#engineering

Elastic Properties and Applications

What characterizes the stress-strain behavior of a material in the region where plastic deformation begins?

When plastic deformation begins, the material experiences permanent deformation, meaning it does not return to its original shape even after the load is removed. As per NCERT, applying relevant law/formula with correct units and sign convention leads to The material experiences permanent deformation. 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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Elastic Properties and Applications

A brass wire of length 2.0m and cross-sectional area 3×10−6m2 is stretched by a force producing a strain of 5×10−4. If t

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×5×10−4 = 4.5×107N/m2. Force: F = Stress×A = 4.5×107×3×10−6 = 135N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 135N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Elastic Properties and Applications

A copper wire of length 2.0m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 1×10−4. If

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 1.1×1011×1×10−4 = 1.1×107N/m2. Force: F = Stress×A = 1.1×107×2×10−6 = 22N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 22N. 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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Elastic Properties and Applications

A copper wire of length 1.2 m and cross-sectional area 1.5 × 10-6 m2 is stretched by a force producing a stress of 2 × 1

Young's modulus: Y = Stress / Strain. Strain: Strain = Stress / Y = (2 × 107) / (1.1 × 1011) ≈ 1.82 × 10-4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82 × 10-4. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Elastic Properties and Applications

A steel rod of length 0.6 m and cross-sectional area 5 × 10-6 m2 is compressed by a force producing a strain of 3 × 10-4

Young's modulus: Y = Stress / Strain. Stress: Stress = Y × Strain = 2 × 1011 × 3 × 10-4 = 6 × 107 N/m2. Force: F = Stress × A = 6 × 107 × 5 × 10-6 = 300 N.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Elastic Properties and Applications

An aluminium wire of length 1.5m and cross-sectional area 2×10−6m2 is stretched by a force of 140N. If the Young's modul

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 140×1.52×10−6×7×1010 = 2101.4×105 = 1.5×10−3m = 1.5mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.5mm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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