Elasticity

If a rod or wire of length $L$ and cross-section area $A$ under the action of stretching force $F$ applied normally to its face suffers a increase $\Delta L$ in length, then

$ Y=\frac{\text { tensile stress }}{\text { longitudinal strain }}=\frac{F / A}{\frac{\Delta L}{L}} $

Hence, Young's modulus is proportionality constant that relates

the force per unit area applied perpendicularly at the surface of an object to the fractional change in length.

For metal wire $A, L_A=L, R_A=R$

For metal wire $B, L_B=3 L, R_B=2 R$ When the ends of the of metal wire $A$ and $B$ are joined and pulled with a constant force $F$ after fixing the one end of combined system, then change in length of wire $A$ and $B$ is same.

$ \begin{array}{ll} \text { i.e., } & \Delta L_A=\Delta L_B \\ \Rightarrow & \frac{F L_A}{A_A Y_A}=\frac{F L_B}{A_B Y_B} \\ \Rightarrow & \frac{F L}{\pi R_A^2 Y_1}=\frac{F \times 3 L}{\pi R_B^2 Y_2} \\ \Rightarrow & \frac{1}{\pi R^2 Y_A}=\frac{3}{\pi(2 R)^2 Y_B} \\ \Rightarrow & \frac{1}{\pi R^2 Y_A}=\frac{3}{4 \pi R^2 Y_B} \\ \Rightarrow & \frac{1}{Y_A}=\frac{3}{4 Y_B} \Rightarrow \frac{Y_B}{Y_A}=\frac{3}{4} \end{array} $

As $Y=$ Young's modulus of material

$ \begin{aligned} & \qquad=\frac{F / A}{\Delta L / L_0} \\ & \Rightarrow \quad \Delta L=\frac{L_0 F}{Y A} \Rightarrow \quad L-L_0=\frac{L_0 F}{Y A} \\ & \Rightarrow \quad L=L_0+\frac{L_0 F}{Y A} \\ & \text { Here, } L_1=L_0+\frac{L_0 T_1}{Y A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \quad L_2=L_0+\frac{L_0 T_2}{Y A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & \left(L_1-L_0\right) T_2=\frac{L_0 T_1 T_2}{Y A} \text { and } \\ & \left(L_2-L_0\right) T_1=\frac{L_0 T_1 T_2}{Y A} \end{aligned} $

So, equating we have

$ \begin{array}{lr} \Rightarrow & \left(L_1-L_0\right) T_2=\left(L_2-L_0\right) T_1 \\ \Rightarrow & L_1 T_2-L_2 T_1=L_0\left(T_2-T_1\right) \\ \Rightarrow & L_0=\frac{L_1 T_2-L_2 T_1}{T_2-T_1} \end{array} $

The given situation is shown in the following figure.

$ \begin{aligned} & \text { Shear strain }=\frac{x}{l}=\frac{0.2 \mathrm{~mm}}{50 \mathrm{~cm}}=\frac{0.2 \times 10^{-3}}{50 \times 10^{-2}} \\ & \text { Shear stress }=\frac{F}{A}=\frac{10^5}{\frac{50}{100} \times \frac{10}{100}} \\ & =\frac{10^5 \times 10^4}{5 \times 10^2}=2 \times 10^6 \\ & \text { Modulus of rigidity }=\frac{\text { Stress }}{\text { Strain }} \\ & =\frac{2 \times 10^6 \times 50 \times 10^{-2}}{0.2 \times 10^{-3}} \\ & =5 \times 10^9 \mathrm{~N} / \mathrm{m}^2=5 \mathrm{GPa} \end{aligned} $

Given, linear dimensions increase by a factor of 8.

So, volume will become (8)³ times and area of cross-section will become (8)² times.

Now, $stress = {{weight} \over {area}}$

Weight = (volume $\times$ density $\times$ g) will also become (8)³ times.

$\therefore$ New stress $ = {{{{(8)}^3} \times {w_0}} \over {{{(8)}^2} \times {A_0}}} = 8\left( {{{{w_0}} \over {{A_0}}}} \right)$

Hence, the stress increases by a factor of 8.