NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Vishal kumarUpdated on 19 Sep 2025, 12:12 AM IST

Have you ever wondered why a bridge can support a big vehicle or how a rubber band that has been extended comes back to its original form? The solutions given in Chapter 8 - Mechanical Properties of Solids of Class 11 covers the behavior of the solids when they are subjected to forces of various kinds like stretching, compression and bending.

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NMechanical Properties of Solids NCERT Solutions: Download PDF
Mechanical Properties of Solids NCERT Solutions: Exercise Questions
Mechanical Properties of Solids NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions
Class 11 Physics Chapter 8 - Mechanical Properties of Solids: Key Topics
Mechanical Properties of Solids NCERT Solutions: Important Formulas
What Extra Should Students Study Beyond the NCERT for JEE/NEET?
NCERT Solutions for Class 11 Physics Chapter-Wise
NCERT Solutions for Class 11 Subject-wise

The NCERT Solutions for Class 11 Physics Chapter 8 - Mechanical Properties of Solids offer step-by-step explanatory features of elasticity, stress, strain, and deformation, which make the students grasp the concepts in a clear way. These NCERT solutions are formulated in accordance with the current CBSE Class 11 Physics syllabus and can, therefore, serve a lot of purpose in the preparation of board exams as well as competitive exams such as JEE Main and NEET. With these well planned NCERT Solutions for Class 11 Physics Chapter 8 - Mechanical Properties of Solids in hand, the students are able to master the difficult concepts with ease, to master their problem solving abilities and to tackle numerical questions with a lot of confidence. Every question of the exercise is answered systematically to ensure that learners not only do well in exams but also build a greater conceptual insight.

NMechanical Properties of Solids NCERT Solutions: Download PDF

One of the most important aspects of Physics is the understanding of the behavior of solids under the influence of forces, and Chapter 8 of Class 11 - Mechanical Properties of Solids describes it most eloquently. Using NCERT solutions, students are in a position to get a step-by-step solution to problems that make difficult topics such as elasticity, stress, strain, and modulus elasticity very easy to acquire. These solutions are in form of a free downloadable PDF, and they assist students to revise faster and be well prepared both in the board examinations and also in competitive exams.

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Mechanical Properties of Solids NCERT Solutions: Exercise Questions

NClass 11 Physics Chapter 8 - Mechanical Properties of Solids question answers gives step by step solutions to every question. The solutions enable the students to become more conceptual and problem-solving. They are organized well and precise in their discussion of relevant topics such as stress, strain, Hooke law, and elastic modulus among others, an ideal mix during exam preparation.

8.1 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 of $4.0 \times 10^{- 5} m^{2}$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

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 $Δ l$ respectively

$Y = \frac{F l}{A Δ l}$

$\frac{F}{Δ l} = \frac{A Y}{l}$

Since F and $Δ l$ is the same for both wires we have

$\frac{A_{S} Y_{S}}{l_{S}} = \frac{A_{C} Y_{C}}{l_{C}}$

$\frac{Y_{S}}{Y_{C}} = \frac{A_{C} l_{S}}{A_{S} l_{C}}$

$\frac{Y_{S}}{Y_{C}} = \frac{4.0 \times 10^{- 5} \times 4.7}{3.0 \times 10^{- 5} \times 3.5}$

$\frac{Y_{S}}{Y_{C}} = 1.79$

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 isYoung’s modulus?

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

$Y = \frac{S t r e s s}{S t r a i n}$

$Y = \frac{150 \times 10^{6}}{0.002}$

$Y = 7.5 \times 10^{2} N m^{- 2}$

8.2(b)What is the approximate yield strength for this material?

Answer:

The Yield Strength is approximately $3 \times 10^{8} N m^{- 2}$ for the given material. We can see above this value of strain, the body stops behaving elastically.

8.3(a) The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale. Which of the materials has the greater Young’s modulus?

1650444808637

Fig 8.10

Answer:

As we can see in the given Stress-Strain graphs that the slope is greater 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 statement 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 to 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 less in the case of steel than that in the case of rubber and therefore the Young's Modulus of rubber is less than that of steel.

8.4 (b) Read the following statement below carefully and state, with reasons, if it is true or false. The stretching of a coil is determined by its shear modulus.

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 its shape would. Therefore the amount of elongation of the coil taking place for corresponding stress depends upon the Shear Modulus of elasticity.

8.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. 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.

1650444888498

Fig 8.11

Answer:

Tension in the steel wire is F₁

$F_{1} = (4 + 6) \times 9.8 F_{1} = 98 N$

Length of steel wire $l_{s}$ = 1.5 m

The diameter of the steel wire, d = 0.25 cm

$A = π {(\frac{d}{2})}^{2}$
$A = π \times {(\frac{0.25 \times 10^{- 2}}{2})}^{2}$
$A = 4.9 \times 10^{- 6} m^{2}$

Area od the steel wire, $A = 4.9 \times 10^{- 6} m^{2}$

Let the elongation in the steel wire be $Δ l_{1}$

Young's Modulus of steel, Y₁ = $2 \times 10^{11} N m^{- 2}$

$Y_{1} = \frac{F_{1} l_{1}}{Δ l_{1} A}$

$Δ l_{1} = \frac{F_{1} l_{1}}{Y_{1} A}$

$Δ l_{1} = \frac{98 \times 1.5}{2 \times 10^{11} \times 4.9 \times 10^{- 6}}$

$Δ l_{1} = 1.5 \times 10^{- 4} m$

Tension in the Brass wire is F₂

$F_{2} = (6) \times 9.8 F_{2} = 58.8 N$

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 $Δ l_{2}$

Young's Modulus of steel, Y₂ = $0.91 \times 10^{11} N m^{- 2}$

$Δ l_{2} = \frac{F_{2} l_{2}}{Y_{2} A}$

$Δ l_{2} = \frac{58.8 \times 1}{0.91 \times 10^{11} \times 4.9 \times 10^{- 6}}$

$Δ l_{2} = 1.32 \times 10^{- 4} m$

8.6 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?

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 m²

Tangential Force is F

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

Tangential Stress is F/A

$\frac{F}{A} = \frac{980}{0.01}$

$\frac{F}{A} = 98000 N$

Shear modulus of aluminium $η = 2.5 \times 10^{10} N m^{- 2}$

$T a n g e n t i a l S t r a i n = \frac{\frac{F}{A}}{η} = \frac{98000}{2.5 \times 10^{10}} = 3.92 \times 10^{- 6}$

Let the Vertical deflection be $Δ l$

$Δ l = T a n g e n t i a l S t r e s s \times S i d e$
$Δ l = 3.92 \times 10^{- 6} \times 0.1$
$Δ l = 3.92 \times 10^{- 7} m$

8.7 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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer:

Inner radii of each column, r₁ = 30 cm = 0.3 m

Outer radii of each column, r₂ = 60 cm = 0.6 m

Mass of the structure, m = 50000 kg

Stress on each column is P

$P = \frac{50000 \times 9.8}{4 \times π \times ({0.6}^{2} - {0.3}^{2})}$

$P = 1.444 \times 10^{5} N m^{- 2}$

Youngs Modulus of steel is $Y = 2 \times 10^{11} N m^{- 2}$

$C o m p r e s s i o n a l s t r a i n = \frac{P}{Y} = \frac{1.444 \times 10^{5}}{2 \times 10^{11}} = 7.22 \times 10^{- 7}$

8.8 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?

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, $η = 42 \times 10^{9} N m^{- 2}$

$S t r a i n = \frac{F}{A η} = \frac{F}{l b η} = \frac{44500}{15.2 \times 10^{- 3} \times 19.1 \times 10^{- 3} \times 42 \times 10^{9}} = 3.65 \times 10^{- 3}$

8.9 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?

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

$T = P π r^{2}$

$T = 10^{8} \times π \times (1.5 \times 10^{- 2})^{2}$

$T = 7.068 \times 10^{4} N$

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

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.

$Y = \frac{F l}{Δ l A}$

$Y = \frac{4 F l}{Δ l π d^{2}}$

$Y d^{2} = k$ As F, l and $Δ l$ are equal for all wires

$Y_{i r o n} = 1.9 \times 10^{11} N m^{- 2}$

$Y_{c o p p e r} = 1.1 \times 10^{11} N m^{- 2}$

$\frac{d_{i r o n}}{d_{c o p p e r}} = \sqrt{\frac{Y_{c o p p e r}}{Y_{i r o n}}}$
$\frac{d_{i r o n}}{d_{c o p p e r}} = \sqrt{\frac{1.1 \times 10^{11}}{1.9 \times 10^{11}}}$
$\frac{d_{i r o n}}{d_{c o p p e r}} = 0.761$

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

Answer:

Mass of the body = 14.5 kg

Angular velocity, $ω$ = 2 rev/s

$ω = 4 π r a d / s$

The radius of the circle, r = 1.0 m

Tension in the wire when the body is at the lowest point is T

$T = m g + m ω^{2} r$
$T = 14.5 \times 9.8 + 14.5 \times (4 π)^{2}$
$T = 2431.84 N$

Cross-Sectional Area of wire, A = 0.065 cm²

Young's Modulus of steel, $Y = 2 \times 10^{11} N m^{- 2}$

$Δ l = \frac{F l}{A Y}$

$Δ l = \frac{2431.84 \times 1}{0.065 \times 10^{- 4} \times 2 \times 10^{11}}$

$Δ l = 1.87 m m$

8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm ( $1 a t m = 1.013 \times 10^{5} P a$ ), 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.

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

$B = \frac{S t r e s s}{V o l u m e t r i c S t r a i n}$

$B = \frac{P}{\frac{Δ V}{V}}$

$B = \frac{100 \times 1.013 \times 10^{5} \times 100 \times 10^{- 3}}{0.5 \times 10^{- 3}}$

$B = 2.026 \times 10^{9} N m^{- 2}$

The bulk modulus of air is $B_{a} = 1.0 \times 10^{5} N m^{- 2}$

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.

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

Answer:

Water at the surface is under 1 atm pressure.

At the depth, the pressure is 80 atm.

Change in pressure is $Δ P = 79 a t m$

Bulk Modulus of water is $B = 2.2 \times 10^{9} N m^{- 2}$

$\begin{aligned} B = - \frac{P}{\frac{Δ V}{V}} \\ \frac{Δ V}{V} = - \frac{P}{B} \\ \frac{Δ V}{V} = - \frac{79 \times 1.013 \times 10^{5}}{2.2 \times 10^{9}} \\ \frac{Δ V}{V} = - 3.638 \times 10^{- 3} \end{aligned}$

The negative sign signifies that for the same given mass the Volume has decreased The density of water at the surface $ρ = 1.03 \times 10^{3} kg m^{- 3}$
Let the density at the given depth be $ρ^{'}$
Let a certain mass occupy V volume at the surface

$\begin{aligned} ρ & = \frac{m}{V} \\ ρ^{'} & = \frac{m}{V + Δ V} \end{aligned}$

Dividing the numerator and denominator of RHS by V we get

$\begin{aligned} ρ^{'} & = \frac{\frac{m}{V}}{1 + \frac{Δ V}{V}} \\ ρ^{'} & = \frac{ρ}{1 + \frac{Δ V}{V}} \\ ρ^{'} & = \frac{1.03 \times 10^{3}}{1 + (- 3.638 \times 10^{- 3})} \\ ρ^{'} & = 1.034 \times 10^{3} kg m^{- 3} \end{aligned}$

The density of water at a depth where pressure is 80.0 atm is $1.034 \times 10^{3} k g m^{- 3}$ .

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

Answer:

Bulk's Modulus of Glass is $B_{G} = 3.7 \times 10^{10} N m^{- 2}$

Pressure is P = 10 atm.

The fractional change in Volume would be given as

$\frac{Δ V}{V} = \frac{P}{B}$

$\frac{Δ V}{V} = \frac{10 \times 1.013 \times 10^{5}}{3.7 \times 10^{10}}$

$\frac{Δ V}{V} = 2.737 \times 10^{- 5}$

The fractional change in Volume is $2.737 \times 10^{- 5}$

8.15 Determine the volume contraction of a solid copper cube, 10 cm on edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} P a$

Answer:

The bulk modulus of copper is $B_{C} = 140 G P a = 1.4 \times 10^{11} N m^{- 2}$

Edge of copper cube is s = 10 cm = 0.1 m

Volume Of copper cube is V = s³

V = (0.1)³

V = 0.001 m³

Hydraulic Pressure applies is $P = 7.0 \times 10^{6} P a$

From the definition of bulk modulus

$\begin{aligned} B_{C} & = \frac{P}{\frac{Δ V}{V}} \\ \frac{Δ V}{V} & = \frac{P}{B_{C}} \\ \frac{Δ V}{V} & = \frac{7 \times 10^{6}}{1.4 \times 10^{11}} \frac{Δ V}{V} = 5 \times 10^{- 5} \end{aligned}$

The volumetric strain is $5 \times 10^{- 5}$
Volume contraction will be

$\begin{aligned} Δ V = Volumetric Strain \times Initial Volume \\ Δ V = 5 \times 10^{- 5} \times 10^{- 3} \\ Δ V = 5 \times 10^{- 8} m^{3} \\ Δ V = 5 \times 10^{- 2} {cm}^{3} \end{aligned}$

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 $Δ V = 0.10 %$

$V o l u m e t r i c S t r a i n = \frac{0.1}{100}$
$\frac{Δ V}{V} = 0.001$

Bulk modulus of water is $B_{w} = 2.2 \times 10^{9} N m^{- 2}$

$\begin{aligned} B_{w} = \frac{P}{\frac{Δ V}{V}} \\ P = \frac{Δ V}{V} \times B_{w} \\ P = 0.001 \times 2.2 \times 10^{9} \\ P = 2.2 \times 10^{6} Pa \end{aligned}$

A pressure of $2.2 \times 10^{6} P a$ 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.

Mechanical Properties of Solids NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions

The Higher Order Thinking Skills (HOTS) Questions of the Class 11 Physics Chapter 8: Mechanical Properties of Solids encourage learners to explain their grasp on the topic in abstract and practical situations. The questions are aimed to improve the critical thinking and understand high-level clarity, which are really helpful in competitive exams such as JEE.

Q1. 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?

Answer: $Y =$ stress/strain

$Y = \frac{F L}{A Δ L} = \frac{4 F L}{π D^{2} Δ L}$

$Δ L$ of copper $= Δ L$ of iron, F is the same in both cases
Hence, $Y \propto \frac{1}{D^{2}}$
So, $\frac{D_{copper}}{D_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{copper}}}$

Q2. What will be the maximum load a wire can withstand without breaking when its length is reduced to half of its original length?

Answer: 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.

Q3. The temperature of a wire is doubled. The Young's modulus of elasticity
a) will also double
b) will become four times
c) will remain same
d) will decrease

Answer: As we learn,

$Y α \frac{1}{Δ T}$

So as the Young’s modulus increases, elasticity decreases.

Q4. A spring is stretched by applying a load to its free end. What is the strain produced in the spring?

Answer: 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.

Q5. 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:
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: 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 $> Y$ rubber
Hence, $\frac{Δ L}{L}$ (rubber) will be larger for the same $\frac{F}{A}$ . In the case of steel $Δ L$ is insignificant. But, in rubber, $Δ L$ is significant as the shape changes.

Class 11 Physics Chapter 8 - Mechanical Properties of Solids: Key Topics

CClass 11 Physics Chapter 8 - Mechanical Properties of Solids, is based on the analysis of solid material behavior in case tension or compression forces, or bending forces, are introduced. Elasticity, stress-strain relationships, modulus of elasticity (Young), modulus of shear, modulus of bulk, and ratio of Poisson are the important concepts of this chapter. Knowledge of these concepts does not only provide a good foundation of board exams, but also provides the basis of higher level studies in engineering and competitive examinations such as JEE and NEET.

8.1 Introduction
8.2 Stress and strain
8.3 Hooke’s law
8.4 Stress-strain curve
8.5 Elastic moduli

8.5.1 Young’s Modulus

8.5.2 Shear Modulus

8.5.3 Bulk Modulus

8.5.4 Poisson's Ratio

8.5.5 Elastic Potential Energy in a Stretched Wire
8.6 Applications of elastic behaviour of materials

Mechanical Properties of Solids NCERT Solutions: Important Formulas

The Mechanical Properties of Solids class 11 question answers: Important Formulae give a concise list of all the important formulae in stress, strain, Young modulus, bulk modulus, shear modulus and elastic potential energy. These formulae will serve as a rapid revision guide, since solutions to problems can be done more quicker and efficiently in terms of exam preparation and competitive examinations like JEE and NEET.

Stress and Strain

1. Stress $= \frac{F}{A}$
(Force per unit area)
2. Longitudinal Strain $= \frac{Δ L}{L}$
(Change in length / Original length)
3. Shear Strain $= \tan θ \approx θ$ (for small angles)
4. Volumetric Strain $= \frac{Δ V}{V}$

Moduli of Elasticity

5. Young's Modulus $(Y) = \frac{Stress}{Longitudinal Strain} = \frac{F / A}{Δ L / L}$
6. Shear Modulus $(n) = \frac{Shearing Stress}{Shearing Strain} = \frac{F / A}{θ}$
7. Bulk Modulus $(K) = \frac{Bulk Stress}{Volumetric Strain} = \frac{Δ P}{Δ V / V}$
8. Relation between $Y, K$ and $η$ :

$Y = 9 K η / (3 K + η)$

Poisson’s Ratio

9. $σ = - \frac{Lateral Strain}{Longitudinal Strain}$

Elastic Energy

10. Elastic Potential Energy per unit volume =

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

11. Work done in stretching a wire =

$W = \frac{1}{2} \frac{F Δ L}{A L} \times A L = \frac{1}{2} F Δ L$

Approach to Solve Questions of Class 11 Physics Chapter 8 - Mechanical Properties of Solids

To answer any questions in Mechanical Properties of Solids, it is important to have a clear understanding of elasticity, stress-strain and the mathematical formulae of elastic constants. The chapter has both theory and numericals hence the students will need a systematic approach when solving problems.

Read the problem - Determine the nature of the question, i.e. tensile stress, compressive stress, shear, or volumetric strain.
Remember the definitions -Always relate the provided data to simple definitions such as stress= Force/ Area or strain= ΔL/L.
Select the appropriate modulus - Select the appropriate modulus depending on the nature of the strain: Youngs modulus (Y), Bulk modulus (K) or Shear modulus (e).
Apply Hooke Law - Use the relationship between strain and stress (inside the elastic limit) i.e. Stress α Strain.
Substitute carefully - Write the known values clearly and substitute them into the correct formula step by step.
Check units and conversions - To avoid calculation errors, pay attention to SI units (e.g., converting cm to m, mm² to m²).
Analyze graphs - In questions involving stress-strain curves, pay attention to such areas as elastic limit, yield point, or breaking point.
Connect to real-life examples - Associating the concept with real-life examples (such as bridges, springs, or wires) aids in consolidating the knowledge as one responds to application-based queries.

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

In Chapter 8: Mechanical Properties of Solids, JEE/NEET students are supposed to learn something extra in NCERT, meaning engaging in application scenarios, conceptual questions, and advanced problems in numerical form. The concepts stress-strain relationships, modulus of elasticity, and practical life should be understood better to score well in the competitive exams

Concept Name	JEE	NCERT
Elasticity	✅	✅
Stress And Strain	✅	✅
Stress-Strain Relationship	✅	✅
Hooke’s Law	✅	✅
Work Done In Stretching A Wire	✅	✅
Relation Between Volumetric Strain, Lateral Strain And Poisson’s Ratio	✅	❌

NCERT Solutions for Class 11 Physics Chapter-Wise

The NCERT Solutions to Class 11 Physics are available in chapter-wise format to have inexpensive, step-by-step solutions to each chapter. The links are useful in revising major concepts, doing numerical problems, and preparing a sound base for exams such as CBSE, JEE, and NEET. All solutions are designed by professionals according to the new syllabus.

NCERT solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT solutions for Class 11 Physics Chapter 2 Motion in a straight line

NCERT solutions for Class 11 Physics Chapter 3 Motion in a Plane

NCERT solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

NCERT solutions for Class 11 Physics Chapter 7 Gravitation

NCERT solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids

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

NCERT solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter

NCERT solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT solutions for Class 11 Physics Chapter 13 Oscillations

NCERT solutions for Class 11 Physics Chapter 14 Wave

NCERT Solutions for Class 11 Subject-wise

NCERT Solutions for Class 11 Biology

NCERT Solutions for Class 11 Maths

NCERT Solutions for Class 11 Chemistry

Also, Check NCERT Books and NCERT Syllabus here

Subject-wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

Q: What is the significance of this chapter to JEE/NEET and other competitive exams?

It forms the basis of the properties of material strength and elasticity, which are of major importance in physics problems in mechanics, and mercy based questions.

Q: What role does Youngs modulus play in engineering?

It facilitates in the design of structures and choice of materials that would sustain a given load without undergoing excessive deformations.

Q: What is Hooke s Law and in what does it apply?

Hooke Law articulates the strain and the stress as being directly proportional to one another and at the elastic limit. It can be used in materials that do not remain in the deformed state.

Q: What does stress-strain curve really represent?

Stress-strain shows the deformation of a material with stress. It assists in finding out elastic limit, yield strength, ultimate strength and breaking strength.

Q: How do I use the Mechanical Properties of Solids Class 11 NCERT PDF for exam preparation?

Use the Mechanical Properties of Solids Class 11 NCERT PDF to revise topics like stress, strain, elasticity, and Hooke’s Law. It's a handy tool for practicing questions before tests and clearing your basics.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties Of Solids

NMechanical Properties of Solids NCERT Solutions: Download PDF

Mechanical Properties of Solids NCERT Solutions: Exercise Questions

Mechanical Properties of Solids NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions

Class 11 Physics Chapter 8 - Mechanical Properties of Solids: Key Topics

Mechanical Properties of Solids NCERT Solutions: Important Formulas

Stress and Strain

Moduli of Elasticity

Poisson’s Ratio

Elastic Energy

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