Question

A steel wire of uniform cross - sectional area $2m{m}^2$ is heated up to ${50}^{\circ }C$ and is stretched by clamping its two ends rigidly. The change in tension when the temperature falls from ${50}^{\circ }$ C to ${30}^{\circ }$C is
$(\text{coefficient of linear expansion}\alpha =1.1\times {10}^{-5}{/}^{\circ }C,\text{Youngs modulus}Y=2.0\times {10}^{11}\frac{N}{m^2})$

• A. $88N$
• B. $1.5\times {2.0}^{10}N$
• C. $5N$
• D. $2.5\times {2.0}^{10}N$

Answer 1

The correct option is A $88N$

As the temperature is reduced from ${50}^{\circ }C$ to ${30}^{\circ }C$ rod should contract. But in this case the rod is clamped at both the ends, so stress is created in the rod.

From Stress strain relationship
$\sigma \propto ϵ$
$\sigma =Yϵ$
$\sigma =Y\frac{l}{\Delta l}=\frac{F}{A}$
Here $\Delta l$ is change in length due to temperature change
Hence $\Delta l=l\alpha \Delta T$
Change in tension:
$F=YA\frac{l}{\Delta l}=YA\alpha \Delta T$

$=2\times {10}^{11}\times 2\times \times {10}^{-6}\times 1.1\times {10}^{-5}\times (50-30)$

$=88N$