Question

Two elastic rods are joined between fixed supports as shown in figure. Condition for no change in the lengths of the individual rods with increase in temperature -
[Given $\alpha \to$ linear expansion coefficient, $A\to$ area of cross-section of rod, $Y\to$ Young's modulus and $L\to$ length of rod]

• A. $\frac{A_1}{A_2}=\frac{{\alpha }_1}{{\alpha }_2}\frac{Y_1}{Y_2}$
• B. $\frac{A_1}{A_2}=\frac{L_1{\alpha }_1{Y}_1}{L_2{\alpha }_2{Y}_2}$
• C. $\frac{A_1}{A_2}=\frac{L_2{\alpha }_2{Y}_2}{L_1{\alpha }_1{Y}_1}$
• D. $\frac{A_1}{A_2}=\frac{{\alpha }_2{Y}_2}{{\alpha }_1{Y}_1}$

Answer 1

The correct option is D $\frac{A_1}{A_2}=\frac{{\alpha }_2{Y}_2}{{\alpha }_1{Y}_1}$

As per question,


For no change in the lengths of individual rods, the compressive forces in both rods due to thermal expansion should be equal. Then, the net force on each rod will be zero.

$A_1{Y}_1{\alpha }_1\Delta T=A_2{Y}_2{\alpha }_2\Delta T$
[ from formula of linear expansion and Young's modulus, also, temperature difference is same ]

$\Rightarrow A_1{Y}_1{\alpha }_1=A_2{Y}_2{\alpha }_2$

$\Rightarrow \frac{A_1}{A_2}=\frac{{\alpha }_2{Y}_2}{{\alpha }_1{Y}_1}$

Hence, option $\left(\text{D}\right)$ is the right choice.