case study based questions class 11 physics chapter 5

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Class 11 Physics Case Study Questions

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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)

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Class 11 Physics Case Study Questions PDF Download

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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.

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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.

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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

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Important Questions for CBSE Class 11 Physics Chapter 5 - Work, Energy and Power

• Class 11 Important Question
• Chapter 6: Work, Energy And Power

CBSE Class 11 Physics Chapter 5 Important Questions - Free PDF Download

Important Questions for Class 11 Physics Chapter 5 - Work, Energy and Power are available in Vedantu. All Questions are designed as per the latest Syllabus of NCERT with reference to frequent questions in exams. We hear the words 'work,' 'energy,' & 'power' all the time. A person carrying materials, a farmer cultivating, a student studying for exams are all said to be performing their work. Work has a specific and definite meaning in physics. In this article we will Solve lot of important questions that will definitely help students to score good marks in final exams. Students can use these solutions for learning the most important concepts and prepare for their board exams.

Download CBSE Class 11 Physics Important Questions 2024-25 PDF

Also, check CBSE Class 11 Physics Important Questions for other chapters:

Topics Covered in Class 11 Physics Chapter 5 - Work, Energy and Power are as follows:

The work-energy theorem

Kinetic energy

Work done by a variable force

The work-energy theorem for a variable force

The concept of potential energy

The conservation of mechanical energy

The potential energy of a spring

Law of conservation of energy

Study Important Questions for Class 11 Physics Chapter 5 - Work, Energy and Power

Very Short Answer Questions                                                                         1 Mark

1. If two bodies stick together after collision, will the collision be elastic or inelastic?

Ans : If two bodies stick together after the collision the collision will be an inelastic collision.

2. When an air bubble rises in water, what happens to its potential energy?

Ans: When an air bubble rises in water, the potential energy of the air bubble decreases because work is done by upthrust on the bubble.

3. A spring is kept compressed by pressing its ends together lightly. It is then placed in a strong acid, and released. What happens to its stored potential energy?

Ans: When spring is kept compressed by pressing its ends together lightly and further placed in strong acid and released, t he loss in potential energy appears as kinetic energy of the molecules of the acid.

4. Define the triple point of water.

Ans: Triple point of water indicates the values of pressure and temperature at which water co-exists in equilibrium in all the three states of matter.

5. State Dulong and Petit law.

Ans: According to Dulong and Petit law, the specific heat of all the solids is constant at room temperature and is equal to \[3R\].

6. Why are the clock pendulums made of invar, a material of low value of coefficient of linear expansion?

Ans: The clock pendulums are made of Inver as it has a low value of α (coefficient of linear expansion) i.e., for a small change in temperature, there won’t be much change in the length of the pendulum.

7. Why is mercury used in making thermometers?

Ans: Mercury is used in making thermometers as it has a wide and useful temperature range and has a uniform rate of expansion.

8. How would a thermometer be different if glass expanded more with increasing temperature than mercury?

Ans: If glass expanded more with increasing temperature than mercury, the scale of the thermometer would be upside down.

9. Show the variation of specific heat at constant pressure with temperature.

Ans: Here is the required variation:

10. Two thermometers are constructed in the same way except that one has a spherical bulb and the other an elongated cylindrical bulb. Which one will respond quickly to temperature change?

Ans: The thermometer with a cylindrical bulb will respond quickly to temperature changes as the surface area of the cylindrical bulb is greater than that of a spherical bulb.

11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \[F=-\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,N\]. Where \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,,\overset{\wedge }{\mathop{k}}\,\] are unit vectors along the x, y and z axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z axis?

Ans: Force exerted on the body is \[F=-\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,N\].

Displacement, \[s\text{ }=4\overset{\wedge }{\mathop{k}}\,\text{ }m\]

Work done, \[W\text{ }=F.s\]

\[=\left( -\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right).\left( 4\overset{\wedge }{\mathop{k}}\, \right)\]

$=12\text{ }J$

Thus, \[12\text{ }J\] of work is done by the force on the body.

12. A molecule in a gas container hits a horizontal wall with speed and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Ans: Yes; the collision is elastic. The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic. The gas molecule moves with a velocity of \[200\text{ }m/s\] and strikes the stationary wall of the container and rebounds with the same speed. This shows that the rebound velocity of the wall remains zero. Thus, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

13. The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Ans: Bob A won’t rise at all. For an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. Here a complete transfer of momentum takes place from the moving mass to the stationary mass. Thus, bob A of mass m, will come to rest after colliding with bob B of equal mass, while bob B will move with the velocity of bob A at the instant of collision.

14. A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg/s. What is the speed of the trolley after the entire sand bag is empty?

Ans: The sand bag is placed on a trolley that is moving with a uniform speed of \[27\text{ }km/h\]. The external forces acting on the system of the sandbag and the trolley is zero. As the leaking action does not produce any external force on the system, the sand starts leaking from the bag with no change in the velocity of the trolley. This is in accordance with Newton's first law of motion. Thus, the speed of the trolley will remain \[27\text{ }km/h\].

15. Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Ans:   From the figure given in the question, (i), (ii), (iii), (iv), and (vi) cannot possibly describe the elastic collision of two billiard balls. We know that the potential energy of a system of two masses is inversely proportional to the separation between them. Here the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero (i.e., \[V\left( r \right)\text{ }=\text{ }0\]) when the two balls touch each other, i.e., at \[r\text{ }=\text{ }2R\], where R is the radius of each billiard ball. The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions. Thus, they do not describe the elastic collisions between them.

Short Answer Questions                                                                                2 Marks

1. A body is moving along Z – axis of a co – ordinate system is subjected to a constant force F is given by $\overset{\to }{\mathop{F}}\,=-\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,N$. Where \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,,\overset{\wedge }{\mathop{k}}\,\] are unit vector along the x, y and z – axis of the system respectively. What is the work done by this force in moving the body a distance of 4m along the Z – axis?

Ans: Force exerted on the body is,

 \[F=-\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,N\].

2. A ball is dropped from the height \[{{h}_{1}}\] and if re-bounces to a height \[{{h}_{2}}\]. Find the value of coefficient of restitution.

Ans: As the ball drops from height \[{{h}_{1}}\] ,

Velocity of approach will be ${{v}_{1}}=\sqrt{2g{{h}_{1}}}$

And the ball rebounds to height \[{{h}_{2}}\] .

Therefore, velocity of separation is ${{v}_{2}}=\sqrt{2g{{h}_{2}}}$

Coefficient of restitution is given by:

$e=\frac{{{v}_{2}}}{{{v}_{1}}}=\frac{\sqrt{2g{{h}_{2}}}}{\sqrt{2g{{h}_{1}}}}$

$\Rightarrow e=\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}$

3. State and prove work energy theorem analytically.

Ans: The work-energy theorem states that work done by force acting on a body is equal to the change produced in its kinetic energy.

If $\vec{F}$ force is applied to move an object through a distance $d\vec{s}$.

Then $dw=\vec{F}.d\vec{s}$

$F=m\vec{a}$

$dw=m\vec{a}.d\vec{s}$

$dw=m\frac{d\vec{v}}{dt}.d\vec{s}$

\[dw=m\frac{d\vec{s}}{dt}.d\vec{v}\]

\[\Rightarrow dw=mv.d\vec{v}\]

On integrating,

\[\sum{dw}=W=\int\limits_{u}^{v}{mv.d\vec{v}}\]

\[W=m\left| \frac{v}{2} \right|_{u}^{v}\]

\[\Rightarrow W=\frac{m{{v}^{2}}}{2}-\frac{m{{u}^{2}}}{2}\]

Hence \[W\text{ }=\text{ }{{K}_{f}}\text{ }\text{ }{{K}_{i}}\] Where \[{{K}_{f}}\] and \[{{K}_{i}}\] are final and initial kinetic energy.

4. An object of mass 0.4kg moving with a velocity of 4m/s collides with another object of mass 0.6kg moving in the same direction with a velocity of 2m/s. If the collision is perfectly inelastic, what is the loss of K.E. due to impact?

Ans: In the above question it is given that:

\[{{m}_{1\text{ }}}=\text{ }0.4kg\],

\[{{u}_{1}}\text{ }=\text{ }4m/s\], 

\[{{m}_{2\text{ }}}=\text{ }0.6kg\] And

\[{{u}_{2\text{ }}}=\text{ }2m/s\].

Total K.E. before collision is given by:

${{K}_{i}}=\frac{1}{2}{{m}_{1}}{{u}_{1}}^{2}+\frac{1}{2}{{m}_{2}}{{u}_{2}}^{2}$

\[\Rightarrow {{K}_{i}}=\frac{1}{2}\left( 0.4 \right){{\left( 4 \right)}^{2}}+\frac{1}{2}\left( 0.6 \right){{\left( 2 \right)}^{2}}\]

\[\Rightarrow {{K}_{i}}=4.4J\]

Since collision is perfectly inelastic,

$v=\frac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}=2.8m/s$

Total K.E. after collision is given by:

${{K}_{f}}=\frac{1}{2}\left( {{m}_{1}}+{{m}_{2}} \right){{v}^{2}}$

$\Rightarrow {{K}_{f}}=\frac{1}{2}\left( 0.4+0.6 \right){{2.8}^{2}}$

$\Rightarrow {{K}_{f}}=3.92J$

Thus, loss in K.E. will be:

$\Delta K={{K}_{i}}-{{K}_{f}}$

$\Rightarrow \Delta K=4.4-3.92=0.48J$

5. Why does the density of solid/liquid decreases with rise in temperature?

Ans: Consider P to be Density of solid/liquid at temperature T.

\[{{P}_{1}}\text{ }=\text{ }Density\text{ }of\text{ }solid/liquid\text{ }at\text{ }Temperature\text{ }T+\Delta T\].

As, \[Density=\frac{Mass}{Volume}\]

\[P=\frac{M}{V}\] …… (1)

\[{{P}_{1}}=\frac{M}{{{V}_{1}}}\] …… (2)

\[{{V}_{1}}=\text{ }Volume\text{ }of\text{ }solid\text{ }at\text{ }temperature\text{ }T\text{ }+\text{ }\Delta T\]

\[V\text{ }=\text{ }Volume\text{ }of\text{ }solid\text{ }at\text{ }temperature\text{ }T\]

As on increasing the temperature, solids/liquids expand, that is their volume increases, so by equation 1) & 2) Density is inversely proportional to volumes. Therefore, if the volume increases on increasing the temperature, Density will decrease.

6. Two bodies at different temperatures \[{{\mathbf{T}}_{\mathbf{1}}}\], and \[{{\mathbf{T}}_{\mathbf{2}}}\] are brought in thermal contact do not necessarily settle down to the mean temperature of \[{{\mathbf{T}}_{\mathbf{1}}}\] and \[{{\mathbf{T}}_{\mathbf{2}}}\] . Why?

Ans: When two bodies at different temperatures \[{{\mathbf{T}}_{\mathbf{1}}}\] and \[{{\mathbf{T}}_{\mathbf{2}}}\] are in thermal contact. They do not always settle at their mean temperature because the thermal capacities of two bodies may not always be equal.

7. The resistance of certain platinum resistance thermometer is found to be \[2.56\text{ }\Omega \] at ${{0}^{\circ }}C$ and \[3.56\text{ }\Omega \] at \[{{1000}^{\circ }}C\]. When the thermometer is immersed in a given liquid, its resistance is observed to \[5.06\text{ }\Omega \]. Determine the temperature of liquid?

\[{{R}_{o}}\text{ }=\text{ }Resistance\]at  ${{0}^{\circ }}C=2.56\Omega $ 

\[{{R}_{t}}\text{ }=\text{ }Resistance\text{ }at\text{ }temperature\text{ }T\text{ }=\text{ }{{1000}^{\circ }}\text{C is 3}.56\text{ }\Omega \]

\[{{R}_{t}}\text{ }=\text{ }Resistance\text{ }at\text{ }unknown\text{ }temperature\text{ }t\];

\[{{R}_{t\text{ }}}=\text{ }5.06\Omega \]

As $t=\frac{{{R}_{t}}-{{R}_{0}}}{{{R}_{100}}-{{R}_{0}}}\times 100$

$\Rightarrow t=\frac{5.06-2.56}{3.56-2.56}\times 100={{250}^{\circ }}C$

8. A ball is dropped on a floor from a height of 2cm. After the collision, it rises up to a height of 1.5m. Assuming that 40% of mechanical energy lost goes to thermal energy into the ball. Calculate the rise in temperature of the ball in the collision. Specific heat capacity of the ball is 800J/k. Take \[g\text{ }=\text{ }10m/{{s}^{2}}\].

\[Initial\text{ }height\text{ }=\text{ }{{h}_{1}}=2m\]

\[Final\text{ }height\text{ }=\text{ }{{h}_{2}}=1.5m\]

\[Since\text{ }potential\text{ }energy\text{ }=\text{ }mechanical\text{ }energy\text{ }for\text{ }a\text{ }body\text{ }at\text{ }rest\text{ }as\text{ }K.E\text{ }=0\]

\[Mechanical\text{ }energy\text{ }lost\text{ }=\left| mg\left( {{h}_{1}}-{{h}_{2}} \right) \right|\]

$=\left| 1\times 10\left( 105-2 \right) \right|=5J$

Now \[\left( mechanical\text{ }energy\text{ }lost \right)\text{ }\times \text{ }40%\text{ }=\text{ }heat\text{ }gained\text{ }by\text{ }ball\]

$\frac{40}{100}\times 5=Cm\Delta T$

$\Rightarrow \frac{40}{100}\times 5=800\times 1\times \Delta T$

Therefore, 

\[\Delta T=\frac{100\times 800}{40\times 5}=2.5\times {{10}^{-3}}{{\text{ }}^{o}}C\], which is the required rise in temperature.

9. A thermometer has wrong calibration. It reads the melting point of ice as \[\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{0}}}\mathbf{C}\]. It reads \[\mathbf{6}{{\mathbf{0}}^{\mathbf{0}}}\mathbf{C}\] in place of \[\mathbf{5}{{\mathbf{0}}^{\mathbf{0}}}\mathbf{C}\]. What is the temperature of boiling point of water on the scale?

Lower fixed point on the wrong scale is \[\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{0}}}\mathbf{C}\].

Let \[n\text{ }=\text{ }no.\text{ }divisions\text{ }between\text{ }upper\text{ }and\text{ }lower\text{ }fixed\text{ }points\text{ }on\text{ }this\text{ }scale\]. 

If \[Q\text{ }=\text{ }reading\text{ }on\text{ }this\text{ }scale\], then

$\frac{C-0}{100}=\frac{Q-\left( -10 \right)}{n}$

Now, \[C\text{ }=\text{ }Incorrect\text{ }Reading\text{ }=\text{ }{{60}^{0}}C\]

\[Q\text{ }=\text{ }Correct\text{ }Reading\text{ }=\text{ }{{50}^{0}}C\]

$\frac{50-0}{100}=\frac{60-\left( -10 \right)}{n}$

$\frac{50}{100}=\frac{70}{n}$

\[n\text{ }=\text{ }140\]

On the Celsius scale, Boiling point of water is \[{{100}^{0}}C\]

$\frac{100-0}{100}=\frac{Q+10}{140}$

\[Q\text{ }=\text{ }{{130}^{0}}C\], which is the temperature of boiling point of water on this scale.

10. Write the advantages and disadvantages of platinum resistance thermometer.

Ans: Advantages of Platinum Resistance thermometer are as follows:

High accuracy of measurement

Measurements of temperature can be made over a wide range of temperature i.e., from \[{{260}^{0}}C\text{ }to\text{ }{{1200}^{0}}C\].

      Disadvantages of Platinum Resistance thermometer are as follows:

Requires additional equipment such as bridge circuit, Power supply etc.

11. If the volume of block of metal changes by 0.12% when it is heated through 200C. What is the co-efficient of linear expansion of the metal?

Ans:   The co-efficient of cubical expansion y of the metal is given by:

$Y=\frac{1}{V}\times \frac{\Delta V}{\Delta T}$

$\Rightarrow Y=\frac{\Delta V}{V}\times \frac{1}{\Delta T}$

Here, $\frac{\Delta V}{V}=\frac{0.12}{100}$

$\Delta T={{20}^{0}}C$

$Y=\frac{0.12}{100}\times \frac{1}{20}$

$\Rightarrow Y=6\times {{10}^{-5}}{{/}^{0}}C$

$\therefore$ Co-efficient of linear expansion of the metal is:

$\alpha =\frac{Y}{3}=\frac{6\times {{10}^{-5}}}{3}=2\times {{10}^{-5}}{{/}^{0}}C$

12. The density of a solid at00C and 5000C is in the ratio 1.027: 1. Find the co-efficient of linear expansion of the solid?

\[Density\text{ }at\text{ }{{0}^{0}}C\text{ }=\text{ }{{S}_{O}}\]

\[Density\text{ }at\text{ }{{500}^{0}}C\text{ }=\text{ }{{S}_{500}}\]

Now, \[{{S}_{O}}\text{ }=\text{ }{{S}_{500}}\left( 1+Y\Delta T \right)\]

Where, \[Y\text{ }=\text{ }Co-efficient\text{ }of\text{ }volume\text{ }expansion\]

\[\Delta T\text{ }=\text{ }Change\text{ }in\text{ }temperature\]

$\frac{{{S}_{O}}}{{{S}_{500}}}=\frac{1.027}{1}$

\[\Delta T\text{ }=\text{ }Final\text{ }Temperature\text{ }\text{ }Initial\text{ }temperature\]

\[\Delta T\text{ }=\text{ }500\text{ }-\text{ }{{0}^{0}}C\]

\[\Delta T\text{ }=\text{ }{{500}^{0}}C\]

Or $1.027=1\times \left( 1+Y\Delta T \right)$

$1.027=1+Y\Delta T$

$\Rightarrow 1.027-1=Y\Delta T$

$\Rightarrow 0.027=Y\Delta T$

$\Rightarrow \frac{0.027}{500}=Y$

$\Rightarrow Y=54\times {{10}^{-5}}{{/}^{0}}C$

Now, the co-efficient of linear expansion \[\left( \alpha  \right)\] is related to co-efficient of volume expansion (Y) as:

\[\alpha =\frac{Y}{3}=\frac{54\times {{10}^{-5}}}{3}=18\times {{10}^{-5}}{{/}^{0}}C\]

13. If one Mole of a monatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular specific heat of the mixture at constant volume?

Ans: We know that:

For, a monatomic gas, \[Specific\text{ }heat\text{ }at\text{ }consent\text{ }volume\text{ }=\text{ }{{C}_{V1}}\text{ }=\frac{3}{2}R\] ; 

\[R\text{ }=\text{ }Universal\text{ }Gas\text{ }Constant\]

\[No.\text{ }of\text{ }moles\text{ }of\text{ }monatomic\text{ }gas\text{ }=\text{ }{{n}_{1}}\text{ }=\text{ }1\text{ }mole\]

\[No.\text{ }of\text{ }moles\text{ }of\text{ }diatomic\text{ }gas\text{ }=\text{ }{{n}_{2}}\text{ }=\text{ }3\text{ }moles\].

For, diatomic gas, specific heat at constant volume, \[{{C}_{V2}}\text{ }=\frac{5}{2}R\]

Applying, conservation of energy.

Let \[{{C}_{V}}\text{ }=\text{ }Specific\text{ }heat\text{ }of\text{ }the\text{ }mixture\];

${{C}_{V}}=\frac{{{n}_{1}}{{C}_{{{v}_{1}}}}+{{n}_{2}}{{C}_{{{n}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}$

$\Rightarrow {{C}_{V}}=\frac{1\times \frac{3}{2}R+3\times \frac{5}{2}R}{1+3}$

$\Rightarrow {{C}_{V}}=\frac{\frac{3}{2}R+\frac{15}{2}R}{4}$

R = Universal Gas constant

$\Rightarrow {{C}_{V}}=\frac{9R}{4}=\frac{9\times 8.31}{4}=18.7J/mol/K$

14. Calculate the difference between two principal specific heats of 1g of helium gas at N. T. P. Given Molecular weight of Helium = 4 and J = 4.186 J/cal and Universal Gas constant, R = 8.314J / mole / K?

\[Molecular\text{ }weight\text{ }of\text{ }Helium\text{ }=\text{ }M\text{ }=\text{ }4\]

\[Universal\text{ }Gas\text{ }Constant,\text{ }R\text{ }=\text{ }8.31J/mole/K\]

\[{{C}_{P}}\text{ }=\text{ }specific\text{ }heat\text{ }at\text{ }constant\text{ }Pressure\]

\[{{C}_{V}}\text{ }=\text{ }specific\text{ }heat\text{ }at\text{ }constant\text{ }Volume\]

Now, ${{C}_{P}}-{{C}_{V}}=\frac{r}{J}$ for 1 mole of gas.

${{C}_{P}}-{{C}_{V}}=\frac{R}{MJ}$

Where, \[R\text{ }=\text{ }Universal\text{ }Gas\text{ }Constant\text{ }=\text{ }8.31J/mole/K\] 

\[J\text{ }=\text{ }4.186\text{ }J/cal\]

\[M\text{ }=\text{ }Molecular\text{ }weight\text{ }of\text{ }Helium\text{ }=\text{ }4\]

$\Rightarrow {{C}_{P}}-{{C}_{V}}=\frac{8.31}{4\times 4.186}$

$\Rightarrow {{C}_{P}}-{{C}_{V}}=0.496cal/g/K$, which is the required difference.

15. Why does heat flow from a body at higher temperature to a body at lower temperature?

Ans: If a body at a higher temperature is in contact with a body at a lower temperature, molecules with a more kinetic energy that are in contact with less energetic molecules give some of their kinetic energy to the less energetic ones.

16. A one litre flask contains some mercury. IT is found that at different temperatures, then volume of air inside the flask remains the same. What is the volume of mercury in the flask? Given the co-efficient of linear expansion of glass \[=\text{ }9\text{ }\times \text{ }{{10}^{-6}}\text{ }/{{\text{ }}^{0}}C\] and co- efficient of volume expansion of mercury \[=\text{ }1.8\text{ }\times \text{ }{{10}^{-4}}\text{ }/{{\text{ }}^{0}}C\].

Ans: It is given that the volume of air in the flask remains the same at different temperatures. This is possible only when the expansion of glass is exactly equal to the expansion of mercury, Co-efficient of cubical expansion of glass is:-

${{\gamma }_{g}}=3{{\alpha }_{g}}=3\times 9\times {{10}^{-5}}=27\times {{10}^{-5}}{{/}^{0}}C$

The co-efficient of cubical expansion of mercury is:

${{\gamma }_{m}}=1.8\times {{10}^{-4}}{{/}^{0}}C$

Volume of flask, \[V\text{ }=\text{ }1\text{ }liter\text{ }=\text{ }1000\text{ }c{{m}^{3}}\].

Let \[{{V}_{m}}\text{ c}{{m}^{3}}\] be the volume of mercury in the flask.

\[Expansion\text{ }of\text{ }flask\text{ }=\text{ }Expansion\text{ }of\text{ }Mercury\]

$C\times {{\gamma }_{g}}\times t={{V}_{m}}\times {{\gamma }_{m}}\times t$

∴ Volume of Mercury, ${{V}_{m}}=\frac{V\times {{\gamma }_{g}}}{{{\gamma }_{m}}}$

$\Rightarrow {{V}_{m}}=\frac{1000\times 27\times {{10}^{-5}}}{1.8\times {{10}^{-4}}}=150c{{m}^{3}}$

17. The potential energy function for a particle executing linear simple harmonic motion is given by \[\mathbf{V}\left( \mathbf{x} \right)\text{ }=k{{x}^{2}}/\mathbf{2}\] , where k is the force constant of the oscillator. For k = 0.5 N/m, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = ± 2 m.

Total energy of the particle, \[E=\text{ }1J\].

Force constant, \[k=\text{ }0.5\text{ }N/m\].

Kinetic energy of the particle, \[K\text{ }=\frac{1}{2}m{{v}^{2}}\]

According to the conservation law:

\[E=\text{ }V\text{ }+\text{ }K\]

\[1=\text{ }\frac{1}{2}k{{x}^{2}}\text{ }+\text{ }\frac{1}{2}m{{v}^{2}}\]

At the moment of ‘turn back', velocity (and hence K) becomes zero.

$\therefore 1=\frac{1}{2}k{{x}^{2}}$

$\frac{1}{2}\times 0.5{{x}^{2}}=1$

${{x}^{2}}=4$

Thus, the particle turns back when it reaches $x=\pm 2$m.

18. State if each of the following statements is true or false. Give reasons for your Answer.

In an elastic collision of two bodies, the momentum and energy of each body is conserved.

Ans: The above statement is false. This is because in an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.

Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

Ans: The above statement is false. This is because although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.

Work done in the motion of a body over a closed loop is zero for every force in nature.

Ans: The above statement is false. This is because the work done in the motion of a body over a closed loop is zero for a conservation force only.

In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Ans: The above statement is true. This is because in an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

19. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

(i) ${{t}^{\frac{1}{2}}}$ 

(ii) t  

(iii) \[{{t}^{\frac{3}{2}}}\] 

(iv) ${{t}^{2}}$

Ans: The power delivered to it at time t is proportional to t i.e., option (ii) t is the correct choice.

\[Mass\text{ }of\text{ }the\text{ }body\text{ }=\text{ }m\]

\[Acceleration\text{ }of\text{ }the\text{ }body\text{ }=\text{ }a\]

Using Newton's second law of motion, the force experienced by the body is given by: \[F\text{ }=\text{ }ma\].

Both m and a are constants. Hence, force F will also be a constant.

\[F\text{ }=\text{ }ma\text{ }=\text{ }Constant\] ..... (i)

For velocity v, acceleration is given as,

$a=\frac{dv}{dt}=c\text{onstant}$

$dv=c\text{onstant}\times \text{dt}$

$v=\alpha t$ …… (ii)

Where, $\alpha $ is another constant.

$v\propto t$ …… (iii)

Power is given by the relation:

\[P\text{ }=\text{ }F.v\]

Using equations (i) and (iii), we have:

$P\propto t$

Hence, power is directly proportional to time.

20. A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

Ans: Its displacement in time t is proportional to \[{{t}^{\frac{3}{2}}}\], $\therefore$ option (iii) is the right answer.

$=mav=mv\frac{dv}{dt}=\text{constant=k}$

$\Rightarrow vdv=\frac{k}{m}dt$

Integrating both sides:

$\frac{{{v}^{2}}}{2}=\frac{k}{m}t$

$v=\sqrt{\frac{2kt}{m}}$

For displacement of the body, we have:

$v=\frac{dx}{dt}=\sqrt{\frac{2k}{m}}{{t}^{\frac{1}{2}}}$

$dx=k'{{t}^{\frac{1}{2}}}dt$

Where $k'=\sqrt{\frac{2k}{3}}=$ New constant

On integrating both sides, we get:

$x=\frac{2}{3}k'{{t}^{\frac{3}{2}}}$

$\Rightarrow x\propto {{t}^{\frac{3}{2}}}$, which is the required proportionality.

21. A pump on the ground floor of a building can pump up water to fill a tank of volume $30{{m}^{3}}$ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Volume of the tank, $V=30{{m}^{3}}$ .

Time of operation, \[t=\text{ }15\text{ }min\text{ }=\text{ }15\text{ }\times 60\text{ }=\text{ }900\text{ }s\].

Height of the tank, \[h=\text{ }40\text{ }m\].

Efficiency of the pump, \[\eta =\text{ }30%\]

Density of water, $\rho ={{10}^{3}}kg/{{m}^{3}}$

Mass of water, \[m=\rho V=30\times {{10}^{3}}kg\]

Output power can be obtained as:

${{P}_{0}}=\frac{Wo\text{rk d}one}{Time}=\frac{mgh}{t}$

$\frac{30\times {{10}^{3}}\times 9.8\times 40}{900}=13.067\times {{10}^{3}}W$

For input power, efficiency is given by the relation:

$\eta =\frac{{{P}_{0}}}{{{P}_{1}}}=30%$

$\Rightarrow {{P}_{1}}=\frac{13.067}{30}\times 100\times {{10}^{3}}$

$\Rightarrow {{P}_{1}}=43.6kW$, which is the required power consumption.

22. A body of mass 0.5 kg travels in a straight line with velocity $v=a{{x}^{\frac{3}{2}}}$ where $a=5m/{{s}^{2}}$. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Mass of the body, \[m=\text{ }0.5\text{ }kg\].

Velocity of the body is governed by the equation,

$v=a{{x}^{\frac{3}{2}}}$ where $a=5m/{{s}^{2}}$.

Initial velocity, \[u\text{ }\left( at\text{ }x\text{ }=\text{ }0 \right)\text{ }=\text{ }0\]

Final velocity \[v\text{ }\left( at\text{ }x\text{ }=\text{ }2\text{ }m \right)=10\sqrt{2}m/s\]

Work done, \[W\text{ }=\text{ }Change\text{ }in\text{ }kinetic\text{ }energy\]

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

$=\frac{1}{2}\times 0.5\times \left( {{\left( 10\sqrt{2} \right)}^{2}}-0 \right)$

23. A family uses 8 kW of power. 

Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? 

Power used by the family, \[P\text{ }=\text{ }8\text{ }kW\text{ }=\text{ 8}\times {{10}^{3}}W\]

Solar energy received per square metre \[=\text{ }200\text{ }W\]

Efficiency of conversion from solar to electricity energy \[=\text{ }20\text{ }%\]

Area required to generate the desired electricity \[=\text{ }A\]

As per the information given in the question, 

\[\text{8}\times {{10}^{3}}=20%\times \left( A\times 200 \right)\]$=\frac{20}{100}\times A\times 200$

$\Rightarrow A=200{{m}^{2}}$

Compare this area to that of the roof of a typical house.

Ans: The area of a solar plate required to generate \[8\text{ }kW\] of electricity is almost equivalent to the area of the roof of a building having dimensions \[14\text{ }m\times 14\text{ }m\].

24. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7m/s. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Mass of the bolt, \[m=\text{ }0.3\text{ }kg\].

\[Speed\text{ }of\text{ }the\text{ }elevator\text{ }=\text{ }7\text{ }m/s\].

Height, \[h\text{ }=\text{ }3\text{ }m\].

As the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.

\[Heat\text{ }produced\text{ }=\text{ }Loss\text{ }of\text{ }potential\text{ }energy\text{ }=\text{ }mgh\text{ }=\text{ }=\text{ }8.82\text{ }J\]

The heat produced will remain the same even if the lift is stationary. This is due to the fact that the relative velocity of the bolt with respect to the lift will remain zero.

25. Consider the decay of a free neutron at rest: \[n\text{ }\to \text{ }p+\text{ }e-\]. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the $\beta $ decay of a neutron or a nucleus (Fig. 6.19).

(Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of $\beta $ decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like e-, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: \[n\text{ }\to \text{ }p+\text{ }e-\])

Ans: The decay process of free neutron at rest is \[n\text{ }\to \text{ }p+\text{ }e-\].

Using Einstein's mass-energy relation, we have the energy of electron as $\Delta m{{c}^{2}}$.

\[\Delta m=\text{ }Mass\text{ }defect\text{ }=\text{ }Mass\text{ }of\text{ }neutron-\left( Mass\text{ }of\text{ }proton\text{ }+\text{ }Mass\text{ }of\text{ }electron \right)\]

\[c\text{ }=\text{ }Speed\text{ }of\text{ }light\]

\[\Delta m\] and c are constants. Thus, the given two-body decay is unable to explain the continuous energy distribution in the $\beta $ decay of a neutron or a nucleus. The presence of neutrino on the LHS of the decay correctly interprets the continuous energy distribution.

Long Answer Questions                                                                                   3 Marks

1. Prove that in an elastic collision in one dimension the relative velocity of approach before impact is equal to the relative velocity of separation after impact.

Ans: Consider the figure given below:

According to law of conservation of linear momentum

${{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$

\[{{m}_{1}}\left( {{u}_{1}}-{{v}_{1}} \right)={{m}_{2}}\left( {{v}_{2}}-{{u}_{2}} \right)\] …… (1)

K.E. also remains conserved.

$\frac{1}{2}{{m}_{1}}{{u}_{1}}^{2}+\frac{1}{2}{{m}_{2}}{{u}_{2}}^{2}=\frac{1}{2}{{m}_{1}}{{v}_{1}}^{2}+\frac{1}{2}{{m}_{2}}{{v}_{2}}^{2}$

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

Dividing (2) by (1)

${{u}_{1}}-{{v}_{1}}={{v}_{2}}+{{u}_{2}}$

${{u}_{1}}-{{u}_{2}}={{v}_{2}}-{{v}_{1}}$

i.e., \[Relative\text{ }velocity\text{ }of\text{ }approach\text{ }=\text{ }Relative\text{ }velocity\text{ }of\text{ }separation\].

2. Calculate \[{{C}_{p}}\] for air, given that \[{{C}_{v}}\text{ }=0.162\text{ }cal\text{ }{{g}^{-1}}\text{ }{{k}^{-1}}\] and density air at N.T. P is \[0.001293\text{ }g/c{{m}^{3}}\].

\[Specific\text{ }heat\text{ }at\text{ }constant\text{ }volume\text{ }=\text{ }{{C}_{v\text{ }}}=\text{ }0.162\text{ }Cal\text{ }{{g}^{-1\text{ }}}{{k}^{-1}}\]

Specific heat at constant pressure = \[{{C}_{p}}=\text{ }?\]

Now, \[{{C}_{p}}\text{ }\text{ }{{C}_{v\text{ }}}=\frac{r}{J}=\frac{PV}{TJ}\left( \because PV=nRT \right)\]

\[{{C}_{p}}\text{ }\text{ }{{C}_{v\text{ }}}=\frac{P\times 1}{s\times TJ}\left( s=Density \right)\]

\[\Rightarrow {{C}_{p}}\text{ }\text{ }{{C}_{v\text{ }}}=\frac{1.01\times {{10}^{5}}}{273\times 4.2\times {{10}^{7}}\times 1.293\times {{10}^{-3}}}\]

\[\Rightarrow {{C}_{p}}\text{ }\text{ }{{C}_{v\text{ }}}=\frac{1.01\times {{10}^{6+3-7}}}{273\times 4.2\times {{10}^{7}}\times 1.293}\]

\[\Rightarrow {{C}_{p}}\text{ }\text{ }{{C}_{v\text{ }}}=\frac{1.01\times {{10}^{2}}}{1482.5}=6.8\times {{10}^{-2}}\]

\[\Rightarrow {{C}_{p}}\text{ }\text{ }{{C}_{v\text{ }}}=0.068\]

\[\Rightarrow {{C}_{p}}\text{ }=\text{ }0.162+0.068\]

\[\Rightarrow {{C}_{p}}\text{ }=\text{ }0.23\text{ c}al\text{ }{{g}^{-1}}\text{ }{{k}^{-1}}\], which is the required value asked to determine.

3. Develop a relation between the co-efficient of linear expansion, co-efficient superficial expansion and coefficient of cubical expansion of a solid.

Ans: As the \[co-efficient\text{ }of\text{ }linear\text{ }expansion\text{ }=\text{ }\alpha \text{ }=\frac{\Delta L}{L\Delta T}\]

\[\Delta L\text{ }=\text{ }change\text{ }in\text{ }length\]

\[L\text{ }=\text{ }length\]

\[\Delta T\text{ }=\text{ }change\text{ }in\text{ }temperature\].

In the same way,

\[Co-efficient\text{ }of\text{ }superficial\text{ }expansion\text{ }=\text{ }\beta \text{ }=\frac{\Delta S}{S\Delta T}\]

\[\Delta S\text{ }=\text{ }change\text{ }in\text{ }area\]

\[S\text{ }=\text{ }original\text{ }area\]

\[\Delta T\text{ }=\text{ }change\text{ }in\text{ }temperature\]

Co-efficient of cubical expansion, \[Y\text{ }=\frac{\Delta V}{V\Delta T}\]

\[\Delta V\text{ }=\text{ }change\text{ }in\text{ }volume\]

\[V\text{ }=\text{ }original\text{ }volume\]

\[\Delta L=\alpha L\text{ }\Delta T\]

\[\Rightarrow L\text{ }+\text{ }\Delta L\text{ }=\text{ }L\text{ }+\text{ }\alpha L\text{ }\Delta T\]

\[\Rightarrow L\text{ }+\text{ }\Delta L\text{ }=\text{ }L\text{ }\left( 1+\alpha \Delta T \right)\] …… (1)

\[V+\text{ }\Delta V\text{ }=\text{ }V\text{ }\left( 1+Y\Delta T \right)\] ……. (2)

\[S+\Delta S=S\text{ }\left( 1+\beta \Delta T \right)\] …… (3)

Also, 

\[\left( V+\Delta V \right)\text{ }=\text{ }{{\left( L+\Delta L \right)}^{3}}\]

\[\Rightarrow V+\Delta V={{L}^{3}}\left( 1+3\alpha \Delta T+3{{\alpha }^{2}}\Delta {{T}^{2}}+{{\alpha }^{3}}{{T}^{3}} \right)\]

Since \[{{\alpha }^{2}},\text{ }{{\alpha }^{3}}\] are negligible, so,

\[V+\gamma \text{ }V\Delta T=\text{ }V\left( 1+3\alpha \Delta T \right)\] \[\left[ as\text{ }{{L}^{3\text{ }}}=\text{ }V \right]\]

\[V+\gamma V\Delta T\text{ }=\text{ }V+V3\alpha \Delta T\]

\[\Rightarrow \gamma V\Delta T\text{ }=\text{ }3\alpha \Delta T\]

\[\Rightarrow \gamma \text{ }=\text{ }3\alpha \]

Similarly, \[\beta \text{ }=\text{ }2\alpha \] (using \[{{L}^{2}}\text{ }=\text{ }S\text{ }\left( Area \right)\])

\[\alpha =\frac{\beta }{2}=\frac{\gamma }{3}\], which is the required relation.

4. Calculate the amount of heat required to convert 1.00kg of ice at \[\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{0}}}C\] into steam at \[\mathbf{10}{{\mathbf{0}}^{\mathbf{0}}}C\] at normal pressure. Specific heat of ice = 2100J/kg/k. Latent heat of fusion of ice = 3.36x105 J/kg, specific heat of water = 4200J/kg/k. Latent heat of vaporization of water \[=\text{ }2.25\times {{10}^{6}}J/kg\].

Heat is required to raise the temperature of ice from \[\text{ }{{10}^{0}}C\text{ }to\text{ }{{0}^{0}}C.\]

Thus, change in temperature \[=\text{ }\Delta T\text{ }=\text{ }T2-T1\text{ }=\text{ }0-\left( -10 \right)\text{ }=\text{ }{{10}^{0}}C\].

So, \[\Delta Q1=cm\Delta T\]

\[C\text{ }=\text{ }specific\text{ }heat\text{ }of\text{ }ice\]

\[M\text{ }=\text{ }Mass\text{ }of\text{ }ice\]

\[\Delta T\text{ }=\text{ }{{10}^{0}}C\]

\[\Delta {{Q}_{1}}\text{ }=\text{ }2100\times 1\times 10=21000J\]

(2) Heat required to melt the ice to \[{{0}^{0}}C\] water:

\[\Delta {{Q}_{2}}=\text{ }mL\]

\[L\text{ }=\text{ }Latent\text{ }heat\text{ }of\text{ }fusion\text{ }of\text{ }ice\text{ }=\text{ }3.36\times 105\text{ }J/kg\]

\[m\text{ }=\text{ }Mass\text{ }of\text{ }ice\]

\[\Rightarrow \Delta {{Q}_{2}}=1\times 3.36\times 105\text{ }J/kg\]

\[\Rightarrow \Delta {{Q}_{2}}=3.36\times 105\text{ }J\]

\[\Rightarrow \Delta {{Q}_{2}}=336000J\]

(3) Heat required to raise the temperature of water from \[{{0}^{0}}C\] to \[{{100}^{0}}C\]:-

\[\Delta T\text{ }=\text{ }T2-T1\text{ }=\text{ }100-0={{100}^{0}}C\]

\[\Delta {{Q}_{3}}=\text{ }cm\Delta Tc\text{ }=\text{ }specific\text{ }heat\text{ }of\text{ }water\]

\[\Rightarrow \Delta {{Q}_{3}}=4200\times 1\times 100\]

\[\Rightarrow \Delta {{Q}_{3}}=420,000J\]

(4) Heat required to convert 1000C water to steam at 1000C

\[\Delta {{Q}_{4}}=\text{ }mL\]

Where, \[L\text{ }=\text{ }Latent\text{ }heat\text{ }of\text{ }vapourisation\text{ }=\text{ }2.25\times {{10}^{6}}J/kg\]

\[\Rightarrow \Delta {{Q}_{4}}=\text{ }1\times 2.25\times 106J/kg\]

\[\Rightarrow \Delta {{Q}_{4}}=2250000J\]

\[Total\text{ }Heat\text{ }required\text{ }=\Delta {{Q}_{1}}+\Delta {{Q}_{2}}+\Delta {{Q}_{3}}+\Delta {{Q}_{4}}\]

\[\Rightarrow \Delta {{Q}_{total}}=\text{ }21000+336000+420000+2250000\]

\[\Rightarrow \Delta {{Q}_{total}}=\text{ }3027000J\]

\[\Rightarrow \Delta {{Q}_{total}}=\text{ }3.027\times {{10}^{6}}J\]

5. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

Ans: The sign of work done is positive. In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

Work done by gravitational force in the above case,

Ans: The sign of work done is negative. In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.

Work done by friction on a body sliding down an inclined plane,

Ans: The sign of work done is negative. Since the direction of frictional force is opposite to the direction of motion, the work done by the frictional force is negative in this case.

Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,

Ans: The sign of work done is positive. Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.

Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Ans: The sign of work done is negative. The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

6. Choose the correct alternative:

When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.

Ans: Decreases, this is because:

A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body. 

Work done by a body against friction always results in a loss of its kinetic/potential energy.

Ans: Kinetic energy, this is because:

The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.

The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

Ans: External force, this is because:

Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system.

In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Ans: Total linear momentum, this is because:

The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

7. Answer carefully, with reasons:

In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact)

Ans: No. This is because in an elastic collision, the total initial kinetic energy of the balls will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.

Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Ans: Yes. This is because in an elastic collision, the total linear momentum of the system always remains conserved.

What are the answers to (a) and (b) for an inelastic collision?

Ans: No; Yes. This is because in an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.

If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Ans: Elastic. This is because in the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

8. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. 

How much work does she do against the gravitational force?

Mass of the weight, \[m\text{ }=\text{ }10\text{ }kg\].

Height to which the person lifts the weight, \[h\text{ }=\text{ }0.5\text{ }m\].

Number of times the weight is lifted, \[n\text{ }=\text{ }1000\]

$\therefore$ Work done against gravitational force is:

$=1000\times 10\times 9.8\times 0.5$

Fat supplies J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

\[Energy\text{ }equivalent\text{ }of\text{ }1\text{ }kg\text{ }of\text{ }fat\text{ }=\text{ 3}\text{.8}\times \text{1}{{\text{0}}^{7}}J\]

\[Efficiency\text{ }rate\text{ }=\text{ }20%\]

Mechanical energy supplied by the person's body will be:

$=\frac{20}{100}\times 3.8\times {{10}^{7}}$

$=\frac{1}{5}\times 3.8\times {{10}^{7}}$

Equivalent mass of fat lost by the dieter will be:

$=\frac{1}{\frac{1}{5}\times 3.8\times {{10}^{7}}}\times 49\times {{10}^{3}}$

$=6.45\times {{10}^{-3}}kg$

Long Answer Questions                                                                                  4 Marks

1. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.

Compute the following:

(a) Work done by the applied force in 10 s,

(b) Work done by friction in 10 s,

(c) Work done by the net force on the body in 10 s,

(d) Change in kinetic energy of the body in 10 s, and interpret your results.

Mass of the body, \[m=\text{ }2\text{ }kg\].

Applied force, \[F\text{ }=\text{ }7\text{ }N\].

Coefficient of kinetic friction, \[\mu =\text{ }0.1\]

Initial velocity, \[u=\text{ }0\]

Time, \[t\text{ }=\text{ }10\text{ }s\].

The acceleration produced in the body by the applied force is given by Newton's second law of motion as:

$a'=\frac{F}{m}=\frac{7}{2}=3.5m/{{s}^{2}}$

Frictional force is given as:

$f=\mu mg=0.1\times 2\times 9.8=-1.96N$

The acceleration produced by the frictional force:

$a''=-\frac{1.96}{2}=0.98m/{{s}^{2}}$

Total acceleration of the body:

$a=a'+a''=3.5-0.98=2.52m/{{s}^{2}}$

The distance travelled by the body is given by the equation of motion:

$s=ut+\frac{1}{2}a{{t}^{2}}$$=0+\frac{1}{2}\times 2.52\times {{10}^{2}}=126m$

Work done by the applied force, ${{W}_{s}}=F\times s=7\times 126=882J$

Work done by the frictional force, ${{W}_{f}}=F\times s=-1.96\times 126=-247J$

Net force \[~=7-1.96=\text{ }5.04\text{ }N\] 

Work done by the net force, \[{{W}_{net}}=\text{ }5.04\text{ }\times 126\text{ }=\text{ }635\text{ }J\].

From the first equation of motion, final velocity can be calculated as:

$v=u+at$$=0+2.52\times 10=25.2m/s$

Change in kinetic energy 

$=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}$

$=\frac{1}{2} 2\left ( v^{2}-u^{2} \right)=\left ( -25.2 \right )^{2}-0^{2}=635 J$

2. A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Mass of the trolley, \[M=\text{ }200\text{ }kg\].

Speed of the trolley, \[v=\text{ }36\text{ }km/h\text{ }=\text{ }10\text{ }m/s\].

Mass of the boy, \[m=\text{ }20\text{ }kg\].

Initial momentum of the system of the boy and the trolley

\[=\text{ }\left( M+\text{ }m \right)v\]

$=\left( 200+20 \right)\times 10$

\[=\text{ }2200\text{ }kg\text{ }m/s\]

Let v' be the final velocity of the trolley with respect to the ground.

Final velocity of the boy with respect to the ground $=v'-4$

Final momentum $=Mv'+m\left( v'-4 \right)$

$=200v'+20v'-80$

$=220v'-80$

As per the law of conservation of momentum:

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

\[2200\text{ }=220v'-80\]

$\Rightarrow v'=\frac{2280}{220}=10.36m/s$

Length of the trolley, \[l=\text{ }10\text{ }m\].

Speed of the boy, \[v''\text{ }=\text{ }4\text{ }m/s\].

Time taken by the boy to run, $t=\frac{10}{4}=2.5s$.

Thus, the distance moved by the trolley $=v''\times t=10.36\times 2.5=25.9m$.

3. A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N/m as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Mass of the block, \[m=\text{ }1\text{ }kg\]

Spring constant, \[k=100N/m\]

Displacement in the block, \[x=\text{ }10\text{ }cm\text{ }=\text{ }0.1\text{ }m\]

The given situation can be shown as in the following figure.

At equilibrium:

Normal reaction, \[R=\text{ }mg\text{ }cos\text{ }37{}^\circ \]

Frictional force, \[f\text{ = }\mu R\text{ }=\text{ }mg\text{ }sin\text{ }37{}^\circ \]

Where, \[\mu \] is the coefficient of friction.

Net force acting on the block \[=mg\sin {{37}^{0}}-f\]

\[=mg\sin {{37}^{0}}-\mu mgco\operatorname{s}{{37}^{0}}\]

\[=mg\left( \sin {{37}^{0}}-\mu co\operatorname{s}{{37}^{0}} \right)\]

At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,

\[mg\left( \sin {{37}^{0}}-\mu mgco\operatorname{s}{{37}^{0}} \right)x=\frac{1}{2}k{{x}^{2}}\]

\[\Rightarrow 1\times 9.8\left( \sin {{37}^{0}}-\mu mgco\operatorname{s}{{37}^{0}} \right)x=\frac{1}{2}100\times {{0.1}^{2}}\]

$\Rightarrow 0.602-\mu \times 0.799=0.510$

$\Rightarrow \mu =\frac{0.092}{0.799}=0.115$, which is the required coefficient of friction.

4. Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

From the given figure: \[x\text{ }>\text{ }a;\text{ 0}\]

Total energy of a system is:\[E\text{ }=\text{ }P.E.\text{ }+\text{ }K.\text{ }E.\]

The kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative. In this case, the potential energy $\left( {{V}_{0}} \right)$ of the particle becomes greater than total energy (E)  for \[x\text{ }>\text{ }a\]. Hence, kinetic energy becomes negative in this region. Therefore, the particle will not exist in this region. The minimum total energy of the particle is zero.

From the given figure, all the regions the particle cannot be found for the given energy. In the given case, the potential energy (PE) is greater than total energy (E) in all regions. Hence, the particle will not exist in this region.

From the given figure: $x>a$ and \[x\text{ }<\text{ }b;-{{V}_{1}}\]

In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between $x>a$ and \[x\text{ }<\text{ }b\].The minimum potential energy in this case is \[{{V}_{1}}\]. Therefore, 

\[KE=\text{ }E-\left( -{{V}_{1}} \right)=E+{{V}_{1}}\].

Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than \[{{V}_{1}}\]. So, the minimum total energy the particle must have is \[{{V}_{1}}\].

From the given figure: $-\frac{b}{2}<x<\frac{a}{2};-\frac{a}{2}<x<\frac{a}{2};-{{V}_{1}}$

In the given case, the potential energy \[\left( \text{ }{{\text{V}}_{0}} \right)\] of the particle becomes greater than the total energy (E) for $-\frac{b}{2}<x<\frac{a}{2}$ and $-\frac{a}{2}<x<\frac{a}{2}$. Therefore, the particle will not exist in these regions.

The minimum potential energy in this case is $-{{V}_{1}}$. Therefore,\[KE=\text{ }E-\left( -{{V}_{1}} \right)=E+{{V}_{1}}\].

Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $-{{V}_{1}}$. So, the minimum total energy the particle must have is $-{{V}_{1}}$.

5. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. 

(electron mass \[=9.11\times {{10}^{-31}}\text{ }kg\], proton mass \[=1.67\times {{10}^{-27}}\text{ }kg\], \[1\text{ }eV\text{ }=\text{ 1}\text{.60}\times \text{1}{{\text{0}}^{-19}}J\]).

Mass of the electron \[=9.11\times {{10}^{-31}}\text{ }kg\]

Mass of the proton \[=1.67\times {{10}^{-27}}\text{ }kg\]

Kinetic energy of the electron, \[{{E}_{Ke}}=\text{ }10\text{ }keV\text{ }=\text{ 1}{{\text{0}}^{4}}eV\]

\[=\text{1}{{\text{0}}^{4}}\times \text{1}\text{.60}\times \text{1}{{\text{0}}^{-19}}\text{=1}\text{.60}\times \text{1}{{\text{0}}^{-15}}J\]

Kinetic energy of the proton, ${{E}_{Kp}}=100keV={{10}^{5}}eV=1.60\times {{10}^{-14}}J$.

For the velocity of an electron ${{v}_{e}}$, its kinetic energy is given by the relation:

${{E}_{Ke}}=\frac{1}{2}m{{v}_{e}}^{2}$

$\Rightarrow {{v}_{e}}=\sqrt{\frac{2\times {{E}_{Ke}}}{m}}$$=\sqrt{\frac{2\times 1.60\times {{10}^{-15}}}{9.11\times {{10}^{-31}}}}=5.93\times {{10}^{7}}m/s$

For the velocity of a proton ${{v}_{p}}$, its kinetic energy is given by the relation:

${{E}_{Kp}}=\frac{1}{2}m{{v}_{p}}^{2}$

$\Rightarrow {{v}_{p}}=\sqrt{\frac{2\times {{E}_{Kp}}}{m}}$$=\sqrt{\frac{2\times 1.60\times {{10}^{-15}}}{1.67\times {{10}^{-27}}}}=4.38\times {{10}^{6}}m/s$

Hence, the electron is moving faster than the proton.

The ratio of their speeds:

$\frac{{{v}_{e}}}{{{v}_{p}}}=\frac{5.93\times {{10}^{7}}}{4.38\times {{10}^{6}}}=13.54:1$

It is clear that electrons are faster; Ratio of speeds is \[13.54:\text{ }1\].

6. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Length of the pendulum, \[l\text{ }=\text{ }1.5\text{ }m\]

Mass of the bob \[=\text{ }m\]

Energy dissipated \[=\text{ }5%\]

Using the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:

Potential energy of the bob, \[EP\text{ }=\text{ }mgl\]

Kinetic energy of the bob, \[EK\text{ }=\text{ }0\]

\[Total\text{ }energy\text{ }=\text{ }mgl\] ... (i)

At the lowermost point (mean position):

Potential energy of the bob, \[EP\text{ }=\text{ }0\]

Kinetic energy of the bob, ${{E}_{k}}=\frac{1}{2}m{{v}^{2}}$.

Total energy, ${{E}_{k}}=\frac{1}{2}m{{v}^{2}}$... (ii)

As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated. The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i.e., 

$\frac{1}{2}m{{v}^{2}}=\frac{95}{100}\times mgl$

$\Rightarrow v=\sqrt{\frac{2\times 95\times 1.5\times 9.8}{100}}=528m/s$, which is the required speed.

7. The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = $30{{m}^{2}}$ , v = 36 km/h and the density of air is \[\mathbf{1}.\mathbf{2}\text{ }\mathbf{kg}/{{m}^{3}}\]. What is the electrical power produced?

\[Area\text{ }of\text{ }the\text{ }circle\text{ }swept\text{ }by\text{ }the\text{ }windmill\text{ }=\text{ }A\]

\[Velocity\text{ }of\text{ }the\text{ }wind\text{ }=\text{ }v\]

\[Density\text{ }of\text{ }air\text{ }=\rho \]

Volume of the wind flowing through the windmill per sec \[=\text{ }Av\]

Mass of the wind flowing through the windmill per sec \[=\text{ }\rho Av\]. Mass m, of the wind flowing through the windmill in time \[t=\rho Avt\]

Kinetic energy of air $=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\left( \rho Avt \right){{v}^{2}}=\frac{1}{2}\rho A{{v}^{3}}t$

Area of the circle swept by the windmill \[=\text{ }A\text{ }=\text{ }30\text{ }{{m}^{2}}\]

Velocity of the wind \[=\text{ }v=\text{ }36\text{ }km/h\]

Density of air, $\rho =1.2kg/{{m}^{3}}$

\[Electric\text{ }energy\text{ }produced\text{ }=\text{ }25%\text{ }of\text{ }the\text{ }wind\text{ }energy\]

$=\frac{25}{100}\times Kinet\text{ic energy of a}ir$

$=\frac{1}{8}\rho A{{v}^{3}}t$

\[Electrical\text{ }power=\frac{Electrica\text{l e}nergy}{Time}\]

$=\frac{1}{8}\frac{\rho A{{v}^{3}}t}{t}=\frac{1}{8}\rho A{{v}^{3}}$

$=\frac{1}{8}\times 1.2\times 30\times {{10}^{3}}=4.5kW$

5 Marks Questions

Define potential energy. Give examples.

Ans: Potential energy is defined as the energy possessed by a body by virtue of its position in a field or due to change in its configuration example – A gas compressed in a cylinder, A wound spring of a water, water raised to the overhead tank in a house etc.

Draw a graph showing the variation of potential energy, kinetic energy and the total energy of a body freely falling on earth from a height h?

Ans: The graph below shows the variation of potential energy, kinetic energy and the total energy of a body freely falling on earth from a height h:

(i) Gravitational potential energy decreases as the body falls downwards and is zero at the earth

(ii) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.

(iii) According to the law of conservation of energy, total mechanical (KE + PE) energy remains constant.

2. Answer the following:

The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

Ans: At rocket’s expense the heat energy required for burning is obtained. The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket. According to the conservation of energy:

\[Total\text{ }Energy\left( T.E. \right)=\text{ }Potential\text{ }enegry\left( P.E. \right)+Kinetic\text{ }energy\left( K.E. \right)\]

$mgh+\frac{1}{2}m{{v}^{2}}$

The reduction in the rocket's mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

Ans: Gravitational force is a conservative force. As the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero.

An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

Ans: When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases due to the reduction in the height. As the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.

In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Ans: In the second case, the work done is greater.

Mass, \[m\text{ }=\text{ }15\text{ }kg\]

Displacement, \[s\text{ }=\text{ }2\text{ }m\]

Work done, \[W=Fs\cos \theta \]

Where, \[\theta =Angle\text{ }between\text{ }force\text{ }and\text{ }displacement\]

\[W=mgs\cos \theta =15\times 2\times 9.8\cos {{90}^{0}}=0\]

Here, the direction of the force applied on the rope and the direction of the displacement of the rope are the same.

Therefore, the angle between them, $\theta ={{0}^{0}}$

As, $\cos {{0}^{0}}=1$ 

Work done, \[W={{F}_{s}}\cos \theta =mgs\]$=15\times 9.8\times 2=294J$

Thus, work done in the second case is greater.

3. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10m/s$?

Radius of the rain drop, \[r=\text{ }2\text{ }mm\text{ }=\text{ 2}\times {{10}^{-3}}m\].

Volume of the rain drop, \[V=\frac{4}{3}\pi {{r}^{3}}\]

\[=\frac{4}{3}\times 3.14\times {{\left( \text{2}\times {{10}^{-3}} \right)}^{3}}\times {{10}^{3}}kg\]

Mass of the rain drop, \[m=\rho V\]

Gravitational force, \[F\text{ }=\text{ }mg\]

\[=\frac{4}{3}\times 3.14\times {{\left( \text{2}\times {{10}^{-3}} \right)}^{3}}\times {{10}^{3}}\times 9.8N\]

The work done by the gravitational force on the drop in the first half of its journey:

${{W}_{1}}=Fs$

\[=\frac{4}{3}\times 3.14\times {{\left( \text{2}\times {{10}^{-3}} \right)}^{3}}\times {{10}^{3}}\times 9.8\times 250\]

\[=\text{ }0.082\text{ }J\]

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i.e., \[{{W}_{II}}=\text{ }0.082\text{ }J\]As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.

Total energy at the top will be:

${{E}_{T}}=mgh+0$

\[=\frac{4}{3}\times 3.14\times {{\left( \text{2}\times {{10}^{-3}} \right)}^{3}}\times {{10}^{3}}\times 9.8\times 500\times {{10}^{-5}}\]

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.

Total energy at the ground will be:

${{E}_{G}}=\frac{1}{2}m=\frac{4}{3}\times 3.14\times {{\left( \text{2}\times {{10}^{-3}} \right)}^{3}}\times {{10}^{3}}\times 9.8\times {{10}^{2}}$

$=1.675\times {{10}^{-3}}J$

\[Work\text{ }done\text{ }by\text{ }resistive\text{ }force\text{ }={{E}_{G}}-{{E}_{T}}=-0.162J\]

4. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figures is a possible result after collision?

Ans:  

Case (i) result is possible after a collision

It is observed that the total momentum before and after collision in each case is constant. For an elastic collision, the total kinetic energy of a system remains conserved before and after the collision. For the mass of each ball bearing m, 

The total kinetic energy of the system before collision:

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

$=\frac{1}{2}m{{v}^{2}}$

Case (ii): Total kinetic energy of the system after collision:

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

$=\frac{1}{4}m{{v}^{2}}$

Hence, the kinetic energy of the system is not conserved in case (i).

Case (iii): Total kinetic energy of the system after collision: 

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

Thus, the kinetic energy of the system is conserved in case (ii).

Case (vi): Total kinetic energy of the system after collision:

$=\frac{1}{2}\left( 3m \right){{\left( \frac{v}{3} \right)}^{2}}$

$=\frac{1}{6}m{{v}^{2}}$

Thus, the kinetic energy of the system is not conserved in case (iii).

5. A bullet of mass 0.012 kg and horizontal speed of $70m/s$ strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Mass of the bullet, \[m=\text{ }0.012\text{ }kg\]

Initial speed of the bullet, \[{{u}_{b}}=\text{ }70\text{ }m/s\].

Mass of the wooden block, \[M=\text{ }0.4\text{ }kg\]

Initial speed of the wooden block, \[{{u}_{g}}=\text{ }0\]

Final speed of the system of the bullet and the block \[=\text{ }v\]

Applying the law of conservation of momentum:

$m{{u}_{b}}+M{{u}_{g}}=\left( m+M \right)v$

$0.012\times 70+0.4\times 0=\left( 0.012+0.4 \right)v$

$\Rightarrow v=\frac{0.84}{0.412}=2.041m/s$

For the system of the bullet and the wooden block:

Mass of the system, \[m'\text{ }=\text{ }0.412\text{ }kg\]

Velocity of the system \[=\text{ }2.04\text{ }m/s\]

Height up to which the system rises \[=\text{ }h\]

Applying the law of conservation of energy to this system:

\[Potential\text{ }energy\text{ }at\text{ }the\text{ }highest\text{ }point\text{ }=\text{ }Kinetic\text{ }energy\text{ }at\text{ }the\text{ }lowest\text{ }point\]

$m'gh=\frac{1}{2}m'{{v}^{2}}$

$\Rightarrow h=\frac{1}{2}\left( \frac{{{v}^{2}}}{g} \right)$$=\frac{1}{2}\times \frac{{{\left( 2.04 \right)}^{2}}}{9.8}$$=0.2123m$

The wooden block will rise to a height of \[0.2123\text{ }m\].

\[Heat\text{ }produced\text{ }=\text{ }Kinetic\text{ }energy\text{ }of\text{ }the\text{ }bullet\text{ }-\text{ }Kinetic\text{ }energy\text{ }of\text{ }the\text{ }system\]

$\Rightarrow \frac{1}{2}m{{u}^{2}}-\frac{1}{2}m'{{v}^{2}}$\[=\frac{1}{2}\times 0.012\times {{70}^{2}}-\frac{1}{2}\times 0.412\times {{2.04}^{2}}=28.54J\]

6. Two inclined frictionless tracks, one gradual and the other steep, meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given \[{{\theta }_{1}}=\text{ }30{}^\circ ,\text{ }{{\theta }_{2}}=\text{ }60{}^\circ ,\]and h = 10 m, what are the speeds and times taken by the two stones?

Ans: No, they won’t reach there with the same speed.

The stone moving down the steep plane will reach the bottom first.

Yes; the stones will reach the bottom with the same speed

\[{{v}_{b}}={{v}_{c}}=\text{ }14\text{ }m/s\]

\[{{t}_{1}}=\text{ }2.86\text{ }s;\text{ }{{\text{t}}_{2}}=\text{ }1.65\text{ }s\]

The given situation can be shown as in the following figure:

Here, the initial height (AD) for both stones is the same (h). Hence, both will have the same potential energy at point A.

As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i.e.,

$\frac{1}{2}m{{v}_{1}}^{2}=\frac{1}{2}m{{v}_{2}}^{2}$

Consider, ${{v}_{1}}={{v}_{2}}=v$

\[m\text{ }=\text{ }Mass\text{ }of\text{ }each\text{ }stone\]

\[v=\text{ }Speed\text{ }of\text{ }each\text{ }stone\text{ }at\text{ }points\text{ }B\text{ }and\text{ }C\]

Hence, both stones will reach the bottom with the same speed, v.

For stone I:

Net force acting on this stone is given by:

${{F}_{net}}=m{{a}_{1}}=mg\sin {{\theta }_{1}}$

$\Rightarrow {{a}_{1}}=g\sin {{\theta }_{1}}$

For stone II:

${{a}_{2}}=g\sin {{\theta }_{2}}$

$\Rightarrow {{\theta }_{2}}>{{\theta }_{1}}$

$\Rightarrow {{a}_{2}}>{{a}_{1}}$

Using the first equation of motion, the time of slide can be obtained as:

$\Rightarrow t=\frac{v}{a}\left( \because u=0 \right)$

${{t}_{1}}=\frac{v}{{{a}_{1}}}$

${{t}_{2}}=\frac{v}{{{a}_{2}}}$

$\Rightarrow {{t}_{2}}<{{t}_{1}}$

Thus, the stone moving down the steep plane will reach the bottom first.

The speed (v) of each stone at points B and C is given by the relation obtained from the law of conservation of energy.

$mgh=\frac{1}{2}m{{v}^{2}}$

$\Rightarrow v=\sqrt{2gh}=\sqrt{2\times 9.8\times 10}=14m/s$

${{t}_{1}}=\frac{v}{{{a}_{1}}}=\frac{v}{g\sin {{\theta }_{1}}}=\frac{14}{9.8\times \sin {{30}^{0}}}=2.86s$

${{t}_{2}}=\frac{v}{{{a}_{2}}}=\frac{v}{g\sin {{\theta }_{2}}}=\frac{14}{9.8\times \sin {{60}^{0}}}=1.65s$, which are the corresponding times taken by both the stones.

Important Questions from Work, Energy and Power (Short, Long & Practice)

Short answer type questions.

1. A spring is stretched. Is the work done by the stretching force positive or negative?

2. What is the work done by the centripetal force? Why?

3. What is the work done by the tension in the string of a simple pendulum?

Long Answer Type Questions

1. Prove that instantaneous power is given, by the dot product of force and velocity i.e. P = F.v.

2. State and prove work-energy theorem or principle.

3. State and prove the law of conservation of energy.

4. A 10 kg block slides without acceleration down a rough inclined plane making an angle of 20° with the horizontal. Calculate the work done over a distance of 1.2 m when the inclination of the plane is increased to 30°.

Practice Questions

1. A bullet of mass 20 g moving with a velocity of 500 ms -1 strikes a tree and goes out from the other end. with a velocity of 400 ms -1 . Calculate the work done in passing through the tree.

2. A stone of mass 5 kg falls from the top of a cliff 30 m high and buries itself one metre deep into the sand. Find the average resistance offered and time taken to penetrate into the sand.

3. A wooden ball is dropped from a height of 2 m. What is the height up to which the ball will rebound if the coefficient of restitution is 0.5?

Download Important Questions For Class 11 Physics Chapter 5

Chapter 5 of Class 11 Physics Work, Power, and Energy is an important topic to be covered by the students for securing good scores in the Board exams as well as in the competitive exams. Students find Chapter 5 Work, Power, and Energy as one of the difficult topics to be covered in Physics, but students will be able to tackle this by practising more and more questions with the help of work energy and power Class 11 important questions with answers Pdf available on the Vedantu platform. Class 11 Physics Chapter 5 Important Questions Pdf is prepared for students in such a way that they will be able to solve the questions effortlessly.

Work Energy And Power Class 11 Important Questions With Answers

Students must be familiar with Chapter 5 Work, Power, and Energy for attempting Class 11 Physics Chapter 5 Important Questions . The chapter includes interesting topics like a work-energy theorem, kinetic energy, potential energy, conservation of energy, power, collisions, etc… 

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Physics Chapter 5 - Work, Energy, and Power

In NCERT Physics syllabus Work, Energy, and Power chapter is of high importance from an examination point of view. Students must be familiar with the terms of work, energy, and power from their elementary classes. In our everyday life, we use these terms so casually but the Physics behind them is interesting. In order to have a hold on any subject, it is only possible with regular studies and practice. Work, Energy, and Power is a chapter that can be understood with real-life exams. Work energy and power Class 11 important questions with answers covered all the questions with suitable examples students can relate it to quickly. 

Students should start studying by questioning, for example, now we are on Chapter 5 Work, Energy, and Power. The very first question that arises in a student's mind is what is work. Work and energy are two terms that we come across most often, work and energy are interrelated concepts. For solving important questions from work energy and power Class 11 , students should understand the complete chapter and remember a few topics such as :

Work: Work is just energy transferred by force.

Work Done: Work done is defined as when an external force acts upon a body in steady motion; it will experience some displacement; this action is termed as the work is done.

Energy: The ability to do work is known as energy. There are several types of energy like electrical energy, solar energy, mechanical energy, etc. In this chapter, we will focus mainly on mechanical energy and its types. There are basically two types of mechanical energy:

Kinetic Energy: The energy gained by an object due to its motion is known as the kinetic energy of that object.

Potential Energy: The energy gained by an object due to its position is known as the potential energy of that object.

Power: The rate of doing work is known as power.

Work Energy Theorem: Work energy theorem states that work done by the external forces on an object is equal to the change in its kinetic energy.

Conservation of Energy: Conservation of energy can be stated as- energy is a conserved quantity it can neither be created nor be destroyed.

Like these, there are many more interesting concepts included in Chapter 5 of Class 11 Physics. Not only these we will also come across topics like momentum, conservation of momentum, collisions, type of collision, etc… There are high chances of getting numerical type problems in competitive exams from this chapter. Not only numerical type questions we also get to see questions that demand for graphical methods, derivations, and multiple choices type questions. 

The work, energy, and power chapter compiled of many more important topics that can be mastered by practicing Class 11 Physics Chapter 5 Important Questions . The work, Energy and Power chapter mainly focus on bringing the idea of how work is defined, the difference between work and work done, Mechanical energy and its types, etc…  Refer to the material available on the Vedantu for a detailed explanation.

Chapter 5 of Class 11 Physics syllabus as prescribed by CBSE, NCERT board have numerous topics, derivations, diagrams, graphs. Students must focus on derivation parts carefully with clear understanding.  Numerical problems, MCQs, Revision notes of Chapter 5 are available on the Vedantu website and they are prepared by experts after deep research and analysis of CBSE, NEET , JEE exam papers and students find them reliable. Before starting with the preparation of the exams, students should go through the chapters thoroughly without skipping any topic as in Physics every topic will be related to another topic.

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FAQs on Important Questions for CBSE Class 11 Physics Chapter 5 - Work, Energy and Power

1. Is Chapter 5-Work, Power, and Energy of Class 11 Physics difficult?

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3. What are the important topics given in Chapter 5-Work, Power, and Energy of Class 11 Physics?

Important topics covered in Chapter 5 of Class 11 Physics include an introduction, the theorem related to work, energy, and power, different types of energies such as kinetic energy, potential energy, and mechanical energy. Students will also learn the concept of conservation of energy and laws related to the conservation of energy. Students can understand all important topics given in Class 11 Physics Chapter 5 by practicing important questions given on Vedantu. 

4. How important is Class 11 Physics Chapter 5 for JEE and NEET Exams?

Class 11 Physics Chapter 5 is very important for JEE and NEET exams. Students should understand the basic concepts of this chapter to score high marks in the board exams and entrance exams. They should practice formulas and numerical for solving numerical problems. Students can get high marks in the board exams and entrance exams.

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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

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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.

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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

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NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

Ncert solutions class 11 physics chapter 5 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion is one of the key tools to prepare Physics efficiently for the examination. The solutions of Chapter 5 Laws of Motion, are given below to help students learn new concepts in an interactive and easy way. Using NCERT Solutions will help students to get an idea about the question paper pattern and the marking scheme of the exam.

In the previous chapter, we mainly focused on the motion of a particle. We saw that uniform motion depends on velocity, whereas non-uniform motion depends on acceleration. So far, we did not give much thought to what governs the motion of bodies. Now, this becomes a basic question in the present chapter. To learn all the basic concepts covered in this chapter, NCERT Solutions for Class 11 Physics are designed with accurate and authentic information strictly based on the latest CBSE Syllabus for 2023-24 and its curriculum. Let’s have a look at NCERT laws of motion Class 11 solutions.

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Access the answers to NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

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 skillfully held stationary in the sky

(d) a car moving with a constant velocity of 30 km/h on a rough road

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

(a) The raindrop is falling at a constant speed. Therefore, acceleration will become zero. When acceleration is zero, the force acting on the drop will become zero since F = ma.

(b) The cork is floating on water, which means the weight of the cork is balanced by the upthrust. Therefore, the net force on the cork will be zero

(c) Since the car moves with a constant velocity, the acceleration becomes zero. Therefore, the force will be zero.

(d) The net force acting on the high-speed electron will be zero since the electron is far from the material objects and free of electric and magnetic fields.

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 Solutions change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

(a) During the upward motion of the pebble, the acceleration due to gravity acts downwards, so the magnitude of the force on the pebble is

F = mg = 0.05 kg x 10 ms -2 = 0.5 N

The direction of the force is downwards

(b) During the downward motion also the magnitude of the force will be equal to 0.5 N and the force acts downwards

(c) If the pebble is thrown at an angle of 45° with the horizontal direction, it will have both horizontal and vertical components of the velocity. At the highest point, the vertical component of velocity will be zero but the horizontal component of velocity will remain throughout the motion of the pebble. This component will not have any effect on the force acting on the pebble. The direction of the force acting on the pebble will be downwards and the magnitude will be 0.5 N because no other force other than acceleration acts on the pebble.

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 with1 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.

(a) Mass of stone = 0.1 kg

Acceleration = 10 ms -2

Net force, F = mg = 0.1 x 10 = 1.0 N

The force acts vertically downwards

(b) The train moves at a constant velocity. Therefore, the acceleration will be equal to zero. So there is no force acting on the stone due to the motion of the train. Therefore, the force acting on the stone will remain the same (1.0 N)

(c) When the train accelerates at 1m/s 2 , the stone experiences an additional force of F’ = ma = 0.1 x 1 = 0.1 N. The force acts in the horizontal direction.

But 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 train floor, its acceleration will be the same as that of the train.

Therefore, the magnitude of the force acting on the stone, F = ma = 0.1 x 1 = 0.1 N.

It acts along 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 – mv 2 /l

(iii) T + mv 2 /l

T is the tension in the string. [Choose the correct alternative].

Solution (i) T

The net force acting on the particle is T, and it is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.

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

Force = – 50 N (since it is a retarding force)

Mass m = 20 kg

u = 15 m s -1

Force F = ma

a = F/m = -50/20 = – 2.5 ms -2

Using the equation, v = u + at

0 = 15 + (- 2.5) t

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

Mass, m = 3.0 Kg

u = 2.0 m/s

v = 3.5 m/s

F = m [(v-u)/t] (since a = (v -u)/t )

F = 3 [ (3.5 – 2)/25] = 0.18 N

The force acts in the direction of motion of the body

7. 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.

Mass, m = 5 kg

Force, F 1 = 8N and F 2 = 6N

The resultant force of the body

Acceleration, a = F/m

a = 10/ 5 = 2m/s in the direction of the resultant force

The direction of the acceleration,

tan β = 6/8 = 0.75

β = tan -1 (0.75)

β = 37 0 with 8N

8. The driver of a three-wheeler moving at 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.

Initial velocity, u= 36 km/h

Final velocity, v = 0

Mass of the three-wheeler, m 1 = 400 Kg

Mass of the driver, m 2 = 65 Kg

Time taken to bring the vehicle to rest = 4.0 s

Acceleration, a = v- u/t = (0 – 10)/ 4 = – 2.5 m/s

Now, F = (m 1 + m 2 )/ a = (400 + 65) x (-2.5)

= – 1162.5 N = – 1.16 x 10 3 N

The negative sign shows that the force is retarding

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

Mass of the rocket, m = 20000 kg = 2 x 10 4 kg

Initial acceleration = 5 ms -2

g = 9.8 m/s 2

The initial thrust (force) should give an upward acceleration of 5 ms -2 and should overcome the force of gravity.

Thus, the thrust should produce a net acceleration of 9.8 + 5.0 = 14.8 ms -2 .

Using Newton’s second law of motion, the initial thrust acting on the rocket

Thrust = force = mass x acceleration

F = 20000 x 14.8 = 2.96 x 10 5 N

10 . A body of mass 0.40 kg moving initially with a constant speed of 10 ms -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.

Mass of the body = 0.40 kg

Initial velocity, u = 10 m/s

Force, f = -8 N (retarding force)

Using the equation S = ut + (½) at 2

(a) Position at the time t = – 5 s

The force starts acting on the body from t = 0 s

So the acceleration of the body when the time is – 5 s is 0

S 1 = (10)(-5) + (½) (0) (-5) 2 = – 50 m

(b) Position at the time t = 25 s

The acceleration of the body due to the force acting in the opposite direction

Acceleration, a = F/a = -8 /0.4 = -20 ms -2

S 2 = (10)(25) + (½) (-20) (25) 2 = – 6000 m

(c) Position at the time t = 100 s

For the first 30 sec, the body will move under the retardation of the force, and after that, the speed will remain constant.

Therefore, distance covered in 30 sec

S 3 = (10)(30) + (½)(-20)(30 2 )

= 300 – 9000 = – 8700m

The speed after 30 sec is

v = 10 – (20 x 30) = 590 m/s

The distance covered in the next 70 sec is

S 4 = – 590 x 70 + (½) (0) (70) 2 = – 41300 m

Therefore the position after 100 sec = S 3 + S 4 = – 8700 – 41300 = – 50000m

11. A truck starts from rest and accelerates uniformly at 2.0 ms -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 (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)

Initial velocity, u = 0

Acceleration, a = 2 ms -2 ,

Using equation, v = u + at, we get

v = 0 + 2 x 10 = 20 m/s

The final velocity, v = 20 m/s

At time, t = 11 sec, the horizontal component of the velocity in the absence of the air resistance remains unchanged

V x =20 m/s

The vertical component of the velocity is given by the equation

V y = u + a y t

Here t = 11 – 10 = 1s and a y = a = 10 m/s

Therefore, the resultant velocity v of the stone is

v = ( v x 2 + v y 2 )½

v = (20 2 + 10 2 ) 1/2

v = 22.36 m/s

tan θ = v y /v x = 10/20 = ½ = 0.5

θ = tan -1 (0.5 ) = 26. 56 0 from the horizontal

(b) When the stone is dropped from the truck, the horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity which is equal to 10 m/s 2 and it acts vertically downwards.

12. 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 ms -1 . What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position?

(a) When the bob is at one of its extreme positions, the velocity is zero. So, if the string is cut the bob will fall vertically downward under the force of its weight F = mg.

(b) At its mean position the bob has a horizontal velocity. If the string is cut, the bob will behave like a projectile and will fall on the ground after taking a parabolic path.

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?

Mass of the man, 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 of 10 m/s, it’s acceleration= 0.

R = mg = 70 x 10 = 700 N

(b) Lift moving downwards with a = 5 ms -2

Using Newton’s second law of motion, the equation of motion can be written as

R = m (g – a) = 70 (10 – 5) = 350 N

(c) Lift moving 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 . a = g

:. R = m(g -a) = m(g – g) = 0

The man will be in a state of weightlessness

14. Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).

When t＜0, the distance covered by the particle is zero. Therefore, the force on the particle is zero.

When 0＜ t ＜4s, the particle is moving with a constant velocity. Therefore, the force will be zero.

When t＞4s, the particle remains at a constant distance. Therefore, the force of the particle will be zero.

Impulse at t = 0.

Here, u = 0

v = ¾ = 0.75 m/s

Impulse= total change in momentum = mv – mu = m (v – u)

= 4 (0 – 0.75) = -3 kg m/s

Impulse at t= 4s

u = 0.75 m/s, v = 0

Impulse = m (v – u) = 4 (0 – 0.75) = -3 kg m/s

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

Mass of the body A, m 1 = 10 kg

Mass of the body B, m 2 = 20 kg

Horizontal force = 600 N

Total mass of the system, m = m 1 + m 2 = 30 kg

Applying Newton’s second law of motion, we have

a = F/m = 600/30 = 20 m/s 2

(i) When the force is applied on A (10 kg)

F – T = m 1 a

T = F – m 1 a

T = 600 – (10 x 20)

= 600 – 200 = 400 N

(ii) When the force is applied on B (20 kg)

F – T = m 2 a

T = F – m 2 a

= 600 – (20 x 20) = 200 N

16. 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.

Smaller mass, m 1 = 8 kg

Larger mass, m 2 = 12 kg

Tension in the string = T

The heavier mass m 2 will move downwards and the smaller mass m 1 will move upwards.

Applying Newton’s second law,

For mass m 1 :

T – m 1 g = m 1 a —– (1)

For mass m2:

m 2 g – T = m 2 a ——(2)

Add (1) and (2)

(m 2 – m 1 ) g = (m 1 + m 2 ) a

a = (m 2 – m 1 ) g/ (m 1 + m 2 )

=[ (12 – 8)/(12 + 8)] x 10 = (4 /20) x 10= 2m/s

Therefore, acceleration of the mass is 2 m/s 2

Substituting this value of acceleration in equation (ii) we get

m 2 g – T = m 2 a

= 2m 1 m 2 g/(m 1 + m 2 )

T = 2 x 12 x 8 x 10/ (12 + 8)

Therefore, the tension on the string is 96 N

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.

Let m 1 , m 2 be the masses of the two daughter nuclei and v 1 ,v 2 be their respective velocities of the daughter nuclei. Let m be the mass of the parent nucleus.

Total linear momentum after disintegration = m 1 v 1 +m 2 v 2 .

Before disintegration, the nucleus is at rest. Therefore, its linear momentum before disintegration is zero.

Applying the law of conservation of momentum,

Total linear momentum before disintegration = Total linear momentum after disintegration

0 = m 1 v 1 + m 2 v 2

v 1 =-m 2 v 2 /m 1

The negative sign indicates v 1 and v 2 are in opposite directions.

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?

Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s

The initial momentum of each ball before the collision

= 0.05 x 6 = 0.3 kg m/s

After the collision, the balls change in the direction of motion without a change in the magnitude of the velocity

Final momentum after collision of the first ball = – 0.05 x 6 = – 0.3 kg m/s

Final momentum after collision of the second ball = 0.3 kg m/s

Impulse imparted to the first ball = (-0.3) – (0.3) = – 0.6 kg m/s

Impulse imparted to the second ball = (0.3) – (-0.3) = 0.6 kg m/s

The two impulses are opposite in direction.

19. 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 ms -1 .What is the recoil speed of the gun?

Mass of the shell, m = 0.020 kg

Mass of the gun, M = 100 Kg

Speed of the shell = 80 m/s

The initial velocity of the shell and the gun is zero. So, the initial momentum of the system is zero.

Applying the law of conservation of momentum, the initial momentum is equal to the final momentum.

0 = mv – MV

Recoil speed of the gun, v = mv/M

v = (0.020 x 80)/100

v = 0.016 m/s

20. 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.)

Velocity of the ball = 54 km/h

The ball is deflected back such that the total angle = 45 0

The initial momentum of the ball is mucosӨ = (0.15 x 54 x 1000 x cos 22. 5)/3600

= 0.15 x 15 x 0.9239 along NO

Final momentum of the ball = mucosӨ along ON

Impulse = change in the momentum = mucosӨ – (-mucosӨ) = 2mucosӨ = 2 x 0.15 x 15 x 0.9239 = 4.16 kg.m/s

21. 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?

Mass of the stone = 0.25 kg

Radius, r = 1.5 m

Number of revolution in a second, n= 40/60 = (⅔) rev/sec

The angular velocity, ω = 2πn = 2 x 3.14 x (⅔)

The tension on the string provides the necessary centripetal force

T = 0.25 x 1.5 x [2 x 3.14 x (⅔)] 2

Maximum tension on the string, T max = 200 N

T max = mv 2 max /r

v 2 max = (T max x r)/m

= ( 200 x 1.5)/0.25 = 1200

v max = 34. 6 m/s

Therefore, the maximum speed of the stone is 34.64 m/s

22. If in problem 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

Solution: (b)

At each point of the circular motion, the velocity will be tangential. If the string breaks suddenly the stone moves in the tangential direction according to Newton’s first law of motion.

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 lawnmower than to push it

(d) a cricketer moves his hands backwards while holding a catch

• The horse pushes the ground with a certain force when it pulls the cart. By applying the third law of motion, the ground will exert an equal and opposite reaction force upon the feet of the horse. This causes the horse to move forward. In empty space, the horse will not experience a reaction force. Therefore, the horse cannot pull the cart in empty space.
• Due to inertia of motion. When a bus stops all of a sudden, the lower part of a person’s body that is in contact with the seat comes to rest suddenly but the upper part will continue to be in motion. As a result, the person’s upper half of the body is thrown forward in the direction of the motion of the bus.
• When the lawnmower is pulled, the vertical component of the applied force acts upwards. This reduces the effective weight of the lawnmower. When the lawn mower is pushed, the vertical component acts in the direction of the weight of the mower. Therefore, there is an increase in the weight of the mower. So, it is easier to pull a lawnmower than to push it.
• When the batsman hits the ball, the ball will have a large momentum. When he moves his hands backwards the time of impact is increased contact, so the force is reduced.

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

This graph could be of a ball rebounding between two walls separated by a distance of 2 cm. The ball rebounds between the walls every 2 seconds with a uniform velocity.

Velocity = displacement/ time = (2 x 10 -2 )/2 = 0.01 m/s

Initial momentum = mu = 0.04 x 0.01 = 4 x 10 -4 kgm/s

Final momentum = mv = 0.04 x (-0.01) = – 4 x 10 -4 kgm/s

Magnitude of impulse = Change in momentum

= (4 x 10 -4 )-(- 4 x 10 -4 ) = 8 x 10 -4 kgm/s

The time between two consecutive impulses is 2 sec, so the ball receives an impulse every 2 seconds.

25. Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms -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.)

Here acceleration of conveyor belt a = 1 ms -2 ,

Coefficient of static friction, μ s = 0.2

mass of man, m = 65 kg

Net Force = ma = 65 x 1 = 65N

This net force is due to the friction between the belt and the man

Maximum static friction = μ s N

Beyond that, the man starts moving relative to the belt as per the pseudo force

At maximum static friction

μ s N = μ s .mg = ma max

a max = μs g = 0.2 x 10 = 2 m/s 2

26. A stone of mass m tied to the end of a string is revolving in a vertical circle of radius R. The net force 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.

Solution: (a)

The net force at the lowest point is (mg – T 1 ) and the net force at the highest point is (mg + T 2 ). Therefore option (a) is correct.

Since mg and T 1 are in mutually opposite directions at the lowest point and mg and T 2 are in the same direction at the highest point.

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) the action of the rotor of the helicopter on surrounding air

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

Mass of helicopter = 1000 kg

Crew and passengers weight = 300 kg

Vertical acceleration, a = 15 ms -2 and g = 10 ms -2

The total mass of the system, m i = 1000 + 300 = 1300 Kg

(a) Force on the floor of the helicopter by the crew and passengers

R – mg = ma

= m (g + a) = 300 (10 + 15) N = 7500 N

(b) Action of the rotor of the helicopter on surrounding air is due to the mass of the helicopter and the passengers.

R’ – m i g = m i a

R’ = m i (g+a)

= 1300 x (10 + 15) = 32500 N

This force acts vertically downwards

(c) Force on the helicopter due to the surrounding air is the reaction of the force applied by the rotor on the air. As action and reaction are equal and opposite, therefore, the force of reaction, F = 32500 N. This force acts vertically upwards.

28. A stream of water flowing horizontally with a speed of 15 m/s 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?

Speed of water flowing, v= 15 m/s

Cross-sectional area of the tube, A= 10 -2 m 2

In one second, the distance travelled is equal to the velocity v

Density of water, ρ = 1000 kgm -3

Mass of water hitting the wall per second = ρ x A x v

= 1000 kgm -3 x 10 -2 m 2 x 15 m/s = 150 kg/s

Force exerted on the wall because of the impact of water = momentum loss of water per second

= Mass x velocity

= 150 kg/s x 15 m/s = 2250 N

29. Ten one rupee coins are put on top of one another on a table. Each coin has a mass of 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 

(c) the reaction of the sixth coin on the seventh coin

(a) The force on the 7th coin is due to the weight of the three coins kept above it.

Weight of one coin is mg. So the weight of three coins is 3mg

Force exerted on the 7th coin is (3mg)N and the 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, the force on the 7th coin due to the 8th coin will be the same as the force on the 7th coin by the three coins above it.

Therefore, the force on the 7th coin by the 8th coin is (3mg)N and the force act vertically downwards.

(c) The sixth coin is under the weight of four coins above it and experiences a downward force due to the four coins.

The total downward force on the 6th coin is 4mg

Applying Newton’s third law of motion, the 6th coin will exert a reaction force upwards. Therefore, the force exerted by the 6th coin on the 7th coin is equal to 4mg and acts in the upward direction.

30. 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?

The speed of the aircraft executing the horizontal loop = 720 km/h = 720 x (5/18) = 200 m/s

The angle of banking = 15°

tanθ= v 2 /rg , we have

r = v 2 /gtanθ = (200 x 200)/(10 x tan 15 0 ) = (200 x 200)/ (10 x 0.2679)= 14931 m

31. A train runs along an unbanked circular track of radius of 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?

Radius of the track = 30 m

Velocity of the train = 54 km/h = 54 x (5/18) = 15 m/s

Mass of the train = 10 6 kg

The required centripetal force is provided by the force of lateral friction due to the rails on the wheels of the train

The angle of banking required to prevent wearing out of the rails

tanθ= v 2 /rg = (15 x 15)/ (30 x 10) = 0.75

θ = tan -1 (0.75) = 37 0

32. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 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?

Mass of the block = 25 kg

Mass of the man = 50 kg

Acceleration due to gravity = 10 m/s 2

Weight of the block, F = 25 x 10 = 250 N

Weight of the man, W = 50 x 10 = 500 N

In the 1st case, the man lifts the block directly by applying an upward force of 250 N (same as the weight of the block)

According to Newton’s third law of motion, there will be a downward reaction on the floor. The action on the floor by the man.

= 500 N + 250 N = 750 N.

In the 2nd case, the man applies a downward force of 25 kg wt. According to Newton’s third law of motion, the reaction is in the upward direction.

In this case, action on the floor by the man

= 500 N – 250 N= 250 N.

Therefore, the man should adopt the second method.

33. A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break, when the monkey

(a) climbs up with an acceleration of 6 ms -2

(b) climbs down with an acceleration of 4 ms -2

(c) climbs up with a uniform speed of 5 ms -1

(d) falls down the rope nearly freely under gravity

(Ignore the mass of the rope).

Mass of the monkey = 40 kg

Maximum tension the rope can withstand, T max = 600 N

(a) When the monkey climbs up with an acceleration of 6m/s2,

Tension T – mg = ma

T = m (g+a)

T = 40 (10 + 6)

Since T > Tmax, the rope will break

(b) When the monkey climbs down with the acceleration of 4m/s 2

mg – T = ma

T = mg – ma = m (g – a)

= 40 (10 – 4) = 240 N

Since T ＜T max , the rope will not break

(c) When the monkey climbs with a uniform speed 5m/s. The acceleration will be zero. The equation of motion is

T – mg = ma

T – mg = 0

T = mg = 40 x 10 = 400 N

(d) When the monkey falls freely, the acceleration of the monkey will be equal to the acceleration due to gravity

The equation of motion is written as

mg + T = mg

T = m(g-g) = 0

34 . 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 (Fig.). 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 Solution to (b) change, when the bodies are in motion? Ignore the difference between μ s and μ k .

Mass of the body, m A = 5 Kg

Mass of the body, m B = 10 kg

Applied Force = 200 N

Coefficient of friction between the bodies and the table μ s = 0.15

(a) The force of friction is given by the relation:

f s = μ(m A + m B )g

= 0.15 (5 + 10) x 10

= 1.5 x 15 = 22.5 N, towards left

Therefore, the net force on the partition is 200 – 22.5 = 177. 5 N rightward

According to Newton’s third law, action and reaction are in the opposite direction

Therefore, the reaction of the partition will be 177.5 N, in the leftward direction

(b) Force of friction on mass A

F A = μm A g

= 0.15 x 5 x 10 = 7.5 N leftward

The net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightwards

An equal amount of reaction force will be applied on mass A by B, i.e., 192.5 N acting leftward

When the wall is removed, the two bodies move in the direction of the applied force

The net force acting on the moving system = 177. 5 N

The equation of motion for the system of acceleration a, can be written as

Net force = (m A + m B ) a

a = Net force/ (m A + m B )

= 177.5/ (5 + 10) = 177. 5/15 = 11. 83 m/s 2

Net force causing mass A to move

F A = m A a = 5 x 11.83 = 59. 15 N

Net force exerted by the mass A on mass B= 192.5 – 59.15= 133. 35 N

This force acts in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3N, acting opposite to the direction of motion.

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 -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.

Mass of the block = 15 kg

Coefficient of static friction between the block and the trolley, p= 0.18

Acceleration of the trolley = 0.5 m/s 2

(a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N, this force acts in the direction of motion of the trolley

Force of friction, F f = p mg = 0.18 x 15 x 10 = 27 N

Force experienced by the block will be less than the friction. So the block will not move. So, for a stationary observer on the ground, the block will be stationary.

(b) The observer moving with the trolley has an accelerated motion. He forms a non-inertial frame in which Newton’s laws of motion are not applicable. The trolley will be at rest for the observer moving with the trolley

36. 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 Fig. 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 ms -2 . At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Force experienced by box, F = ma = 40 x 2 = 80 N

Frictional force F f = μmg = 0.15 x 40 x 10 = 60 N

Net force = F – F f = 80 – 60 = 20 N

Backward acceleration produced in the box, a =20/40(Net Force/m)

a = 0.5 ms -2

If t is time taken by the box to travel s =5 metre and fall off the truck then from

S = ut + 1/2 at 2 `

5 = 0 x t + (½) x 0.5 t 2

t= √(5×2/0.5) = 4.47 s`

If the truck travels a distance x during this time, then again from

S = ut + 1/2 at 2

x = 0 x 4.47 + (½) x 2(4.47) 2 = 19.98 m

37. A disc revolves with a speed of 33⅓ rpm 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 coefficient of friction between the coins and record is 0.15, which of the coins will revolve with the record?

Speed of revolution of the disc = 33 ⅓ rpm= 100/3 rpm= 100/ (3 x 60) rps = 5/9 rps

ω = 2πv = 2 x (22/7)x (5/9) = 220/63 rad/s

The coins revolve with the disc, the centripetal force is provided by the frictional force mv 2 /r ≤ μmg ——–(1)

As v = rω, equation (1) becomes mrω 2 /r ≤ μmg

r ≤ μg/ω 2 = (0.15 x 10)/(220/63) 2 = 12 cm

For coin A, r = 4 cm

The condition (r≤ 12) is satisfied for coin placed at 4cm, so coin A will revolve with the disc

For coin B, r = 14 cm

The condition (r≤ 12) is not satisfied for the coin placed at 14cm, so coin B will not revolve with the disc.

38. 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?

When the motorcyclist is at the uppermost point of the death-well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. The outward centrifugal force acting on the motorcyclist is balanced by these two forces.

R + mg = mv 2 /r ———(1)

Here, v is the velocity of the motorcyclist

m is the mass of the motorcyclist and the motorcycle

Because of the balance between the forces, the motorcyclist does not fall

The minimum speed required at the uppermost position to perform a vertical loop is given by the equation (1) when R = 0

mg = mv 2 min /r

V 2 min = gr

V min = √10 x 25 = 15.8 m/s

39. 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?

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction between the wall and his clothing, μ = 0.15

Number of revolution of hollow cylindrical drum = 200 rev/min= 200/60 = 10/3 rev/s

The centripetal force required is provided by the normal N of the wall on the man

N = mv 2 /R = mω 2 R

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force acting vertically upwards.

The man will not fall if mg ≤ limiting frictional force f e (μN)

mg ≤ μ (mω 2 R)

Therefore, for minimum rotational speed of the cylinder

ω 2 = g/Rμ = 10/(0.15 x 3) = 22. 2

ω =√22.2 = 4.7 rad/s

40. 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 ω ≤ √g / R . What is the angle made by the radius vector joining the centre to the bead with the vertically downward direction for ω = √2g/ R? Neglect friction.

Let θ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction

mg = N cosθ —–(1)

mrω 2 = Nsinθ —– (2)

(or) m (Rsinθ) ω 2 = Nsinθ

Substituting the value of N in (1)

mg = mRω 2 cosθ

(or) cos θ = g/Rω 2 ———(3)

As l cosθ I ≤ 1 the bead will remain at the lowermost point

g/Rω 2 ≤ 1 or ω ≤ √g/R

For ω = √2g/ R , equation (3) becomes

cos θ = g/Rω 2

cos θ = (g/R) (R/2g) = ½

Chapter 5 Solutions for Class 11 Physics PDF given here has answers to the questions in the textbook, along with NCERT Exemplar Problems , Worksheets, Extra questions, Short answer questions MCQs, HOTS (High Order Thinking Skills) and Important questions.

Subtopics of Class 11 Physics Chapter 5 Laws of Motion

Some of the topics and subtopics covered in the NCERT Solutions for Class 11 Physics Chapter 5 are listed below.

Chapter 5 of Class 11 Physics is an important chapter in mechanics. Numericals on the conservation of momentum are frequently asked in the exams. If all three laws of motion are well understood, students will be able to relate these laws to day-to-day activities. An ample amount of everyday practice will help you in solving the numericals easily. While practising, one can refer to the NCERT Solutions for Class 11 Laws of Motion to cross-check the right answers.

BYJU’S provides advanced study materials, CBSE notes , videos and animation that help you to understand the topic thoroughly. Our interactive teaching method helps you to remember the topic for a significant period. This will help you to score good marks in the Class 11 examinations as well as in the entrance examinations.

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Important Questions Class 11 Physics Chapter 5

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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.

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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

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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.

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

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Solutions for chapter 5: laws of motion.

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

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

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

it is at rest.

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

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

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

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

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

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

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

neither the force nor the torque need to be zero.

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

`-(0.45 hati + 0.6 hatj)`

`-(0.9 hati + 1.2 hatj)`

`-5 (hati + hatj)`

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

0.75 kg ms –1

1.5 kg ms –1

14 kg ms –1

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

Conservation of energy.

Newton’s first law only.

Newton’s second law only.

Both Newton’s second and third law.

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

frictional force along westward.

muscle force along southward.

frictional force along south-west.

muscle force along south-west.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 A cricket bowler releases the ball in two different ways

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

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

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

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

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

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

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

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

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

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

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

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

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M-Physics Tutorial

Class 11 physics chapter 5 important questions laws of motion.

It is important for the students that all the concepts should be very clear for better marks in future. Here, we are providing important conceptual questions and answers for class 11 physics chapter 5 Laws of Motion. In this lesson, students will learn about Laws of Motion . This will not only help the students to know the important questions but will also help them during revision.

Q.1. Is force needed to keep a body moving with uniform velocity? Ans. No.

Q.2. What is inertia? What gives the measure of inertia? Ans. The inherent property of the bodies that they do not change their state unless acted upon by an external force is called inertia. Mass of the body give the measure of its inertia.

Q.3. How is inertia related to mass of a body? Ans. More is the mass of a body, more is its inertia.

Q.4. If you jerk a piece of paper under a book quick enough the book will not move. Why? Ans. The book remains there as much due to the inertia of rest.

Q.5. Define the term momentum. Give its SI unit. Ans. Momentum of a body is defined as the total quantity of motion contained in a body and is measured as the product of the mass of body and its velocity. The SI unit of momentum is kg ms -1 .

Q.6. A bullet fired from a rifle is more dangerous than an air molecule hitting a person, though both of them have almost the same speed. Explain. Ans. It is because mass of the bullet is about 1025 times that of the air molecule. As such the momentum of bullet is very large.

Q.7. From which Newton's law of motion the definition of force comes? Ans. First law of motion.

Q.8. Is Newton's second law (F = m.a) always valid. Give an example in support of your answer. Ans. It is valid in an inertial frame of reference. In a non inertial frame of reference (such as a car moving along a circular path) Newton's second law does not hold apparently.

Q.9. the rate of change of momentum of a body is 5 kg m s -1 . What is the force acting on the body? Ans. 5 N.

Read also: Laws of Motion Class 11 Physics Notes Chapter 5

Q.10. Give and state SI unit of force. Ans. The SI unit of force is Newton(N). Force is said to be 1 N, if it produces an acceleration of 1m s -2 in a body of mass 1 kg.

Q.11. What is the difference between mN and nm? Ans. The symbol mN stands for millinewton (10 -3 N) and nm stands for nanometre (10 -9 m).

Q.12. Action and reaction forces do not balance each other. Why? Ans. When one body exerts a force on another body, it gets an equal and opposite reaction. The action and reaction forces do not cancel each other, as they do not act on the same body.

Q.13. What is an impulsive force? Ans. A force which acts for a small time and also varies with time is called an impulsive force.

Q.14. What is the net force on a cork floating on water? Ans. Zero.

Q.15. Which is greater -- the attraction of 10 kg mass for earth or the earth's attraction for 10 kg mass?

Ans. Both are equal.

Q.16. A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body? Ans. Yes, the force is needed to change the direction of motion of the body i.e. to deflect the body from its straight path to circular path.

Q.17. State principle of conservation of momentum. Ans. The principle of conservation of momentum states that if no external force acts on a system the momentum of the system remains constant.

Read also: Class 11 Physics Chapter 5 MCQs with Answer Laws of Motion

Q.18. State Newton's first law of motion. Ans. Newton's first law of motion states that every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled by some external force to change that state.

Q.19. State Newton's second law of motion. Ans. Newton's second law of motion states that the time rate of change of momentum of a body is directly proportional to the external force applied on it and the change in momentum takes place in the direction of force.

Q.20. State Newton's third law of motion. Ans. Newton's third law of motion states that to every action there is an equal and opposite reaction.

Q.21. What is an impulse? Ans. Impulse received during an impact is defined as the product of the average force and the time for which the force acts.

Q.22. Can a body remain in state of rest when external forces acting on it. Explain your answer. Ans. Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Q.23. If the net force acting on a body be zero, will the body remain necessarily in rest position? Explain your answer. Ans. In case the body is in uniform motion along a straight line, it will continue to do so, if the net force acting on a body be zero.

Q.24. Name physical situation, where the mass of a body changes with time. Ans. When the body moves with a speed comparable to the speed of light. For example, the mass of a charged particle increases, when it is accelerated in a cyclotron.

Q.25. When are the two bodies said to possess equal masses? Ans. If the same force acting on the two bodies produces the same acceleration, then the two bodies must be of equal masses.

Q.26. Is a bus moving along a circular track, an inertial frame of reference? Ans. No. It is a non-inertial frame of reference.

Q.27. What can be said about the motion of the vehicle, if a plumb line hanging from its roof drops vertically? Ans. Either the vehicle is at rest or it is moving with constant velocity.

Q.28. The assertion made by the Newton's first law of motion that everybody continues in a state of uniform motion in the absence of external force appears to be contradicted in everyday experience. Why? Ans. When we roll a ball on the floor, it ultimately stops because of the force of friction of the ground. Thus, the state of uniform motion of the object changes due to the external force (friction). On the earth, every change in uniform motion of a body can be connected with an external force acting on it. However, in free space, when no external force acts, the state of motion described by the Newton's first law of motion can be obtained and experienced.

Q.29. According to Newton's first law of motion, a body moving with a uniform speed along a straight line should continue moving. In practice a body in motion stops after some time. Explain the reason. Ans. The force of friction acts as the external retarding force. It brings the body to rest.

Q.30. An astronaut accidentally gets thrown out of his small space ship accelerating in interstellar space at a constant rate of 100 ms -2 . What is the acceleration of the astronaut, the instant after he is outside the spaceship? Assume that there are no nearby stars to exert gravitational force on him. Ans. In the interstellar space, moment the astronaut is out of the spaceship, no external force acts on him. It is because, there are no nearby stars to exert gravitational force on him. The spaceship too can not exert any appreciable gravitational force on him due to its small size. Therefore, once the astronaut is out of the spaceship, his acceleration will be zero.

Q.31. According to Newton's third law, every force is accompanied by an equal and opposite force. How can a movement ever take place? Ans. Since action and reaction do not act on the same body, they do not cancel each other. Therefore, a body may move either under the effect of the action of force or the other body may move under the effect of reaction on it.

Q.32. It is easier to pull than to push a body. Explain. Ans. When we pull a body, the vertical component of the applied pull acts opposite to the weight of the body and it reduces its effective weight. On the other hand, when a body is pushed, the vertical component of the applied push adds to the weight of the body and hence its effective weight increases. Thus, the effective weight is lesser, when the body is pulled. Hence, it is easier to pull than to push a body.

Q.33. A stone when thrown on a glass window smashes the window pane into pieces, but a bullet from the gun passes through making a clean hole. Why? Ans. The velocity of the stone is much less than that of the bullet fired from a gun. Due to its low speed, the stone remains in contact contact with the window pane for a longer time and its motion is shared by whole of the window pane. As a result, it smashes into pieces. On the other hand, the bullet fired from the gun remains in contact with the window pane for such a small time that it can share its motion only with the portion of the window pane, it comes in contact with. As such, it makes a clean hole in the window pane.

Q.34. What is friction? Ans. The opposing force that is set up between the surfaces of contact, when one body slides or rolls or tends to do so on the surface of another body is called friction.

Q.35. Define limiting friction. Ans. The maximum value of the force of friction which comes into play before a body just begins to slide over the surface of another body is called limiting value of static friction.

Q.36. Why do we call friction a self adjusting force? Ans. When applied force is zero, friction is zero. As the applied force is increased, friction also increases and becomes equal to the applied force. It happens so, till the body does not start moving. That is why, friction is called a self adjusting force.

Q.37. What is the unit of coefficient of limiting friction? Ans. It has no unit.

Q.38. What are the factors on which the coefficient of friction between two surfaces depend? Ans. The coefficient of friction between two surfaces depends upon the nature of the two surfaces and their state of roughness.

Q.39. What is the relation between angle of repose and angle of friction? Ans. Angle of friction and angle of repose are equal.

Q.40. Why is friction a non-conservative force? Ans. It is because work done against friction along a closed path is non zero.

Q.41. What happens to limiting friction, when a wooden block is moved with increasing speed on a horizontal surface? Ans. The limiting friction decreases as the wooden block is moved with increasing speed on the horizontal surface.

Q.42. Why are tyres made of rubber and not iron? Ans. It is because, the coefficient of friction between rubber and concrete (material of the road) is less than that between iron and the road.

Q.43. Give any three advantages of friction. Ans.  

• Friction helps us to walk.
• Brakes make use of friction to stop the vehicles.
• Friction helps to transmit power from the motors and engines to other machines by making use of belts and clutches.

Q.44. Why are wheels made circular? Ans. The rolling friction is less than the sliding friction. The wheels are made circular so as to convert the sliding friction into the rolling friction.

Q.45. It is easier to roll a barrel than to slide it along the road. Why? Ans. The rolling friction is lesser as compared to the sliding friction.

Q.46. What happens to the fluid friction, as speed of the object moving through it is increased? Ans. The fluid friction increases, as the speed of the object moving through it, is increased.

Q.47. smoother the surface lesser is friction. Comment. Ans. When the surfaces are made smoother, the size of the irregularities in the surfaces decreases. As a result, the area of actual contact decreases. As the number of atoms in contact will also decrease due to decrease in area of contact, the force of molecular attraction and hence the force of friction decreases.

Q.48. Polishing a surface beyond a certain limit may increase friction. Why? Ans. When the surfaces are polished beyond a certain limit, the area at each point of contact becomes very small. However, the actual area of contact between the two surfaces increases appreciably. It is because, the number of points of contact becomes very large on making the surface is highly polished. Since number of atoms (or molecules) of the two surfaces in contact is proportional to their area in contact, the force of friction increases due to the greater force of molecular attraction between the two surfaces.

Q.49. A horse has to apply more force to start a cart than to keep it moving. Why? Ans. When the cart is at rest, the friction between the wheels of the cart and the road is static in nature and once the cart starts moving, it is kinetic in nature. The kinetic friction is always less than the static friction. Due to this reason, a horse has to apply more force to start a cart than to keep it moving.

Q.50. Why do we slip on a rainy day? Ans. On a rainy day, the wet ground becomes very smooth. As a result, the coefficient of friction between our feet and the wet ground gets much reduced. Consequently, the friction between the feet and the ground becomes very small. It may cause us to slip.

Q.51. Explain how friction helps in walking? Ans. When we walk, with push the ground in backward direction. Our foot will get equal and opposite reaction from the ground only, if the food does not slip i.e. there is adequate friction between the food and the ground. The forward horizontal component of the reaction helps us to move forward.

Q.52. How can proper inflation of tyres save fuel? Ans. Under the weight of the vehicle, the road surface gets depressed and a small hump is created just ahead of the wheel. In case the tyres are not properly inflated, the tyres require a large effort to climb the hump. In case the tires are properly inflated, the compression of the tyres as much smaller and hence they experience less resistance (rolling friction). In other words, it will make the vehicle to cover greater distance for the same quantity of fuel spent.

Q.53. What furnishes centripetal force for earth to go round the sun? Ans. The gravitational pull of the sun on the earth.

Q.54. What provides the centripetal force to satellite revolving round the earth? Ans. Gravitational force of attraction on the satellite due to the earth.

Q.55. What furnishes the centripetal force for the electrons to go round the nucleus? Ans. The electrostatic force of attraction on the electrons due to the nucleus.

Q.56. For uniform circular motion, does the direction of the centripetal force depends upon the sense of rotation? Ans. The direction of the centripetal force does not depend, whether the body is moving in clockwise or anticlockwise direction. It is always directed along the radius and towards the centre of the circle.

Q.57. Can centripetal force produce rotation? Ans. No. A centripetal force can move a body along a circular path but it can not produce rotational motion.

Q.58. What provides the centripetal force to a car taking a turn on a level road? Ans. The force of friction between the tyres and the road provides the necessary centripetal force to the car to take a turn on a level road.

Q.59. The speed of the stone is increased beyond the maximum permissible value (the value of speed at which the tension in the string becomes equal to the maximum tension the string can stand) the string breaks suddenly. What will be the trajectory of the stone after the string breaks? Ans. At the instant, the string breaks, the stone will fly off tangentially from its position at that instant. It is because, the speed of the stone at any instant is directed along tangent to the circular path at that point.

Q.60. A train moves on an unbanked circular band of rails. Which rail will wear out faster? Ans. Inner rail. It is because, the inward pressure on the inner rail is more than that on the outer rail.

Q.61. What are the advantages of banking? Ans. The vehicle can be moved along a circular track at a reasonable speed without the fear of skidding. Further, while taking a turn, one may not need to decrease the speed of the vehicle.

Q.62. A stone is moved along a vertical circle so as to just loop the loop. What is the tension in the string, when the stone is at the highest point? Ans. The tension is equal to six times the weight of the stone.

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CBSE Class 11 Physics Laws Of Motion MCQ with Answers PDF

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Class 11 Physics Chapter 5 Laws Of Motion MCQs

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Chapter 5 Laws of Motion

Answer The Following Question Answer

Question 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 skillfully held stationary in the sky, d. a car moving with a constant velocity of 30 km/h on a rough road, e. a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Solution : a. As the rain drop is flling with a constant speed, its accleration, a = 0. Hence net force F= ma  = 0. b. As the cork is floating on water, its weight is balanced by the upthrust due to water. Therefore, the net force on the cork is 0. c. As the kite is held stationery, in accordance with the first law of motion, the net force on the kite is 0. d. Force is being applied to overcome the force of friction. But as velocity of the car is constant, its accleleration, a = 0. Hence net force on the car F = ma = 0. e. As the high speed electron in space is far away from all gravitating objects and free of electric and magnetic fields, the net force on electron is 0.

Question 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? Ignore air resistance.

Solution : 0.5 N, in vertically downward direction, in all cases Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as: F = m × a Where, F = Net force m = Mass of the pebble = 0.05 kg a = g = 10 m/s 2 ∴F = 0.05 × 10 = 0.5 N The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction. If the pebble is thrown at an angle of 45° with the horizontal direction, 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.

Question 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 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.

Solution : a. Here, m = 0.1 Kg, a = + g = 10 m/s 2 Net force, F = ma = 0.1 × 10 = 1.0 N This forcer acts vertically downwards. b. When the train is running at a constant velocity, its acceleration = 0, No force acts on the stone due to this motion. Therefore, force on the stone F = weight of stone = mg = 0.1 × 10 = 1.0 N This force also acts vertically downwards. c. When the train is accelerating with 1 m s -2 , an additional force F' = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F' becomes zero and the net force on the stone is F = mg = 0.1 × 10 = 1.0 N, acting vertically downwards. d. As the stone is lying on the floor of the trin, its acceleration is same as that of the train. ∴ force acting on stone, F = ma = 0.1 × 1 = 0.1 NThis force is along the horizontal direction of motion of the train. Note that in each case, the weight of the stone is being balanced by the normal reaction.

Question 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].

Solution : (i) T When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e., F = T = mv 2 / l Where F is the net force acting on the particle.

Question 5. 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?

Solution : Retarding force, F = –50 N Mass of the body, m = 20 kg Initial velocity of the body, u = 15 m/s Final velocity of the body, v = 0 Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as: F = ma –50 = 20 × a ∴ a = -50/20 = -2.5 ms -2 Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as: v = u + at ∴ t = -u / a = -15 / -2.5 = 6 s

Question 6. 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?

Solution : Mass of the body, m = 3 kg Initial speed of the body, u = 2 m/s Final speed of the body, v = 3.5 m/s Time, t = 25 s Using the first equation of motion, the acceleration (a) produced in the body can be calculated as: v = u + at ∴ a = (v - u) / t = (3.5 - 2) / 25 = 0.06 ms -2 As per Newton’s second law of motion, force is given as: F = ma = 3 × 0.06 = 0.18 N Since 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.

Question 7. 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.

Solution : Mass of the body, m = 5 kg The given situation can be represented as follows

The resultant of two forces is given as:

Question 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 4.0 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.

Solution : Initial speed of the three-wheeler, u = 36 km/h = 10 m/s Final speed of the three-wheeler, v = 0 m/s Time, t = 4 s Mass of the three-wheeler, m = 400 kg Mass of the driver, m' = 65 kg Total mass of the system, M = 400 + 65 = 465 kg Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as: v = u + at ∴ a = (v - u) / t = (0 - 10) / 4 = -2.5 ms -2 The negative sign indicates that the velocity of the three-wheeler is decreasing with time. Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as: F = Ma = 465 × (–2.5) = –1162.5 N The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

Question 9. 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.

Solution : Mass of the rocket, m = 20,000 kg Initial acceleration, a = 5 m/s 2 Acceleration due to gravity, g = 10 m/s 2 Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation: F – mg = ma F = m (g + a) = 20000 × (10 + 5) = 20000 × 15 = 3 × 10 5 N

Question 10. 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

Solution : Mass of the body, m = 0.40 kg Initial speed of the body, u = 10 m/s due north Force acting on the body, F = –8.0 N Acceleration produced in the body, a = F / m = -8.0 / 0.40 = -20 ms -2 (i) At t = –5 s Acceleration, a' = 0 and u = 10 m/s s = ut + (1/2) a' t 2 = 10 × (–5) = –50 m (ii) At t = 25 s Acceleration, a'' = –20 m/s 2 and u = 10 m/s s' = ut' + (1/2) a" t 2 = 10 × 25 + (1/2) × (-20) × (25) 2 = 250 - 6250 = -6000 m (iii) At t = 100 s For 0 ≤ t ≤ 30 s a = -20 ms -2 u = 10 m/s s 1 = ut + (1/2)a"t 2 = 10 × 30 + (1/2) × (-20) × (30) 2 = 300 - 9000  =  -8700 m For 30 < t ≤ 100 s As per the first equation of motion, for t = 30 s, final velocity is given as: v = u + at = 10 + (–20) × 30 = –590 m/s Velocity of the body after 30 s = –590 m/s For motion between 30 s to 100 s, i.e., in 70 s: s 2 = vt + (1/2) a" t 2 = -590 × 70 = -41300 m ∴ Total distance, s" = s 1 + s 2 = -8700 -41300 = -50000 m = -50 km.

Question 11. 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 (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)

v = (v x 2 + v y 2 ) 1/2 = (20 2 + 10 2 ) 1/2 = 22.36 m/s Let θ be the angle made by the resultant velocity with the horizontal component of velocity, v x ∴ tan θ = (v y / v x ) θ = tan -1 (10 / 20) = 26.57 0 (b) 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. Hence, the acceleration of the stone is 10 m/s 2 and it acts vertically downward.

Question 12. 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 (a) at one of its extreme positions, (b) at its mean position.

Solution : (a) Vertically downward At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground. (b) Parabolic path At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Question 13. A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s –1 , (b) downwards with a uniform acceleration of 5 m s –2 , (c) upwards with a uniform acceleration of 5 m s –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?

Solution : (a) Mass of the man, m = 70 kg Acceleration, a = 0 Using Newton’s second law of motion, we can write the equation of motion as: R – mg = ma Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration a = 0 ∴ R = mg = 70 × 10 = 700 N ∴ Reading on the weighing scale = 700 / g = 700 / 10 = 70 kg (b) Mass of the man, m = 70 kg Acceleration, a = 5 m/s 2 downward Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = 70 (10 – 5) = 70 × 5 = 350 N ∴ Reading on the weighing scale = 350 g = 350 / 10 = 35 kg (c) Mass of the man, m = 70 kg Acceleration, a = 5 m/s 2 upward Using Newton’s second law of motion, we can write the equation of motion as: R – mg = ma R = m(g + a) = 70 (10 + 5) = 70 × 15 = 1050 N ∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg (d) When the lift moves freely under gravity, acceleration a = g Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = m(g – g) = 0 ∴ Reading on the weighing scale = 0 / g = 0 kg The man will be in a state of weightlessness.

Question 14. Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s,0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).

Solution : (a) For t < 0 It can be observed from the given graph 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. Hence, the force acting on the particle is zero. For t > 4 s It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle. For 0 < t < 4 It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero. (b) At t = 0 Impulse = Change in momentum = mv – mu Mass of the particle, m = 4 kg Initial velocity of the particle, u = 0 Final velocity of the particle, v = 3 / 4 m/s ∴ Impulse = 4 ( 3/4 - 0) = 3 kg m/s At t = 4 s Initial velocity of the particle, u = 3 / 4 m/s Final velocity o9f the particle, v = 0 ∴ Impulse = 4 (0 - 3/4) = -3 kg m/s

Question 15. 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 (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

The equation of motion can be written as: F – T = m 2 a T = F – m 2 a ∴T = 600 – 20 × 20 = 200 N … (ii) which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.

Question 16. 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.

Smaller mass, m 1 = 8 kg Larger mass, m 2 = 12 kg Tension in the string = T Mass m 2 , owing to its weight, moves downward with acceleration a,and mass m 1 moves 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 … (i) For mass m 2 : The equation of motion can be written as: m 2 g – T = m 2 a … (ii) Adding equations (i) and (ii), we get: (m 2 - m 1 )g = (m 1 + m 2 )a ∴ a = ( (m 2 - m 1 ) / (m 1 + m 2 ) )g    ....(iii) = (12 - 8) / (12 + 8) × 10  =  4 × 10 / 20  =  2 ms -2 Therefore, the acceleration of the masses is 2 m/s 2 . Substituting the value of a in equation (ii), we get: m 2 g - T = m 2 (m 2 - m 1 )g / (m 1 + m 2 ) T = (m 2 - (m 2 2 - m 1 m 2 ) / (m 1 + m 2 ) )g = 2m 1 m 2 g / (m 1 + m 2 ) = 2 × 12 × 8 × 10 / (12 + 8) = 96 N Therefore, the tension in the string is 96 N.

Question 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.

Solution : Let m, m 1 , and m 2 be 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 According to the law of conservation of momentum: Total initial momentum = Total final momentum 0 = m 1 v 1 + m 2 v 2 v 1 = -m 2 v 2 / m 1 Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

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

Solution : Mass of each ball = 0.05 kg Initial velocity of each ball = 6 m/s Magnitude of the initial momentum of each ball, p i = 0.3 kg m/s After collision, the balls change their directions of motion without changing the magnitudes of their velocity. Final momentum of each ball, p f = –0.3 kg m/s Impulse imparted to each ball = Change in the momentum of the system = p f – p i = –0.3 – 0.3 = –0.6 kg m/s The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Question 19. 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?

Solution : Mass of the gun, M = 100 kg Mass of the shell, m = 0.020 kg Muzzle speed of the shell, v = 80 m/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. According to the law of conservation of momentum: Final momentum = Initial momentum mv – MV = 0 ∴ V = mv / M = 0.020 × 80 / (100 × 1000) = 0.016 m/s

Question 20. 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.)

Where, AO = Incident path of the ball OB = Path followed by the ball after deflection ∠AOB = Angle between the incident and deflected paths of the ball = 45° ∠AOP = ∠BOP = 22.5° = θ Initial and final velocities of the ball = v Horizontal component of the initial velocity = vcos θ along RO Vertical component of the initial velocity = vsin θ along PO Horizontal component of the final velocity = vcos θ along OS Vertical component of the final velocity = vsin θ along OP The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions. ∴ Impulse imparted to the ball = Change in the linear momentum of the ball = mvCosθ - (-mvCosθ)  =  2mvCosθ Mass of the ball, m = 0.15 kg Velocity of the ball, v = 54 km/h = 15 m/s ∴ Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s

Question 21. 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?

Solution : Mass of the stone, m = 0.25 kg Radius of the circle, r = 1.5 m Number of revolution per second, n = 40 / 60 = 2 / 3 rps Angular velocity, ω = v / r = 2πn The centripetal force for the stone is provided by the tension T, in the string, i.e., T = F Centripetal = mv 2 / r = mrω = mr(2πn) 2 = 0.25 × 1.5 × (2 × 3.14 × (2/3) ) 2 = 6.57 N Maximum tension in the string, T max = 200 N T max = mv 2 max / r ∴ v max = (T max  × r  / m) 1/2 = (200 × 1.5 / 0.25) 1/2 = (1200) 1/2 = 34.64 m/s Therefore, the maximum speed of the stone is 34.64 m/s.

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

Solution : b.  the stone flies off tangentially from the instant the string breaks, 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.

Question 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.

Solution : a. While trying to pull a cart, ahorse 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. Therefore, a horse cannot pull a cart and run in empty space. b. This is due to inertia of motion. When a speeding bus stops suddenly, the lower part of a passenger's body, which is in contact with the seat, suddenly comes to rest. However, the upper part tends to remain in motion (as per the first law of motion). As a result, the passenger's upper body is thrown forward in the direction in which the bus was moving. c. When the lawnmower is pulled, the vertical component of the applied force acts upwards. This reduces the effective weight of the lawnmower. When the lawn mower is pushed, the vertical component acts in the direction of the weight of the mower. Therefore, there is an increase in the weight of the mower. So, it is easier to pull a lawnmower than to push it. d. According to Newton’s second law of motion, we have the equation of motion: F = ma = m ∆v /∆t      ...(i) Where, F = Stopping force experienced by the cricketer as he catches the ball m = Mass of the ball Δt = Time of impact of the ball with the hand It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e., F  ∝1/ Δt      ....(ii) Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa. While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (Δt). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

Additional Excercises

Question 24. Figure 5.17 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?

Question 25.Figure 5.18 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.)

Solution : Mass of the man, m = 65 kg Acceleration of the belt, a = 1 m/s 2 Coefficient of static friction, μ = 0.2 The net force F, acting on the man is given by Newton’s second law of motion as: F net = ma = 65 × 1 = 65 N The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force f s , exerted by the belt, i.e., F' net = f s ma' = μmg ∴ a' = 0.2 × 10 = 2 m/s 2 Therefore, the maximum acceleration of the belt up to which the man can stand stationary is 2 m/s 2 .

Question 26. 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.

Using Newton’s second law of motion, we have: T + mg = mv 2 2 / R    ...(ii) Where, v 2 = Velocity at the highest point It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively (T – mg) and (T + mg).

Question 27. 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.

Solution : a. Mass of the helicopter, m h = 1000 kg Mass of the crew and passengers, m p = 300 kg Total mass of the system, m = 1300 kg Acceleration of the helicopter, a = 15 m/s 2 Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as: R – m p g = ma = m p (g + a) = 300 (10 + 15) = 300 × 25 = 7500 N Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward. b. Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as: R' - mg = ma = m(g + a) = 1300 (10 + 15) = 1300 × 25 = 32500 N The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward. c.  The force on the helicopter due to the surrounding air is 32500 N, directed upward.

Question 28. 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?

Solution : Speed of the water stream, v = 15 m/s Cross-sectional area of the tube, A = 10 –2 m 2 Volume of water coming out from the pipe per second, V = Av = 15 × 10 –2 m 3 /s Density of water, ρ = 10 3 kg/m 3 Mass of water flowing out through the pipe per second = ρ × V = 150 kg/s The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as: F = Rate of change of momentum = ∆P / ∆t = mv / t = 150 × 15 = 2250 N

Question 29. 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.

Solution : a. Force on the seventh coin is exerted by the weight of the three coins on its top. Weight of one coin = mg Weight of three coins = 3mg Hence, the force exerted on the 7 th coin by the three coins on its top is 3mg. This force acts vertically downward. b. Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top. Weight of the eighth coin = mg Weight of the ninth coin = mg Weight of the tenth coin = mg Total weight of these three coins = 3mg Hence, the force exerted on the 7 th coin by the eighth coin is 3mg. This force acts vertically downward. c. The 6 th coin experiences a downward force because of the weight of the four coins (7 th , 8 th , 9 th , and 10 th ) on its top. Therefore, the total downward force experienced by the 6 th coin is 4mg. As per Newton’s third law of motion, the 6 th coin will produce an equal reaction force on the 7 th coin, but in the opposite direction. Hence, the reaction force of the 6 th coin on the 7 th coin is of magnitude 4mg. This force acts in the upward direction.

Question 30. 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?

Solution : Speed of the aircraft, v = 720 km/h = 720 × 5 / 18  =  200 m/s Acceleration due to gravity, g = 10 m/s 2 Angle of banking, θ = 15° For radius r, of the loop, we have the relation: tan θ = v 2 / rg r = v 2 / g tan θ = 200 2  / (10 × tan 15) = 4000 / 0.26 = 14925.37 m = 14.92 km

Question 31. 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?

Solution : Radius of the circular track, r = 30 m Speed of the train, v = 54 km/h = 15 m/s Mass of the train, m = 10 6 kg The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail The angle of banking θ, is related to the radius (r) and speed (v) by the relation: tan θ = v 2 / rg = 15 2 / (30 × 10) = 225 / 300 θ = tan -1 (0.75) = 36.87 0 Therefore, the angle of banking is about 36.87°.

Question 32. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. 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?

750N and 250 N in the respective cases; Method (b) Mass of the block, m = 25 kg Mass of the man, M = 50 kg Acceleration due to gravity, g = 10 m/s 2 Force applied on the block, F = 25 × 10 = 250 N Weight of the man, W = 50 × 10 = 500 N Case (a): When the man lifts the block directly In this case, the man applies a force in the upward direction. This increases his apparent weight. ∴Action on the floor by the man = 250 + 500 = 750 N Case (b): When the man lifts the block using a pulley In this case, the man applies a force in the downward direction. This decreases his apparent weight. ∴Action on the floor by the man = 500 – 250 = 250 N If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Question 33. A monkey of mass 40 kg climbs on a rope (Fig. 5.20) 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).

Solution : Case (a) Mass of the monkey, m = 40 kg Acceleration due to gravity, g = 10 m/s Maximum tension that the rope can bear, T max = 600 N Acceleration of the monkey, a = 6 m/s 2 upward Using Newton’s second law of motion, we can write the equation of motion as: T – mg = ma ∴T = m(g + a) = 40 (10 + 6) = 640 N Since T > T max , the rope will break in this case. Case (b) Acceleration of the monkey, a = 4 m/s 2 downward Using Newton’s second law of motion, we can write the equation of motion as: mg – T = ma ∴T = m (g – a) = 40(10 – 4) = 240 N Since T < T max , the rope will not break in this case. Case (c) The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0. Using Newton’s second law of motion, we can write the equation of motion as: T – mg = ma T – mg = 0 ∴T = mg = 40 × 10 = 400 N Since T < T max , the rope will not break in this case. Case (d) When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g Using Newton’s second law of motion, we can write the equation of motion as: mg – T = mg ∴T = m(g – g) = 0 Since T < T max , the rope will not break in this case.

Question 34.  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 (Fig. 5.21). 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 .

Solution : a.  Mass of body A, m A = 5 kg Mass of body B, m B = 10 kg Applied force, F = 200 N Coefficient of friction, μ s = 0.15 The force of friction is given by the relation: f s = μ (m A + m B )g = 0.15 (5 + 10) × 10 = 1.5 × 15 = 22.5 N leftward Net force acting on the partition = 200 – 22.5 = 177.5 N rightward As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force. Hence, the reaction of the partition will be 177.5 N, in the leftward direction. b. Force of friction on mass A: f A = μm A g = 0.15 × 5 × 10 = 7.5 N leftward Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5 N acting leftward. When the wall is removed, the two bodies will move in the direction of the applied force. Net force acting on the moving system = 177.5 N The equation of motion for the system of acceleration a,can be written as: Net force = (m A + m B ) a ∴ a = Net force / (m A + m B ) = 177.5 / (5 + 10) = 177.5 / 15 = 11.83 ms -2 Net force causing mass A to move: F A = m A a = 5 × 11.83 = 59.15 N Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

Question 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 m s –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.

Solution : a.  Mass of the block, m = 15 kg Coefficient of static friction, μ = 0.18 Acceleration of the trolley, a = 0.5 m/s 2 As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation: F = ma = 15 × 0.5 = 7.5 N This force is acted in the direction of motion of the trolley. Force of static friction between the block and the trolley: f = μmg = 0.18 × 15 × 10 = 27 N The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest. When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation. b.  An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Question 36. 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 Fig. 5.22. 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).

Solution : Mass of the box, m = 40 kg Coefficient of friction, μ = 0.15 Initial velocity, u = 0 Acceleration, a = 2 m/s 2 Distance of the box from the end of the truck, s' = 5 m As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by: F = ma = 40 × 2 = 80 N As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck. This force is given by: f = μmg = 0.15 × 40 × 10 = 60 N ∴Net force acting on the block: F net = 80 – 60 = 20 N backward The backward acceleration produced in the box is given by: a back = F net / m  =  20 / 40 = 0.5 ms -2 Using the second equation of motion, time t can be calculated as: s' = ut + (1/2)a back t 2 5 = 0 + (1/2) × 0.5 × t 2 ∴ t = √20 s Hence, the box will fall from the truck after √ 20 s from start. The distance s, travelled by the truck in √ 20 s is given by the relation: s = ut + (1/2)at 2 = 0 + (1/2) × 2 × (√20) 2 = 20 m

Question 37. A disc revolves with a speed of 100 / 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?

Solution : Coin placed at 4 cm from the centre Mass of each coin = m Radius of the disc, r = 15 cm = 0.15 m Frequency of revolution, ν = 100 / 3 rev/min = 100 / (3 × 60) = 5 / 9 rev/s Coefficient of friction, μ = 0.15 In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc. Coin placed at 4 cm: Radius of revolution, r' = 4 cm = 0.04 m Angular frequency, ω = 2πν = 2 × (22/7) × (5/9) = 3.49 s -1 Frictional force, f = μmg = 0.15 × m × 10 = 1.5m N Centripetal force on the coin: F cent. = mr'ω 2 = m × 0.04 × (3.49) 2 = 0.49m N Since f > F cent , the coin will revolve along with the record. Coin placed at 14 cm: Radius, r" = 14 cm = 0.14 m Angular frequency, ω = 3.49 s –1 Frictional force, f' = 1.5m N Centripetal force is given as: F cent. = mr"ω 2 = m × 0.14 × (3.49) 2 = 1.7m N Since f < F cent. , the coin will slip from the surface of the record.

Question 38. 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?

Solution : In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

Question 39. 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?

Solution : Mass of the man, m = 70 kg Radius of the drum, r = 3 m Coefficient of friction, μ = 0.15 Frequency of rotation, ν = 200 rev/min = 200 / 60 = 10 / 3 rev/s The necessary centripetal force required for the rotation of the man is provided by the normal force (F N ). When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force (f = μF N ) acting upward. Hence, the man will not fall until: mg < f mg < μF N = μmrω 2 g < μrω 2 ω > (g / μr) 1/2 The minimum angular speed is given as: ω min = (g / μr) 1/2 = ( 10 / (0.15 X 3) ) 1/2   =  4.71 rad s -1

Question 40. 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 ω ≤ √g/R. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = √2g/R? Neglect friction.

Solution : Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

OP = R = Radius of the circle N = Normal reaction The respective vertical and horizontal equations of forces can be written as: mg = N Cosθ ... (i) mlω 2 = N Sinθ … (ii) In ΔOPQ, we have: Sin θ = l / R l = R Sinθ … (iii) Substituting equation (iii) in equation (ii), we get: m(R Sinθ) ω 2 = N Sinθ mR ω 2 = N ... (iv) Substituting equation (iv) in equation (i), we get: mg = mR ω 2 Cosθ Cosθ = g / Rω 2     ...(v) Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω 2 ≤ 1, i.e., for ω ≤ (g / R) 1/2 For ω = (2g / R) 1/2   or ω 2 = 2g / R   .....(vi) On equating equations (v) and (vi), we get: 2g / R = g / RCos θ Cos θ = 1 / 2 ∴ θ = Cos -1 (0.5)  =  60 0

NCERT Solutions For Class-11 Physics Chapter Wise

Chapter 1 Physical World

Chapter 2 Units and Measurements

Chapter 3 Motion In A Straight Line

Chapter 4 Motion In A Plane

Chapter 6 Work, Energy and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solid

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

Related 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

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Case Study Questions Class 11 Physics Motion in a Straight Line

Case study questions class 11 physics chapter 3 motion in a straight line.

CBSE Class 11 Case Study Questions Physics Motion in a Straight Line. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Motion in a Straight Line .

CBSE Case Study Questions Class 11 Physics Motion in a Straight Line

Case study – 1.

If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of actual path length travelled by object. It is scalar quantity having SI unit of metre while displacement refers to the shortest distance between initial and final position of object. It is vector quantity. The magnitude of the displacement for a course of motion may be zero but the corresponding path length is not zero. using this data answer following questions.

1) Can path length be zero for motion of body from one point to other point?

d) All of the above

4) Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of “how much distance an object has covered during its motion” while displacement refers to the measure of “how far the abject actually from initial place.”

Case Study – 2 

v = u + at           s = ut + ½ at 2            2a s = v 2 – u 2

1) equation of motions are applicable to motion with

3) The brakes applied to a car produce an acceleration of 10 m/s 2 in the opposite direction to the motion. If the car takes 1 s to stop after the application of brakes, calculate the distance traveled during this time by car.

4) An object is dropped from a tower falls with a constant acceleration of 10 m/s2. Find its speed 10 s after it was dropped.

a = -10 m/s 2 (as acceleration is retarding)

s = ut + ½ at 2

v = 0+ 10*10

Case Study – 3

a) Displacement over given time interval

2) Slope of velocity time graph gives

3) The change in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x- axis and the velocity is represented along the y -axis. slope of velocity time graph represents acceleration of object The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement and

Case Study – 4

x A (t) = x A (0) + v A t

b) V BA = v B – v A

c) None of these

Case Study – 5

c) Instantaneous speed

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

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

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CBSE Class 11th Physics Value Based Questions Chapter 5 Laws of Motion PDF Download

CBSE Class 11th Physics Value Based Questions Chapter 5 Laws of Motion are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Physics Value Based Questions for Class 11th are available here at Free of cost. These questions are expected to be asked in the Class 11th board examination. These Physics Value Based Questions are from complete CBSE Syllabus.

CBSE Class 11th Physics Value Based Questions Chapter 5 Laws of Motion

Most of these Physics Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Physics Value Based Questions for board examinations. These Physics Value Based Questions are prepared by Directorate of Education, Delhi.

CBSE Physics Value Based Questions Class 11th Chapter 5 Laws of Motion PDF

The purpose of the Physics Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Physics Value Based Questions for Class 11th Chapter 5 Laws of Motion download here in PDF format. The most CBSE Physics Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 11th exam are collected from various sources. It covers questions asked in previous year examinations.

CBSE Physics Value Based Questions for Class 11th Chapter 5 Laws of Motion Free PDF

Class 11th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Physics Value Based Questions Chapter 5 Laws of Motion certainly help students to obtain good marks in the examinations. 

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