case study questions class 11 physics chapter 9

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CBSE Case Study Questions Class 11 Physics PDF Download

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

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

Class 11 Physics Case Study Questions

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

Chapter-wise Solved Case Study Questions for Class 11 Physics

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

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

Class 11 Books for Boards

Why Case Study Questions Matter

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

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

1. Motion and Gravitation

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

2. Work, Energy, and Power

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

3. Mechanical Properties of Solids and Fluids

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

4. Thermodynamics

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

5. Oscillations and Waves

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

6. Kinetic Theory and Laws of Motion

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

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

8. Electrical Effects of Current

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

9. Magnetic Effects of Current

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

10. Electromagnetic Induction

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

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

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

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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.

Table of Contents

Class 11th PHYSICS: Chapterwise Case Study Question & Solution

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

Case Study-Based Questions for Class 11 Physics

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

Class 11 Physics MCQ Questions

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

Class 11 Books for Boards

Class 11 Physics Syllabus 2024

Unit I: Physical World and Measurement 08 Periods

Chapter–2: Units and Measurements

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

Unit II: Kinematics 24 Periods

Chapter–3: Motion in a Straight Line

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

Chapter–4: Motion in a Plane

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

Unit III: Laws of Motion 14 Periods

Chapter–5: Laws of Motion

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

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

Unit IV: Work, Energy and Power 14 Periods

Chapter–6: Work, Energy and Power

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

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

Chapter–7: System of Particles and Rotational Motion

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

Unit VI: Gravitation 12 Periods

Chapter–8: Gravitation

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

Unit VII: Properties of Bulk Matter 24 Periods

Chapter–9: Mechanical Properties of Solids

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

Chapter–10: Mechanical Properties of Fluids

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

Chapter–11: Thermal Properties of Matter

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

Unit VIII: Thermodynamics 12 Periods

Chapter–12: Thermodynamics

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

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

Chapter–13: Kinetic Theory

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

Unit X: Oscillations and Waves 26 Periods

Chapter–14: Oscillations

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

Chapter–15: Waves

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

FAQs about Class 11 Physics Case Studies

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

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

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

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

Are the case study questions based on the latest syllabus?

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

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Class 11 Physics Case Study Questions

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

What is the purpose of physics?

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

What are case study questions in physics?

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

CBSE Case Study Questions in Class 11 Physics

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

Sample Class 11 Physics Case Study Questions

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

Class 11 Physics Case Study Question 1

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

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

Answer Key:

Class 11 Physics Case Study Question 2

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

  Class 11 Physics Case Study Question 3

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

Class 11 Physics Case Study Question 4 

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

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

Examining Class 11 Physics syllabus

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

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

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

CBSE Class 11 Physics (Code No. 042)

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

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

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids.

Topics and Subtopics in  NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Solids :

NCERT Solutions Class 11 Physics Physics Sample Papers

QUESTIONS FROM TEXTBOOK

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• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
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• NCERT Solutions Class 11 Business Studies
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• NCERT Solutions Class 11 Computer Science

Question 9. 4. Read the ‘allowing two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus. Answer:   (a) False. The-Young’s modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel. (b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance, as well as shape of helicals of the coil change and it, involves shear modulus.

Question 9. 20. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 x 10 7 Pa? Assume that each rivet is to carry one-quarter of the load. Answer:   Diameter = 6mm; Radius, r = 3 x 10 -3 m; Maximum stress = 6.9 x 10 7 Pa Maximum load on a rivet = Maximum stress x cross-sectional area = 6.9 x 10 7 x 22/7 (3 x 10 -3 ) 2 N = 1952 N Maximum tension = 4 x 1951.7 N = 7.8 x 10 3 N.

NCERT Solutions for Class 11 Physics All Chapters

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

We hope the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids, drop a comment below and we will get back to you at the earliest.

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NCERT Solutions For Class 11 Physics chapter-9 Mechanical Properties Of Solids

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NCERT Solutions for class-11 Physics Chapter 9 Mechanical Properties of Solid is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-11 in NCERT textbook, also do read theory of this Chapter 9 Mechanical Properties of Solid while going before solving the NCERT questions. You can download NCERT solution of all chapters from Physics Wallah in PDF.

Chapter 9 Mechanical Properties of Solid

Answer The Following Question Answer

Question 1. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10 –5 m 2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10 –5 m 2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Solution : Length of the steel wire, L 1 = 4.7 m Area of cross-section of the steel wire, A 1 = 3.0 × 10 –5 m 2 Length of the copper wire, L 2 = 3.5 m Area of cross-section of the copper wire, A 2 = 4.0 × 10 –5 m 2 Change in length = ΔL 1 = ΔL 2 = ΔL Force applied in both the cases = F Young’s modulus of the steel wire: Y 1 = (F 1 / A 1 ) (L 1 / ΔL 1 ) = (F / 3 X 10 -5 ) (4.7 / ΔL)     ....(i) Young’s modulus of the copper wire: Y 2 = (F 2 / A 2 ) (L 2 / ΔL 2 ) = (F / 4 × 10 -5 ) (3.5 / ΔL)     ....(ii) Dividing (i) by (ii), we get: Y 1 / Y 2   =  (4.7 × 4 × 10 -5 ) / (3 × 10 -5  × 3.5) = 1.79 : 1 The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

Question 2. Figure 9.11 shows the strain-stress curve for a given material. What are

a. Young’s modulus and

b. approximate yield strength for this material?

Solution : (a) It is clear from the given graph that for stress 150 × 10 6 N/m 2 , strain is 0.002. ∴Young’s modulus, Y = Stress / Strain = 150 × 10 6 / 0.002  =  7.5 × 10 10 Nm -2 Hence, Young’s modulus for the given material is 7.5 ×10 10 N/m 2 . (b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit. It is clear from the given graph that the approximate yield strength of this material is 300 × 10 6 Nm/ 2 or 3 × 10 8 N/m 2 .

Question 3. The stress-strain graphs for materials A and B are shown in Fig. 9.12. The graphs are drawn to the same scale. (a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material?

Solution : (a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence, Young's modulus (=stress/strain) is greater for A than that of B. (b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question 4. Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus. Solution : (a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain. (b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elsticity is involved.

Question 5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution : Elongation of the steel wire = 1.49 × 10 –4 m Elongation of the brass wire = 1.3 × 10 –4 m Diameter of the wires, d = 0.25 m Hence, the radius of the wires, r = d/2  =  0.125 cm Length of the steel wire, L 1 = 1.5 m Length of the brass wire, L 2 = 1.0 m Total force exerted on the steel wire: F 1 = (4 + 6) g = 10 × 9.8 = 98 N Young’s modulus for steel: Y 1 = (F 1 /A 1 ) / (ΔL 1 / L 1 ) Where, ΔL 1 = Change in the length of the steel wire A 1 = Area of cross-section of the steel wire = πr 1 2 Young’s modulus of steel, Y 1 = 2.0 × 10 11 Pa ∴ ΔL 1 = F 1   × L 1 / (A 1   × Y 1 ) = (98  × 1.5) /[ π(0.125  × 10 -2 ) 2   × 2  × 10 11 ]   =   1.49  × 10 -4 m Total force on the brass wire: F 2 = 6 × 9.8 = 58.8 N Young’s modulus for brass: Y 2 = (F 2 /A 2 )/ (ΔL 2 / L 2 ) Where, ΔL 2 = Change in the length of the brass wire A 1 = Area of cross-section of the brass wire = πr 1 2 ∴ ΔL 2 = F 2   × L 2 / (A 2   × Y 2 ) = (58.8 X 1) / [ (π  × (0.125  × 10 -2 ) 2   × (0.91  × 10 11 )] = 1.3  × 10 -4 m Elongation of the steel wire = 1.49 × 10 –4 m Elongation of the brass wire = 1.3 × 10 –4 m.

Question 6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Solution : Edge of the aluminium cube, L = 10 cm = 0.1 m The mass attached to the cube, m = 100 kg Shear modulus (η) of aluminium = 25 GPa = 25 × 10 9 Pa Shear modulus, η = Shear stress / Shear strain  =  (F/A) / (L/ΔL) Where, F = Applied force = mg = 100 × 9.8 = 980 N A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m 2 ΔL = Vertical deflection of the cube ∴ ΔL = FL / Aη = 980 × 0.1/ [ 10 -2  × (25 × 10 9 ) ] = 3.92 × 10 –7 m The vertical deflection of this face of the cube is 3.92 ×10 –7 m.

Question 7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.< /p>

Solution : Mass of the big structure, M = 50,000 kg Inner radius of the column, r = 30 cm = 0.3 m Outer radius of the column, R = 60 cm = 0.6 m Young’s modulus of steel, Y = 2 × 10 11 Pa Total force exerted, F = Mg = 50000 × 9.8 N Stress = Force exerted on a single column = 50000 × 9.8 / 4  =  122500 N Young’s modulus, Y = Stress / Strain Strain = (F/A) / Y Where, Area, A = π (R 2 – r 2 ) = π ((0.6) 2 – (0.3) 2 ) Strain = 122500/ [ π ((0.6) 2 – (0.3) 2 ) × 2 × 10 11 ]  =  7.22 × 10 -7 Hence, the compressional strain of each column is 7.22 × 10 –7 .

Question 8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Solution : Length of the piece of copper, l = 19.1 mm = 19.1 × 10 –3 m Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10 –3 m Area of the copper piece: A = l × b = 19.1 × 10 –3 × 15.2 × 10 –3 = 2.9 × 10 –4 m 2 Tension force applied on the piece of copper, F = 44500 N Modulus of elasticity of copper, η = 42 × 10 9 N/m 2 Modulus of elasticity, η = Stress / Strain = (F/A) / Strain ∴ Strain = F / Aη = 44500/ (2.9 × 10 -4  × 42 × 10 9 ) = 3.65 × 10 –3 .

Question 9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10 8 N m –2 , what is the maximum load the cable can support?

Solution : Radius of the steel cable, r = 1.5 cm = 0.015 m Maximum allowable stress = 10 8 N m –2 Maximum stress = Maximum force / Area of cross-section ∴ Maximum force = Maximum stress × Area of cross-section = 10 8 × π (0.015) 2 = 7.065 × 10 4 N Hence, the cable can support the maximum load of 7.065 × 10 4 N.

Question 10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

Solution : The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same. The relation for Young’s modulus is given as: Y = Stress / Strain = (F/A)/ Strain  =  (4F/πd 2 )/ Strain     ....(i) Where, F = Tension force A = Area of cross-section d = Diameter of the wire It can be inferred from equation (i) that Y ∝ (1/d 2 ) Young’s modulus for iron, Y 1 = 190 × 10 9 Pa Diameter of the iron wire = d 1 Young’s modulus for copper, Y 2 = 120 × 10 9 Pa Diameter of the copper wire = d 2 Therefore, the ratio of their diameters is given as:

Question 11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm 2 . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Solution : Mass, m = 14.5 kg Length of the steel wire, l = 1.0 m Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s Cross-sectional area of the wire, a = 0.065 cm 2 = 0.065 × 10 -4 m 2 Let Δl be the elongation of the wire when the mass is at the lowest point of its path. When the mass is placed at the position of the vertical circle, the total force on the mass is: F = mg + mlω 2 = 14.5 × 9.8 + 14.5 × 1 × (12.56) 2 = 2429.53 N Young's modulus = Strss / Strain Y = (F/A) / (∆l/l) ∴ ∆l = Fl / AY Young’s modulus for steel = 2 × 10 11 Pa ∆l = 2429.53 × 1 / (0.065 × 10 -4  × 2 × 1011)   =   1.87 × 10-3 m Hence, the elongation of the wire is 1.87 × 10 –3 m.

Question 12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10 5 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Solution : Initial volume, V 1 = 100.0l = 100.0 × 10 –3 m 3 Final volume, V 2 = 100.5 l = 100.5 ×10 –3 m 3 Increase in volume, ΔV = V 2 – V 1 = 0.5 × 10 –3 m 3 Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10 5 Pa Bulk modulus = Δp/ (ΔV/V 1 )  =  Δp × V 1 / ΔV = 100 × 1.013 × 10 5  × 100 × 10 -3 / (0.5 × 10 -3 ) = 2.026 × 10 9 Pa Bulk modulus of air = 1 × 10 5 Pa ∴ Bulk modulus of water / Bulk modulus of air  =  2.026 × 10 9 / (1 × 10 5 )  =  2.026 × 10 4 This ratio is very high because air is more compressible than water.

Question 13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10 3 kg m –3 ?

Solution : Let the given depth be h. Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10 5 Pa Density of water at the surface, ρ 1 = 1.03 × 10 3 kg m –3 Let ρ 2 be the density of water at the depth h. Let V 1 be the volume of water of mass m at the surface. Let V 2 be the volume of water of mass m at the depth h. Let ΔV be the change in volume. ΔV = V 1 - V 2 = m [ (1/ρ 1 ) - (1/ρ 2 ) ] ∴ Volumetric strain = ΔV / V 1 = m [ (1/ρ 1 ) - (1/ρ 2 ) ] × (ρ 1 / m) ΔV / V 1 = 1 - (ρ 1 /ρ 2 )     ......(i) Bulk modulus, B = pV 1 / ΔV ΔV / V 1 = p / B Compressibility of water = (1/B) = 45.8 × 10 -11 Pa -1 ∴ ΔV / V 1 = 80 × 1.013 × 10 5  × 45.8 × 10 -11   =  3.71 × 10 -3    ....(ii) For equations (i) and (ii), we get: 1 - (ρ 1 /ρ 2 )   =   3.71 × 10 -3 ρ 2 = 1.03 × 10 3 / [ 1 - (3.71 × 10 -3 ) ] = 1.034 × 10 3 kg m -3 Therefore, the density of water at the given depth (h) is 1.034 × 10 3 kg m –3 .

Question 14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Solution : Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 10 5 Pa Bulk modulus of glass, B = 37 × 10 9 Nm –2 Bulk modulus, B = p / (∆V/V) Where, ∆V/V = Fractional change in volume ∴ ∆V/V = p / B = 10 × 1.013 × 105 / (37 × 109) = 2.73 × 10 -5 Hence, the fractional change in the volume of the glass slab is 2.73 × 10 –5 .

Question 15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×10 6 Pa.

Solution : Length of an edge of the solid copper cube, l = 10 cm = 0.1 m Hydraulic pressure, p = 7.0 × 10 6 Pa Bulk modulus of copper, B = 140 × 10 9 Pa Bulk modulus, B = p / (∆V/V) Where, ∆V/V = Volumetric strain ΔV = Change in volume V = Original volume. ΔV = pV / B Original volume of the cube, V = l 3 ∴ ΔV = pl 3 / B = 7 × 10 6  × (0.1) 3 / (140 × 10 9 ) = 5 × 10 -8 m 3   =  5 × 10 -2 cm -3 Therefore, the volume contraction of the solid copper cube is 5 × 10 –2 cm –3 .

Question 16. How much should the pressure on a litre of water be changed to compress it by 0.10%?

Solution : Volume of water, V = 1 L It is given that water is to be compressed by 0.10%. ∴ Fractional change, ∆V / V = 0.1 / (100 × 1)  =  10 -3 Bulk modulus, B = ρ / (∆V/V) ρ = B × (∆V/V) Bulk modulus of water, B = 2.2 × 10 9 Nm -2 ρ = 2.2 × 10 9  × 10 -3   =  2.2 × 10 6 Nm -2 Therefore, the pressure on water should be 2.2 ×10 6 Nm –2 .

Question 17. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Solution : Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10 –3 m Radius, r = d/2  =  0.25 × 10 -3 m Compressional force, F = 50000 N Pressure at the tip of the anvil: P = Force / Area  =  50000/ π(0.25 × 10 -3 ) 2 = 2.55  × 10 11 Pa Therefore, the pressure at the tip of the anvil is 2.55 × 10 11 Pa.

Question 18. A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm 2 and 2.0 mm 2 , respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution : Cross-sectional area of wire A, a 1 = 1.0 mm 2 = 1.0 × 10 –6 m 2 Cross-sectional area of wire B, a 2 = 2.0 mm 2 = 2.0 × 10 –6 m 2 Young’s modulus for steel, Y 1 = 2 × 10 11 Nm –2 Young’s modulus for aluminium, Y 2 = 7.0 ×10 10 Nm –2 (a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached. Stress in the wire = Force / Area  =  F / a If the two wires have equal stresses, then: F 1 / a 1   =  F 2 / a 2 Where, F 1 = Force exerted on the steel wire F 2 = Force exerted on the aluminum wire F 1 / F 2 = a 1 / a 2   =  1 / 2    ....(i) The situation is shown in the following figure:

Question 19. A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10 –2 cm 2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Let x be the depression at the mid point

i.e. CD = x.

AC= CB = l = 0.5 m ;

m = 100 g = 0.100 Kg

AD= BD = (l 2 + x 2 ) 1/2

Increase in length, ∆l = AD + DB - AB = 2AD - AB

Question 20. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10 7 Pa? Assume that each rivet is to carry one quarter of the load.

Solution : Diameter of the metal strip, d = 6.0 mm = 6.0 × 10 –3 m Radius, r = d/2 = 3 × 10 -3 m Maximum shearing stress = 6.9 × 10 7 Pa Maximum stress = Manimum load or force / Area Maximum force = Maximum stress × Area = 6.9 × 10 7 × π × (r) 2 = 6.9 × 10 7 × π × (3 ×10 –3 ) 2 = 1949.94 N Each rivet carries one quarter of the load. ∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N.

Question 21. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10 8 Pa. A steel ball of initial volume 0.32 m 3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Solution : Water pressure at the bottom, p = 1.1 × 10 8 Pa Initial volume of the steel ball, V = 0.32 m 3 Bulk modulus of steel, B = 1.6 × 10 11 Nm –2 The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface. Let the change in the volume of the ball on reaching the bottom of the trench be ΔV. Bulk modulus, B = p / (∆V/V) ∆V  =  B / pV = 1.1 × 10 8  × 0.32 / (1.6 × 10 11 )  =  2.2 × 10 -4 m 3 Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10 –4 m 3 .

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 5 Laws of Motion

Chapter 6 Work, Energy and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

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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Mechanical Properties of Solids Class 11 Notes CBSE Physics Chapter 9 (Free PDF Download)

• Revision Notes
• Chapter 9 Mechanical Properties Of Solids

Revision Notes for CBSE Class 11 Physics Chapter 9 (Mechanical Properties of Solids) - Free PDF Download

The notes of mechanical properties of solids class 11 are available on Vedantu in PDF format and students can download it for free. Mechanical properties of Solids come under Unit VII- Properties of Bulk Matter. This entire unit is very important from an examination point of view and carries a total weightage of 20 marks.

This chapter talks about various laws and concepts around solid bodies. The chapter is theoretically dense, so to memorize them well students should understand the concepts first. To strengthen the understanding, physics class 11 chapter 9 notes include examples to explain the concepts clearly and logically. Students are advised to go through the notes of mechanical properties of solids class 11 regularly to maximize retention of the concepts and examples.

Download CBSE Class 11 Physics Revision Notes 2024-25 PDF

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Mechanical Properties of Solids Chapter-Related Important Study Materials It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Mechanical Properties of Solids Class 11 Notes Physics - Basic Subjective Questions

Section – a (1 mark questions).

1. Why does spring balance show wrong readings after they have been used for a long time?

Ans. Because of elastic fatigue.

2. What is an elastomer?

Ans. It is a substance that can be elastically stretched to large values of strain.

3. Give two examples which are nearly perfectly plastic material.

Ans. Putty and paraffin wax.

4. What is the Yield point on a graph drawn between stress-strain?

Ans. Yield point is the point, beyond which the wire starts showing increase in strain without any increase in stress.

5. The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If Y A and Y B are the Young’s modulii of the materials, then find the relation between Y A  and Y B .

Ans. $\dfrac{Y_{A}}{Y_{B}}=\dfrac{tan\;60^{\circ}}{tan\;30^{\circ}}=\dfrac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=3$

$Y_{A}=3Y_{B}$

Section – B (2 Marks Questions)

6. A body of mass 1 kg is attached to one end of a wire and rotated in a horizontal circle of diameter 40 cm with a constant speed of 2 m/s. What is the area of cross-section of the wire if the stress developed in the wire is $5\times 10^{6}N/m^{2}$ ?

Ans. $F=T=\dfrac{mv^{2}}{r}=\dfrac{1\times 4}{0\cdot 2}=20N$

$Stress=\dfrac{F}{A}$

$\therefore A=\dfrac{F}{Stress}=\dfrac{20}{5\times 10^{6}}=4\times 10^{-6}m^{2}=4mm^{2}$

7. The radii of two wires of the same material are in the ratio 2 : 1. If the wires are stretched by equal forces, find the ratio of the stresses produced in them.

Ans. $Stress=\dfrac{F}{A}$

$\dfrac{S_{1}}{S_{2}}=\dfrac{\dfrac{F}{A_{1}}}{\dfrac{F}{A_{2}}}=\dfrac{A_{2}}{A_{1}}$

$\Rightarrow \dfrac{S_{1}}{S_{2}}=\left ( \dfrac{r_{2}}{r_{1}} \right )^{2}=\left ( \dfrac{1}{2} \right )^{2}$

$\therefore \dfrac{S_{1}}{S_{2}}=\dfrac{1}{4}$

8. The face EFGH of the cube shown in the figure is displaced 2 mm parallel to itself when forces of 5 × 10 5 N each are applied on the lower and upper faces. The lower face is fixed. Then find the strain produced in the cube.

Ans. Shear strain $=\dfrac{X}{l}=\dfrac{2\times 10^{-3}}{4\times 10^{-2}}=0\cdot 05$

9. For a given material the Young’s modulus is 2.4 times that of rigidity modulus, then find the Poisson’s ratio.

Ans. $Y=2\cdot 4\eta$ 

$Y=2\eta (1+\sigma )$

$2\cdot 4\eta =2\eta (1+\sigma )$

$1\cdot 2=(1+\sigma )$

$\sigma =0\cdot 2$

10. A wire is stretched by 5 mm when it is pulled by a certain force. If the wire of same material but of double the length and double the diameter be stretched by the same force, then find the elongation in wire.

Ans. $dl=\dfrac{Fl}{AY}$

$dl\varpropto\dfrac{l}{r^{2}}$

$\dfrac{dl_{1}}{dl_{2}}=\dfrac{l_{1}r_{2}^{2}}{l_{2}r_{1}^{2}}$

$\dfrac{5}{dl_{2}}=\dfrac{l(2r)^{2}}{2lr^{2}}$

$\dfrac{5}{dl_{2}}=\dfrac{4}{2}=2$

$dl_{2}=\dfrac{5}{2}=2\cdot 5mm$

PDF Summary - Class 11 Physics Mechanical Properties of Solids Notes (Chapter 9)  

1. introduction: .

A rigid body refers to a hard solid object having a definite shape and size. However, in reality, bodies can be stretched, compressed and bent. Even the strongest rigid steel bar can be deformed when a sufficiently large external force is applied on it. This suggests that solid bodies are not perfectly rigid. Solids have a definite shape and size. In order to make a change (or deform) their shapes or sizes, a force is always required.

2. Deforming Force: 

A deforming force can be defined as a force that produces a change in the configuration (size or shape) of the object on application.

3. Elasticity: 

Elasticity refers to the property of an object by virtue of which it regains its original configuration after having the deforming force removed. For instance, when we stretch a rubber band and release it, it snaps back to its original shape and length. 

4. Perfectly Elastic Body: 

The bodies which have the capability to regain their original configuration immediately and completely after having the deforming force removed are termed perfectly elastic bodies. Quartz fibre can be considered as a perfectly elastic body. 

5. Plasticity:

When a body does not have the capability to regain its original size and shape completely and immediately after having the deforming force removed, it is called a plastic body and this property is termed as plasticity. 

6. Perfectly Plastic Body: 

A body that does not regain its original configuration at all on the removal of deforming force is known as a perfectly plastic body. Putty and paraffin wax can be considered nearly perfectly plastic bodies. 

7. Stress: 

When an object gets deformed under the action of an external force, then at each section of the object, stress (an internal reaction force) is produced, which tends to restore the body into its original state.

7.1 Definition: 

The internal restoring force produced per unit area of the cross-section of the deformed object is termed stress. 

7.2  Mathematical Form: 

$\text{Stress}=\frac{\text{Applied Force}}{\text{Area}}$ 

Its unit is $\text{N/}{{\text{m}}^{2}}$ or pascal (Pa).

Its dimensional formula is \[\left[ \text{M}{{\text{L}}^{-1}}{{\text{T}}^{-2}} \right]\].

7.3 Types of Stress: 

Three different types of stress are known. They are:

1.Longitudinal Stress:

When a deforming force is applied normal to the area of a cross-section, then the stress is termed as longitudinal stress or normal stress. It is further differentiated into two kinds: 

Tensile Stress: When there is an increase in length of the object under the effect of applied force, then the stress is termed as tensile stress. 

 Compressional Stress: When there is a decrease in the length of the object under the effect of applied force, then the stress is termed as compression stress.

2. Tangential or Shearing Stress: 

When the deforming force acts tangentially to the surface of a body, it generates a change in the shape of the body. This tangential force applied per unit area is termed as tangential stress or shearing stress. 

3. Hydraulic Stress: 

When the applied force is due to a liquid uniformly from all sides, then the corresponding stress is termed as hydrostatic stress. 

8. Strain: 

When a deforming force gets applied on an object, the object undergoes a change in its shape and size. The fractional change in their setup is termed a strain. 

8.1  Mathematical Equation: 

$\text{Strain}=\frac{\text{change in dimension}}{\text{original dimension}}$ 

It is a dimensionless quantity and has no unit. 

According to the change in setup, the strain is differentiated into three types:  

$\text{a) Longitudinal strain}=\frac{\text{change in length}}{\text{original length}}$

$\text{b) Volumetric strain}=\frac{\text{change in volume}}{\text{Original volume}}$

\[c)\text{ }Shearing\text{ }strain=\frac{tangential\text{ }applied\text{ }force}{Area\text{ }of\text{ }force}\]  

9. Hooke’s Law 

Robert Hook observed that within the elastic limit, the stress turns out to be directly proportional to the strain. i.e.,\[stress\propto strain\Rightarrow stress=K.strain\] 

where $K$ is the constant of proportionality known as the ‘Elastic Modulus’ of the material. 

Here, it is to be noted that there are some materials that do not obey Hooke’s law like rubber, human’s muscle, etc. 

9.1 Types of Modulus of Rigidity: 

9.1.1  young’s modulus of rigidity \[\left( y \right)\]:  .

It refers to the ratio of normal(longitudinal) stress to the longitudinal strain within the elastic limit. 

$Y=\frac{\text{longitudinal stress}}{\text{Longitudinal strain}}$ 

It has the same unit as stress because strain does not have any unit. Clearly, $Y$ is measured in $N/{{m}^{2}}$ or Pa. 

Metals usually have high values of Young’s modulus compared to other materials. Scientifically, the higher Young’s modulus of the material, the more elastic it is.

9.1.2  Bulk Modulus of Rigidity: 

It refers to the ratio of direct stress to the volumetric strain within the elastic limit. 

\[\kappa =\frac{\text{direct stress}}{\text{Volumetric strain}}\] 

\[\kappa =\frac{\frac{-F}{A}}{\frac{\Delta V}{V}}=\frac{-PV}{\Delta V}\] 

The SI unit of bulk modulus is $N/{{m}^{2}}$.

Compressibility: 

The compressibility of a material refers to the reciprocal of its bulk modulus of elasticity. Mathematically, it is given by

$C=\frac{1}{\kappa }$  

Its SI unit is ${{N}^{-1}}{{m}^{2}}$ and CGS unit is $dyn{{e}^{-1}}c{{m}^{2}}$. 

9.1.3  Modulus of Rigidity or Shear Modulus \[(\eta )\]: 

It refers to the ratio of tangential stress to the shear strain within the elastic limit. Mathematically,

$\eta =\frac{\text{tangential stress}}{\text{shear strain}}$ 

\[\eta =\frac{\frac{F}{A}}{Y}=\frac{F}{AY}\] 

$\eta =\frac{F}{AY}$ 

The SI unit of shear modulus is $N/{{m}^{2}}$.

Here, it is to be noted that the shear modulus of a material is always considerably smaller than the Young’s modulus 

10. Limit of Elasticity: 

The maximum value of deforming force for which elasticity is experienced in the body is known as its limit of elasticity.

11.Stress-Strain Curve:

Above graph shows the stress-strain curve for a metal wire which is gradually being loaded.  

The initial part OA of the graph is a straight line expressing that stress is proportional to strain. Up to the point A, Hooke’s law is obeyed. Point A is known as the proportional limit. In this region, the wire is perfectly elastic. 

After the point A, the stress is not proportional to strain and a curved portion AB is generated. But, if the load is removed at any point between O and B, the curve is retraced along BAO and the wire regains its original length. The portion OB of the graph is known as the elastic region and the point B is termed the elastic limit or yield point. The stress corresponding to B is known as yield strength. 

Beyond the point B, the strain increases more rapidly than stress. When the load is removed at any point C, the wire cannot come back to its original length but follows the dashed line. Even on decreasing the stress to zero, a residual strain same as OE is left in the wire. Here, the material acquires a permanent set. The fact that the stress-strain curve is not retraced on reversing the strain is termed elastic hysteresis. 

When the load is increased beyond the point C, there is a large increase in the strain or the length of the wire. In this region, the constrictions (termed necks and waists) develop at some points along the length of the wire and the wire breaks finally at the point D, termed as the fracture point.

In the region between B and D, the length of the wire keeps on increasing even without any addition of load. This region is known as the plastic region and the material undergoes plastic flow or plastic deformation here. The stress corresponding to the breaking point is termed the ultimate strength or tensile strength of the material. 

12. Elastic after Effect: 

Objects return to their original state when deforming force is removed. Certain objects return to their original state immediately after the removal of the deforming force, whereas some objects take longer to do so. The delay in attaining back the original state by an object on the removal of the deforming force is termed an elastic after effect.

13. Elastic Fatigue: 

The property of an elastic body by virtue of which its behaviour becomes less elastic under the application of repeated alternating deforming force is known as elastic fatigue. 

14. Ductile Materials: 

The materials that have a large plastic range of extension are known as ductile materials. These materials undergo an irreversible rise in their lengths before snapping. Thus, they can be drawn into thin wires. Some examples of ductile materials are copper, silver, iron and aluminium. 

15. Brittle Materials: 

The materials that have a very small range of plastic extension are known as brittle materials. These materials break as soon as the stress is increased beyond the elastic limit. Some examples of brittle materials are cast iron, glass and ceram. ics

16. Elastomers: 

The materials for which the strain produced is much larger than the stress applied, within the limit of elasticity, are termed elastomers. Some examples of elastomers are rubber, the elastic tissue of the aorta and the large vessel carrying blood from the heart. They have no plastic range. 

17. Elastic Potential Energy of Stretched Wire: 

When a wire is made to stretch, interatomic forces come into play, which opposes the change. Work has to be done against these restoring forces. The work done in stretching the wire is stored in it as its elastic potential energy.

18. Poisson’s Ratio: 

On the application of a deforming force at the free end of a suspended wire of length \[l\] and diameter \[D\], its length increases by \[\Delta l\] but its diameter decreases by \[\Delta D\]. 

Now, two kinds of strains are produced by a single force: 

Longitudinal strain \[=\frac{\Delta l}{l}\] 

Lateral strain \[=\frac{-\Delta D}{D}\]

Mathematically, 

Poisson’s Ratio \[\left( \sigma  \right)=\frac{Lateral\text{ }strain}{longitudinal\text{ }strain}=\frac{\left( \frac{-\Delta D}{D} \right)}{\left( \frac{\Delta l}{l} \right)}=-\frac{l\Delta D}{D\Delta l}\]

The negative sign shows that longitudinal and lateral strains are opposite in nature.

Because Poisson’s ratio is the ratio of two strains, it has no units and dimensions. 

The theoretical value of Poisson’s ratio lies between $-1\text{ and }0.5$  whereas its practical value lies between $0\text{ and }0.5$.

19. Applications of Elasticity: 

The elastic behavior of materials plays a major role in everyday life. All engineering designs need precise knowledge of the elastic behavior of materials. For instance, while designing a building, the structural design of the columns, beams and supports need ample knowledge of the strength of materials used. 

A bridge has to be designed such that it should withstand the load of the flowing traffic, the force of winds as well as its own weight. Likewise, in the design of buildings, usage of beams and columns is popular. In both these cases, the overcoming of the problem of bending of beam under a load is of utmost priority. The beam must not bend too much or break. Now, let us consider the case of a beam loaded at the centre and supported near its ends as shown in the following diagram.

A bar of length \[l\], breadth \[b\], and depth \[d\], when loaded at the centre by a load \[W\] sinks by an amount given by \[\delta =\frac{W{{I}^{3}}}{\left( 4b{{d}^{3}}Y \right)}\]

Bending can be limited by using a material with a large Young’s modulus \[Y\]. Depression can be reduced more effectively by increasing the depth d rather than the breadth \[b\]. However, a deep bar has a tendency to bend under the weight of moving traffic, thus a better choice is to have a bar of I-shaped cross-section. Such an arrangement gives a large load-bearing surface and ample depth to prevent bending. Also, such a shape reduces the weight of the beam without any sacrifice in its strength and thus reduces the cost.

FAQs on Mechanical Properties of Solids Class 11 Notes CBSE Physics Chapter 9 (Free PDF Download)

Question 1. Briefly Explain the Different Types of Strain.

Ans: Strain is defined as the change in the shape or size of the body after a deforming force is applied to the body.

Longitudinal Strain - Any change in the length of the body is defined as a longitudinal or tensile strain.

Strain = Change in length/Original length = △l/l

Volumetric Strain - If the deforming force brings about a change in the volume of the body it acts upon it is defined as the volumetric strain.

Strain = Change in volume/Original volume = △V/V

Shear Strain - The tilt in angle caused due to the tangential stress on the body.

Strain = θ = △L/L

Question 2. Explain Poisson’s Ratio.

Ans: Poisson’s ratio talks about the change in the width of the material to the change in the length of the material per unit width and length respectively. Mathematically it can be defined as the negative ratio of a transverse strain to longitudinal strain.

The change in the diameter of the body due to external force applied is the transverse strain and the change in the length of the body is a longitudinal strain.

Poisson’s ratio = σ = -Transverse Strain/Longitudinal strain

σ = -(△D/D)/(△l/l)

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NCERT Solutions For Class 11 Physics Chapter 9

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The main motive of NCERT Solutions is to make understanding the subject easier for the students. This also helps in performing better in the exams and scoring high marks. Class 11 Physics introduces students to topics like Thermodynamics, Laws of Motion, etc. These topics are extremely important for the students, as many times, these topics are also asked in competitive exams. It is extremely important that the students are thorough and well prepared with Physics for Class 11. 

NCERT Solutions for Class 11 Physics Chapter 9  

Mechanical Properties of Solid explains the well-known feature displayed by matters in this stage and goes on to spot the laws governing these situations. It offers a simple answer to complex questions. 

NCERT Solutions For Class 11 Physics Chapter 9 Mechanical Properties Of Solids (Include Download Button)

NCERT Solutions is a one-in-all study material prepared for helping the students to improve their scores and understanding of the subjects. NCERT book solution focuses on the NCERT pattern of questions and answers and provides the students with all the solutions to the problems asked in the textbooks. NCERT solutions are the product of multiple efforts from teachers with years of experience in teaching students. The entire NCERT book solution is a complete guide for anyone looking to sharpen their knowledge of any topic, concept, or even if they want to go back and use the study material as a reference. 

NCERT Solutions for Class 11 Physics Chapter 9 prepared by subject matter experts makes all the concepts appear very clear and simple. The complex problems are broken down into simple steps so that students can grasp all the concepts involved in solving problems in all chapters of Physics Class 11 NCERT. NCERT for Physics Chapter 9 introduces the Mechanical Properties of Solids. The chapter on Mechanical Properties of solids Class 11 explains the prominent characteristics displayed by matters in this stage and goes on to pinpoint the laws. This chapter talks about various laws and concepts around solid bodies. The chapter is theoretically dense, so to memorise them well students should understand the concepts first physics class 11 chapter 9 notes include examples to explain the concepts clearly and logically. Students are advised to go through the notes of mechanical properties of solids in class 11 regularly to maximise the keeping of the concepts and examples.

NCERT Solution Of Physics Class 11 Chapter 9 

The Physics NCERT solutions Class 11 prepared by expert teachers help students to understand the topics well. The chapters are laid out in such a way that they can be downloaded directly to your phone. You have the option of downloading either the entire syllabus or each chapter separately. With the Class 11 Physics NCERT Solutions, our teachers give you important step-by-step advice on how to approach a certain problem and reach the solution. The solutions contain all the chapters laid out and answered by our panel of expert teachers. With many years of expertise on board-issued textbooks, they mark out important questions and give you tips on how to solve them.

With NCERT Solutions for Physics Class 11 , it becomes easier for the students to learn the subject. Having a solutions book also makes it easier for the students to revise before the exam as all the syllabus is consolidated in one place for the students to quickly go through. With the NCERT Physics Class 11 solutions, you can perform better in your class and gain the necessary confidence to tackle your final year in school.

NCERT Solutions Of Physics Class 11 Chapter 9 (Free Download)

The NCERT Books for Class 11 are used extensively by students across the nation for a thorough revision and refresher of key concepts that appear in the Mathematics and Science subject components in the Class 11 syllabus. Class 11 is the stepping stone to the Class 12 board exams and beyond and doing well in one’s academics at this point sets up the foundation for the remaining years to come. Chapters related to science solutions cover topics related to Tissues, Motion, Atoms and Molecules, Natural Resources, etc. The NCERT Solutions for class 11 physics chapter 9 have been designed keeping this purpose in mind that students will need all the access and the concepts related to the topics and other materials. NCERT solutions are the product of multiple efforts from the teachers with years of experience in teaching students actively involved in making this solution.

The NCERT book solutions are a complete guide for anyone looking to sharpen his knowledge of any subject, topic, concept, or even if you want to go back and use the study material as a reference. The focus is also on the overall development of the student, also keeping in mind understanding the important topics. NCERT Solutions for Class 11 Physics Chapter 9 prepared by subject matter experts makes all the concepts appear very clear and simple.

NCERT Solutions For Class 11 Physics Chapter 9 Topics

Class 11 Physics Chapter 9 – Mechanical Properties of Solids. The notes of mechanical properties of solids class 11 are available in NCERT solutions by Extramarks. Mechanical properties of Solids come under Unit VII- Properties of Bulk Matter. This entire unit is very important from an examination point of view. This chapter talks about various laws and concepts around solid bodies. The chapter is theoretically dense, so to memorise them well students should understand the concepts first. To strengthen the understanding, Physics Class 11 Chapter 9 notes include examples to explain the concepts clearly and logically.

Elastic Behaviour Of Solid

Elasticity is the ability of a body to resist a distorting influence or stress and to return to its original size and shape when the stress is removed. Solid objects will deform when forces are applied to them. If the material is elastic, the object will return to its initial shape and size when these forces are removed.  To a greater or lesser extent, most solid materials exhibit elastic behaviour, but there is a limit to the magnitude of the force and the accompanying deformation within which elastic recovery is possible for any given material. Stresses beyond the elastic limit cause material to yield or flow. For such materials, the elastic limit marks the end of elastic behaviour and the beginning of plastic behaviour.

Stress And Strain

Stress and Strain are the two terms in Physics that describe the forces causing the deformation of objects. Deformation is known as the change of the shape of an object by applications of force. The object experiences it due to external force, for example, the forces might be like squeezing, squashing, twisting, shearing, ripping, or pulling the objects apart.  

Types of Strain

A person’s body can be stressed in one of two ways, depending on how much stress is applied

Tensile Strain: Tensile strain is defined as a change in the length of a body caused by tensile tension. Compressive Strain: The change in length of a body caused by compressive strain is known as compressive strain.

Hooke’s Law

It is named after the scientist Robert Hooke. Hooke’s Law states that stress developed is directly proportional to the strain produced in an object, within the elastic limit. An object that can come back to its original shape is its elasticity. Therefore, Hooke’s law applies to elastic objects. It doesn’t apply to the plasticity property of solids. It can be, therefore, represented as Stress = k * Strain Where k is the modulus of elasticity

Stress-Strain Curve

A curve drawn between stress and strain is called the stress-strain curve. When stress and stress are drawn along the y-axis and x-axis respectively, a linear graph is formed in the ideal situation of Hooke’s law. However, when actual experiments are drawn, a curve is formed known as the stress-strain curve. Stress and strain have a straight proportional relationship up to an elastic limit. The relationship between stress and strain is explained by Hooke’s law.

Hooke’s law states that the strain in a solid is proportional to the applied stress, which must be within the solid’s elastic limit. 

Elastic Modulus

Elastic modulus measures the resistance of the material to elastic—or “springy”— deformation.

Low modulus materials are floppy and stretch a lot when they are pulled. High modulus materials are the opposite—they stretch very little when pulled. Elastic modulus is another key property that determines the deformation of the structural element under flexure. The compressive strength of concrete is closely related to its compressive strength.

Application Of Elastic Behaviour Of Materials

This theory of elasticity is used to design safe and stable man-made structures such as skyscrapers and bridges to make life suitable. Cranes used to lift loads use ropes that are designed so that the stress due to maximum load does not exceed the breaking stress. It is also found that a collection of thinner wire strands when compacted together makes the rope stronger than a solid rope of the same cross-section.  That is the reason, crane ropes are made of several strands instead of one. The beams used in buildings and bridges should have to be carefully designed so that they do not bend excessively and break under the stress of the load on them. Structures such as bridges and tall buildings that have to support static or dynamic loads are generally constructed using pillars and beams to support them. Beams and pillars are designed to remain stable and safe within the range of the maximum load they are designed to carry.

Benefits Of NCERT Solutions For Class 11 Physics Chapter Mechanical Properties Of Solids

Mechanical properties of solids elaborate the characteristics such as the resistance to deformation and their strength. Strength is the ability of an object to withstand the applied stress, to what extent can it bear the stress. Resistance to deformation is how resistant an object is to the change of shape. If the resistance to deformation is less, the object can easily change its shape and vice versa. So, NCERT Solution gives benefits of focusing on the NCERT pattern of Questions and answers and providing the students with all the solutions to the problems asked in the textbooks. Our teachers and experts give you important step-by-step advice on how to approach a certain problem and get a solution. Complex problems are broken down into simple steps so that students can grasp all the concepts involved in solving problems in all chapters of Physics Class 11 NCERT. NCERT book solution is a complete guide for anyone looking to shape his/her knowledge of any concepts taught in class.

What Are Mechanical Properties?

Mechanical properties are physical properties that a material exhibits upon the application of forces. The mechanical properties of a material are defined as those properties that influence the material’s reaction to applied loads. Mechanical properties are used to determine how a material would behave in a given application and are helpful during the material selection and coating specification process. The mechanical properties of materials are crucial in many applications. In the coating industry, properties such as abrasion resistance, impact resistance, toughness, chemical resistance and tensile strength determine how well a particular coating will perform in a specific environment to protect against corrosion. Examples of mechanical properties include modulus of elasticity, tensile strength, hardness, ductility, impact resistance, compression modulus and fatigue limit.

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Q.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10 –5 m 2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10 –5 m 2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Here,   length of steel wire,   l s = 4 .7 m Area of cross-section   of steel wire,   a s = 3 .0×10 –5   m 2 Length of copper wire, l c = 3 .5 m Area of cross-section of copper wire,   a c = 4 .0 × 10 –5   m 2 Change in length = Δl s = Δl c = Δl Let   force exerted in both the cases = F Young’s modulus of steel wire   is   given   as:     Y s = F s a s × l s Δl     Y s =   F × 4 .7 m 3 .0×10 –5 m 2 × Δl           … ( i ) Young’s modulus of copper wire   is   given   as: Y c = F c a c × l c Δl   Y c = F×3 .5 m 4 .0×10 –5 m 2 ×Δl           … ( ii ) Dividing equation   ( i ) by equation   ( ii ) , we   obtain: Y s Y c = 4 .7 m × 4 .0 × 10 –5 m 2 3 .0 × 10 –5 m 2 × 3 .5 m = 1 .79 ∴ Ratio of the   Young’s modulus of steel to that of copper =1 .79

Q.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

(a)From the given graph, for stress =150×10 6 Nm -2 , Corresponding   strain   =   0 .002 ∴ Young’s modulus,   Y   = Stress Strain =   150   ×   10 6 Nm -2 0 .002 Y =   7 .5   × 10 10   Nm -2 ∴ Young’s modulus for the given material   =   7 .5   ×   10 10   Nm -2 (b)The approximate   yield strength   of a material is the maximum stress it can   bear without crossing the elastic limit . ∴ From the given graph, the approximate   yield strength of the   given material =   300 × 10 6   Nm -2 =   3×10 8   Nm -2

Q.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12. The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material?

( a ) From the given graphs, it is clear that for a given   strain, the stress for material A is more than that for material B . Since,   Young’s modulus,   Y   =   Stress Strain ∴ For a given value   of   strain,   if the stress for a material is greater,   then Young’s modulus will also   be greater for that material .

Hence, Young’s modulus for material A is greater in comparison to it is for material B.

(b) Strength of a material is determined by the amount of stress required for fracturing a material, corresponding to its fracture point. The extreme point in a stress-strain curve is called as the Fracture point.

From the given graphs, it is clear that material A can bear more strain than material B.

Therefore, material A is stronger in comparison to material B.

Q.4 Read the following two statements below carefully and state, with reasons, if it is true or false.

(a) The Young’s modulus of rubber is greater than that of steel;

(b) The stretching of a coil is determined by its shear modulus.

(a) The   given   statement   is   false,   because   for a given stress   there   is   more   strain in rubber than steel . Young’s modulus   is   given   by   the   relation: Y   =   Stress Strain ∴ For a constant   value   of   stress:   Y   ∝   1 Strain   ∴ The   Young’s   modulus   of rubber is less as   compared   to it   is   for steel . (b) Shear modulus is given by the ratio of the applied stress to the change in the shape of a body . The stretching of a coil changes its shape without any change in the length of the wire used in the coil . Therefore, shear modulus of elasticity is involved in it .

Q.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Here,   diameter of wires,   d   =   0 .25 m ∴ Radius of wires,   r = d 2 = 0 .125 cm Length of steel wire,   L 1 = 1 .5 m Length of brass wire,   L 2 = 1 .0 m Total force applied   on the steel wire   is   given   as: F s = (4 kg + 6 kg) × g F s = 10 kg × 9 .8 ms -2 = 98 N Young’s modulus for steel,   Y s = 2 .0 ×10 11   Pa Let change in the length of the steel wire = ΔL s Area   of   cross-section   of   steel   wire = A s = πr 2 ∴ ΔL 1 = F s × L s A s × Y s = F s × L s πr 2 × Y s ΔL 1 = 98 N × 1 .5 m π(0 .125 ×10 -2 m 2 ) 2 × (2 × 10 11 Pa) ΔL 1 = 1 .49 ×10 -4   m Total force applied   on brass wire,   F b = 6 × 9 .8 = 58 .8   N Young’s modulus for brass,Y b = 0 .91×10 11   Pa Let change in the length of the brass wire = ΔL b Area   of   cross-section   of   brass   wire =   A b = πr 2 ∴ ΔL 2 = F b × L b A b × Y b = F 2 × L 2 πr 2 2 × Y 2 ΔL 2 = 58 .8 N × 1 .0 m π(0 .125 × 10 -2 m) 2 × (0 .91 × 10 11 Pa) = 1 .3×10 -4   m ∴ Elongation   of   steel   wire = ΔL s = 1 .49 ×10 -4   m Elongation   of   brass   wire = ΔL b = 1 .3 ×10 -4   m

Q.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Here,   edge of aluminium cube,   L   =   10 cm =   0 .1 m Mass attached   to aluminium cube, m   =   100 kg Shear modulus of aluminium,   η   =   25 GPa =   25   ×   10 9   Pa Shear modulus   is   given   as: η = Shear   stress Shear   strain =   F/A L/ΔL (i) Here, Applied force,   F   =   mg F = 100 N × 9 .8 ms -2 = 980 N Area of one of faces of the cube,   A = 0 .1 m × 0 .1m A = 0 .01 m 2 ΔL = Vertical deflection of the cube From   equation   (i),   we   have   ΔL= FL Aη ΔL= 980 N × 0 .1 m 10 –2 m 2 × ( 25 × 10 9 Pa ) = 3 .92 × 10 –7   m ∴ Vertical deflection of this face of the cube = 3 .92 × 10 –7   m

Q.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Here,   mass of the big structure, M   =   50,000 kg Inner radius of cylinderical   column, r   =   30 cm =   0 .3 m Outer radius of cylinderical   column, R   =   60 cm=   0 .6 m Young’s modulus of steel, Y   =   2   ×   10 11   Pa Total force applied, F   =   Mg =   50000 kg   ×   9 .8 ms − 2 Stress   =   Force   applied on each   column = 50000 × 9 .8 N 4   =   122500 N Young’s modulus   is   given   as: Y   =   Stress Strain ∴ Strain   =   F/A Y   ( i ) Here, Area,   A   =   π ( R 2 –r 2 ) = π [ ( 0 .6 m ) 2 – ( 0 .3 m ) 2 ] Using   equation   ( i ) ,   we   have Strain   =   122500 N π [ ( 0 .6 m ) 2 – ( 0 .3 m ) 2 ]   ×   2   ×   10 11 P a Strain = 7 .22 × 10 –7 ∴ Compressional   strain   of   each   column   =   7 .22   ×   10 –7

Q.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Here,   length of the copper   piece,l   =   19 .1 mm =   19 .1   ×   10 –3 m Breadth of the copper   piece,b   =   15 .2 mm =   15 .2   ×   10 –3 m Area of the piece   of   copper,   A   =   l   ×   b A   = 19 .1×10 –3 m ×15 .2×10 –3 m = 2 .9×10 –4 m 2                             Tension force   exerted on the copper   piece,   F   =   44500 N Modulus of elasticity of copper, η = 42 ×10 9   Nm -2 Modulus of elasticity   is   given   by   the   relation: η = Stress Strain = F/A Strain ∴ Strain= F Aη Strain= 44500 N 2 .9 × 10 –4 m 2 × 42 × 10 9 Nm 2 Strain = 3 .65×10 -3

Q.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10 8 Nm –2 , what is the maximum load the cable can support?

Here,   radius of the steel cable,   r = 1 .5 cm = 0 .015 m         Area   of   cross-section   of the steel cable = π r 2 = π (0 .015 m) 2 Maximum stress = 10 8   Nm –2 Maximum stress = Maximum force Area   of   cross-section ∴ Maximum force = Maximum stress ×Area of cross-section F = 10 8 × π (0 .015 m) 2 = 7 .065 × 10 4   N ∴ The steel   cable can carry the maximum load   of 7 .065 ×10 4   N .  

Q.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

As   each   wire   has   same   tension   force,   so   extension   is   the   same   in   each   case . The relation for Young’s modulus is given   as: Y   =   Stress Strain = F/A Strain Y   = 4F/ π d 2 Strain (i) Here,   F   =   Force   of   tension A=   Cross-sectional   area d=   Diameter of wire From   equation   (i),   it   is   clear   that: Y   ∝   1 d 2 Young’s modulus of iron, Y iron = 190 × 10 9   Pa Young’s modulus of copper,   Y cu = 110 × 10 9   Pa Let   diameter of the iron wire = d iron Let   diameter of the copper wire = d cu ∴ d cu d iron = Y iron Y cu = 190 × 10 9 110 × 10 9 = 19 11 d cu d iron =1 .31

Q.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev s -1 at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm 2 . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Here,   mass, m = 14 .5 kg Length of steel wire, l = 1 .0 m Angular velocity, ω = 2 revs -1 Area of Cross-section   of the wire, a = 0 .065 cm 2 = 0 .065 ×10 -4   m 2 When the mass is at the lowest point of its path: Let   elongation of the wire = ΔL Total force exerted   on the   mass,   when it is placed at the lowest   position of the vertical circleis   given   as: F = mg + mlω 2 F =14 .5 kg × 9 .8 ms − 2 +14 .5 kg ×1 m× ( 2 revs -1 ) 2 F = 200 .1 N The   relation   for   Young’s modulus is   given   as,   Y = Stress Strain = F A Δl l = Fl AΔl ∴ Δl = Fl AY (i) Young’s modulus   for   Steel =       2 × 10 11   Pa Applying   equation   (i),   we   obtain: Δl = 200 .1 N ×1 m 0 .065×10 -4 ×2×10 11 Pa Δl = 1 .539× 10 –4   m ∴ Elongation of the wire = 1 .539 × 10 –4   m

Q.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10 5 Pa), Final volume =100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Here,   initial volume   of   water,V 1 = 100 .0   L = 100 .0 × 10 –3 m 3   Final volume   of   water,   V 2 = 100 .5 L = 100 .5 × 10 –3 m 3 Rise in volume, ΔV = V 2 – V 1 = 0 .5 × 10 –3 m 3 Rise in pressure, Δp = 100 .0 atm Δp = 100 × 1 .013 × 10 5   Pa Bulk   modulus   of   water = ΔpV 1 ΔV = 100 × 1 .013 × 10 5 Pa × 100 × 10 –3 m 3 0 .5 × 10 –3 m 3 = 2 .026 ×10 9   Pa Bulk   modulus   of   air = 1 .0×10 5   Pa ∴ Bulk   modulus   of   water Bulk   modulus   of   air = 2 .026 × 10 9   Pa 1 .0 ×10 5   Pa = 2 .026 × 10 4   This ratio is very large because air is more compressible than water .

Q.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 ×10 3 kg m –3 ?

Pressure at the given depth,   p = 80 .0 atm p = 80 × 1 .013 × 10 5   Pa Density of water at surface,   ρ 1 = 1 .03 × 10 3   kgm –3 Let   ρ 2   = density of water at the given   depth   . Let V 1 and   V 2 be   the volume of water of mass m at the surface   and   at the   given depth   respectively . ∴ Change in volume,   ΔV = V 2 – V 1 ΔV = m ( 1 ρ 1 – 1 ρ 2 ) ∴ Volumetric   strain   =   ΔV V 1   =   m( 1 ρ 1 – 1 ρ 2 )   ×   ρ 1 m ∴ ΔV V 1 =1   –   ρ 1 ρ 2     (i) Bulk   modulus   is   given   as:   B   =   pV 1 ΔV ∴ ΔV V 1 = p B Since   compressibility   of   water   = 1 B = 45 .8×10 -11   Pa -1 ∴ ΔV V 1 = 80 atm ×1 .013 × 10 5 kgm –3 ×45 .8×10 -11 Pa -1 ΔV V 1 = 3 .712×10 -3 (ii) From   equation   (i)   and   (ii),   we   obtain: 1- ρ 1 ρ 2 = 3 .712×10 -3 ρ 2 = 1 .03 × 10 3 1-(3 .712×10 -3 ) =1 .034 × 10 3   kgm -3 ∴ Density   of   water   at   the   given   depth   = 1 .034 ×10 3   kgm -3

Q.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Given,   hydraulic pressure applied on the glass slab,     p = 10 atm = 10 × 1 .013 × 10 5   Pa Bulk modulus of glass,   B = 37 × 10 9   Nm –2 Bulk   modulus   is   given   by   the   relation:     B = p ΔV V Here,   ΔV V = Fractional change in volume ∴ ΔV V = p B = 10 × 1 .013 × 10 5 Pa   37 × 10 9 Pa ΔV V = 2 .73× 10 -5 ∴ Fractional change in the volume of the glass slab = 2 .73 × 10 –5

Q.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×10 6 Pa.

Here,   length of an edge of the solid copper cube, l   =   10 cm = 0 .1 m Hydraulic pressure   on   the solid copper cube,   p =7 .0 × 10 6 Pa Bulk modulus of copper,   B = 140 × 10 9   Pa Bulk   modulus   is   given   by   the   relation:   B = p ΔV V (i) Here,   ΔV V = Volumetric strain ΔV = Change in volume V = Original volume From   equation   (i),   we   obtain: ΔV = pV B Since volume of the cube   is   given   as:   V = l 3 ∴ ΔV= pl 3 B = 7 .0 × 10 6 Pa × ( 0 .1 m 3 ) 3 140 × 10 9 Pa ΔV = 5 ×10 -8   m 3 = 5×10 -2   cm 3 Thus, volume contraction of the solid copper cube, ΔV = 5 × 10 –2   cm –3

Q.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?

Here,   volume of water,   V = 1 L Bulk   modulus   of   water,   B = 2 .2×10 9   Nm -2 Given that, water is to be compressed by 0 .10% .                   ∴ Fractional   change,   ΔV V = 0 .1 L 100×1 L = 10 -3 Bulk   modulus   is   given   by   the   relation: B = p ΔV V ∴ p = B × ΔV V ∴ p = ( 2 .2×10 9 Nm -2 ) ×10 -3 = 2 .2 ×10 6   Pa ∴ The   required   pressure   on   water   =   2 .2 × 10 6   Pa

Q.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Here,   diameter of cones at narrow ends,   d   =   0 .50 mm         d =   0 .5   ×   10 –3   m ∴ Radius, r   =   d 2   = 0 .50× 10 –3 2 = 0 .25 × 10 -3   m Compressional force   at   wide   ends,   F = 50000 N Pressure at the tip of the anvil   is   given   as: P = Force Area = F π r 2 P = 50000 N π ( 0 .25 × 10 -3 m ) 2 = 2 .55×10 11   Pa ∴ Pressure at the tip of the anvil   =   2 .55   ×   10 11   Pa .

Q.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm 2 and 2.0 mm 2 respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Here,   area of cross-section   of   wire   A,   a A = 1 .0 mm 2   = 1 .0 × 10 –6   m 2 Area of cross-section of wire B,   a B = 2 .0 mm 2 = 2 .0 × 10 –6 m 2 Young’s modulus of steel,Y s = 2 × 10 11   Nm –2 Young’s modulus of aluminium,Y a = 7 .0 × 10 10   Nm –2 ( a )   Let mass m be suspended to the rod at a distance y from the end where wire A is suspended . Stress = Force Area = F a If stress   is   equal   in   the   two   wires,   then: F 1 a A = F 2 a B Here, F 1 = Force on the steel wire F 2 = Force on the aluminum wire F 1 a A = F 2 a B = 1 2 (i) The situation of   the   problem   is shown in the figure   (a) . Taking torque about the point of suspension   of   mass   from   the   rod,   we obtain: F 1 y = F 2 ( 1 .05 – y ) F 1 F 2 = ( 1 .05 – y ) y (ii) From equations ( i ) and ( ii ) , we have:   ( 1 .05 – y ) y = 1 2 2 ( 1 .05 – y ) = y 2 .1 – 2y = y 3y = 2 .1 ∴ y = 0 .7   m ∴ To produce an equal stress in both   of   the wires, the mass must be suspended at a distance of 0 .7 m from the   end   where   A   is   attached . (b)   The   relation   for   Young’s   modulus   is   given   as: Young’s   modulus   =   Stress Strain ∴ Strain   =   Stress Young’s   modulus = a Y If   the   strain   in   both   of   the   wires   is   equal,   then F 1 a A Y s = F 2 a B Y a F 1 F 2 = a A a B Y s Y a = 1 2 × 2 × 10 11 Nm –2 7 × 10 10 Nm –2 = 10 7 (iii) Taking torque about the point of suspension   of   mass   m, we obtain: F 1 y s = F 2 ( 1 .05 – y s ) F 1 F 2 = ( 1 .05 – y s ) y s ( iv ) From equations ( iii ) and ( iv ) , we obtain: ( 1 .05 – y s ) y s = 10 7 7 ( 1 .05 – y s ) = 10   y s 17   y s = 7 .35 ∴   y s = 0 .432 In order to produce an equal strain in both   of   the wires, the mass must be hanged at a distance of 0 .432 m from the   end where wire A attached .

Q.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10 –2 cm 2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Here,   length of the steel wire = 1 .0 m Cross-sectional   area, A = 0 .50 × 10 –2 cm 2 =0 .50 × 10 –6 m 2 A mass 100 g is suspended from midpoint   of   the   wire . m= 100 g = 0 .1 kg ∴ The wire get   depressed as shown in the given figure . Original length of   wire = AC Depression in   the   wire = l The length of   the   wire   after mass m is attached to the wire = AO + OC Increase in length of the   wire,   Δl = ( AO + OC ) –AC Here, AO = OC = [ ( 0 .5 ) 2 + ( l ) 2 ] 1 2 ∴ Δl = [ ( 0 .5 m ) 2 + ( l ) 2 ] 1 2 – 1 .0 Expanding   the   expression   and   neglecting   higher   terms, we   obtain: Δl = l 2 0 .5 Strain= Increase   in   length Original   length Let T = Tension in the wire . ∴ mg = 2Tcosθ From the given   figure,we   obtain: cosθ   =   l [ ( 0 .5 m ) 2 + ( l ) 2 ] 1 2   =   l ( 0 .5 m ) [ 1 + ( l 0 .5 m ) 2 ] 1 2 Expanding the   above   expression and neglecting higher terms,   we   obtain: cos   θ   =   l ( 0 .5 m ) [ 1   +   l 2 2 ( 0 .5 m ) 2 ] [ 1   +   l 2 2 ( 0 .5 m ) 2 ]   ;   1   for   small   value   of   l ∴ cos   θ   =   l 0 .5 ∴ T   =   mg 2 ( l 0 .5 )   =   mg 4l Stress   =   Tension Area   =   mg 4l   ×   A   Young’s modulus   is   given   as:   Y   =   Stress Strain ∴ Y = mg × 0 .5 4l × A×l 2 ∴ l = mg × 0 .5 4YA 3 Young’s modulus of steel, Y = 2×10 11   Pa ∴ l = 0 .1 kg × 9 .8 ms − 2 ×0 .5 m 4 × 2 ×10 11 Pa × 0 .50 ×10 -6   m 2 3 = 0 .0106   m ∴ The depression at the midpoint = 0 .0106 m

Q.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 ×10 7 Pa? Assume that each rivet is to carry one quarter of the load.

Here,   diameter of the metal strip, d   =   6 .0 mm d = 6 .0   ×   10 –3   m ∴ Radius   of the metal strip,   r   =   d 2   = 3 .0 × 10 –3   m Maximum shearing stress   on   the   riveted   strip = 6 .9   ×   10 7   Pa Maximum   stress   =   Maximum   load   on   a   rivet Area   ∴ Maximum load   on   a   rivet   =   Maximum stress × Area = 6 .9 × 10 7 Pa × π ×(r) 2 = 6 .9 × 10 7 Pa × π × (3 ×10 –3 m) 2 = 1949 .94 N Since   each rivet supports one quarter of the load, ∴ Maximum tension   on   each   rivet = 4 × 1949 .94 N T = 7799 .76 N

Q.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10 8 Pa. A steel ball of initial volume 0.32 m 3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Here, water pressure at the bottom   of the trench, P =   1 .1 × 10 8   Pa Initial volume of the   ball,   V = 0 .32 m 3 Bulk modulus for steel,   B = 1 .6 × 10 11   Nm –2 The ball is   located in the Pacific Ocean,   nearly   11 km below the surface . Let the change in the volume of the ball on arriving   at   the bottom of the trench = ΔV Bulk   modulus   is   given   by   the   relation:   B = p ΔV V ∴ ΔV = B pV = 1 .1 × 10 8 Pa × 0 .32 m 3 1 .6 × 10 11 Nm –2 ΔV = 2 .2 × 10 -4   m 3 ∴ Change in volume of the ball on reaching the bottom of   the trench =   2 .2   ×   10 –4   m 3

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• After Completing the Chapter: Students can prefer utilising the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids revision after completing the chapter thoroughly. 
• To Identify Gaps in Knowledge: It is advisable for students to utilise the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids theory if they want to identify the gaps in their knowledge. After identification of gaps, students can also eliminate those gaps by solving the Chapter 9 Mechanical Properties of Solids questions on a regular basis. 
• To Verify the Answers: Once students have attempted the questions of Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids then they can prefer utilising the NCERT Solutions so that they can match their own answers. 
• To Practise More Questions: If students want to practise more questions on Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids then they can prefer using the NCERT Solutions for Class 11 Physics. By practising more Chapter 9 Mechanical Properties of Solids questions, students can easily enhance their problem solving skills. 
• To Revise: The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids PDF can be utilised as good study material as through it, students can easily revise the topics and concepts. 
• To Build Strong Foundation: If students want to build a strong foundation for the Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids, then they can prefer to utilise the NCERT Solutions for Class 11 Physics. By building a strong foundation, students can score good marks in questions related to Chapter 9 Mechanical Properties of Solids. 

What are the Challenges Faced While Referring to the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids?

While referring to the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids, students may face challenges, those challenges are discussed below: 

• Lack of Understanding: Students may struggle to understand the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids revision if they don’t have a clear understanding of the concepts and topics. In this case, students need to have a clear understanding of the Chapter 9 Mechanical Properties of Solids concepts. 
• Lengthy Solutionss: Some answers in the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids theory can be lengthy so in this case students may face difficulty in understanding the Solutions. To overcome this, students can break the answers of Chapter 9 Mechanical Properties of Solids questions into smaller parts. 
• Technical Language: If the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids PDF contains any kind of technical language then some students may face difficulty in understanding those words. 
• Lack of Visual Aids: If there are not enough visual diagrams of Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids questions, then they may face difficulty in understanding the NCERT Class 11 Physics Solutionss. To overcome this, students may refer to the Selfstudys website as there are enough visual diagrams related to Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids. 
• Limited Scope: The NCERT Solutionss for Class 11 Physics provides a limited number of Chapter 9 Mechanical Properties of Solids questions to solve; for this students need to refer to other study materials which can be available on the website. 
• Inadequate Explanations: Some answers of Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids may not provide adequate explanations in the NCERT Solutionss; so in this case students may face difficulty understanding the questions. 

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

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids in Hindi and English Medium with additional exercises for session 2024-25 along with MCQ extra question answers. Focus on understanding the problem-solving techniques specific to each topic. This might involve understanding how to approach numerical problems, using formulas correctly, and applying appropriate units.

Class 11 Physics Chapter 9 Mechanical Properties of Fluids Question Answers

• Class 11 Physics Chapter 9 NCERT Solutions
• Class 11 Physics Chapter 9 in Hindi Medium
• Class 11 Physics Chapter 9 NCERT Book
• Class 11 Physics all Chapters NCERT Solutions
• Class 11 all Subjects NCERT Solutions

The height of a liquid in a fine capillary tube

As per the expression: h=2 T cos rpg, we can say that height climbed by the liquid depends upon the surface tension (T) of the liquid, the angle of contact (8) between surface of tube and liquid, radius of tube (r), density of liquid (p) and net value of acceleration due to gravity (g).

• View Answer

If the surface of a liquid is plane, then the angle of contact of the liquid with the walls of container is

The angle of contact with wall of container is angle b/w liquid surface and normal to the wall of container. Hence for a plane liquid angle of contact is 0 degree.

One end of a towel dips into a bucket full of water and other end hangs over the bucket. It is found that after some time the towel becomes fully wet. It happens

One end of a towel dips into a bucket full of water and other end hangs over the bucket. It is found that after some time the towel becomes fully wet. It happens because of the capillary action of cotton threads.

The surface of water in contact with glass wall is

Capillary action occurs because water is sticky in nature due to which the water modules get attracted to the glass and stick to it. Complete answer: When the lower end of the glass wall is placed in the surface of liquid like water then a concave meniscus is formed.

Surface tension is the ability of a fluid surface to shrink to the smallest possible area. Have you ever noticed that even with a tall glass of water, you can only add a few drops before it overflows? Have you ever lost a thermometer and watched how mercury reacts when it falls? All of these are caused by the surface tension of the surface. Its insulated surface behaves like a solid rubber membrane due to the inhibitory force of fluid molecules.

As a result, the different surfaces of the fluid remain under stress and tend to have minimum fields. So, surface tension refers to the tension on a single surface of a fluid. Surface tension is described as the phenomenon that occurs when the surface of a liquid comes into contact with another phase (which can also be a liquid). Liquids seem to have the smallest surface area. The surface of the liquid looks like an elastic sheet.

When the angle of contact between a solid and a liquid is 90, then

When the angle of contact b/w a solid and a liquid is 900 then it shows that they are at equilibrium and cohesive forces are equal to the adhesive forces.

At critical temperature, the surface tension of a liquid

At critical temperature, the surface tension of a liquid is zero.

Choose the wrong statement from the following

In drinking the cold drinks through a straw, we use to phenomenon of capillarity.

Plants get water through the roots because of

The transfer of water using xylem in plants occurs using the capillary theory Hence plants get water through the roots because of capillary action.

Bernoulli’s principle is an important formula used in fluid mechanics. Bernoulli’s principle is also known as Bernoulli’s equation or Bernoulli’s theorem. Bernoulli’s principle provides the relationship between pressure (P) of fluid flow and height (h) of a vessel with kinetic energy and gravitational potential vitality. This principle was first proposed by Daniel Bernoulli and then formulated in Bernoulli’s equations by Leonard Euler in 1752.

Through the statement of Bernoulli’s principle, conservation of energy was found to be true for flowing fluids. It may sound contradictory, but Bernoulli’s principle describes how the velocity and pressure of a fluid are related to each other. The phenomenon of Bernoulli’s principle can be observed in rivers. In some places, the width of the river turned out to have changed. As the width of a river widens, the speed of the water flowing through it decreases. In narrower areas, the water velocity increases. Therefore, Bernoulli’s equation can be derived using the law of conservation of energy.

Property of a liquid, when there is a relative movement between the different layers of the liquid, the different layers of the liquid play a role thanks to this property (internal friction), which is called viscosity. Moving parallel until it is in a higher horizontal plane, a fluid can be considered as composed of several layers. If the fluid flow is vigorous, the velocity of the layer in contact with the horizontal plane is close to zero. But as the distance of the layer from the horizontal increases, the velocity of the laminar flow increases. Therefore, the uppermost fluid layer has the maximum velocity, so there is always a velocity difference between two adjacent addresses of the fluid.

One layer flows faster than the other, as in solid objects, and due to their relative motion, opposing frictional forces act between their entire planes. Likewise, due to the relative motion between adjacent layers in the fluid, they also create friction against the layers to resist movement. In other words, the slow-flowing layer can be said to try to reduce the velocity of the fast-flowing layer, which is known as viscosity.

Velocity gradient is defined as the ratio of change in velocity (dv) to distance (dx). The direction of the velocity gradient is perpendicular to the direction of flow, pointing in the direction of the velocity increase. Mathematically, it is given by. Velocity gradient = dv / dx.

How is the NCERT Solutions for Class 11 Physics Chapter 9 useful for students?

Studying the NCERT solutions as reference material with a manual can help a grade 11 student to learn more about the topics of chapter 9 in Physics. These solutions answer all the exercises as well as additional exercises questions and clarify all the doubts, which can help users perform better in your exams. Class 11 Physics chapter 9 solutions are prepared by a team of academics, guaranteeing the quality of the study material that all will refer for further study.

What are the core properties of the fluid discussed in Class 11 Physics Chapter 9?

Class 11 Physics Chapter 9 comprise of the concepts of motion of fluid. It stats that a fluid is a substance that can flow under the action of an external force. Gases and liquids belong to the category of fluids because they can flow. The study of hydrodynamic properties is called hydrostatics. When external pressure is applied, the fluid can change shape. The student must know that the fluids have special physical properties that help to understand how they behave when exposed to outside forces.

What is the significance of Reynolds number in Chapter 9 of 11th Physics?

The Reynolds number plays an important role to describe a streamline flow of fluid. It has no dimension. It is used to determine if the type of flow pattern when flowing through a pipe is laminar or turbulent. The Reynolds number can be determined by the ratio of internal force to viscous force. An 11th standard student must know that if the Reynolds number is high, the flow through the pipe is called turbulent flow, and it is laminar only if the Reynolds number is low.

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Case Study Questions Class 11 Physics Work, Energy And Power

Case study questions class 11 physics chapter 6 work, energy and power, cbse case study questions class 11 physics work, energy and power, case study – 2.

Kinetic energy is a scalar quantity. The kinetic energy of an object is a measure of the work and The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule (J).

Case Study – 3

Case study – 4.

Answer key – 4

Case Study – 5

1 hp = 746 W

4) The instantaneous power is defined as the limiting value of the average power as time interval approaches zero.

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• Class 11 Physics

CBSE Important Questions for Class 11 Physics

CBSE Important Questions for Class 11 Physics provide you with excellent strategies to prepare for the Physics examination. Questions can be framed from any corner of the book or outside the textbook for the final exams. Thus, you need advanced knowledge rather than practising only through textbooks. This page provides you with CBSE Important Questions Class 11 with medium to advance level Physics questions, helping you to become well-prepared for the competitive examinations.

CBSE Class 11 Physics Important Questions

Important questions for Class 11 Physics are framed according to the NCERT syllabus. Students can click the link below to access the chapter-wise important questions of CBSE Class 11 Physics.

The repeatedly asked questions and the questions of high priority according to the syllabus are provided with answers, at BYJU’S for free. Thus, CBSE Important Questions are essential in building clear concepts and excelling in the Class 11 examination. It plays a major role in helping students face exams confidently and achieve good grades.

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NCERT Solutions for Class 11 Physics: Download PDF

NCERT Solutions for Class 11 Physics 2023-24: The class 11 syllabus is important to build a strong base for certain topics in Class 12. Thus, students must understand the topics thoroughly so that they can grasp more complex concepts in the higher classes. The NCERT Solutions for Class 11 Physics help students comprehend the topics so they can write comprehensive answers in the final exam.

The CBSE Class 11 Physics NCERT Solutions on Embibe are prepared by subject specialists to make studying easier and help students build a strong base in the concepts. Students can download the NCERT Solutions for all the chapters in CBSE Class 11 Physics. They can refer to the same while studying the chapters and ace the subject.

NCERT Solutions for Class 11 Physics 2023-24

NCERT Solutions for Class 11 Physics is the best study material for students as it helps them in their studies while they advance through the syllabus. We have arranged the solutions in a chapter-wise format so that students can download the solutions for the chapter they are preparing or solving questions on.

NCERT Solutions for 11th class Physics  is provided ahead on this page, which students can download for. To access these solutions, all they need to do is click on the links provided above and get on to downloading chapter-wise Class 11 Physics NCERT solutions. They can also attempt mock tests for CBSE Class 11 Physics topics or chapters for.

Students can also check the important chapters in NCERT Class 11 Physics from the table below:

NCERT Solutions for Class 11 Physics: Points to Remember

We have provided some important points that are covered in NCERT Class 11 Physics to help students in their exam preparations. Refer to the list below:

• Physics is a branch of science which explains nature and natural phenomena. 
• The basic laws of physics are universal and apply in widely different contexts and conditions.
• Branches of Classical physics are Mechanics, Thermodynamics, Electromagnetism, Optics and Acoustics.
• Branches of Modern Physics are Relativity, Quantum Mechanics, Atomic and Nuclear Physics.
• According to Newton’s first law of motion, everybody continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.
• According to Newton’s second law of motion, 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. F = dP/dt = ma
• Energy transferred from one body to another body due to their temperature difference is known as heat.
• The temperature of a body is a physical quantity which determines whether the heat energy flows from one body to another or not. The temperature of a substance is the measure of the average kinetic energy of its molecules.
• A transverse wave is a wave motion in which the particles of the medium vibrate about their mean positions at right angles to the direction of the propagation of the wave.
• A longitudinal wave is a wave motion in which the particles of the medium vibrate about their mean positions along the direction of the propagation of the wave.

Students can visit Embibe to get access to all the important points for NCERT Class 11 Physics.

NCERT Solutions for Class 11 Physics: All Chapters

Students can download the NCERT Solutions for Class 11 Physics from the following list. They can practice questions on the NCERT Class 11 Physics chapters on Embibe.

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

NCERT Class 11 Physics Reference Books

Apart from the NCERT Class 11 Physics textbooks, students can also refer to the books available on the Embibe app. These books are prepared by subject-matter experts, as per the NCERT syllabus. The NCERT Class 11 Physics chapter topics can be understood with the help of Embibe’s 3D videos. Click on the chapter links to refer to the NCERT Class 11 Physics reference books on Embibe.

FAQs on NCERT Solutions for Class 11 Physics 2023-24

Below are some of the most frequently asked questions on NCERT Solutions for Class 11 Physics 2024-25:

Ans: The NCERT Physics Class 11 book is divided into two parts with a total of 15 chapters. Part 1 contains 8 chapters, and Part 2 contains 7 chapters.

Ans. Concepts of Physics by HC Verma is the best book for Class 11 CBSE Physics.

Ans. No, for the academic year, the officials of CBSE have decided to omit the topic Doppler Effect.

Ans: Students looking for NCERT solutions can download the chapter-wise solutions PDF from this article.

Ans: There might be moments when you feel stuck while revising or preparing for a chapter. At times like this, you can refer to these solutions and know exactly what to do or how to approach a question. Since the solutions are provided with detailed diagrams, you will be able to understand the concepts more effectively and excel in the subject.

We hope this article on CBSE NCERT Solutions for Class 11 Physics is helpful to you. Stay tuned to Embibe for the latest CBSE Class 11 NCERT Solutions exam updates.

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Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy

• Last modified on: 3 years ago
• Reading Time: 4 Minutes

Case Study Questions for Class 9 Science Chapter 11 Work and Energy

In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on  case studies and passage based  as well. In that, a paragraph will be given, and then questions based on it will be asked.

Here, we have provided case based/passage based questions for Class 9 Science  Chapter 11 Work and Energy . Students can practice these questions for their exam.

Case Study/Passage Based Questions

Question 1:

Read the following and answer any four questions from (i) to (iii).

Figure shows a watch glass embedded in clay. A tiny spherical ball is placed at the edge B at a height h above the centre A.

(i) The kinetic energy of ball, when it reaches at point A is (a) zero (b) maximum (c) minimum (d) can’t say.

(ii) The ball comes to rest because of (a) frictional force (b) gravitational force (c) both (a) and (b) (d) none of these.

(iii) The energy possessed by ball at point C is (a) potential energy (b) kinetic energy (c) both potential and kinetic energy (d) heat energy.

You may also like:

Case study and passage based questions for other chapters of class 9 science is given below.

Chapter 3 Atoms and Molecules

Chapter 4 Structure of Atom

Chapter 10 Gravitation

Chapter 11 Work and Energy

Chapter 13 Why Do We Fall Ill?

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