#engineering mechanics

Bulk Modulus and Compressibility

A brass rod of radius 0.008m and length 1.0m is subjected to a tensile force producing a stress of 5×107N/m2. What is th

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.008)2 = 3.14×6.4×10−5≈2.01×10−4m2. Force: F = Stress×A = 5×107×2.01×10−4≈1.005×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.005×104N. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bulk Modulus and Compressibility

An aluminium block of dimensions 0.6m×0.4m×0.2m is subjected to a shearing force of 8×104N. If the shear modulus of alum

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.6×0.4 = 0.24m2, L = 0.2m. Substitute: Δx = 8×104×0.20.24×2.5×1010 = 160006×109≈2.67×10−6m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.67×10−6m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bulk Modulus and Compressibility

A rectangular slab of length 0.4 m, width 0.2 m, and thickness 0.1 m is subjected to a shearing force of 8 × 104 N. If t

Shear modulus: G = (Shear stress) / (Shear strain) = (F / A) / (Δx / L). Rearrange: Δx = (F L) / (A G). Area A = 0.4 × 0.2 = 0.08 m2, L = 0.1 m. Substitute: Δx = (8 × 104 × 0.1) / (0.08 × 4 × 109) = 8000 / (3.2 × 108) = 2.5 × 10-5 m.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bulk Modulus and Compressibility

A copper wire of length 2.0m and cross-sectional area 2.5×10−6m2 is stretched by a force of 250N. If the Young's modulus

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 250×2.02.5×10−6×1.1×1011 = 5002.75×105≈1.82×10−3m = 1.82mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82mm. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bulk Modulus and Compressibility

A steel cable of radius 0.02 m is used to lift a load. If the maximum stress it can withstand is 1 × 108 N/m2, what is t

Stress: Stress = F / A. Area: A = π r2 = 3.14 × (0.02)2 = 3.14 × 0.0004 = 1.256 × 10-3 m2. Maximum force: F = Stress × A = 1 × 108 × 1.256 × 10-3 = 1.256 × 105 N.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Bulk Modulus and Compressibility

A steel wire of length 2 m and cross-sectional area 2 × 10-6 m2 is stretched by a force of 200 N. If the Young's modulus

Using Young's modulus formula: Y = (F L) / (A ΔL). Rearrange for elongation: ΔL = (F L) / (A Y). Substitute values: ΔL = (200 × 2) / (2 × 10-6 × 2 × 1011). Calculate: ΔL = 400 / (4 × 105) = 10-3 m = 1 mm.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Bulk Modulus and Compressibility

An aluminium rod of length 1.4m and cross-sectional area 2×10−6m2 is compressed by a force producing a stress of 3×107N/

Young's modulus: Y = StressStrain. Strain: Strain = StressY = 3×1077×1010≈4.29×10−4. Compression: ΔL = Strain×L = 4.29×10−4×1.4≈6×10−4m = 0.6mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.6mm. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bulk Modulus and Compressibility

What is the key difference between elastic and plastic deformation in a material?

Elastic deformation is reversible, meaning the material returns to its original shape after the load is removed, while plastic deformation results in permanent changes. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Elastic deformation is reversible, plastic is not. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Stress, Strain and Hooke's Law

In the stress-strain behavior of a material, what occurs after the maximum stress point when the material is ductile?

For a ductile material, after the maximum stress point (ultimate tensile strength), the stress decreases with increasing strain as the material necks, leading to fracture. As per NCERT, applying relevant law/formula with correct units and sign convention leads to The stress decreases leading to fracture. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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