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

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.

This Story also Contains

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

Download Solution PDF

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 3.0×105m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0×105m2 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 YS and YC, respectively.

Length of the steel wire ls= 4.7 m

Length of the copper wire lc = 4.7 m

The cross-sectional area of the steel wire AS = 3.0×105m2

The cross-sectional area of the Copper wire AC = 4.0×105m2

Let the load and the change in the length be F and Δl respectively

Y=FlAΔl


FΔl=AYl

Since F and Δl is the same for both wires we have

ASYSlS=ACYClC


YSYC=AClSASlC


YSYC=4.0×105×4.73.0×105×3.5


YSYC=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=StressStrain


Y=150×1060.002


Y=7.5×102Nm2

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

Answer:

The Yield Strength is approximately 3×108Nm2 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

F1=(4+6)×9.8F1=98N

Length of steel wire ls = 1.5 m

The diameter of the steel wire, d = 0.25 cm

A=π(d2)2
A=π×(0.25×1022)2
A=4.9×106m2

Area od the steel wire, A=4.9×106 m2

Let the elongation in the steel wire be Δl1

Young's Modulus of steel, Y1 = 2×1011Nm2

Y1=F1l1Δl1A


Δl1=F1l1Y1A


Δl1=98×1.52×1011×4.9×106


Δl1=1.5×104 m

Tension in the Brass wire is F2

F2=(6)×9.8F2=58.8N

Length of Brass wire l2 = 1.5 m

Area of the brass wire, A=4.9×106 m2

Let the elongation in the steel wire be Δl2

Young's Modulus of steel, Y2 = 0.91×1011Nm2

Δl2=F2l2Y2A


Δl2=58.8×10.91×1011×4.9×106


Δl2=1.32×104 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 = l2 = 0.01 m2

Tangential Force is F

F=100×9.8F=980 N

Tangential Stress is F/A

FA=9800.01


FA=98000N

Shear modulus of aluminium η=2.5×1010Nm2

Tangential Strain=FAη=980002.5×1010=3.92×106

Let the Vertical deflection be Δl

Δl=Tangential Stress×Side
Δl=3.92×106×0.1
Δl=3.92×107m

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=50000×9.84×π×(0.620.32)


P=1.444×105Nm2

Youngs Modulus of steel is Y=2×1011Nm2

Compressional strain=PY=1.444×1052×1011=7.22×107

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=42×109Nm2

Strain=FAη=Flbη=4450015.2×103×19.1×103×42×109=3.65×103

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 108Nm2 , 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=Pπr2


T=108×π×(1.5×102)2


T=7.068×104N

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=FlΔlA


Y=4FlΔlπd2


Yd2=k As F, l and Δl are equal for all wires

Yiron=1.9×1011Nm2

Ycopper=1.1×1011Nm2

dirondcopper=YcopperYiron
dirondcopper=1.1×10111.9×1011
dirondcopper=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 0.065cm2 . 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=4π 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+mω2r
T=14.5×9.8+14.5×(4π)2
T=2431.84N

Cross-Sectional Area of wire, A = 0.065 cm2

Young's Modulus of steel, Y=2×1011Nm2

Δl=FlAY


Δl=2431.84×10.065×104×2×1011


Δl=1.87mm

8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm ( 1atm=1.013×105Pa ), 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=StressVolumetric Strain


B=PΔVV


B=100×1.013×105×100×1030.5×103


B=2.026×109Nm2

The bulk modulus of air is Ba=1.0×105Nm2

The Ratio of the Bulk Modulus of water to that of air is

BBa=2.026×1091.0×105


BBa=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 1.03×103 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 ΔP=79 atm

Bulk Modulus of water is B=2.2×109Nm2

B=PΔVVΔVV=PBΔVV=79×1.013×1052.2×109ΔVV=3.638×103

The negative sign signifies that for the same given mass the Volume has decreased The density of water at the surface ρ=1.03×103 kg m3
Let the density at the given depth be ρ
Let a certain mass occupy V volume at the surface

ρ=mVρ=mV+ΔV

Dividing the numerator and denominator of RHS by V we get

ρ=mV1+ΔVVρ=ρ1+ΔVVρ=1.03×1031+(3.638×103)ρ=1.034×103 kg m3

The density of water at a depth where pressure is 80.0 atm is 1.034×103 kg m3 .

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 BG=3.7×1010Nm2

Pressure is P = 10 atm.

The fractional change in Volume would be given as

ΔVV=PB


ΔVV=10×1.013×1053.7×1010


ΔVV=2.737×105

The fractional change in Volume is 2.737×105

8.15 Determine the volume contraction of a solid copper cube, 10 cm on edge, when subjected to a hydraulic pressure of 7.0×106Pa

Answer:

The bulk modulus of copper is BC=140GPa=1.4×1011Nm2

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=7.0×106Pa

From the definition of bulk modulus

BC=PΔVVΔVV=PBCΔVV=7×1061.4×1011ΔVV=5×105

The volumetric strain is 5×105
Volume contraction will be

ΔV= Volumetric Strain × Initial Volume ΔV=5×105×103ΔV=5×108 m3ΔV=5×102 cm3

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 ΔV=0.10%

Volumetric Strain=0.1100
ΔVV=0.001

Bulk modulus of water is Bw=2.2×109 Nm2

Bw=PΔVVP=ΔVV×BwP=0.001×2.2×109P=2.2×106 Pa

A pressure of 2.2×106Pa 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=FLAΔL=4FLπD2ΔL

ΔL of copper =ΔL of iron, F is the same in both cases
Hence, Y1D2
So, Dcopper Diron =Yiron Ycopper

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,

Yα1ΔT

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, ΔLL (rubber) will be larger for the same FA. In the case of steel ΔL is insignificant. But, in rubber, ΔL 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 =FA
(Force per unit area)
2. Longitudinal Strain =ΔLL
(Change in length / Original length)
3. Shear Strain =tanθθ (for small angles)
4. Volumetric Strain =ΔVV

Moduli of Elasticity

5. Young's Modulus (Y)= Stress Longitudinal Strain =F/AΔL/L
6. Shear Modulus (n)= Shearing Stress Shearing Strain =F/Aθ
7. Bulk Modulus (K)= Bulk Stress Volumetric Strain =ΔPΔV/V
8. Relation between Y,K and η :

Y=9Kη/(3K+η)

Poisson’s Ratio

9. σ= Lateral Strain Longitudinal Strain

Elastic Energy

10. Elastic Potential Energy per unit volume =

U=12× Stress × Strain

11. Work done in stretching a wire =

W=12FΔLAL×AL=12FΔL

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

Frequently Asked Questions (FAQs)

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

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?
A:

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?
A:

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?
A:

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?
A:

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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