Concept Card: Bulk Modulus, Modulus of Rigidity & Elastic Inter-Relations
1. Architectural Scope & Hydrostatic/Shear States:
While Young’s modulus ($Y$) describes 1D longitudinal deformation under uniaxial tension or compression, general stress states involve multi-dimensional changes:
- Bulk Modulus ($B$): Resistance of a substance (solid, liquid, or gas) to change in volume under uniform all-around hydrostatic pressure, without change in shape.
- Modulus of Rigidity ($\eta$): Resistance of a solid to change in shape (angular skew) under tangential shear forces, without change in volume.
- Poisson’s Ratio ($\nu$): Coupling ratio connecting lateral contraction to longitudinal elongation.
2. Bulk Modulus ($B$) & Compressibility ($k$)
(A) Definition and Formulation:
When a body of initial volume $V_0$ is subjected to a uniform hydraulic stress (excess pressure $\Delta P$), its volume contracts by $-\Delta V$.
$$\mathbf{B = \frac{\text{Volumetric / Hydraulic Stress}}{\text{Volumetric Strain}} = \frac{\Delta P}{-\frac{\Delta V}{V_0}} = -V_0 \frac{\Delta P}{\Delta V}}$$
- Negative Sign Significance: An increase in pressure ($\Delta P > 0$) causes a decrease in volume ($\Delta V < 0$). The negative sign ensures that $B$ remains a positive quantity.
- Differential Form:
$$B = -V \frac{dP}{dV} = \rho \frac{dP}{d\rho}$$
(since mass $m = \rho V = \text{constant} \implies \frac{d\rho}{\rho} = -\frac{dV}{V}$). - SI Units & Dimensions:
- SI Unit: $\text{N/m}^2$ or $\text{Pascal (Pa)}$.
- Dimensional Formula: $[B] = \mathbf{[M L^{-1} T^{-2}]}$.
- Comparative Hierarchy:
$$B_{\text{solids}} \ (\sim 10^{11}\text{ Pa}) \ > \ B_{\text{liquids}} \ (\sim 10^9\text{ Pa}) \ > \ B_{\text{gases}} \ (\sim 10^5\text{ Pa})$$
Solids are nearly incompressible, liquids are slightly compressible at extreme depths, and gases are highly compressible. For an idealized perfectly incompressible fluid, $\Delta V = 0 \implies \mathbf{B = \infty}$.
(B) Compressibility ($k$):
The fractional decrease in volume per unit increase in hydrostatic pressure:
$$\mathbf{k = \frac{1}{B} = -\frac{1}{V_0} \frac{\Delta V}{\Delta P}}$$
- SI Unit: $\text{Pa}^{-1}$ or $\text{m}^2/\text{N}$.
- Dimensions: $[k] = \mathbf{[M^{-1} L^1 T^2]}$.
(C) Isothermal vs. Adiabatic Bulk Modulus of an Ideal Gas:
- Isothermal Bulk Modulus ($B_{\text{iso}}$):
$$P V = \text{constant} \implies P dV + V dP = 0 \implies -V \frac{dP}{dV} = P \implies \mathbf{B_{\text{iso}} = P}$$ - Adiabatic Bulk Modulus ($B_{\text{adia}}$):
$$P V^\gamma = \text{constant} \implies \gamma P V^{\gamma-1} dV + V^\gamma dP = 0 \implies -V \frac{dP}{dV} = \gamma P \implies \mathbf{B_{\text{adia}} = \gamma P}$$ - Ratio:
$$\mathbf{\frac{B_{\text{adia}}}{B_{\text{iso}}} = \gamma = \frac{C_p}{C_v} > 1}$$
Acoustic Significance: Laplace corrected Newton’s formula for the speed of sound in air by recognizing that rapid sound compressions are adiabatic, replacing $B_{\text{iso}} = P$ with $B_{\text{adia}} = \gamma P$:
$$v_{\text{sound}} = \sqrt{\frac{B_{\text{adia}}}{\rho}} = \sqrt{\frac{\gamma P}{\rho}}$$
3. Modulus of Rigidity / Shear Modulus ($\eta$)
(A) Definition and Formulation:
When tangential deforming forces are applied parallel to opposite faces of a body of cross-sectional area $A$, the body undergoes angular deformation without change in volume:
$$\mathbf{\eta = \frac{\text{Tangential / Shearing Stress}}{\text{Shearing Strain}} = \frac{F_\parallel / A}{\theta} = \frac{F_\parallel}{A \theta}}$$
where $\theta$ (in radians) is the angle of shear:
$$\theta \approx \tan\theta = \frac{\Delta x}{h}$$
($\Delta x$ is the relative lateral displacement of the top face, and $h$ is the height perpendicular to the faces).
- SI Unit: $\text{N/m}^2$ or $\text{Pascal (Pa)}$.
- Dimensions: $[\eta] = \mathbf{[M L^{-1} T^{-2}]}$.
- Fluids Have Zero Shear Modulus: Liquids and gases cannot support static tangential shear stress without continuous flow; hence for all fluids:
$$\mathbf{\eta_{\text{fluid}} = 0}$$
Only solids possess non-zero shear moduli ($\eta \ne 0$).
4. Poisson’s Ratio ($\nu$) & Volumetric Strain Under Tension
When a cylindrical wire of original length $L_0$ and diameter $d_0$ is pulled longitudinally:
- Longitudinal strain: $\varepsilon_l = \frac{\Delta L}{L_0} > 0$.
- Lateral strain: $\varepsilon_d = -\frac{\Delta d}{d_0} < 0$ (cross-section shrinks).
Poisson’s Ratio ($\nu$):
$$\mathbf{\nu = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta d / d_0}{\Delta L / L_0}}$$
As a ratio of two identical strains, $\nu$ is dimensionless and unitless ($[M^0 L^0 T^0]$).
Fractional Change in Volume Under Tensile Strain:
Volume of cylindrical wire: $V = \frac{\pi}{4} d^2 L$.
Taking the logarithm and differentiating:
$$\ln V = \ln\left(\frac{\pi}{4}\right) + 2 \ln d + \ln L \implies \frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L}$$
Substitute $\frac{\Delta d}{d} = -\nu \frac{\Delta L}{L}$:
$$\mathbf{\frac{\Delta V}{V} = \frac{\Delta L}{L} (1 – 2\nu) = \varepsilon_l (1 – 2\nu)}$$
Theoretical & Practical Limits of Poisson’s Ratio:
1. If volume does not decrease when pulled in tension ($\Delta V \ge 0$), then $1 – 2\nu \ge 0 \implies \mathbf{\nu \le 0.5}$.
2. For a perfectly incompressible substance ($\Delta V = 0$), $1 – 2\nu = 0 \implies \mathbf{\nu = 0.5}$ (e.g., vulcanized rubber, water).
3. Theoretical bounds: $\mathbf{-1 \le \nu \le 0.5}$. Practical range for real solids: $\mathbf{0 \le \nu \le 0.5}$ (cork: $\nu \approx 0$; metals: $\nu \approx 0.25 – 0.35$).
5. Inter-Relationships Among Elastic Moduli ($Y, B, \eta, \nu$)
For homogeneous, isotropic elastic solids, the four elastic parameters are connected by fundamental continuum mechanics relations:
- $$\mathbf{Y = 3 B (1 – 2\nu)}$$
- $$\mathbf{Y = 2 \eta (1 + \nu)}$$
- $$\mathbf{\frac{9}{Y} = \frac{1}{B} + \frac{3}{\eta}}$$
- $$\mathbf{\nu = \frac{3 B – 2 \eta}{6 B + 2 \eta}}$$
6. Strain Energy Densities in Bulk and Shear Deformations
- Volumetric Compression Energy Density:
$$\mathbf{u_{\text{volume}} = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} \Delta P \left(-\frac{\Delta V}{V_0}\right) = \frac{(\Delta P)^2}{2 B} = \frac{1}{2} B \varepsilon_v^2}$$ - Shear Strain Energy Density:
$$\mathbf{u_{\text{shear}} = \frac{1}{2} \times \text{Shear Stress} \times \text{Shear Strain} = \frac{1}{2} \sigma_s \theta = \frac{\sigma_s^2}{2 \eta} = \frac{1}{2} \eta \theta^2}$$
Solved Examples
Example 1 (Hydrostatic Bulk Compression & Ocean Trench Density – JEE Main):
At the deepest point of the Mariana Trench, the water depth is $h = 11.0\text{ km} = 1.10 \times 10^4\text{ m}$. Assume the surface density of sea water is $\rho_0 = 1000\text{ kg/m}^3$, surface gravity is $g = 9.8\text{ m/s}^2$, and the bulk modulus of water is $B = 2.20 \times 10^9\text{ N/m}^2$.
(a) Determine the hydrostatic gauge pressure $\Delta P$ at this depth.
(b) Calculate the percentage decrease in volume of water at the trench floor.
(c) Determine the actual density $\rho$ of sea water at the bottom of the trench.
Solution:
(a) Hydrostatic Gauge Pressure:
$$\Delta P = \rho_0 g h = (1000\text{ kg/m}^3)(9.8\text{ m/s}^2)(1.10 \times 10^4\text{ m}) = 1.078 \times 10^8\text{ Pa} = \mathbf{107.8\text{ MPa}}$$
(This corresponds to approximately $1064\text{ atmospheres}$ of pressure!).
(b) Fractional and Percentage Decrease in Volume:
$$-\frac{\Delta V}{V_0} = \frac{\Delta P}{B} = \frac{1.078 \times 10^8\text{ Pa}}{2.20 \times 10^9\text{ Pa}} = 0.0490$$
$$\% \text{ volume decrease} = 0.0490 \times 100\% = \mathbf{4.90\%}$$
(c) Density at Depth:
Since mass is conserved: $m = \rho_0 V_0 = \rho V \implies \rho = \rho_0 \left(\frac{V_0}{V}\right) = \rho_0 \left(\frac{V_0}{V_0 + \Delta V}\right) \approx \rho_0 \left(1 – \frac{\Delta V}{V_0}\right)$.
$$\rho \approx \rho_0 \left(1 + \frac{\Delta P}{B}\right) = (1000\text{ kg/m}^3)(1 + 0.0490) = \mathbf{1049.0\text{ kg/m}^3}$$
Example 2 (Shear Deformation of a Cemented Metal Cube – JEE Main):
A solid structural metal cube of edge length $L = 20.0\text{ cm} = 0.20\text{ m}$ has its bottom face firmly cemented to the ground. A horizontal tangential force $F = 4.0 \times 10^5\text{ N}$ is applied uniformly across its top face. The modulus of rigidity of the metal is $\eta = 8.0 \times 10^{10}\text{ N/m}^2$.
(a) Find the shear stress developed on the top face.
(b) Calculate the angle of shear strain $\theta$ in radians.
(c) Calculate the lateral horizontal displacement $\Delta x$ of the top face.
(d) Compute the total elastic strain energy stored in the sheared cube.
Solution:
Area of the top face: $A = L^2 = (0.20\text{ m})^2 = 0.040\text{ m}^2$.
(a) Shear Stress:
$$\sigma_s = \frac{F}{A} = \frac{4.0 \times 10^5\text{ N}}{0.040\text{ m}^2} = \mathbf{1.0 \times 10^7\text{ N/m}^2 = 10.0\text{ MPa}}$$
(b) Angle of Shear Strain ($\theta$):
$$\theta = \frac{\sigma_s}{\eta} = \frac{1.0 \times 10^7\text{ Pa}}{8.0 \times 10^{10}\text{ Pa}} = \mathbf{1.25 \times 10^{-4}\text{ radians}}$$
(c) Lateral Displacement ($\Delta x$):
$$\Delta x = L \theta = (0.20\text{ m})(1.25 \times 10^{-4}\text{ rad}) = 2.50 \times 10^{-5}\text{ m} = \mathbf{25.0\text{ }\mu\text{m}}$$
(d) Total Elastic Strain Energy:
$$U = \frac{1}{2} F \Delta x = \frac{1}{2}(4.0 \times 10^5\text{ N})(2.50 \times 10^{-5}\text{ m}) = \mathbf{5.00\text{ J}}$$
Example 3 (Thermodynamic Bulk Moduli of Gases – JEE Advanced):
An ideal diatomic gas ($\gamma = 1.40$) is held at standard atmospheric pressure $P_0 = 1.013 \times 10^5\text{ Pa}$.
(a) Derive the isothermal bulk modulus $B_{\text{iso}}$ and evaluate its numerical value.
(b) Derive the adiabatic bulk modulus $B_{\text{adia}}$ and evaluate its numerical value.
(c) If the gas density is $\rho = 1.29\text{ kg/m}^3$, compare the speed of sound computed using Newton’s isothermal assumption with Laplace’s adiabatic formula.
Solution:
(a) Isothermal Bulk Modulus:
For an isothermal process, $T = \text{constant} \implies P V = \text{constant}$.
Differentiating: $P dV + V dP = 0 \implies \frac{dP}{dV} = -\frac{P}{V}$.
$$B_{\text{iso}} = -V \frac{dP}{dV} = -V \left(-\frac{P}{V}\right) = \mathbf{P_0 = 1.013 \times 10^5\text{ Pa}}$$
(b) Adiabatic Bulk Modulus:
For a reversible adiabatic process, $P V^\gamma = \text{constant}$.
Differentiating: $\gamma P V^{\gamma – 1} dV + V^\gamma dP = 0 \implies \gamma P dV + V dP = 0 \implies \frac{dP}{dV} = -\frac{\gamma P}{V}$.
$$B_{\text{adia}} = -V \frac{dP}{dV} = -V \left(-\frac{\gamma P}{V}\right) = \mathbf{\gamma P_0 = (1.40)(1.013 \times 10^5\text{ Pa}) = 1.418 \times 10^5\text{ Pa}}$$
(c) Speed of Sound Comparison:
Newton’s Isothermal formula:
$$v_{\text{Newton}} = \sqrt{\frac{B_{\text{iso}}}{\rho}} = \sqrt{\frac{1.013 \times 10^5}{1.29}} \approx \mathbf{280.2\text{ m/s}}$$
Laplace’s Adiabatic formula:
$$v_{\text{Laplace}} = \sqrt{\frac{B_{\text{adia}}}{\rho}} = \sqrt{\frac{\gamma P_0}{\rho}} = \sqrt{1.40} \times 280.2 \approx 1.1832 \times 280.2 \approx \mathbf{331.5\text{ m/s}}$$
Laplace’s adiabatic correction precisely accounts for the experimentally observed velocity of sound in air ($332\text{ m/s}$).
Example 4 (Elastic Inter-Relations & Incompressibility – JEE Advanced):
A structural metallic alloy has Young’s modulus $Y = 1.20 \times 10^{11}\text{ Pa}$ and modulus of rigidity $\eta = 4.80 \times 10^{10}\text{ Pa}$.
(a) Determine the Poisson’s ratio $\nu$ of the alloy.
(b) Calculate the bulk modulus $B$ of the alloy.
(c) Deduce whether the material is compressible and calculate its compressibility $k$.
Solution:
(a) Poisson’s Ratio ($\nu$):
Using the relation $Y = 2 \eta (1 + \nu)$:
$$1 + \nu = \frac{Y}{2 \eta} = \frac{1.20 \times 10^{11}}{2(4.80 \times 10^{10})} = \frac{1.20 \times 10^{11}}{0.96 \times 10^{11}} = 1.25$$
$$\mathbf{\nu = 1.25 – 1.00 = 0.25}$$
(b) Bulk Modulus ($B$):
Using the relation $Y = 3 B (1 – 2\nu)$:
$$B = \frac{Y}{3 (1 – 2\nu)} = \frac{1.20 \times 10^{11}}{3 [1 – 2(0.25)]} = \frac{1.20 \times 10^{11}}{3(1 – 0.50)} = \frac{1.20 \times 10^{11}}{1.50} = \mathbf{8.00 \times 10^{10}\text{ Pa} = 80.0\text{ GPa}}$$
(c) Compressibility ($k$):
Since $B$ is finite, the material is compressible. Its compressibility is:
$$k = \frac{1}{B} = \frac{1}{8.00 \times 10^{10}\text{ Pa}} = \mathbf{1.25 \times 10^{-11}\text{ Pa}^{-1}}$$
Practice Worksheet: Bulk Modulus, Rigidity & Elastic Moduli
Section A: Single Correct Questions (Q1 to Q5)
Q1. The ratio of the adiabatic bulk modulus ($B_{\text{adia}}$) to the isothermal bulk modulus ($B_{\text{iso}}$) for an ideal gas having ratio of molar heat capacities $\gamma = C_p / C_v$ is:
(A) $\gamma$
(B) $\frac{1}{\gamma}$
(C) $\gamma – 1$
(D) $\frac{\gamma + 1}{2}$
Q2. The compressibility of water is $k = 4.5 \times 10^{-10}\text{ Pa}^{-1}$. If a volume $V_0 = 100\text{ litres}$ of water is subjected to an increase in hydrostatic pressure $\Delta P = 2.0 \times 10^7\text{ Pa}$, the decrease in volume $-\Delta V$ is:
(A) $0.90\text{ litres}$
(B) $0.45\text{ litres}$
(C) $0.09\text{ litres}$
(D) $0.18\text{ litres}$
Q3. A square metal plate of side length $50\text{ cm}$ and thickness $10\text{ cm}$ is subjected to a shearing force $F = 9.0 \times 10^4\text{ N}$ applied tangentially on its narrow upper face ($50\text{ cm} \times 10\text{ cm}$), while the lower face is rigidly fixed. If the shear modulus of the metal is $\eta = 3.0 \times 10^{10}\text{ Pa}$, the lateral displacement of the top edge is:
(A) $3.0 \times 10^{-5}\text{ m}$
(B) $1.5 \times 10^{-5}\text{ m}$
(C) $6.0 \times 10^{-5}\text{ m}$
(D) $9.0 \times 10^{-6}\text{ m}$
Q4. If the density of a liquid at the surface is $\rho_0$ and its bulk modulus is $B$, the density $\rho$ at depth $h$ below the surface (where $g$ is acceleration due to gravity) is given by:
(A) $\rho_0 \left(1 + \frac{\rho_0 g h}{B}\right)$
(B) $\rho_0 \left(1 – \frac{\rho_0 g h}{B}\right)$
(C) $\frac{\rho_0}{1 + \frac{B}{\rho_0 g h}}$
(D) $\rho_0 \left(1 + \frac{B}{\rho_0 g h}\right)$
Q5. A solid sphere of bulk modulus $B$ is submerged in a liquid where it experiences a uniform hydrostatic pressure increase $\Delta P$. The elastic potential energy stored per unit volume of the sphere is:
(A) $\frac{(\Delta P)^2}{2 B}$
(B) $\frac{(\Delta P)^2}{B}$
(C) $\frac{\Delta P}{2 B}$
(D) $\frac{B (\Delta P)^2}{2}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Regarding the Modulus of Rigidity ($\eta$) of different states of matter, which of the following statements is/are correct?
(A) Solids possess a definite non-zero modulus of rigidity because they resist changes in shape.
(B) Ideal liquids and gases have zero modulus of rigidity ($\eta = 0$) because they cannot sustain static shear stress.
(C) In pure shear deformation, the volume of the body remains strictly constant ($\Delta V = 0$).
(D) The shear modulus $\eta$ of an isotropic solid is generally smaller than its Young’s modulus $Y$.
Q7. A cylindrical metal wire of original length $L_0$ and radius $r_0$ is subjected to longitudinal tensile stress producing longitudinal strain $\varepsilon_l$. If $\nu$ is the Poisson’s ratio of the material, which of the following statements is/are correct?
(A) The lateral strain developed in the wire is $\varepsilon_r = -\nu \varepsilon_l$.
(B) The fractional change in cross-sectional area is $\frac{\Delta A}{A_0} \approx -2\nu \varepsilon_l$.
(C) The fractional change in volume is $\frac{\Delta V}{V_0} \approx \varepsilon_l (1 – 2\nu)$.
(D) For any material whose volume does not decrease when pulled in tension, $\nu \le 0.5$.
Q8. Which of the following inter-relationships among the elastic constants ($Y, B, \eta, \nu$) are valid for an isotropic elastic material?
(A) $Y = 3 B (1 – 2\nu)$
(B) $Y = 2 \eta (1 + \nu)$
(C) $\frac{9}{Y} = \frac{1}{B} + \frac{3}{\eta}$
(D) $\nu = \frac{3 B – 2 \eta}{6 B + 2 \eta}$
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A material has the property that its volume remains strictly invariant ($\Delta V = 0$) when subjected to longitudinal tensile strain. The Poisson’s ratio of this material is $\nu$. Determine the integer value of $10\nu$.
Q10. For a certain metallic alloy, the Young’s modulus is $2.5$ times its modulus of rigidity ($Y = 2.5\eta$). The Poisson’s ratio of the alloy is $\nu$. Determine the integer value of $100\nu$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Ideal Gas Bulk Modulus Ratio: $B_{\text{adia}} / B_{\text{iso}} = \gamma$ |
| Q2 | (A) | Compressibility Relation: $-\Delta V = k V_0 \Delta P = 0.90\text{ litres}$ |
| Q3 | (A) | Shear Strain & Lateral Displacement: $\Delta x = h (F / A\eta) = 3.0 \times 10^{-5}\text{ m}$ |
| Q4 | (A) | Density Variation with Depth: $\rho = \rho_0 (1 + \rho_0 g h / B)$ |
| Q5 | (A) | Volumetric Strain Energy Density: $u = \frac{(\Delta P)^2}{2 B}$ |
| Q6 | (A, B, C, D) | Fundamental Properties of Shear Modulus in Solids vs. Fluids |
| Q7 | (A, B, C, D) | Poisson’s Ratio & Incompressibility Boundary Limit ($\nu \le 0.5$) |
| Q8 | (A, B, C, D) | Continuum Elastic Constant Inter-Relations |
| Q9 | 5 | Incompressible Material Poisson’s Ratio: $\nu = 0.5 \implies 10\nu = 5$ |
| Q10 | 25 | Elastic Ratio: $Y = 2.5\eta \implies \nu = 0.25 \implies 100\nu = 25$ |
Detailed Explanations
Solution 1:
For an ideal gas:
– Isothermal bulk modulus: $B_{\text{iso}} = -V \frac{dP}{dV} = P$.
– Adiabatic bulk modulus: $B_{\text{adia}} = -V \frac{dP}{dV} = \gamma P$.
Taking the ratio:
$$\frac{B_{\text{adia}}}{B_{\text{iso}}} = \frac{\gamma P}{P} = \mathbf{\gamma}$$
Correct Answer: (A)
Solution 2:
Compressibility is defined as $k = -\frac{1}{V_0} \frac{\Delta V}{\Delta P}$.
Rearranging for volume reduction $-\Delta V$:
$$-\Delta V = k V_0 \Delta P$$
Given $k = 4.5 \times 10^{-10}\text{ Pa}^{-1}$, $V_0 = 100\text{ litres}$, and $\Delta P = 2.0 \times 10^7\text{ Pa}$:
$$-\Delta V = (4.5 \times 10^{-10}\text{ Pa}^{-1})(100\text{ L})(2.0 \times 10^7\text{ Pa}) = 9.0 \times 10^{-8} \times 10^7 = \mathbf{0.90\text{ litres}}$$
Correct Answer: (A)
Solution 3:
Area of the face subjected to tangential force:
$$A = 50\text{ cm} \times 10\text{ cm} = 0.50\text{ m} \times 0.10\text{ m} = 0.050\text{ m}^2$$
Shear stress developed:
$$\sigma_s = \frac{F}{A} = \frac{9.0 \times 10^4\text{ N}}{0.050\text{ m}^2} = 1.80 \times 10^6\text{ Pa}$$
Height perpendicular to fixed base: $h = 50\text{ cm} = 0.50\text{ m}$.
Shear strain angle $\theta$:
$$\theta = \frac{\sigma_s}{\eta} = \frac{1.80 \times 10^6\text{ Pa}}{3.0 \times 10^{10}\text{ Pa}} = 6.0 \times 10^{-5}\text{ rad}$$
Lateral displacement of top edge:
$$\Delta x = h \theta = (0.50\text{ m})(6.0 \times 10^{-5}\text{ rad}) = \mathbf{3.0 \times 10^{-5}\text{ m}}$$
Correct Answer: (A)
Solution 4:
Gauge pressure at depth $h$: $\Delta P = \rho_0 g h$.
From the definition of bulk modulus:
$$B = -V_0 \frac{\Delta P}{\Delta V} \implies -\frac{\Delta V}{V_0} = \frac{\Delta P}{B} = \frac{\rho_0 g h}{B}$$
Since mass is constant ($m = \rho V = \rho_0 V_0$):
$$\frac{\Delta \rho}{\rho_0} = -\frac{\Delta V}{V_0} = \frac{\rho_0 g h}{B} \implies \mathbf{\rho = \rho_0 \left(1 + \frac{\rho_0 g h}{B}\right)}$$
Correct Answer: (A)
Solution 5:
Volumetric strain energy density is the work done per unit volume during hydrostatic compression:
$$u = \frac{1}{2} \times (\text{Hydraulic Stress}) \times (\text{Volumetric Strain}) = \frac{1}{2} \Delta P \left(-\frac{\Delta V}{V_0}\right)$$
Since $-\frac{\Delta V}{V_0} = \frac{\Delta P}{B}$:
$$u = \frac{1}{2} \Delta P \left(\frac{\Delta P}{B}\right) = \mathbf{\frac{(\Delta P)^2}{2 B}}$$
Correct Answer: (A)
Solution 6:
– (A) True: Solids maintain a crystalline lattice that opposes tangential skew deformation.
– (B) True: Fluids cannot resist static shear; any non-zero tangential stress produces continuous shear flow.
– (C) True: In pure shear, opposite faces translate laterally without altering height or base area; hence $\Delta V = 0$.
– (D) True: From $Y = 2\eta(1 + \nu)$, since $\nu > 0$, $Y = 2\eta(1 + \nu) > 2\eta \implies \eta < Y/2$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: By definition, $\nu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} \implies \varepsilon_r = -\nu \varepsilon_l$.
– (B) True: Area $A = \pi r^2 \implies \frac{\Delta A}{A} = 2 \frac{\Delta r}{r} = -2\nu \varepsilon_l$.
– (C) True: Volume $V = A L \implies \frac{\Delta V}{V} = \frac{\Delta A}{A} + \frac{\Delta L}{L} = -2\nu \varepsilon_l + \varepsilon_l = \varepsilon_l (1 – 2\nu)$.
– (D) True: For volume not to decrease under tension ($\frac{\Delta V}{V} \ge 0$), $1 – 2\nu \ge 0 \implies \nu \le 0.5$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
All four equations represent standard, mathematically proven inter-relations among isotropic elastic moduli derived from Hooke’s law generalized to three dimensions.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
The fractional change in volume under longitudinal strain is:
$$\frac{\Delta V}{V_0} = \varepsilon_l (1 – 2\nu)$$
For an incompressible material ($\Delta V = 0$):
$$1 – 2\nu = 0 \implies \nu = 0.50$$
Multiplying by 10:
$$10\nu = 10 \times 0.50 = \mathbf{5}$$
Correct Answer: 5
Solution 10:
Using the relation between Young’s modulus and shear modulus:
$$Y = 2 \eta (1 + \nu)$$
Given $Y = 2.5 \eta$:
$$2.5 \eta = 2 \eta (1 + \nu) \implies 1 + \nu = \frac{2.5}{2} = 1.25 \implies \nu = 0.25$$
Multiplying by 100:
$$100\nu = 100 \times 0.25 = \mathbf{25}$$
Correct Answer: 25