Question

Two rods of different materials and of equal cross-sectional areas and length $1\text{m}$ are joined face to face at one end and their free ends are fixed to rigid walls. If the temperature of the surrounding is increased by $30{}^{\circ }\text{C}$, the magnitude of the displacement of the joint of the rods is :
$[{\alpha }_1=2\times {10}^{-5}{}^{\circ }{\text{C}}^{-1},{\alpha }_2=5\times {10}^{-5}{}^{\circ }{\text{C}}^{-1},$
$Y_1=2\times {10}^{10}{\text{N/m}}^2,Y_2={10}^{10}{\text{N/m}}^2]$

• A. $15\times {10}^{-5}\text{m}$
• B. $12\times {10}^{-5}\text{m}$
• C. $10\times {10}^{-5}\text{m}$
• D. $18\times {10}^{-5}\text{m}$

Answer 1

The correct option is C $10\times {10}^{-5}\text{m}$

Given:
Length of each rod, $L=1\text{m}$
Change in temperature, $\Delta T=30{}^{\circ }\text{C}$
Coefficient of expansion of rod $1$, ${\alpha }_1=2\times {10}^{-5}/{}^{\circ }\text{C}$
Young's modulus of rod $1$, $Y_1=2\times {10}^{10}{\text{N/m}}^2$
Coefficient of expansion of rod $2$, ${\alpha }_2=5\times {10}^{-5}/{}^{\circ }\text{C}$
Young's modulus of rod $2$, $Y_2={10}^{10}{\text{N/m}}^2$
To find:
Magnitude of displacement of the joint when temperature is increased, $x=?$


Net expansion of rod $1=$ expansion due to heating $+$ displacement of joint
$\Delta L_1=L{\alpha }_1\Delta T+x$ .........(1)
[ from formula of linear expansion ]
Similarly, net expansion of rod $2=$ expansion due to heating $-$ displacement of joint
$\Delta L_2=L{\alpha }_2\Delta T-x$ .........(2)

We know that for equilibrium of joint, stress due to rod $1$ on joint should be equal to stress due to rod $2$ on joint.
$\Rightarrow Y_1{ϵ}_1=Y_2{ϵ}_2$
where $ϵ$ is the strain.
[ from formula of Young's modulus ]
$\Rightarrow Y_1\times \frac{\Delta L_1}{L}=Y_2\times \frac{\Delta L_2}{L}$
[ from formula of strain ]
$\Rightarrow Y_1\times (L{\alpha }_1\Delta T+x)=Y_2\times (L{\alpha }_2\Delta T-x)$
[ from (1) and (2) ]
$\Rightarrow x=\left[\frac{Y_2{\alpha }_2-Y_1{\alpha }_1}{Y_1+Y_2}\right]L\Delta T$
$\Rightarrow x=\left[\frac{{10}^{10}\times 5\times {10}^{-5}-2\times {10}^{10}\times 2\times {10}^{-5}}{{10}^{10}+2\times {10}^{10}}\right]\times 1\times 30$
$\Rightarrow x=10\times {10}^{-5}\text{m}$