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**NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Solids **is an invaluable resource that aids students in comprehending the topics and preparing effectively for their exams. class 11 Physics Chapter 9 exercise solutions chapter consists of a total of twenty-one questions, each of which is explained in great detail and in a straightforward manner in Class 11 Physics Ch 9 NCERT solutions. It is crucial to take comprehensive notes while studying each topic. Solving textbook questions and gaining a strong grasp of the subject matter are equally vital. The experienced faculty has meticulously crafted the NCERT solutions in alignment with the latest CBSE Syllabus for 2023, with the aim of enhancing students' conceptual understanding.

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This Story also Contains

- NCERT Solutions Class 11 Physics Chapter 9 – Free PDF Download
- NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids
- NCERT Solutions for Class 11 Physics Chapter Wise
- Importance of Mechanical Properties Of Solids Class 11 Solutions
- Key features of mechanical properties of solids exercise solutions
- Subject wise NCERT Exemplar solutions

Class 11 Physics Chapter 9 NCERT solutions are part of NCERT Solutions for Class 11 Physics. If a rubber band is stretched by applying a variable force after a particular amount of force the rubber band breaks. The breakage happens when the applied force is beyond the elastic limit. To compare the elastic limit we use the term modulus of elasticity. The Mechanical Properties Of Solids Class 11 solutions mainly deal with questions related to the concepts of stress, strain, and modulus of elasticity. If these concepts are clear then it is very easy to understand the CBSE NCERT solutions for Class 11 Physics chapter 9 Mechanical Properties Of Solids. The chapter is small in length and easy to score. Class 11 Physics Chapter 9 question answer helps students in their preparations. NCERT is considered the bible for many competitive exams.

**Mechanical properties of solids class 11 exercise pdf download**** **for free.

****According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 9, " Mechanical properties of solids," has been renumbered as Chapter 8.**

**Access NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Exercise **

** 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 } =

The cross-sectional area of the Copper wire A _{ C } =

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

Since F and is the same for both wires we have

The ratio of Young’s modulus of steel to that of copper is 1.79.

** Q2 (a) ** Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus

** Answer: **

Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.

For the given material

** Answer: **

The Yield Strength is approximately for the given material. We can see above this value of strain, the body stops behaving elastically.

** 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.

** Q3 (b) ** The stress-strain graphs for materials A and B are shown in Fig. 9.12. The graphs are drawn to the same scale. Which of the two is the stronger material?

Fig. 9.12

** 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.

** Q4 (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.

** Answer: **

Tension in the steel wire is F _{ 1 }

Length of steel wire l _{ 1 } = 1.5 m

The diameter of the steel wire, d = 0.25 cm

Area od the steel wire,

Let the elongation in the steel wire be

Young's Modulus of steel, Y _{ 1 } =

Tension in the Brass wire is F _{ 2 }

Length of Brass wire l _{ 2 } = 1.5 m

Area od the brass wire,

Let the elongation in the steel wire be

Young's Modulus of steel, Y _{ 2 } =

** Answer: **

Edge of the aluminium cube, l = 10 cm = 0.1 m

Area of a face of the Aluminium cube, A = l ^{ 2 } = 0.01 m ^{ 2 }

Tangential Force is F

Tangential Stress is F/A

Shear modulus of aluminium

Let the Vertical deflection be

** Answer: **

Inner radii of each column, r _{ 1 } = 30 cm = 0.3 m

Outer radii of each colum, r _{ 2 } = 60 cm = 0.6 m

Mass of the structure, m = 50000 kg

Stress on each column is P

Youngs Modulus of steel is

** 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,

** 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

** 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.

As F, l and are equal for all wires

** Answer: **

Mass of the body = 14.5 kg

Angular velocity, = 2 rev/s

The radius of the circle, r = 1.0 m

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

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

Young's Modulus of steel,

** 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

The bulk modulus of air is

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

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

Bulk Modulus of water is

The negative sign signifies that for the same given mass the Volume has decreased

The density of water at the surface

Let the density at the given depth be

Let a certain mass occupy V volume at the surface

Dividing the numerator and denominator of RHS by V we get

The density of water at a depth where pressure is 80.0 atm is .

** Answer: **

Bulk's Modulus of Glass is

Pressure is P = 10 atm.

The fractional change in Volume would be given as

The fractional change in Volume is

** Answer: **

The bulk modulus of copper is

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

From the definition of bulk modulus

The volumetric strain is

Volume contraction will be

The volume contraction has such a small value even under high pressure because of the extremely large value of bulk modulus of copper.

** Q16 ** How much should the pressure on a litre of water be changed to compress it by 0.10%?

** Answer: **

Change in volume is

Bulk modulus of water is

A pressure of 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.

** Answer: **

The diameter of at the end of the anvil, d = 0.50 mm

Cross-sectional area at the end of the anvil is A

Compressional Force applied, F = 50000N

The pressure at the tip of the anvil is P

The pressure at the tip of the anvil is .

** Answer: **

Cross-Sectional Area of wire A is A _{ A } = 1 mm ^{ 2 }

Cross-Sectional Area of wire B is A _{ B } = 2 mm ^{ 2 }

Let the Mass m be suspended at x distance From the wire A

Let the Tension in the two wires A and B be F _{ A } and F _{ B } respectively

Since the Stress in the wires is equal

Equating moments of the Tension in the wires about the point where mass m is suspended we have

The Load should be suspended at a point 70 cm from a wire A such that there are equal stresses in the two wires.

** Answer: **

Cross-Sectional Area of wire A is A _{ A } = 1 mm ^{ 2 }

Cross-Sectional Area of wire B is A _{ B } = 2 mm ^{ 2 }

Let the Mass m be suspended at y distance From the wire A

Let the Tension in the two wires A and B be F _{ A } and F _{ B } respectively

Since the Strain in the wires is equal

Equating moments of the Tension in the wires about the point where mass m is suspended we have

The Load should be suspended at a point 43.2 cm from the wire A such that there is an equal strain in the two wires.

** Answer: **

Let the ends of the steel wire be called A and B.

length of the wire is 2l = 1 m.

The cross-sectional area of the wire is

Let the depression at the midpoint due to the suspended 100 g be y.

Change in the length of the wire is

The strain is

The vertical components of the tension in the arms balance the weight of the suspended mass, we have

The stress in the wire will be

The Young's Modulus of steel is

The depression at the mid-point of the steel wire will be 1.074 cm.

** Answer: **

Diameter of each rivet, d = 6.0 mm

Maximum Stress

The number of rivets, n = 4.

The maximum tension that can be exerted is T

** Answer: **

The pressure at the bottom of the trench,

The initial volume of the steel ball, V = 0.32 m ^{ 3 }

Bulk Modulus of steel,

The change in the volume of the ball when it reaches the bottom of the trench is .

Class 11 Physics Chapter 9 exercise solutions hold significant importance for students. Questions related to the Mechanical Properties of Solids are commonly featured in Class 11 examinations, making it a crucial topic for those aiming to achieve high marks in both school exams and entrance exams. To enhance their understanding of this chapter, students are encouraged to solve the mechanical properties of solids exercise provided at the end of the chapter.

Mechanical properties of solids class 11 exercise explore the fundamental concept of determining a solid's mechanical properties, a central theme in Physics. Students will delve into topics such as Young's modulus for materials like copper and steel. They will learn to extract valuable information, including yield strength, from graph plots and compare stress-strain relation graphs to determine Young's modulus and strength values.

By mastering the content in this class 11 physics ch 9 ncert solutions, students will not only excel in their academic examinations but also gain a deeper insight into the mechanical behaviour of materials, a skill that can be valuable in various real-world applications.

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 |

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

Hooke's law

longitudinal strain

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Download EBookThe 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

r = Radius of wire

L = Original length

= Change in length

#### Pressure

Pressure applied by a liquid column, p = hρg

#### Viscous Force

(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

n_{t}= n_{o}/(1+αt+βt^{2})

where n_{o }and n_{t} 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 of the important formulas that play a crucial role in understanding the concepts covered in NCERT Solutions class 11 physics chapter 9. Utilize them effectively for your studies and problem-solving.

By practising NCERT solutions for class 11, you will be able to answer most of the numerical problems asked in final exams for Class 11 and other competitive exams like NEET and JEE mains. The chapter Mechanical Properties of Solids come under the unit properties of bulk matter for NEET and properties of solids and liquids for JEE Mains. Two or three questions are expected for NEET from the unit properties of bulk matter out of which one question is expected from the chapter Mechanical Properties of Solids and in JEE Mains one question may be asked from Mechanical Properties of Solids.

**Comprehensive Coverage:** NCERT class 11 physics chapter 9 question answer cover all the topics and concepts present in the chapter, ensuring a thorough understanding of mechanical properties of solids.

**Step-by-Step Solutions:** Detailed step-by-step numericals on mechanical properties of solids class 11 for each question are provided, making it easy for students to follow and grasp the concepts.

**Clarity and Simplicity: **The mechanical properties of solids numericals are explained in a clear and simple language, making complex concepts easy to understand.

**Diagrams and Graphs**: Wherever necessary, diagrams and graphs are included to enhance visual understanding and clarify concepts.

**Concise Summaries**: NCERT mechanical properties of solids exercise solutions often include concise summaries of the main points in each section, helping students revise and retain important information.

**Examination-Oriented:** The mechanical properties of solids class 11 numericals are designed to help students prepare for exams, including CBSE board exams and various competitive exams.

**NCERT Solutions for Class 11 ** **Subject wise**

- NCERT Books Class 11 Physics
- NCERT Syllabus Class 11 Physics
- NCERT Books Class 11
- NCERT Syllabus Class 11

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.

3. Can using NCERT Solutions for Class 11 Physics Chapter 9 improve a student's performance and lead to a higher score in CBSE exams?

The chapter 9 physics class 11 ncert solutions are designed to facilitate easy comprehension of the concepts by students. The solutions are prepared by the faculty at careers360, keeping in mind the latest CBSE syllabus and guidelines. The language used in the solutions is simple and clear to enable students to understand the method of solving complex problems effortlessly.

4. is mechanical properties of solids class 11 exercise solutions important chapter for board ?

Yes, the chapter "Mechanical Properties of Solids" in Class 11 is important for board exams as it covers fundamental concepts frequently tested in physics examinations.

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