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Shear Modulus and Rigidity

Covers shear modulus and rigidity with problems on deformation under shear stress. Helps students understand how materials respond to twisting or sliding forces.

21 questions

Shear Modulus and Rigidity

An aluminium wire of length 1.7m and cross-sectional area 2×10−6m2 is stretched by 0.34mm. If the Young's modulus of alu

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.34×10−3m. F = 7×1010×2×10−6×0.34×10−31.7 = 47.61.7≈28N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 28N. 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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Shear Modulus and Rigidity

A glass slab of volume 0.025m3 is subjected to a hydraulic pressure of 4×106N/m2. If the bulk modulus of glass is 3.7×10

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −4×1063.7×1010≈−1.08×10−4. Change in volume: ΔV = ΔVV×V = −1.08×10−4×0.025≈−2.7×10−6m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.7×10−6m3. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

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Shear Modulus and Rigidity

A steel wire of length 3.0m and cross-sectional area 2×10−6m2 is stretched by a force of 200N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 200×3.02×10−6×2×1011 = 6004×105 = 1.5×10−3m = 1.5mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.5mm. 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 water sample of volume 1.5litres is compressed by a pressure of 3×106N/m2. If the bulk modulus of water is 2.2×109N/m2

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −3×1062.2×109≈−1.36×10−3. Magnitude: 1.36×10−3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.36×10−3. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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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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Shear Modulus and Rigidity

A glass slab of volume 0.015m3 is subjected to a hydraulic pressure of 2×106N/m2. If the bulk modulus of glass is 3.7×10

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −2×1063.7×1010≈−5.41×10−5. Change in volume: ΔV = ΔVV×V = −5.41×10−5×0.015≈−8.11×10−7m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.11×10−7m3. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Shear Modulus and Rigidity

A steel wire of length 3.2m and cross-sectional area 5×10−6m2 is stretched by 0.8mm. If the Young's modulus of steel is

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.8×10−3m. F = 2×1011×5×10−6×0.8×10−33.2 = 8003.2 = 250N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 250N. 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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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

Which type of stress causes a change in the shape of a body without altering its volume?

Shearing stress causes a change in the shape of a body by producing relative displacement of opposite faces (pure shear) without changing its volume. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Shearing stress. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Shear Modulus and Rigidity

A steel wire of length 2.5m and cross-sectional area 3.5×10−6m2 is stretched by a force of 350N. If the elongation is 0.

Young's modulus: Y = FLAΔL. Substitute: ΔL = 0.2×10−3m. Y = 350×2.53.5×10−6×0.2×10−3 = 8757×10−10≈1.25×1012N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.25×1012N/m2. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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