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#elongation

18 public questions tagged with this topic.

Stress, Strain and Hooke's Law

A copper wire of length 2.8m and cross-sectional area 2×10−6m2 is stretched by a force of 160N. If the Young's modulus o

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 160×2.82×10−6×1.1×1011 = 4482.2×105≈2.04×10−3m = 2.04mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.04mm. 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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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 steel wire of length 2.0m and cross-sectional area 3×10−6m2 is stretched by 0.4mm. If the Young's modulus of steel is

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.4×10−3m. F = 2×1011×3×10−6×0.4×10−32 = 2402 = 120N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 120N. 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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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 3.0m and cross-sectional area 2×10−6m2 is stretched by a force of 200N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 200×3.02×10−6×2×1011 = 6004×105 = 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 A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Bulk Modulus and Compressibility

A steel wire of length 2.6m and cross-sectional area 2×10−6m2 is stretched by a force of 400N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 400×2.62×10−6×2×1011 = 10404×105 = 2.6×10−3m = 2.6mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6mm. 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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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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Young's Modulus and Elasticity

An aluminium wire of length 2.0m and cross-sectional area 1.5×10−6m2 is stretched by a force of 150N. If the Young's mod

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 150×2.01.5×10−6×7×1010 = 3001.05×105≈2.86×10−3m = 2.86mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.86mm. 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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Young's Modulus and Elasticity

A brass wire of length 2.6m and cross-sectional area 4×10−6m2 is stretched by a force of 400N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 400×2.64×10−6×9×1010 = 10403.6×105≈2.89×10−3m = 2.89mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.89mm. 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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Young's Modulus and Elasticity

A steel wire of length 1.5m and cross-sectional area 1×10−6m2 is stretched by a force of 100N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 100×1.51×10−6×2×1011 = 1502×105 = 7.5×10−4m = 0.75mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.75mm. 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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