case study questions class 11 physics mechanical properties of solids

• Lakhmir Singh

Case Study Questions Class 11 Physics Mechanical Properties of Solid

Case study questions class 11 physics chapter 9 mechanical properties of solid.

CBSE Class 11 Case Study Questions Physics Mechanical Properties of Solid. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Mechanical Properties of Solid.

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

CBSE Case Study Questions Class 11 Physics Mechanical Properties of Solid

Case study – 1.

The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity and the deformation caused is known as elastic deformation. However, if you apply force to a lump of putty or mud, they have no gross tendency to regain their previous shape, and they get permanently deformed. Such substances are called plastic and this property is called plasticity. Putty and mud are close to ideal plastics. We know that in a solid, each atom or molecule is surrounded by neighboring atoms or molecules. These are bonded together by interatomic or intermolecular forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic (or intermolecular) distances. When the deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size. Answer the following

1) Putty and mud are example of

a) Ideal plastic

b) Ideal elastic

c) Pseudo plastic

d) None of these

2) The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as

a) Elasticity

b) Plasticity

3) Define elasticity

4) Define plasticity

5) Explain elastic behavior of solid

Answer key – 1

3) The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity and the deformation caused is known as elastic deformation.

4) The property of a body, in which body does not regain its original size and shape when the applied force is removed and get permanently deformed, is known as plasticity.

5 We know that in a solid, each atom or molecule is surrounded by neighboring atoms or molecules. These are bonded together by interatomic or intermolecular forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic (or intermolecular) distances. When the deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size.

Case Study – 2

2) When a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force is equal in magnitude but opposite in direction to the applied force. The restoring force per unit area is known as stress. If F is the force applied normal to the cross–section and A is the area of cross section of the body.

Magnitude of the stress = F/A

The SI unit of stress is N-m -2 or Pascal (Pa) and its dimensional formula is [ML -1 T -2 ]. The restoring force per unit area in this case is called tensile stress. If the cylinder is compressed under the action of applied forces, the restoring force per unit area is known as compressive stress. Tensile or compressive stress can also be termed as longitudinal stress. In both the cases, there is a change in the length of the cylinder. The change in the length ΔL to the original length L of the body is known as longitudinal strain.

The restoring force per unit area developed due to the applied tangential force is known as tangential or shearing stress.

1) Restoring force per unit area is called as

c) Modulus of elasticity

2) Ratio of change in dimension to original dimension is called

3) Define shear stress.

4) Define stress. Give its SI unit and dimension

5) Define strain. Give its SI unit and dimension

Answer key – 2

3) The tangential restoring force per unit area developed known as tangential or shearing stress.

4) When a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force is equal in magnitude but opposite in direction to the applied force. The restoring force per unit area is known as stress.

If F is the force applied normal to the cross–section and A is the area of cross section of the body.

The SI unit of stress is N-m -2 or Pascal (Pa) and its dimensional formula is [ML -1 T -2 ].

5) Ratio of change in dimension to original dimension is called strain. As it is ratio of similar quantities so it carries no unit and hence no dimensions.

Case Study – 3

For small deformations within elastic limit the stress and strain are proportional to each other. This is known as

Hooke’s law. Thus, stress α strain

Stress = k × strain

Where k is the proportionality constant and is known as modulus of elasticity. Hooke’s law is an empirical law and is found to be valid for most materials. However, there are some materials which do not exhibit this linear relationship.

In the region from A to B, stress and strain are not proportional. Nevertheless, the body still returns to its original dimension when the load is removed. The point B in the curve is known as yield point (also known as elastic limit) and the corresponding stress is known as yield strength (σ y ) of the material.

If the load is increased further, the stress developed exceeds the yield strength and strain increases rapidly even for a small change in the stress. The portion of the curve between B and D shows this. When the load is removed, say at some point C between B and D, the body does not regain its original dimension. In this case, even when the stress is zero, the strain is not zero. The material is said to have a permanent set. The deformation is said to be plastic deformation. The point D on the graph is the ultimate tensile strength (σ u ) of the material. Beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at point E. If the ultimate strength and fracture points D and E are close, the material is said to be brittle. If they are far apart, the material is said to be ductile.

1) Stress is directly proportional to strain this is valid

a) Above elastic limit

b) Within elastic limit

c) Above plastic limit

2) SI unit of modulus of elasticity is

3) Define modulus of elasticity.

4) State hooks law

5) Write note on stress strain curve for ductile material

Answer key – 3

3) Modulus of elasticity is defined as ration of stress to strain.

4) For small deformations within elastic limit the stress and strain are proportional to each other. This is known as

If the load is increased further, the stress developed exceeds the yield strength and strain increases rapidly even for a small change in the stress. The portion of the curve between B and

D shows this. When the load is removed, say at some point C between B and D, the body does not regain its original dimension. In this case, even when the stress is zero, the strain is not zero. The material is said to have a permanent set. The deformation is said to be plastic deformation. The point D on the graph is the ultimate tensile strength (σ u ) of the material.

Beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at point E. If the ultimate strength and fracture points D and E are close, the material is said to be brittle. If they are far apart, the material is said to be ductile

Case Study – 4

The proportional region within the elastic limit of the stress-strain curve is of great importance for structural and manufacturing engineering designs. The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material.

Experimental observation show that for a given material, the magnitude of the strain produced

is same whether the stress is tensile or compressive. The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ϵ) is defined as Young’s modulus and is denoted by the symbol Y.

Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress i.e., N-m -2 or Pascal (Pa). As steel has more modulus of elasticity than copper brass and aluminum hence steel is preferred in heavy-duty machines and in structural designs. Wood, bone, concrete and glass have rather small Young’s moduli. Answer the following.

1) If stress strain changes then young’s modulus is

a) Also changes

b) Remains constant

c) Either changes or remains constant depends on amount of stress and strain

2) SI unit of young’s modulus is

b) Pascal (Pa).

c) N-m -2 or Pascal (Pa).

3) Which of the following is more elastic

a) Aluminum

4) Defines young’s modulus. Give its SI unit and dimensions

5) Why steel is more preferred in heavy industries than copper and brass?

Answer key – 4

4) The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ϵ) is defined as Young’s modulus and is denoted by the symbol Y.

the unit of Young’s modulus is the same as that of stress i.e., N-m -2 or Pascal (Pa) and its dimensional formula is [ML -1 T -2 ]

5) Steel is more preferred in heavy industries than copper and brass because steel has more modulus of elasticity that is higher young’s modulus than copper and brass. In short steel is more elastic than copper and brass.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

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

We have a strong team of experienced Teachers who are here to solve all your exam preparation doubts

Jkbose class 5 hindi chapter 17 lohri solution, duff and dutt class 10 2024 chapter 7 transformation of sentences solution, dav public school new punaichak, patna admission 2024 – 2025 details, st karens secondary school khagaul road, patna admission 2024 – 2025 details.

Sign in to your account

Username or Email Address

Remember Me

Class 11 Physics Case Study Questions PDF Download

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

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

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

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

Table of Contents

Class 11th PHYSICS: Chapterwise Case Study Question & Solution

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

Case Study-Based Questions for Class 11 Physics

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

Class 11 Physics MCQ Questions

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

Class 11 Books for Boards

Class 11 Physics Syllabus 2024

Unit I: Physical World and Measurement 08 Periods

Chapter–2: Units and Measurements

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

Unit II: Kinematics 24 Periods

Chapter–3: Motion in a Straight Line

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

Chapter–4: Motion in a Plane

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

Unit III: Laws of Motion 14 Periods

Chapter–5: Laws of Motion

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

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

Unit IV: Work, Energy and Power 14 Periods

Chapter–6: Work, Energy and Power

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

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

Chapter–7: System of Particles and Rotational Motion

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

Unit VI: Gravitation 12 Periods

Chapter–8: Gravitation

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

Unit VII: Properties of Bulk Matter 24 Periods

Chapter–9: Mechanical Properties of Solids

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

Chapter–10: Mechanical Properties of Fluids

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

Chapter–11: Thermal Properties of Matter

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

Unit VIII: Thermodynamics 12 Periods

Chapter–12: Thermodynamics

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

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

Chapter–13: Kinetic Theory

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

Unit X: Oscillations and Waves 26 Periods

Chapter–14: Oscillations

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

Chapter–15: Waves

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

FAQs about Class 11 Physics Case Studies

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

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

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

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

Are the case study questions based on the latest syllabus?

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

You Might Also Like

50+ jee mains mcq questions dual nature of radiation and matter with solutions, 50+ neet mcq questions work, energy and power with solutions.

Units and Measurement Notes Class 11 Physics Notes- Download PDF for JEE/NEET

Leave a reply cancel reply.

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

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

The Topper Combo Flashcards

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

No thanks, I’m not interested!

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• DK Goel Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• ML Aggarwal Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Sandeep Garg Textbook Solution
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• HOTS Question
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• JEE Crash Course
• JEE Previous Year Paper
• Important Info
• JEE Mock Test
• JEE Sample Papers
• SRM-JEEE Mock Test
• VITEEE Mock Test
• BITSAT Mock Test
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• COMEDK Previous Year Paper
• GUJCET Previous Year Paper
• KCET Previous Year Paper
• KEAM Previous Year Paper
• Manipal Previous Year Paper
• MHT CET Previous Year Paper
• WBJEE Previous Year Paper
• AMU Previous Year Paper
• TS EAMCET Previous Year Paper
• SRM-JEEE Previous Year Paper
• VITEEE Previous Year Paper
• BITSAT Previous Year Paper
• UPSEE Previous Year Paper
• CGPET Previous Year Paper
• CUSAT Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

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

The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids contains a vast collection of questions to practise. By practising these questions of Chapter 9 Mechanical Properties of Solids, students can strengthen their problem solving skills as well as reasoning skills; these skills can be used in further chapters and real-life problems. These questions of Chapter 9 Mechanical Properties of Solids can also help students to build a strong foundation for the chapter. 

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

The Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids can be a bit confusing if students are not able to understand the concepts properly. For that purpose students need to start practising Chapter 9 Mechanical Properties of Solids questions from the NCERT textbook. After practising those questions, students can prefer referring to the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids PDF which is available in the Selfstudys website. 

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

In the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids, a plethora of exercises are given so that students can understand the concepts as well as can solve confusions then and there. By practising exercise wise questions, students can learn to approach different questions of Chapter 9 Mechanical Properties of Solids in different and creative ways. 

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

The questions in the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids are arranged according to each and every formula. By practising the questions of Chapter 9 Mechanical Properties of Solids formula wise, students can become more confident while applying the formulas. By applying the right formulas in the right questions of Chapter 9 Mechanical Properties of Solids, students can score good marks in those questions.

Where can Students Find the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids?

Students can find the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids, steps to attempt are clearly explained below: 

• Visit the Selfstudys website. 

• Bring the arrow towards NCERT Books & Solutionss which can be seen in the navigation bar. 
• A drop down menu will appear, select NCERT Solutionss from the list. 

• A new page will appear, select Class 11 from the list of classes. 

• Now select the subject Physics from the list. 
• Again a new page will appear, select the chapter Chapter 9 Mechanical Properties of Solids. 

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

The features of NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids is considered to be distinctive trait, some important features are discussed below: 

• Based on NCERT Syllabus: The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids revision are based on the NCERT syllabus so that by referring to it students can have an updated knowledge. 
• Different Levels of Questions: In the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids theory, different levels of questions are asked; that is easy to difficult.
• Hints and Solutionss are Given: In the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids PDF, hints and Solutionss are given so that students can easily solve all their doubts and confusions. 
• All Exercises are Covered: Inside the Class 11 Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids, all the exercises are covered in the NCERT Solutions; accordingly students can learn to approach in different ways. 
• Diagrams are Given: Diagrams are generally considered to be visual representation of the Class 11 Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids questions and concepts; the same is followed in NCERT Solutions. 
• Available in the PDF: The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids is available in the PDF so that students can access answers whenever they want to. 

What Are the Advantages of Using NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids?

The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids provides plenty of advantages, some of the advantages are discussed below: 

• Improves Performance: By solving questions in the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids revision, students can improve their problem solving skills and can increase the chances of getting good marks in the test. The Solutions helps students understand the concepts of Chapter 9 Mechanical Properties of Solids as well as questions effectively and efficiently.
• Builds Confidence: By solving questions from NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids revision, students can build their confidence as it provides a variety of questions. Being confident while attempting Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids questions can help one to get right and accurate answers. 
• Enhances the Learning Process: By using the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids theory, students can enhance their learning process and can reinforce the learning strategies. 
• Saves Time and Effort: Students don’t need to search for answers as it is already available in the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids PDF as it saves both time and effort of students. 
• Easy to Understand: As the answers of Class 11 Physics Chapter 9 Mechanical Properties of Solids are very simple this makes students understand all levels of questions easily. 
• Helps to Maintain Accuracy: Accuracy is considered to be the freedom from making mistakes; students can maintain the accuracy level by solving Class 11 Physics Chapter 9 Mechanical Properties of Solids questions from NCERT Solutions. 

Is NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Right for Students?

The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids is right for students as it is arranged according to the latest syllabus,some of the key factors are discussed below: 

• To Understand Concepts: The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids revision is considered to be the right one as it allows students to understand the concepts easily. 
• Comprehensive Coverage: The topics are covered in a comprehensive way in the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids theory so it is right for students to refer. 
• Reliable: Students can be dependent on the NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids PDF as the questions and answers are accurate and reliable. Accordingly, students don’t need to check for the right answers of Chapter 9 Mechanical Properties of Solids questions from other reference materials. 
• Improves Problem Solving Skills: It is a skill in which students tend to look up to questions of Chapter 9 Mechanical Properties of Solids Chapter 9 Mechanical Properties of Solids in a logical way and solve those questions in a logical way. So to improve problem solving skills while attempting questions of Chapter 9 Mechanical Properties of Solids, students can refer to the NCERT Solutions of Class 11 Physics. 
• Follows the Latest Syllabus: The Chapter 9 Mechanical Properties of Solids questions in the NCERT Solutions Class 11 Physics follows the latest syllabus so it is right for students to choose it. 
• Can Solve Questions with Ease: Students can choose the NCERT Solutions for Class 11 Physics so that they can solve the Chapter 9 Mechanical Properties of Solids questions with ease. 

When Is the Best Time to Use NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids?

The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids can be used at any time while preparing; but is it better to follow the given tips: 

• 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. 

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences

One Last Step...

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.

• NCERT Solutions
• NCERT Class 11
• NCERT 11 Physics
• Chapter 9: Mechanical Properties Of Solids

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

Ncert solutions class 11 physics chapter 9 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

NCERT Solutions Class 11 Physics Chapter 9 Mechanical Properties of Solids are important resources that help students understand the topics and prepare well for the exam. It is very important to make notes while studying each topic thoroughly. It is important to solve the questions from the textbook and get well-versed in the topics. The experienced faculty drafted the NCERT Solutions , according to the latest CBSE Syllabus 2022-23, with the aim of improving conceptual knowledge among students.

We all know that a rigid body is something that has a definite size and shape, but in reality, it can be compressed, stretched and bent. In order to change the shape of a solid, a force is needed. Elasticity and plasticity are the main concepts of concern in this chapter. To help students with their exam preparations, we at BYJU’S have provided NCERT Solutions for Class 11 Physics in an easily downloadable PDF format from the link below.

carouselExampleControls112

Previous Next

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

Q1. 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 Young’s modulus of steel to that of copper?

Length of the steel wire, l 1 = 4.7 m

Cross-sectional area of the steel wire, a 1 = 3.0 × 10 –5 m 2

Length of the copper wire, l 2 = 3.5 m

Cross-section area of the copper wire, a 2 = 4.0 × 10 –5 m 2

Change in length = Δl 1 = Δl 2 = Δl

Force applied in both cases = F

Young’s modulus of the steel wire

Young’s modulus of the copper wire

Dividing (1) by (2), we get

Q2. The figure below shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

Young’s modulus ,Y =Stress/Strain =150 x 10 6 /0.002 = 150 x 10 6 /2 x 10 -3 = 75 x 10 9  Nm -2 =75 x 10 10  Nm -2

(a)Yield strength of a material is the maximum stress that the material can sustain and retain its elastic property. From the graph, the approximate yield strength of the given material = 300 x 10 6  Nm -2 = 3 x 10 8  Nm -2  .

Q3. The stress-strain graphs for materials A and B are shown in the figure below. 

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?

Young’s modulus = Stress/Strain

(a) From the graphs, we can see that for the given strain, stress for A is greater than that of B. Therefore, Young’s modulus of A is greater than B. (b) Young’s modulus is also a measure of the strength of the material. Young’s modulus is greater for A; therefore, material A is stronger than B.

Q4. 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 ) True. Stretching a coil does not change its length; only its shape is altered, and this involves shear modulus.

( b ) False. This is because, for the same value of stress, there is more strain in rubber than in steel. And as Young Modulus is an inverse of strain, it is greater in steel.

Q5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass, are  loaded, as shown in Fig. 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. [Young’s modulus of steel is 2.0 x 10 11  Pa. (1 Pa = 1 N m 2 )]

Diameter of the two wires, d =0.25m Radius of the wires, r= d/2 =0.125cm Unloaded length of the steel wire, l 1 =1.5m Unloaded length of the brass wire, l 2 =1.0m

Force exerted on the steel wire F 1 =(4+6)g=10×9.8=98N

Cross-section area of the steel wire, a 1  =πr 1 2

Change in length of the steel wire = Δl 1

Young’s modulus for steel= 2.0 x 10 11  Pa

= 1.49 x 10 -4 m

Force of the brass wire ,F 2 = 6 x 9.8 = 58. 8 N

Cross-section area of the brass wire, a 2  =πr 2 2

Change in length of the brass wire = Δl 2

Young’s modulus of the brass wire = 0.91 x 10 11 Pa

= 1.3 x 10 -4 m

The elongation of the steel wire is 1.49 x 10 -4 m, and that of brass is 1.3 x 10 -4 m.

Q6. 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?

Edge of the aluminium cube, L = 10 cm =10/100= 0.1 m Area of each face, A = (0.1) 2  = 0.01 m 2

Mass attached to the opposite face of the cube = 100 kg

Tangential force acting on the face, F = 100 kg = 100 x 9.8 = 980 N

Shear modulus, η = Tangential stress/Shearing strain

Shearing strain = Tangential stress/ Shear modulus

= F/Aη = 980/(0.01 x 25 x 10 9 ) = 3.92 x 10 -6

Shearing strain = Lateral strain/Side of the cube

Lateral strain = Shearing strain x Side of the cube = 3.92 x 10 -6 x 0.1

= 3.92 x 10 -7 m ≈ 4 x 10 -7 m

Q7. Four identical hollow cylindrical columns of mild steel support a big structure with a mass of 50,000 kg. The inner and outer radii of each column are 30 and 60 cm, respectively. Assuming the load distribution to be uniform, calculate the compressional strain of  each column.

Mass of the big structure, M = 50,000 kg

Total force exerted on the four columns= Total weight of the structure=50000×9.8N

The compressional force on each column = Mg/4 = (50000×9.8)/4 N= 122500 N Therefore, Stress = 122500 N Young’s modulus of steel, Y=2×10 11 Pa

Young’s modulus, Y= Stress/Strain

Strain = Young’s modulus/Stress

Strain = (F/A)/Y

Inner radius of the column, r = 30 cm = 0.3 m Outer radius of the column, R = 60 cm = 0.6 m Where, Area, A=π(R 2 −r 2 )=π((0.6) 2 −(0.3) 2 ) = 0.27 π m 2

Strain =122500/[0.27 x 3.14×2×10 11 ]=7.22×10 −7

Hence, the compressional strain of each column is 7.22×10 −7 .

Q8. 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.

Area of the copper piece, A=19.1×10 −3 ×15.2×10 −3 =2.9×10 −4 m 2

Tension force applied on the piece of copper, F=44,500 N

Modulus of elasticity of copper, Y=42×10 9  Nm −2

Modulus of elasticity (Y) = Stress / Strain

=(F/A) / Strain Strain = F/(YA) = 44500/(2.9×10 −4 ×42×10 9 ) = 3.65×10 −3

Q9. A steel cable with a radius of 1.5 cm supports a chairlift in a ski area. If the maximum stress is not to exceed 10 8  Nm -2 .What is the maximum load the cable can support?

Radius of the steel cable, r=1.5 cm = 0.015 m

Cross-sectional area of the cable = πr 2 = 3.14 x ( 0.015) 2

= 7.06 x 10 -4 m

Maximum stress allowed on the steel cable = 10 8 N/m 2

Maximum load the cable can support = Maximum stress × Area of cross-section

= 10 8  x 7.06 x 10 -4

= 7.065×10 4  N Hence, the cable can support the maximum load of 7.065×10 4 N.

Q10. 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 ratios of their diameters if each is to have the same tension.

As the tension on the wires is the same, the extension of each wire will also be the same. Now, as the length of the wires is the same, the strain on them will also be equal.

Now, we know

Y = Stress / Strain = (F/A) / Strain  =  (4F/πd 2 ) / Strain     . . . . . . . . . . ( 1 ) Where, A = Area of cross-section F = Tension force d = Diameter of the wire We can conclude from equation ( 1 ) that Y ∝ (1/d 2 ). We know that Young’s modulus for iron, Y 1  = 190 × 10 9  Pa Let the diameter of the iron wire = d 1 Also, Young’s modulus for copper, Y 2  = 120 × 10 9  Pa Let the diameter of the copper wire = d 2 Thus, the ratio of their diameters can be given as

= \(\begin{array}{l}\sqrt{\frac{190×10^{9}}{120×10^{9}}}\end{array} \) =1 : 25 : 1

Q11. 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.

Mass, m = 14.5 kg

Length of the steel wire, l = 1 m

Angular velocity, v = 2 rev/s

Cross-sectional area of the wire, A = 0.065 x 10 -4  m 2

Total pulling force on the steel wire when the mass is at the lowest point of the vertical circle, F = mg + mr ω 2

= 14.5×9.8+14.5×1×(12.56) 2 =2429.53 N

Young’s modulus = Stress / Strain \(\begin{array}{l}Y=\frac{F}{A}\frac{l}{\Delta l}\end{array} \)

Δl=(2429.53×1)/(0.065×10 −4 ) × (2×10 11 ) = 1.87×10 −3 m Hence, the elongation of the wire when the mass is at the lowest is 1.87×10 −3 m.

Q12. Compute the bulk modulus of water from the following data: Initial volume = 100.0  litres, Pressure increase = 100.0 atm (1 atm = 1.013 × 10 5   Pa), Final volume = 100.5  litres. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Initial volume, V 1 =100.0 litre =100.0×10 −3 m 3 Final volume, V 2 =100.5 litre =100.5×10 −3 m 3

Change in the volume, ΔV=V 2 −V 1 =0.5×10 −3 m 3

Pressure increase, p=100.0atm=100×1.013×10 5 Pa

= 101.3 x 10 5 Pa Bulk modulus of water= p/(ΔV/V 1 )=pV 1 /ΔV = 101.3×10 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

The intermolecular force in liquids is much larger than in air as the distance between the molecules is much lesser in liquid than in air. Therefore, at the same temperature, the strain for water is much more than for air.

Q13. 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 ?

Let the depth be the alphabet ‘d’.

Given, Pressure at the given depth, p = 60.0 atm = 60 × 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 d. V 1  be the volume of water of mass m at the surface. Then, let V 2 be the volume of water of mass m at the depth h. And ΔV is 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 )        . .  . . .  . . ( 1 ) We know, 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  = 60 × 1.013 × 10 5  × 45.8 × 10 -11   =  2.78 × 10 -3     . . . . . ( 2 )

Using equation ( 1 ) and equation ( 2 ), we get 1 – (ρ 1 /ρ 2 )   =   2.78 × 10 -3 ρ 2  = 1.03 × 10 3  / [ 1 – (2.78 × 10 -3 ) ] = 1.032 × 10 3  kg m -3 Therefore, at depth d, water has a density of 1.034 × 10 3  kg m –3 .

Q14. Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm.

Pressure acting on the glass plate, 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 × 10 5 ] / (37 × 10 9 ) = 2.73 × 10 -4 Therefore, the fractional change in the volume of the glass plate is 2.73 × 10 –4 .

Q15. Determine the volume contraction of a solid copper cube, 10 cm on edge, when  subjected to a hydraulic pressure of 7.0 × 10 6 Pa.

Side of the copper cube, a = 10 cm

Therefore, the volume of the copper cube, V = a 3  = 10 -3 m 3

Hydraulic pressure, p = 7.0 x 10 6  Pa

Bulk modulus of copper B = 140 G Pa = 140 x 10 9  Pa.

Bulk modulus, K=−P/(ΔV/V) ​

We get the value of volume contraction as ΔV = – PV/K =−(7×10 6  ×0.001)/(140×10 9 )

=−0.05×10 −6 m 3

Q16.How much should the pressure on a litre of water be changed to compress it by 0.10%?

Volume of water, V=1 litre

Water should be compressed by 0.10% The fractional change in volume, △V/V=(0.1/100)×1=10 −3

Bulk modulus, B =P/(△V/V) = PV/△V P=B×(△V/V)

Bulk modulus of water, B = 2.2×10 9 Nm −2

Pressure on water, P=2.2×10 9 ×10 −3 =2.2×10 6  Pa

Q17. Anvils made of single crystals of diamond, with the shape as shown in the f igure, are used to investigate the 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?

Flat faces at the narrow end of the anvil have a diameter, d=0.50mm=0.5×10 −3 m Radius, r=d/2=0.25×10 −3 m Compressional force, F=50000N Pressure at the tip of the anvil P=Force/Area

Area = πr 2  = 3.14 x (0.25×10 −3  ) 2 = 0.1925 x 10 -6 m 2

Pressure at the tip of the anvil = F/A

= 50000/0.1925 x 10 -6 =2.59×10 11  Pa

Q18. 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 the figure. 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?

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 mm 2  = 2 × 10 –6  m 2

We know 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 ( i ) Let a mass m be hung on the stick at a distance y from the end where wire A is attached. Stress in the wire = Force / Area  =  F / a Now, it is given that the two wires have equal stresses F 1  / a 1   =  F 2  / a 2

Where, F 1  = Force acting on wire A and F 2  = Force acting on wire B F 1  / F 2  = a 1  / a 2   =  1 / 2    . . . . . . . . . . . . ( 1 ) The above situation can be represented as

Moment of forces about the point of suspension, we have F 1 y = F 2  (1.5 – y) F 1  / F 2  = (1.5 – y) / y    . . . . . . . . . . ( 2 ) Using equation ( 1 ) and equation ( 2 ), we can write (1.5 – y) / y  = 1 / 2 2 (1.5 – y)  =  y y = 1 m

Therefore, the mass needs to be hung at a distance of 1m from the end where wire A is attached in order to produce equal stress in the two wires.

( ii ) We know,

Young’s modulus = Stress / Strain => Strain = Stress / Young’s modulus  =  ( F/a)/ Y It is given that the strain in the two wires is equal ( F 1 /a 1 ) / Y 1   =  ( F 2 /a 2 ) / Y 2 F 1  / F 2  = a 1 Y 1  / a 2 Y 2 a 1  / a 2  = 1 / 2 F 1  / F 2  = (1 / 2) (2 × 10 11  / 7 × 10 10 )  =  10 / 7      . . . . . . . . . . . ( 3 )

Let the mass m be hung on the stick at a distance y 1 from the end where the steel wire is attached in order to produce equal strain.

Taking the moment of the force about the point where mass m is suspended

F 1 y 1  = F 2  (1.5 – y 1 ) F 1  / F 2   =  (1.5 – y 1 ) / y 1                                     . . . . . . . . . . . ( 4 )

From  equations ( 3 ) and ( 4 ), we get (1.05 – y 1 ) / y 1   =  10 / 7 7(1.05 – y 1 )  =  10y 1 y 1  = 0.432 m

Therefore, the mass needs to be hung at a distance of 0.432 m from the end where wire A is attached in order to produce equal strain in the two wires.

Q19. 9 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.

Steel wire length = 1.0 m

Area of cross-section, A = 0.50 \(\begin{array}{l}\times 10^{-2} cm^2\end{array} \)

Mass of the given object, m = 100 g = 0.1 kg

From the given diagram, We can deduce that \(\begin{array}{l}x \end{array} \) is the depression at the midpoint, that is, CD = \(\begin{array}{l}x \end{array} \) It is given that, AD = BD = \(\begin{array}{l}(I^2 + x^2)^\frac{1}{2}\end{array} \) AC = CB = I = 0.5m m = 100g = 0.100kg

Gain in length, \(\begin{array}{l}\bigtriangleup I\end{array} \) = AD + DB – AB = 2AD – AB = \(\begin{array}{l}2(I^2 + x^2)^\frac{1}{2} – 2I = \left ( 1 + \frac{x^2}{I^2} \right )2I -(2I)\end{array} \) = \(\begin{array}{l}2I \left (1 + \frac{x^2}{2I^2} \right ) – 2I = \frac{x^2}{I}\end{array} \) Thus, strain = \(\begin{array}{l}\frac{\bigtriangleup I}{2I} = \frac{x^2}{2I^2} \end{array} \) If the tension of the wire is \(\begin{array}{l}T\end{array} \) , then \(\begin{array}{l}2 T cos\theta = mg\end{array} \) ∴ \(\begin{array}{l}T = \frac{mg}{2cos\theta }\end{array} \) Now, \(\begin{array}{l}cos\theta\end{array} \) = \(\begin{array}{l}\frac{x}{(I^2 + x^2)^\frac{1}{2}} = \frac{x}{I\left ( 1 + \frac{1}{2}\frac{x^2}{I^2} \right )^\frac{1}{2}} = \frac{x}{\left ( 1 + \frac{1}{2}\frac{x^2}{I^2} \right )}\end{array} \) Since, \(\begin{array}{l}x \end{array} \)   << 1, then 1 >> \(\begin{array}{l}\frac{1x^2}{2I^2} \end{array} \)   &  \(\begin{array}{l}1 + \frac{1}{2} \frac{x^2}{I^2} \approx 1\end{array} \) Therefore, \(\begin{array}{l}T = \frac{mg}{2\left ( \frac{x}{1} \right )} = \frac{mgI}{2x}\end{array} \) Stress = \(\begin{array}{l}\frac{T}{A} = \frac{mgl}{2Ax} \end{array} \) Young’s modulus \(\begin{array}{l}Y = \frac{stress}{strain}\end{array} \) = \(\begin{array}{l}\frac{mgI}{2Ax}\times \frac{2I^2}{x^2} = \frac{mgI^3}{Ax^3} \end{array} \) Therefore, \(\begin{array}{l}x = I \left ( \frac{mg}{YA} \right )^\frac{1}{3} = 0.5 \left ( \frac{0.1 \times 10}{20 \times 10^{11} \times 10^{-6}} \right )\end{array} \) = \(\begin{array}{l}1.074 \times 10^{-2}m\end{array} \)

Q20. 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.

Diameter of the metal strips= 6mm = 6 x  10 -3  m

Radius, r = 3 x 10 -3  m; Shearing stress on the rivet= 6.9 x 10 7  Pa Maximum load or force on a rivet = Maximum stress x cross-sectional area = 6.9 x 10 7  x 3.14 x (3 x 10 -3 ) 2  N = 1950 N Maximum tension = 4 x 1950 N = 7800 N

Q21. The Marina trench is located in the Pacific Ocean, and in one place, it is nearly eleven km beneath the surface of the 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 the bottom?

Water pressure at the bottom of the trench, p=1.1×10 8 Pa Initial volume of the steel ball, V=0.32m 3 Bulk modulus of steel, B=1.6×10 11 Nm −2

The ball falls at the bottom of the trench, which is nearly 11 km beneath the surface of the water.

The volume change of the ball after reaching the bottom of the trench is  △V. Bulk modulus, B=p/(△V/V) = pV/△V △V= pV/B =(1.1×10 8 ×0.32)/(1.6×10 11 ) = 0.352 ×10 8 /1.6×10 11

= 0.22 x 10 -3 m 3

The change in volume of the ball on reaching the bottom of the trench is 0.22 x 10 -3 m 3

Q22.  A mild steel wire of cross-sectional area 0.60 x 10 -2 cm 2 and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.

Let YZ be the mild steel wire of length 2l = 2m and cross-sectional area A = 0.60 x 10 -2 cm 2 . Let the mass of m = 100 g = 0.1 kg be hung from the midpoint O, as shown in the figure. And let x be the depression at the midpoint, i.e., OD

From the figure,

ZO =YO = l = 1 m

ZD = YD =  (l 2  + x 2 ) 1/2

Increase in length, ∆l = YD + DZ – ZY

= 2YD – YZ                           ( As DZ = YD)

= 2(l 2  + x 2 ) 1/2 – 2l

∆l = 2l( x 2 /2l 2 ) = x 2 / l

Therefore, longitudinal strain = ∆l / 2l = x 2 /2l 2   . . . . . . . . ( i )

If T is the tension in the wires, then in equilibrium 2Tcosθ = 2mg

Or,          T = mg / 2cos θ

= [ mg (l 2  + x 2 ) 1/2 ] / 2x =mgl / 2x

Therefore, Stress = T / A = mgl / 2Ax   . . . . . . . . . . . . ( ii )

= \(\begin{array}{l}\frac{mgl^{3}}{2Ax^{3}}\end{array} \)

x = \(\begin{array}{l}l[\frac{mg}{YA}]^{\frac{1}{3}}\end{array} \) = \(\begin{array}{l}1[\frac{0.1 \times 10}{20 \times 10^{11} \times 0.6 \times 10^{-6}}]^{\frac{1}{3}}\end{array} \)

= 9.41 x 10 -3 m.

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids is one of the most important chapters for the students. Questions from Mechanical Properties of Solids are asked in most Class 11 examinations. Students who aim to score good marks in these exams and entrance exams should try to solve NCERT questions given at the end of the chapter. Solving NCERT questions will help them to understand the chapter in a better way.

Finding out a solid’s mechanical properties is one of the core concepts in Physics. Here, students will be covering topics such as Young’s modulus of copper and steel. We will derive information, such as yield strength, from a plotted graph and even compare between stress-strain relation graph and find out the values in Young’s modulus and strength in this chapter.

Check out Physics NCERT Solutions for Class 11 for more information.

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

Have you ever stretched a coil, do you want to know what factors determine the stretch of a coil? Find out why Young’s modulus is greater in steel than in rubber. We will be finding the compression strain of each column of a cylinder, and we are finding the ratio of the diameter of three wires if tension is the same in all of them. We will find the density of water when the pressure is at the bottom, and we will also find the fractional change in the volume of a glass plate when pressure is applied. We will be calculating the pressure on a litre of water if it is compressed; along with that, we will be seeing what will happen when a ball is dropped into the Marina Trench, the deepest point in the planet’s ocean. Similar to this, we will see many other examples in this chapter whose sole purpose is to make students understand the concepts clearly, and if practised effectively, they will be able to score excellent results.

Subtopics of C lass 11 Physics Chapter 9 Mechanical Properties of Solids

• Introduction
• Elastic behaviour of solids
• Stress and strain
• Hooke’s law
• Stress-strain curve
• Elastic Moduli
• Applications of elastic behaviour of materials.

NCERT Solutions for Class 11 Physics Chapter 9 is prepared by the subject experts by verifying different textbooks, previous years’ question papers and sample papers. In order to get a good score in Class 11 examinations and entrance exams, it is very important for the students to study these solutions repeatedly. BYJU’S provide students with the finest of study materials, notes, sample papers, MCQs (multiple choice questions), short and long answer questions, exemplary problems and worksheets. These will assist them in equipping you better to face Class 11 examinations and all-important engineering and medical entrance examinations. These are prepared as per the latest CBSE Syllabus 2023-24 .

Disclaimer – 

Dropped Topics – 

9.2 Elastic Behaviour of Solids 9.6.2 Determination of Young’s Modulus of the Material of a Wire

Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 9

What is biofortification in chapter 9 mechanical properties of solids of ncert solutions for class 11 physics, what are the topics covered in chapter 9 of ncert solutions for class 11 physics, do the ncert solutions for class 11 physics chapter 9 help students to score well in the cbse exams, leave a comment cancel reply.

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

Request OTP on Voice Call

Post My Comment

I’m happy with this aap

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Counselling

• CBSE Class 11 Physics Chapter 9 – Mechanical Properties of Solids Class 11 Notes

Mechanical Properties of Solids Class 11 CBSE Revision Notes

Mechanical Properties of Solids Class 11 Notes – First of all the chapter explains stress. Also, it tells that stress is the restoring force per unit area. On the other hand, the strain is the fractional change in dimension. Generally, there are three types of stresses namely: tensile stress, shearing stress, and hydraulic stress. Moreover, the tensile stress is further classified as longitudinal stress or compressive stress. In addition, for small deformation stress, is directly proportional to the strain for many materials. Most noteworthy, this is known as Hooke’s Law and the constant of proportionality is the modulus of elasticity.

Elastomers are a class of solid that does not obey Hooke’s Law. Moreover, when we apply a parallel pair of forces to a solid’s upper and lower faces the solid deforms. Furthermore, the upper face of the solid moves sideways with respect to the lower face. Also, the horizontal displacement of the upper face is perpendicular to the vertical height. In addition, this deformation is known as shear and the stress is shearing stress. Besides, stress is not a vector quantity as unlike a force, stress cannot be assigned in a specific direction.

Download Toppr app for Android and iOS or signup for free.

Sub-topics covered under Mechanical Properties of Solids

• Elasticity and Plasticity – This topic explains elasticity, plasticity, stress, types of stress, and Hooke’s Law.
• Applications of Elastic Behaviour of Materials – The topic overviews the behavior of materials on the application of elasticity.
• Stress and Strain – This topic defines solids, stress, and strain.
• Elastic Moduli – The topic highlights elastic moduli, and various moduli of Young, Shear, etc.
• Hooke’s Law and Stress-strain Curve – This topic teaches about Hooke’s law stress, and strain curve, and analysis of the curve.

You can download CBSE Class 11 Physics Chapter 9  Mechanical Properties of Solids Revision Notes by clicking on the download button below

Customize your course in 30 seconds

Which class are you in.

CBSE Class 11 Physics Revision Notes

• CBSE Class 11 Physics Chapter 5 – Law of Motion Class 11 Notes
• CBSE Class 11 Physics Chapter 10 – Mechanical Properties of Fluids Class 11 Notes
• CBSE Class 11 Physics Chapter 8 – Gravitation Class 11 Notes
• CBSE Class 11 Physics Chapter 7 – Systems of Particles and Rotational Motion Class 11 Notes
• CBSE Class 11 Physics Chapter 6 – Work, Energy and Power Class 11 Notes
• CBSE Class 11 Physics Chapter 2 – Units and Measurement Class 11 Notes
• CBSE Class 11 Physics Chapter 1 Physical World Class 11 Notes
• CBSE Class 11 Physics Chapter 11 – Thermal Properties of Matter Class 11 Notes
• CBSE Class 11 Physics Chapter 12 – Thermodynamics Class 11 Notes

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

Talk to our experts

1800-120-456-456

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

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 2023-24 PDF

Also, check CBSE Class 11 Physics revision notes for other chapters:

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)

CBSE Study Materials for Class 11

CBSE Expert

CBSE Class 11 Physics Mechanical Properties of Solids MCQ with Answers PDF

CBSE MCQ Questions for Class 11 Physics Chapter 9 Mechanical Properties of Solids with Answers Pdf free download. MCQ Questions for Class 11 Physics with Answers were prepared based on the latest exam pattern. This MCQ will help you score good marks in the final exam. Mechanical Properties of Solids Class 11 Physics MCQs are prepared for a better understanding of the concept. Mechanical Properties of Solids Class 11 Physics MCQ is prepared by Experts of CBSE.

Class 11 Physics Chapter 9 Mechanical Properties of Solids MCQs

Check the multiple-choice questions for the 11th Class Physics Chapter 9 Mechanical Properties of Solids. Select one answer out of 4 options.

Mechanical Properties of Solids Class 11 MCQs Questions with Answers

Out of the following materials, whose elasticity is independent of temperature? (a) Copper (b) Invar steel (c) Brass (d) Silver

Answer: (b) Invar steel

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is (a) 10 cm (b) 15 cm (c) 2.5 cm (d) 5 cm

Answer: (d) 5 cm

The ratio of the change in dimension at right angles to the applied force to the initial dimension is known as (a) Youngs modulus (b) Poissons ratio (c) Lateral strain (d) Shearing strain

Answer: (c) Lateral strain

A rubber cord of cross sectional area 1 mm² and unstretched length 10 cm is stretched to 12 cm and then released to project a stone of mass 5 gram. If Y for rubber = 5 ? 108N/m², then the tension in the rubber cord is (a) 25 N (b) 50 N (c) 100 N (d) 200 N

Answer: (c) 100 N

If a glass prism is dipped in water, its dispersive power (a) increases (b) decreases (c) does not change (d) may increase or decrease depending on whether the angle of the prism is less than or greater than 60º

Answer: (b) decreases

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods. (a) both the rods will elongate but there shall be no perceptible change in shape (b) the steel rod will elongate and change shape but the rubber rod will only elongate (c) the steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse (d) the steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre

Answer: (d) the steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre

A spring is stretched by applying a load to its free end. The strain produced in the spring is (a) volumetric (b) shear (c) longitudinal and shear (d) longitudinal

Answer: (c) longitudinal and shear

A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1 cm. The two wires will have (a) the same stress (b) different stress (c) the same strain (d) different strain

Answer: (a) the same stress and (d) different strain

The maximum load a wire can withstand without breaking when its length is reduced to half of its original length, will (a) be doubled (b) be half (c) be four times (d) remain same

Answer: (d) remains same

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight. (a) tensile stress at any cross-section A of the wire is F/A (b) tensile stress at any cross-section is zero (c) tensile stress at any cross-section A of the wire is 2F/A (d) tension at any cross-section A of the wire is FA

Answer: (a) tensile stress at any cross-section A of the wire is F/A

Hookes law essentially defines (a) Stress (b) Strain (c) Yield point (d) Elastic limit

Answer: (d) Elastic limit

Two wires have the same material and length, but their masses are in the ration of 4 : 3. If they are stretched by the same force, their elongations will be in the ratio of (a) 2 : 3 (b) 3 : 4 (c) 4 : 3 (d) 9 : 16

Answer: (b) 3 : 4

Four wires whose lengths and diameter respectively are given below are made of the same material. Which of these will have the largest extension when same tension is applied? (a) 0.50 m, 0.50 mm (b) 1.00 mm, 1.00 mm (c) 2.00 m, 2.00 mm (d) 4.00 m, 4.00 mm

Answer: (a) 0.50 m, 0.50 mm

An iron bar of length l m and cross section A m² is pulled by a force of F Newton from both ends so as to produce and elongation in meters. Which of the following statement statements is correct (a) Elongation is inversely proportional to length l (b) Elongation is directly proportional to cross section A (c) Elongation is inversely proportional to A (d) Elongation is directly proportional to Youngs modulus

Answer: (c) Elongation is inversely proportional to A

A rubber cord of cross sectional area 1 mm² and unstretched length 10 cm is stretched to 12 cm and then released to project a stone of mass 5 gram. If Y for rubber = 5 X 10 8 N/m², then the tension in the rubber cord is (a) 25 N (b) 50 N (c) 100 N (d) 200 N

You are given two wires W 1  and W 2 . Both are made of the same material and are of the same length. The radius of cross-section of W 2  is twice that of W 1 . Same load is suspended from both of them. If the strain in W 1  be 4, then calculate the strain in W 2 . (a) 8 (b) 4 (c) 2 (d) 1

Answer: (d) 1

A steel wire is loaded by 2 kg weight. If the radius of the wire is doubled, then its extension will become (a) half (b) four times (c) one-fourth (d) double

Answer: (c) one-fourth

The energy per unit volume of a strained wire is given by (a) 1/2 stress × strain (b) 1/2 strain × load (c) 1/2 extension × stress (d) 1/2 load × extension

Answer: (a) 1/2 stress × strain

In the following data, the four wires made of the same material have length L and radius r. Which 6f these will have the maximum extension when the same force is applied? (a) L = 50 cm, r = 0.4 mm (b) L = 100 cm, r = 1 mm (c) L = 200 cm, r = 2 mm (d) L = 300 cm, r = 3 mm

Answer: (a) L = 50 cm, r = 0.4 mm

A cable which can support a load W is cut into two equal parts. The maximum load that can be supported by either part is (a) W (b) 2W (c) W/2 (d) W/4

Answer: (a) W

Which of the following affects the elasticity of a substance? (a) Hammering and annealing (b) Change in temperature (c) Impurity in substance (d) All of the above

Answer: (d) All of the above

A body of mass 1 kg is attached to one end of a wire and rotated in 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 × 10 6  N/m²? (a) 2 mm² (b) 3 mm² (c) 4 mm² (d) 5 mm²

Answer: (c) 4 mm²

The upper end of wire 1 m long and 2 mm radius is clamped. The lower end is twisted through an angle of 45°. The angle of shear is (a) 0.09° (b) 0.9° (c) 9° (d) 90°

Answer: (a) 0.09°

The magnitude of the force developed by raising the temperature from 0°C to 100°C of the iron bar of 1.00 m long and 1 cm² cross-section when it is held so that it is not permitted to expand or bend is (Temperature Co-efficient = 10 -5 / o C and Y = 10 11  N/m²) (a) 10 3  N (b) 10 4  N (c) 10 5  N (d) 10 9  N

Answer: (b) 104 N

Longitudinal strain is possible in the case of (a) Gases (b) Liquid (c) Only solids (d) Only gases & liquids

Answer: (c) Only solids

In a wire, when elongation is 2 cm energy stored is E. if it is stretched by 10 cm, then the energy stored will be (a) E (b) 2 E (c) 4 E (d) 25 E

Answer: (d) 25 E

When the intermolecular distance increases due to tensile force, then (a) There is no force between the molecules (b) There is a repulsive force between the molecules (c) There is an attractive force between the molecules (d) There is zero resultant force between the molecules

Answer: (c) There is an attractive force between the molecules

The magnitude of the force developed by raising the temperature from 0°C to 100°C of the iron bar of 1.00 m long and 1 cm² cross-section when it is held so that it is not permitted to expand or bend is (a = 10 -5 / o C and Y = 10 11 N/m²) (a) 10 3 N (b) 10 4 N (c) 10 5 N (d) 10 9 N

Which of the following phenomena is used in optical fibres ? (a) Total internal reflection (b) Scattering (c) Diffraction (d) Refraction

Answer: (a) Total internal reflection

Critical angle of light passing from glass to water is minimum for (a) red colour (b) green colour (c) yellow colour (d) violet colour

Answer: (d) violet colour

The substance which shows no elastic after effect practically is (a) steel (b) rubber (c) copper (d) quartz

Answer: (d) quartz

The breaking stress of a material is defined as (a) breaking load per unit area (b) breaking load per unit volume (c) breaking load per unit length (d) total breaking load

Answer: (a) breaking load per unit area

The maximum length of the steel wire (of density ρ = 7.8 × 10³ kg/m³) which when fixed at one end can hang freely without breaking (breaking stress for steel is 7.8 × 10 8 Nm -2 ) is (take g = 10 ms -2 ): (a) 1 km (b) 0.01 km (c) 0.1 km (d) 10 km

Answer: (d) 10 km

Girders are made in I shape to (a) increase its strength (b) make it appear more elegant (c) reduce the quantity of material used (d) none of the above

Answer: (c) reduce the quantity of material used

Two wires A and B are of the same material, but A is half as long and has diameter three times the diameter of wire B. If they are stretched by the same amount, then the required force in wire A must be (a) three times that on B (b) one third that on B (c) nine times that on B (d) eighteen times that on B

Answer: (d) eighteen times that on B

One end of a steel wire of area of cross-section 3 mm² is attached to the ceiling of an elevator moving up with an acceleration of 2.2 m/s². if a load of 8 kg is attached at its free end, then the stress developed in the wire will be (a) 8 × 10 6  N/m² (b) 16 × 10 6  N/m² (c) 20 × 10 6  N/m² (d) 32 × 10 6  N/m²

Answer: (d) 32 × 106 N/m²

What is the phenomenon of temporary delay in regaining the original configuration by an elastic body, after the removal of a deforming force? (a) Elastic fatigue (b) Elasticity (c) Plasticity (d) Elastic after effect

Answer: (d) Elastic after effect

In magnitude hydraulic stress is equal to (a) hydraulic force (b) hydraulic pressure (c) restoring force (d) hydraulic strain

Answer: (b) hydraulic pressure

The property of a body by virtue of which it tends to regain its original size and shape when the applied force is removed is called (a) elasticity (b) plasticity (c) rigidity (d) compressibility

Answer: (a) elasticity

Which of the following statement is False? (a) The elastomers are the elastic substances which can be subjected to large value of strain i.e. greatly stretched e.g.- rubber (b) The work done in stretching the wire is stored in it in the form of elastic potential energy (c) The modulus of rigidity of water is zero (d) None of these

Answer: (d) None of these

How does the Young’s modulus vary with the increase of temperature? (a) increases (b) decreases (c) remains constant (d) first increases and then decreases

For obtaining appreciable extension, the wire must be (a) short and thin (b) long and thick (c) long and thin (d) short and thick

Answer: (c) long and thin

The bulk modulus of a perfectly rigid body is (a) one (b) infinity (c) zero (d) none of the above

Answer: (b) infinity

Which of the following statement is False? (a) Steel is preferred for making spring over copper because Young’s modulus of steel is more than, that of copper (b) Breaking load for unit area of cross-section of a wire is called tensile strength (c) The dimensional formula for stress is same as that for pressure (d) None of these

Answer: (b) Breaking load for unit area of cross-section of a wire is called tensile strength

Hope these MCQs help you in your preparation for your final exams. Leave your comment below, if you have any questions regarding Class 11 Physics Chapter 9 Mechanical Properties of Solids MCQ with Answers .

Leave a Comment Cancel reply

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

Download India's best Exam Preparation App Now.

Key Features

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

No thanks, I’m not interested!

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

• Last modified on: 9 months ago
• Reading Time: 6 Minutes

Case Study Questions:

Question 1:

The property due to which the free surface of liquid tends to have minimum surface area and behaves like a stretched membrane is called surface tension. It is a force per unit length acting in the plane of interface between the liquid and the bounding surface i.e ., S = F/L, where F = force acting on either side of imaginary line on surface and L = length of imaginary line. Surface tension decreases with rise in temperature. Highly soluble impurities increases surface tension and sparingly soluble impurities decreases surface tension.

1. The excess pressure inside a soap bubble is three times than excess pressure inside a second soap bubble, then the ratio of their surface area is

2. Which of the following statements is not true about surface tension?

(a) A small liquid drop takes spherical shape due to surface tension.

(b) Surface tension is a vector quantity.

(c) Surface tension of liquid is a molecular phenomenon.

(d) Surface tension of liquid depends on length but not on the area.

3. Which of the following statement is not true about angle of contact?

(a) The value of angle of contact for pure water and glass is zero.

(b) Angle of contact increases with increase in temperature of liquid.

(c) If the angle of contact of a liquid and a solid surface is less than 90°, then the liquid spreads on the surface of solid.

(d) Angle of contact depend upon the inclination of the solid surface to the liquid surface.

4. Which of the following statements is correct?

(a) Viscosity is a vector quantity.

(c) Reynolds number is a dimensionless quantity.

(d) Angle of contact is a vector quantity.

5. A liquid does not wet the solid surface if the angle of contact is

(b) equal to 45°

(c) equal to 90°

(d) greater than 90°

Related Posts

You may also like:, category lists (all posts).

All categories of this website are listed below with number of posts in each category for better navigation. Visitors can click on a particular category to see all posts related to that category.

• Full Form (1)
• Biography of Scientists (1)
• Assertion Reason Questions in Biology (37)
• Case Study Questions for Class 12 Biology (14)
• DPP Biology for NEET (12)
• Blog Posts (35)
• Career Guidance (1)
• Assertion Reason Questions for Class 10 Maths (14)
• Case Study Questions for Class 10 Maths (15)
• Extra Questions for Class 10 Maths (12)
• Maths Formulas for Class 10 (1)
• MCQ Questions for Class 10 Maths (15)
• NCERT Solutions for Class 10 Maths (4)
• Quick Revision Notes for Class 10 Maths (14)
• Assertion Reason Questions for Class 10 Science (16)
• Case Study Questions for Class 10 Science (14)
• Evergreen Science Book Solutions for Class 10 (17)
• Extra Questions for Class 10 Science (23)
• HOTS for Class 10 Science (17)
• Important Questions for Class 10 Science (10)
• Lakhmir Singh Class 10 Biology Solutions (4)
• Lakhmir Singh Class 10 Chemistry Solutions (5)
• Lakhmir Singh Class 10 Physics Solutions (5)
• MCQ Questions for Class 10 Science (20)
• NCERT Exemplar Solutions for Class 10 Science (16)
• NCERT Solutions for Class 10 Science (15)
• Quick Revision Notes for Class 10 Science (4)
• Study Notes for Class 10 Science (17)
• Assertion Reason Questions for Class 10 Social Science (14)
• Case Study Questions for Class 10 Social Science (24)
• MCQ Questions for Class 10 Social Science (3)
• Topicwise Notes for Class 10 Social Science (4)
• CBSE CLASS 11 (1)
• Assertion Reason Questions for Class 11 Chemistry (14)
• Case Study Questions for Class 11 Chemistry (11)
• Free Assignments for Class 11 Chemistry (1)
• MCQ Questions for Class 11 Chemistry (8)
• Very Short Answer Questions for Class 11 Chemistry (7)
• Assertion Reason Questions for Class 11 Entrepreneurship (8)
• Important Questions for CBSE Class 11 Entrepreneurship (1)
• Assertion Reason Questions for Class 11 Geography (24)
• Case Study Questions for Class 11 Geography (24)
• Assertion Reason Questions for Class 11 History (12)
• Case Study Questions for Class 11 History (12)
• Assertion and Reason Questions for Class 11 Maths (16)
• Case Study Questions for Class 11 Maths (16)
• Formulas for Class 11 Maths (6)
• MCQ Questions for Class 11 Maths (17)
• NCERT Solutions for Class 11 Maths (8)
• Case Study Questions for Class 11 Physical Education (11)
• Assertion Reason Questions for Class 11 Physics (15)
• Case Study Questions for Class 11 Physics (12)
• Class 11 Physics Study Notes (5)
• Concept Based Notes for Class 11 Physics (2)
• Conceptual Questions for Class 11 Physics (10)
• Derivations for Class 11 Physics (3)
• Extra Questions for Class 11 Physics (13)
• MCQ Questions for Class 11 Physics (16)
• NCERT Solutions for Class 11 Physics (16)
• Numerical Problems for Class 11 Physics (4)
• Physics Formulas for Class 11 (7)
• Revision Notes for Class 11 Physics (11)
• Very Short Answer Questions for Class 11 Physics (11)
• Assertion Reason Questions for Class 11 Political Science (20)
• Case Study Questions for Class 11 Political Science (20)
• CBSE CLASS 12 (8)
• Extra Questions for Class 12 Biology (14)
• MCQ Questions for Class 12 Biology (13)
• Case Studies for CBSE Class 12 Business Studies (13)
• MCQ Questions for Class 12 Business Studies (1)
• Revision Notes for Class 12 Business Studies (10)
• Assertion Reason Questions for Class 12 Chemistry (15)
• Case Study Based Questions for Class 12 Chemistry (14)
• Extra Questions for Class 12 Chemistry (5)
• Important Questions for Class 12 Chemistry (15)
• MCQ Questions for Class 12 Chemistry (8)
• NCERT Solutions for Class 12 Chemistry (16)
• Revision Notes for Class 12 Chemistry (7)
• Assertion Reason Questions for Class 12 Economics (9)
• Case Study Questions for Class 12 Economics (9)
• MCQ Questions for Class 12 Economics (1)
• MCQ Questions for Class 12 English (2)
• Assertion Reason Questions for Class 12 Entrepreneurship (7)
• Case Study Questions for Class 12 Entrepreneurship (7)
• Case Study Questions for Class 12 Geography (18)
• Assertion Reason Questions for Class 12 History (8)
• Case Study Questions for Class 12 History (13)
• Assertion Reason Questions for Class 12 Informatics Practices (13)
• Case Study Questions for Class 12 Informatics Practices (11)
• MCQ Questions for Class 12 Informatics Practices (5)
• Assertion and Reason Questions for Class 12 Maths (14)
• Case Study Questions for Class 12 Maths (13)
• Maths Formulas for Class 12 (5)
• MCQ Questions for Class 12 Maths (14)
• Problems Based on Class 12 Maths (1)
• RD Sharma Solutions for Class 12 Maths (1)
• Assertion Reason Questions for Class 12 Physical Education (11)
• Case Study Questions for Class 12 Physical Education (11)
• MCQ Questions for Class 12 Physical Education (10)
• Assertion Reason Questions for Class 12 Physics (16)
• Case Study Based Questions for Class 12 Physics (14)
• Class 12 Physics Conceptual Questions (16)
• Class 12 Physics Discussion Questions (1)
• Class 12 Physics Latest Updates (2)
• Derivations for Class 12 Physics (8)
• Extra Questions for Class 12 Physics (4)
• Important Questions for Class 12 Physics (8)
• MCQ Questions for Class 12 Physics (14)
• NCERT Solutions for Class 12 Physics (18)
• Numerical Problems Based on Class 12 Physics (16)
• Physics Class 12 Viva Questions (1)
• Revision Notes for Class 12 Physics (7)
• Assertion Reason Questions for Class 12 Political Science (16)
• Case Study Questions for Class 12 Political Science (16)
• Notes for Class 12 Political Science (1)
• Assertion Reason Questions for Class 6 Maths (13)
• Case Study Questions for Class 6 Maths (13)
• Extra Questions for Class 6 Maths (1)
• Worksheets for Class 6 Maths (1)
• Assertion Reason Questions for Class 6 Science (16)
• Case Study Questions for Class 6 Science (16)
• Extra Questions for Class 6 Science (1)
• MCQ Questions for Class 6 Science (9)
• Assertion Reason Questions for Class 6 Social Science (1)
• Case Study Questions for Class 6 Social Science (26)
• NCERT Exemplar for Class 7 Maths (13)
• NCERT Exemplar for Class 7 Science (19)
• NCERT Exemplar Solutions for Class 7 Maths (12)
• NCERT Exemplar Solutions for Class 7 Science (18)
• NCERT Notes for Class 7 Science (18)
• Assertion Reason Questions for Class 7 Maths (14)
• Case Study Questions for Class 7 Maths (14)
• Extra Questions for Class 7 Maths (5)
• Assertion Reason Questions for Class 7 Science (18)
• Case Study Questions for Class 7 Science (17)
• Extra Questions for Class 7 Science (19)
• Assertion Reason Questions for Class 7 Social Science (1)
• Case Study Questions for Class 7 Social Science (30)
• Assertion Reason Questions for Class 8 Maths (7)
• Case Study Questions for Class 8 Maths (17)
• Extra Questions for Class 8 Maths (1)
• MCQ Questions for Class 8 Maths (6)
• Assertion Reason Questions for Class 8 Science (16)
• Case Study Questions for Class 8 Science (11)
• Extra Questions for Class 8 Science (2)
• MCQ Questions for Class 8 Science (4)
• Numerical Problems for Class 8 Science (1)
• Revision Notes for Class 8 Science (11)
• Assertion Reason Questions for Class 8 Social Science (27)
• Case Study Questions for Class 8 Social Science (23)
• CBSE Class 9 English Beehive Notes and Summary (2)
• Assertion Reason Questions for Class 9 Maths (14)
• Case Study Questions for Class 9 Maths (14)
• MCQ Questions for Class 9 Maths (11)
• NCERT Notes for Class 9 Maths (6)
• NCERT Solutions for Class 9 Maths (12)
• Revision Notes for Class 9 Maths (3)
• Study Notes for Class 9 Maths (10)
• Assertion Reason Questions for Class 9 Science (16)
• Case Study Questions for Class 9 Science (14)
• Evergreen Science Book Solutions for Class 9 (15)
• Extra Questions for Class 9 Science (22)
• MCQ Questions for Class 9 Science (11)
• NCERT Solutions for Class 9 Science (15)
• Revision Notes for Class 9 Science (1)
• Study Notes for Class 9 Science (15)
• Topic wise MCQ Questions for Class 9 Science (2)
• Topicwise Questions and Answers for Class 9 Science (15)
• Assertion Reason Questions for Class 9 Social Science (15)
• Case Study Questions for Class 9 Social Science (19)
• CHEMISTRY (8)
• Chemistry Articles (2)
• Daily Practice Problems (DPP) (3)
• Books for CBSE Class 9 (1)
• Books for ICSE Class 10 (3)
• Editable Study Materials (8)
• Exam Special for CBSE Class 10 (3)
• H. C. Verma (Concepts of Physics) (13)
• Study Materials for ICSE Class 10 Biology (14)
• Extra Questions for ICSE Class 10 Chemistry (1)
• Study Materials for ICSE Class 10 Chemistry (5)
• Study Materials for ICSE Class 10 Maths (16)
• Important Questions for ICSE Class 10 Physics (13)
• MCQ Questions for ICSE Class 10 Physics (4)
• Study Materials for ICSE Class 10 Physics (8)
• Study Materials for ICSE Class 9 Maths (7)
• Study Materials for ICSE Class 9 Physics (10)
• Topicwise Problems for IIT Foundation Mathematics (4)
• Challenging Physics Problems for JEE Advanced (2)
• Topicwise Problems for JEE Physics (1)
• DPP for JEE Main (1)
• Integer Type Questions for JEE Main (1)
• Integer Type Questions for JEE Chemistry (6)
• Chapterwise Questions for JEE Main Physics (1)
• Integer Type Questions for JEE Main Physics (8)
• Physics Revision Notes for JEE Main (4)
• JEE Mock Test Physics (1)
• JEE Study Material (1)
• JEE/NEET Physics (6)
• CBSE Syllabus (1)
• Maths Articles (2)
• NCERT Books for Class 12 Physics (1)
• NEET Chemistry (13)
• Important Questions for NEET Physics (17)
• Topicwise DPP for NEET Physics (5)
• Topicwise MCQs for NEET Physics (32)
• NTSE MAT Questions (1)
• Physics (1)
• Alternating Current (1)
• Electrostatics (6)
• Fluid Mechanics (2)
• PowerPoint Presentations (13)
• Previous Years Question Paper (3)
• Products for CBSE Class 10 (15)
• Products for CBSE Class 11 (10)
• Products for CBSE Class 12 (6)
• Products for CBSE Class 6 (2)
• Products for CBSE Class 7 (5)
• Products for CBSE Class 8 (1)
• Products for CBSE Class 9 (3)
• Products for Commerce (3)
• Products for Foundation Courses (2)
• Products for JEE Main & Advanced (10)
• Products for NEET (6)
• Products for ICSE Class 6 (1)
• Electrostatic Potential and Capacitance (1)
• Topic Wise Study Notes (Physics) (2)
• Topicwise MCQs for Physics (2)
• Uncategorized (138)

Test series for students preparing for Engineering & Medical Entrance Exams are available. We also provide test series for School Level Exams. Tests for students studying in CBSE, ICSE or any state board are available here. Just click on the link and start test.

Download CBSE Books

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online JEE Test Series for 499/- Only (Web + App) for 1 Year

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Leave a Reply Cancel reply

Join our Online Test Series for CBSE, ICSE, JEE, NEET and Other Exams

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

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

• New QB365-SLMS
• NEET Materials
• JEE Materials
• Banking first yr Materials
• TNPSC Materials
• DIPLOMA COURSE Materials
• 5th Standard Materials
• 12th Standard Materials
• 11th Standard Materials
• 10th Standard Materials
• 9th Standard Materials
• 8th Standard Materials
• 7th Standard Materials
• 6th Standard Materials
• 12th Standard CBSE Materials
• 11th Standard CBSE Materials
• 10th Standard CBSE Materials
• 9th Standard CBSE Materials
• 8th Standard CBSE Materials
• 7th Standard CBSE Materials
• 6th Standard CBSE Materials
• Tamilnadu Stateboard
• Scholarship Exams
• Scholarships

Class 11th Physics - Mechanical Properties of Fluids Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Physics Subject - Mechanical Properties of Fluids, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Mechanical properties of fluids case study questions with answer key.

11th Standard CBSE

Final Semester - June 2015

Fluid is the name given to a substance which begins to flow when external force is applied on it. Shape and volume of the fluid changes by the application of very small shear stress. The branch of physics which deals the study of fluids at rest is called hydrostatics and that branch of physics which. deals with study of fluid in motion is called hydrodynamics when liquid is at rest in a container, it exerts a force on the surface of object in contact with liquid, which is always normal to the surface of object. The total normal force exerted by liquid at rest on a given surface in contact with it is called thrust of liquid on that surface. (i) What is the cause of thrust of liquid on surface in contact? (ii) Define hydrostatic pressure. Write its S.I. unit (iii) Is hydrostatic pressure a vector quantity or scalar quantity? Give reason. (iv) Define relative density of a substance (v) What will be the measure of total preassure at a depth h below the liquid surface? (vi) What is gauge pressure? On what factors does it depend? (vii) What is hydrostatic paradox?

It is observed that if gravity effect is neglected the pressure at every point of liquid in equilibrium of rest is same and the increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points of the liquid. This is accounted to Pascal's law. Hydraulic lift and hydraulic brakes working is based on the Pascal's law, in which a small force applied on the smaller piston will appear as a very large force on the large piston. (i) State Pascal's law (ii) A bottle full of a liquid is fitted with a tight cork. Explain why a slight blow on the cork may be sufficient to break the bottle? (iii) Two Pistons of hydraulic press have diameter of 30.0 cm and 2.5 cm. Find the force exerted by longer piston when 50.0 kg wt is placed on smaller Piston. (iv) In the above question, find the distance through which the longer piston would move after 10 strokes if the stroke of the smaller piston is 40 cm. (v) In a car lift, compressed air exerts a force F 1  on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 10.0 cm, If the mass of the car to be lifted is 1300 kg. Calculate F 1 . What is pressure necessary to accomplish the task?

It has been found that a liquid in small quantity at rest, free from external force like gravity, always tends to have a spherical shape. Since for a given volume, a sphere has the least surface area, hence it shows that the free surface of every liquid at rest has a tendency to have a least surface area. The free surface of liquid behaves as if covered by a stretched membrane, having tension in all directions parallel to the surface. This tension in the free surface of liquid at rest is called the surface tension. It arises due to the fact that the free surface of liquid at rest has some additional potential energy. (i) What is surface tension and its origin? (ii) Why does oil spread over the surface of water? (iii) At what temperature the surface tension of a liquid is zero? (iv) Surface tension of all lubricating oils and paints is kept low. Why? (v) What is the effect of impurities on the surface tension of liquid? (vi) What is work done in blowing a soap bubble of radius r and surface tension S? (vii) Define surface energy of liquid.

Stokes' Law: A body falling through a viscous medium experiences a retarding force resulting in absorption of energy by the medium in the form of heat .:The motion of the body produces a relative motion between the different layers of the fluid. Consequently, it experiences a force which tends to retard its motion. When a small spherical body is dropped in a viscous liquid such as glycerine, it accelerates first, but soon begins to experience a retarding force. When the retarding force becomes equal to the effective weight of the body in the fluid, the body experiences no net force and falls with a constant velocity known as the terminal velocity. George stokes found that a small spherical body of radius r moving with a uniform velocity v in a fluid of coefficient of viscosity \(\eta\)  experiences a retarding force F given by F =  \(6 \pi \eta r v\)   (i) Define viscosity. (ii) Define S.I unit of coefficient of viscosity of a liquid. (iii) Name the forces which act on the small spherical body falling freely through a viscous medium. (iv) Write the dimenslonal formula of coefficient of viscosity  \(\eta\)  . (v) What would be the terminal velocity of the body if the upthrust on the body is negligible to its weight? (vi) Write the expression for terminal velocity of spherical body of radius r and density \(\rho\)  falling freely through a liquid of density \(\sigma\)  and coefficient of viscosity  \(\eta\) .

Equation of continuity is a fundamental equation of liquid flow and is a special case of the general law of conservation of mass. Consider an incompressible and non viscous liquid flowing slowly and steadily through a pipe of non-uniform cross-section. Let A and B be two different sections of a pipe having cross-sectional area Q 1  and Q 2 respectively. Let v 1  and v 2 be the respecstive velocities of the liquid flow through these cross sections. According to the equation of continuity of flow Q 1  v 1  = Q 2 V 2 or QV = constant i.e., the velocity of liquid flow at any section of the pipe is inversely proportional to area of cross-section of the pipe at that section. (i) Water flows through a horizontal pipe of nonuniform cross-section at the rate of 31.4 litre per minute. Determien the velocity of flow of water at the section of the pipe where diameter is 2 cm. (ii) Water flows through a horizontal pipe of diameter 2 cm at a speed of 3 cm s -1 . The pipe has a nozzle of diameter 5 mm at its end. Determine the speed of water emerging from the nozzle. (iii) What is meant by streamline flow? (iv) Distinguish between laminar flow and turbulent flow. (v) Still water runs deep, why?

*****************************************

Mechanical properties of fluids case study questions with answer key answer keys.

(i) The molecules of liquid are in random motion due to their thermal velocity they collide with surface and rebound that result in the change in momentum which is transferred to surface in contact. (ii) It is thrust exerted by 1iquid at rest per unit area of the surface in contact. The S.I unit of pressure is Nm -2 or Pascal. (iii) Hydrostatic pressure is a scalar quantity, because it is exerted in all direction equally. It has arbitrary direction. (iv) Relative density of a substance is defined as the ratio of its density to the density of water at 4° C i.e.. \(\text { Relative density }=\frac{\text { density of substance }}{\text { density of water at } 4^{\circ} \mathrm{C}}\) (v) It is equal to P 0 = Po + h \(\rho\) g where P 0  = atmospheric Pressure g = acceleration due to gravity. \(\rho\)  = density of liquid (vi) Gauge pressure at a point in a liquid is the difference of total pressure at that point and atmospheric pressure. It is independent of area of cross-section A but depends upon the height h of the liquid column and density of the liquid. (vii) The liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as hydrostatic paradox.

(i) Pascal's law states that the increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container provided the effect of gravity is neglected. (ii) It is because the increase in pressure at one part of liquid is communicated equally at the other parts of liquid that result in large force on walls of bottle to break the bottle. (iii) Here  \(A_{1}=\pi\left(\frac{2.5}{2}\right)^{2} \mathrm{~cm}^{2}, A_{2}=\pi\left(\frac{30}{2}\right)^{2} \mathrm{~cm}^{2}\) F 1  = 50 kg wt Now  \(F_{2}=\frac{F_{1}}{A_{1}} \times A_{2}=\frac{50 \times \pi \times\left(\frac{30}{2}\right)^{2}}{\pi\left(\frac{2.5}{2}\right)^{2}}\) = 7200 kg wt (iv) In one stroke, Input workdone = Output workdone So F 1 l 1  = F 2 l 2 or  \(l_{2}=\frac{F_{1} l_{1}}{F_{2}}=\frac{50 \times 4}{7200}=0.028 \mathrm{~cm}\) Distance covered after 10 strokes = 0.028 x 10= 0.28 cm (v) Here,  \(F_{1}=?, r_{1}=\frac{5}{100} \mathrm{~m}\) F 2 = 1300 Kg wt = 1300 x 9.8 N \(r_{2}=\frac{10}{100} \mathrm{~m}\) As  \(\frac{F_{1}}{a_{1}}=\frac{F_{2}}{a_{2}} \text { or } F_{1}=\frac{a_{1}}{a_{2}} F_{2}=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}} F_{2}\) or  \(F_{1}=\frac{r_{1}^{2}}{r_{2}^{2}} F_{2}=\frac{\left(\frac{5}{100}\right)^{2}}{\left(\frac{10}{100}\right)^{2}} \times 1300 \times 9.8\) = 3185 N Pressure  \(P=\frac{F_{1}}{a_{1}}=\frac{3185}{\frac{22}{7} \times\left(\frac{5}{100}\right)^{2}}=4.05 \times 10^{5} \mathrm{~Pa}\)  

(i) Surface tension is the property of liquid by virtue of which the free surface of liquid at rest tends to have minimum surface area and as such it behaves as if covered with stretched membrane. Surface tension is due to force of cohesion. (ii) Surface tension of oil is less than that of water. When oil is dropped on surface of water, water stretches the oil drops on all sides. Hence the oil spreads over the surface of water. (iii) At critical temperature. (iv) In order to have low value of surface tension so that it can spread over larger area. (v) A highly soluble substance like sodium chloride when dissolved in water, increases the surface tension of water (liquid). But the sparingly substance like phenol when dissolved in water, reduces the surface tension of water. (vi) Work done = Surface tension x area of soap bubble \(=S \times\left(4 \pi r^{2}\right) \times 2=8 \pi S r^{2}\) (vii) Surface energy of a given liquid surface is the amount of work done against the force of surface tension, in forming the liquid surface of given area at a constant temperature.

(i) Viscosity is the property of the fluid by virtue of which an internal frictional force comes into play when the fluid is in motion in the form of layers having relative motion of the different layers. (ii) Force  \(F=\frac{\eta A d v}{d x}\) \(\hat{\eta}=\frac{N \times m}{m^{2} \times m s^{-1}}=\mathrm{Nm}^{-2} \mathrm{~s}\) The SI unit of   \(\eta=\mathrm{Nm}^{-2} \mathrm{~s}\) (iii) Forces acting on the small spherical body falling freely through a viscous medium are the following (i) Weight of the body acting vertically downwards. (ii) Upward thrust due to buoyancy equal to weight of body. (iii) Viscous drag acting in the direction opposite to the direction of motion of body. (iv)  \(\text { Here } \eta=\frac{F}{6 \pi r v}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\) (v) When upthrust acting on the body is negligible then terminal velocity of the body \(v_{t}=\frac{m g}{6 \pi \eta r}\)   (vi)  \(v_{\mathbf{T}}=\frac{2 r^{2}(\rho-\sigma) g}{9 \eta}\) (viii) Viscosity of liquid decreases with temperature due to decrease in intermolecular force of attraction. \(\eta \alpha \sqrt{\mathrm{T}}\)  for gases viscosity increases with temperature. Viscosity of liquid increases with pressure whereas that of gases remain independent.

(i) Volume of water flowing per second V = 31.4 litre/min  \(=\frac{31.4 \times 10^{-3}}{60} \mathrm{~m}^{3} / \mathrm{s}\) Area of cross-section  \(a=\pi r^{2}\)   = 3.14 x 10 - 2 m 2 , since  \(r=\frac{d}{2}=\frac{2}{2}=1 \mathrm{~cm}=10^{-2} \mathrm{~m}\)   velocity of flow  \(=\frac{V}{a}=\frac{5}{3} \mathrm{~ms}^{-1}\)   (ii) a 1 v 1 = a 2 v 2 therefore  \(v_{2}=\left(\frac{a_{1}}{a_{2}}\right) v_{1}=\left(\frac{r_{1}}{r_{2}}\right)^{2} v_{1}\) \(=\left(\frac{1}{0.25}\right)^{2} \times 3=48 \mathrm{~cm} \mathrm{~s}^{-1}\)   (iii) It is that flow in which every particle of the liquid follows exactly the path of its preceding particle and has the same velocity in magnitude and direction as that of its preceding particle while crossing through that point. (iv) In a laminar flow the liquid moves in layers and one layer slides over the other layer of liquid. While in turbulent flow liquid moves with a velocity greater than its critical velocity, the motion of the particles of liquid becomes disorderly. (v) According to equation of continuity av = constant At high depth area of cross-section becomes very large therefore velocity of flow becomes negligible, this is reason behind the still water run deep.

Related 11th Standard CBSE Physics Materials

11th standard cbse syllabus & materials, cbse 11th chemistry structure of atom chapter case study question with answers, cbse 11th chemistry some basic concept of chemistry chapter case study questions with answers, 11th biology biological classification chapter case study question with answers cbse, 11th biology the living world chapter case study question with answers cbse, class 11th business studies - internal trade case study questions and answers 2022 - 2023, class 11th business studies - social responsibilities of business and business ethics case study questions and answers 2022 - 2023, class 11th business studies - emerging modes of business case study questions and answers 2022 - 2023, class 11th business studies - business service case study questions and answers 2022 - 2023, class 11th business studies - private, public and global enterprises case study questions and answers 2022 - 2023, class 11th business studies - business, trade and commerce case study questions and answers 2022 - 2023.

Class 11th Applied Mathematics - Coordinate Geometry Case Study Questions and Answers 2022 - 2023

Class 11th applied mathematics - basics of financial mathematics case study questions and answers 2022 - 2023, class 11th applied mathematics - descriptive statistics case study questions and answers 2022 - 2023, class 11th applied mathematics - probability case study questions and answers 2022 - 2023, class 11th applied mathematics - calculus case study questions and answers 2022 - 2023.

Class VI to XII

Tn state board / cbse, 3000+ q&a's per subject, score high marks.

11th Standard CBSE Study Materials

11th Standard CBSE Subjects