocer’s box $=2.300kg$ 

Mass of gold piece $I=20.15g=0.02015kg$ 

Mass of gold piece $II=20.17g=0.02017kg$ 

Total mass of the box $=2.3+0.02015+0.02017=2.34032kg$ 

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is $2.3kg$.

b) The Difference in the Masses of the Pieces to Correct Significant Figures?       

Difference in masses $=20.17-20.15=0.02g$ 

While subtracting, the final result should retain as many decimal places as there are in the number with the least decimal places.

13. A Famous Relation in Physics Relates ‘Moving Mass’ M to the ‘Rest Mass’ ${{m}_{0}}$of a Particle in Terms of Its Speed v and Speed of Light c. (This Relation First Arise 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:

$m=\frac{{{m}_{0}}}{{{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}}$

We are given the following relation:

 $m=\frac{{{m}_{0}}}{{{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}}$

Dimension of m, ${{M}^{1}}{{L}^{0}}{{T}^{0}}$

Dimension of ${{m}_{0}}$, ${{M}^{1}}{{L}^{0}}{{T}^{0}}$

Dimension of v, ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$

Dimension of ${{v}^{2}}$, ${{M}^{0}}{{L}^{2}}{{T}^{-2}}$

Dimension of c, ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$

For the formula to be dimensionally correct, the dimensions on the LHS should be the same as those on the RHS. In order to satisfy this condition, ${{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}$should be dimensionless and for that we require ${{v}^{2}}$ be divided by ${{c}^{2}}$. So, the dimensionally correct version of the above relation would be,

$m=\frac{{{m}_{0}}}{{{\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)}^{\frac{1}{2}}}}$

14. The Unit of Length Convenient on the Atomic Scale is Known as an Angstrom and is Denoted By $\overset{{}^\circ }{\mathop{A}}\,:1\overset{{}^\circ }{\mathop{A}}\,={{10}^{-10}}m$. The Size of a Hydrogen Atom Is About 0.5a. What is the Total Atomic Volume In ${{m}^{3}}$of a Mole of Hydrogen Atoms?

Radius of hydrogen atom is given to be, 

$r=0.5\overset{{}^\circ }{\mathop{A}}\,=0.5\times {{10}^{-10}}m$

The expression for the volume is,

$V=\frac{4}{3}\pi {{r}^{3}}$

Now on substituting the given values, 

$V=\frac{4}{3}\pi {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

But we know that 1 mole of hydrogen would contain Avogadro number of hydrogen atoms, so volume of 1 mole of hydrogen atoms would be, 

$V'={{N}_{A}}V=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$

Therefore, we found the required volume to be $3.16\times {{10}^{-7}}{{m}^{3}}$.

15.One Mole of an Ideal Gas at Standard Temperature and Pressure Occupies $22.4L$ (molar Volume). What is the Ratio of Molar Volume to the Atomic Volume of a Mole of Hydrogen? (Take the Size of a Hydrogen Molecule to Be About $1\overset{{}^\circ }{\mathop{\text{A}}}\,$). Why is This Ratio So Large?

Radius of hydrogen atom, \[r=0.5\overset{{}^\circ }{\mathop{\text{A}}}\,=0.5\times {{10}^{-10}}m\] 

Volume of hydrogen atom, $V=\frac{4}{3}\pi {{r}^{3}}$ 

$\Rightarrow V=\frac{4}{3}\times \frac{22}{7}\times {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

Now, 1 mole of hydrogen contains $6.023\times {{10}^{23}}$ hydrogen atoms. 

Volume of 1 mole of hydrogen atoms,${{V}_{a}}=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$. 

Molar volume of 1 mole of hydrogen atoms at STP, ${{V}_{m}}=22.4L=22.4\times {{10}^{-3}}{{m}^{3}}$ 

So, the required ratio would be, 

$\frac{{{V}_{m}}}{{{V}_{a}}}=\frac{22.4\times {{10}^{-3}}}{3.16\times {{10}^{-7}}}=7.08\times {{10}^{4}}$

Hence, we found that the molar volume is $7.08\times {{10}^{4}}$ times higher than the atomic volume. 

For this reason, the interatomic separation in hydrogen gas is much larger than the size of a hydrogen atom. 

16. Explain This Common Observation Clearly: If You Look Out of the Window of a Fast-Moving Train, the Nearby Trees, Houses Etc. Seems to Move Rapidly in a Direction Opposite to the Train's Motion, but the Distant Objects (hill Tops, the Moon, the Stars Etc.) Seems to Be Stationary. (In Fact, Since You Are Aware That You Are Moving, These Distant Objects Seem to Move With You).

Ans: Line-of-sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc., while sitting in a moving train, they appear to move rapidly in the opposite direction because the line-of-sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc., appear stationary because of the large distance. As a result, the line-of-sight does not change its direction rapidly.

17. The Sun Is a Hot Plasma (ionized Matter) With Its Inner Core at a Temperature Exceeding ${{10}^{7}}K$ and Its Outer Surface at a Temperature of About 6000k. 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$=2.0\times {{10}^{30}}kg$, Radius of the Sun$=7.0\times {{10}^{8}}m$ .

We are given the following:

Mass of the sun, $M=2.0\times {{10}^{30}}kg$

Radius of the sun, $R=7.0\times {{10}^{8}}m$

Now we find the volume of the sun to be, 

$V=\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\pi {{\left( 7.0\times {{10}^{8}} \right)}^{3}}=1437.3\times {{10}^{24}}{{m}^{3}}$

Density of the sun is found to be, 

$\rho =\frac{M}{V}=\frac{2.0\times {{10}^{30}}}{1437.3\times {{10}^{24}}}$

$\therefore \rho \sim 1.4\times {{10}^{3}}kg/{{m}^{3}}$

So, we found the density of the sun to lie in the density range of solids and liquids. 

Clearly, the high intensity is attributed to the intense gravitational attraction of the inner layers on the outer layer of the sun.

Dimensional Formulae of Physical Quantities

Overview of Deleted Syllabus for CBSE Class 11 Physics Units and Measurements

NCERT Class 11 Physics Chapter 1 Solutions on Units And Measurements provided by Vedantu serves as the foundation for all scientific study and experimentation. This chapter introduces students to the fundamental concepts of measuring physical quantities, ensuring consistency and accuracy in their scientific observations and calculations. Understanding units and measurements is critical for conducting experiments, interpreting results, and communicating scientific findings effectively. These concepts are essential for mastering the topic and are often tested in exams. From previous year's question papers, typically around 2-3 questions are asked from this chapter. These questions test students' theoretical concepts as well as their problem-solving skills. 

Other Study Material for CBSE Class 11 Physics Chapter 1

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

1. What are the topics that are covered in class 11 physics chapter 2?

The concepts that are covered in chapter 2 are:

Introduction to units and measurement. 

The international system of the units.

SI Base unit

SI Derived unit

Advantages of learning SI units and CGS units  

Significant figures. 

Applications of significant figures

Exact Number

Dimensions of physical quantities. 

Dimensional formulae and dimensional equations. 

Dimensional analysis and its applications.  

Unit conversion and dimensional analysis

Using dimensional analysis to check the correctness of physical equation. 

Homogeneity principle of dimensional analysis. 

Applications of dimensional analysis. 

Limitations of dimensional analysis.

2. What is dimensional analysis?

We quantify the size and shape of things using Dimensional Analysis. It helps us study the nature of objects mathematically. It involves lengths and angles as well as geometrical properties such as flatness and straightness. The basic concept of dimension is that we can add and subtract only those quantities that have the same dimensions. Similarly, two physical quantities are equal if they have the same dimensions.

The main benefits of the dimensional analysis of a problem have reduced the number of variables in the issue by combining dimensional variables to one form of non-dimensional parameters. By far the easiest and most useful method in the analysis of any fluid problem is that of direct mathematical solution.

3. What are gross errors?

This section basically takes into account human oversight and other mistakes while reading, recording and readings. The most common errors, the human error in the measurement fall which is under this category of errors in measurement. For instance, a person taking the reading from the meter of the instrument and he may read 33 as 38. Gross errors could be avoided by using two suitable measures and they are mentioned below:

A proper care should always be taken while reading and recording the data. Also, calculation of error should be done accurately.

By increasing the total number of experimenters we can reduce the gross errors. If each experimenter takes different reading at various points, then by taking an average of more readings we can reduce the gross errors.

4. How Vedantu will help me in exam preparation?

Our NCERT solutions are prepared by our maths experts with various real-life examples. These examples will make you understand the concept quickly and memorise them for a longer time. Solutions provided to the questions are 100% accurate in the exercises which are crisp and concise to the point.

Our solutions are the best study guides, which help you in smart learning and efficient answering of questions. These solutions will also help you in improving a strong conceptual base with all the important concepts in a very easy and understandable language. You will also enjoy learning from our solutions which are really fun and interactive.

5. Are NCERT Solutions important for Class 11 Physics Chapter 2?

Referring to NCERT Solutions is as important as referring to the questions while preparing for your Class 11 Physics Chapter 2. Only reading the questions that are provided in the NCERT is not going to be helpful if students are not able to answer them correctly. NCERT Solutions for Class 11 Physics Chapter 2 available on Vedantu provide students the correct step-by-step solutions for all NCERT questions so that students do not lose any marks in their exams.

6. Do I need to practice all the questions provided in Class 11 Physics Chapter 2 NCERT Solutions?

Questions provided in Class 11 Physics NCERT Solutions for Chapter 2 are to be considered crucial when preparing for your Class 11 Physics exams. The exam question papers always include questions that have been provided in the Physics NCERT for Class 11. Practicing all the questions will only help you increase your understanding of all the concepts taught in Chapter 2 and also the possibility of scoring well in the questions framed from the chapter.

7. How can I understand the Class 11 Physics Chapter 2?

Chapter 2 in Class 11 Physics NCERT is called “Units and Measurement”. This chapter talks in detail about various units used for determining the measurement of different physical quantities, instruments used for such measurements and their accuracy, etc. Students can easily understand this chapter by indulging in regular reading of the chapter and solving the questions provided in the NCERT. For more help, students can also refer to NCERT Solutions for Class 11 Physics Chapter 2 on the official website of Vedantu or download the Vedantu app where these resources are available at free of cost.

8. What is the marks distribution for Class 11 Physics Chapter 2?

Chapter 2 - Units and Measurement in Class 11 Physics is a part of Unit - I along with Chapter 1 - Physical World. According to the marks distribution provided by CBSE for Class 11 Physics, Unit - I consisting of both the chapters, carries a total of 23 marks. Hence, preparation from both chapters should be given equal priority to avoid losing any marks in questions framed from them in the exam.

9. What are the important topics covered in NCERT Solutions for Class 11 Physics Chapter 2?

NCERT Solutions for Class 11 Physics  Chapter 2 - Units and Measurements includes various topics like the International System of Units, Measurement of length, mass, and time, Application of Significant Figures, etc. Among these, the most important topics from the chapter include SI Units, Absolute Errors, Dimensional Analysis, and Significant Figures. Short-answer and numerical-based questions can also be asked from the topic of evaluating errors during the measurement of quantities. 

10. What is the concept of units and measurements in class 11 exercise solutions?

In units and measurements class 11 exercise solutions in Physics Chapter 1 involves quantifying physical quantities in a standardized manner. A unit is a definite magnitude of a quantity, defined and adopted by convention or law, used as a standard for measuring the same kind of quantity. Measurement is the process of determining the size, length, or amount of something, typically using standard units.

11. Which topics are important in units and measurements class 11 NCERT solutions?

Key topics in Physics Class 11 Chapter 1 Exercise Solutions include:

Physical Quantities and Units: Base and derived quantities, SI units, and other systems of units.

Measurement Techniques: Direct and indirect measurement methods.

Significant Figures: Rules for determining significant figures in measurements.

Errors in Measurement: Types of errors (systematic and random) and methods to minimize them.

Dimensional Analysis: Checking the dimensional consistency of equations and converting units.

12. What is the direct method of measurement of length class 11 physics chapter 1 NCERT solutions?

In units and measurements class 11 exercise solutions, the direct method of measurement of length involves using instruments that directly provide the measurement of length, such as rulers, measuring tapes, vernier calipers, and micrometer screw gauges. These tools are used to measure the length of objects by comparing them directly against a standard unit of length.

13. What is the full form of SI unit?

According to physics class 11 chapter 1, the full form of SI unit is "Système International 'Unités", which translates to the International System of Units. It is the modern form of the metric system and the most widely used system of measurement.

14. What are the three main units of measurement in Class 11 Physics Ch 1 NCERT Solutions?

The three main base units and measurements class 11 ncert solutions in the SI system are:

Meter (m) for length

Kilogram (kg) for mass

Second (s) for time

15. What is a measurement of time in units and measurements class 11 solutions?

The measurement of time is the process of quantifying the duration or interval between two events. The standard unit of time in the SI system is the second (s). Time can be measured using various devices such as clocks, stopwatches, and atomic clocks.

NCERT Solutions for Class 11 Physics

Elasticity Numericals Class 11 Physics

Important formulae for solving numericals of elasticity :.

(i)   Stress =  $\frac{Force}{area}$ 

(ii)   Strain = $\frac{Change\,\,in\,\,configuration}{Original\,\,configuration}$

(iii)   Longitudinal strain = $\frac{Change\,\,in\,\,length}{Original\,\,length}$

(iv)   Volumetric strain = $\frac{Change\,\,in\,\,volume}{Original\,\,volume}$

(v)   Modulus of elasticity (E) = $\frac{Stress}{Strain}$

(vi)   Young’s modulus of elasticity, Y = $\frac{Fl}{eA}$

(vii)   Elastic potential energy stored, E = $\frac{1}{2}$ .F .e

(viii)   Work done to stretch the wire = Energy stored

(ix)   Energy density, ${{\rho }_{E}}$ = $\frac{1}{2}$ stress × strain

(x)   Bulk Modulus, K = $\frac{Normal\,\,stress}{Volumetric\,\,strain}$

(xi)   Modulus of rigidity ($\eta $) = $\frac{Tangential\,\,stress}{Shear\,\,strain}$

(xii)   Poisson’s ratio ($\sigma $) = $\frac{Lateral\,\,strain}{Longitudinal\,\,strain}$

(xiii)   Shearing strength = $\frac{Force}{area}$

(xiv)   Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

1. What force is required to stretch a steel wire of cross sectional area 1 cm 2 to double its length?

[Y Steel = 2 × 10 11 N/m ]

Force, F = ?

Area of cross-section, A = 1 cm 2  

A = (1×10 -2 ) 2 m 2 = 1×10 -4 m 2

Let initial length, l 1 = l

Then final length, l 2 = 2 l

Elongation, e = l 2 – l 1 = 2 l – l = l

$\therefore $ Elongation, e = l

Y Steel = 2×10 11 N/m 2

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$

Or, F = $\frac{YAe}{l}$

Or, F = $\frac{2\times {{10}^{11}}\times l\times 1\times {{10}^{-4}}}{l}$

$\therefore $ F = 2×10 7 N

2. Calculate the work done in stretching a steel wire 100 cm in length and of cross sectional area 0.030 cm 2 when a load of 100 N is slowly applied before the elastic limit is reached.  [Y Steel = 2 × 10 11 N/m ]

Work done, W = ? 

Length of wire, l = 100 cm = 1m

Cross-sectional area, A = 0.03 cm 2 

                           A   = 0.03× (1×10 –2 ) 2 m 2

                 A = 0.03×10 –4 m 2

Force, F = 100 N

Y Steel = 2×10 11 Nm –2 

Young’s modulus of elasticity

        Y = $\frac{Fl}{eA}$

Or,   e = $\frac{Fl}{YA}$

Or,   e = $\frac{100\times 1}{2\times {{10}^{11}}\times 0.03\times {{10}^{-4}}}$

Elongation, e = $\frac{1}{6000}$ m

Work done (W) = Energy stored (E)

Or,   W = $\frac{1}{2}$ .F .e

Or,   W = $\frac{1}{2}$ × 100  × $\frac{1}{6000}$

$\therefore $    W = 8.33×10 -3 J

3. Find the work done in stretching a wire of cross sectional area 10 –2 cm 2 and 2m long through 0.1 mm, If Y for the material of wire is 2 × 10 11 Nm –2 .

Work done, W = ?

Cross-sectional area, A = 10 –2 cm 2 

A   = 10 –2 × (1×10 –2 ) 2 m 2

A = 1×10 –6 m 2

Length of wire, l = 2 m

Elongation, e = 0.1 mm = 0.1×10 –3 m

    Y = $\frac{Fl}{eA}$

    F = $\frac{YeA}{l}$

    F = $\frac{2\times {{10}^{11}}\times 0.1\times {{10}^{-3}}\times 1\times {{10}^{-6}}}{2}$

    F = 10 N

Now, work done, (W) = Energy stored (E)

      W = $\frac{1}{2}$ .F .e

      W = $\frac{1}{2}$ × 10  × 0.1×10 –3

$\therefore $  W = 5×10 -4 J

4. A vertical brass rod of circular section is loaded by placing of 5 kg weight on top of it.   If its length is 50 cm and radius of cross section is 1 cm, find the contraction of the rod and the energy stored in it. 

[Y Brass = 3.5 × 10 10 Nm –2 ]

Mass on top, m  = 5 kg

Length of rod, l = 50 cm = 0.5 m

Radius, r = 1 cm = 1×10 –2 m  

(i)  Contraction, e (or  $\Delta l$ or l 1 – l 1 ) = ?

(ii)  Energy stored, E = ?

Y Brass = 3.5×10 10 Nm –2 

We know, 

e = $\frac{Fl}{YA}$

e = $\frac{mgl}{Y\times \pi {{r}^{2}}}$

e = $\frac{5\times 10\times 0.5}{3.5\times {{10}^{10}}\times \pi {{(1\times {{10}^{-2}})}^{2}}}$

$\therefore $ Contraction, e = 2.27×10 –6 m

 Energy stored, E = $\frac{1}{2}$ .F .e

E = $\frac{1}{2}$ × mg × e 

E = $\frac{1}{2}$ × 5 × 10 × 2.27×10 –6

$\therefore $ E = 5.68×10 -5 J

 5. Calculate the work done in stretching a steel wire 100 cm in length and of cross sectional area 0.03 cm 2 when a load of 100 N is slowly applied without the elastic limit being reached.  [Y Steel = 2 × 10 11 N/m ]

A   = 0.03× (1×10 –2 ) 2 m 2

A = 0.03×10 –4 m 2

   Y = $\frac{Fl}{eA}$

   e = $\frac{Fl}{YA}$

   e = $\frac{100\times 1}{2\times {{10}^{11}}\times 0.03\times {{10}^{-4}}}$

Now, work done (W) = Energy stored (E)

     W = $\frac{1}{2}$ .F .e

     W = $\frac{1}{2}$ × 100  × $\frac{1}{6000}$

$\therefore $ W = 8.33×10 -3 J

6. A uniform steel wire of density 8000 Kgm –3 weight 20 g and is 2.5 m long. It lengthens by 1 mm when trenched by a force of 80 N. Calculate the value of the Young’s modulus of steel and the energy stored in the wire.

Density of steel wire, $\rho $ = 8000 kgm -3

Mass of wire, m = 20g = 20×10 -3 kg

Length of wire l = 2.5m

Elongation, e = 1mm = 1×10 -3 m

Force, F = 80 N 

Young’s modulus of steel, Y = ?

Energy stored in the wire, E = ?

Y = $\frac{Fl}{eA}$ …….(i)

Again, 

Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

A = $\frac{m}{\rho l}$ = $\frac{20\times {{10}^{-3}}}{8000\times 2.5}$ 

A = 1×10 -6 m 2

From equation (i),

Y = $\frac{Fl}{eA}$= $\frac{80\times 2.5}{1\times {{10}^{-3}}\times 1\times {{10}^{-6}}}$  

$\therefore $ Y = 2×10 11 Nm –2

Energy stored in the stretched wire,

      E = $\frac{1}{2}$ .F .e

Or, E = $\frac{1}{2}$ × 80 × 1×10 -3

$\therefore $ E = 40×10 -3 J

7. A uniform Steel wire of density 7800 Kgm –3 weights 16 gm and is 250 cm long. It lengthens by 1.2 mm when is stretched by a force of 80 N. Calculate the Young’s modulus and energy stored in the wire.

Density of steel wire, $\rho $ = 7800 kgm -3

Mass of wire, m = 16 g = 16×10 -3 kg

Length of wire l = 250 cm = 2.5 m

Elongation, e = 1.2 mm = 1.2×10 -3 m

Force, F = 80 N

Density, $\rho $ = $\frac{mass}{volume}$ =  $\frac{m}{A.l}$

A = $\frac{m}{\rho l}$

       Y = $\frac{Fl}{e}$× $\frac{\rho l}{m}$

Or,  Y = $\frac{80\times 2.5}{1.2\times {{10}^{-3}}}$×  $\frac{7800\times 2.5}{16\times {{10}^{-3}}}$

$\therefore $    Y = 2.03×10 11 Nm –2

Or, E = $\frac{1}{2}$ × 80 × 1.2×10 -3

$\therefore $ E = 4.8×10 -2 J

8. A steel wire of density 8000 Kgm –3 weights 24 g and is 250 cm long. It lengthens by 1.2 mm when stretched by a force of 80 N. Calculate the Young’s modulus of Steel and the energy stored in the wire.

Mass of wire, m = 24 g = 24×10 -3 kg

Length of wire l = 250 cm = 2.5m

Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$ 

 A = $\frac{m}{\rho l}$ 

   Y = $\frac{Fl}{e}$× $\frac{\rho l}{m}$

Or,  Y = $\frac{80\times 2.5}{1.2\times {{10}^{-3}}}$×  $\frac{8000\times 2.5}{24\times {{10}^{-3}}}$

$\therefore $    Y = 1.389×10 11 Nm –2

9. A uniform steel wire of density 7800 Kgm –3 weighs 16 gram and is 250 cm. It lengthens by 1.2 mm when a load of 8 kg is applied. Calculate the value of Young’s modulus for the steel and the energy stored in the wire.

Hanging mass, M = 8 Kg

Y = $\frac{Mg\times l}{eA}$ …….(i)     [$\because $ F = Mg]

A = $\frac{m}{\rho l}$ 

  Y = $\frac{Fl}{e}$× $\frac{\rho l}{m}$

Or,  Y = $\frac{8\times 10\times 2.5}{1.2\times {{10}^{-3}}}$×  $\frac{7800\times 2.5}{16\times {{10}^{-3}}}$

10. A wire of length 2.5 m and area of cross section 1 × 10 –6 m 2 has a mass of 15 kg hanging on it. What is the extension produced? How much is the energy stored in the standard wire if Young’s modulus of wire is 2 × 10 11 Nm –2 .

Area of cross-section, A = 1×10 –6 m 2

Hanging mass, m = 15 kg

Elongation, e = ?

Young’s modulus of steel, Y = 2×10 11 Nm –2 .

Y = $\frac{Fl}{eA}$ 

e = $\frac{Fl}{YA}$    [here, F = mg]

e = $\frac{15\times 10\times 2.5}{2\times {{10}^{11}}\times 1\times {{10}^{-6}}}$

e = 1.875 ×10 -3 m

Or, E = $\frac{1}{2}$ × mg × e 

Or, E = $\frac{1}{2}$ × 15×10 ×1.875 ×10 -3

$\therefore $ E = 0.141 J

11. A steel cable with cross sectional area 3 cm 2 has an elastic limit of 2.40 × 10 8 pa. Find the maximum upward acceleration that can be given a 1200 kg elevator supported by the cable if the stress does not exceed one third of the elastic limit.

Cross-sectional area, A = 3 cm 2  

Or,                            A = 3×(1×10 –2 ) 2  

Or,                            A = 3×10 –4 m 2

Elastic limit = 2.4×10 8 pa (i.e. Nm –2 )

Maximum upward acceleration, a max = ?

Mass of elevator, m = 1200 kg

Maximum stress = $\frac{1}{3}$ of elastic limit

Or,  $\frac{Force\,\,(ma{{x}^{m}})}{area}$ = $\frac{1}{3}$× elastic limit

Or,  $\frac{m\times {{a}_{max}}}{A}$ = $\frac{1}{3}$× elastic limit

Or,  a max = $\frac{1}{3}$× $\frac{A}{m}$× elastic limit

Or,  a max = $\frac{1}{3}$× $\frac{3\times {{10}^{-4}}}{1200}$×2.4×10 8 

$\therefore $  a max = 20 ms -2

12. How much force is required to punch a hole 1 cm in diameter in a steel sheet 5mm thick whose shearing strength is 2.76 × 10 8 Nm –2 .

Diameter of hole, d = 1 cm

$\therefore $ Radius of hole, r = 0.5 cm 

r = 0.5 × 10 –2 m

Thickness of steel sheet, t = 5 mm 

t = 5 × 10 –3 m

Shearing strength = 2.76 ×10 8 Nm –2

    Shearing strength = $\frac{Force}{area}$

Or, Shearing strength = $\frac{F}{C\times t}$    here, C is circumference.

Or, F = Shearing strength × C × t

Or, F = Shearing strength × 2$\pi $r × t

Or, F = 2.76 ×10 8 × 2$\pi $× 0.5 × 10 –2 × 5 × 10 –3

$\therefore $ F = 43353.98 N

13. A copper wire and a steel wire of the same cross sectional area and of length 1 m and 2 m respectively are connected end to end. A force is applied, which stretches their combined length by 1 cm. Find how much each wire is elongated. [Y copper = 1.2 × 10 11 N/m 2 and Y steel = 2 × 10 11 N/m 2 ]

Length of copper wire, l cu = 1 m 

Length of steel wire, l cu = 2 m Total elongation, e cu + e s = 1 cm = 1×10 –2 m……(i)

(i)  Elongation in copper wire, e cu = ?

(ii)  Elongation in steel wire, e s = ?

For copper, Young’s modulus of elasticity,

Y cu = $\frac{F\,\,{{l}_{cu}}}{{{e}_{cu}}A}$

e cu = $\frac{F\,\,{{l}_{cu}}}{{{Y}_{cu\,\,}}A}$….(ii)

For steel, Young’s modulus of elasticity,

Y s = $\frac{F{{l}_{s}}}{{{e}_{s}}A}$

e s = $\frac{F\,{{l}_{S}}}{{{Y}_{S}}A}$….(iii)

Dividing equation (ii) by (iii)

Or,  $\frac{{{e}_{cu}}}{{{e}_{s}}}$= $\frac{{{l}_{cu}}}{{{Y}_{cu}}}$×$\frac{{{Y}_{S}}}{{{l}_{S}}}$

Or,  $\frac{1\times {{10}^{-2}}-{{e}_{S}}}{{{e}_{S}}}$= $\frac{1}{1.2\times {{10}^{11}}}$× $\frac{2\times {{10}^{11}}}{2}$ 

Or,  $\frac{1\times {{10}^{-2}}-{{e}_{S}}}{{{e}_{S}}}$= $\frac{5}{6}$

Or,  6×1×10 –2 – 6 e s = 5 e s

Or,  e s = $\frac{6\times 1\times {{10}^{-2}}}{11}$

 ∴   Elongation in steel wire, e s  = 5.45×10 –3 m.

 and elongation in copper wire, e cu   =  1×10 –2 – 5.45×10 –3 = 4.54 ×10 –3 m

14. A rubber cord of a catapult has a cross-sectional area 1.0 mm 2 and total un-stretched length 10 cm. It is stretched to 12 cm and then released to project a missile of mass 5.0 g. Calculate the velocity of projection. [Y rubber = 5 × 10 8 N/m 2 ]

Area of cross-section, A = 1 mm 2  

A = (1×10 -3 ) 2 m 2 = 1×10 -6 m 2

Un-stretched (original) length, l 1 = 10 cm = 10 × 10 -2 m

Stretched (final) length, l 2 = 12 cm = 12 × 10 -2 m

Elongation, e = l 2 – l 1  

e = 2×10 -2 m

Mass of missile, m = 5g 

m = 5×10 -3 kg

Velocity projection, v = ?

Y rubber = 5×10 8 N/m 2

We have, 

F = $\frac{YAe}{l}$

F = $\frac{5\times {{10}^{8}}\times 2\times {{10}^{-2}}\times 1\times {{10}^{-6}}}{10\times {{10}^{-2}}}$

Elastic energy stored in a stretched rubber

E = $\frac{1}{2}$ .F .e

E = $\frac{1}{2}$ × 100 × 2 × 10 -2

According to work energy theorem,

Elastic potential energy, E = K.E 

Or, K.E = $\frac{1}{2}$ .m . v 2

Or, 1 = $\frac{1}{2}$ × 5×10 -3 v 2

Or, v = $\sqrt{\frac{2}{5\times {{10}^{-3}}}}$

$\therefore $ v = 20 m/s

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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements - PDF Download

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements are the best study resources. You can get to understand the main topics and to score well in your examination. This solution provides appropriate answers to the textbook questions. To get a good grip on this chapter, you can make use of the NCERT Solutions for Class 11 Physics available at our website for free in a pdf form, you can download them and use them offline.

Chapter 1 of NCERT Solutions for Class 11 Physics mainly helps understand the fundamentals of units and measurements. In our daily lives, most of the activities depend on this, and it is very important for us to learn it effectively. Everything around us depends on units and measurements, from buying milk in the morning to the pounds of bread needed for breakfast or from buying sugar for milk to the kilograms of rice needed for lunch. You can access the Physics NCERT Solutions for Class 11 to comprehend the key concepts present in this chapter.

Topics Covered in Class 11 Chapter 2 Physics Units and Measurement

Introduction

Measurement of any kind of physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard we called that standard as Unit. The result of a measurement of any physical quantity is expressed by a numerical measure accompanied by a unit. However, if the number of physical quantities appears to be very large, we use only a limited number of units for expressing all the physical quantities, since all the units are interrelated with one another. The units for the fundamental or base quantities are called fundamental or base units.

The International System of Units

In past times, the scientists of different countries were using different systems of units for measurement. There are basically 3 such systems like CGS, the FPS (or British) system and the MKS system were in use extensively till recently.

 In CGS(Centimeter Gram Second system) system they were centimetre, gram and second respectively

In the FPS(foot-pound-second system of units) system they were foot, pound and second respectively.

In MKS system they were meter, kilogram and second respectively. 

Significant figure

Following are the rules for significant figure

All the non-zero digits are significant.

All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.

If the number is less than 1, the zero(s) on the right of the decimal point but to the left of the first non-zero digit are not significant. 

The terminal or trailing zero(s) in a number without a decimal point are not significant.

The trailing zero(s) in a number with a decimal point are significant. 

Rules for Arithmetic Operations with Significant Figures

The result of a calculation involving approximate measured values of quantities (i.e. values with limited number of significant figures) must reflect the uncertainties in the original measured values. It cannot be more accurate than the original values that we measured themselves on which the result is based. In common, the final result should not have more significant figures than the original data from which it was picked up.

The given rules for arithmetic operations with significant figures ensure that the final result of a calculation is shown with the precision that is consistent with the precision of the input measured values :

 In multiplication or division, the final result should retain as many significant figures as there are in the original result with the least significant figures.

 In subtraction or  addition, the final result should retain as many decimal places as are there in the number with the least decimal places.

Rounding off the Uncertain Digits

The result of calculation with approximate numbers, which contain more than one uncertain digit, should be rounded off. The rules for rounding off numbers to the appropriate significant figures are very obvious in most cases. Let's consider a number 1.746 rounded off to three significant figures is 1.75, while the number 2.743 would be 2.74. The rule by convention is that the preceding digit is raised by 1 if the insignificant digit to be dropped is more than 5, and is left unchanged if the digit is less than 5. But what if the number is 3.745 in which the insignificant digit is 5. Here, the old method is that if the preceding digit is even, the insignificant digit is dropped and, if it is odd, the preceding digit is raised by 1.

Rules for Determining the Uncertainty in the Results of Arithmetic Calculations

The rules for determining the uncertainty or error in the number/measured quantity in arithmetic operations can be understood from the below examples. 

(1) If the length and breadth of a thin rectangular sheet are measured, using a meter scale as 17.2 cm and 10.1 cm respectively, there are three significant figures in each measurement. It means that the length l can be written as l = 17.2 ± 0.1 cm 

And breadth b can be written as b = 10.1 ± 0.1 cm

                                                       = 10.1 cm ± 1 %

(2) If a set of experimental data is specified to n significant figures, a result gained by combining the data will also be valid to n significant figures.

(3) The relative error of a value of number specified to significant figures depends not only on n but on the number itself too. 

Dimensions of Physical Quantities

The nature of all physical quantities is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental quantities or base quantities. We can call these base quantities as the seven dimensions of the physical world, which are represented in square brackets [ ]. So , length of dimension [L], , time [T], mass [M], electric current [A], thermodynamic temperature [K], luminous intensity [cd], and amount of substance [mol]. 

Dimensional Formulae and Dimensional Equations

The expression which shows how and which of the base quantities represent the dimensions of a physical quantity are known as the dimensional formula of the given physical quantity. For example, the dimensional formula of the volume is [M° L3 T°], and that of speed or velocity is [M° L T-1]. Similarly,  [M L–3 T°] that of mass density and  [M° L T–2] is the dimensional formula of acceleration.

 An equation which is obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity. So, the dimensional equations are the equations, which represent the dimensions of a physical quantity in terms of the base quantities. 

Dimensional Analysis and Its Applications

The recognition of concepts of dimensions, which guide the description of physical behaviour is of basic importance as only those physical quantities can be added or subtracted which have the similar dimensions. A thorough understanding of dimensional analysis helps you in deducing certain relations among different physical quantities and checking the derivation, accuracy and dimensional consistency or homogeneity of various mathematical expressions.

Checking the Dimensional Consistency of Equations 

 if an equation fails this consistency test, it is proved wrong, but if it passes, it is not proved right. Thus, a dimensionally correct equation need not be actually an exact (correct) equation, but a dimensionally wrong (incorrect) or inconsistent equation must be wrong.

Benefits of the Class 11 Physics Chapter 1 Solution

The benefits of using the NCERT Solutions for Class 11 Physics Chapter 1 are –

1. Completely solved answers for all the questions given in the NCERT textbook are freely available in PDF format.

2. Simple and easy-to-understand language is used to make learning fun for you.

3. Our Subject teacher experts prepare the solutions after conducting vast research on each concept.

4. The solutions not only help you with your exam preparation but also for various competitive exams like JEE, NEET, etc.

5. PDF format of solutions is available in chapter-wise and exercise-wise formats to help students learn the concepts in a better way.

Frequently Asked Questions

Question 1 : What are the topics in chapter 1 in physics class 11th?

Answer : Following are the topics in chapter 1 physics class 11th

1.1 Introduction

1.2 The International System of Units

1.3 Significant figure

1.3.1 Measurement of Length

1.3.2 Rounding off the uncertain digits

1.3.3 Rules for determining the uncertainty in the result of Arithmetic calculation

1.4 Dimensions of Physical Quantities

1.5 Dimensional Formulae and Dimensional Equations

1.6 Dimensional Analysis and Its Applications

1.6.1 Checking the Dimensional Consistency of Equations

1.6.2 Deducing Relation among the physical quantities 

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NCERT Solutions for Class 11 Physics Chapter 1 Free PDF Download

Ncert solutions for class 11 physics chapter 1 – physical world.

Physics is a subject that studies everything present in the universe. In addition, how everything works and what are their characteristics. But class 11 physics is pretty tough to understand for students. That’s why we prepared these NCERT Solutions for Class 11 Physics Chapter 1 which will guide you through the chapter thoroughly. Also, our video tutorial explains each topic in simple language.

These solutions are pretty easy to understand as our expert teacher has designed them to convey a proper and complete meaning of each topic and equations of the chapter.

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Download NCERT Solutions for Class 11 here.

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CBSE Class 11 Physics Chapter 1 Physical World NCERT Solutions

The world that we see, feel, and experience around us is the physical world. In this NCERT Solutions for Class 11 Physics Chapter 1, we are going to discuss what is physical world from the viewpoint of physics. Moreover, this NCERT Solution clear all your doubts and queries related to this chapter. Also, it explains each and every topic of the chapter in detail.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 1

1.1 what is physics.

Physics refers to that branch of science that is related to the nature and properties of energy and substance. Likewise, its subject matter includes mechanics, sound, gravity, electricity, heat, light and radiation, magnetism, and the structure of atoms. Besides, this topic discusses how this science work and its origin. And various terms related to physics are discussed in the topic.

• Natural science- the science that relates to nature.

1.2 Scope And Excitement Of Physics

This topic describes how interesting physics is and its scope. Physics is fun or exciting because you can learn the concepts by performing them. Moreover, the scope of physics is very wide that the whole universe comes in it. In addition, the topic describes the scope of physics in gravity, motion, electricity, light, radiation, heat and many others.

1.3 Physics, Technology, and Society

This topic overviews the various scientist and philosopher who contributed to the development of physics and technology. Likewise, the topic contains a chart that contains the name of those scientists and philosophers whom contribution in physics we can’t deny.

1.4 Fundamental Forces in Nature

This topic describes all those forces that work on every object or body. Besides, the topic has a chart that contains the link between physics and technology.

1.4.1 Gravitational Force- It refers to the force that attracts an object towards the earth.

1.4.2 Electromagnetic Force- It refers to the force that every object produce which attract or repel other objects.

1.4.3 Strong Nuclear Force- It refers to the force that binds the protons and electrons with the nucleus of the atom.

1.4.4 Weak Nuclear Force- It refers to that force, which is not strong enough to bind the protons and electrons in an element. Likewise, it only appears in β-decay of the nucleus.

1.4.5 Towards Unification of Forces- This topic defines how various scientists grouped different forces of nature among a common force.

1.5 Nature of Physical Laws

This topic talks about the nature of laws of physics and how different scientist theories are categorized in one group of category. Besides, this topic contains a chart that shows the unification of different forces.

You can download NCERT Solutions for Class 11 Physics Chapter 1 PDF by clicking on the button below.

Solved Questions for You

Question 1: Physics involves the study of the

• Birds and animals
• Nature and natural phenomena

Answer :  Physics is a branch of science which deals with the study of nature and natural phenomena.

Question 2: Which year was declared as International Year of Physics. ?

Answer: The year 2005 was declared as the International Year of Physics.

Question 3: Lightning was discovered by the

Answer: Lightning was discovered by Franklin.

Question 4: What is the full form of GMRT?

• Ground Mobile Receive Terminal
• Geometric Mean Reciprocal Titer
• Giant Metrewave Radio Telescope
• General Maintenance and Repair Technician

Answer: Giant Metrewave Radio Telescope (GMRT).

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NCERT Solutions for Class 11

• NCERT Solutions for Class 11 Chemistry Chapter 7 Free PDF Download
• NCERT Solutions for Class 11 Physics Chapter 6
• NCERT Solutions for Class 11 Biology Chapter 6 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 2 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 18 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 10 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 22 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 16 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 20 Free PDF Download
• NCERT Solutions for Class 11 Biology Chapter 19 Free PDF Download

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NCERT Solutions for Class 11th: Ch 5 Laws Of Motion Physics

Ncert solutions for class 11th: ch 5 laws of motion physics science.

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NCERT Exemplar solutions for Physics Class 11 chapter 5 - Laws of Motion [Latest edition]

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Solutions for chapter 5: laws of motion.

Below listed, you can find solutions for Chapter 5 of CBSE NCERT Exemplar for Physics Class 11.

NCERT Exemplar solutions for Physics Class 11 Chapter 5 Laws of Motion Exercises [Pages 29 - 37]

A ball is travelling with uniform translatory motion. This means that ______.

it is at rest.

the path can be a straight line or circular and the ball travels with uniform speed.

all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.

the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

A metre scale is moving with uniform velocity. This implies ______.

the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.

the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.

the total force acting on it need not be zero but the torque on it is zero.

neither the force nor the torque need to be zero.

A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s −1 and a final velocity `v = - (3hati + 4hatj)` m s −1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s 1 )

`-(0.45 hati + 0.6 hatj)`

`-(0.9 hati + 1.2 hatj)`

`-5 (hati + hatj)`

In the previous problem (5.3), the magnitude of the momentum transferred during the hit is ______.

0.75 kg ms –1

1.5 kg ms –1

14 kg ms –1

Conservation of momentum in a collision between particles can be understood from ______.

Conservation of energy.

Newton’s first law only.

Newton’s second law only.

Both Newton’s second and third law.

A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is ______.

frictional force along westward.

muscle force along southward.

frictional force along south-west.

muscle force along south-west.

A body of mass 2 kg travels according to the law x(t) = pt + qt 2 + rt 3 where p = 3 ms −1 , q = 4 ms −2 and r = 5 ms −3 . The force acting on the body at t = 2 seconds is ______.

A body with mass 5 kg is acted upon by a force F = `( –3hati + 4hatj)` N. If its initial velocity at t = 0 is v = `(6hati - 12hatj)` ms –1 , the time at which it will just have a velocity along the y-axis is ______.

A car of mass m starts from rest and acquires a velocity along east `v = vhati (v > 0)` in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is ______.

`(mv)/2` eastward and is exerted by the car engine.

`(mv)/2` eastward and is due to the friction on the tyres exerted by the road.

more than `(mv)/2` eastward exerted due to the engine and overcomes the friction of the road.

`(mv)/2` exerted by the engine.

The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?

• The force at t = (1/8) s on the particle is – 16π 2 Am.
• The particle is acted upon by on impulse of magnitude 4π 2 A m at t = 0 s and t = (1/4) s.
• The particle is not acted upon by any force.
• The particle is not acted upon by a constant force.
• There is no impulse acting on the particle.

In Figure, the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is m /2 and of B is m. Which of the following statements are true?

• The bodies will move together if F = 0.25 mg.
• The body A will slip with respect to B if F = 0.5 mg.
• The bodies will move together if F = 0.5 mg.
• The bodies will be at rest if F = 0.1 mg.
• The maximum value of F for which the two bodies will move together is 0.45 mg.

Mass m 1 moves on a slope making an angle θ with the horizontal and is attached to mass m 2 by a string passing over a frictionless pulley as shown in figure. The coefficient of friction between m 1 and the sloping surface is µ.

• If m 2 > m 1 sin θ, the body will move up the plane.
• If m 2 > m 1 (sin θ + µ cos θ), the body will move up the plane.
• If m 2 < m 1 (sin θ + µ cos θ), the body will move up the plane.
• If m 2 < m 1 (sin θ − µ cos θ), the body will move down the plane.

In figure, a body A of mass m slides on plane inclined at angle θ 1 to the horizontal and µ 1 is the coefficent of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle θ 2 to the horizontal. Which of the following statements are true?

• A will never move up the plane.
• A will just start moving up the plane when `µ = (sin  θ_2 - sin  θ_1)/(cos  θ_1)`
• For A to move up the plane, θ 2 must always be greater than θ 1 .
• B will always slide down with constant speed.

Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 ms –1 each, collide and rebound with the same speed. If the collision lasts for 10 –3 s, which of the following statements are true?

• The impulse imparted to each ball is 0.25 kg ms –1 and the force on each ball is 250 N.
• The impulse imparted to each ball is 0.25 kg ms –1 and the force exerted on each ball is 25 × 10 –5 N.
• The impulse imparted to each ball is 0.5 Ns.
• The impulse and the force on each ball are equal in magnitude and opposite in direction.

A body of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is ______.

• 1 m s –2 at an angle of tan −1 `(4/3)` w.r.t 6 N force.
• 0.2 m s –2 at an angle of tan −1 `(4/3)` w.r.t 6 N force.
• 1 m s –2 at an angle of tan −1 `(3/4)` w.r.t 8 N force.
• 0.2 m s –2 at an angle of tan −1 `(3/4)` w.r.t 8 N force.

A girl riding a bicycle along a straight road with a speed of 5 ms –1 throws a stone of mass 0.5 kg which has a speed of 15 ms –1 with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 ms –2 , what would be the reading of the weighing scale? (g = 10 ms –2 )

The position time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 0 s and t = 4 s.

A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?

The velocity of a body of mass 2 kg as a function of t is given by `v(t) = 2t  hati + t^2hatj`. Find the momentum and the force acting on it, at time t = 2s.

A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.

Why are porcelain objects wrapped in paper or straw before packing for transportation?

Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?

A woman throws an object of mass 500 g with a speed of 25 ms 1 .

• What is the impulse imparted to the object?
• If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?

Why are mountain roads generally made winding upwards rather than going straight up?

A mass of 2 kg is suspended with thread AB (Figure). Thread CD of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward directon so as to apply force on AB. Which of the threads will break and why?

In the above given problem if the lower thread is pulled with a jerk, what happens?

Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in figure. Calculate T 1 and T 2 when whole system is going upwards with acceleration = 2 ms 2 (use g = 9.8 ms –2 ).

Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° (figure). A flexible cord attached to A passes over a frictonless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.

A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is µ and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to hold the block against the wall?

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 ms –2 )

Figure shows (x, t), (y, t ) diagram of a particle moving in 2-dimensions.

If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.

A person in an elevator accelerating upwards with an acceleration of 2 ms –2 , tosses a coin vertically upwards with a speed of 20 ms 1 . After how much time will the coin fall back into his hand? ( g = 10 ms –2 )

There are three forces F 1 , F 2 and F 3 acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.

• Show that the forces are coplanar.
• Show that the torque acting on the body about any point due to these three forces is zero.

When a body slides down from rest along a smooth inclined plane making an angle of 45° with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the co-efficient of friction between the body and the rough plane.

Figure shows (vx, t) and (vy, t) diagrams for a body of unit mass. Find the force as a function of time.

A racing car travels on a track (without banking) ABCDEFA (Figure). ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The co-efficient of friction on the road is µ = 0.1. The maximum speed of the car is 50 ms –1 . Find the minimum time for completing one round.

The displacement vector of a particle of mass m is given by `r(t) = hati` A cos ωt + `hatj` B sin ωt. Show that the trajectory is an ellipse.

The displacement vector of a particle of mass m is given by r(t) = `hati` A cos ωt + `hatj` B sin ωt. Show that F = − mω 2 r.

 A cricket bowler releases the ball in two different ways

• giving it only horizontal velocity, and
• giving it horizontal velocity and a small downward velocity. The speed v s at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

There are four forces acting at a point P produced by strings as shown in the figure, which is at rest. Find the forces F 1 and F 2 .

A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is µ. Let the mass of the box be m.

• At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane?
• What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to α > θ ?
• What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
• What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a?

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 ms –2 . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the (g = 10 ms –2 )

• force on the floor of the helicopter by the crew and passengers.
• action of the rotor of the helicopter on the surrounding air.
• force on the helicopter due to the surrounding air.

NCERT Exemplar solutions for Physics Class 11 chapter 5 - Laws of Motion

Shaalaa.com has the CBSE Mathematics Physics Class 11 CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT Exemplar solutions for Mathematics Physics Class 11 CBSE 5 (Laws of Motion) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT Exemplar textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Physics Class 11 chapter 5 Laws of Motion are Aristotle’s Fallacy, The Law of Inertia, Newton's First Law of Motion, Newton’s Second Law of Motion, Newton's Third Law of Motion, Conservation of Momentum, Equilibrium of a Particle, Common Forces in Mechanics, Circular Motion and Its Characteristics, Solving Problems in Mechanics, Static and Kinetic Friction, Laws of Friction, Inertia, Intuitive Concept of Force, Dynamics of Uniform Circular Motion - Centripetal Force, Examples of Circular Motion (Vehicle on a Level Circular Road, Vehicle on a Banked Road), Lubrication - (Laws of Motion), Law of Conservation of Linear Momentum and Its Applications, Rolling Friction, Introduction of Motion in One Dimension.

Using NCERT Exemplar Physics Class 11 solutions Laws of Motion exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Exemplar Solutions are essential questions that can be asked in the final exam. Maximum CBSE Physics Class 11 students prefer NCERT Exemplar Textbook Solutions to score more in exams.

Get the free view of Chapter 5, Laws of Motion Physics Class 11 additional questions for Mathematics Physics Class 11 CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

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

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane.

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

Topics and Subtopics in  NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane :

QUESTIONS FROM TEXTBOOK

Question 4. 1. State, for each of the following physical quantities, if it is a scalar or a vector:  volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. Answer:   Scalars: Volume, mass, speed, density, number of moles, angular frequency. Vectors: Acceleration, velocity, displacement, angular velocity.

Question 4. 2. Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. Answer:  Work and current are the scalar quantities in the, given list.

More Resources for CBSE Class 11

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• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
• NCERT Solutions Class 11 Biology
• NCERT Solutions Class 11 Hindi
• NCERT Solutions Class 11 English
• NCERT Solutions Class 11 Business Studies
• NCERT Solutions Class 11 Accountancy
• NCERT Solutions Class 11 Psychology
• NCERT Solutions Class 11 Entrepreneurship
• NCERT Solutions Class 11 Indian Economic Development
• NCERT Solutions Class 11 Computer Science

Question 4. 3. Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. Answer:   Impulse.

Question 4. 4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful: (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. Answer:  (a) No, because only the scalars of same dimensions can be added. (b) No, because a scalar cannot be added to a vector. (c) Yes, multiplying a vector with a scalar gives the scalar (number) times the vector quantity which makes sense and one gets a bigger vector. For example, when acceleration A is multiplied by mass m, we get a force F = ml (d) Yes, two scalars multiplied yield a meaningful result, for example multiplication of rise in temperature of water and its mass gives the amount of heat absorbed by that mass of water. (e) No, because the two vectors of same dimensions can be added. (f) Yes, because both are vectors of the same dimensions.

Question 4.5. Read each statement below carefully and state with reasons, if it is true or false: (a) The magnitude of a vector is always a scalar. (b) Each component of a vector is always a scalar. (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. Answer:  (a) True, magnitude of the velocity of a body moving in a straight line may be equal to the speed of the body. (b) False, each component of a vector is always a vector, not scalar. (c) False, total path length can also be more than the magnitude of displacement vector of a particle. (d) True, because the total path length is either greater than or equal to the magnitude of the displacement vector. (e) True, this is because the resultant of two vectors will not lie in the plane of third vector and hence cannot cancel its effect to give null vector.

Question 4. 19. Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre. (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point. (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector. Answer:  (a) False, the net acceleration of a particle in circular motion is along the radius of the circle towards the centre only in uniform circular motion. (b) True, because while leaving the circular path, the particle moves tangentially to the circular path. (c) True, the direction of acceleration vector in a uniform circular motion is directed towards the centre of circular path. It is constantly changing with time. The resultant of all these vectors will be a zero vector.

Question 4. 24. Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (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. Answer:  (a) False, because kinetic energy is a scalar but does not remain conserved in an inelastic collision. (b) False, because potential energy in a gravitational field may have negative values. (c) False, because mass, length, time, speed, work etc., all have dimensions. (d) False, because speed, energy etc., vary from point to point in space. (e) True, because a scalar quantity will have the same value for observers with different orientations of axes since a scalar has no direction of its own.

Question 4. 26. A vector has magnitude and direction. (i) Does it have a location in the space? (ii) Can it vary with time? (iii) Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer. Answer:  (i) Besides having magnitude and direction, each vector has also a location in space. (ii) A vector can vary with time. As an example, velocity and acceleration vectors may vary with time. (iii) Two equal vectors a and b having different locations may not have same physical effect. As an example, two balls thrown with the same force, one from earth and the other from moon will attain different ‘maximum heights’.

Question 4. 27. A vector has both magnitude and direction. Does that mean anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation and the angle of rotation about the axis. Does that make any rotation a vector? Answer:   No. Finite rotation of a body about an axis is not a vector because finite rotations do not obey the laws of vector addition.

Question 4. 28. Can you associate vectors with (a) the length of a wire bent into a loop (b) a plane area (c) a sphere? Explain. Answer:  (a) We cannot associate a vector with the length of a wire bent into a loop. This is because the length of the loop does not have a definite direction. (b) We can associate a vector with a plane area. Such a vector is called area vector and its direction is represented by a normal drawn outward to the area. (c) The area of a sphere does not point in any difinite direction. However, we can associate a null vector with the area of the sphere. We cannot associate a vector with the volume of a sphere.

NCERT Solutions for Class 11 Physics All Chapters

• Chapter 1 Physical World
• Chapter 2 Units and Measurements
• Chapter 3 Motion in a Straight Line
• Chapter 4 Motion in a plane
• Chapter 5 Laws of motion
• Chapter 6 Work Energy and power
• Chapter 7 System of particles and Rotational Motion
• Chapter 8 Gravitation
• Chapter 9 Mechanical Properties Of Solids
• Chapter 10 Mechanical Properties Of Fluids
• Chapter 11 Thermal Properties of matter
• Chapter 12 Thermodynamics
• Chapter 13 Kinetic Theory
• Chapter 14 Oscillations
• Chapter 15 Waves

We hope the NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane, drop a comment below and we will get back to you at the earliest.

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