Question

Three rods of equal length of $1\text{m}$ and equal cross-sectional areas are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at $0{}^{\circ }\text{C}$. Find the length of the system when the temperature is raised to $500{}^{\circ }\text{C}$.
$[{\alpha }_A=2\times {10}^{-5}{}^{\circ }{\text{C}}^{-1},{\alpha }_B=3\times {10}^{-5}{}^{\circ }{\text{C}}^{-1},$
$Y_A={10}^{10}{\text{N/m}}^2,Y_B=2\times {10}^{10}{\text{N/m}}^2]$

• A. $1.0125\text{m}$
• B. $1.125\text{m}$
• C. $1.25\text{m}$
• D. $1.00125\text{m}$

Answer 1

The correct option is A $1.0125\text{m}$

Given:
Length of each rods, $l_0=1\text{m}$
Change in temperature, $\Delta T=500-0=500{}^{\circ }\text{C}$
Coefficient of linear expansion of rod $A$, ${\alpha }_A=2\times {10}^{-5}{}^{\circ }{\text{C}}^{-1}$
Young's modulus of rod $A$, $Y_A={10}^{10}{\text{N/m}}^2$
Coefficient of linear expansion of rod $B$, ${\alpha }_B=3\times {10}^{-5}{}^{\circ }{\text{C}}^{-1}$
Young's modulus of rod $B$, $Y_B=2\times {10}^{10}{\text{N/m}}^2$
To find:
Length of the system, $l=?$


We know that,
Total strain $=$ Total stress $/$ Total Young's modulus
$\Rightarrow \frac{\Delta l}{l_0}=\frac{2\times Y_A\times \frac{\Delta l_A}{l_0}+Y_B\times \frac{\Delta l_B}{l_0}}{2Y_A+Y_B}$
[ from formula of Young's modulus ]

$\Rightarrow \frac{\Delta l}{l_0}=\frac{2\times Y_A\times \frac{l_0{\alpha }_A\Delta T}{l_0}+Y_B\times \frac{l_0{\alpha }_B\Delta T}{l_0}}{2Y_A+Y_B}$
[ from formula of linear expansion ]

$\Rightarrow \frac{\Delta l}{l_0}=\frac{2Y_A{\alpha }_A\Delta T+Y_B{\alpha }_B\Delta T}{2Y_A+Y_B}$
$\Rightarrow \Delta l=\left[\frac{2Y_A{\alpha }_A\Delta T+Y_B{\alpha }_B\Delta T}{2Y_A+Y_B}\right]l_0$
$\therefore l=l_0+\Delta l=[1+\frac{2Y_A{\alpha }_A\Delta T+Y_B{\alpha }_B\Delta T}{2Y_A+Y_B}]l_0$
$=[1+\frac{2\times {10}^{10}\times 2\times {10}^{-5}\times 500+2\times {10}^{10}\times 3\times {10}^{-5}\times 500}{2\times {10}^{10}+2\times {10}^{10}}]\times 1$
$=1.0125\text{m}$