Test: Mechanical Properties of Solids - JEE MCQ

Explanation:A perfectly rigid body is hypothetical in nature but for some phenomena ( in rotational bodies ) we assume bodies are perfectly rigid i.e. the intermolecular forces are always in equilibrium irrespective of the external forces due to which their shape and size are constant.

Stress = F⟂/A = (Mass x Acceleration⟂)/Cross-sectional Area

= (550 kg x 9.8 m/s²)/ (0.30×10−4 m²)

= 1.8×10⁸ Pa

Young's ModulusSteel (Y) = 20×10¹⁰ Pa = Stress/Strain

=> Strain = Stress/Y = 1.8×108 Pa / 20×10¹⁰ Pa

= 9.0×10⁻⁴ 

Given Data,
Length of the piece of copper = l = 19.1 mm = 19.1 × 10^-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10^-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10^-4 m²
Modulus of elasticity of copper from standard list, η = 42× 10⁹ N/m2
By definition, Modulus of elasticity, η = stress/strain

⇒ Strain = F/Aη

⇒ Strain = 3.65 × 10^-3
Hence, the resulting strain is 3.65 × 10^-3

Y=F x l/A x Δ l
Δ l=0.5cm=0.5x10^-2m, l=10M, F=940N
Y=20x10¹⁰pa
20x10¹⁰=940x10/πr²x0.5x10^-10
πr²=94x100/5x10^-3x2x10¹¹=94x10²/10x10⁸
r²=94/π x 10^-7 =2.99 x 10^-6
r2 ≅3x10^-6
r=1.13x10^-10 m
diameter=2r=3.6mm

Explanation:When external force is applied on the solid bodies, the solid bodies get deformed. The atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic ( or intermolicular ) distances. When the deforming foce is removed, the interatomic forces tend to drive them back to their original postions. Thus the body regains its original shape and size.

An elastomer is a polymer with viscoelasticity (i. e., both viscosity and elasticity) and very weak intermolecular forces, and generally low Young's modulus and high failure strain compared with other materials. Elastomer rubber compounds are made from five to ten ingredients, each ingredient playing a specific role. Polymer is the main component, and determines heat and chemical resistance, as well as low- temperature performance. Reinforcing filler is used, typically carbon black, for strength properties.

σ=Stress and ε=strain
σ=F/A= (550kg) × (9.81m/s²)3×10-5m²/=0.18GPA
ε=Δl/l0=σ/Υ​=0.18×10⁹/200×109​=9×10^-4
Δl=εl₀= (9×10^-4) (2m) = 0.0018m=1.8mm

Explanation:More the elastic a material is , more it has the property to regain its original position which is required in construction works.

Shear modulus or modulus of rigidity is the ratio of shearing stress to the corresponding shearing strain, by the definition of shear modulus.

• Hydraulic pressure = 10 atm


• Bulk modulus of glass = 37 GPa

• Given formula for bulk modulus: \(B = \frac{\Delta P}{\frac{\Delta V}{V}}\)


• Given pressure = 10 atm = 10 x 1.013 x \(10^5\) Pa


• Substitute the values: 37 x \(10^9\) Pa = \(\frac{10 x 1.013 x 10^5}{\frac{\Delta V}{V}}\)


• Solving for \(\frac{\Delta V}{V}\): \(\frac{\Delta V}{V} = \frac{10 x 1.013 x 10^5}{37 x 10^9}\)


• Calculating the numerical value: \(\frac{\Delta V}{V} = 0.0027\)

The fractional change in volume of the glass slab when subjected to a hydraulic pressure of 10 atm is 0.0027.

Explanation:External force permanently distubed the equilibrium position of the interatomic ( or intermolecular ) forces between the particles of solid bodies.

I section is generally used as a beam because of its high section modulus as its most of the area is situated away from its neutral axis hence it has high moment of inertia i.e. high section modulus i.e. high moment carrying capacity which is the major requirement for a good beam section.

F/A = e/L
F/A = 0.18/l ( let the length be l)
F/A = x/2l
0.18/L = x/2l
0.18×2l = xl
0.36l = xl
x = 0.36

• Stress is defined as the force per unit area applied to an object, typically measured in units of Pascals (Pa) or Newtons per square meter (N/m^2).

• Stress is a measure of how much force is applied to a certain area of an object.


• It is important in engineering and physics as it helps determine how materials will react under different forces.


• Stress can be tensile (pulling force), compressive (pushing force), or shear (sliding force).


• Stress is calculated by dividing the force applied to an object by the area over which the force is distributed.


• Therefore, stress is commonly expressed in units of force per unit area, such as N/m^2 or Pa.

• Therefore, the correct definition of stress in the given options is force per unit area (B).

Correct Answer :- d

Explanation : The load applied to a column would place the column in compression; conversely, a load hanging from a rod would place the rod in tension. Strain is the deformation of a structural member because of stress within the member.

Stress has its own SI unit called the Pascal. 1 Pascal (Pa) is equal to 1 N/m². In imperial units stress is measured in pound force per square inch which is often shortened to "psi". The dimension of stress is same as that of pressure.

A ductile material is one that can withstand a large amount of plastic deformation between the elastic limit and the fracture point.
A material that breaks suddenly when elongated or fracture occurs in it soon after the elastic limit is crossed is called a brittle material.
A ductile material that exhibits extra elongation or deformation and does not fracture is also referred as superplastic material.

Hence 1.6 mm upper, 1.0 mm lower is correct.

• Definition: Volume strain is defined as the ratio of change in volume (ΔV) to the original volume V.

• Volume strain is a measure of how much a material changes in volume under stress.


• It is calculated by taking the change in volume and dividing it by the original volume of the material.


• Volume strain is usually expressed as a decimal or a percentage.


• It is an important concept in physics and engineering, especially in the study of materials and structures.


• Understanding volume strain is crucial for predicting how materials will behave under different conditions and loads.

Hook’s Law states that for small deformations the stress and strain are proportional to each other.

• Original stretch in wire: 0.18 mm


• Stress is proportional to strain

• When the diameter of the wire is doubled, the cross-sectional area of the wire will increase by a factor of 4 (since area is proportional to the square of the diameter).


• Since stress is proportional to strain, the new stretch in the wire will be 4 times the original stretch.


• Therefore, the new stretch in the wire with double the diameter will be 4 * 0.18 mm = 0.72 mm.

• Original stretch in wire: 0.18 mm


• New stretch in wire with double the diameter: 0.72 mm


• Difference in stretch: 0.72 mm - 0.18 mm = 0.54 mm

• The wire would have stretched 0.54 mm more if it had the same length but twice the diameter.


• Therefore, the correct answer is 0.54 mm.

Hooke’s law: a law stating that the strain in a solid is proportional to the applied stress within the elastic limit of that solid.
In the OA line Hooke’s law is valid because stress is directly proportional to strain.

• Steel wire stretches by 0.18 mm


• Stress is proportional to strain

• Since stress is proportional to strain, we can use the formula: Stress = Young's Modulus x Strain


• For steel wire, the strain is 0.18 mm


• For the copper wire, we need to find the strain


• Young's Modulus for steel is greater than that of copper


• Therefore, for the same stress applied, the strain in copper wire will be greater


• Let the strain in copper wire be x


• Since stress is proportional to strain, we have: Stress_steel = Stress_copper


• Therefore, Young's Modulus_steel x Strain_steel = Young's Modulus_copper x Strain_copper


• Using this equation, we can find the strain in the copper wire

• 0.18 mm = Young's Modulus_steel x Strain_steel


• Let x be the strain in copper wire


• 0.18 mm = Young's Modulus_copper x x


• Since Young's Modulus_steel > Young's Modulus_copper, x > 0.18 mm


• Therefore, the copper wire would have stretched more than 0.18 mm

• The copper wire would have stretched by approximately 0.33 mm