NCERT Solutions For Class 11 Physics Chapter 9
The Class 11 Physics NCERT solutions help you understand the basic concepts and master the subject. The concepts taught are completely new to what the students have previously learned in Class 10. Preparing for assignments and understanding what is being taught in class in college becomes easier if you have a strong understanding of the basic concepts.
The main motive of NCERT Solutions is to make understanding the subject easier for the students. This also helps in performing better in the exams and scoring high marks. Class 11 Physics introduces students to topics like Thermodynamics, Laws of Motion, etc. These topics are extremely important for the students, as many times, these topics are also asked in competitive exams. It is extremely important that the students are thorough and well prepared with Physics for Class 11.
NCERT Solutions for Class 11 Physics Chapter 9
Mechanical Properties of Solid explains the well-known feature displayed by matters in this stage and goes on to spot the laws governing these situations. It offers a simple answer to complex questions.
NCERT Solutions For Class 11 Physics Chapter 9 Mechanical Properties Of Solids (Include Download Button)
NCERT Solutions is a one-in-all study material prepared for helping the students to improve their scores and understanding of the subjects. NCERT book solution focuses on the NCERT pattern of questions and answers and provides the students with all the solutions to the problems asked in the textbooks. NCERT solutions are the product of multiple efforts from teachers with years of experience in teaching students. The entire NCERT book solution is a complete guide for anyone looking to sharpen their knowledge of any topic, concept, or even if they want to go back and use the study material as a reference.
NCERT Solutions for Class 11 Physics Chapter 9 prepared by subject matter experts makes all the concepts appear very clear and simple. The complex problems are broken down into simple steps so that students can grasp all the concepts involved in solving problems in all chapters of Physics Class 11 NCERT. NCERT for Physics Chapter 9 introduces the Mechanical Properties of Solids. The chapter on Mechanical Properties of solids Class 11 explains the prominent characteristics displayed by matters in this stage and goes on to pinpoint the laws. This chapter talks about various laws and concepts around solid bodies. The chapter is theoretically dense, so to memorise them well students should understand the concepts first physics class 11 chapter 9 notes include examples to explain the concepts clearly and logically. Students are advised to go through the notes of mechanical properties of solids in class 11 regularly to maximise the keeping of the concepts and examples.
NCERT Solution Of Physics Class 11 Chapter 9
The Physics NCERT solutions Class 11 prepared by expert teachers help students to understand the topics well. The chapters are laid out in such a way that they can be downloaded directly to your phone. You have the option of downloading either the entire syllabus or each chapter separately. With the Class 11 Physics NCERT Solutions, our teachers give you important step-by-step advice on how to approach a certain problem and reach the solution. The solutions contain all the chapters laid out and answered by our panel of expert teachers. With many years of expertise on board-issued textbooks, they mark out important questions and give you tips on how to solve them.
With NCERT Solutions for Physics Class 11, it becomes easier for the students to learn the subject. Having a solutions book also makes it easier for the students to revise before the exam as all the syllabus is consolidated in one place for the students to quickly go through. With the NCERT Physics Class 11 solutions, you can perform better in your class and gain the necessary confidence to tackle your final year in school.
NCERT Solutions Of Physics Class 11 Chapter 9 (Free Download)
The NCERT Books for Class 11 are used extensively by students across the nation for a thorough revision and refresher of key concepts that appear in the Mathematics and Science subject components in the Class 11 syllabus. Class 11 is the stepping stone to the Class 12 board exams and beyond and doing well in one’s academics at this point sets up the foundation for the remaining years to come. Chapters related to science solutions cover topics related to Tissues, Motion, Atoms and Molecules, Natural Resources, etc. The NCERT Solutions for class 11 physics chapter 9 have been designed keeping this purpose in mind that students will need all the access and the concepts related to the topics and other materials. NCERT solutions are the product of multiple efforts from the teachers with years of experience in teaching students actively involved in making this solution.
The NCERT book solutions are a complete guide for anyone looking to sharpen his knowledge of any subject, topic, concept, or even if you want to go back and use the study material as a reference. The focus is also on the overall development of the student, also keeping in mind understanding the important topics. NCERT Solutions for Class 11 Physics Chapter 9 prepared by subject matter experts makes all the concepts appear very clear and simple.
NCERT Solutions For Class 11 Physics Chapter 9 Topics
Class 11 Physics Chapter 9 – Mechanical Properties of Solids. The notes of mechanical properties of solids class 11 are available in NCERT solutions by Extramarks. Mechanical properties of Solids come under Unit VII- Properties of Bulk Matter. This entire unit is very important from an examination point of view. This chapter talks about various laws and concepts around solid bodies. The chapter is theoretically dense, so to memorise them well students should understand the concepts first. To strengthen the understanding, Physics Class 11 Chapter 9 notes include examples to explain the concepts clearly and logically.
Elastic Behaviour Of Solid
Elasticity is the ability of a body to resist a distorting influence or stress and to return to its original size and shape when the stress is removed. Solid objects will deform when forces are applied to them. If the material is elastic, the object will return to its initial shape and size when these forces are removed. To a greater or lesser extent, most solid materials exhibit elastic behaviour, but there is a limit to the magnitude of the force and the accompanying deformation within which elastic recovery is possible for any given material. Stresses beyond the elastic limit cause material to yield or flow. For such materials, the elastic limit marks the end of elastic behaviour and the beginning of plastic behaviour.
Stress And Strain
Stress and Strain are the two terms in Physics that describe the forces causing the deformation of objects. Deformation is known as the change of the shape of an object by applications of force. The object experiences it due to external force, for example, the forces might be like squeezing, squashing, twisting, shearing, ripping, or pulling the objects apart.
Types of Strain
A person’s body can be stressed in one of two ways, depending on how much stress is applied
Tensile Strain: Tensile strain is defined as a change in the length of a body caused by tensile tension. Compressive Strain: The change in length of a body caused by compressive strain is known as compressive strain.
It is named after the scientist Robert Hooke. Hooke’s Law states that stress developed is directly proportional to the strain produced in an object, within the elastic limit. An object that can come back to its original shape is its elasticity. Therefore, Hooke’s law applies to elastic objects. It doesn’t apply to the plasticity property of solids. It can be, therefore, represented as Stress = k * Strain Where k is the modulus of elasticity
A curve drawn between stress and strain is called the stress-strain curve. When stress and stress are drawn along the y-axis and x-axis respectively, a linear graph is formed in the ideal situation of Hooke’s law. However, when actual experiments are drawn, a curve is formed known as the stress-strain curve. Stress and strain have a straight proportional relationship up to an elastic limit. The relationship between stress and strain is explained by Hooke’s law.
Hooke’s law states that the strain in a solid is proportional to the applied stress, which must be within the solid’s elastic limit.
Elastic modulus measures the resistance of the material to elastic—or “springy”— deformation.
Low modulus materials are floppy and stretch a lot when they are pulled. High modulus materials are the opposite—they stretch very little when pulled. Elastic modulus is another key property that determines the deformation of the structural element under flexure. The compressive strength of concrete is closely related to its compressive strength.
Application Of Elastic Behaviour Of Materials
This theory of elasticity is used to design safe and stable man-made structures such as skyscrapers and bridges to make life suitable. Cranes used to lift loads use ropes that are designed so that the stress due to maximum load does not exceed the breaking stress. It is also found that a collection of thinner wire strands when compacted together makes the rope stronger than a solid rope of the same cross-section. That is the reason, crane ropes are made of several strands instead of one. The beams used in buildings and bridges should have to be carefully designed so that they do not bend excessively and break under the stress of the load on them. Structures such as bridges and tall buildings that have to support static or dynamic loads are generally constructed using pillars and beams to support them. Beams and pillars are designed to remain stable and safe within the range of the maximum load they are designed to carry.
Benefits Of NCERT Solutions For Class 11 Physics Chapter Mechanical Properties Of Solids
Mechanical properties of solids elaborate the characteristics such as the resistance to deformation and their strength. Strength is the ability of an object to withstand the applied stress, to what extent can it bear the stress. Resistance to deformation is how resistant an object is to the change of shape. If the resistance to deformation is less, the object can easily change its shape and vice versa. So, NCERT Solution gives benefits of focusing on the NCERT pattern of Questions and answers and providing the students with all the solutions to the problems asked in the textbooks. Our teachers and experts give you important step-by-step advice on how to approach a certain problem and get a solution. Complex problems are broken down into simple steps so that students can grasp all the concepts involved in solving problems in all chapters of Physics Class 11 NCERT. NCERT book solution is a complete guide for anyone looking to shape his/her knowledge of any concepts taught in class.
What Are Mechanical Properties?
Mechanical properties are physical properties that a material exhibits upon the application of forces. The mechanical properties of a material are defined as those properties that influence the material’s reaction to applied loads. Mechanical properties are used to determine how a material would behave in a given application and are helpful during the material selection and coating specification process. The mechanical properties of materials are crucial in many applications. In the coating industry, properties such as abrasion resistance, impact resistance, toughness, chemical resistance and tensile strength determine how well a particular coating will perform in a specific environment to protect against corrosion. Examples of mechanical properties include modulus of elasticity, tensile strength, hardness, ductility, impact resistance, compression modulus and fatigue limit.
Q.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Q.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Q.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12.
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Hence, Young’s modulus for material A is greater in comparison to it is for material B.
(b) Strength of a material is determined by the amount of stress required for fracturing a material, corresponding to its fracture point. The extreme point in a stress-strain curve is called as the Fracture point.
From the given graphs, it is clear that material A can bear more strain than material B.
Therefore, material A is stronger in comparison to material B.
Q.4 Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Q.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Q.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Q.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Q.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Q.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 Nm–2, what is the maximum load the cable can support?
Q.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
Q.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev s-1 at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Q.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume =100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Q.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 ×103 kg m–3?
Q.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Q.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×106 Pa.
Q.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?
Q.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?
Q.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2 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.
Q.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 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.
Q.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 ×107 Pa? Assume that each rivet is to carry one quarter of the load.
Q.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
FAQs (Frequently Asked Questions)
1. Where to find Class 11 Physics Chapter 9 Solutions?
You can easily find NCERT solutions class 11 physics chapter 9 on Extramarks. You can download the solutions and access them anywhere, anytime.
2. How can I score well in Physics Class 11, Chapter 9?
Many students find Physics difficult to crack but if the concepts are clear, it becomes a lot easier to score in physics. Extramarks offers you a perfect solution set solved by expert teachers in simple explanations.
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