Class 11 Physics (Physics Part-II) Chapter 8 Mechanical Properties Of Solids

Key Concept: Basic Definition

d) Pascal (Pa)

[Solution Description]

The unit of Young’s Modulus can be derived from its formula $Y = \frac{\sigma}{\epsilon}$ where $\sigma$ is stress with a unit of Pascal (Pa), and $\epsilon$ is strain, which is dimensionless. Thus, the unit of Young’s modulus is Pascal (Pa).

Key Concept: Basic Definition

b) They can be stretched to large strains and return to original shape

[Solution Description] Elastomers have the unique ability to be stretched to large strains and return to their original shape after the force is removed. This property differentiates them from other materials such as metals or plastics.

Key Concept: Safety Margin

c) Higher safety margin

[Solution Description]

A thicker rope provides a higher safety margin, which accounts for unexpected increases in load or potential defects in the material. It ensures reliability and reduces the risk of failures.

Key Concept: Fracture Point

d) Point E

[Solution Description] Fracture occurs at the fracture point on the stress-strain curve, which is designated as Point E.

Key Concept: Conceptual Understanding

c) Due to rock’s elastic limit

[Solution Description] Mountains cannot exceed a certain height because rocks will reach their elastic limit under too much stress, causing deformation or failure.

Key Concept: Buckling Concept

b) Buckling

[Solution Description] Buckling refers to the sudden bending or warping of beams and structures, typically occurring when compressive loads are applied off-center or if the structure is very slender compared to its length.

Key Concept: Advanced Elasticity Concepts, Material Behavior

c) Lower density and higher plasticity

[Solution Description] Polymers often exhibit greater plasticity and lower density compared to metals, making them ideal for applications requiring materials that can flex significantly without breaking while remaining lightweight. Metals could offer higher rigidity but at the cost of increased weight.

Key Concept: Structural Design Challenges, Elastic Moduli

c) Shear modulus

[Solution Description]

For materials enduring shear force, the relevant modulus is the Shear modulus $G$, which measures the material’s response to shear stresses. It is different from Young’s modulus, which is concerned with tensile and compressive loadings, or the Bulk modulus, related to volumetric changes under pressure.

Therefore, minimizing deformation under shear would involve choosing a material with a high Shear modulus $G$.

Key Concept: Concept Identification

b) Elasticity

[Solution Description] Elasticity is the property of a material that allows it to regain its original shape and size after the removal of applied force.

Key Concept: Real-World Problem Solving, Advanced Reasoning

c) To maintain structural integrity under load changes

[Solution Description]

Considering the elastic potential energy stored in overhead transmission cables is critical because it determines how much the cables can stretch before becoming permanently deformed or breaking. It affects the durability and safety of the system, especially under varying loads due to weather conditions or mechanical stresses.

Key Concept: Poisson’s Ratio, Advanced Reasoning

c) Material B expands more laterally.

[Solution Description] Poisson’s ratio ($\nu$) measures the rate of lateral expansion relative to axial compression. A higher Poisson’s ratio indicates greater lateral expansion when compressed. Thus, since material B has a higher Poisson’s ratio than material A, material B will experience larger lateral expansion when subjected to the same axial force.

Key Concept: Conceptual Understanding

b) Elastic potential energy

[Solution Description]

When a wire is stretched, elastic potential energy is stored in it due to its deformation within the elastic limit.

Key Concept: Strain Types

c) Volume strain

[Solution Description] Volume strain refers to the fractional change in a body’s volume due to stress, represented mathematically as $\frac{\Delta V}{V}$, where $\Delta V$ is the change in volume and $V$ is the original volume.

Key Concept: Strain Definition

a) Change in length to original length

[Solution Description] Longitudinal strain is defined as the change in length ($\Delta L$) divided by the original length ($L$).

Key Concept: Basic Formula Application

c) 40,000 N/m$^{2}$

[Solution Description]

The formula for shear modulus $G$ is given by

$G = \frac{\sigma_s}{\text{shearing strain}}$. Substituting the given values, we have:

$G = \frac{200}{0.005} = 40,000 \, \text{N/m}^2$.

Key Concept: Modulus of Elasticity

b) Ratio of stress to strain

[Solution Description] The modulus of elasticity measures the ratio of stress to strain in the linear portion of the stress-strain curve, indicating how much a material will deform under a given load.

Key Concept: Elastic Potential Energy, Advanced Reasoning

c) 12500 J

[Solution Description] The elastic potential energy per unit volume is given by $U/V = \frac{1}{2} \sigma \epsilon$. Thus, the total energy $U = V \times \left(\frac{1}{2} \times 250000 \times 0.001\right)$. Substituting the values gives $U = 100 \times 125 = 12500 \, \text{J}$.

Key Concept: Basic Definition

a) Ratio of tensile stress to longitudinal strain

[Solution Description]

Young’s Modulus is defined as the ratio of tensile stress to longitudinal strain. It measures a material’s stiffness.

Key Concept: Basic Definition

a) The ratio of lateral strain to longitudinal strain

[Solution Description]

Poisson’s ratio is defined as the ratio of lateral strain to longitudinal strain in a stretched wire.

Key Concept: Elastic Limit

a) Elastic Limit

[Solution Description]

The elastic limit is the point on the stress-strain curve beyond which the material will not return to its original shape and will have permanent deformation.

Key Concept: Elastic Potential Energy, Real-World Application

c) $2 \times 10^6 J/m^3$

[Solution Description] Using the formula $u = \frac{1}{2} \sigma \epsilon$, substitute:

$u = \frac{1}{2} \times 5 \times 10^6 \, \text{Pa} \times 0.8 = 2 \times 10^6 \, \text{J/m}^3$

Key Concept: Formula Application

d) 0.02

[Solution Description] Shearing strain is calculated using the formula $\Delta x / L$. Here, $\Delta x = 3 \, \text{cm}$ and $L = 150 \, \text{cm}$.

Calculating:

$\frac{\Delta x}{L} = \frac{3}{150} = 0.02$

Thus, the shearing strain is 0.02.

Key Concept: Material Behavior

c) An elastomer

[Solution Description] Materials like elastomers exhibit a large elastic region but do not follow Hooke’s Law, implying they can stretch significantly and still return to their original shape.

Key Concept: Basic Definition

b) $B = – p/(\Delta V / V)$

[Solution Description]

The bulk modulus $B$ is defined as the ratio of hydraulic stress to the corresponding hydraulic strain, given by the formula:

$$B = – \frac{p}{\Delta V / V}$$

Therefore, option b) is correct.

Key Concept: Material Behavior

d) Steel

[Solution Description] Materials with higher Young’s modulus are considered more elastic because they have a greater ability to return to their original shape after deformation. Steel usually has a high Young’s modulus compared to other materials like copper or aluminium.

Key Concept: Young’s Modulus

b) Stiffness of the material

[Solution Description] Young’s modulus is defined as the ratio of stress to strain in the elastic region of the stress-strain curve. It measures the stiffness of a material.