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#cross-sectional area

6 public questions tagged with this topic.

Specific Heat and Calorimetry

A brass rod of length 2.5m at 20∘C is heated to 220∘C. If its cross-sectional area increases by 0.018cm2, what was its o

Given: ΔT = 220−20 = 200∘C, ΔA = 0.018cm2, αl = 1.8×10−5K−1. Area expansion: ΔA = A0×2αlΔT. 0.018 = A0×2×1.8×10−5×200. 0.018 = A0×7.2×10−3⇒A0 = 0.0187.2×10−3 = 2.5cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 cm². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Temperature Scales and Triple Point

A silver rod of length 1m at 10∘C is heated to 110∘C. If its cross-sectional area increases by 0.0076cm2, what was its o

Given: ΔT = 110−10 = 100∘C, ΔA = 0.0076cm2, αl = 1.9×10−5K−1. ΔA = A0×2αlΔT. 0.0076 = A0×2×1.9×10−5×100. 0.0076 = A0×3.8×10−3⇒A0 = 0.00763.8×10−3 = 2cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2 cm². This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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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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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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Shear Modulus and Rigidity

A cylindrical rod of length 0.5 m and radius 0.01 m is compressed by a force of 5000 N. If the compressive stress is 5 ×

Stress: Stress = F / A. Rearrange: A = F / Stress. Substitute: A = 5000 / (5 × 106) = 10-3 m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10-3 m2. 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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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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