#shear displacement

Stress, Strain and Hooke's Law

A brass block of dimensions 0.4m×0.3m×0.1m is subjected to a shearing force of 3×104N. If the shear modulus of brass is

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.4×0.3 = 0.12m2, L = 0.1m. Substitute: Δx = 3×104×0.10.12×3.6×1010 = 30004.32×109≈6.94×10−7m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.94×10−7m. 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 block of dimensions 0.5m×0.3m×0.1m is subjected to a shearing force of 6×104N. If the shear modulus of copper i

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.5×0.3 = 0.15m2, L = 0.1m. Substitute: Δx = 6×104×0.10.15×4.2×1010 = 60006.3×109≈9.52×10−7m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.52×10−7m. 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 copper block of dimensions 0.4 m × 0.3 m × 0.1 m is subjected to a shearing force of 3 × 104 N. If the shear modulus o

Shear modulus: G = (F / A) / (Δx / L). Rearrange: Δx = (F L) / (A G). Area: A = 0.4 × 0.3 = 0.12 m2, L = 0.1 m. Substitute: Δx = (3 × 104 × 0.1) / (0.12 × 4.2 × 1010) = 3000 / (5.04 × 109) ≈ 5.95 × 10-7 m.

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

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Bulk Modulus and Compressibility

An aluminium block of dimensions 0.6m×0.4m×0.2m is subjected to a shearing force of 8×104N. If the shear modulus of alum

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.6×0.4 = 0.24m2, L = 0.2m. Substitute: Δx = 8×104×0.20.24×2.5×1010 = 160006×109≈2.67×10−6m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.67×10−6m. 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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Bulk Modulus and Compressibility

A rectangular slab of length 0.4 m, width 0.2 m, and thickness 0.1 m is subjected to a shearing force of 8 × 104 N. If t

Shear modulus: G = (Shear stress) / (Shear strain) = (F / A) / (Δx / L). Rearrange: Δx = (F L) / (A G). Area A = 0.4 × 0.2 = 0.08 m2, L = 0.1 m. Substitute: Δx = (8 × 104 × 0.1) / (0.08 × 4 × 109) = 8000 / (3.2 × 108) = 2.5 × 10-5 m.

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

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Young's Modulus and Elasticity

An aluminium block of dimensions 0.5m×0.3m×0.1m is subjected to a shearing force of 7×104N. If the shear modulus of alum

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.5×0.3 = 0.15m2, L = 0.1m. Substitute: Δx = 7×104×0.10.15×2.5×1010 = 70003.75×109≈1.87×10−6m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.87×10−6m. 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 block of dimensions 0.3 m × 0.2 m × 0.1 m is subjected to a shearing force of 5 × 104 N. If the shear modulus of

Shear modulus: G = (F / A) / (Δx / L). Rearrange: Δx = (F L) / (A G). Area: A = 0.3 × 0.2 = 0.06 m2, L = 0.1 m. Substitute: Δx = (5 × 104 × 0.1) / (0.06 × 3.6 × 1010) = 5000 / (2.16 × 109) ≈ 2.31 × 10-6 m.

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

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