NIOS Class 12 Physics Chapter 8 Elastic Properties of Solids

NIOS Class 12 Physics Chapter 8 Elastic Properties of Solids Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 8 Elastic Properties of Solids Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.

NIOS Class 12 Physics Chapter 8 Elastic Properties of Solids

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.

Elastic Properties of Solids

Chapter: 8

Module – II: Mechanics of Solids and Fluids

INTEXT QUESTIONS 8.1

1. What will be the nature of inter-atomic forces when deforming force applied on an object (i) increases, (ii) decreases the inter-atomic separation?

Ans: From the inter-molecular force theory, we know that at equilibrium separation R₀, the net inter-atomic force is zero.

(i) When deforming force increases the inter-atomic separation (R > R₀):

• The inter-atomic forces become attractive in nature

• These attractive forces try to bring the atoms back to their equilibrium positions

• The atoms resist the increase in separation

(ii) When deforming force decreases the inter-atomic separation (R < R₀):

• The inter-atomic forces become repulsive in nature

• These repulsive forces try to push the atoms back to their equilibrium positions

• The atoms resist being compressed further

2. If we clamp a rod rigidly at one end and a force is applied normally to its cross section at the other end, name the type of stress and strain?

Ans: When a rod is clamped at one end and force is applied normally at the other end:

(i) Type of Stress: Longitudinal stress if pulling, compressive stress if pushing)

The stress acts along the length of the rod.

Stress = F/A,

Where F is the applied force and A is the cross-sectional area.

Type of Strain: Linear strain 

• The strain represents the change in length per unit original length.

• Strain = ΔL/L, where ΔL is change in length and L is original length.

3. The ratio of stress to strain remains constant for small deformation of a metal wire. For large deformations what will be the changes in this ratio?

Ans: For small deformations, the ratio of stress to strain remains constant according to Hooke’s Law. This ratio is called the modulus of elasticity.

For large deformations:

(i) The ratio of stress to strain decreases.

(ii) This happens because the wire crosses its elastic limit.

(iii) Beyond the elastic limit, the material enters the plastic region.

(iv) In the plastic region, stress increases more slowly compared to strain.

(v) The stress-strain relationship becomes non-linear.

(vi) Eventually, the wire may reach the breaking point where it fractures

4. Under what conditions, a stress is known as breaking stress?

Ans: A stress is known as breaking stress under the following conditions:

(i) At the fracture point: When the applied stress reaches the maximum value that the material can withstand before breaking

(ii) Beyond ultimate tensile strength: It represents the stress at which the material actually breaks or fractures

(iii) Point F on stress-strain curve: It corresponds to the breaking point on the stress-strain diagram

(iv) Material failure: At this stress, the material loses its structural integrity completely

(v) No recovery possible: Beyond this point, the material cannot be restored to its original form Breaking stress is also called fracture stress or ultimate breaking strength.

5. If mass of 4 kg is attached to the end of a vertical wire of length 4 m with a diameter 0.64 mm, the extension is 0.60 mm. Calculate the tensile stress and strain?

Ans Given: Mass, m = 4 kg

Length, L = 4 m

Diameter, d = 0.64 mm 

= 0.64 × 10⁻³ m

Extension, ΔL = 0.60 mm 

= 0.60 × 10⁻³ m

g = 9.8 m/s²

Tensile Stress:

Force, F = mg = 4 × 9.8 

= 39.2 N

Area, A = πd²/4 

= π × (0.64 × 10⁻³)²/4

= π × 0.4096 × 10⁻⁶/4 

= 3.22 × 10⁻⁷ m²

Tensile Stress = F/A 

= 39.2/(3.22 × 10⁻⁷)

= 1.22 × 10⁸ N/m²

Tensile Strain:

Strain = ΔL/L

= (0.60 × 10⁻³)/4 

= 1.5 × 10⁻⁴

Tensile stress = 1.22 × 10⁸ N/m²,

Tensile strain = 1.5 × 10⁻⁴