Question

Two elastic rods are joined between fixed supports as shown in the figure. If there is no change in the lengths of the individual rods with increase in temperature, then choose the correct option.
[${\alpha }_1=2{\alpha }_2,Y_1=Y_2$ ($Y$ is Young's modulus, $\alpha$ is the linear expansion coefficient and $A$ is the cross-sectional area of rod)]

• A. $\frac{A_1}{A_2}=\frac{2}{1}$
• B. $\frac{A_1}{A_2}=\frac{1}{2}$
• C. $\frac{A_1}{A_2}=\frac{1}{4}$
• D. $\frac{A_1}{A_2}=4$

Answer 1

The correct option is B $\frac{A_1}{A_2}=\frac{1}{2}$

Young's modulus of elasticity is
$Y=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$
Strain produced due to thermal expansion
$\frac{\Delta L}{L}=\alpha \Delta T$
$\Rightarrow Y=\frac{F}{A\alpha \Delta T}\Rightarrow F=YA\alpha \Delta T$

Condition for no change in length of the rod is
Tension in rod 1 = Tension in rod 2
i.e $Y_1{A}_1{\alpha }_1\Delta T=Y_2{A}_2{\alpha }_2\Delta T$
$[Y_1=Y_2,{\alpha }_1=2{\alpha }_2]$
$\Rightarrow \frac{A_1}{A_2}=\frac{Y_2}{Y_1}\frac{{\alpha }_2}{{\alpha }_1}$
$\Rightarrow \frac{A_1}{A_2}=\frac{Y_2}{Y_2}\times \frac{{\alpha }_2}{2{\alpha }_2}=\frac{1}{2}$
Thus, option (b) is the correct answer.