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# NCERT Solutions Class 11 Physics Chapter 8 Mechanical Properties of Solids

Last updated date: 11th Sep 2024
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## NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids - FREE PDF Download

Physics Chapter 8 Mechanical Properties of Solids NCERT solutions by Vedantu explains the behaviour of solid materials under various types of forces and deformations. It covers essential concepts such as stress and strain, Hooke’s Law, and different types of elastic moduli, including Young’s modulus, shear modulus, and bulk modulus. The chapter also explores Poisson’s ratio and the distinctions between elastic and plastic behaviour. This foundational knowledge is critical for advanced studies and practical applications in physics and engineering disciplines. Understanding these properties is crucial for applications in engineering, construction, and materials science. With Vedantu's Class 11 Physics NCERT Solutions, you'll find step-by-step explanations of all the exercises in your textbook, ensuring you understand the concepts thoroughly.

Table of Content
1. NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids - FREE PDF Download
2. Glance on Class 11 Physics Mechanical Properties of Solids NCERT Solutions
3. Access NCERT Solutions for Class 11 Physics Chapter 8 – Mechanical Properties of Solids
4. Mechanical Properties of Solids Chapter Summary - Class 11 NCERT Solutions
5. Overview of Deleted Syllabus for CBSE Class 11 Physics Mechanical Properties of Solids
6. Other Study Material for CBSE Class 11 Physics Chapter Mechanical Properties of Solids
7. Chapter-Specific NCERT Solutions for Class 11 Physics
FAQs

## Glance on Class 11 Physics Mechanical Properties of Solids NCERT Solutions

• Chapter 8 of Class 11 Physics introduces the concept of Mechanical Properties of Solids, focusing on how materials deform and return to their original shape under various forces.

• The chapter explains stress and strain, Hooke's Law, Young's modulus, Poisson's ratio, and Stress-strain curves.

• The chapter also covers shear modulus and bulk modulus, which describe a material's response to shear stress and volumetric stress, respectively.

• The concept of elasticity and plasticity is explained, highlighting the difference between reversible and permanent deformation.

• The importance of thermal stress is highlighted, explaining how temperature changes can induce stress in materials due to expansion or contraction.

• Class 11 physics Ch 8 NCERT solutions, explores the applications of mechanical properties in real-life scenarios, such as in construction, manufacturing, and material selection for various engineering projects.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 8 - Mechanical Properties of Solids, which you can download as PDFs.

• There are 16 fully solved questions in the exercise of class 11th Physics Chapter 8 Mechanical Properties of Solids.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Physics Chapter 8 – Mechanical Properties of Solids

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

Ans: In the above question it is given that:

The length of the steel wire is ${{L}_{1}}=4.7m$.

Area of cross-section of the steel wire is ${{A}_{1}}=3.0\times {{10}^{-5}}{{m}^{2}}$.

The length of the copper wire is ${{L}_{2}}=3.5m$.

Area of cross-section of the copper wire is ${{A}_{2}}=4.0\times {{10}^{-5}}{{m}^{2}}$ .

Now,

The change in length is given by:

$\Delta L={{L}_{1}}-{{L}_{2}}=4.7-3.5=1.2m$

Let the force applied in both the cases be $F$ .

Therefore, Young’s modulus of the steel wire is given by:

${{Y}_{1}}=\frac{F}{{{A}_{1}}}\times \frac{{{L}_{1}}}{\Delta L}=\frac{F\times 4.7}{3.0\times {{10}^{-5}}\times 1.2}$ …… (1)

And Young’s modulus of the copper wire is given by:

${{Y}_{2}}=\frac{F}{{{A}_{2}}}\times \frac{{{L}_{2}}}{\Delta L}=\frac{F\times 3.5}{4.0\times {{10}^{-5}}\times 1.2}$ …… (2)

Therefore,

$\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{F\times 4.7\times 4.0\times {{10}^{-5}}\times 1.2}{3.0\times {{10}^{-5}}\times 1.2\times F\times 3.5}=\frac{1.79}{1}$

Hence the ratio of Young’s modulus of steel to that of copper is $1.79:1$ .

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

a) Young’s modulus?

Ans: From the graph given in the above question it is clear that:

Stress for a given material is $150\times {{10}^{6}}N/{{m}^{2}}$ and strain is $0.002$.

Young’s modulus is given by:

$Y=\frac{Stress}{Strain}$

$\Rightarrow Y=\frac{150\times {{10}^{6}}N/{{m}^{2}}}{0.002}=7.5\times {{10}^{10}}N/{{m}^{2}}$

Therefore, Young’s modulus for the given material is $7.5\times {{10}^{10}}N/{{m}^{2}}$ .

b) Approximate yield strength for this material?

Ans: The yield strength of the material is the maximum stress it sustains without crossing the elastic limit.

From the graph given in the above question, it is clear that approximate yield strength for this material is $300\times {{10}^{6}}N/{{m}^{2}}$ or $3\times {{10}^{8}}N/{{m}^{2}}$ .

3. The stress-strain graphs for materials A and B are shown in figure.

The graphs are drawn to the same scale.

a) Which of the materials has the greater Young’s modulus?

Ans:  In the two graphs it is that given that stress for A is more than that of B.

As,

$Young's\text{ }modulus=\frac{Stress}{Strain}$,

Therefore, material A has greater Young's modulus.

b) Which of the two is the stronger material?

Ans: The strength of a material is determined by the amount of stress required for fracturing a material, corresponding to its fracture point.

Fracture point is defined as the extreme point in a stress-strain curve.

From the graph it is clear that material A can withstand more strain than material B.

Therefore, material A is stronger than material B.

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.

Ans: The given statement is false.

As there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain. Therefore, the Young’s modulus of steel is greater than that of rubber.

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

Ans: The given statement is true.

As the shear modulus of a coil relates with the change in shape of the coil and the stretching of coil changes its shape without any change in the length. Therefore, the shear modulus of elasticity is involved. Hence the stretching of a coil is determined by its shear modulus.

5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. 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.

Ans: In the above question it is given that:

Diameter of the wires is $d=0.25m$.

Hence $r=0.125m$.

Length of the steel wire is ${{L}_{1}}=1.5m$.

Length of the brass wire is ${{L}_{2}}=1.0m$.

Total force exerted on the steel wire is ${{F}_{1}}=\left( 4+6 \right)g=10g$

$\therefore {{F}_{1}}=10\times 9.8=98N$.

Young’s modulus for steel is given by

${{Y}_{1}}=\frac{{{F}_{1}}}{{{A}_{1}}}\times \frac{{{L}_{1}}}{\Delta {{L}_{1}}}$

Where,

$\Delta {{L}_{1}}$ is the change in the length of the steel wire.

And ${{A}_{1}}$ is the area of cross-section of the steel wire.

$\therefore {{A}_{1}}=\pi {{r}_{1}}^{2}$

We have,

Young’s modulus of steel is ${{Y}_{1}}=2.0\times {{10}^{11}}Pa$.

$\Rightarrow \Delta {{L}_{1}}=\frac{{{F}_{1}}\times {{L}_{1}}}{{{A}_{1}}\times {{Y}_{1}}}$

$\Rightarrow \Delta {{L}_{1}}=\frac{98\times 1.5}{\pi {{\left( 0.125 \right)}^{2}}\times 2.0\times {{10}^{11}}}=1.49\times {{10}^{-4}}m$.

Total force on the brass wire is ${{F}_{2}}=6\times 9.8=58.8N$.

Young’s modulus for brass is given by

${{Y}_{2}}=\frac{{{F}_{2}}}{{{A}_{2}}}\times \frac{{{L}_{2}}}{\Delta {{L}_{2}}}$

Where,

$\Delta {{L}_{2}}$ is the change in the length of the brass wire.

And ${{A}_{2}}$ is the area of cross-section of the brass wire.

$\therefore {{A}_{2}}=\pi {{r}_{2}}^{2}$

We have,

Young’s modulus of brass is ${{Y}_{2}}=0.91\times {{10}^{11}}Pa$.

$\Rightarrow \Delta {{L}_{2}}=\frac{{{F}_{2}}\times {{L}_{2}}}{{{A}_{2}}\times {{Y}_{2}}}$

$\Rightarrow \Delta {{L}_{2}}=\frac{58.8\times 1}{\pi {{\left( 0.125 \right)}^{2}}\times 0.91\times {{10}^{11}}}=1.3\times {{10}^{-4}}m$.

Clearly, the elongation of the steel wire is $1.49\times {{10}^{-4}}m$ and that of the brass wire is $1.3\times {{10}^{-4}}m$.

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?

Ans: In the above question it is given that:

Edge of the aluminium cube is $L=10cm=0.1m$.

The mass attached to the cube is $m=100kg$.

Shear modulus $\eta$ of aluminium is $25GPa=25\times {{10}^{10}}Pa$.

We know that:

$Shear\text{ }modulus\left( \eta \right)=\frac{\text{Shear stress}}{\text{Shear strain}}$

$\Rightarrow \eta =\frac{\left( \frac{F}{A} \right)}{\left( \frac{L}{\Delta L} \right)}$

Where,

$F$ is the applied force.

$\therefore F=mg=100\times 9.8=980N$

Area of one of the faces of the cube is $A=0.1\times 0.1=0.01{{m}^{2}}$.

Vertical deflection of the cube is $\Delta L$.

$\Delta L=\frac{FL}{A\eta }$

$\Rightarrow \Delta L=\frac{980\times 0.1}{0.01\times 25\times {{10}^{9}}}=3.92\times {{10}^{-7}}m$

Clearly, the vertical deflection of this face of the cube is $3.92\times {{10}^{-7}}m$.

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.

Ans: In the above question it is given that:

Mass of the big structure is $M=50000kg$.

Inner radius of the column is $r=30cm=0.3m$.

Outer radius of the column is $R=60cm=0.6m$

Young’s modulus of steel is $Y=2\times {{10}^{11}}Pa$

The total force exerted is $F=Mg=50000\times 9.8N$.

$\text{Stress = Force exerted on a single column}$

$\Rightarrow Stress=\frac{50000\times 9.8}{4}=122500N$

Young’s modulus is given by:

$Y=\frac{Stress}{Strain}$

$\Rightarrow Strain=\frac{\left( \frac{F}{A} \right)}{Y}$

Where,

Area is given by

$A=\pi \left( {{R}^{2}}-{{r}^{2}} \right)=\pi \left( {{\left( 0.6 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right)$.

$\Rightarrow Strain=\frac{\left( \frac{50000\times 9.8}{\pi \left( {{\left( 0.6 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right)} \right)}{2\times {{10}^{11}}}=7.22\times {{10}^{-7}}$

Therefore, the compressional strain of each column is $7.22\times {{10}^{-7}}$.

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?

Ans: In the above question it is given that:

Length of the piece of copper is $l=19.1mm=19.1\times {{10}^{-3}}m$.

Breadth of the piece of copper is $b=15.2mm=15.2\times {{10}^{-3}}m$

Area of the copper piece will be:

$A=l\times b$

$\Rightarrow A=19.1\times {{10}^{-3}}\times 15.2\times {{10}^{-3}}=2.9\times {{10}^{-4}}{{m}^{2}}$

Tension force applied on the piece of copper is $F=44500N$.

Modulus of elasticity of copper is $\eta =42\times {{10}^{9}}N/{{m}^{2}}$.

We know that :

$\text{Modulus of elasticity}\left( \eta \right)=\frac{Stress}{Strain}$

$\Rightarrow \eta =\frac{\left( \frac{F}{A} \right)}{Strain}$

$\Rightarrow Strain=\frac{F}{A\eta }$

$\Rightarrow Strain=\frac{44500}{2.9\times {{10}^{-4}}\times 42\times {{10}^{9}}}=3.65\times {{10}^{-3}}$

Hence, the resulting strain is $3.65\times {{10}^{-3}}$.

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

Ans: In the above question it is given that:

Radius of the steel cable is $r=1.5cm=0.015m$.

Maximum allowable stress is ${{10}^{8}}N/{{m}^{2}}$.

We know that:

$\text{Maximum force = Maximum stress }\times \text{Area of cross-section}$

$\Rightarrow \text{Maximum force =1}{{\text{0}}^{8}}\times \pi {{\left( 0.015 \right)}^{2}}=7.065\times {{10}^{4}}N$

Therefore, the cable can support the maximum load of $7.065\times {{10}^{4}}N$.

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.

Ans: In the above question it is given that:

Tension force acting on each wire is the same.

Therefore the extension produced in each wire is the same.

As the length of both wires is the same, the strain in both wires is also the same.

Young’s modulus is given by:

$Y=\frac{Stress}{Strain}$

$\Rightarrow Y=\frac{\left( \frac{F}{A} \right)}{Strain}=\frac{\frac{4F}{\pi {{d}^{2}}}}{Strain}$ …… (1)

Where,

$F$ is the Tension force,

$A$ is the area of cross-section and

$d$ is the diameter of the wire

From equation (1), it is clear that $Y\propto \frac{1}{{{d}^{2}}}$.

Young’s modulus for iron is ${{Y}_{1}}=190\times {{10}^{9}}Pa$.

Let the diameter of the iron wire be ${{d}_{1}}$.

Young’s modulus for copper is ${{Y}_{2}}=100\times {{10}^{9}}Pa$.

Let the diameter of the copper wire be ${{d}_{2}}$.

Therefore, the ratio of their diameters is given as:

$\frac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\frac{{{Y}_{1}}}{{{Y}_{2}}}}=\sqrt{\frac{190\times {{10}^{9}}}{100\times {{10}^{9}}}}=\frac{1.31}{1}$

Therefore, the ratio of diameters of copper wire to iron wire is $1.31:1$.

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

Ans: In the above question it is given that:

Mass is $m=14.5\text{ }kg$.

Length of the steel wire is $l=1.0\text{ }m$

Angular velocity is $\omega =2rev/s$

Cross-sectional area of the wire is $a=0.065c{{m}^{2}}=0.065\times {{10}^{-4}}{{m}^{2}}$.

Consider the elongation of the wire when the mass is at the lowest point of its path to be $\Delta l$.

The total force on the mass when the mass is placed at the position of the vertical circle is given by:

$F=mg+ml{{\omega }^{2}}$

$\Rightarrow F=14.5\times 9.8+14.5\times 1\times {{2}^{4}}=200.1N$

Young’s modulus is given by:

$Y=\frac{Stress}{Strain}$

$\Rightarrow Y=\frac{\left( \frac{F}{A} \right)}{\left( \frac{\Delta l}{l} \right)}$

$\Rightarrow \Delta l=\frac{Fl}{AY}$

We know that Young’s modulus for steel is $2\times {{10}^{11}}Pa$.

Therefore,

$\Delta l=\frac{220.1\times 1}{0.065\times {{10}^{-4}}\times 2\times {{10}^{11}}}=1.53\times {{10}^{-4}}m$

Thus, the elongation of the wire is $1.53\times {{10}^{-4}}m$.

12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm$\left( 1atm=1.013\times {{10}^{5}}Pa \right)$. 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.

Ans: In the above question it is given that:

Initial volume is ${{V}_{1}}=100.0l=100\times {{10}^{-3}}{{m}^{3}}$.

Final volume is ${{V}_{2}}=100.5l=100.5\times {{10}^{-3}}{{m}^{3}}$.

Thus, the increase in volume is ${{V}_{2}}-{{V}_{1}}=0.5\times {{10}^{-3}}{{m}^{3}}$.

Increase in pressure is $\Delta p=100atm=100\times 1.013\times {{10}^{5}}Pa$.

The formula for bulk modulus is

$\text{Bulk Modulus}=\frac{\Delta p}{\left( \frac{\Delta V}{{{V}_{1}}} \right)}=\frac{\Delta p{{V}_{1}}}{\Delta V}$

$\Rightarrow \text{Bulk Modulus=}\frac{100\times 1.013\times {{10}^{5}}\times 100\times {{10}^{-3}}}{0.5\times {{10}^{-3}}{{m}^{3}}}=2.026\times {{10}^{9}}Pa$

We know that Bulk modulus of air is $1\times {{10}^{5}}Pa$.

$\Rightarrow \frac{Bulk\text{ }modulus\text{ }of\text{ }water}{Bulk\text{ }modulus\text{ }of\text{ }air}=\frac{2.026\times {{10}^{9}}}{1\times {{10}^{5}}}=2.026\times {{10}^{4}}$

This ratio is very high because air is more compressible than water.

13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03\times {{10}^{3}}kg/{{m}^{3}}$ ?

Ans: In the above question it is given that:

Pressure at the given depth is $p=80.0atm=80\times 1.01\times {{10}^{5}}Pa$.

Consider the given depth to be h.

Density of water at the surface is ${{\rho }_{1}}=1.03\times {{10}^{3}}kg/{{m}^{3}}$

Consider ${{\rho }_{2}}$ to be the density of water at the depth h.

Consider ${{V}_{1}}$ to be the volume of water of mass m at the surface.

Consider ${{V}_{2}}$ to be the volume of water of mass m at the depth h.

Consider $\Delta V$ to be the change in volume.

$\Delta V={{V}_{1}}-{{V}_{2}}$

$\Rightarrow \Delta V=m\left[ \left( \frac{1}{{{\rho }_{1}}} \right)-\left( \frac{1}{{{\rho }_{2}}} \right) \right]$

Now,

$Volumetric\text{ }strain=m\left[ \left( \frac{1}{{{\rho }_{1}}} \right)-\left( \frac{1}{{{\rho }_{2}}} \right) \right]\times \left( \frac{{{\rho }_{1}}}{m} \right)$

$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=1-\left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)$ …… (1)

Bulk modulus is given by:

$Bulk\text{ }modulus=\frac{p{{V}_{1}}}{\Delta V}$

$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=\frac{p}{B}$

Compressibility of water is given by:

$\frac{1}{B}=45.8\times {{10}^{-11}}P{{a}^{-1}}$

$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=80\times 1.013\times {{10}^{5}}\times 45.8\times {{10}^{-11}}=3.71\times {{10}^{-3}}$ …… (2)

From equations (1) and (2) we get:

$1-\left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)=3.71\times {{10}^{-3}}$

$\Rightarrow {{\rho }_{2}}=\frac{1.03\times {{10}^{3}}}{\left[ 1-\left( 3.71\times {{10}^{-3}} \right) \right]}=1.034\times {{10}^{3}}kg{{m}^{-3}}$

Clearly, the density of water at the given depth (h) is $1.034\times {{10}^{3}}kg{{m}^{-3}}$.

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

Ans: In the above question it is given that:

The hydraulic pressure exerted on the glass slab is $p=10atm=10\times 1.013\times {{10}^{5}}Pa$.

Also, we know that the bulk modulus of glass is $B=37\times {{10}^{9}}N/{{m}^{2}}$.

Bulk modulus is given by the relation:

$B=\frac{p}{\left( \frac{\Delta V}{V} \right)}$

Where,

$\frac{\Delta V}{V}$ is the fractional change in volume.

$\Rightarrow \left( \frac{\Delta V}{V} \right)=\frac{p}{B}$

$\Rightarrow \left( \frac{\Delta V}{V} \right)=\frac{10\times 1.013\times {{10}^{5}}}{37\times {{10}^{9}}}=2.73\times {{10}^{-5}}$

Clearly, the fractional change in the volume of the glass slab is $2.73\times {{10}^{-5}}$.

15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7\times {{10}^{6}}Pa$.

Ans: In the above question it is given that:

The length of an edge of the solid copper cube is $l=10cm=0.1m$.

Hydraulic pressure is $p=7\times {{10}^{6}}Pa$.

Bulk modulus of copper is $B=140\times {{10}^{9}}Pa$.

Bulk modulus is given by the relation:

$B=\frac{p}{\left( \frac{\Delta V}{V} \right)}$

Where,

$\frac{\Delta V}{V}$ is the volumetric strain

$\Delta V$ is the change in volume.

$V$ is the original volume.

$\Rightarrow \Delta V=\frac{pV}{B}$

The original volume of the cube is $V={{l}^{3}}$.

$\Rightarrow \Delta V=\frac{p{{l}^{3}}}{B}$

$\Rightarrow \Delta V=\frac{7\times {{10}^{6}}\times {{\left( 0.1 \right)}^{3}}}{140\times {{10}^{9}}}=5\times {{10}^{-8}}{{m}^{3}}=5\times {{10}^{-2}}c{{m}^{3}}$

Clearly, the volume contraction of the solid copper cube is $5\times {{10}^{-2}}c{{m}^{3}}$.

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

Ans: In the above question it is given that:

Volume of water is $V=1L$.

The water is to be compressed by $0.10%$.

$\therefore Fractional\text{ }change=\frac{\Delta V}{V}=\frac{0.1}{100\times 1}={{10}^{-3}}$

Bulk modulus is given by the relation:

$B=\frac{p}{\left( \frac{\Delta V}{V} \right)}$

$\Rightarrow p=B\times \left( \frac{\Delta V}{V} \right)$

We know that, bulk modulus of water is $B=2.2\times {{10}^{9}}N/{{m}^{2}}$

$\Rightarrow p=2.2\times {{10}^{9}}\times {{10}^{-3}}=2.2\times {{10}^{6}}N/{{m}^{2}}$.

Clearly, the pressure on water should be $2.2\times {{10}^{6}}N/{{m}^{2}}$.

## Mechanical Properties of Solids Chapter Summary - Class 11 NCERT Solutions

### Elastic and Plastic Behaviour of Materials

• Whenever a force is applied on a body, then it tends to change the size or shape of the body.

• 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.

• Those substances which do not have a tendency to regain their shape and hence gets permanently deformed are called Plastic and the property is called Plasticity.

#### Stress

• The restoring force per unit cross-sectional are set up within the body is called stress.

• $Stress\frac{Restoring\; force }{Area\;of\;cross\;section}=\frac{F}{A}$

In SI system, unit of stress is N/m2 or pascal (Pa). In general there are three types of stresses

(a) Longitudinal Stress – Tensile stress (associated with stretching or compressive stress (associated with compression).

(b) Shearing Stress.

(c) Bulk Stress.

#### Breaking Stress

• The minimum stress after which the wire breaks is called breaking stress.

• A wire of length l will break due to its own weight, when $l=\frac{Breaking\;stress}{dg}$

where, d = density of material of wire and g = acceleration due to gravity.

#### Strain

• Strain is defined as the ratio of change in dimension of an object to the original dimension.

$Strain=\frac{Change\;in\;configuration\;of\;the\;Object}{Original\;configuration\;of\;the\;Object}$

It is a pure number and has no unit.

#### Hooke’s Law

• This law states that, for small deformations, the stress and strain are proportional to each other.

Thus, Stress\propto Strain

$Stress=k\times\;strain$

The SI unit of modulus of elasticity is Nm-2

A class of solids called elastomers do not obey Hooke’s law.

#### Moduli of Elasticity

• Three elastic moduli viz. Young’s Modulus, shear modulus and bulk modulus are used to describe the elastic behavior of objects as they respond to deforming forces that act on them.

When an object is under tension or compression, the Hooke’s law takes the form:

$\frac{F}{A}=\frac{Y\Delta L}{L}$

where $\frac{\Delta L}{L}$ is the tensile or compressive strain of the object, F is the magnitude of the applied fore causing the strain. A is the cross-sectional area over which F is applied (perpendicular to A) and Y is the Young’s modulus for the object.

A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement $\Delta L$ of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. This type of stress is possible only in solids.

In this kind of deformation the Hooke’s law takes the form :

$\frac{F}{A}=G\times\frac{\Delta L}{L}$ where $\Delta L$ is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus.

When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke’s law takes the form $P= -B\left (\frac{\Delta V}{V} \right )$. where p is the pressure (hydraulic stress) on the object due to the fluid, $\frac{\Delta V}{V}$ (the volume strain) is the absolute fractional change in the object’s volume due to that pressure and B is the bulk modulus of the object.

• The Young’s modulus and shear modulus are relevant only for solids since only solids have lengths and shapes.

• Bulk modulus is relevant for solids, liquid and gases. it refers to the change in volume when every part of the body is under the uniform stress so that the shape of the body remains unchanged.

• Metal shave larger values of Young’s modulus than alloys and elastomers. A material with large value of Young’s modulus require a large force to produce small changes in its length.

#### Poisson’s Ratio

• It is the ratio of transverse or lateral strain to longitudinal strain in the direction of stretching force. It is expressed as

$\sigma =\frac{Lateral\;Contraction\;strain}{Longitudinal\;Contraction\;strain}=\frac{-d/D}{l/L}$

where, d = change in diameter & D = original diameter,  l change in length   &  L = original length.

Theoretical limits of Poisson’s ratio are -1 and 0.5, while the practical limits are 0 and 0.5.

Relation between  Y,K,G and 𝝈 as given below

1. $L = \frac{y}{{3(1 - 2\sigma )}}$

2. $G = \frac{y}{{2{\mkern 1mu} (1 + \sigma )}}$

3. $\sigma = {\mkern 1mu} \frac{{(3B - 2G)}}{{2(3B + G)}}$

4. $Y = \frac{{9GB}}{{3B + G}}$

5. $\frac{9}{Y} = \frac{3}{G} + \frac{1}{B}$

#### Stress-strain Curve

• A typical stress-strain curve for a metal is shown in figure

• In the region O to A, stress is found proportional to strain. Thus, Hooke’s law is fully obeyed in this region and body regain its original shape.

• Point A is known as point of proportional limit.

• In the region from A to B, stress and strain are not proportional, but the body still returns to its original shape and size.

• The point B is yield point (also called elastic limit) and corresponding stress is yield stress (\sigma y).

• If stress increases beyond point B, the strain further increases, but on removing the strain wire does not regain its original length.

• Beyond point C, for a small stress, the strain produced is large upto point D. The wire will break at point D called fracture point of wire.

#### Elastic Potential Energy Stored in a Stretched Wire

• The work done in stretching a wire against the interatomic forces is stored as the elastic potential energy.

• Potential energy of work done per unit is given by

$U=\frac{W}{V}=\frac{1}{2}\times\;Stress\times\;Strain$

$=\frac{1}{2}\times\;Y\;\times\;\left ( Strain \right )^{2}=\frac{1}{2Y}\times\;\left ( Strain \right )^{2}\left [ \vec{} Y=\frac{Stress}{Strain}\right ]$

• Potential energy stored in stretching wire by restoring force is expressed as

$W=\frac{1}{2}\times\;F\;\times\;l$

where, F = restoring force and l= elongation produced.

• If the force acting on the body is increased from F1 to F2 within the elastic limit, then

$W=\frac{\left ( F_{1}+F_{2} \right )}{2}\times\;extension$

## Overview of Deleted Syllabus for CBSE Class 11 Physics Mechanical Properties of Solids

 Chapter Dropped Topics Mechanical Properties of Solids 9.2 Elastic Behaviour of Solids 9.6.2 Determination of Young’s Modulus of the Material of a Wire Exercises 9.17 – 9.21

## Conclusion

Class 11 Physics chapter 8 NCERT solutions on Mechanical Properties of Solids provided by Vedantu provides an in-depth understanding of how solids behave under various forces and deformations. By studying stress, strain, elastic moduli, and the distinctions between elastic and plastic behaviour, you gain valuable insights into the material properties that are crucial for engineering, construction, and material science applications. This knowledge equips students with the foundational principles needed to analyse and design structures and materials that can withstand real-world stresses, laying the groundwork for further studies in physics and engineering disciplines. From previous year's question papers, typically around 3–4 questions are asked from this chapter.

## Other Study Material for CBSE Class 11 Physics Chapter Mechanical Properties of Solids

 S.No. Important Links for Chapter 8 Mechanical Properties of Solids 1 Class 11 Mechanical Properties of Solids Important Questions 2 Class 11 Mechanical Properties of Solids Revision Notes 3 Class 11 Mechanical Properties of Solids Important Formulas 4 Class 11 Mechanical Properties of Solids NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 11 Physics

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 11 Physics Chapter 8 Mechanical Properties of Solids

1. What are Some Mechanical Properties of Solids?

Mechanical Properties of Solids Class 11 NCERT PDF go into the depths of the topic. It also describes the various mechanical properties of solids, some of which are as follows:

1. Brittleness: Brittleness is the property of solids due to which they break into smaller pieces when enough force is applied.

2. Compressive Strength: Due to their compressive strength, solids resist being pushed together.

3. Ductility: It is the property of solids that results in their deformation under tensile strength.

4. Elasticity: An elastic substance is capable of resisting distortions or deformations under an applied force. These materials return to their original shape when the force is removed.

Other than that, solids also show plasticity, which is the ability to undergo permanent deformation when a force is applied.

2. What is Elastic Moduli?

Most solids resist elastic deformation when stress is applied. Elastic modulus is the quantity that measures this resistance. It is an important inclusion of Class 11 Physics Chapter 9.

3. What is Hooke’s Law?

Hooke’s law states that the size of deformation of an object is directly proportional to the force of deformation.

4. What is an intermolecular force?

As discussed in the chapter, the force of attraction or repulsion acting between the particles (atoms, molecules and ions) neighbouring each other, is referred to as the intermolecular force of attraction or repulsion. Intermolecular forces are considered weaker in comparison to the intramolecular (covalent and ionic bonds) force acting between the atoms and molecules. Examples of intermolecular force are the London dispersion force, ion-dipole interaction etc.

5. What are the practical applications of the Mechanical Properties of Solids Class 11?

Solid mechanics has a plethora of real-life applications. It is one of the most fundamental applied engineering sciences which plays a crucial role in explaining a lot of physical phenomena around us and in explaining human anatomy. It is also used extensively in surgical implants and the design of dental prostheses. The principles and properties of solid mechanics have also been used in the Euler–Bernoulli beam equation.

6. Where can I get the NCERT Solutions for Class 11 Physics, Chapter 9?

To avail the NCERT Solutions for Class 11 Physics, Chapter 9, refer to Vedantu's NCERT Solutions for this chapter. After clicking on the given link, a web page will open, click on the download option to download the PDF of Vedantu's NCERT Solutions for this chapter or you can also continue your studies online. It is available free of cost and can be assessed online anytime, anywhere. These solutions will clear all your conceptual doubts and will help you to fetch good marks in this Physics chapter.

7. What do I do to score well in Class 11 Physics, Chapter 9?

First of all, to study Class 11 Physics, Chapter 9, you must carefully read from the NCERT textbook to understand all basic concepts well. Once you have learnt and understood the chapter basics, proceed to the NCERT back exercise questions for this chapter. If you get stuck in between, refer to Vedantu's NCERT Solutions for this chapter to clear all doubts as soon as possible. Last but not the least, practice additional questions available on the Vedantu app and website and consistently revise this chapter to fetch a high score in the Physics exam.

8. Is Class 11 Physics Chapter 9 difficult?

Class 11 Physics, Chapter 9-Mechanical Properties of Solids can be made easy if you follow the right strategy and approach while preparing this chapter. After studying this chapter from the NCERT and solving the back exercise questions, you should solve the previous year questions from this chapter in the Physics exam. Focus more on numerical problems and prepare a list of all the formulae from this chapter. Through consistent practice and optimism, you too can master this chapter.

9. What is the stress in the Mechanical Properties of Solids NCERT solutions?

In class 11 physics ch 8 NCERT  solutions, stress is defined as the force applied per unit area on a material. It is a measure of the internal forces that particles within the material exert on each other when subjected to external forces.

10. What are the important topics in Mechanical Properties of Solids in mechanical properties of fluids class 11 NCERT solutions?

Important topics in class 11 physics Mechanical Properties of Solids NCERT  solutions include:

• Stress and strain

• Hooke's Law

• Young's modulus

• Shear modulus and bulk modulus

• Elasticity and plasticity

• Poisson's ratio

• Stress-strain curves

• Thermal stress

11. According to class 11 physics chapter 8 NCERT  solutions, does stress depend on material?

Yes, stress depends on the material's properties as mentioned in mechanical properties of fluids class 11 NCERT solutions. Different materials have different stress responses to the same applied force, influenced by factors such as composition, structure, and temperature.

12. What is the symbol of stress mentioned in class 11 Mechanical Properties of Solids NCERT solutions?

The symbol for stress is typically denoted by the Greek letter sigma (σ).

13. What is another name for the Mechanical Properties of Solids?

Another term for the Mechanical Properties of Solids is "material properties" or "mechanical behaviour of materials."

14. What are the applications of Mechanical Properties of Solids NCERT solutions?

Applications of class 11 physics chapter 8 exercise solutions include:

• Construction and civil engineering

• Material selection in manufacturing

• Designing structures and machinery

• Understanding failure mechanisms in materials

• Enhancing product durability and performance in various industries