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A brass rod of length $50 \mathrm{~cm}$ and diameter $3.0 \mathrm{~mm}$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} \mathrm{C}$, if the original lengths are at $40.0^{\circ} \mathrm{C}$ ? Is there a thermal stress developed at the function ? The ends of the rod are free to expand (coefficient of linear expansion of brass
$=2.0 \times 10^{-5} \mathbf{K}^{-1}$, steel $\left.=1.2 \times 10^{-5} \mathbf{K}^{-1}\right)$
$=2.0 \times 10^{-5} \mathbf{K}^{-1}$, steel $\left.=1.2 \times 10^{-5} \mathbf{K}^{-1}\right)$
Solution:
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Verified Answer
$\Delta l_1=l_2 \alpha_1 \Delta T=50 \times\left(2.1 \times 10^{-5}\right) \times(250-40)$ $=0.2205 \mathrm{~cm}$
$$
\begin{aligned}
&\Delta l_2=l_2 \alpha_2 \Delta \mathrm{T}=50 \times\left(1.2 \times 10^{-5}\right) \times(250-40) \\
&=0.126 \mathrm{~cm}
\end{aligned}
$$
$\therefore$ Change in length of the combined rod $=\Delta l_1+\Delta l_2=0.2205+0.126=0.3465 \mathrm{~cm}$.
$$
\begin{aligned}
&\Delta l_2=l_2 \alpha_2 \Delta \mathrm{T}=50 \times\left(1.2 \times 10^{-5}\right) \times(250-40) \\
&=0.126 \mathrm{~cm}
\end{aligned}
$$
$\therefore$ Change in length of the combined rod $=\Delta l_1+\Delta l_2=0.2205+0.126=0.3465 \mathrm{~cm}$.
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