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#copper wire

13 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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Universal Law of Gravitation

A steel wire and a copper wire have the same length and cross-sectional area. Both are stretched by the same force. If Y

Elongation: ΔL = (F L) / (A Y). Ratio: (ΔLsteel)/(ΔLcopper) = Ycopper / Ysteel = (1.1 × 1011) / (2 × 1011) = 11/20 = 0.55. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.55. 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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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 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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Shear Modulus and Rigidity

A copper wire of length 2.2m and cross-sectional area 1.8×10−6m2 is stretched by a force of 180N. If the Young's modulus

Stress: Stress = FA = 1801.8×10−6 = 1×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1×1081.1×1011≈9.09×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.09×10−4. 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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Bulk Modulus and Compressibility

A copper wire of length 2.4m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 1.5×10−4. I

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 1.1×1011×1.5×10−4 = 1.65×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.65×107N/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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Bulk Modulus and Compressibility

A copper wire of length 2.3m 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. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1×1081.1×1011≈9.09×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.09×10−4. 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 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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Stress, Strain and Hooke's Law

A copper wire of length 1.6m and cross-sectional area 1.8×10−6m2 is stretched by a force of 180N. If the Young's modulus

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 180×1.61.8×10−6×1.1×1011 = 2881.98×105≈1.45×10−3m = 1.45mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.45mm. 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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