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27 public questions tagged with this topic.

Stress, Strain and Hooke's Law

An aluminium wire of length 1.8m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 2×10−4.

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 7×1010×2×10−4 = 1.4×107N/m2. Force: F = Stress×A = 1.4×107×2×10−6 = 28N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 28N. 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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Stress, Strain and Hooke's Law

A steel wire of length 2.6m and cross-sectional area 2×10−6m2 is stretched by 0.52mm. If the Young's modulus of steel is

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.52×10−3m. F = 2×1011×2×10−6×0.52×10−32.6 = 2082.6 = 80N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 80N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Stress, Strain and Hooke's Law

A steel 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

Stress: Stress = FA = 2502.5×10−6 = 1×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1×1082×1011 = 5×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Stress, Strain and Hooke's Law

A copper wire of length 1.8m and cross-sectional area 3×10−6m2 is stretched by a force producing a strain of 3×10−4. If

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 1.1×1011×3×10−4 = 3.3×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.3×107N/m2. 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 steel wire of length 2.8m and cross-sectional area 4×10−6m2 is stretched by a force producing a strain of 1.5×10−4. If

Strain: Strain = ΔLL. Rearrange: ΔL = Strain×L = 1.5×10−4×2.8 = 4.2×10−4m = 0.42mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.42mm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Elastic Properties and Applications

A brass wire of length 1.7m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 4×10−4. If t

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×4×10−4 = 3.6×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6×107N/m2. 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 brass wire of length 1.9m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 2×10−4. If t

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×2×10−4 = 1.8×107N/m2. Force: F = Stress×A = 1.8×107×2×10−6 = 36N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 36N. 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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