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Ever been curious about why bridges don't break under weight? Or why does a rubber band expand and come back to shape? Well, the Mechanical Properties of Solids has all the answers! This chapter in the NCERT Class 11 Physics goes deep into the way to let students understand all this.
Mechanical Properties of Solids is a crucial topic in covering fundamental concepts that help explain why objects are attracted to the Earth. If you struggle to grasp the concept of the mechanical properties of solids or solve NCERT textbook questions, Careers360 provides detailed NCERT solutions to assist you in your studies.
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This chapter primarily deals with questions related to elasticity and deformation, making it easier to understand the fundamental concepts, which are crucial in competitive exams such as NEET and JEE Main. The NCERT solutions for Class 11 Physics Chapter 8 help students strengthen their understanding and improve their problem-solving skills.
According to the CBSE Syllabus for the academic year 2025-26, the chapter you previously referred to as Chapter 9, " Physics class 11 Mechanical Properties of Solids, has been renumbered as Chapter 8.
Answer:
Let the Young's Modulus of steel and copper be Y S and Y C respectively.
Length of the steel wire l S = 4.7 m
Length of the copper wire l C = 4.7 m
The cross-sectional area of the steel wire A S = $3.0 \times 10^{-5} m^2$
The cross-sectional area of the Copper wire A C = $4.0 \times 10^{-5} m^2$
Let the load and the change in the length be F and $\Delta l$ respectively
$\\Y=\frac{Fl}{A\Delta l}\\ \frac{F}{\Delta l}=\frac{AY}{l}$
Since F and $\Delta l$ is the same for both wires we have
$\\\frac{A_{S}Y_{S}}{l_{S}}=\frac{A_{C}Y_{C}}{l_{C}}\\ \frac{Y_{S}}{Y_{C}}=\frac{A_{C}l_{S}}{A_{S}l_{C}}\\ \frac{Y_{S}}{Y_{C}}=\frac{4.0\times 10^{-5}\times 4.7}{3.0\times 10^{-5}\times3.5}\\ \frac{Y_{S}}{Y_{C}}=1.79$
The ratio of Young’s modulus of steel to that of copper is 1.79.
8.2 (a) Figure 8.9 shows the strain-stress curve for a given material. What are (a)
Fig. 8.9
Answer:
Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.
For the given material
$\\Y=\frac{Stress}{Strain}\\ Y=\frac{150\times 10^{6}}{0.002}\\ Y=7.5\times 10^{2}Nm^{-2}$
8.2(b)What are approximate yield strength for this material?
Answer:
The Yield Strength is approximately $3\times 10^{8}Nm^{-2}$ for the given material. We can see above this value of strain, the body stops behaving elastically.
Fig 8.10
Answer:
As we can see in the given Stress-Strain graphs that the slope is more in the graph corresponding to material A. We conclude that A has a greater Young's Modulus.
8.3(b) Which of the two is the stronger material?
Answer: The material which fractures at higher stress is said to be stronger. As we can see in the given Stress-Strain graphs the stress at which the material fractures is higher in A than that in B we conclude that A is the stronger material.
8.4(a) Read the following two statements below carefully and state, with reasons, if it is true or false.
The Young’s modulus of rubber is greater than that of steel;
Answer:
False: Young's Modulus is defined as the ratio of the stress applied on a material and the corresponding strain that occurs. As for the same amount of pressure applied on a piece of rubber and steel, the elongation will be much lesser in case of steel than that in the case of rubber and therefore the Young's Modulus of rubber is lesser than that of steel.
Answer:
True: As the force acts Normal to the parallel planes in which helical parts of the wire lie, the actual length of the wire would not change but it's shape would. Therefore the amount of elongation of the coil taking place for corresponding stress depends upon the Shear Modulus of elasticity.
Fig 8.11
Answer:
Tension in the steel wire is F 1
$\\F_{1}=(4+6)\times 9.8\\ F_{1}=98N$
Length of steel wire l 1 = 1.5 m
The diameter of the steel wire, d = 0.25 cm
$\\A=\pi \left ( \frac{d}{2} \right )^{2}\\ A=\pi \times \left ( \frac{0.25\times 10^{-2}}{2} \right )^{2}\\ A=4.9\times 10^{-6}m^{2}$
Area od the steel wire, $A=4.9\times 10^{-6}\ m^{2}$
Let the elongation in the steel wire be $\Delta l_{1}$
Young's Modulus of steel, Y 1 = $2\times 10^{11}Nm^{-2}$
$\\Y_{1}=\frac{F_{1}l_{1}}{\Delta l_{1}A }\\ \Delta l_{1}=\frac{F_{1}l_{1}}{Y_{1}A}\\ \Delta l_{1}=\frac{98\times 1.5}{2\times 10^{11}\times 4.9\times 10^{-6}}\\ \Delta l_{1}=1.5\times 10^{-4}\ m$
Tension in the Brass wire is F 2
$\\F_{2}=(6)\times 9.8\\ F_{2}=58.8N$
Length of Brass wire l 2 = 1.5 m
Area of the brass wire, $A=4.9\times 10^{-6}\ m^{2}$
Let the elongation in the steel wire be $\Delta l_{2}$
Young's Modulus of steel, Y 2 = $0.91\times 10^{11}Nm^{-2}$
$\\ \Delta l_{2}=\frac{F_{2}l_{2}}{Y_{2}A}\\ \Delta l_{2}=\frac{58.8\times 1}{0.91\times 10^{11}\times 4.9\times 10^{-6}}\\ \Delta l_{2}=1.32\times 10^{-4}\ m$
Answer:
Edge of the aluminium cube, l = 10 cm = 0.1 m
Area of the face of the Aluminium cube, A = l 2 = 0.01 m 2
Tangential Force is F
$\\F=100\times 9.8\\ F=980\ N$
Tangential Stress is F/A
$\\\frac{F}{A}=\frac{980}{0.01}\\ \frac{F}{A}=98000N$
Shear modulus of aluminium $\eta =2.5\times 10^{10}Nm^{-2}$
$\\Tangential\ Strain=\frac{\frac{F}{A}}{\eta }\\ =\frac{98000}{2.5\times 10^{10}}\\ =3.92\times 10^{-6}$
Let the Vertical deflection be $\Delta l$
$\\\Delta l=Tangential\ Stress\times Side \\\Delta l=3.92\times 10^{-6}\times 0.1\\ \Delta l=3.92\times 10^{-7}m$
Answer:
Inner radii of each column, r 1 = 30 cm = 0.3 m
Outer radii of each column, r 2 = 60 cm = 0.6 m
Mass of the structure, m = 50000 kg
Stress on each column is P
$\\P=\frac{50000\times 9.8}{4\times \pi \times (0.6^{2}-0.3^{2})}\\ P=1.444\times 10^{5}Nm^{-2}$
Youngs Modulus of steel is $Y=2\times 10^{11}Nm^{-2}$
$\\Compressional\ strain=\frac{P}{Y} \\=\frac{1.444\times 10^{5}}{2\times 10^{11}}\\ =7.22\times 10^{-7}$
Answer:
Length of the copper piece, l = 19.1 mm
The breadth of the copper piece, b = 15.2 mm
Force acting, F = 44500 N
Modulus of Elasticity of copper, $\eta =42\times 10^{9}Nm^{-2}$
$\\Strain=\frac{F}{A\eta }\\ =\frac{F}{lb\eta }\\ =\frac{44500}{15.2\times 10^{-3}\times 19.1\times 10^{-3}\times 42\times 10^{9}}\\ =3.65\times 10^{-3}$
Answer:
Let the maximum Load the Cable Can support be T
Maximum Stress Allowed, P = 10 8 N m -2
Radius of Cable, r = 1.5 cm
$\\T=P\pi r^{2}\\ T=10^{8}\times \pi \times (1.5\times 10^{-2})^{2}\\ T=7.068\times 10^{4}N$
Answer:
Each wire must support the same load and are of the same length and therefore should undergo the same extension. This, in turn, means they should undergo the same strain.
$\\Y=\frac{Fl}{\Delta lA}\\ Y=\frac{4Fl}{\Delta l\pi d^{2}}\\ Yd^{2}=k$ As F, l and $\Delta l$ are equal for all wires
$Y_{iron}=1.9\times 10^{11}Nm^{-2}$
$Y_{copper}=1.1\times 10^{11}Nm^{-2}$
$\\\frac{d_{iron}}{d_{copper}}=\sqrt{\frac{Y_{copper}}{Y_{iron}}}\\ \frac{d_{iron}}{d_{copper}}=\sqrt{\frac{1.1\times 10^{11}}{1.9\times 10^{11}}}\\ \frac{d_{iron}}{d_{copper}}=0.761$
Answer:
Mass of the body = 14.5 kg
Angular velocity, $\omega$ = 2 rev/s
$\omega =4\pi \ rad/s$
The radius of the circle, r = 1.0 m
Tension in the wire when the body is at the lowest point is T
$\\T=mg+m\omega ^{2}r\\ T=14.5\times 9.8+14.5\times (4\pi )^{2}\\ T=2431.84N$
Cross-Sectional Area of wire, A = 0.065 cm 2
Young's Modulus of steel, $Y=2\times 10^{11}Nm^{-2}$
$\\\Delta l=\frac{Fl}{AY}\\ \Delta l=\frac{2431.84\times1}{0.065\times 10^{-4}\times 2\times 10^{11}}\\ \Delta l=1.87 mm$
Answer:
Pressure Increase, P = 100.0 atm
Initial Volume = 100.0 l
Final volume = 100.5 l
Change in Volume = 0.5 l
Let the Bulk Modulus of water be B
$\\B=\frac{Stress}{Volumetric\ Strain}\\ B=\frac{P}{\frac{\Delta V}{V}}\\ B=\frac{100\times 1.013\times 10^{5}\times 100\times 10^{-3}}{0.5\times 10^{-3}}\\ B=2.026\times 10^{9}Nm^{-2}$
The bulk modulus of air is $B_{a}=1.0\times 10^{5}Nm^{-2}$
The Ratio of the Bulk Modulus of water to that of air is
$\\\frac{B}{B_{a}}=\frac{2.026\times 10^{9}}{1.0\times 10^{5}}\\ \frac{B}{B_{a}}=20260$
This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water.
Answer:
Water at the surface is under 1 atm pressure.
At the depth, the pressure is 80 atm.
Change in pressure is $\Delta P=79\ atm$
Bulk Modulus of water is $B=2.2\times 10^{9}Nm^{-2}$
$\\B=-\frac{P}{\frac{\Delta V}{V}}\\ \frac{\Delta V}{V}=-\frac{P}{B} \\\frac{\Delta V}{V}=-\frac{79\times 1.013\times 10^{5}}{2.2\times 10^{9}}\\ \frac{\Delta V}{V}=-3.638\times 10^{-3}$
The negative sign signifies that for the same given mass the Volume has decreased
The density of water at the surface $\rho=1.03\times 10^{3}\ kg\ m^{-3}$
Let the density at the given depth be $\rho '$
Let a certain mass occupy V volume at the surface
$\\\rho=\frac{m}{V}\\$
$\\\rho'=\frac{m}{V+\Delta V}\\$
Dividing the numerator and denominator of RHS by V we get
$\\\rho'=\frac{\frac{m}{V}}{1+\frac{\Delta V}{V}}\\ \rho'=\frac{\rho }{1+\frac{\Delta V}{V}}\\ \rho'=\frac{1.03\times 10^{3}}{1+(-3.638\times 10^{-3})}\\ \rho'=1.034\times 10^{3}\ kg\ m^{-3}$
The density of water at a depth where pressure is 80.0 atm is $1.034\times 10^{3}\ kg\ m^{-3}$ .
Answer:
Bulk's Modulus of Glass is $B_{G}=3.7\times 10^{10}Nm^{-2}$
Pressure is P = 10 atm.
The fractional change in Volume would be given as
$\\\frac{\Delta V}{V}=\frac{P}{B}\\ \frac{\Delta V}{V}=\frac{10\times 1.013\times 10^{5}}{3.7\times 10^{10}}\\ \frac{\Delta V}{V}=2.737\times 10^{-5}$
The fractional change in Volume is $2.737\times 10^{-5}$
Answer:
The bulk modulus of copper is $B_{C}=140GPa=1.4\times 10^{11}Nm^{-2}$
Edge of copper cube is s = 10 cm = 0.1 m
Volume Of copper cube is V = s 3
V = (0.1) 3
V = 0.001 m 3
Hydraulic Pressure applies is $P=7.0\times 10^{6}Pa$
From the definition of bulk modulus
$\\B_{C}=\frac{P}{\frac{\Delta V}{V}}\\ \frac{\Delta V}{V}=\frac{P}{B_{C}}\\ \frac{\Delta V}{V}=\frac{7\times 10^{6}}{1.4\times 10^{11}} \frac{\Delta V}{V}=5\times 10^{-5}$
The volumetric strain is $5\times 10^{-5}$
Volume contraction will be
$\\\Delta V=Volumetric\ Strain\times Initial\ Volume\\ \Delta V=5\times 10^{-5}\times 10^{-3}\\ \Delta V=5\times 10^{-8}m^{3}\\ \Delta V=5\times 10^{-2}cm^{3}$
The volume contraction has such a small value even under high pressure because of the extremely large value of the bulk modulus of copper.
8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer:
Change in volume is $\Delta V=0.10 \%$
$\\Volumetric\ Strain =\frac{0.1}{100}\\ \frac{\Delta V}{V}=0.001$
Bulk modulus of water is $B_{w}=2.2\times 10^{9}\ Nm^{-2}$
$\\B_{w}=\frac{P}{\frac{\Delta V}{V}}\\ P=\frac{\Delta V}{V}\times B_{w}\\ P=0.001\times 2.2\times 10^{9}\\ P=2.2\times 10^{6}Pa$
A pressure of $2.2\times 10^{6}Pa$ is to be applied so that a litre of water compresses by 0.1%.
Note: The answer is independent of the volume of water taken into consideration. It only depends on the percentage change.
1. A rigid bar of mass M is supported symmetrically by three wires, each of length L. Those at each end are of copper, and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to?
Solution: $\mathrm{Y}=$ stress/strain
$
Y=\frac{F L}{A \Delta L}=\frac{4 F L}{\pi D^2 \Delta L}
$
$\Delta L$ of copper $=\Delta L$ of iron, F is the same in both cases
Hence, $Y \propto \frac{1}{D^2}$
So, $\frac{D_{\text {copper }}}{D_{\text {iron }}}=\sqrt{\frac{Y_{\text {iron }}}{Y_{\text {copper }}}}$
2. What will be the maximum load a wire can withstand without breaking when its length is reduced to half of its original length?
Solution: Stress = force/area
The breaking stress is not dependent on the length, and hence, if the cross-sectional area is changed, it will not affect the breaking force.
3. What happens to the Young’s modulus of elasticity when the temperature of a wire is doubled?
Solution: As we learn,
$Y \alpha \frac{1}{\Delta T}$
So as the Young’s modulus increases, elasticity decreases.
4. A spring is stretched by applying a load to its free end. What is the strain produced in the spring?
Solution: The strain produced in a material can be shearing or longitudinal, depending on the type of force applied. In the case of a spring stretched by a load, the shape and length of the spring both change. When a longitudinal strain is produced, the length of the spring increases or decreases in the direction of the applied force.
5. Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both rods are fixed rigidly at one end of the roof. A mass M is attached to each of the free ends at the centre of the rods. Then:
Solution: 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.
$M$ is the mass which is being positioned in the middle of the rods made of rubber and steel.
Y steel $>\mathrm{Y}$ rubber
Hence, $\frac{\Delta L}{L}$ (rubber) will be larger for the same $\frac{F}{A}$. In the case of steel $\Delta L$ is insignificant. But, in rubber, $\Delta L$ is significant as the shape changes.
Understand the Basic Terms and Definitions
Know these key concepts first:
$
\text { Stress }=\frac{\text { Force }}{\text { Area }}
$
Types: Tensile, Compressive, Shearing
$
\text { Strain }=\frac{\text { Change in dimension }}{\text { Original dimension }}
$
Types: Longitudinal, Volumetric, Shearing
$
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F \cdot L}{A \cdot \Delta L}
$
$
B=-\frac{\Delta P}{\Delta V / V}
$
Understand key points of the curve:
Proportional Limit
Elastic Limit
Yield Point
Ultimate Stress
Breaking Point
Know the difference between elastic and plastic behaviour
Choose the Right Formula Based on the Type of Deformation
For stretching/compression:
$
Y=\frac{F L}{A \Delta L}
$
For volumetric strain under pressure:
$
B=-\frac{P}{\Delta V / V}
$
For shear deformation:
$
\eta=\frac{F / A}{\Delta x / L}
$
NCERT – Mechanical Properties of Solids |
JEE - Mechanical Properties of Solids |
Concept Name |
Concept Name |
Stress and Strain | |
Hooke's Law | |
Young’s Modulus | |
Elasticity | |
Elastic Limit | |
Stress-Strain Graph | |
Types of Stress (Tensile, Compressive, Shear) |
Types of Stress (Tensile, Compressive, Shear) |
Applications of Mechanical Properties in Real Life |
Numerical Problems on Stress and Strain |
Plastic and Elastic Deformation |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 |
Mechanical Properties of Solids |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
NCERT Solutions for Class 11 Subject-wise
The topics covered in the NCERT book Class 11 physics chapter 9 are
One question may be asked from mechanical properties of solids for JEE Main exam. To practice more questions refer to previous year JEE main papers, NCERT book and NCERT exemplar problems.
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