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#bulk modulus

24 public questions tagged with this topic.

Stress, Strain and Hooke's Law

A water sample of volume 3litres is compressed by a pressure of 4×106N/m2. If the bulk modulus of water is 2.2×109N/m2,

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −4×1062.2×109≈−1.82×10−3. Volume: V = 3litres = 3×10−3m3. Change in volume: ΔV = ΔVV×V = −1.82×10−3×3×10−3≈−5.45×10−6m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.45×10−6m3. 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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Stress, Strain and Hooke's Law

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. Volume: V = 1.5litres = 1.5×10−3m3. Change in volume: ΔV = ΔVV×V = −1.36×10−3×1.5×10−3≈−2.04×10−6m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.04×10−6m3. 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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Elastic Properties and Applications

A glass slab of volume 0.02m3 is subjected to a hydraulic pressure of 5×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −5×1063.7×1010≈−1.35×10−4. Magnitude: 1.35×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.35×10−4. 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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Elastic Properties and Applications

A glass slab of volume 0.03m3 is subjected to a hydraulic pressure of 5×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −5×1063.7×1010≈−1.35×10−4. Change in volume: ΔV = ΔVV×V = −1.35×10−4×0.03≈−4.05×10−6m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.05×10−6m3. 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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Elastic Properties and Applications

A glass slab of volume 0.04m3 is subjected to a hydraulic pressure of 7×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −7×1063.7×1010≈−1.89×10−4. Magnitude: 1.89×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.89×10−4. 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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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 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 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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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.

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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