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

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

Edited By Vishal kumar | Updated on Mar 16, 2025 07:58 PM IST

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 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.

This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids
  2. Mechanical Properties of Solids: Important Formulas + eBook link
  3. Importance of Mechanical Properties Of Solids Class 11 Solutions
  4. Mechanical Properties of Solids Exercise Solutions: Features
  5. NCERT Solutions for Class 11 Physics Chapter Wise
  6. Subject wise NCERT Exemplar solutions

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.

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

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

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 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 = 3.0×105m2

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

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

Y=FlAΔlFΔl=AYl

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

ASYSlS=ACYClCYSYC=AClSASlCYSYC=4.0×105×4.73.0×105×3.5YSYC=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)

Young’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=StressStrainY=150×1060.002Y=7.5×102Nm2

8.2(b)What are 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 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.

8.4 (b) Read the following two statements 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 it's 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 F 1

F1=(4+6)×9.8F1=98N

Length of steel wire l 1 = 1.5 m

The diameter of the steel wire, d = 0.25 cm

A=π(d2)2A=π×(0.25×1022)2A=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, Y 1 = 2×1011Nm2

Y1=F1l1Δl1AΔl1=F1l1Y1AΔl1=98×1.52×1011×4.9×106Δl1=1.5×104 m

Tension in the Brass wire is F 2

F2=(6)×9.8F2=58.8N

Length of Brass wire l 2 = 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, Y 2 = 0.91×1011Nm2

Δl2=F2l2Y2AΔl2=58.8×10.91×1011×4.9×106Δl2=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 = l 2 = 0.01 m 2

Tangential Force is F

F=100×9.8F=980 N

Tangential Stress is F/A

FA=9800.01FA=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×SideΔl=3.92×106×0.1Δl=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, 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=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=Pπr2T=108×π×(1.5×102)2T=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=FlΔlAY=4FlΔlπd2Yd2=k As F, l and Δl are equal for all wires

Yiron=1.9×1011Nm2

Ycopper=1.1×1011Nm2

dirondcopper=YcopperYirondirondcopper=1.1×10111.9×1011dirondcopper=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+mω2rT=14.5×9.8+14.5×(4π)2T=2431.84N

Cross-Sectional Area of wire, A = 0.065 cm 2

Young's Modulus of steel, Y=2×1011Nm2

Δl=FlAYΔl=2431.84×10.065×104×2×1011Δl=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 StrainB=PΔVVB=100×1.013×105×100×1030.5×103B=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=2.026×1091.0×105BBa=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 pressure is 80.0 atm, given that its density at the surface is 1.03×103Kg ?

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=mV

ρ=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=PBΔVV=10×1.013×1053.7×1010ΔVV=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 = s 3

V = (0.1) 3

V = 0.001 m 3

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 VolumeΔV=5×105×103ΔV=5×108m3ΔV=5×102cm3

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=0.1100ΔVV=0.001

Bulk modulus of water is Bw=2.2×109 Nm2

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

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 upon the percentage change.

Mechanical Properties of Solids: Important Formulas + eBook link

Here are some important formulas for the mechanical properties of solids class 11 exercise. You can also access the eBook link for comprehensive study materials.

- Stress = Force  Area 

- Hooke's law E= Stress  Strain 

E= modulus of elasticity 

- longitudinal strain

The ratio of change in length to the original length.

- A wire of length L and radius r is clamped to a rigid support and a mass m is attached to the other end, then Young's modulus is

Y=mgLπr2Δl

r = Radius of wire

L= Original length

Δl= Change in length

  • Pressure

Pressure applied by a liquid column, p = hρg

  • Viscous Force

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(F)= -η A (dv/dx)

here,

(dv/dx) = rate of change of the velocity with distance and velocity gradient, A = area of cross-section and coefficient of viscosity

  • Variation in Viscosity

nt= no/(1+αt+βt2)

where no and nt are coefficients of viscosities at 0°C and t°C, α and β are constants

  • Poiseuille’s Formula

ν= π/2(pr4/ηl)

Where:

p = pressure difference across the two ends of the tube, r = radius of the tube,n = coefficient of viscosity and l = length of the tube

These are some important formulas that play a crucial role in understanding the concepts covered in NCERT Solutions class 11 physics chapter 8. Utilize them effectively for your studies and problem-solving.
Download- Formula Sheet for Physics Class 11: Chapterwise Important Formulas

Importance of Mechanical Properties Of Solids Class 11 Solutions

Practising NCERT solutions for Class 11 is highly beneficial, especially for tackling numerical problems commonly asked in final exams and competitive tests like NEET and JEE Main.

In NEET, around 2–3 questions are typically asked from Properties of Bulk Matter, with at least one question focused on Mechanical Properties of Solids. Likewise, in JEE Main, this chapter frequently contributes one question, highlighting its importance in effective exam preparation.

Mechanical Properties of Solids Exercise Solutions: Features

  • Comprehensive Coverage: The NCERT solutions for Class 11 Physics (Mechanical Properties of Solids) cover all essential topics, ensuring a deep understanding of the chapter.
  • Step-by-Step Solutions: Each numerical problem is solved with a detailed, structured approach, making it easier for students to grasp the concepts.
  • Clear and Simple Explanations: Complex topics related to the mechanical properties of solids are presented in easy-to-understand language to enhance comprehension.
  • Diagrams and Graphs: Visual aids such as graphs and diagrams are included wherever necessary to improve conceptual clarity.
  • Concise Summaries: Each section includes summaries of key points to help students revise efficiently and retain important information.
  • Exam-Focused Approach: The mechanical properties of solids numerical are structured to assist students in preparing for CBSE board exams and various competitive exams.

NCERT Solutions for Class 11 Physics Chapter Wise

Chapter 1

Physical world

Chapter 2

Units and Measurement

Chapter 3

Motion in a straight line

Chapter 4

Motion in a Plane

Chapter 5

Laws of Motion

Chapter 6

Work, Energy and Power

Chapter 7

System of Particles and Rotational motion

Chapter 8

Gravitation

Chapter 9

Mechanical Properties of Solids

Chapter 10

Mechanical Properties of Fluids

Chapter 11

Thermal Properties of Matter

Chapter 12

Thermodynamics

Chapter 13

Kinetic Theory

Chapter 14

Oscillations

Chapter 15

Waves

NCERT Solutions for Class 11 Subject-wise

Also, Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What are the topics covered in the Class 11 chapter Mechanical Properties of Solids

The topics covered in the NCERT book Class 11 physics chapter 9 are

  • Elastic behaviour of solids
  • Stress and strain
  • Hooke’s law
  • Stress-strain curve
  • Elastic moduli
  • Applications of elastic behaviour of materials
2. What is the weightage of the chapter mechanical properties of solids for JEE Main exam

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.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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