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#brass wire

8 public questions tagged with this topic.

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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Shear Modulus and Rigidity

A brass wire of length 1.8m and cross-sectional area 2.5×10−6m2 is stretched by a force producing a strain of 2×10−4. If

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×2×10−4 = 1.8×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.8×107N/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 brass wire of length 2.4m and cross-sectional area 3×10−6m2 is stretched by a force producing a stress of 5×107N/m2. I

Young's modulus: Y = StressStrain. Strain: Strain = StressY = 5×1079×1010≈5.56×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.56×10−4. 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 brass wire of length 1.6m and cross-sectional area 2×10−6m2 is stretched by a force producing a stress of 4×107N/m2. I

Young's modulus: Y = StressStrain. Strain: Strain = StressY = 4×1079×1010≈4.44×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.44×10−4. 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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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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Stress, Strain and Hooke's Law

A brass wire of length 2.0m and cross-sectional area 2×10−6m2 is stretched by a force of 300N. If the Young's modulus of

Stress: Stress = FA = 3002×10−6 = 1.5×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1.5×1089×1010≈1.67×10−3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.67×10−3. 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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