Bulk Modulus and Compressibility

Bulk Modulus and Compressibility

A steel rod of radius 0.01m and length 1.5m is subjected to a tensile force producing a stress of 4×107N/m2. What is the

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.01)2 = 3.14×10−4m2. Force: F = Stress×A = 4×107×3.14×10−4 = 1.256×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.256×104N. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details

Bulk Modulus and Compressibility

A brass rod of radius 0.008m and length 1.0m is subjected to a tensile force producing a stress of 5×107N/m2. What is th

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.008)2 = 3.14×6.4×10−5≈2.01×10−4m2. Force: F = Stress×A = 5×107×2.01×10−4≈1.005×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.005×104N. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

View Details

Bulk Modulus and Compressibility

A steel wire of length 2.6m and cross-sectional area 2×10−6m2 is stretched by a force of 400N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 400×2.62×10−6×2×1011 = 10404×105 = 2.6×10−3m = 2.6mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6mm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details

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.

View Details

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.

View Details

Bulk Modulus and Compressibility

A glass slab of volume 0.01m3 is subjected to a hydraulic pressure of 2×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −2×1063.7×1010≈−5.41×10−5. Change in volume: ΔV = ΔVV×V = −5.41×10−5×0.01≈−5.41×10−7m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.41×10−7m3. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

View Details

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.

View Details

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.

View Details

Bulk Modulus and Compressibility

A brass block of dimensions 0.6m×0.4m×0.2m is subjected to a shearing force of 9×104N. If the shear modulus of brass is

Shear modulus: G = Shear stressShear strain. Shear stress: Shear stress = FA, A = 0.6×0.4 = 0.24m2. Shear stress: 9×1040.24 = 3.75×105N/m2. Shear strain: Shear strain = Shear stressG = 3.75×1053.6×1010≈1.04×10−5.

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

View Details

Bulk Modulus and Compressibility

A steel wire of length 2.7m and cross-sectional area 2×10−6m2 is stretched by 0.54mm. If the Young's modulus of steel is

Strain: Strain = ΔLL = 0.54×10−32.7 = 2×10−4. Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 2×1011×2×10−4 = 4×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4×107N/m2. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details

Bulk Modulus and Compressibility

A steel cable of radius 0.015 m is used to support a load. If the maximum stress it can withstand is 2 × 108 N/m2, what

Stress: Stress = F / A. Area: A = π r2 = 3.14 × (0.015)2 = 3.14 × 0.000225 = 7.065 × 10-4 m2. Maximum force: F = Stress × A = 2 × 108 × 7.065 × 10-4 = 1.413 × 105 N.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details

Bulk Modulus and Compressibility

A water sample of volume 1litre is compressed by a pressure of 1×106N/m2. If the bulk modulus of water is 2.2×109N/m2, w

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −1×1062.2×109≈−4.55×10−4. Magnitude: 4.55×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.55×10−4. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

View Details