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#stress

23 public questions tagged with this topic.

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 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 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 copper wire of length 2.5m and cross-sectional area 2×10−6m2 is stretched by a force of 200N. If the Young's modulus o

Stress: Stress = FA = 2002×10−6 = 1×108N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1×108N/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 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

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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Shear Modulus and Rigidity

A steel wire of length 2.5m and cross-sectional area 3.5×10−6m2 is stretched by a force of 350N. If the elongation is 0.

Young's modulus: Y = FLAΔL. Substitute: ΔL = 0.2×10−3m. Y = 350×2.53.5×10−6×0.2×10−3 = 8757×10−10≈1.25×1012N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.25×1012N/m2. 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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Bulk Modulus and Compressibility

A steel rod of radius 0.01m and length 1.5m is subjected to a tensile force producing a stress of 4×107N/m2. What is the

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.01)2 = 3.14×10−4m2. Force: F = Stress×A = 4×107×3.14×10−4 = 1.256×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.256×104N. 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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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

A steel wire of length 2.7m and cross-sectional area 2×10−6m2 is stretched by 0.54mm. If the Young's modulus of steel is

Strain: Strain = ΔLL = 0.54×10−32.7 = 2×10−4. Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 2×1011×2×10−4 = 4×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4×107N/m2. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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