Poisson's Ratio and relation between Modulus Questions in English

(A) The relationship between the elastic moduli is derived from the Poisson's ratio $\sigma$.
We have the following standard relations:
$Y = 3K(1 - 2\sigma)$
$Y = 2\eta(1 + \sigma)$
From the first equation,$1 - 2\sigma = \frac{Y}{3K} \implies 2\sigma = 1 - \frac{Y}{3K} \implies \sigma = \frac{1}{2} - \frac{Y}{6K}$.
From the second equation,$1 + \sigma = \frac{Y}{2\eta} \implies \sigma = \frac{Y}{2\eta} - 1$.
Equating the two expressions for $\sigma$:
$\frac{Y}{2\eta} - 1 = \frac{1}{2} - \frac{Y}{6K}$
$\frac{Y}{2\eta} + \frac{Y}{6K} = \frac{3}{2}$
Multiplying by $6\eta K$:
$3YK + Y\eta = 9\eta K$
$Y(3K + \eta) = 9\eta K$
$Y = \frac{9\eta K}{3K + \eta}$
Thus,the correct option is $A$.

(C) The relationship between Young's modulus $(Y)$,Bulk modulus $(K)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 3K(1 - 2\sigma)$.
Rearranging the formula to solve for $\sigma$:
$1 - 2\sigma = \frac{Y}{3K}$
$2\sigma = 1 - \frac{Y}{3K}$
$\sigma = \frac{3K - Y}{6K}$
Substituting the given values $Y = 7.25 \times 10^{10} \ N/m^2$ and $K = 11 \times 10^{10} \ N/m^2$:
$\sigma = \frac{3(11 \times 10^{10}) - 7.25 \times 10^{10}}{6(11 \times 10^{10})}$
$\sigma = \frac{33 \times 10^{10} - 7.25 \times 10^{10}}{66 \times 10^{10}}$
$\sigma = \frac{25.75}{66} \approx 0.39$.

(A) Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain for a material under tension or compression.
This definition is applicable only to solids that possess a definite shape and can sustain shear stress.
Liquids do not have a definite shape and cannot sustain shear stress; they only resist changes in volume (bulk modulus).
Therefore,the concept of lateral and longitudinal strain does not apply to liquids,meaning they have no Poisson's ratio.

(D) The relations between elastic constants are given by:
$Y = 3K(1 - 2\sigma)$ --- $(1)$
$Y = 2\eta(1 + \sigma)$ --- $(2)$
From $(1)$,we have: $\frac{Y}{3K} = 1 - 2\sigma$ --- $(3)$
From $(2)$,we have: $\frac{Y}{2\eta} = 1 + \sigma$ --- $(4)$
To eliminate $\sigma$,multiply equation $(4)$ by $2$:
$\frac{Y}{\eta} = 2 + 2\sigma$ --- $(5)$
Adding equations $(3)$ and $(5)$:
$\frac{Y}{3K} + \frac{Y}{\eta} = (1 - 2\sigma) + (2 + 2\sigma)$
$\frac{Y}{3K} + \frac{Y}{\eta} = 3$
Dividing both sides by $3Y$:
$\frac{1}{9K} + \frac{1}{3\eta} = \frac{1}{Y}$
Thus,the correct relation is $\frac{1}{Y} = \frac{1}{3\eta} + \frac{1}{9K}$.

(D) Poisson's ratio $(\sigma)$ is defined as the ratio of lateral strain to longitudinal strain.
For most materials,the theoretical range of Poisson's ratio is between $-1.0$ and $0.5$.
Among the given options,$0.4$ falls within this range,while $1$,$0.9$,and $0.8$ are outside the physically possible range for stable,isotropic materials.
Therefore,the correct option is $D$.

(B) Young's modulus $(Y)$ represents the stiffness of a material,which is a measure of the stress required to produce a unit strain. Its value is typically in the order of $10^9$ to $10^{11} \text{ Pa}$ for metals.
Poisson's ratio $(\sigma)$ is a dimensionless quantity representing the ratio of lateral strain to longitudinal strain. For most stable,isotropic materials,the value of Poisson's ratio lies between $-1.0$ and $0.5$.
Since $Y$ is a physical quantity with units of pressure (Pascal) and $\sigma$ is a dimensionless ratio,comparing their magnitudes directly depends on the units used. However,in standard physical contexts where $Y$ is expressed in Pascals,$Y$ is numerically much larger than $\sigma$. Thus,$Y > \sigma$ is the correct relation.

(D) Poisson's ratio $(
u)$ is defined as the ratio of lateral strain to longitudinal strain.
For most metals,when a force is applied in the longitudinal direction,the lateral dimension decreases,resulting in a positive value for Poisson's ratio.
The theoretical limits for Poisson's ratio for stable,isotropic materials are between $-1.0$ and $0.5$.
However,for most practical metals,the value is positive and typically ranges from $0$ to $0.5$.
Therefore,the correct range for metals is $0$ to $0.5$.

(A) The relationship between Young's modulus $(Y)$,Bulk modulus $(K)$,and Modulus of rigidity $(\eta)$ with Poisson's ratio $(\sigma)$ is given by:
$Y = 3K(1 - 2\sigma)$
$Y = 2\eta(1 + \sigma)$
For a stable material,the moduli of elasticity must be positive $(Y, K, \eta > 0)$.
From $Y = 3K(1 - 2\sigma) > 0$,we get $1 - 2\sigma > 0$,which implies $\sigma < \frac{1}{2}$.
From $Y = 2\eta(1 + \sigma) > 0$,we get $1 + \sigma > 0$,which implies $\sigma > -1$.
Therefore,the theoretical range for Poisson's ratio is $-1 < \sigma < \frac{1}{2}$.

(A) The Poisson's ratio $(\sigma)$ is defined as the ratio of lateral strain to longitudinal strain.
For most materials,the theoretical range of Poisson's ratio is between $-1.0$ and $0.5$.
Specifically,for stable,isotropic,linear elastic materials,the value must be greater than $-1.0$ and less than or equal to $0.5$.
Since $0.7$ is greater than the maximum theoretical limit of $0.5$,it cannot be a valid value for Poisson's ratio.
Therefore,the correct option is $A$.

(B) The volume $V$ of a wire of radius $r$ and length $l$ is given by $V = \pi r^2 l$.
Since the volume remains unchanged during stretching,$dV = 0$.
Differentiating the volume equation,we get $dV = 2\pi rl dr + \pi r^2 dl = 0$.
Rearranging the terms,$2\pi rl dr = -\pi r^2 dl$.
This simplifies to $\frac{dr}{r} = -\frac{1}{2} \frac{dl}{l}$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{dr/r}{dl/l}$.
Substituting the value from our equation,$\sigma = -(-\frac{1}{2}) = 0.5$.

(D) The fractional change in volume $\frac{dV}{V}$ for a rod under longitudinal strain $\epsilon_l = \frac{dL}{L}$ is given by the formula: $\frac{dV}{V} = (1 - 2\sigma) \frac{dL}{L}$,where $\sigma$ is the Poisson's ratio.
Given $\sigma = 0.50$ and $\frac{dL}{L} = 2 \times 10^{-3}$.
Substituting these values: $\frac{dV}{V} = (1 - 2 \times 0.50) \times (2 \times 10^{-3})$.
$\frac{dV}{V} = (1 - 1) \times (2 \times 10^{-3}) = 0$.
Therefore,the percentage change in volume is $0\%$.

(C) The relationship between Young's modulus $(Y)$,modulus of rigidity $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
For any metal,the Poisson's ratio $(\sigma)$ lies between $0$ and $0.5$ (typically around $0.3$).
Since $(1 + \sigma) > 1$,it follows that $Y = 2\eta(1 + \sigma) > 2\eta$.
Therefore,$Y > \eta$ is always true for metals.

(A) The relation between Young's modulus $(Y)$,modulus of rigidity $(\eta)$,and Poisson's ratio $(\sigma)$ is given by $Y = 2\eta(1 + \sigma)$.
Given that there is no transverse strain,the Poisson's ratio $\sigma = 0$.
Substituting the values,we get $Y = 2\eta(1 + 0) = 2\eta$.
Therefore,$\eta = Y / 2$.
Given $Y = 6 \times 10^{12} \ N/m^2$,we have $\eta = (6 \times 10^{12}) / 2 = 3 \times 10^{12} \ N/m^2$.

(B) Given that Young's modulus $Y = 3\eta$,where $\eta$ is the modulus of rigidity.
We know the relation $Y = 2\eta(1 + \sigma)$,where $\sigma$ is the Poisson's ratio.
Substituting $Y = 3\eta$ into the relation: $3\eta = 2\eta(1 + \sigma) \implies 1.5 = 1 + \sigma \implies \sigma = 0.5$.
Now,we use the relation between Young's modulus $Y$,bulk modulus $K$,and Poisson's ratio $\sigma$: $Y = 3K(1 - 2\sigma)$.
Substituting $\sigma = 0.5$ into the equation: $Y = 3K(1 - 2(0.5)) = 3K(1 - 1) = 3K(0) = 0$.
However,solving for $K$: $K = \frac{Y}{3(1 - 2\sigma)}$.
As $\sigma \to 0.5$,the denominator $3(1 - 2(0.5)) = 0$.
Therefore,$K = \frac{Y}{0} = \infty$.

(D) The relationship between Young's modulus $(Y)$,modulus of rigidity $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Given that $Y = 2.4\eta$,we substitute this into the formula:
$2.4\eta = 2\eta(1 + \sigma)$.
Dividing both sides by $2\eta$,we get:
$1.2 = 1 + \sigma$.
Solving for $\sigma$:
$\sigma = 1.2 - 1 = 0.2$.
Therefore,the Poisson's ratio of the material is $0.2$.

(A) The volume $V$ of a wire of length $L$ and radius $r$ is given by $V = \pi r^2 L$.
Taking the logarithmic derivative,we get $\frac{dV}{V} = 2\frac{dr}{r} + \frac{dL}{L}$.
Since the volume remains unchanged,$\frac{dV}{V} = 0$,so $2\frac{dr}{r} = -\frac{dL}{L}$,which implies $\frac{dr/r}{dL/L} = -\frac{1}{2}$.
The Poisson's ratio $\sigma$ is defined as $\sigma = -\frac{\text{lateral strain}}{\text{longitudinal strain}} = -\frac{dr/r}{dL/L}$.
Substituting the value,we get $\sigma = -(-0.5) = 0.5$.

(C) The relationship between Young's modulus $(Y)$,bulk modulus $(K)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 3K(1 - 2\sigma)$.
Given:
$Y = 8 \times 10^{10} \ N/m^2$
$K = 10 \times 10^{10} \ N/m^2$
Substituting the values into the formula:
$8 \times 10^{10} = 3 \times (10 \times 10^{10}) \times (1 - 2\sigma)$
Divide both sides by $10^{10}$:
$8 = 30 \times (1 - 2\sigma)$
$8/30 = 1 - 2\sigma$
$0.2667 = 1 - 2\sigma$
Rearranging for $\sigma$:
$2\sigma = 1 - 0.2667$
$2\sigma = 0.7333$
$\sigma = 0.7333 / 2$
$\sigma \approx 0.3666 \approx 0.37$.

(A) Poisson's ratio $(\sigma)$ is defined as the ratio of lateral strain to longitudinal strain.
$\sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}}$
Given:
Longitudinal strain $(\frac{\Delta L}{L})$ = $0.05\% = 0.0005$
Poisson's ratio $(\sigma)$ = $0.4$
Lateral strain = $\sigma \times \text{Longitudinal strain}$
Lateral strain = $0.4 \times 0.05\% = 0.02\%$
Since the wire is stretched,the length increases,which causes the diameter to decrease.
Therefore,the diameter reduces by $0.02\%$.

(A) The fractional change in volume $\frac{dV}{V}$ is related to the longitudinal strain $\frac{dL}{L}$ and Poisson's ratio $\sigma$ by the formula: $\frac{dV}{V} = (1 - 2\sigma) \frac{dL}{L}$.
Given $\sigma = -\frac{1}{2}$,substituting this value into the equation gives: $\frac{dV}{V} = (1 - 2(-\frac{1}{2})) \frac{dL}{L} = (1 + 1) \frac{dL}{L} = 2 \frac{dL}{L}$.
Wait,let us re-evaluate the standard relation for volume change: $\frac{dV}{V} = (1 - 2\sigma) \frac{dL}{L}$.
If $\sigma = 0.5$,then $\frac{dV}{V} = 0$,which implies the material is incompressible.
However,if $\sigma = -0.5$,then $\frac{dV}{V} = (1 - 2(-0.5)) \frac{dL}{L} = 2 \frac{dL}{L}$.
Actually,for most materials,$-1 < \sigma < 0.5$. If $\sigma = -0.5$,the material is highly compressible. Given the options provided and the standard physics context where $\sigma = 0.5$ represents incompressibility,there is a common confusion in literature. If the question implies $\sigma = 0.5$,the answer is Incompressible. If the question strictly asks for $\sigma = -0.5$,it is Compressible.

(C) For a unit cube (side length $L = 1$),the tensile stress on each face is $\sigma_{stress} = \frac{F}{A} = \frac{F}{1^2} = F$.
The longitudinal strain along one axis due to the force on that axis is $\epsilon_{long} = \frac{F}{Y}$.
Due to the tensile forces on the other two perpendicular faces,there will be lateral strains. The lateral strain due to one perpendicular force is $-\sigma \epsilon_{long} = -\sigma \frac{F}{Y}$.
Since there are two such perpendicular faces,the total lateral strain is $2 \times (-\sigma \frac{F}{Y}) = -\frac{2\sigma F}{Y}$.
The net strain (change in length $\Delta L$ for $L=1$) is the sum of the longitudinal strain and the lateral strains:
$\Delta L = \frac{F}{Y} - \frac{2\sigma F}{Y} = \frac{F(1 - 2\sigma)}{Y}$.

(D) Given: Radius $r = 1\, mm = 10^{-3}\, m$,Length $L = 1\, m$,Force $F = 100\, N$,Young's modulus $Y = 2 \times 10^{11}\, N/m^2$,Poisson's ratio $\mu = \pi / 10$,Area $A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6}\, m^2$.
$1$. Calculate longitudinal strain $(\epsilon_L)$:
$\epsilon_L = \frac{\Delta L}{L} = \frac{F}{AY} = \frac{100}{(\pi \times 10^{-6}) \times (2 \times 10^{11})} = \frac{100}{2\pi \times 10^5} = \frac{1}{2\pi} \times 10^{-3}$.
$2$. Calculate lateral strain $(\epsilon_d)$:
Poisson's ratio $\mu = - \frac{\text{lateral strain}}{\text{longitudinal strain}} = - \frac{\Delta r / r}{\Delta L / L}$.
$\frac{\Delta r}{r} = - \mu \times \epsilon_L = - (\frac{\pi}{10}) \times (\frac{1}{2\pi} \times 10^{-3}) = - \frac{1}{20} \times 10^{-3} = - 0.05 \times 10^{-3} = - 5 \times 10^{-5}$.
$3$. Calculate change in radius $(\Delta r)$:
$\Delta r = - 5 \times 10^{-5} \times r = - 5 \times 10^{-5} \times 1\, mm = - 0.00005\, mm$.
$4$. New radius $(r')$:
$r' = r + \Delta r = 1\, mm - 0.00005\, mm = 0.99995\, mm$.

(A) Let the initial length be $L$ and the cross-sectional area be $A$. The volume $V = A \times L$.
Since the volume remains constant during stretching,$dV = 0$.
Taking the derivative,$d(AL) = A \cdot dL + L \cdot dA = 0$.
This implies $\frac{dA}{A} = -\frac{dL}{L}$.
The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain:
$\sigma = -\frac{dA/A}{dL/L}$.
Substituting the relation $\frac{dA}{A} = -\frac{dL}{L}$ into the formula:
$\sigma = -\frac{-dL/L}{dL/L} = 1$.
Wait,re-evaluating: For an incompressible material,the Poisson's ratio is exactly $0.5$.
Let longitudinal strain be $\epsilon_l = \frac{dL}{L}$ and lateral strain be $\epsilon_d = \frac{dr}{r}$.
Volume $V = \pi r^2 L$. Taking log and differentiating: $\frac{dV}{V} = 2\frac{dr}{r} + \frac{dL}{L} = 0$.
Thus,$\frac{dr}{r} = -\frac{1}{2} \frac{dL}{L}$.
Poisson's ratio $\sigma = -\frac{\epsilon_d}{\epsilon_l} = -\frac{-0.5 \epsilon_l}{\epsilon_l} = 0.5$.

(C) The volume $V$ of a wire of length $\ell$ and diameter $D$ is given by $V = \ell \cdot \frac{\pi D^2}{4}$.
Taking the differential,we get $dV = \frac{\pi D^2}{4} d\ell + \ell \cdot \frac{\pi}{2} D dD$.
Since the volume remains constant,$dV = 0$.
Therefore,$\frac{\pi D^2}{4} d\ell = -\ell \cdot \frac{\pi}{2} D dD$.
Rearranging the terms,we get $\frac{d\ell}{\ell} = -2 \frac{dD}{D}$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{dD/D}{d\ell/\ell}$.
Substituting the relation $\frac{dD}{D} = -\frac{1}{2} \frac{d\ell}{\ell}$,we get $\sigma = -\frac{-\frac{1}{2} \frac{d\ell}{\ell}}{\frac{d\ell}{\ell}} = 0.5$.

(C) For isotropic materials, the elastic moduli are related by Poisson's ratio $\sigma$ as follows:
$Y = 2n(1 + \sigma)$
$Y = 3k(1 - 2\sigma)$
Where $Y$ is Young's modulus, $n$ is shear modulus (rigidity), and $k$ is bulk modulus.
For most metals like aluminium and copper, the Poisson's ratio $\sigma$ lies between $0$ and $0.5$ (typically around $0.3$).
Since $0 < \sigma < 0.5$, we can derive the relationship:
From $Y = 2n(1 + \sigma)$, since $(1 + \sigma) > 1$, we get $Y > n$.
From $Y = 3k(1 - 2\sigma)$, since $(1 - 2\sigma) < 1$, we get $Y < 3k$, which implies $k > Y/3$.
Comparing the magnitudes for typical metals, the relationship is $n < Y < k$.
Therefore, the correct order is shear modulus < Young's modulus < bulk modulus.

(D) The relationship between Young's modulus $(Y)$,Bulk modulus $(K)$,and Modulus of rigidity $(\eta)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Rearranging this equation to solve for Poisson's ratio $(\sigma)$:
$Y = 2\eta + 2\eta\sigma$
$Y - 2\eta = 2\eta\sigma$
$\sigma = \frac{Y - 2\eta}{2\eta}$
$\sigma = \frac{Y}{2\eta} - \frac{2\eta}{2\eta}$
$\sigma = \frac{0.5Y}{\eta} - 1$
Alternatively,expressing it as $\sigma = \frac{0.5Y - \eta}{\eta}$ matches option $D$.

(N/A) Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain produced in a material when it is subjected to a tensile or compressive force.
When a tensile force is applied to a wire,its length increases (longitudinal strain),while its cross-sectional diameter decreases (lateral strain).
Conversely,when a compressive force is applied,the length decreases,and the diameter increases.
Mathematically,Poisson's ratio $(\mu)$ is given by:
$\mu = -\frac{\text{lateral strain}}{\text{longitudinal strain}}$
If the original diameter is $d$ and the change in diameter is $\Delta d$,the lateral strain is $\frac{\Delta d}{d}$. If the original length is $L$ and the change in length is $\Delta L$,the longitudinal strain is $\frac{\Delta L}{L}$.
Therefore,$\mu = -\frac{(\Delta d / d)}{(\Delta L / L)}$.
Since it is a ratio of two strains,it is a dimensionless and unitless quantity.
The magnitude of Poisson's ratio depends solely on the nature of the material of the body.

(N/A) Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain.
Let a cylindrical rod have length $l$ and radius $r$. When the rod is stretched,its length increases by $\Delta l$ and its radius decreases by $\Delta r$.
The lateral strain is $\epsilon_{lat} = -\frac{\Delta r}{r}$ and the longitudinal strain is $\epsilon_{long} = \frac{\Delta l}{l}$.
Poisson's ratio $\mu$ is given by $\mu = \frac{\epsilon_{lat}}{\epsilon_{long}} = -\frac{\Delta r / r}{\Delta l / l}$.
Thus,$\frac{\Delta r}{r} = -\mu \frac{\Delta l}{l} \quad \dots (1)$
The volume $V$ of the rod is $V = \pi r^2 l$.
Differentiating to find the change in volume:
$\frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta l}{l} \quad \dots (2)$
Substituting equation $(1)$ into $(2)$:
$\frac{\Delta V}{V} = 2(-\mu \frac{\Delta l}{l}) + \frac{\Delta l}{l} = \frac{\Delta l}{l} (1 - 2\mu)$
Since the volume of a material cannot increase when it is stretched $(\Delta V \ge 0)$,we must have $(1 - 2\mu) \ge 0$.
Therefore,$1 \ge 2\mu$,which implies $\mu \le 0.5$.

(D) For an isotropic elastic material,the relationship between Young's modulus $(Y)$,shear modulus $(G)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2G(1 + \sigma)$.
Rearranging this formula to solve for the shear modulus $(G)$,we get: $G = \frac{Y}{2(1 + \sigma)}$.
Therefore,both expressions represent the correct physical relationship between these elastic constants.

(B) Lateral strain is defined as the ratio of the change in the lateral dimension (such as diameter or width) of a body to its original lateral dimension when it is subjected to a longitudinal stress.
If a wire of diameter $D$ undergoes a change in diameter $\Delta D$ due to an applied force,the lateral strain is given by the formula: $\text{Lateral Strain} = \frac{\Delta D}{D}$.
It is dimensionless and is typically associated with the Poisson's ratio,which relates lateral strain to longitudinal strain.

(C) The relationship between Young's modulus $(Y)$,Bulk modulus $(K)$,and modulus of rigidity $(\eta)$ is derived using Poisson's ratio $(\sigma)$.
We know the relations:
$Y = 3K(1 - 2\sigma) \implies 1 - 2\sigma = \frac{Y}{3K} \implies 2\sigma = 1 - \frac{Y}{3K} \implies \sigma = \frac{1}{2} - \frac{Y}{6K} \dots (i)$
$Y = 2\eta(1 + \sigma) \implies 1 + \sigma = \frac{Y}{2\eta} \implies \sigma = \frac{Y}{2\eta} - 1 \dots (ii)$
Equating $(i)$ and $(ii)$:
$\frac{1}{2} - \frac{Y}{6K} = \frac{Y}{2\eta} - 1$
$1 + \frac{1}{2} = \frac{Y}{2\eta} + \frac{Y}{6K}$
$\frac{3}{2} = \frac{3KY + Y\eta}{6K\eta}$
$9K\eta = 3KY + Y\eta$
$9K\eta = Y(3K + \eta)$
$Y = \frac{9K\eta}{3K + \eta}$
Thus,the correct relation is $Y = \frac{9K\eta}{3K + \eta}$.

(A) The relationship between Young's modulus $(Y)$, modulus of rigidity $(\eta)$, and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Given that the Young's modulus is $2.4$ times its modulus of rigidity, we have: $Y = 2.4\eta$.
Equating the two expressions for $Y$:
$2.4\eta = 2\eta(1 + \sigma)$
Divide both sides by $2\eta$:
$1.2 = 1 + \sigma$
Solving for $\sigma$:
$\sigma = 1.2 - 1 = 0.2$.
Therefore, the Poisson's ratio is $0.2$.

(D) The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{\Delta r/r}{\Delta l/l} = 0.5$.
Given longitudinal strain $\frac{\Delta l}{l} = 3 \times 10^{-3}$.
Therefore, lateral strain $\frac{\Delta r}{r} = -\sigma \times \frac{\Delta l}{l} = -0.5 \times 3 \times 10^{-3} = -1.5 \times 10^{-3}$.
The volume of a cylindrical rod is $V = \pi r^2 l$.
Taking the logarithmic derivative, the fractional change in volume is $\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$.
Substituting the values: $\frac{\Delta V}{V} = 2(-1.5 \times 10^{-3}) + (3 \times 10^{-3}) = -3 \times 10^{-3} + 3 \times 10^{-3} = 0$.
Thus, the percentage increase in volume is $0\%$.

(D) The relationship between Young's modulus $(Y)$ and the modulus of rigidity (shear modulus,$\eta$) is given by the formula:
$Y = 2\eta(1 + \sigma)$
where $\sigma$ is the Poisson's ratio.
The practical range for Poisson's ratio $(\sigma)$ for most materials is $0 < \sigma \leq 0.5$.
Substituting the minimum value of $\sigma$ (which is $0$):
$Y = 2\eta(1 + 0) = 2\eta$
Substituting the maximum value of $\sigma$ (which is $0.5$):
$Y = 2\eta(1 + 0.5) = 2\eta(1.5) = 3\eta$
Since $Y$ ranges from $2\eta$ to $3\eta$ for valid materials,it is clear that $Y > \eta$ always holds true for elastic materials.

(A) Given:
Poisson's ratio $(\sigma)$ = $1/4 = 0.25$
Young's modulus $(Y)$ = $8 \times 10^{10} \, N/m^2$
Lateral strain = $0.02 \% = 0.02 / 100 = 2 \times 10^{-4}$
Formula for Poisson's ratio:
$\sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}}$
Calculating longitudinal strain $(\epsilon)$:
$\epsilon = \frac{\text{Lateral strain}}{\sigma} = \frac{2 \times 10^{-4}}{0.25} = 8 \times 10^{-4}$
Elastic potential energy per unit volume $(u)$:
$u = \frac{1}{2} \times Y \times \epsilon^2$
Substituting the values:
$u = \frac{1}{2} \times (8 \times 10^{10}) \times (8 \times 10^{-4})^2$
$u = 4 \times 10^{10} \times 64 \times 10^{-8}$
$u = 256 \times 10^2 = 2.56 \times 10^4 \, J/m^3$

(A) The relationships between Young's modulus $(Y)$,bulk modulus $(K)$,modulus of rigidity $(\eta)$,and Poisson's ratio $(\sigma)$ are given by:
$Y = 2\eta(1+\sigma)$ --- $(1)$
$Y = 3K(1-2\sigma)$ --- $(2)$
Equating the two expressions for $Y$:
$2\eta(1+\sigma) = 3K(1-2\sigma)$
$2\eta + 2\eta\sigma = 3K - 6K\sigma$
$2\eta\sigma + 6K\sigma = 3K - 2\eta$
$\sigma(6K + 2\eta) = 3K - 2\eta$
$\sigma = \frac{3K - 2\eta}{6K + 2\eta}$

(B) Given: Length $L = 2 \ m$,Young's modulus $Y = 2.0 \times 10^{11} \ N/m^2$,Poisson's ratio $\mu = 0.2$,and transverse strain $\epsilon_t = \frac{\Delta r}{r} = 10^{-3}$.
Poisson's ratio is defined as $\mu = -\frac{\text{transverse strain}}{\text{longitudinal strain}} = -\frac{\epsilon_t}{\epsilon_l}$.
Taking the magnitude,$\epsilon_l = \frac{\epsilon_t}{\mu} = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}$.
The elastic potential energy density $u$ is given by $u = \frac{1}{2} Y \epsilon_l^2$.
Substituting the values: $u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (5 \times 10^{-3})^2$.
$u = 1.0 \times 10^{11} \times 25 \times 10^{-6} = 25 \times 10^5 \ J/m^3$.
Thus,the value is $25$.

(B) The relationship between Young's modulus $(Y)$,rigidity modulus $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Given that $Y = 2.6\eta$,we substitute this into the equation:
$2.6\eta = 2\eta(1 + \sigma)$.
Dividing both sides by $2\eta$,we get:
$1.3 = 1 + \sigma$.
Solving for $\sigma$:
$\sigma = 1.3 - 1 = 0.3$.
Therefore,the Poisson's ratio is $0.3$.

(D) The relationship between Young's modulus $(Y)$,modulus of rigidity $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula:
$Y = 2 \eta (1 + \sigma)$
Therefore,the correct relation is $Y = 2 \eta (1 + \sigma)$.

(D) The relationship between Young's modulus $(Y)$,rigidity modulus $(\eta)$,and Poisson's ratio $(\sigma)$ is given by the formula: $Y = 2\eta(1 + \sigma)$.
Given that $Y = 2.4\eta$,we substitute this into the equation:
$2.4\eta = 2\eta(1 + \sigma)$.
Dividing both sides by $2\eta$,we get:
$1.2 = 1 + \sigma$.
Solving for $\sigma$:
$\sigma = 1.2 - 1 = 0.2$.
Therefore,the Poisson's ratio is $0.2$.

(B) The longitudinal strain is given by $\epsilon_L = \frac{\Delta L}{L} = \frac{F}{AY} = \frac{Mg}{\pi r^2 Y}$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{\epsilon_D}{\epsilon_L} = -\frac{\Delta r / r}{\Delta L / L}$.
Therefore,the lateral strain is $\frac{\Delta r}{r} = -\sigma \epsilon_L$.
The magnitude of the decrease in radius is $\Delta r = r \sigma \epsilon_L$.
Substituting the value of $\epsilon_L$: $\Delta r = r \sigma \left( \frac{Mg}{\pi r^2 Y} \right) = \frac{Mg \sigma}{\pi r Y}$.

(D) The Poisson’s ratio $(\sigma)$ for a homogeneous isotropic material is theoretically constrained by the requirement that the Young’s modulus $(Y)$, shear modulus $(G)$, and bulk modulus $(K)$ must all be positive values.
This leads to the theoretical range for Poisson’s ratio being $-1.0 < \sigma < 0.5$.
Most common materials have a Poisson’s ratio between $0$ and $0.5$.
Since $0.8$ lies outside this valid range, it cannot be the value of Poisson’s ratio for a homogeneous isotropic material.

(B) The Poisson's ratio $\sigma$ is defined as the negative ratio of transverse strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L}$.
For a wire of length $L$ and diameter $D$,the volume $V = \frac{\pi D^2 L}{4}$.
Since the volume $V$ remains constant,$\frac{\Delta V}{V} = 2\frac{\Delta D}{D} + \frac{\Delta L}{L} = 0$.
This implies $2\frac{\Delta D}{D} = -\frac{\Delta L}{L}$,or $\frac{\Delta D/D}{\Delta L/L} = -0.5$.
Substituting this into the definition of Poisson's ratio: $\sigma = -(-0.5) = 0.5$.
Therefore,for a material with constant volume under stretching,the Poisson's ratio is $0.5$.
Thus,option $(B)$ is correct.

(B) The longitudinal strain is given by $\epsilon_L = 0.2 \% = 0.002$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$ to longitudinal strain $\epsilon_L$,so $\epsilon_d = -\sigma \cdot \epsilon_L$.
Here,$\sigma = 0.3$,so the lateral strain is $\epsilon_d = -0.3 \times 0.002 = -0.0006$.
The volume strain $\frac{\Delta V}{V}$ for a wire is given by the sum of longitudinal strain and two lateral strains: $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_d$.
Substituting the values: $\frac{\Delta V}{V} = 0.002 + 2(-0.0006) = 0.002 - 0.0012 = 0.0008$.
Converting to percentage: $0.0008 \times 100 \% = 0.08 \%$.

(C) Length of steel wire,$l = 3 \ m$.
Increased length,$\Delta l = 0.3 \ cm = 0.3 \times 10^{-2} \ m = 3 \times 10^{-3} \ m$.
Poisson's ratio $(\sigma)$ $= 0.26$.
By definition,Poisson's ratio is the ratio of lateral strain to longitudinal strain:
$\sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = 0.26$.
Longitudinal strain $= \frac{\Delta l}{l} = \frac{3 \times 10^{-3} \ m}{3 \ m} = 10^{-3}$.
Therefore,Lateral strain $= 0.26 \times \text{Longitudinal strain} = 0.26 \times 10^{-3}$.

(D) Given,Poisson's ratio,$\sigma = 0.5$.
Longitudinal strain,$\frac{\Delta l}{l} = 2 \times 10^{-3}$.
The volumetric strain $\left(\frac{\Delta V}{V}\right)$ is related to the longitudinal strain $\left(\frac{\Delta l}{l}\right)$ by the formula:
$\frac{\Delta V}{V} = (1 - 2\sigma) \frac{\Delta l}{l}$
Substituting the given values:
$\frac{\Delta V}{V} = (1 - 2 \times 0.5) \times 2 \times 10^{-3}$
$\frac{\Delta V}{V} = (1 - 1) \times 2 \times 10^{-3} = 0 \times 2 \times 10^{-3} = 0$
Therefore,the percentage change in volume is $\frac{\Delta V}{V} \times 100 = 0 \times 100\% = 0\%$.

(C) Given: Length of the wire,$L = 10 \ m$,diameter,$D = 0.6 \times 10^{-3} \ m$,Poisson's ratio,$\sigma = 0.3$,and change in wire length,$\Delta L = 6 \times 10^{-3} \ m$.
The Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain:
$\sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{\Delta D / D}{\Delta L / L}$
Rearranging the formula to solve for the change in diameter $\Delta D$:
$\Delta D = \sigma \times D \times \frac{\Delta L}{L}$
Substituting the given values:
$\Delta D = 0.3 \times (0.6 \times 10^{-3} \ m) \times \frac{6 \times 10^{-3} \ m}{10 \ m}$
$\Delta D = 0.3 \times 0.6 \times 10^{-3} \times 6 \times 10^{-4} \ m$
$\Delta D = 1.08 \times 10^{-7} \ m = 10.8 \times 10^{-8} \ m$
Therefore,the change in the diameter of the wire is $10.8 \times 10^{-8} \ m$. The correct option is $C$.

(D) The fractional change in volume $\frac{\Delta V}{V}$ is given by the formula: $\frac{\Delta V}{V} = \epsilon_l (1 - 2\sigma)$,where $\epsilon_l$ is the longitudinal strain and $\sigma$ is the Poisson's ratio.
Given,$\epsilon_l = 2 \times 10^{-3}$ and $\sigma = 0.5$.
Substituting these values into the formula:
$\frac{\Delta V}{V} = (2 \times 10^{-3}) \times (1 - 2 \times 0.5)$
$\frac{\Delta V}{V} = (2 \times 10^{-3}) \times (1 - 1)$
$\frac{\Delta V}{V} = (2 \times 10^{-3}) \times 0 = 0$.
Therefore,the percentage change in volume is $0 \times 100 = 0\%$.

(A) Given: Tension $F = 22 \,N$,Area $A = 0.02 \,cm^2 = 0.02 \times 10^{-4} \,m^2$,Young's modulus $Y = 1.1 \times 10^{11} \,N/m^2$,Poisson's ratio $\sigma = 0.32$.
Longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{22}{(0.02 \times 10^{-4}) \times (1.1 \times 10^{11})} = \frac{22}{2.2 \times 10^4} = 10^{-4}$.
Poisson's ratio is defined as $\sigma = -\frac{\Delta r/r}{\Delta l/l}$. The lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 \times 10^{-4}$.
The area is $A = \pi r^2$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Substituting the values,$\frac{\Delta A}{A} = 2 \times (-0.32 \times 10^{-4}) = -0.64 \times 10^{-4}$.
The decrease in area is $\Delta A = |\frac{\Delta A}{A}| \times A = (0.64 \times 10^{-4}) \times (0.02 \,cm^2) = 1.28 \times 10^{-6} \,cm^2$.

(B) The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L}$.
Since the area $A = \pi r^2$,the fractional change in area is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Given $\frac{\Delta A}{A} = -2\% = -0.02$,we have $2 \frac{\Delta r}{r} = -0.02$,so $\frac{\Delta r}{r} = -0.01$.
The lateral strain is $\frac{\Delta r}{r} = -0.01$.
Using the Poisson's ratio formula: $\sigma = -\frac{\Delta r/r}{\Delta L/L}$.
$0.4 = -\frac{-0.01}{\Delta L/L}$.
$\frac{\Delta L}{L} = \frac{0.01}{0.4} = 0.025$.
Therefore,the percentage increase in length is $0.025 \times 100 = 2.5 \%$.