NCERT Solutions for class 11 Physics Chapter 9: Mechanical properties of solids

NCERT Solutions for Class 11 Physics Chapter 9: Mechanical Properties of Solids are provided in this article below. Mechanical properties of solids define the various characteristics of solids such as their resistance to deformation and their strength.

Chapter 9 Mechanical Properties of Solids belongs to Unit 7 Properties of Bulk Matter. Unit 7 along with Unit 8 Thermodynamics and Unit 9 Behaviour of Perfect Gases and Kinetic Theory of Gases has a weightage of 20 marks in the Class 11 Examination. The Class 11 NCERT Solutions for Mechanical Properties of Solids discusses concepts such as Elastic behaviour of solids, Hooke’s law, and the Stress-strain curve.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 9

NCERT Solutions for Class 11 Physics Chapter 9

Class 11 Physics Chapter 9 – Concepts Covered

Stress is the ratio of the internal force F, produced when a substance is deformed, to the area A over which the force acts.

\(Stress = {F \over A}\)

Strain is the ratio of the change in shape or size to the original shape or size. It is just a number and has no dimensions.

Strain is of 3 types – Longitudinal strain, Volumetric strain, and Shear strain

As per Hooke’s law, the ratio of stress to the corresponding strain produced within elastic limits is a constant. This constant is known as the modulus of elasticity.

\(F_s = -kx\)

For a solid, in wire or a thin rod form, Young’s modulus of elasticity is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit.

Young's Modulus

Poisson’s Ratio – The ratio of change in diameter (ΔD) to the original diameter (D) is known as lateral strain. The ratio of change in length (Δl) to the original length (l) is known as longitudinal strain. The ratio of lateral strain to the longitudinal strain is called Poisson’s ratio.

\(Poisson's\ Ratio\ = {Lateral\ strain \over Longitudinal\ Strain}\)

Class 11 Physics Chapter 9 Related Guides –

Stress and Strain	Elastic Moduli	Shear Modulus
Bulk Modulus	Shearing Stress	Yield Strength
Elastic Limit	Tensile Stress	Elastomers
Beam Deflection Formula	Solid Deformation	Stress Formula

Class 11 Physics Study Guides:

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Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.
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NCERT Solutions for class 11 Physics Chapter 9: Mechanical properties of solids

NCERT Solutions for Class 11 Physics Chapter 9

Class 11 Physics Chapter 9 – Concepts Covered

CBSE CLASS XII Related Questions

1.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

4.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

5.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

6.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

Similar Physics Concepts

Comments

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NCERT Solutions for class 11 Physics Chapter 9: Mechanical properties of solids

NCERT Solutions for Class 11 Physics Chapter 9

Class 11 Physics Chapter 9 – Concepts Covered

CBSE CLASS XII Related Questions

1.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

4.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

5.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

6.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

4.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

5.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

6.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.