Poisson's Ratio and relation between Modulus Questions in English

(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 $0.025 \times 100 = 2.5 \%$.

(B) The relationship between the elastic moduli is derived from the theory of elasticity. For an isotropic material,the Young's modulus $(Y)$,bulk modulus $(K)$,and modulus of rigidity $(\eta)$ are related by the equation:
$\frac{9}{Y} = \frac{3}{\eta} + \frac{1}{K}$.
This formula allows us to calculate one elastic constant if the other two are known.

(D) Young's modulus is given by $Y = \frac{F/A}{\Delta l/l}$, so the longitudinal strain is $\frac{\Delta l}{l} = \frac{F}{YA}$.
Given $F = 22 \,N$, $Y = 1.1 \times 10^{11} \,N/m^2$, and $A = 0.01 \,cm^2 = 10^{-6} \,m^2$.
Calculating longitudinal strain: $\frac{\Delta l}{l} = \frac{22}{1.1 \times 10^{11} \times 10^{-6}} = 2 \times 10^{-4}$.
Poisson's ratio $\sigma$ is defined as $\sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = \frac{\Delta r/r}{\Delta l/l}$.
Thus, $\frac{\Delta r}{r} = \sigma \cdot \frac{\Delta l}{l} = 0.32 \times 2 \times 10^{-4} = 6.4 \times 10^{-5}$.
The area of the cross-section is $A = \pi r^2$. Taking the derivative, the fractional change in area is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Substituting the values: $\frac{\Delta A}{A} = 2 \times 6.4 \times 10^{-5} = 12.8 \times 10^{-5}$.
Therefore, the decrease in area is $\Delta A = (12.8 \times 10^{-5}) \times (0.01 \,cm^2) = 1.28 \times 10^{-6} \,cm^2$.

(C) The relationship between Young's modulus $(Y)$, rigidity modulus $(\eta)$, and Poisson's ratio $(\sigma)$ is given by $Y = 2\eta(1 + \sigma)$.
Given $Y / \eta = 2.8$, we have $2(1 + \sigma) = 2.8$, which implies $1 + \sigma = 1.4$, so $\sigma = 0.4$.
The volume of the wire $V = A \cdot L$ remains constant, so $dV/V = dA/A + dL/L = 0$, which means $dL/L = -dA/A$.
Given the decrease in area is $2 \%$, $dA/A = -0.02$, so $dL/L = 0.02$ (a $2 \%$ increase in length).
However, considering the lateral strain $\epsilon_l = -\sigma \cdot \epsilon_L$, where $\epsilon_l = (dA/A)/2 = -0.01$.
Thus, $-0.01 = -0.4 \cdot \epsilon_L$, which gives $\epsilon_L = 0.01 / 0.4 = 0.025$.
Therefore, the percentage change in length is $2.5 \%$.

(A) Given:
Area of cross-section $A = 0.01 \ cm^2 = 10^{-6} \ m^2$.
Tension $T = 22 \ N$.
Young's modulus $Y = 1.1 \times 10^{11} \ N \ m^{-2}$.
Poisson ratio $\sigma = 0.32$.
Longitudinal strain $\frac{\Delta l}{l} = \frac{T}{AY} = \frac{22}{10^{-6} \times 1.1 \times 10^{11}} = \frac{22}{1.1 \times 10^5} = 20 \times 10^{-5}$.
Lateral strain $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 \times 20 \times 10^{-5} = -6.4 \times 10^{-5}$.
Since $A = \pi r^2$,the change in area is $\Delta A = 2\pi r \Delta r$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Percentage change in area $= |2 \times (-6.4 \times 10^{-5})| \times 100 = 12.8 \times 10^{-3} \%$.

(C) Let the original radius of the wire be $r$ and its length be $L$. The area of cross-section is $A = \pi r^2$.
Taking the derivative,$\Delta A = 2 \pi r \Delta r$.
Poisson's ratio $\sigma$ is defined as $\sigma = -\frac{\Delta r / r}{\Delta L / L}$,so $\frac{\Delta r}{r} = -\sigma \frac{\Delta L}{L}$.
Substituting this into the area change equation: $\Delta A = 2 \pi r^2 (-\sigma \frac{\Delta L}{L}) = -2 A \sigma \frac{\Delta L}{L}$.
Taking the magnitude,$|\Delta A| = 2 A \sigma \frac{\Delta L}{L}$,which gives $\frac{\Delta L}{L} = \frac{\Delta A}{2 A \sigma}$.
Young's modulus $Y$ is defined as $Y = \frac{F/A}{\Delta L/L}$,so $F = Y A \frac{\Delta L}{L}$.
Substituting $\frac{\Delta L}{L}$: $F = Y A (\frac{\Delta A}{2 A \sigma}) = \frac{Y \Delta A}{2 \sigma}$.

(A) Given: Tension $F = 20 \,N$,Area $A = 0.01 \,cm^2 = 10^{-6} \,m^2$,Young's modulus $Y = 1.1 \times 10^{11} \,N/m^2$,Poisson's ratio $\sigma = 0.32$.
The longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{20}{10^{-6} \times 1.1 \times 10^{11}} = \frac{20}{1.1 \times 10^5} \approx 1.818 \times 10^{-4}$.
Poisson's ratio is defined as $\sigma = -\frac{\Delta r/r}{\Delta l/l}$,so the lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 \times 1.818 \times 10^{-4} \approx -0.5818 \times 10^{-4}$.
The area is $A = \pi r^2$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
The decrease in area is $\Delta A = 2 \times A \times \sigma \times \frac{\Delta l}{l} = 2 \times 10^{-6} \times 0.32 \times 1.818 \times 10^{-4} \approx 1.16 \times 10^{-10} \,m^2$.
Converting to $cm^2$: $1.16 \times 10^{-10} \times (10^2 \,cm)^2 = 1.16 \times 10^{-6} \,cm^2$.

(A) Poisson's ratio $(\sigma) = \frac{\text{lateral strain}}{\text{longitudinal strain}}$.
Given lateral strain $= 0.01 \times 10^{-3} \ m$ (Note: Strain is dimensionless,assuming the value represents the change in diameter $\Delta D$ or similar,but here we use it directly as the numerator for the ratio).
$\sigma = \frac{\text{lateral strain}}{\Delta L / L} = 0.4$.
$\frac{\Delta L}{L} = \frac{0.01 \times 10^{-3}}{0.4} = 0.025 \times 10^{-3} = 2.5 \times 10^{-5}$.
Young's modulus $Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}$.
$Y = \frac{100 \ N}{0.025 \ m^2 \times (2.5 \times 10^{-5})} = \frac{100}{0.0625 \times 10^{-5}} = \frac{100}{6.25 \times 10^{-7}} = 16 \times 10^7 = 1.6 \times 10^8 \ N/m^2$.

(C) Given: The fractional change in length is $\frac{\Delta l}{l} = -0.01$ (since it decreases by $1 \%$).
Poisson's ratio, $\sigma = 0.2$.
For a rod, the volume $V = A \times l$, where $A$ is the cross-sectional area.
The fractional change in volume is given by $\frac{\Delta V}{V} = \frac{\Delta A}{A} + \frac{\Delta l}{l}$.
Since $A = w \times t$ (width $\times$ thickness), $\frac{\Delta A}{A} = \frac{\Delta w}{w} + \frac{\Delta t}{t}$.
By definition of Poisson's ratio, $\sigma = -\frac{\Delta w / w}{\Delta l / l} = -\frac{\Delta t / t}{\Delta l / l}$.
Thus, $\frac{\Delta w}{w} = -\sigma \frac{\Delta l}{l}$ and $\frac{\Delta t}{t} = -\sigma \frac{\Delta l}{l}$.
Substituting these into the volume equation:
$\frac{\Delta V}{V} = -\sigma \frac{\Delta l}{l} - \sigma \frac{\Delta l}{l} + \frac{\Delta l}{l} = \frac{\Delta l}{l} (1 - 2\sigma)$.
Substituting the values: $\frac{\Delta V}{V} = -0.01 \times (1 - 2 \times 0.2) = -0.01 \times (1 - 0.4) = -0.01 \times 0.6 = -0.006$.
Therefore, the volume decreases by $0.6 \%$.

(D) The longitudinal strain is given by $\epsilon_L = 2 \times 10^{-3}$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$ to longitudinal strain $\epsilon_L$,i.e.,$\sigma = -\frac{\epsilon_d}{\epsilon_L}$.
Therefore,the lateral strain is $\epsilon_d = -\sigma \epsilon_L = -0.50 \times (2 \times 10^{-3}) = -1 \times 10^{-3}$.
The volumetric strain $\frac{\Delta V}{V}$ is given by the sum of the longitudinal strain and two lateral strains: $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_d$.
Substituting the values: $\frac{\Delta V}{V} = (2 \times 10^{-3}) + 2(-1 \times 10^{-3}) = 2 \times 10^{-3} - 2 \times 10^{-3} = 0$.
Since the volumetric strain is $0$,the percentage change in volume is $0$.

(D) The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{dD/D}{dL/L}$.
Given $\sigma = 0.5$,we have $0.5 = -\frac{dD/D}{dL/L}$,which implies $\frac{dD}{D} = -0.5 \frac{dL}{L}$.
The cross-sectional area $A$ is given by $A = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Taking the logarithmic derivative,$\frac{dA}{A} = 2 \frac{dD}{D}$.
Substituting the expression for $\frac{dD}{D}$,we get $\frac{dA}{A} = 2 (-0.5 \frac{dL}{L}) = -\frac{dL}{L}$.
Given that the area decreases by $4 \%$,we have $\frac{dA}{A} = -0.04$.
Therefore,$-0.04 = -\frac{dL}{L}$,which means $\frac{dL}{L} = 0.04$ or $4 \%$.
Thus,the percentage increase in length is $4 \%$.

(D) Given that stress $\sigma_{s} = \frac{1}{100} Y$,where $Y$ is the Young's modulus.
We know that $Y = \frac{\text{stress}}{\text{longitudinal strain}} = \frac{\sigma_{s}}{\Delta l / l}$.
Substituting the value of stress: $Y = \frac{Y / 100}{\Delta l / l} \implies \frac{\Delta l}{l} = \frac{1}{100} = 0.01$.
Poisson's ratio $\nu = -\frac{\Delta w / w}{\Delta l / l} = -\frac{\Delta t / t}{\Delta l / l} = 0.3$.
Thus,the lateral strain $\frac{\Delta w}{w} = \frac{\Delta t}{t} = -\nu \left( \frac{\Delta l}{l} \right) = -0.3 \times 0.01 = -0.003$.
The volume of a rectangular rod is $V = l \times w \times t$.
The fractional change in volume is $\frac{\Delta V}{V} \approx \frac{\Delta l}{l} + \frac{\Delta w}{w} + \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta V}{V} = 0.01 + (-0.003) + (-0.003) = 0.01 - 0.006 = 0.004$.
Therefore,the percentage change in volume is $0.004 \times 100 \% = 0.4 \%$.