laws of motion case study questions class 11

Case Study Questions Class 11 Physics Laws Of Motion

Case study questions class 11 physics chapter 5 laws of motion.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 11 Physics Laws Of Motion

Case study – 1, case study – 2.

4) If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.

Case Study – 3

Answer key -3, case study – 4, answer key – 4, case study – 5.

Friction: Let us return to the example of a body of mass m at rest on a horizontal table. The force of gravity (mg) is cancelled by the normal reaction force (N) of the table. Now suppose a force F is applied horizontally to the body. We know from experience that a small applied force may not be enough to move the body. But if the applied force F were the only external force on the body, it must move with acceleration F/m, however small. Clearly, the body remains at rest because some other force comes into play in the horizontal direction and opposes the applied force F, resulting in zero net force on the body. This force f s parallel to the surface of the body in contact with the table is known as frictional force, or simply friction. When there is no applied force, there is no static friction. It comes into play the moment there is an applied force. As the applied force F increases, fs also increases, remaining equal and opposite to the applied force (up to a certain limit), keeping the body at rest. Hence, it is called static friction. Static friction opposes impending motion. The term impending motion means motion that would take place (but does not actually take place) under the applied force, if friction were absent. It is found experimentally that the limiting value of static friction (f s )max f is independent of the area of contact and varies with the normal force(N) approximately as :

Answer key-5

Leave a reply cancel reply, we have a strong team of experienced teachers who are here to solve all your exam preparation doubts, pseb class 6 agriculture chapter 8 major vegetables of punjab solution, up scert solutions class 8 english chapter 9 – where do all the teachers go, up scert solutions class 6 english chapter 15 – yes, we can, telangana scert class 7 math solution chapter 13 area and perimeter.

CBSE Expert

CBSE Case Study Questions Class 11 Physics PDF Download

Are you a Class 11 Physics student looking to enhance your understanding and prepare effectively for your exams? Look no further! In this comprehensive guide, we present a curated collection of CBSE Case Study Questions Class 11 Physics that will help you grasp the core concepts of Physics while reinforcing your problem-solving skills.

CBSE 11th Standard CBSE Physics question papers, important notes, study materials, Previous Year Questions, Syllabus, and exam patterns. Free 11th Standard CBSE Physics books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Physics Case Study Questions

CBSE Class 11 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 11 Physics

• 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

Class 11 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 11 Physics examinations. Our expert faculty for standard 11 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 11 students understand the concepts and also easy-to-learn solutions.

Class 11 Books for Boards

Why Case Study Questions Matter

Case study questions are an invaluable resource for Class 11 Physics students. Unlike traditional textbook exercises, these questions simulate real-life scenarios, challenging students to apply theoretical knowledge to practical situations. This approach fosters critical thinking and helps students build a deep understanding of the subject matter.

Let’s delve into the different topics covered in this collection of case study questions:

1. Motion and Gravitation

In this section, we explore questions related to motion, velocity, acceleration, and the force of gravity. These questions are designed to test your grasp of the fundamental principles governing motion and gravitation.

2. Work, Energy, and Power

This set of questions delves into the concepts of work, energy, and power. You will encounter scenarios that require you to calculate work done, potential and kinetic energy, and power in various contexts.

3. Mechanical Properties of Solids and Fluids

This section presents case study questions about the mechanical properties of solids and fluids. From stress and strain calculations to understanding the behavior of fluids in different situations, these questions cover a wide range of applications.

4. Thermodynamics

Thermodynamics can be a challenging topic, but fear not! This part of the guide offers case study questions that will clarify the laws of thermodynamics, heat transfer, and thermal expansion, among other concepts.

5. Oscillations and Waves

Get ready to explore questions related to oscillations, simple harmonic motion, and wave characteristics. These questions will deepen your understanding of wave propagation and the behavior of oscillatory systems.

6. Kinetic Theory and Laws of Motion

Kinetic theory and the laws of motion can be complex, but with our case study questions, you’ll find yourself mastering these topics effortlessly.

Discover a wide array of questions dealing with light, lenses, and mirrors. This section will improve your problem-solving skills in optics and enhance your ability to analyze optical phenomena.

8. Electrical Effects of Current

Electricity and circuits are fundamental to physics. The case study questions in this section will challenge you to apply Ohm’s law, Kirchhoff’s laws, and other principles in various electrical circuits.

9. Magnetic Effects of Current

Delve into the fascinating world of magnets and magnetic fields. This set of questions will strengthen your understanding of magnetic effects and their applications.

10. Electromagnetic Induction

The final section covers electromagnetic induction, Faraday’s law, and Lenz’s law. You’ll be presented with scenarios that test your ability to predict induced electromotive forces and analyze electromagnetic phenomena.

In conclusion, mastering Class 11 Physics requires a thorough understanding of fundamental concepts and their practical applications. The case study questions provided in this guide will undoubtedly assist you in achieving a deeper comprehension of the subject.

Remember, practice is key! Regularly attempt these case study questions to strengthen your problem-solving abilities and boost your confidence for the exams. Happy studying, and may you excel in your Physics journey!

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Download India's best Exam Preparation App Now.

Key Features

• Revision Notes
• Important Questions
• Previous Years Questions
• Case-Based Questions
• Assertion and Reason Questions

No thanks, I’m not interested!

myCBSEguide

• Class 11 Physics Case...

Class 11 Physics Case Study Questions

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Looking for complete and comprehensive case study questions for class 11 Physics? myCBSEguide is just a click away! With extensive study materials, sample papers, case study questions and mock tests, myCBSEguide is your one-stop solution for class 11 Physics exam preparation needs. So, what are you waiting for? Log on to myCBSEguide and get started today!

What is the purpose of physics?

Physics is the study of the fundamental principles governing the natural world. It is a vital part of the scientific enterprise, providing the foundation on which other sciences are built. Physics is essential for understanding how the world works, from the smallest particles to the largest structures in the Universe. In class 11 Physics, students are introduced to the basic concepts of physics and learn about the fundamental principles governing the natural world. Class 11 Physics concepts are essential for understanding the world around us and for further study in physics and other sciences.

What are case study questions in physics?

In physics, case study questions are intended to evaluate a student’s ability to apply theoretical principles to real-life situations. These questions usually ask the student to assess data from a specific experiment or setting in order to discover what physical principles are at play. Problem-solving and critical-thinking skills are developed through case study questions, which are an important aspect of physics education.

CBSE Case Study Questions in Class 11 Physics

CBSE Class 11 Physics question paper pattern includes case study questions. Class 11 Physics case study questions assess a student’s ability to apply physics principles to real-world environments. The questions are usually focused on a situation provided in the Class 11 Physics question paper, and they demand the student to answer the problem using their physics knowledge. Class 11 Physics case study questions are an important aspect of the CBSE physics curriculum. Class 11 Physics case study questions are a useful way to assess a student’s expertise in the subject.

Sample Class 11 Physics Case Study Questions

Expert educators at myCBSEguide have created a collection of Class 11 physics case study questions. The samples of Class 11 physics case study questions are given below. Class 11 physics case study questions are designed to test your understanding of the concepts and principles of physics. They are not meant to be easy, but they should be done if you have a good grasp of the subject. So, take a look at the questions and see how you fare. Good luck!

Class 11 Physics Case Study Question 1

Read the case study given below and answer any four subparts: Potential energy is the energy stored within an object, due to the object’s position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy. Potential energy depends on the force acting on the two objects.

• kinetic energy
• potential energy
• mechanical energy
• none of these
• potential energy decreases
• potential energy increases
• kinetic energy decreases
• kinetic energy increases
• only when spring is stretched
• only when spring is compressed
• both a and b
• 5  ×  10 4  J
• 5  ×  10 5  J

Answer Key:

Class 11 Physics Case Study Question 2

• distance between body
• source of heat
• all of the above
• convection and radiation
• (b) convection
• (d) all of the above
• (a) convection
• (a) increase
• (c) radiation

  Class 11 Physics Case Study Question 3

• internal energy.
• 1 +(T 2 /T 1 )
• (T 1 /T 2 )+1
• (T 1  /T 2 )- 1
• 1 – (T 2  / T 1 )
• increase or decrease depending upon temperature ratio
• first increase and then decrease
• (d) 1- (T 2 / T 1 )
• (b) increase
• (c) constant

Class 11 Physics Case Study Question 4 

• It is far away from the surface of the earth
• Its surface temperature is 10°C
• The r.m.s. velocity of all the gas molecules is more than the escape velocity of the moon’s surface
• The escape velocity of the moon’s surface is more than the r.m.s velocity of all molecules
• T(H 2 ) = T(N 2 )
• T(H 2 ) < T(N 2 )
• T(H 2 ) > T(N 2 )

The given samples of Class 11 Physics case study questions will help Class 11 Physics students to get an idea on how to solve it. These Class 11 Physics case study questions are based on the topics covered in the Class 11 Physics syllabus and are designed to test the student’s conceptual understanding. The questions are of varying difficulty levels and cover a wide range of topics. By solving these Class 11 Physics case study questions, students will be able to develop their problem-solving skills and improve their understanding of the concepts.

Examining Class 11 Physics syllabus

Senior Secondary school education is a transitional step from general education to a discipline-based curriculum concentration. The current curriculum of Class 11 Physics takes into account the rigour and complexity of the disciplinary approach, as well as the learners’ comprehension level. Class 11 Physics syllabus has also been carefully crafted to be similar to international norms.

The following are some of the Class 11 Physics syllabus’s most notable features:

• Emphasis is placed on gaining a fundamental conceptual knowledge of the material.
• Use of SI units, symbols, naming of physical quantities, and formulations in accordance with international standards are emphasised.
• For enhanced learning, provide logical sequencing of subject matter units and suitable placement of concepts with their links.
• Eliminating overlapping concepts/content within the field and between disciplines to reduce the curricular load.
• Process skills, problem-solving ability, and the application of Physics principles are all encouraged.

CBSE Class 11 Physics (Code No. 042)

myCBSEguide: A true saviour for many students

myCBSEguide has been a true saviour for many students who struggle to find resources elsewhere. It is a reliable source of information and provides students with everything they need to excel in their academics. myCBSEguide has helped many students score high marks in their exams and has been a valuable resource for their studies. Teachers recommend myCBSEguide to all CBSE students. And best of all, it’s available 24/7, so you can study at your own pace, anytime, anywhere. So why wait? Get started today and see the difference myCBSEguide can make to your studies.

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

• Competency Based Learning in CBSE Schools
• Class 11 Physical Education Case Study Questions
• Class 11 Sociology Case Study Questions
• Class 12 Applied Mathematics Case Study Questions
• Class 11 Applied Mathematics Case Study Questions
• Class 11 Mathematics Case Study Questions
• Class 11 Biology Case Study Questions
• Class 12 Physical Education Case Study Questions

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

Class 11 Physics Case Study Questions PDF Download

• Post author: studyrate
• Post published:
• Post category: Physics
• Post comments: 0 Comments

Class 11 Physics Case Study Questions are available here. You can read these Case Study questions by chapter for your final physics exam. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve high marks on your Board exams, practice these questions.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

We are providing Case Study questions for Class 11 Physics based on the Latest syllabus. There is a total of 14 chapters included in the CBSE Class 11 physics exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

Class 11th PHYSICS: Chapterwise Case Study Question & Solution

Case study questions play a crucial role in the Class 11 Physics curriculum. They are designed to assess your understanding of various concepts and principles in real-life scenarios. These questions help you apply theoretical knowledge to practical situations, enhancing your problem-solving skills.

Case Study-Based Questions for Class 11 Physics

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

Class 11 Physics MCQ Questions

Before the exams, students in class 11 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be offered in Physics exams for Grade 11. These questions were created by our highly qualified faculty for standard 11 Physics based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 11 in understanding the topics.

Class 11 Books for Boards

Class 11 Physics Syllabus 2024

Unit I: Physical World and Measurement 08 Periods

Chapter–2: Units and Measurements

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. significant figures. Dimensions of physical quantities, dimensional analysis and its applications.

Unit II: Kinematics 24 Periods

Chapter–3: Motion in a Straight Line

The frame of reference, Motion in a straight line, Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, and instantaneous velocity, uniformly accelerated motion, velocity-time and position-time graphs. Relations for uniformly accelerated motion (graphical treatment).

Chapter–4: Motion in a Plane

Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors. Motion in a plane, cases of uniform velocity and uniform acceleration projectile motion, uniform circular motion.

Unit III: Laws of Motion 14 Periods

Chapter–5: Laws of Motion

Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.

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

Unit IV: Work, Energy and Power 14 Periods

Chapter–6: Work, Energy and Power

Work done by a constant force and a variable force; kinetic energy, workenergy theorem, power. Notion of potential energy, potential energy of a spring, conservative forces: non- conservative forces, motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.

Unit V: Motion of System of Particles and Rigid Body 18   Periods

Chapter–7: System of Particles and Rotational Motion

Centre of mass of a two-particle system, momentum conservation and Centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod. Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications. Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions. Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation).

Unit VI: Gravitation 12 Periods

Chapter–8: Gravitation

Kepler’s laws of planetary motion, universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite.

Unit VII: Properties of Bulk Matter 24 Periods

Chapter–9: Mechanical Properties of Solids

Elasticity, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear modulus of rigidity (qualitative idea only), Poisson’s ratio; elastic energy.

Chapter–10: Mechanical Properties of Fluids

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its simple applications. Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.

Chapter–11: Thermal Properties of Matter

Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv – calorimetry; change of state – latent heat capacity. Heat transfer-conduction, convection and radiation, thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law .

Unit VIII: Thermodynamics 12 Periods

Chapter–12: Thermodynamics

Thermal equilibrium and definition of temperature zeroth law of thermodynamics, heat, work and internal energy. First law of thermodynamics, Second law of thermodynamics: gaseous state of matter, change of condition of gaseous state -isothermal, adiabatic, reversible, irreversible, and cyclic processes.

Unit IX:   Behavior of Perfect Gases and Kinetic Theory of Gases 08   Periods

Chapter–13: Kinetic Theory

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

Unit X: Oscillations and Waves 26 Periods

Chapter–14: Oscillations

Periodic motion – time period, frequency, displacement as a function of time, periodic functions and their application. Simple harmonic motion (S.H.M) and its equations of motion; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period.

Chapter–15: Waves

Wave motion: Transverse and longitudinal waves, speed of traveling wave, 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.

FAQs about Class 11 Physics Case Studies

What is the best website for a  case   study  of physics  class   11 .

studyrate.in is the best website for Class 11 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

How do you write a case study question for Class 11?

The CBSE will ask two Case Study Questions in the CBSE Class 11th Maths Question Paper. Question numbers 15 and 16 will be case-based questions where 5 MCQs will be asked based on a paragraph.

Are the case study questions based on the latest syllabus?

Yes, the case study questions are curated to align with the latest Class 11 Physics syllabus.

You Might Also Like

Download disha ncert xtract physics pdf for neet/aiims latest edition, 50+ neet mcq questions semiconductor electronics with solutions, 50+ neet mcq questions mechanical properties of solids with solutions, leave a reply cancel reply.

Save my name, email, and website in this browser for the next time I comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed .

The Topper Combo Flashcards

• Contains the Latest NCERT in just 350 flashcards.
• Colourful and Interactive
• Summarised Important reactions according to the latest PYQs of NEET(UG) and JEE

No thanks, I’m not interested!

CBSE Class 11 Physics Chapter 5 Laws of motion Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Laws of Motion : Notes and Study Materials -pdf

• Concepts of Laws of Motion
• Laws of Motion Master File
• Laws of Motion Revision Notes
• Laws of Motion MindMap
• NCERT Solution Laws of Motion
• NCERT Exemplar Solution Laws of Motion
• Laws of Motion : Solved Example 1
• Laws of Motion: Solved Example 2

Law of Motion Class 11 Notes Physics Chapter 5

• Dynamics  is the branch of physics in which we study the motion of a body by taking into consideration the cause i.e., force which produces the motion. • Force Force is an external cause in the form of push or pull, which produces or tries to produce motion in a body at rest, or stops/tries to stop a moving body or changes/tries to change the direction of motion of the body. • The inherent property, with which a body resists any change in its state of motion is called inertia. Heavier the body, the inertia is more and lighter the body, lesser the inertia. • Law of inertia states that a body has the inability to change its state of rest or uniform motion (i.e., a motion with constant velocity) or direction of motion by itself. • Newton’s Laws of Motion Law 1.  A body will remain at rest or continue to move with uniform velocity unless an external force is applied to it. First law of motion is also referred to as the ‘Law of inertia’. It defines inertia, force and inertial frame of reference. I here is always a need of ‘frame of reference’ to describe and understand the motion of particle, lhc simplest ‘frame of reference’ used are known as the inertial frames. A frame of referent, e is known as an inertial frame it, within it, all accelerations of any particle are caused by the action of ‘real forces’ on that particle. When we talk about accelerations produced by ‘fictitious’ or ‘pseudo’ forces, the frame of reference is a non-inertial one. Law 2. When an external force is applied to a body of constant mass the force produces an acceleration, which is directly proportional to the force and inversely proportional to the mass of the body. Law 3.  “To every action there is equal and opposite reaction force”. When a body A exerts a force on another body B, B exerts an equal and opposite force on A. • Linear Momentum The linear momentum of a body is defined as the product of the mass of the body and its velocity. • Impulse Forces acting for short duration are called impulsive forces. Impulse is defined as the product of force and the small time interval for which it acts. It is given by Impulse of a force is a vector quantity and its SI unit is 1 Nm. — If force of an impulse is changing with time, then the impulse is measured by finding the area bound by force-time graph for that force. — Impulse of a force for a given time is equal to the total change in momentum of the body during the given time. Thus, we have • Law of Conservation of Momentum The total momentum of an isolated system of particles is conserved. In other words, when no external force is applied to the system, its total momentum remains constant. • Recoiling of a gun, flight of rockets and jet planes are some simple applications of the law of conservation of linear momentum. • Concurrent Forces and Equilibrium “A group of forces which are acting at one point are called concurrent forces.” Concurrent forces are said to be in equilibrium if there is no change in the position of rest or the state of uniform motion of the body on which these concurrent forces are acting. For concurrent forces to be in equilibrium, their resultant force must be zero. In case of three concurrent forces acting in a plane, the body will be in equilibrium if these three forces may be completely represented by three sides of a triangle taken in order. If number of concurrent forces is more than three, then these forces must be represented by sides of a closed polygon in order for equilibrium. • Commonly Used Forces (i) Weight of a body.  It is the force with which earth attracts a body towards its centre. If M is mass of body and g is acceleration due to gravity, weight of the body is Mg in vertically downward direction. (ii) Normal Force. If two bodies are in contact a contact force arises, if the surface is smooth the direction of force is normal to the plane of contact. We call this force as Normal force. Example.  Let us consider a book resting on the table. It is acted upon by its weight in vertically downward direction and is at rest. It means there is another force acting on the block in opposite direction, which balances its weight. This force is provided by the table and we call it as normal force. (iii) Tension in string. Suppose a block is hanging from a string. Weight of the block is acting vertically downward but it is not moving, hence its weight is balanced by a force due to string. This force is called ‘Tension in string’. Tension is a force in a stretched string. Its direction is taken along the string and away from the body under consideration. • Simple Pulley Consider two bodies of masses m 1  and m 2  tied at the ends of an in extensible string, which passes over a light and friction less pulley. Let m 1  > m 2 . The heavier body will move downwards and the lighter will move upwards. Let a be the common acceleration of the system of two bodies, which is given by • Apparent Weight and Actual Weight — ‘Apparent weight’ of a body is equal to its ‘actual weight’ if the body is either in a state of rest or in a state of uniform motion. — Apparent weight of a body for vertically upward accelerated motion is given as Apparent weight =Actual weight + Ma = M (g + a) — Apparent weight of a body for vertically downward accelerated motion is given as Apparent weight = Actual weight Ma = M (g – a). • Friction The opposition to any relative motion between two surfaces in contact is referred to as friction. It arises because of the ‘inter meshing’ of the surface irregularities of the two surfaces in contact. • Static and Dynamic (Kinetic) Friction The frictional forces between two surfaces in contact (i) before and (ii) after a relative motion between them has started, are referred to as static and dynamic friction respectively. Static friction is always a little more than dynamic friction. The magnitude of kinetic frictional force is also proportional to normal force. • Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. We calculate its value using laws of friction. Laws of Friction: (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. (ii) The magnitude of limiting frictional force is independent of area of contact between the surfaces. • Coefficient of Friction The coefficient of friction (μ) between two surfaces is the ratio of their limiting frictional force to the normal force between them, i.e., • Angle of Friction It is the angle which the resultant of the force of limiting friction F and the normal reaction R makes with the direction of the normal reaction. If θ is the angle of friction, we have • Angle of Repose Angle of repose (α) is the angle of an inclined plane with the horizontal at which a body placed over it just begins to slide down without any acceleration. Angle of repose is given by α = tan-1 (μ) • Motion on a Rough Inclined Plane Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane. • Centripetal Force Centripetal force is the force required to move a body uniformly in a circle. This force acts along the radius and towards the centre of the circle. It is given by where, v is the linear velocity, r is the radius of circular path and ω is the angular velocity of the body. • Centrifugal Force Centrifugal force is a force that arises when a body is moving actually along a circular path, by virtue of tendency of the body to regain its natural straight line path. The magnitude of centrifugal force is same as that of centripetal force. • Motion in a Vertical Circle The motion of a particle in a horizontal circle is different from the motion in vertical circle. In horizontal circle, the motion is not effected by the acceleration due to gravity (g) whereas in the motion of vertical circle, the motion is not effected by the acceleration due to gravity (g) whereas in the motion of vertical circle, the value of ‘g’ plays an important role, the motion in this case does not remain uniform. When the particle move up from its lowest position P, its speed continuously decreases till it reaches the highest point of its circular path. This is due to the work done against the force of gravity. When the particle moves down the circle, its speed would keep on increasing. Let us consider a particle moving in a circular vertical path of radius V and centre o tide with a string. L be the instantaneous position of the particle such that Here the following forces act on the particle of mass ‘m’. (i) Its weight = mg (verticaly downwards). (ii) The tension in the string T along LO. We can take the horizontal direction at the lowest point ‘p’ as the position of zero gravitational potential energy. Now as per the principle of conservation of energy, From this relation, we can calculate the tension in the string at the lowest point P, mid-way point and at the highest position of the moving particle. Case (i) : At the lowest point P, θ = 0° When the particle completes its motion along the vertical circle, it is referred to as “Looping the Loop” for this the minimum speed at the lowest position must be √5gr • IMPORTANT TABLES

CBSE Class 11 Physics Chapter-5 Important Questions

1 Marks Questions

1.What is the unit of coefficient of friction?

Ans:  It has no unit.

2.Name the factor on which coefficient of friction depends?

Ans: Coefficient of friction  depends on the nature of surfaces in contact and nature of motion.

3.What provides the centripetal force to a car taking a turn on a level road?

Ans:  Centripetal force is provided by the force of friction between the tyres and the road.

4.Why is it desired to hold a gun tight to one’s shoulder when it is being fired?

Ans:  Since the gun recoils after firing so it must be held lightly against the shoulder because gun and the shoulder constitute one system of greater mass so the back kick will be less.

5.Why does a swimmer push the water backwards?

Ans:  A swimmer pushes the water backwards because due to reaction of water he is able to swim in the forward direction

6.Friction is a self adjusting force. Justify.

Ans:  Friction is a self adjusting force as its value varies from zero to the maximum value to limiting friction.

7.A thief jumps from the roof of a house with a box of weight W on his head. What will be the weight of the box as experienced by the thief during jump?

Ans:  Weight of the box W = m (g – a) = m (g – g) = 0.

8.Which of the following is scalar quantity? Inertia, force and linear momentum.

Ans: Inertia and linear momentum is measured by mass of the body and is a vector quantity and mass is a scalar quantity.

9.Action and reaction forces do not balance each other. Why?

Ans: Action and reaction do not balance each other because a force of action and reaction acts always on two different bodies.

10.If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?

Ans: No change in speed, but there can be change in the direction of motion.

11.The two ends of spring – balance are pulled each by a force of 10kg.wt. What will be the reading of the balance?

Ans: The reading of the balance will be 10kgwt.

12.A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?

Ans: The apparent weight will increase. If the lift is going with uniform speed, then the apparent weight will remain the same as the real weight.

14.  If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

Ans.(b) When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

2 Marks Questions

1.Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with constant speed.

(b) A kite skillfully held stationary in the sky.

Ans: (1)  According to first law of motion F = 0 as a = 0 (particle moves with constant speed)

(2)  Since kite is stationary net force on the kite is also zero.

4.A force is being applied on a body but it causes no acceleration. What possibilities may be considered to explain the observation?

Ans: (1)  If the force is deforming force then it does not produce acceleration.

 (2)  The force is internal force which cannot cause acceleration.

5.Force of 16N and 12N are acting on a mass of 200kg in mutually perpendicular directions. Find the magnitude of the acceleration produced?

6.An elevator weighs 3000kg. What is its acceleration when the in the tension supporting cable is 33000N. Given that g = 9.8m/s 2 .

Ans: Net upward force on the

ElevatorF = T – mg

7.Write two consequences of Newton’s second law of motion?

Ans: (1) It  shows that the motion is accelerated only when force is applied.

(2)  It gives us the concept of inertial mass of a body.

8.A bird is sitting on the floor of a wire cage and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?

Ans: When the bird starts flying inside the cage the weight of bird is no more experienced as air inside is in free contact with atmospheric air hence the cage will appear lighter.

9.Why does a cyclist lean to one side, while going along curve? In what direction does he lean?

Ans: A cyclist leans while going along curve because a component of normal reaction of the ground provides him the centripetal force he requires for turning.

He has to lean inwards from his vertical position i.e. towards the centre of the circular path.

10.How does banking of roads reduce wear and tear of the tyres?

Ans: When a curved road is unbanked force of friction between the tyres and the road provides the necessary centripetal force. Friction has to be increased which will cause wear and tear. But when the curved road is banked, a component of normal reaction of the ground provides the necessary centripetal force which reduces the wear and tear of the tyres.

12.A soda water bottle is falling freely. Will the bubbles of the gas rise in the water of the bottle?

Ans: bubbles will not rise in water because water in freely falling bottle is in the state of weight – lessens hence no up thrust force acts on the bubbles.

13.Two billiard balls each of mass 0.05kg moving in opposite directions with speed 6m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to other.

Ans: Initial momentum to the ball A = 0.05(6) = 0.3 kg m/s

As the speed is reversed on collision,

final momentum of ball A = 0.05(-6) = -0.3 kg m/s

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = -0.3 -0.3 = -0.6 kg m/s.

15.Explain why passengers are thrown forward form their seats when a speeding bus stops suddenly.

Ans: When the speeding bus stops suddenly, lower part of the body in contact with the seat comes to rest but the upper part of the body of the passengers tends to maintain its uniform motion. Hence the passengers are thrown forward.

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 5 Laws of motion are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 5 Laws of motion.

NCERT Solutions for Class 11 Physics Chapter 5 Laws of motion

Topics and Subtopics in  NCERT Solutions for Class 11 Physics Chapter 5 Laws of motion :

QUESTIONS FROM TEXTBOOK

Question 5. 1. Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skilfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. Answer:  (a) As the drop of rain is falling with constant speed, in accordance with first law of motion, the net force on the drop of rain is zero. (b) As the cork is floating on water, its weight is being balanced by the upthrust (equal to.weight of water displaced). Hence net force on the cork is zero. (c) Net force on a kite skilfully held stationary in sky is zero because it is at rest. (d) Since car is moving with a constant velocity, the net force on the car is zero. (e) Since electron is far away from all material agencies producing electromagnetic and gravitational forces, the net force on electron is zero.

Question 5. 2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, . (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction 1 Ignore air resistance. Answer:  (a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms -2 = 0.5 N. (b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards). (c) The pebble is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at 45° because no other acceleration except ‘g’ is acting on pebble.

More Resources for CBSE Class 11

NCERT Solutions

• 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 5. 3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/ h, (c) just after it is dropped from the window of a train accelerating with 1 ms -2 , (d) lying on the floor of a train which is accelerating with 1 m s~2, the stone being at rest relative to the train.Neglect air resistance throughout. Answer:  (a) Mass of stone = 0.1 kg Net force, F = mg = 0.1 x 10 = 1.0 N. (vertically downwards). (b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.). (c) The stone will experience an additional force F’ (along horizontal) i.e.,F = ma = 0.1 x l = 0.1 N As the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (vertically downwards). (d) As the stone is lying on the floor of the train, its acceleration is same as that of the train. .•. force acting on the stone, F = ma = 0.1 x 1 = 0.1 N. It acts along the direction of motion of the train.

Question 5. 4. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: (i) T, (ii) T – mv 2 /l, (iii) T +mv 2 /l, (iv) 0 T is the tension in the string. [Choose the correct alternative]. Answer:  (i) T The net force T on the particle is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.

Question 5. 9. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms -2 . Calculate the initial thrust (force) of the blast. Answer:  Here, m = 20000 kg = 2 x 10 4  kg  Initial acceleration = 5 ms -2 Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms -2 . Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms -2 . Since, thrust = force = mass x acceleration F = 2 x 10 4 x 14.8 = 2.96 x 10 5 N.

Question 5. 13. A man of mass 70 kg, stands on a weighing machine in a lift, which is moving (a) upwards with a uniform speed of 10 ms -1 . (b) downwards with a uniform acceleration of 5 ms -2 . (c) upwards with a uniform acceleration of 5 ms -2 . What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? Answer:  Here, m = 70 kg, g = 10 m/s 2 The weighing machine in each case measures the reaction R i.e., the apparent weight. (a) When the lift moves upwards with a uniform speed, its acceleration is zero. R = mg = 70 x 10 = 700 N (b) When the lift moves downwards with a = 5 ms -2 R = m (g – a) = 70 (10 – 5) = 350 N (c) When the lift moves upwards with a = 5 ms -2 R = m (g + a) = 70 (10 + 5) = 1050 N (d) If the lift were to come down freely under gravity, downward acc. a = g :. R = m(g -a) = m(g-g) = Zero.

Question 5. 18. Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms -1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other? Answer:  Initial momentum of each ball before collision = 0.05 x 6 kg ms -1 = 0.3 kg ms -1 Final momentum of each ball after collision = – 0.05 x 6 kg ms -1 = – 0.3 kg ms -1 Impulse imparted to each ball due to the other = final momentum – initial momentum = 0.3 kg m s-1 – 0.3 kg ms -1 = – 0.6 kg ms -1 = 0.6 kg ms -1  (in magnitude) The two impulses are opposite in direction.

Question 5. 22. If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stoneflies off at an angle with the tangent whose magnitude depends on the speed of the particle? Answer:  (b) The velocity is tangential at each point of circular motion. At the time the string breaks, the particle continues to move in the tangential direction according to Newton’s first law of motion.

Question 5. 23. Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch. Answer:  (a) A horse by itself cannot move in space due to law of inertia and so cannot pull a cart in space. (b) The passengers in a speeding bus have inertia of motion. When the bus is suddenly stopped the passengers are thrown forward due to this inertia of motion. (c) In the case of pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight. (d) The ball comes with large momentum after being hit by the batsman. When the player takes catch it causes large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of ball and hand is increased so the force is reduced.

Question 5. 27. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms -2 . The crew and the passengers weigh 300 kg. Give the magnitude and direction of (a) force on the floor by the crew and passengers, (b) action of the rotor of the helicopter on surrounding air, (c) force on the helicopter due to the surrounding air, Answer:  Here, mass of helicopter, m 1 = 1000 kg Mass of the crew and passengers, m 2 = 300 kg upward acceleration, a = 15 ms -2 and g = 10 ms -2 (a)Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers = m 2 (g + a) = 300 (10 + 15) N = 7500 N (b)Action of rotor of helicopter on surrounding air is obviously vertically downwards, because helicopter rises on account of reaction to this force. Thus, force of action F = (m 1 + m 2 ) (g + a) = (1000 + 300) (10 + 15) = 1300 x 25 = 32500 N (c)Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction, F = 32500 N, vertically upwards.

5.28.A stream of water flowing horizontally with a speed of 15 ms -1 pushes out of a tube of cross sectional area 10 -2 m 2 , and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming that it does not rebound? Ans. In one second, the distance travelled is equal to the velocity v. Volume of water hitting the wall per second, V = av where a is the cross-sectional area of the tube and v is the speed of water coming out of the tube. V = 10 -2 m 2 x 15 ms -1 = 15 x 10 -2 m 3 s -1 Mass of water hitting the wall per second = 15 x 10 -2 x 10 3 kg s -1 = 150 kg s -1 [v density of water = 1000 kg m -3 ] Initial momentum of water hitting the wall per second = 150 kg s -1 x 15 ms -1 = 2250 kg ms -2 or 2250 N Final momentum per second = 0 Force exerted by the wall = 0 – 2250 N = – 2250 N Force exerted on the wall = – (- 2250) N = 2250 N.

Question 5. 29. Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of (a) the force on the 7th coin (counted from the bottom) due to all coins above it. (b) the force on the 7th coin by the eighth coin and (c) the reaction of the sixth coin on the seventh coin. Answer:  (a) The force on 7th coin is due to weight of the three coins lying above it. Therefore, F = (3 m) kgf = (3 mg) N where g is acceleration due to gravity. This force acts vertically downwards. (b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e. F = 2 m + m = (3 m) kg f = (3 mg) N The force acts vertically downwards. (c) The sixth coin is under the weight of four coins above it. Reaction, R = – F = – 4 m (kg) = – (4 mgf) N Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 5. 35. A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms -1 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley. Answer:   (a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N Force of friction,F f = p mg = 0.18 x 15 x 10 = 27 N. i.e., force experienced by block will be less than the friction. So the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground. (b) The observer moving with trolley has an accelerated motion i.e., he forms non-inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.

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 5 Laws of motion help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 5 Laws of motion, drop a comment below and we will get back to you at the earliest.

Free Resources

Quick Resources

• Physics Important Questions
• Class 11 Physics

Important Questions For Class 11 Physics Chapter 4 - Laws of Motion

Class 11 is one of the most crucial stages in students academic life, and for Science stream students,  Physics  plays a major role. Students must prepare well in class 11 to get conceptual clarity on topics. They must be thorough with the concepts of physics to excel in the exam. The Important Questions For Class 11 Physics Chapter 4 Laws of Motion are provided here to help students prepare effectively for the exam.

Important Questions For Class 11 Physics Chapter 4 Laws of Motion

Also Access  CBSE Class 11 Physics Sample Papers 

Understand the Laws of Motion and the concepts behind these theories by watching this intriguing video.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

• School Solutions
• Star Program
• NCERT Solutions Class 12 Maths
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Commerce
• NCERT Solutions Class 12 Economics
• NCERT Solutions Class 12 Accountancy
• NCERT Solutions Class 12 English
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
• NCERT Solutions Class 11 Biology
• NCERT Solutions Class 11 Commerce
• NCERT Solutions Class 11 Accountancy
• NCERT Solutions Class 11 English
• NCERT Solutions Class 11 Hindi
• NCERT Solutions Class 11 Statistics
• NCERT Solutions Class 10 Maths
• NCERT Solutions Class 10 Science
• NCERT Solutions Class 10 English
• NCERT Solutions Class 10 Hindi
• NCERT Solutions Class 10 Social Science
• NCERT Solutions Class 9 Maths
• NCERT Solutions Class 9 Science
• NCERT Solutions Class 9 English
• NCERT Solutions Class 9 Hindi
• NCERT Solutions Class 9 Social Science
• NCERT Solutions Class 8 Maths
• NCERT Solutions Class 8 Science
• NCERT Solutions Class 8 English
• NCERT Solutions Class 8 Hindi
• NCERT Solutions Class 8 Social Science
• NCERT Solutions Class 7 Maths
• NCERT Solutions Class 7 Science
• NCERT Solutions Class 7 English
• NCERT Solutions Class 7 Hindi
• NCERT Solutions Class 7 Social Science
• NCERT Solutions Class 6 Maths
• NCERT Solutions Class 6 Science
• NCERT Solutions Class 6 English
• NCERT Solutions Class 6 Hindi
• NCERT Solutions Class 6 Social Science
• NCERT Solutions Class 5 Maths
• NCERT Solutions Class 5 English
• NCERT Solutions Class 5 EVS
• NCERT Solutions Class 4 Maths
• NCERT Solutions Class 4 English
• NCERT Solutions Class 4 EVS
• NCERT Solutions Class 4 Hindi
• NCERT Solutions Class 3 Maths
• NCERT Solutions Class 3 English
• NCERT Solutions Class 3 EVS
• NCERT Solutions Class 3 Hindi
• NCERT Solutions Class 2 Maths
• NCERT Solutions Class 2 English
• NCERT Solutions Class 2 Hindi
• NCERT Solutions Class 1 Maths
• NCERT Solutions Class 1 English
• NCERT Solutions Class 1 Hindi
• NCERT Books Class 12
• NCERT Books Class 11
• NCERT Books Class 10
• NCERT Books Class 9
• NCERT Books Class 8
• NCERT Books Class 7
• NCERT Books Class 6
• NCERT Books Class 5
• NCERT Books Class 4
• NCERT Books Class 3
• NCERT Books Class 2
• NCERT Books Class 1
• Important Questions Class 12
• Important Questions Class 11
• Important Questions Class 10
• Important Questions Class 9
• Important Questions Class 8
• Important Questions Class 7
• important questions class 6
• CBSE Class 12 Revision Notes
• CBSE Class 11 Revision Notes
• CBSE Class 10 Revision Notes
• CBSE Class 9 Revision Notes
• CBSE Class 8 Revision Notes
• CBSE Class 7 Revision Notes
• CBSE Class 6 Revision Notes
• CBSE Class 12 Syllabus
• CBSE Class 11 Syllabus
• CBSE Class 10 Syllabus
• CBSE Class 9 Syllabus
• CBSE Class 8 Syllabus
• CBSE Class 7 Syllabus
• CBSE Class 6 Syllabus
• CBSE Class 5 Syllabus
• CBSE Class 4 Syllabus
• CBSE Class 3 Syllabus
• CBSE Class 2 Syllabus
• CBSE Class 1 Syllabus
• CBSE Sample Question Papers For Class 12
• CBSE Sample Question Papers For Class 11
• CBSE Sample Question Papers For Class 10
• CBSE Sample Question Papers For Class 9
• CBSE Sample Question Papers For Class 8
• CBSE Sample Question Papers For Class 7
• CBSE Sample Question Papers For Class 6
• CBSE Sample Question Papers For Class 5
• CBSE Sample Question Papers For Class 4
• CBSE Sample Question Papers For Class 3
• CBSE Sample Question Papers For Class 2
• CBSE Sample Question Papers For Class 1
• CBSE Previous Year Question Papers Class 12
• CBSE Previous Year Question Papers Class 10
• Extra Questions For Class 8 Maths
• Extra Questions For Class 8 Science
• Extra Questions For Class 9 Maths
• Extra Questions For Class 9 Science
• Extra Questions For Class 10 Maths
• Extra Questions For Class 10 Science
• NEET 2021 Question Paper
• NEET 2020 Question Paper
• NEET 2019 Question Paper
• NEET 2018 Question Paper
• NEET 2017 Question Paper
• NEET 2016 Question Paper
• NEET 2015 Question Paper
• NEET Physics Questions
• NEET Chemistry Questions
• NEET Biology Questions
• NEET Sample Papers
• NEET Physics Syllabus
• NEET Chemistry Syllabus
• NEET Biology Syllabus
• NEET Mock Test
• NEET Eligibility Criteria
• JEE Main 2021 Question Paper
• JEE Main 2020 Question Paper
• JEE Main 2019 Question Paper
• JEE Main 2018 Question Paper
• JEE Main 2017 Question Paper
• JEE Main 2016 Question Paper
• JEE Main 2015 Question Paper
• JEE Main Sample Papers
• JEE Main Physics Syllabus
• JEE Main Chemistry Syllabus
• JEE Main Maths Syllabus
• JEE Main Physics Questions
• JEE Main Chemistry Questions
• JEE Main Maths Questions
• JEE main revision notes
• JEE Main Mock Test
• JEE Advanced Physics Questions
• JEE Advanced Chemistry Questions
• JEE Advanced Maths Questions
• JEE Advanced 2021 Question Paper
• JEE Advanced 2020 Question Paper
• JEE Advanced 2019 Question Paper
• JEE Advanced 2018 Question Paper
• JEE Advanced 2017 Question Paper
• JEE Advanced 2016 Question Paper
• JEE Advanced 2015 Question Paper
• JEE Advanced Physics Syllabus
• JEE Advanced Chemistry Syllabus
• JEE Advanced Maths Syllabus
• JEE Advanced Mock Test
• ISC Class 12 Syllabus
• ISC Class 11 Syllabus
• ICSE Class 10 Syllabus
• ICSE Class 9 Syllabus
• ICSE Class 8 Syllabus
• ICSE Class 7 Syllabus
• ICSE Class 6 Syllabus
• ISC Sample Question Papers for Class 12
• ISC Sample Question Papers for Class 11
• ICSE Sample Question Papers for Class 10
• ICSE Sample Question Papers for Class 9
• ICSE Sample Question Papers for Class 8
• ICSE Sample Question Papers for Class 7
• ICSE Sample Question Papers for Class 6
• ICSE Class 10 Revision Notes
• ICSE Class 9 Revision Notes
• ISC Important Questions for Class 12
• ISC Important Questions for Class 11
• ICSE Important Questions for Class 10
• ICSE Important Questions for Class 9
• ICSE Important Questions for Class 8
• ICSE Important Questions for Class 7
• ICSE Important Questions for Class 6
• ISC Class 12 Question Paper
• ICSE Class 10 Question Paper
• Maharashtra Board Syllabus
• Maharashtra Board Sample Question Paper
• Maharashtra Board Previous Year Question Paper
• AP Board Syllabus
• AP Board Sample Question Paper
• AP Board Previous Year Question Paper
• Tamilnadu Board Syllabus
• Tamilnadu Board Sample Question Paper
• Tamilnadu Board Previous Year Question Paper
• Telangana Board Syllabus
• Telangana Board Sample Question Paper
• Telangana Board Previous Year Question Paper
• Karnataka Board Syllabus
• Karnataka Board Sample Question Paper
• Karnataka Board Previous Year Question Paper
• Examination Full Forms
• Physics Full Forms
• Chemistry Full Forms
• Biology Full Forms
• Educational Full Form
• CUET Eligibility Criteria
• CUET Exam Pattern
• CUET Cutoff
• CUET Syllabus
• CUET Admit Card
• CUET Counselling
• CUET Previous Year Question Papers
• CUET Application Form
• CUET Sample Papers
• CUET Exam Centers
• CUET Exam Dates
• CUET Results
• Physics Formulas
• Chemistry Formulas
• Math Formulas
• Algebra Formulas
• Geometry Formulas
• Trigonometry Formulas
• Subscription

Important Questions Class 11 Physics Chapter 5

Home » CBSE » Important Questions Class 11 Physics Chapter 5

• CBSE Important Questions
• Important Questions Class 6
• CBSE Previous Year Question Papers
• CBSE Revision Notes
• CBSE Syllabus
• CBSE Extra Questions
• CBSE Sample Papers
• ISC & ICSE Syllabus
• ICSE Syllabus Class 9
• ICSE Syllabus Class 8
• ICSE Syllabus Class 7
• ICSE Syllabus Class 6
• ICSE Syllabus Class 10
• ICSE Question Paper
• ICSE Sample Question Papers
• ISC Sample Question Papers For Class 12
• ISC Sample Question Papers For Class 11
• ICSE Sample Question Papers For Class 10
• ICSE Sample Question Papers For Class 9
• ICSE Sample Question Papers For Class 8
• ICSE Sample Question Papers For Class 7
• ICSE Sample Question Papers For Class 6
• ICSE Revision Notes
• ICSE Important Questions
• ISC Important Questions For Class 12
• ISC Important Questions For Class 11
• ICSE Important Questions For Class 10
• ICSE Important Questions For Class 9
• ICSE Important Questions For Class 8
• ICSE Important Questions For Class 7
• ICSE Important Questions For Class 6
• Maharashtra board
• Rajasthan-Board
• Andhrapradesh Board
• AP Board syllabus
• Telangana Board
• Tamilnadu Board
• Tamilnadu Sample Question Paper
• Tamilnadu Syllabus
• Tamilnadu Previous Year Question Paper
• NCERT Solutions Class 12
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 9
• NCERT Solutions Class 8
• NCERT Solutions Class 7
• NCERT Solutions Class 6
• NCERT Solutions Class 5
• NCERT Solutions Class 4
• NCERT Solutions Class 3
• NCERT Solutions Class 2
• NCERT Solutions Class 1
• JEE Main Question Papers
• JEE Main Syllabus
• JEE Main Questions
• JEE Main Revision Notes
• JEE Advanced Question Papers
• JEE Advanced Syllabus
• JEE Advanced Questions
• JEE Advanced Sample Papers
• NEET Question Papers
• Neet 2021 Question Paper
• Neet 2020 Question Paper
• Neet 2019 Question Paper
• Neet 2018 Question Paper
• Neet 2017 Question Paper
• Neet 2016 Question Paper
• Neet 2015 Question Paper
• NEET Syllabus

Important Questions for CBSE Class 11 Physics Chapter 5 – Law of Motion

You can find the best revision questions for Laws of Motion Class 11 Physics Chapter 5 Important Questions here at Extramarks. The questions contain step-by-step solutions that are chosen by subject matter experts. The following Class 11 Physics Chapter 5 Important Questions will help students score well in your final exams. 

All of the important topics of the Chapter Law of Motion are covered by the Important Questions and Answers. This set of questions is prepared in such a way that they will aid students in studying the chapter while keeping the important questions and answers in mind. The study material has been created with the types of questions asked in the CBSE Class 11 examination in mind.

This article also includes many important concepts and formulas for Class 11 Physics Chapter 5 that are explained in detail for better understanding.

CBSE Class 11 Physics Chapter 5 Important Questions

These are examples of Chapter 5 Class 11 Physics Important Questions, click here to get the full set of Important Questions for Class 11 Physics Chapter 5.

1 Mark Answers and Questions

Q1. Name the factor on which the coefficient of friction depends.

Ans: The coefficient of friction will mainly depend on two factors, which are as follows:

• The materials of the surfaces in contact.
• The characteristics of the surfaces.

Q2. What provides the centripetal force to a car taking a turn on a level road?

Ans: Centripetal force is provided by the frictional contact between the tyres and the road.

Q3. Why does a swimmer push the water backwards?

Ans: From Newton’s 3rd Law of Motion, we know that “when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted”. As a result, the swimmer pushes water backward with his hands in order to swim ahead.

Q4. Action and reaction forces do not balance each other. Why?

Ans: Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

Q5. The two ends of a spring-balance are pulled by a force of 10 kg each. What will be the reading of the balance?

Ans: As the spring balancing is based on the tension in the spring, it gauges weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg, and the reading will be 10kg.

Q6. A lift is an acceleration upward. Will the apparent weight of a person inside the lift increase, decrease, or remain the same relative to its real weight? What happens if the lift is going at a uniform speed?

Ans: There will be an increase in perceived weight. The apparent weight will stay the same as the true weight if the lift moves at a constant pace.

Q7.  Why is it desirable to hold a gun tight to one’s shoulder when it is being fired?

Ans. Because the gun recoils after being fired, it must be held softly on the shoulder. Because the gun and the shoulder are combined into one mass system, the back kick is reduced. When shooting, a gunman must keep his weapon securely against his shoulder.

Q8. Justify that friction is a self-adjusting force. 

Ans: Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit friction.

2 Marks Answers and Questions

Q1. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with constant speed.

(b) a kite skillfully held stationary in the sky.

Ans: (a) As the raindrop is falling at a constant speed, its acceleration will be 0. The net force acting on the raindrop will be 00 because the force acting on a particle is given by.

(b) As the kite is held stationary, by Newton’s first law of motion, the algebraic sum of forces acting on the kite is zero.

Q2. Write two consequences of Newton’s second law of motion.

Ans: The two consequences of Newton’s Second Law of Motion are as follows.

1.It demonstrates that the motion is only accelerated when force is applied to it.

2. It introduces the notion of a body’s inertial mass.

Q3. A bird is sitting on the floor of a wire cage, and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?

Ans: When the bird begins to fly within the cage, the weight of the bird is no longer felt since the air inside is in direct touch with ambient air, making the cage look lighter.

Q4. Why does a cyclist lean to one side, while going along a curve? In what direction does he lean?

Ans: A cyclist leans while riding along a curve because a component of the ground’s natural response supplies him with the centripetal force he needs to turn.

He must lean inward from his vertical posture, towards the circular path’s centre.

Q5. A soda water bottle is falling freely. Will the gas bubbles rise to the surface of the water in the bottle?

Ans: As the water in a freely falling bottle is in a state of weightlessness, As a result, there is no upthrust force on the bubbles, and the bubbles do not ascend in the water.

Q6. Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly.

Ans: When a fast bus comes to a complete stop, the bottom half of the body in touch with the seat comes to a complete halt, while the upper section of the passengers’ bodies prefer to retain their uniform motion. As a result, the passengers are pushed forward.

Q7. How does road banking reduce tyre wear and tear?

Ans: When a curving road is not banked, friction between the tyres and the road provides centripetal force.

Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground’s natural response supplies the necessary centripetal force, reducing tyre wear and tear.

Q8. A force is being applied to a body, but it causes no acceleration. What possibilities might be considered to explain the observation?

Ans: (1) If the force is a deforming force, no acceleration is produced.

(2) Internal force is incapable of causing acceleration.

3 Marks Answers and Questions

Q1. In which of the following cases is centripetal force provided?

(i) Motion of planet around the sun

(ii) Motion of moon around the Earth

(iii) Motion of an electron around the nucleus in an atom

Ans: (i) The centripetal force is provided by the gravitational force acting on the Earth and the sun.

(ii) The centripetal force is provided by the Earth’s gravitational attraction to the moon.

(iii) The centripetal force is provided by the electrostatic attraction between the electron and the proton.

Q2. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g10 g floating on water,

(c) a kite skillfully held stationary in the sky,

(d) a car moving with a constant velocity of 30 km/h−130 km/h−1 on a rough road, and

(e) a high-speed electron in space, far from all material objects, and free of electric and magnetic fields.

Ans: (a) Zero net force

The raindrops are falling at a steady rate. As a result, the acceleration is zero. The net force acting on the raindrop is zero, according to Newton’s second law of motion.

(b) Zero net force

The cork’s weight is acting downward. The buoyant force exerted by the water in an upward direction balances it. As a result, there is no net force acting on the floating cork.

(c) Zero net force

In the sky, the kite is stationary, i.e. it is not moving at all. As a result, according to Newton’s first law of motion, there is no net force acting on the kite.

(d) Zero net force

The car is going at a constant speed down a bumpy route. As a result, it has no acceleration. There is no net force operating on the car, according to Newton’s second law of motion.

(e) Zero net force

All fields have no effect on the high-speed electron. As a result, there is no net force acting on the electron.

Q3. A train runs along an unbanked circular bend of radius 30 m at a speed of 54 km/hr. The mass of the train is 106 kg. What provides the necessary centripetal force required for this purpose, the engine or the rails? What is the angle of banking required to prevent the rail from wearing out? Ans: From the question, we have the radius of a circular bend given as, r=30 m.

Speed of train = v = 54 km h−1 = 54×518 = 15 ms−1

Mass of train given as, m = 106 kg

Then we need to find the angle of banking θ.

(1) The centripetal force is generated by the lateral force exerted by rails on the train’s wheels.

(2) The centripetal force is provided by the lateral thrust of the outer rail.

(3) According to Newton’s third law of motion, the train exerts (i.e., causes) an equal and opposite thrust on the outer rail, causing its wear and tear.

Therefore, the angle of baking:

tanθ = v 2 rg = 15 2 13×19.8

4 Marks Answers and Questions

Q1. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg

(a) just after it is dropped from the window of a stationary train,

(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,

(c) just after it is dropped from the window of a train accelerating with 1 m s −2 ,

(d) lying on the floor of a train which is accelerating with 1 m s−2, the stone being at rest relative to the train. Neglect air resistance throughout.

Ans: (a) 1 N; vertically downward

From the question, we have the mass of the stone given as, m=0.1 kg

The acceleration of the stone is given as, a = g = 10 m/s2

The net force exerted on the stone, according to Newton’s second law of motion, is

F = ma = m g

= 0.1×10 = 1 N

Gravitational acceleration always works in the downward direction.

(b) 1 N; vertically downward

The train is travelling at a constant speed. As a result, its acceleration in the horizontal direction, where it is moving, is zero. As a result, there is no horizontal force acting on the stone.

The net force acting on the stone is due to gravity’s acceleration, and it is always vertically downward. This force has a magnitude of 1 N.

(c) 1 N; vertically downward

It is given that the train is accelerating at the rate of 1 m/s2.

Hence, the net force acting on the stone will be equal to, F′ = ma = 0.1×1 = 0.1 N

This force has a horizontal component to it. The horizontal force F′, no longer acts on the stone when it is dropped. This is due to the fact that the force acting on a body at any one time is determined by the current circumstances rather than previous ones.

As a result, the net force acting on the stone is determined only by gravity’s acceleration.

F = mg = 1 N

This force acts vertically downward.

(d) 0.1 N; in the direction of motion of the train

The typical reaction of the floor balances the weight of the stone. The train’s horizontal motion is the only source of acceleration.

Acceleration of the train, a = 0.1 m/s2

The net force acting on the stone will be directed in the train’s direction of travel. Its magnitude is given by:

= 0.1×1 = 0.1 N

5 Marks Answers and Questions

Q1. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms−2. The crew and the passengers each weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

Ans: (a) Mass of the helicopter is given as, mh= 1000 kg

Mass of the crew and passengers is given as, mp= 300 kg

Therefore, the total mass of the system, m = 1300 kg

As the acceleration of the helicopter is given as, a = 15 m/s2

The reaction force R exerted on the system by the floor may be computed using Newton’s second equation of motion.

= 300(10+15) = 300×25

The response force will likewise be directed upward because the helicopter is accelerating vertically.As a result, the force exerted on the floor by the crew and passengers is 7500 N, directed downward, according to Newton’s third law of motion.

(b) The reaction force R experienced by the helicopter may be computed using Newton’s second equation of motion as follows = 32500 N

The helicopter is being pushed higher by the reaction force of the surrounding air. As a result, the rotor’s action on the surrounding air will be 32500N, directed downward, according to Newton’s third law of motion.

(c) The surrounding air exerts a force of 32500N on the helicopter, which is directed upward.

• Click to share on Facebook (Opens in new window)
• Click to share on Twitter (Opens in new window)
• Click to share on LinkedIn (Opens in new window)
• Click to share on WhatsApp (Opens in new window)

Q.1 Read the assertion and reason carefully to mark the correct option out of the options given below. Assertion: Motion of the rocket is not projectile motion. Reason: An object is in projectile motion only under the force of gravity. However, a rocket is acted upon by a propulsive force which acts against gravity to provide an upward lift to the rocket.

Assertion is true but reason is false., assertion and reason both are false., both assertion and reason are true and the reason is the correct explanation of the assertion., both assertion and reason are true but reason is not the correct explanation of the assertion..

When a body is thrown, shot or launched through the air with some initial velocity we call it as a projectile. A projectile moves due to a force which is applied to it initially without any propulsion. During motion the only force acted upon projectile is the force of gravity.

A rocket is not a projectile. In rocket propulsion, the pressure inside the rocket pushes the gas or liquid downward, produces equal and opposite reaction pushes the rocket upward.

Q.2 A lift of mass 1000 kg, which is moving with an acceleration of 1 m/s 2 in upward direction, has tension developed in its string equal to

T = m (g + a)

= 1000 (10 + 1)

Q.3 A box is lying on an inclined plane. If the box starts sliding when the angle of inclination is 60°, then the coefficient of static friction of the box and plane is

Q.4 A stone tied at the end of a string is whirled in a horizontal circle .When the string breaks, the stone flies away tangentially. Explain why.

When the stone moves in a circular path, its velocity is always tangential to the point of circle. When the string breaks, the force (centripetal) ceases to act on the stone. Hence, according to Newtons first law of motion, the stone flies away in the direction of its motion.

Q.5 A body of mass 10 kg initially at rest explodes and breaks into three fragments of masses in the ratio 1:1:3.  The two equal pieces fly away in the direction perpendicular to each other. What will be the velocity of heaviest fragment?

m 1 + m 2 + m 3 = 10 kg m 1 : m 2 : m 3 = 1 : 1 : 3. m 1 = m 2 = 2 kg m 3 = 6 kg v 1 = v 2 = 30 m/s, v 3 = ?

According to law of conservation of momentum.

m 3 v 3 = m 1 v 1 cos 45 0 + m 2 v 2 cos 45 0

Please register to view this section

Faqs (frequently asked questions), 1. who discovered the three laws of motion.

Isaac Newton, who was a brilliant mathematician and physicist, discovered the three laws of motion. The following are the three laws:

Objects at rest stay at rest, and objects in motion stay in motion unless acted upon by an external force.

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

For every action, there is an equal and opposite reaction.

2. Why are the laws of motion important?

Newton’s Laws of Motion form a very important part of physics. They explain almost all the types of movements that we experience in our daily lives. The laws pose an explanation for simple things like why we do not fall out of chairs when sitting, how a car moves, what role friction plays in a moving bus. All of these movements can be easily explained using Newton’s laws of motion.

3. How is centripetal force provided for an electron moving around the nucleus in an atom?

Centripetal force is the force required to make an object move in a circular path. In the case of an electron moving around the nucleus in an atom, the required centripetal force is provided by the electrostatic force of attraction between the electron and proton. Extramarks’ Class 11 Physics Chapter 5 Important Questions discusses many such important questions that are carefully selected by subject matter experts.

Fill this form to view question paper

Otp verification.

NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion in Hindi and English Medium exercises and additional exercises question answers for session 2024-25 with Extra multiple choice questions. Begin by thoroughly understanding the concepts of each important topic given in 11th Physics chapter 4. Read the NCERT textbook chapters related to these topics and take notes. Focus on grasping the fundamental principles and key equations.

Class 11 Physics Chapter 4 Laws of Motion Question Answers

• Class 11 Physics Chapter 4 NCERT Solutions
• Class 11 Physics Chapter 4 in Hindi Medium
• Class 11 Physics Chapter 4 NCERT Book
• Class 11 Physics all Chapters NCERT Solutions
• Class 11 all Subjects NCERT Solutions

A block of mass M is placed on a flat surface. A force is applied to move it parallel to the surface. The frictional force f developed is proportion to the

The friction force is directly proportional to the mass of the body. The frictional force is a constant force that resist sliding between the surfaces.

• View Answer

What is the frame reference attached to a satellite of the earth is

Earth can not be inertial frame of reference because it is revolving around the sun and also on its own axis. A accelerated object can’t be taken an initial frame of reference frame attached to an non-inertial frame also can’t be taken as inertial frame of reference.

A block of wood is placed on a surface. A force is applied parallel to the surface to move the body. The frictional force developed acts

Frictional force always oppose the applied force. Hence frictional force acts opposite to the direction of the applied force.

Inside the nucleus, two protons are held together by force which overcomes the repulsion. This force is called

Inside the nucleus, two protons are held together by a force which overcomes the repulsion. This force is called strong force. A strong force inside a nucleus hold two protons or proton and neutron and can overcome the action of repulsion. A nucleus of an atom is formed by the combination of both protons and neutrons. The nuclear forces keeps the molecules together and prevents them from any repulsive action. This force is a limited distant force which acts on close molecules or the molecules which are at a limited nuclear distance from each other.

Newton’s laws are usually expressed in terms of the mass of a point or a particle, that is, an object of negligible volume. It is a reasonable approximation of real objects when the movement of internal parts is negligible and when the distance between objects is much larger than the size of each object. For example, the Earth and the Sun can be approximated as points when the orbits of the former around the latter are taken into account, but not when animals active on their surfaces are taken into account.

The mathematical description of motion or kinematics is based on the idea of specifying a position using numerical coordinates. The simplest case is one-dimensional, that is, when the object is constrained to move only in a straight line. Its position can then be given by a number indicating its position relative to a chosen reference point.

For example, if an object can slide freely along a track that runs left to right, its position can be specified by its distance to a convenient zero point, or by a negative number for position relative to the elevator and a positive number for the position at the elevator origin. Motion is represented by a function that assigns a values of all position coordinates to each value of the time variable.

A passenger sitting in a bus moving at uniform speed, feels pushed backward whenever the bus is accelerated forward. This type of force is called

It is due to inertia of motion. The force really doesn’t exist so it a pseudo force.

A passenger in a moving bus is thrown forward when the bus is suddenly stopped. This is explained

When the speeding bus stops suddenly, the lower part of the passenger’s body in contact with the seat mains at rest whereas the upper part of the body of the passengers continues to be in state of motion due to inertia. Hence, the passengers are thrown forward.

A machine gun fires a bullet of mass 40g with a velocity of 1200 ms-1. The man holding it can exert a maximum force on 144N on the gum. How many bullets can he fire per second at the most?

Let n be the total number of bullets fired. The force exerted by man must balanced the change in momentum of the bullets per second. P= nmv F = d/dt (nmv) Hence, total number of bullets fired per second i.e dn/dt = 3

The mass of a body which is equal to the ratio of the force acting on a body to the acceleration produced in the body is

Inertial mass represents the mass discovered by merely applying a load and calculating its acceleration. Gravitational mass is defined as the mass calculated by dividing the wight of an object by g. The mass of something like a body may be computed by dividing the load exerted on it by its acceleration. Inertial mass is the name given to this mass.

Newton’s second law of motion applies to the behavior of all existing force imbalances. The second law states that the acceleration of an object depends on two variables, the resultant force acting on the object increases and the acceleration of the object increases. Increasing the mass of the object with decrease the acceleration.

The momentum of an object is equal to the product of its mass and its velocity, and momentum-like velocity is a vector that has both magnitude and direction. The force exerted on an object can change the amplitude of the momentum or its direction. Newton’s second law is important tools for almost all the chapters of class 11 physics . For objects of constant mass m, it can be written in the form F = ma, where F and a are both vector quantities. If an object is subjected to a resultant force, it will accelerate according to the equation. Conservatively, if an object is not accelerated, no net force acts on it.

A force is a pushing or pulling force acting on an object due to its interaction with another object. Force results from interactions. According to Newton’s third law of motion, whenever bodies A and B interact, they exert a force on each other. If we sit somewhere on any object, our body exerts a downward force. At the same time the object exerts an upward force on our body.

This law represents a certain symmetry in nature. Forces always come in pairs, and one object cannot exert a force on another without feeling the force itself. We sometimes loosely call this law the action-reaction force, where the force applied is the force of action and the force experienced as a result is the force of reaction.

For example, a swimmer uses his feet to accelerate the wall of the pool in the opposite direction he is pushing. The wall exerts equal and opposite forces on the swimmer. You might think that two equal and opposite forces would cancel each other out, but they don’t because they act on different systems. In this case, two types of systems can be studied: bathers or walls.

What main concepts are covered in Chapter 4 of NCERT Class 11 Physics Solutions?

Class 11 Physics Chapter 4 solutions covers basic concepts based on Newton’s Laws. These concepts can be associated with almost all activities of our daily life. 11th Physics chapter 4 addresses one of the most fundamental and important concepts taught in grade 11 CBSE. The NCERT solution for class 11 Physics Chapter 4 will help students answer questions based on the laws of motion and other related concepts in this chapter.

How to perform well in 11th Physics Chapter 4?

A good grade in Class 11 Physics Chapter 4 simply means having a good understanding of all the concepts taught throughout the chapter and being able to use that knowledge to solve any problems based on the chapter. Students should regularly practice and revise everything they learn in school. While preparing for the Grade 11 Physics exam, it can be very helpful to write down important details and formulas. Numerical covers about 20% of the question papers, so must practice numerical questions given in chapter 4 of 11th Physics.

Which one is the most important topic of Chapter 4 of 11th Physics?

Class 11 Physics Chapter 4 deals with Laws of Motion, which is considered as one of the most important topics in mechanics. It is comprise of Newton’s three laws, which are certainly the most important topics covered in chapter 4 of 11th Physics. However, there are many other topics that are very important and should be focused on while preparing for the 11th grade physics exam. The main topics are momentum and its conservation, circular motion and its variations and problem solving in other sections of mechanics.

Copyright 2024 by Tiwari Academy | A step towards Free Education

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 11 Physics

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

Share this:

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

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

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

Contact Form

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

Online mock tests.

Advertisements

Solutions for chapter 5: laws of motion.

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

NCERT solutions for Physics Class 11 Chapter 5 Laws of Motion Exercises [Pages 109 - 113]

Give the magnitude and direction of the net force acting on a drop of rain falling down with a constant speed.

Give the magnitude and direction of the net force acting on a cork of mass 10 g floating on water.

Give the magnitude and direction of the net force acting on a kite skillfully held stationary in the sky.

Give the magnitude and direction of the net force acting on a car moving with a constant velocity of 30 km/h on a rough road.

Give the magnitude and direction of the net force acting on a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

• during its upward motion,
• during its downward motion,
• at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, just after it is dropped from the window of a stationary train. Neglect air resistance.

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, just after it is dropped from the window of a train running at a constant velocity of 36 km/h. Neglect air resistance.

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, just after it is dropped from the window of a train accelerating with 1 m s –2 . Neglect air resistance.

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, lying on the floor of a train which is accelerating with 1 m s –2 , the stone being at rest relative to the train. Neglect air resistance.

One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed the net force on the particle (directed towards the centre) is:

T is the tension in the string.

`"T" - ("mv"^2)/"l"`

`"T" + ("mv"^2)/"l"`

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

A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s –1  to 3.5 m s –1  in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

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

The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s –2 . Calculate the initial thrust (force) of the blast.

A body of mass 0.40 kg moving initially with a constant speed of 10 m s –1  to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

A truck starts from rest and accelerates uniformly at 2.0 m s –2 . At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground).

• What are the velocity, and
• acceleration of the stone at t = 11?(Neglect air resistance.)

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s –1 . What is the trajectory of the bob if the string is cut when the bob is

• At one of its extreme positions,
• At its mean position.

A man of mass 70 kg stands on a weighing scale in a lift which is moving

• upwards with a uniform speed of 10 m s- 1
• downwards with a uniform acceleration of 5 m s –2
• upwards with a uniform acceleration of 5 m s –2 . What would be the readings on the scale in each case?
• What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

The below figure shows the position-time graph of a particle of mass 4 kg.

• What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
• What is the impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to

• B along the direction of string. What is the tension in the string in each case?

Two masses 8 kg and 12 kg are connected at the two ends of a light, inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms -1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s –1 , what is the recoil speed of the gun?

A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

If the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks?

The stone moves radially outwards.

The stone flies off tangentially from the instant the string breaks.

The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

Explain why a horse cannot pull a cart and run in empty space.

Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly.

Explain why it is easier to pull a lawn mower than to push it.

Explain why a cricketer moves his hands backwards while holding a catch.

Figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of each impulse ?

Figure  shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s –2 . What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

A stone of mass  m  tied to the end of a string revolves in a vertical circle of radius  R . The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

  T 1  and  v 1  denote the tension and speed at the lowest point.  T 2  and  v 2  denote corresponding values at the highest point.

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s –2 . The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

A stream of water flowing horizontally with a speed of 15 m s –1  gushes out of a tube of cross-sectional area 10 –2  m 2 , and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Ten one-rupee coins are put on top of each other on a table. Each coin has a mass  m . Give the magnitude and direction of

(a) the force on the 7 th  coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7 th  coin by the eighth coin,

(c) the reaction of the 6th coin on the 7 th  coin.

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10 6  kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail

A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Figure. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

A monkey of mass 40 kg climbs on a rope in given Figure which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s –2

(b) climbs down with an acceleration of 4 m s –2

(c) climbs up with a uniform speed of 5 m s –1

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Two bodies  A  and  B  of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Figure). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to  A . What are (a) the reaction of the partition (b) the action-reaction forces between  A  and  B ?What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ s  and μ k

A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms –2  for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Figure. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s –2 . At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

A disc revolves with a speed of `33 1/3` rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

A thin circular loop of radius  R  rotates about its vertical diameter with an angular frequency ω .  Show that a small bead on the wire loop remains at its lowermost point for `omega <= sqrt(g/R)` .What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for `omega = sqrt("2g"/R)` ?Neglect friction.

NCERT 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 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 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, 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 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 Solutions are essential questions that can be asked in the final exam. Maximum CBSE Physics Class 11 students prefer NCERT 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.

• Maharashtra Board Question Bank with Solutions (Official)
• Balbharati Solutions (Maharashtra)
• Samacheer Kalvi Solutions (Tamil Nadu)
• NCERT Solutions
• RD Sharma Solutions
• RD Sharma Class 10 Solutions
• RD Sharma Class 9 Solutions
• Lakhmir Singh Solutions
• TS Grewal Solutions
• ICSE Class 10 Solutions
• Selina ICSE Concise Solutions
• Frank ICSE Solutions
• ML Aggarwal Solutions
• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Chemistry
• NCERT Solutions for Class 12 Biology
• NCERT Solutions for Class 11 Maths
• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Biology
• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Science
• NCERT Solutions for Class 9 Maths
• NCERT Solutions for Class 9 Science
• CBSE Study Material
• Maharashtra State Board Study Material
• Tamil Nadu State Board Study Material
• CISCE ICSE / ISC Study Material
• Mumbai University Engineering Study Material
• CBSE Previous Year Question Paper With Solution for Class 12 Arts
• CBSE Previous Year Question Paper With Solution for Class 12 Commerce
• CBSE Previous Year Question Paper With Solution for Class 12 Science
• CBSE Previous Year Question Paper With Solution for Class 10
• Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Arts
• Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Commerce
• Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Science
• Maharashtra State Board Previous Year Question Paper With Solution for Class 10
• CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Arts
• CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Commerce
• CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Science
• CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 10
• Entrance Exams
• Video Tutorials
• Question Papers
• Question Bank Solutions
• Question Search (beta)
• More Quick Links
• Privacy Policy
• Terms and Conditions
• Shaalaa App
• Ad-free Subscriptions

Select a course

• Class 1 - 4
• Class 5 - 8
• Class 9 - 10
• Class 11 - 12
• Search by Text or Image
• Textbook Solutions
• Study Material
• Remove All Ads
• Change mode

Talk to our experts

1800-120-456-456

NCERT Solutions Class 11 Physics Chapter 4 Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion - FREE PDF Download

NCERT for Chapter 4 Laws of Motion Class 11 Solutions by Vedantu, explores the principles governing the motion of objects. This chapter builds on the concepts of force and motion that explains the laws formulated by Sir Isaac Newton.

The chapter covers the practical applications of these laws through free-body diagrams, the equilibrium of forces, motion on inclined planes, and circular motion. These concepts are fundamental for solving real-world problems in engineering, design, and everyday activities, making the chapter essential for students pursuing further studies in physics and engineering. Also, understanding these laws is crucial for analysing and predicting the motion of objects in various physical situations. With Vedantu's Class 11 Physics NCERT Solutions , you'll find step-by-step explanations of all the exercises in your textbook, ensuring that you understand the concepts thoroughly. 

Glance on Physics Chapter 4 Class 11 - Laws of Motion

Chapter 4 of Class 11 Physics likely covers Newton's Laws of Motion, which are the foundation of classical mechanics.

The chapter begins with Newton's First Law, which states that an object will remain at rest or continue to move at a constant velocity unless acted upon by an external force. This law introduces the concept of inertia, which is the tendency of objects to resist changes in their state of motion.

The chapter emphasizes the use of free-body diagrams to visually represent the forces acting on an object. These diagrams are essential tools for analysing problems involving multiple forces and predicting the resulting motion.

The chapter includes a detailed analysis of forces acting on objects on slopes, considering both gravitational components and frictional forces. It addresses the forces involved in circular motion, particularly centripetal force, which keeps an object moving in a curved path.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 4 - Laws of Motion, which you can download as PDFs.

There are 23 fully solved questions in the exercise of class 11th Physics chapter 4 Laws of Motion.

Access NCERT Solutions for Class 11 Physics Chapter 4 – Laws of Motion

1. Give the magnitude and direction of the net force acting on 

a drop of rain falling down with a constant speed, 

Ans: The net force is zero.

As the speed of the rain drop falling down is constant, its acceleration is zero.

Therefore, from Newton’s second law of motion, the net force acting on the rain drop is zero.

a cork of mass $10g$ floating on water, 

It is known that the weight of a cork floating on water acts downward. 

The weight of the cork is balanced by buoyant force exerted by the water in the upward direction.

Therefore, no net force acts on the floating cork.

a kite skilfully held stationary in the sky, 

The kite is stationary in the sky indicates that it is not moving.

Therefore, from Newton’s first law of motion, the net force acting on the kite is zero.

a car moving with a constant velocity of $30{km}/{h}\;$ on a rough road, 

As the car is moving with constant velocity, its acceleration is zero.

Therefore, from Newton’s second law of motion, net force acting on the car is equal to zero.

a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 

As the high speed electron is free from the influence of all the fields, no net force acts on the electron.

2. A pebble of mass \[\mathbf{0}.\mathbf{05kg}\] is thrown vertically upwards. Ignore air resistance and give the direction and magnitude of the net force on the pebble,

during its upward motion, 

Ans: It is known that,

Acceleration due to gravity always acts downward irrespective of the direction of motion of an object. The only force that acts on the pebble thrown vertically upward during its upward motion is the gravitational force.

From Newton’s second law of motion: $F=m\times a$ 

$F$ is the net force 

$m$ is the mass of the pebble, $m=0.05kg$ 

a is the acceleration due to gravity, $a=g=10{m}/{{{s}^{-2}}}\;$ 

$\Rightarrow F=0.05\times 10=0.5N$ 

Therefore, the net force on the pebble is $0.5N$ and this force acts in the downward direction. 

during its downward motion, 

Ans: The only force that acts on the pebble during its downward motion is the gravitational force.

Therefore, the net force on the pebble in its downward direction is same as in upward direction i.e., $0.5N$ and this force acts in the downward direction. 

at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of ${{45}^{\circ }}$ with the horizontal direction? 

Ans: When the pebble is thrown at an angle of ${{45}^{\circ }}$with the horizontal, it will have both the horizontal and vertical components of velocity. 

At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble. 

Therefore, the net force on the pebble is $0.5N$.

3. Neglect air resistance throughout and give the magnitude and direction of the net force acting on a stone of mass $0.1kg$ ,

just after it is dropped from the window of a stationary train, 

Ans: It is given that,

Mass of the stone, \[\mathbf{m}=\mathbf{0}.\mathbf{1kg}\] 

Acceleration of the stone, \[\mathbf{a}=g=10{m}/{{{s}^{2}}}\;\] 

From Newton’s second law of motion, 

The net force acting on the stone is $F=ma=mg$ 

$\Rightarrow F=0.1\times 10=1N$ 

It is known that acceleration due to gravity always acts in the downward direction. 

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

just after it is dropped from the window of a train running at a constant velocity of $36{km}/{h}\;$, 

The train is moving with a constant velocity. 

Therefore, its acceleration is zero in the direction of its motion, i.e. in the horizontal direction. 

Thus, no force is acting on the stone in the horizontal direction. 

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. 

just after it is dropped from the window of a train accelerating with $1m{{s}^{-2}}$ 

Ans: It is given that, 

The train is accelerating at the rate of $1m{{s}^{-2}}$.

Therefore, the net force acting on the stone is $F'=ma=0.1\times 1=1N$ 

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. 

Therefore, the net force acting on the stone is given only by acceleration due to gravity i.e., $F=mg=1N$.

lying on the floor of a train which is accelerating with $1m{{s}^{-2}}$, the stone being at rest relative to the train. 

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. 

Acceleration of the train, $a=1m{{s}^{-2}}$

The net force acting on the stone will be in the direction of motion of the train. 

Magnitude: $F=ma=0.1\times 1=0.1N$ 

Therefore, the magnitude of force is $0.1N$ and its direction is in the direction of motion of the train.

4. One end of a string of length $l$ is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$ the net force on the particle (directed towards the centre) is: 

ii. $T-\frac{m{{v}^{2}}}{l}$ ,

iii. $T+\frac{m{{v}^{2}}}{l}$,

iv. $0$ 

$T$  is the tension in the string. (Choose the correct alternative). 

Ans: (i) $T$ 

The centripetal force of a particle connected to a string revolving in a circular path around a centre is provided by the tension produced in the string.

Therefore, the net force on the particle is the tension $T$ , i.e., 

$F=T=\frac{m{{v}^{2}}}{l}$ 

Where $F$ is the net force acting on the particle. 

5. A constant retarding force of $50N$ is applied to a body of mass $20kg$ moving initially with a speed of $15m{{s}^{-1}}$ . How long does the body take to stop?

Retarding force, \[F=50N\] 

Mass of the body, $m=20kg$ 

Initial velocity of the body, \[u=15m/s\] 

Final velocity of the body, $v=0$ 

From Newton’s second law of motion,

The acceleration $(a)$ produced in the body: $F=ma$ 

$\Rightarrow -50=20\times a$ 

$\Rightarrow a=\frac{-50}{20}=-2.5m{{s}^{-2}}$ 

From the first equation of motion, 

The time $(t)$ taken by the body to come to rest: $v=u+at$ 

$\Rightarrow 0=15+(-2.5)t$ 

$\Rightarrow t=\frac{-15}{-2.5}=6s$ 

Therefore, the time taken by the body to stop is $6s$.

6. A constant force is acting on a body of mass $3.0kg$ changes its speed from $2.0m{{s}^{-1}}$ to $3.0m{{s}^{-1}}$  in $25s$ . The direction of the motion of the body remains unchanged.  What is the magnitude and direction of the force? 

Mass of the body, $m=3kg$ 

Initial speed of the body, \[u=2m/s\] 

Final speed of the body, \[v=3.5m/s\]

Time, $t=25s$ 

The acceleration $(a)$ produced in the body: $v=u+at$

$\Rightarrow a=\frac{v-u}{t}$ 

$\Rightarrow a=\frac{3.5-2}{25}=\frac{1.5}{25}=0.6m{{s}^{-2}}$ 

Force, $F=ma$ 

\[\Rightarrow F=3\times 0.06=0.18N\] 

As the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion. 

Therefore, the magnitude of force is $0.18N$ and direction is along the direction of motion.

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

Mass of the body, $m=5kg$ 

Representation of given data:

Law of Vectors

Resultant of two forces $8N$ and $6N$, $R=\sqrt{{{\left( 8 \right)}^{2}}+{{\left( -6 \right)}^{2}}}$ 

$\Rightarrow R=\sqrt{64+36}$ 

$\Rightarrow R=10N$ 

Angle made by $R$ with the force of $8N$ 

$\theta ={{\tan }^{-1}}\left( \frac{-6}{8} \right)=-{{36.87}^{\circ }}$ 

The negative sign indicates that $\theta $ is in the clockwise direction with respect to the force of magnitude $8N$. 

$a=\frac{F}{m}=\frac{10}{5}=2m{{s}^{-2}}$ 

Therefore, the magnitude of acceleration is $2m{{s}^{-2}}$ and direction is ${{37}^{\circ }}$ with a force of $8N$.

8. The driver of a three-wheeler moving with a speed of $36{km}/{h}\;$ sees a child standing in the middle of the road and brings his vehicle to rest in $4.0s$  just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is $400kg$ and the mass of the driver is $65kg$.

Initial speed of the three-wheeler, \[u=36\text{ }km/h\] 

Final speed of the three-wheeler, \[v=0m/s\] 

Time, $t=4s$ 

Mass of the three-wheeler, $m=400kg$ 

Mass of the driver, $m'=65kg$ 

Total mass of the system, \[M=400+65=465kg\] 

From the first law of motion, 

The acceleration $(a)$ of the three-wheeler can be calculated from: $v=u+at$ 

$\Rightarrow a=\frac{v-u}{t}=\frac{0-10}{4}$ 

$\Rightarrow a=-2.5{m}/{{{s}^{2}}}\;$ 

The negative sign indicates that the velocity of the three-wheeler is decreasing with time. 

The net force acting on the three-wheeler can be calculated as: $F=ma$ 

$\Rightarrow F=465\times (-2.5)$ 

$\Rightarrow F=-1162.5N$ 

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler. 

Therefore, the average retarding force on the vehicle is $-1162.5N$.

9. A rocket with a lift-off mass $20,000kg$ is blasted upwards with an initial acceleration of $5.0m{{s}^{-2}}$. Calculate the initial thrust (force) of the blast.

Mass of the rocket, $m=20,000kg$ 

Initial acceleration, $a=5m{{s}^{-2}}$ 

Acceleration due to gravity, $g=10m{{s}^{-2}}$

By using Newton’s second law of motion, 

The net force (thrust) acting on the rocket can be written as:

$F-mg=ma$ 

$\Rightarrow F=m(g+a)$ 

$\Rightarrow F=20000(10+5)=20000\times 15$ 

$\Rightarrow F=3\times {{10}^{5}}N$  

Therefore, the initial thrust(force) of the blast is $3\times {{10}^{5}}N$.

10. A body of mass \[\mathbf{0}.\mathbf{40kg}\] moving initially with a constant speed of $10m{{s}^{-1}}$ subject to a constant force of $8.0N$ directed towards the south for $30s$. Take the instant the force is applied to be $t=0$ , the position of the body at that time to be predict its position at \[\mathbf{t}=-\mathbf{5}\text{ }\mathbf{s},\text{ }\mathbf{25}\text{ }\mathbf{s},\text{ }\mathbf{100}\text{ }\mathbf{s}\].

Mass of the body, $m=0.40kg$ 

Initial speed of the body, \[u=10m/s\] due north 

Force acting on the body, $F=-8.0N$ 

Acceleration produced in the body, $a=\frac{F}{m}$

At $t=0$ 

$\Rightarrow a=\frac{-8}{0.4}=-20m{{s}^{-2}}$ 

At $t=-5s$ 

Acceleration, $a'=0$ and $u=10m/s$ 

$s=ut+\frac{1}{2}a'{{t}^{2}}$ 

$\Rightarrow s=10\times (-5)=-50m$ 

Acceleration, $a=-20m{{s}^{-2}}$ and $u=10m/s$ 

$s'=ut'+\frac{1}{2}a{{(t')}^{2}}$ 

$\Rightarrow s'=10\times (25)+\frac{1}{2}\times (-20){{(25)}^{2}}$ 

$\Rightarrow s'=250+(-6250)$ 

$\Rightarrow s'=-6000m$ 

At $t=100s$

For $0\le t\le 30s$, $a=-20m{{s}^{-2}}$ and $u=10m/s$

${{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow {{s}_{1}}=10\times 30+\frac{1}{2}\times (-20)\times {{(30)}^{2}}$

$\Rightarrow {{s}_{1}}=300-9000$

$\Rightarrow {{s}_{1}}=8700m$

For $30s\le t\le 100s$

First equation of motion: $v=u+at$ 

$\Rightarrow v=10+(-20)\times 30$ 

$\Rightarrow v=-590m{{s}^{-1}}$ 

Velocity of body after $30s=-590m/s$ 

For motion between $30s$ to $100s$ i.e., in $70s$ 

${{s}_{2}}=vt+\frac{1}{2}a'{{t}^{2}}$

${{s}_{2}}=-590\times 70=-41300m$

Total distance, $s''={{s}_{1}}+{{s}_{2}}$ 

$\Rightarrow s''=-8700+(-41300)=-50000m$ 

Therefore, the position of the body at $t=-5s$ is $-50m$ at $t=25s$ is  $-6000m$ and at $t=100s$ is$-50,000m$.

11. A truck starts from rest and accelerates uniformly at $2.0m{{s}^{-2}}$. At $t=10s$, a stone is dropped by a person standing on the top of the truck ($6m$ high from the ground). Neglect air resistance. What are the

velocity, and 

Initial velocity of the truck, $u=0$ (Initially at rest)

Acceleration, $a=2m{{s}^{-2}}$ 

Time, $t=10s$ 

From first equation of motion: $v=u+at$ 

$\Rightarrow v=0+2\times 10=20{m}/{s}\;$ 

Therefore, the final velocity of the truck and the stone is$20{m}/{s}\;$.

At $t=11s$:

The horizontal component $({{v}_{x}})$  of velocity, in the absence of air resistance, remains unchanged, i.e. \[{{v}_{x}}=20m/s\].

The vertical component of velocity $({{v}_{y}})$ of the stone is given by the first equation of motion as: 

${{v}_{y}}=u+{{a}_{y}}\delta t$ 

Where, 

$\delta t=11-10=1s$ 

$a=g=10m/{{s}^{2}}$ 

$\Rightarrow {{v}_{y}}=0+10\times 1=10{m}/{{{s}^{2}}}\;$ 

The resultant velocity $(v)$  of the stone is:

Two Velocities of Different Magnitude

$\Rightarrow v=\sqrt{v_{x}^{2}+v_{y}^{2}}$ 

$\Rightarrow v=\sqrt{{{20}^{2}}+{{10}^{2}}}=\sqrt{400+100}$

$\Rightarrow v=\sqrt{500}=22.36m/s$

Consider $\theta $  as the angle made by the resultant velocity with the horizontal component of velocity, ${{v}_{x}}$.

$\Rightarrow \tan \theta =\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{10}{20} \right)$

$\Rightarrow \theta ={{\tan }^{-1}}\left( 0.5 \right)$

$\Rightarrow \theta ={{26.57}^{\circ }}$

Therefore, the magnitude of resultant velocity is $22.36{m}/{s}\;$ making an angle of ${{26.57}^{\circ }}$ with the horizontal component of velocity.

acceleration of the stone at $t=11s$?

Ans: When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. 

Therefore, the acceleration of the stone is $10m{{s}^{-2}}$  and it acts vertically downward. 

12. A bob of mass \[\mathbf{0}.\mathbf{1kg}\] hung from the ceiling of a room by a string $2m$ long is set into oscillation. The speed of the bob at its mean position is $1m{{s}^{-2}}$. What is the trajectory of the bob if the string is cut when the bob is

at one of its extreme positions, 

Ans: If the string is cut when the bob is at one of its extremes then the bob will fall vertically on the ground.

Therefore, at the extreme position, the velocity of the bob becomes zero.

at its mean position. 

Ans: If the string is cut when the bob is at its mean position then the bob will trace a projectile path having the horizontal components of velocity only.

The direction of this velocity is tangential to the arc formed by the oscillating bob. At the mean position, the velocity of the bob is $1m/s$. 

Therefore, it will follow a parabolic path. 

13. What would be the readings on the scale of a man of mass \[\mathbf{70kg}\] stands on a weighing scale in a lift which is moving

upwards with a uniform speed of $10m{{s}^{-1}}$,

Mass of the man, $m=70kg$ 

Acceleration, $a=0$(uniform speed)

From Newton’s second law: $R-mg=ma$ 

$ma$ is the net force acting on the man.

$\Rightarrow R-70\times 10=0$ 

$\Rightarrow R=700N$ 

Reading on the weighing scale$=\frac{700}{g}=\frac{700}{10}=70kg$ 

Therefore, the mass of the man, $m=70kg$ 

downwards with a uniform acceleration of $5m{{s}^{-2}}$,

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ downward

From Newton’s second law: $R=m(g-a)$

$\Rightarrow R=70\left( 10-5 \right)=70\times 5$ 

$\Rightarrow R=350N$ 

Reading on the weighing scale$=\frac{350}{g}=\frac{350}{10}=35kg$ 

Therefore, the mass of the man, $m=35kg$ 

upwards with a uniform acceleration of $5m{{s}^{-2}}$.

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ upward

From Newton’s second law: $R=m(g+a)$

$\Rightarrow R=70\left( 10+5 \right)=70\times 15$ 

$\Rightarrow R=1050N$ 

Reading on the weighing scale$=\frac{1050}{g}=\frac{1050}{10}=105kg$ 

Therefore, the mass of the man, $m=105kg$ 

What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? 

Ans: When the lift moves freely under gravity,

Acceleration, $a=g=10m{{s}^{-2}}$ 

$\Rightarrow R=70\left( 10-10 \right)=0$ 

Reading on the weighing scale$=\frac{0}{g}=\frac{0}{10}=0kg$ 

Therefore, the man will be in a state of weightlessness.

14. Figure shows the position-time graph of a particle of mass $4kg$ . Consider one-dimensional motion only and find the 

Position- Time Graph for a Particle

force on the particle for $t<0,t>4s$,$0<t<4s$ ?

Ans:  For $t<0$ :

From the given graph, it is observed that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. 

Therefore, the force acting on the particle is zero. 

For $t>4s$ :

From the given graph, it is observed that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of $3m$ from the origin.

Therefore, no force acts on the particle. 

For $0<t<4$ :

From the given position-time graph, it is observed that it has a constant slope. Thus, the acceleration produced in the particle is zero. 

impulse at $t=0$ and $t=4s$ ? 

Ans: At $t=0$:

\[Impulse=Change\text{ }in\text{ }momentum=mv-mu\] 

Mass of the particle, $m=4kg$

Initial velocity of the particle, $u=0$ 

Final velocity of the particle, $v=\frac{3}{4}m/s$ 

Impulse $=4\left( \frac{3}{4}-0 \right)=3kgm{{s}^{-1}}$ 

Initial velocity of the particle, $u=\frac{3}{4}m/s$ 

Final velocity of the particle, $v=0$ 

Impulse\[=4\left( 0-\frac{3}{4} \right)=-3kgm{{s}^{-1}}\]

Therefore, the impulse at $t=0$ is $3kgm{{s}^{-1}}$ and at $t=4s$is $-3kgm{{s}^{-1}}$.

15. Two bodies of masses $10kg$  and $20kg$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. What is the tension in the string if a horizontal force $F=600N$ is applied along the direction of string to a) A, b) B along the direction of the string. What is the tension in the string in each case?

Horizontal force, $F=600N$ 

Mass of body A, ${{m}_{1}}=10kg$ 

Mass of body B, ${{m}_{2}}=20kg$

Total mass of the system, $m={{m}_{1}}+{{m}_{2}}=30kg$

The acceleration $(a)$ produced in the system is: $F=ma$ 

$\Rightarrow a=\frac{F}{m}=\frac{600}{30}=20m{{s}^{-2}}$

When force $F$ is applied on body A: 

Two Bodies of Masses 

The equation of motion can be written as: $F-T=ma$

$\Rightarrow T=F-{{m}_{1}}a$ 

\[\Rightarrow T=600-10\times 20=400N\]

Therefore, the tension in the string is $400N$.

B along the direction of the string

Ans: When force $F$  is applied on body B: 

The equation of motion can be written as:  $F-T={{m}_{2}}a$ 

$\Rightarrow T=F-{{m}_{2}}a$ 

\[\Rightarrow T=600-20\times 20=200N\]

Therefore, the tension in the string is $200N$.

16. Two masses $8kg$ and $12kg$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Ans: The given system of two masses and a pulley are represented in the following figure: 

Smaller mass, ${{m}_{1}}=8kg$ 

Larger mass, ${{m}_{2}}=12kg$

Tension in the string $=T$ 

Mass \[{{m}_{2}}\], owing to its weight, moves downward with acceleration $a$, and mass moves ${{m}_{1}}$upward.

Applying Newton’s second law of motion to the system of each mass: 

For mass ${{m}_{1}}$: 

The equation of motion can be written as: \[T-{{m}_{1}}g=ma\]    .......$(1)$ 

For mass ${{m}_{2}}$:

The equation of motion can be written as: \[{{m}_{2}}g-T={{m}_{2}}a\]   ........$(2)$ 

Adding equations $(1)$ and $(2)$, we get:

$({{m}_{2}}-{{m}_{1}})g=({{m}_{1}}+{{m}_{2}})a$ 

$\Rightarrow a=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g$        ........$(3)$ 

$\Rightarrow a=\left( \frac{12-8}{12+8} \right)\times 10=\frac{4}{20}\times 10$

$\Rightarrow a=2m{{s}^{-2}}$ 

Thus, the acceleration of the masses is $2m{{s}^{-2}}$ . Substituting the value of  $a$ in equation $(2)$:

\[\Rightarrow {{m}_{2}}g-T={{m}_{2}}\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( {{m}_{2}}-\frac{m_{2}^{2}-{{m}_{1}}{{m}_{2}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( \frac{2{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)g\]

\[\Rightarrow T=\left( \frac{2\times 12\times 8}{12+8} \right)\times 10\]

\[\Rightarrow T=96N\]

Thus, the tension in the string is $96N$.

17. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions. 

Ans: Consider $m$, ${{m}_{1}}$ and ${{m}_{2}}$ as the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest. 

Initial momentum of the system (parent nucleus) $=0$ 

Let ${{v}_{1}}$ and ${{v}_{2}}$  be the respective velocities of the daughter nuclei having masses ${{m}_{1}}$ and ${{m}_{2}}$.

Total linear momentum of the system after disintegration$={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

From the law of conservation of momentum: 

\[\mathbf{Total}\text{ }\mathbf{initial}\text{ }\mathbf{momentum}=\mathbf{Total}\text{ }\mathbf{final}\text{ }\mathbf{momentum}\] 

 $\Rightarrow 0={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

$\Rightarrow {{m}_{1}}{{v}_{1}}=-{{m}_{2}}{{v}_{2}}$

$\Rightarrow {{v}_{1}}=\frac{-{{m}_{2}}{{v}_{2}}}{{{m}_{1}}}$

The negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

18. Two billiard balls each of mass $0.05kg$ moving in opposite directions with speed $6m{{s}^{-1}}$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Mass of each ball$=0.05kg$ 

Initial velocity of each ball$=6m/s$ 

Magnitude of the initial momentum of each ball, ${{p}_{i}}=0.3kgm{{s}^{-1}}$ 

After collision, the balls change their directions of motion without changing the 

magnitudes of their velocity. 

Final momentum of each ball, ${{p}_{f}}=-0.3kgm{{s}^{-1}}$

Impulse imparted to each ball$=$ Change in the momentum of the system 

$\Rightarrow \operatorname{Im}pulse={{p}_{f}}-{{p}_{i}}$ 

$\Rightarrow \operatorname{Im}pulse=-0.3-0.3=-0.6kgm{{s}^{-1}}$

The negative sign indicates that the impulses imparted to the balls are opposite in direction. 

19. A shell of mass \[\mathbf{0}.\mathbf{020kg}\] is fired by a gun of mass \[\mathbf{100kg}\]. If the muzzle speed of the shell is $80m{{s}^{-1}}$ what is the recoil speed of the gun?

Mass of the gun, $M=100kg$ 

Mass of the shell, $m=0.020kg$ 

Muzzle speed of the shell, $v=80m/s$ 

Recoil speed of the gun  $=V$.

Both the gun and the shell are at rest initially. 

Initial momentum of the system$=0$

Final momentum of the system$=mv-MV$ 

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other. 

From the law of conservation of momentum:

\[Final\text{ }momentum=Initial\text{ }momentum\] 

$\Rightarrow V=\frac{mv}{M}$ 

$\Rightarrow V=\frac{0.020\times 80}{100\times 1000}=0.016m/s$ 

Therefore, the recoil speed of the gun is $0.016m/s$.

20. A batsman deflects a ball by an angle of ${{45}^{\circ }}$ without changing its initial speed which is equal to $54{km}/{h}\;$ . What is the impulse imparted to the ball? (Mass of the ball is $0.15kg$) 

Ans: The given situation can be represented as:

Deflection of a Ball Hit by a Batsman

$AO=$ Incident path of the ball 

$OB=$ Path followed by the ball after a deflection 

$\angle AOB=$ Angle between the incident and deflected paths of the ball$={{45}^{\circ }}$ 

$\angle AOB=\angle BOP={{22.5}^{\circ }}=\theta $

Initial and final velocities of the ball$=v$ 

The horizontal component of the initial velocity$=v\cos \theta $ along $RO$ 

Vertical component of the initial velocity$=v\sin \theta $ along $PO$ 

The horizontal component of the final velocity$=v\cos \theta $ along $OS$ 

The vertical component of the final velocity $=v\sin \theta $ along $OP$ 

The horizontal components of velocities suffer no change. The vertical components of velocities are in opposite directions. 

It is known that Impulse imparted to the ball$=$ Change in the linear momentum of the ball.

$\operatorname{Im}pulse=mv\cos \theta -(-mv\cos \theta )=2mv\cos \theta $ 

It is given that,

Mass of the ball, $m=0.15kg$ 

Velocity of the ball, $v=54km/h=54\times \frac{5}{18}=15m/s$ 

Impulse$=2\times 0.15\times 15\cos {{22.5}^{\circ }}=4.16kgm{{s}^{-1}}$ 

Therefore, impulse imparted to the ball is $4.16kgm{{s}^{-1}}$.

21. A stone of mass $0.25kg$  tied to the end of a string is whirled round in a circle of radius $1.5m$ with a speed of $40rev./\min $ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200N$?

Mass of the stone, $m=0.25kg$ 

Radius of the circle, $r=1.5m$ 

Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3}rps$ 

Angular velocity, $\omega =\frac{v}{r}=2\pi n$     ........$(1)$ 

The centripetal force for the stone is provided by the tension $T$ , in the string, i.e., $T={{F}_{Centripetal}}$

$\Rightarrow \frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}=mr{{\left( 2\pi n \right)}^{2}}$ 

$\Rightarrow {{F}_{Centripetal}}=0.25\times 1.5\times {{\left( 2\times 3.14\times \frac{2}{3} \right)}^{2}}$

$\Rightarrow {{F}_{Centripetal}}=6.57N$

Maximum tension in the string, ${{T}_{\max }}=200N$ 

${{T}_{\max }}=\frac{mv_{\max }^{2}}{r}$

$\Rightarrow {{v}_{\max }}=\sqrt{\frac{{{T}_{\max }}\times r}{m}}$

$\Rightarrow {{v}_{\max }}=\sqrt{\frac{200\times 1.5}{0.25}}$

\[\Rightarrow {{v}_{\max }}=\sqrt{1200}=34.64m/s\]

Thus, the maximum speed of the stone is \[34.64m/s\].

22. If, in Exercise 21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: 

the stone moves radially out wards, 

the stone flies off tangentially from the instant the string breaks, 

the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle? 

Ans: $(ii)$ 

From the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. So, if the string breaks, the stone will move in the direction of the velocity at that instant. 

Therefore, the stone will fly off tangentially from the instant the string breaks. 

23. Explain why

a horse cannot pull a cart and run in empty space, 

Ans: In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. 

This reaction force causes the horse to move forward. An empty space is devoid of any such reaction force. 

Hence, a horse cannot pull a cart and run in empty space.

passengers are thrown forward from their seats when a speeding bus stops suddenly, 

Ans: If a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion).

So, the passenger’s upper body is thrown forward in the direction in which the bus was moving. 

it is easier to pull a lawn mower than to push it, 

Ans: While pulling a lawn mower, a force at an angle $\theta $  is applied on it, as shown in the following figure:

Resolution of Force while Pulling a Lawn Mower

The vertical component of this applied force acts upward. This reduces the effective weight of the mower. 

While pushing a lawn mower, a force at an angle $\theta $ is applied on it, as shown in the following figure:

Resolution of Force while pushing a Lawn Mower

In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower. 

As the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it. 

From Newton’s second law of motion: $F=ma=m\frac{\Delta v}{\Delta t}$  ........$(1)$ 

$F=$ Stopping force experienced by the cricketer as he catches the ball 

$m=$ Mass of the ball 

$t=$ Time of impact of the ball with the hand 

From equation $(1)$it can be observed that the impact force is inversely proportional to the impact time, i.e., $F<\frac{1}{\Delta t}$       ........$(2)$ 

Equation $(2)$  shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa. 

Therefore, it is easier to pull a lawn mower than to push it.

a cricketer moves his hands backwards while holding a catch.

Ans: While taking a catch, a cricketer moves his hand backward so as to increase the time of impact $\Delta t$. This in turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

Therefore, a cricketer moves his hands backwards while holding a catch.

Laws of Motion Chapter Summary - Class 11 NCERT Solutions

Force .

A force is something which changes or tends to change the state of rest or motion of a body. It causes a body to start moving if it is at rest or stop it, if it is in motion, or deflect it from its initial path of motion.

Force is a vector quantity having SI unit Newton (N) and dimension [MLT –2 ].

Types of Force 

There are, basically, four forces, which are commonly encountered in mechanics.

(a) Weight : Weight of an object is the force with which earth attracts it. It is also called the force of gravity or the gravitational force.

(b) Contact Force : When two bodies come in contact, they exert forces on each other that are called contact forces.

(i) Normal Force (N): It is the component of contact force normal to the surface. It measures how strongly the surfaces in contact are pressed together.

(ii)  Frictional Force (f) : It is the component of contact force parallel to the surface. It opposes the relative motion (or attempted motion) of the two surfaces in contact.

(c) Tension: The force exerted by the ends of a taut string, rope or chain is called the tension. The direction of tension is so as to pull the body while that of normal reaction is to push the body.

(d) Spring Force: Every spring resists any attempt to change its length; the more you alter its length the harder it resists. The force exerted by a spring is given by F = –kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm –1 ).

Newton’s Laws of Motion 

First Laws of Motion

(a) Everybody continues in its state of rest or of uniform motion in a straight line unless it is compelled by a resultant force to change that state

(b) A frame of reference in which Newton’s first law is valid is called an inertial frame, i.e., if a frame of reference is at rest or in uniform motion it is called inertial, otherwise non-inertial.

Second Laws of Motion

(a) This law gives the magnitude of force.

(b) According to second Laws of Motion, the rate of change of momentum of a body is directly proportional to the resultant force acting on the body, i.e.,

$\vec{F}\alpha \left ( \frac{d\vec{p}}{dt} \right )$

$\vec{F}=K\dfrac{d\vec{p}}{dt}$

Here, the change in momentum takes place in the direction of the applied resultant force. Momentum, $\vec{p}=m\vec{v}$  is a measure of the sum of the motion contained in the body.

Third Laws of Motion

(a) According to this law, for every action, there is an equal and opposite reaction. When two bodies A and B exert force on each other, the force by A on B (i.e., action represented by $\vec{F}_{BA}$ ), is always equal and opposite to the force by B on A (i.e., reaction represented $\vec{F}_{BA}$ ). Thus, $\vec{F}_{BA}=-\vec{F}_{BA}$ .

(b) The two forces involved in any interaction between two bodies are called action and reaction. But we cannot say that a particular force is action and the other one is reaction. Action and Reaction force always acts on different bodies.

Applications of Newton’s Laws of Motion

There are two kinds of problems in classical mechanics :

(a) To find unknown forces acting on a body, given the body’s acceleration.

(b) To predict the future motion of a body, given the body’s initial position and velocity and the forces acting on it. For either kind of problem, we use Newton’s second law . The following general strategy is useful for solving such problems :

(i) Draw a simple, neat diagram of the system.

(ii) Isolate the object of interest whose motion is being analysed. Draw a free body diagram for this object, that is, a diagram showing all external forces acting on the object. For systems containing more than one object, draw separate diagrams for each object. Do not include forces that the object exerts on its surroundings.

(iii) Establish convenient coordinate axes for each body and find the components of the forces along these axes. Now, apply Newton’s second law, $\sum \vec{F}=m\vec{a}$ , in component form. Check your dimensions to make sure that all terms have units of force.

(iv) Solve the component equations for the unknowns. Remember that you must have as many independent equations as you have unknowns in order to obtain a complete solution.

(v) It is a good idea to check the predictions of your solutions for extreme values of the variables. You can often detect errors in your results by doing so.

Friction is an opposing force that comes into play when one body actually moves (slides or rolls) or even tries to move over the surface of another body.

Thus, the force of friction is the force that develops at the surfaces of contact of two bodies and impedes (opposes) their relative motion.

Static Friction, Limiting Friction and Kinetic Friction

The opposing force that comes into play when one body tends to move over the surface of another, but the actual relative motion has yet not started is called Static friction.

Limiting friction is the maximum opposing force that comes into play, when one body is just at the verge of moving over the surface of the other body.

Kinetic friction or dynamic friction is the opposing force that comes into play when one body is actually moving over the surface of another body.

Coefficient of Static Friction

We know that, $f_{ms}\alpha N$ or $f_{ms}=\mu _{s}N$ or $\mu _{s}=\frac{f_{ms}}{N}$  ...(2)

Here, μ s is a constant of proportionality and is called the coefficient of static friction. Thus : Coefficient of static friction for any pair of surfaces in contact is equal to the ratio of the limiting friction and the  normal reaction. μ s , being a pure ratio, has got no units and its value depends upon the nature of the surfaces in contact. Further, μ s , is usually less than unity and is never equal to zero.

Since the force of static friction (f s ) can have any value from zero to maximum (f ms ), i.e. f s < f ms , eqn. (2) is generalised to f s < μ s N

Kinetic Friction

The laws of kinetic friction are exactly the same as those for static friction. Accordingly, the force of kinetic friction is also directly proportional to the normal reaction, i.e.,

$f_{k}\alpha N$ or $f_{k}=\mu _{k}N$ ...(4)

μ k is coefficient of kinetic friction.  μ k < μ s .

Circular Motion

It is the movement of particles along the circumference of a circle.

The uniform circular motion is that in which the particle is moving at a constant speed on a circular path.

The non-uniform circular motion is that in which the particles move with variable speed on its circular path.

Variables in Circular Motion

Angular Displacement: It is the angle subtended by the position vector at the centre of the circular path. Angular displacement, $\Delta \theta =\Delta s/r$ where, $\Delta s$ is the arc length and r is the radius

Angular Velocity: The time rate of change of angular displacement $(\Delta \theta )$ is called angular velocity.

Angular velocity, $\omega =\Delta \theta /\Delta t$

Angular velocity is a vector quantity

Relation between linear velocity (v) and angular velocity $(\omega )$ is given by

$v=r\omega $

Angular Acceleration: The rate of change of angular velocity is called angular acceleration.

Angular acceleration,

$\alpha =\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \omega }{\Delta t}=\dfrac{d\omega }{dt}=\dfrac{d^{2}\theta }{dt^{2}}$

Its SI unit is rad/s 2 and dimensional formula is [T -2 ]

Acceleration in a circular motion has two components, as given below:

(a) Tangential acceleration is the change in magnitude of linear velocity and act along tangent to the circular path. It is given by:

$\alpha _{T}=r\alpha $

(b) Radial Acceleration is the change in direction of linear velocity and acts along the radius towards the centre of circle. It is given by $\alpha _{r}=\dfrac{v^{2}}{r}=\omega ^{2}r$

It is also called centripetal acceleration.

Relation between linear acceleration (a) and angular acceleration $(\alpha )$ , 

$a=r\alpha$ , where r = radius

Relation between angular acceleration $(\alpha )$ and linear velocity (v)

$\alpha =\dfrac{v^{2}}{r}$

Centripetal and Centrifugal Force

Centripetal Force: In uniform circular motion, the force acting on the particle along the radius and towards the centre keeps the body moving along the circular path. This force is called centripetal force.

Centrifugal Force: The pseudo force experienced by a particle performing uniform circular motion due to an accelerated frame of reference which is along the radius and directed way from the centre is called centrifugal force.

Motion of a Car on a Plane  Circular Road

For motion without skidding

$\dfrac{Mv^{2}max}{r}=\mu M_{g}$

$\Rightarrow v_{max}\sqrt{\mu rg}$

Motion on a Banked Road

Angle of banking $=\theta$

$tan\theta =\dfrac{h}{b}$ 

Maximum safe speed at the bend $v_{max}=\left [ \dfrac{rg(\mu +tan\theta )}{1-(\mu tan\theta )} \right ]^{1/2}$

If friction is negligible $v_{max}=\sqrt{rg \tan\theta }=\sqrt{\frac{rhg}{b}}$ and $tan\theta =\dfrac{v^{2}_{max}}{rg}$

Motion of Cyclist on a Curve

In equilibrium angle with vertical is $\theta $ , then $tan\theta =\dfrac{v^{2}}{rg}$

Maximum safe speed $=v_{max}=\sqrt{\mu rg}$

Overview of Deleted Syllabus for CBSE Class 11 Physics Laws of Motion

NCERT Class 11 Physics Chapter 4 Solutions on Laws of Motion provided by Vedantu provide a comprehensive understanding of the principles that govern the motion of objects. The solution cover exercise in the NCERT textbook offers clear explanations and step-by-step methods for solving problems. Important points to focus on include Newton's First Law or the Law of Inertia, Newton's Second Law, Newton's Third Law, the importance of free-body diagrams for visualizing forces, conditions for equilibrium of forces, and the analysis of motion on inclined planes and circular paths. From previous year's question papers, typically around 6–7 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills.

Other Study Material for CBSE Class 11 Physics Chapter 4

Chapter-Specific NCERT Solutions for Class 11 Physics

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 Class 11 Physics Chapter 4 Laws of Motion

1. Why should I refer to the NCERT Solutions for Chapter 5 Laws of Motion of Class 11 Physics?

NCERT Solutions for Class 11 Physics Chapter 5- Laws of Motion will help students understand and answer correctly the questions provided at the end of the chapter by NCERT. These questions and exercises are important since many of them can be asked in the exams. NCERT Solutions are very useful in case you face difficulty in grasping the concept behind any questions. Referring to the solutions will provide you with an idea as to how to solve any given question.

2. What are the concepts covered in Chapter 5 of NCERT Solutions for Class 11 Physics?

Chapter 5 covers the fundamental concepts based on Newton’s laws. These concepts can be correlated to nearly all activities in our daily lives. This chapter talks about one of the most basic and important concepts that is taught in Class 11 Physics. The NCERT Solutions of chapter 5 in Class 11 Physics will help students answer questions that are based on the laws of motion and other related concepts in the chapter.

3. What are the most important topics in Class 11 Physics Chapter 5?

Chapter 5 - Laws of Motion is one of the most important chapters in Mechanics. It talks about the three of Newton’s Laws which are certainly the most important topic that has been covered in this chapter. However, there are more topics that hold high importance and need focus when preparing for your Class 11 Physics exam. These topics are Conservation of Momentum, Problem-Solving in Mechanics, and Circular Motion.

4. Where can I get the NCERT Solutions for Class 11 Physics Chapter 5?

You can find NCERT Solutions for Class 11 Physics Chapter 5 - Laws of Motion on  Vedantu  app or website. Experts at Vedantu have carefully designed these solutions to help students achieve in-depth knowledge while answering the questions provided in the NCERT for Class 11 Physics. These solutions have been framed based on the latest syllabus provided by CBSE. Access and download the NCERT Solutions for Class 11 Physics Chapter 5 free of cost.

5. How do I score well in Class 11 Physics Chapter 5?

Scoring well in Class 11 Physics Chapter 5 simply means having proper knowledge of all concepts that have been taught throughout the chapter and being able to use this knowledge in solving all the questions based on this chapter. Students must regularly practice and revise everything they study in school. Making notes of important details and formulas can be very helpful during your preparation for the Class 11 Physics exams.

6. What are the important questions in Laws of Motion class 11 NCERT Solutions?

The important questions in the laws of motion class 11 ncert solutions often revolve around the fundamental principles and applications of Newton's laws. Key questions include:

Deriving and explaining Newton's three laws of motion.

Problems involving free-body diagrams and calculating net forces.

Questions on equilibrium of forces, including both static and dynamic situations.

Analysing motion on inclined planes and circular motion.

Real-life applications and conceptual questions about action-reaction pairs.

7. What are the two marks questions in Class 11 Physics Laws of Motion NCERT Solutions?

Two-mark questions in the Laws of Motion Class 11 Solutions typically involve concise explanations or simple problem-solving exercises. Examples include:

Stating and explaining each of Newton's laws.

Drawing and interpreting free-body diagrams for simple systems.

Short problems involving the calculation of force, mass, or acceleration using $F=ma$.

Explain the concept of inertia and give examples.

Briefly discussing applications of Newton’s third law in everyday scenarios.

8. How many types of Newton's law are there in Class 11 Physics Ch 4 NCERT Solutions?

There are three types of Newton's laws of motion mentioned in laws of motion class 11 ncert solutions:

Newton's First Law (Law of Inertia): An object will remain at rest or in uniform motion unless acted upon by an external force.

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass $F=ma$.

Newton's Third Law: For every action, there is an equal and opposite reaction.

9. How is the concept of inertia explained in Class 11 Physics Chapter 4 Exercise Solutions?

According to class 11 physics laws of motion ncert solutions, Inertia is the tendency of an object to resist changes in its state of motion. It is directly related to the mass of an object; the greater the mass, the greater the inertia. The concept is explained through examples such as why passengers lurch forward in a sudden stop or why a heavy object is harder to move than a lighter one.

10. What are some real-life applications of Newton’s Third Law discussed in Class 11 Physics Chapter Laws Of Motion NCERT Solutions?

Real-life applications of Newton’s Third Law in laws of motion class 11 solutions include:

The propulsion of rockets and jet engines, where exhaust gases are expelled backward, results in a forward thrust.

The recoil experienced when firing a gun.

Swimming, where the swimmer pushes water backward, propelling themselves forward.

11. How do the Laws of Motion NCERT Solutions for Class 11 Physics explain the use of free-body diagrams?

Free-body diagrams are visual tools used to represent the forces acting on an object. The Ch 4 Physics Class 11 NCERT Solutions explains how to draw these diagrams, identify forces such as tension, normal force, gravitational force, and friction, and use them to solve problems involving equilibrium and motion.

12. What is the importance of equilibrium in the Laws of Motion Class 11 Physics Chapter 4 NCERT Solutions?

Equilibrium is a state where the net force acting on an object is zero. The class 11 physics ch 4 ncert solutions discusses two types of equilibrium:

Static Equilibrium: Where an object is at rest and remains at rest.

Dynamic Equilibrium: Where an object is moving with constant velocity. Understanding these concepts is essential for analysing situations where forces balance each other.

NCERT Solutions for Class 11 Physics

• connect with us
• 1800-572-9877
• [email protected]
• We’re on your favourite socials!

Frequently Search

Couldn’t find the answer? Post your query here

• Engineering Exams
• JEE Main Exam
• JEE Main News & Updates
• Laws of Motion JEE Main Questions 2025: Important Practice Questions with PYQs

Updated On: September 16, 2024 11:00 am IST | JEE Main

Laws of Motion carries an average weightage of 6.6% in the JEE Mains. Candidates can expect 2-3 questions from this chapter. Students can access solved problems, and previous years' Laws of Motion JEE Main questions.  

What To Study in Laws of Motion?

• JEE Main Laws of Motion Important Questions 2025 

Laws of Motion JEE Main Questions 2025: The Laws of Motion chapter is one of the important topics of the JEE Main Physics syllabus. This chapter has a decent weightage in the JEE Mains 2025 exam as per last year's trends. Candidates should solve the questions and evaluate their performance. Preparing all the chapters with high weightage properly will help the candidate score well on the JEE Main exam. The authorities will soon release the syllabus for the JEE Main 2025. Some of the Laws of Motion topics from which questions can be asked are Force & Inertia, Newton’s first/second/third Law of Motion, the Law of Conservations of Linear Momentum (its applications), Concurrent forces and Equilibrium, Kinetic and Static Friction, and Laws of Fiction, and others.

Students should solve all the questions provided here to practice the Laws of Motion chapters from the Physics Syllabus of JEE Main 2025 . Before we move on to the important questions for this chapter, have a look at the topics that need to be focused on while studying this chapter.

Here are the topics that applicants should not miss while their preparation for this chapter.

• Force & interia
• First/second/third law of Newton
• Law of conservation of linear momentum (its applications)
• Concurrent forces and equilibrium
• Laws of friction
• Kinetic and static friction
• Rolling friction
• Uniform circular motion dynamics
• Centripetal force (its applications)

JEE Main Laws of Motion Important Questions 2025

Check the practice important questions from previous years for the chapters ‘laws of motion.’

• While peddling a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
• in the backward direction on the front wheel and in the forward direction on the rear wheel
• in the forward direction on the front wheel and in the backward direction on the rear wheel
• in the backward direction on both the front and rear wheels
• in the forward direction on both the front and rear wheels

Correct Answer: (a)

• A car is moving in a circular horizontal track of radius 10 m with a constant speed of lOm/s. A plumb bob is suspended from the roof of the car by a light, rigid rod. The angle made by the rod with the track is (g = 10 m/s2).

Correct Answer: (c)

• A block of mass 0.1 kg is held against a wall, applying a horizontal force of 5N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is
• A block of mass 2 kg rests on a rough inclined plane, making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is
• 0.7 x 9.8 x√3N
• 0.7 x 9.8 N
• A ship of mass 3 x 107 kg initially at rest is pulled by a force of 5 x 104 N through a distance of 3 m. Assuming that the resistance due to water is negligible, the speed of the ship is
• If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
• linear momentum of the system does not change in time
• the kinetic energy of the system does not change in time
• angular momentum of the system does not change in time
• the potential energy of the system does not change in time
• A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

Correct Answer: (b)

• Two blocks of mass m1 = 10kg and m2 = 5kg, connected to each other by a massless, inextensible string of length 0.3 m, are placed along the diameter of a turntable. The coefficient of friction between the table is 0.5, while there is no friction between rr^ and the table. The table is rotating with an angular velocity of lOrad/s about a vertical axis passing through its center O. The masses are placed along the diameter of the table on either side of the center O such that the mass m1 is at a distance of 0.124m from O.The masses are observed to be at rest with respect to an observer on the turntable.
• Calculate the frictional force on m1.
• What should be the minimum angular speed of the turntable so that the masses will slip from this position?
• How should the masses be placed with the string remaining taut so that there is no frictional force acting on the mass m1?
• As per the given figure, a weightless pulley P is attached to a double-inclined frictionless surface. The tension is the string (massless) will be (if g = 10m/s2)

• Two bodies of mass m1=5kg and m2=3Kg are connected by a light string going on a smooth light pulley a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m1 will be:

• Two masses of m1=5kg and m2=10kg connected via an inextensible string over a frictionless pulley are moving as shown in the figure. The coefficient of friction of a horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop motion is:

• A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocity vs time?

Applicants can judge their performance before the JEE Main 2025 by solving these questions above. For more such updates, stay tuned with CollegeDekho. Also Check:

For more articles and updates on Laws and Motion JEE Main Questions 2025, stay tuned with Collegedekho !

Are you feeling lost and unsure about what career path to take after completing 12th standard?

Say goodbye to confusion and hello to a bright future!

Although maximum candidates usually require one or two years to prepare for the JEE Main and the JEE Advanced exam, candidates can also ace these exams with a tight study routine of 6 months.

HC Verma is considered to be a good book for the JEE Main preparation. It comprises good explanations and many practice problems. However, it is not enough for the preparations and candidates are required to study the NCERT textbooks as well.

The physics section in the JEE Main exam is typically of moderate level of difficulty. It comprises a mix of numerical and concept-based questions.

Candidates must keep in mind that there is no requirement of 75% marks in class for appearing in the JEE Main exam. The JEE Main eligibility criteria of minimum 75% marks in class 12 is required at the time of admission in NITs, IIITs and GFTIs. Candidates can apply and appear in the JEE Main exam irrespective of their class 12 marks.

The JEE Main exam has a total of 90 questions from all three subjects. The JEE Main exam has 30 questions each from the subjects of Physics, Chemistry and Mathematics. Out of 30 questions, there are 20 multiple choice questions and 10 are questions with numerical value answers.

The official syllabus of JEE Main entrance exam and NCERT books cover almost all topics, and it is preferred by the candidates for their preparation. The syllabus of JEE Main and that of CBSE's Class 11 and 12 board is almost the same. Candidates can consider NCERT books as one of the best reference materials for the JEE Main exam.

Some NITs and IIITs which candidates can get with 35000 rank in the JEE Main entrance exam are Indian Institute of Information Technology Kalyani, National Institute of Technology Mizoram,  Indian Institute of Information Technology Dharwad, National Institute of Technology Nagaland, and National Institute of Technology Manipur.

The exact marks to be obtained for getting the 99 percentile in the JEE Main exam depends on the difficulty level of the paper. If the JEE Main question paper turns out to be easy, then marks secured by most of the candidates are good and the higher percentile is relative and gets pushed for higher marks.

No, 40 is regarded as  a very good score in the JEE Main exam. Since 85 to 95 percentile is considered to be good for admission into NITs, and IITs, securing anything less than that is not a good score for the General, OBC, and EWS candidates. With this score, such candidates can barely make it through a few engineering colleges of India.

Was this article helpful?

Related questions, cut off mark for b tech computer science, be computer engg. admission.

Hi there, admission for most of the program has closed right now. You can visit website or reach out to LPU officials for more details. LPUU is one of he top ranking university with NAAC A ++ accreditation. GOod Luck

Vacant post for spot round of VIT

The vacant seat matrix with reference to the spot round at Vishwakarma Institute of Technology VIT Pune , was updated on September 11, 2024. The spot round is conducted for filling the remaining vacant seats after the CAP (Centralized Admission Process) rounds, and the exact number of vacancies for the various branches can be found on their official website www.vit.edu.

Do you have a question? Ask us.

Typical response between 24-48 hours

Get personalized response

Free of Cost

Access to community

Similar Articles

• Rotational Motion Weightage in JEE Main 2025
• Work Energy and Power JEE Main Questions 2025: Important Practice Questions with PYQs
• Laws of Motion Weightage in JEE Mains 2025
• Units and Measurement JEE Main Questions 2025: Important Practice Questions with PYQs
• Units and Dimensions Weightage in JEE Main 2025
• Kinematics JEE Main Questions 2025: Important Practice Questions with PYQs

Recent Articles

• Kinematics Weightage in JEE Mains 2025
• Work Energy and Power Weightage in JEE Main 2025
• Matrices Weightage Marks in TG EAMCET 2025
• REAP (Rajasthan) B.Tech Admission 2024: Admission (Closed), Merit List, Selection Process
• Polytechnic Colleges in Bihar
• Polytechnic Colleges in Chennai
• Polytechnic Colleges in Coimbatore
• GUJCET 2025: Exam Date (March 23), Syllabus, Eligibility, Exam Pattern
• GUJCET 2025 Syllabus: Check Subject Wise, Download PDF
• GUJCET 2025 Exam Date (Released)
• How to Become AI Engineer
• Polytechnic Colleges in Bangalore
• Polytechnic Colleges in Lucknow: Govt Colleges, Private Colleges, Top Courses
• Algebra Weightage in TG EAMCET 2025
• BTech Admission through COMEDK Counselling 2024: Seat Allotment Round 3 (Out), Pay Fees (Till Sept 13), Reporting (Till Sept 14)
• List of Documents Required for COMEDK Counselling 2024
• Polytechnic Colleges in Hyderabad
• List of Algebra Topics in TG EAMCET 2025
• Tier 2 Engineering Colleges in India
• Tier 1 Engineering Colleges in India
• Polytechnic College in Patna
• Autonomous engineering colleges in Pune
• Autonomous Engineering Colleges In Bangalore

Recent News

• Karnataka PGCET 2024 on September 18: Exam day instructions, bell timings
• BCECE Engineering Round 1 Seat Allotment Result 2024 Download Link Activated
• TS DOST Special Drive Seat Allotment Result 2024 (Released): Download link, reporting dates
• Karnataka PGCET Admit Card 2024 (Released): ME, M.Tech, M.Arch hall ticket download link
• Maharashtra DSE CAP Round 3 Cutoff 2024: Final admission closing ranks
• TS DOST Special Drive Allotment Expected Release Time 2024
• TS PGECET First Phase Allotment Expected Release Time 2024
• BCECE Engineering Revised Counselling Dates 2024 Released: 1st and 2nd round for remaining vacant seats
• BITSAT Cutoff 2024 Released: Check campus and course-wise final cutoff scores
• COMEDK Round 3 Seat Allotment Result 2024 (Released): Final allotment download link, reporting dates
• COMEDK Round 3 Cutoff 2024 (Released): GM, KKR engineering final closing ranks
• KCET Round 2 Vacant Seats 2024 (Released): Vacancy matrix after first allotment
• KCET Round 1 DSCE Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 SJCE Mysore Cutoff 2024 Course-Wise
• KCET Round 1 MSRIT Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 PES Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 BMS Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 KLE Technological Institute Cutoff 2024 Course-Wise
• KCET Round 1 RVCE Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 Result 2024 (Released): First seat allotment out at cetonline.karnataka.gov.in
• KCET Round 1 BIT Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 NIE Mysore Cutoff 2024 Course-Wise
• KCET Round 1 SIT Tumkur Cutoff 2024 Course-Wise
• KCET Round 1 NMIT Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 NHCE Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 Dr AIT Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 RNSIT Bangalore Cutoff 2024 Course-Wise
• KCET Round 1 Cutoff 2024 (Available): Course-wise GM and HKR closing ranks
• KCET B.Tech Round 1 Cutoff 2024 (Released): Engineering closing ranks for all institutes
• KCET Round 2 Option Entry 2024 (Released): Direct link to edit options

Trending Now

Subscribe to CollegeDekho News

Top 10 engineering colleges in india.

• Approved by: AICTE, BCI, NCTE, Government of Gujarat
• Type: Private UnAided
• Get Free Counselling
• Approved by: AICTE
• Type: Local Body
• Download Brochure
• Approved by: NAAC
• Type: Government
• Approved by: AICTE, NBA
• Approved by: UGC, AICTE
• Approved by: AICTE, NAAC
• Approved by: UGC, NBA, AICTE

Popular Degrees

• B.Arch. (Bachelor of Architecture)
• B.Tech. (Bachelor of Technology)
• M.Tech. (Master of Technology)
• B.Tech. + M.Tech.

CollegeDekho's expert counsellors can help you with all your doubts

• Enter a Valid Name
• Enter a Valid Mobile
• Enter a Valid Email
• By proceeding ahead you expressly agree to the CollegeDekho terms of use and privacy policy

Tell us your JEE Main score & access the list of colleges you may qualify for!

Details Saved