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There is some change in length when a $33000 \mathrm{~N}$ tensile force is applied on a steel rod of area of cross-section $10^{-3} \mathrm{~m}^2$. The change of temperature required to produce the same elongation, if the steel rod is heated, is (The modulus of elasticity is $3 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and the coefficient of linear expansion of steel is $\left.1.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}\right)$.
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$10^{\circ} \mathrm{C}$
$\begin{aligned} & \text { Modulus of elasticity }=\frac{\text { Force }}{\text { Area }} \times \frac{l}{\Delta l} \\ & \qquad \begin{aligned} 3 \times 10^{11} & =\frac{33000}{10^{-3}} \times \frac{l}{\Delta l} \\ \frac{\Delta l}{l} & =\frac{33000}{10^{-3}} \times \frac{1}{3 \times 10^{11}} \\ & =11 \times 10^{-5}\end{aligned}\end{aligned}$
Change in length, $\frac{\Delta l}{l}=\alpha \Delta T$
$$
\begin{aligned}
11 \times 10^{-5} & =1.1 \times 10^{-5} \times \Delta T \\
\Rightarrow \quad \Delta T & =10 \mathrm{~K} \text { or } 10^{\circ} \mathrm{C}
\end{aligned}
$$
Change in length, $\frac{\Delta l}{l}=\alpha \Delta T$
$$
\begin{aligned}
11 \times 10^{-5} & =1.1 \times 10^{-5} \times \Delta T \\
\Rightarrow \quad \Delta T & =10 \mathrm{~K} \text { or } 10^{\circ} \mathrm{C}
\end{aligned}
$$
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