Young's Modulus and Elasticity

Young's Modulus and Elasticity

A brass block of dimensions 0.5m×0.3m×0.2m is subjected to a shearing force of 6×104N. If the shear modulus of brass is

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.5×0.3 = 0.15m2, L = 0.2m. Substitute: Δx = 6×104×0.20.15×3.6×1010 = 120005.4×109≈2.22×10−6m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.22×10−6m. 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 copper block of dimensions 0.3m×0.2m×0.05m is subjected to a shearing force of 3×104N. If the shear modulus of copper

Shear modulus: G = Shear stressShear strain. Shear stress: Shear stress = FA, A = 0.3×0.2 = 0.06m2. Shear stress: 3×1040.06 = 5×105N/m2. Shear strain: Shear strain = Shear stressG = 5×1054.2×1010≈1.19×10−5.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Young's Modulus and Elasticity

A glass slab of volume 0.05m3 is subjected to a hydraulic pressure of 8×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −8×1063.7×1010≈−2.16×10−4. Magnitude: 2.16×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.16×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Young's Modulus and Elasticity

A copper rod of length 2.0m and cross-sectional area 2.5×10−6m2 is subjected to a tensile force of 500N. If the Young's

Stress: Stress = FA = 5002.5×10−6 = 2×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 2×1081.1×1011≈1.82×10−3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82×10−3. 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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Young's Modulus and Elasticity

A glass slab of volume 0.03m3 is subjected to a hydraulic pressure of 6×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −6×1063.7×1010≈−1.62×10−4. Magnitude: 1.62×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.62×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Young's Modulus and Elasticity

What property of a material primarily determines its resistance to uniform compression?

The bulk modulus determines a material’s resistance to uniform compression by measuring how much it resists volume change under pressure applied in all directions. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Bulk modulus. 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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Young's Modulus and Elasticity

Why do materials like steel require a significantly larger force to produce small deformations compared to materials lik

Steel has a much larger Young’s modulus compared to rubber, meaning it has greater stiffness and requires more force to produce the same strain. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Due to larger Young’s modulus. 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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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 solid aluminium sphere is subjected to a hydraulic pressure of 1 × 107 N/m2. If the bulk modulus of aluminium is 7.2 ×

Bulk modulus: B = -p / (ΔV / V). Rearrange: ΔV / V = -p / B. Substitute: ΔV / V = -(1 × 107) / (7.2 × 1010) ≈ -1.39 × 10-4. Magnitude: 1.39 × 10-4.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Young's Modulus and Elasticity

A copper wire of length 1.5 m and diameter 1 mm is subjected to a tensile force of 100 N. If the Young's modulus of copp

Stress is given by: Stress = F / A. Area A = π r2, where r = (1/2) × 10-3 m = 5 × 10-4 m. So, A = 3.14 × (5 × 10-4)2 = 3.14 × 25 × 10-8 = 7.85 × 10-7 m2. Stress: 100 / (7.85 × 10-7) ≈ 1.27 × 108 N/m2.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Young's Modulus and Elasticity

A copper block of dimensions 0.5 m × 0.3 m × 0.2 m is subjected to a shearing force of 7 × 104 N on its top face. If the

Shear modulus: G = (Shear stress) / (Shear strain). Shear stress: Shear stress = F / A, A = 0.5 × 0.3 = 0.15 m2. Shear stress = (7 × 104) / 0.15 = 4.67 × 105 N/m2. Shear strain: Shear strain = Shear stress / G = (4.67 × 105) / (4.2 × 1010) ≈ 1.11 × 10-5.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Young's Modulus and Elasticity

A steel wire of length 2.4m and cross-sectional area 4×10−6m2 is stretched by 0.6mm. If the Young's modulus of steel is

Strain: Strain = ΔLL = 0.6×10−32.4 = 2.5×10−4. Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 2×1011×2.5×10−4 = 5×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5×107N/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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