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A rail track made of steel having length $10 \mathrm{~m}$ is clamped on a railway line at its two ends (figure). On a summer day due to rise in temperature by $20^{\circ} \mathrm{C}$. It is deformed as shown in figure. Find $x$ (displacement of the centre) if $\alpha_{\text {steel }}$ $=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$.
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Verified Answer
As given that, $\alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
$$
L=10 \mathrm{~m}, \Delta t=20^{\circ} \mathrm{C}
$$
Let us consider the diagram.
By Pythagoras theorem in right angled triangle in figure.
$$
\begin{aligned}
&\left(\frac{L+\Delta L}{2}\right)^2=\left(\frac{L}{2}\right)^2+x^2 \Rightarrow x=\sqrt{\left(\frac{L+\Delta L}{2}\right)^2-\left(\frac{L}{2}\right)^2} \\
&=\frac{1}{2} \sqrt{(L+\Delta L)^2-L^2}=\frac{1}{2} \sqrt{\left(L^2+\Delta L^2+2 L \Delta L\right)-L^2} \\
&=\frac{1}{2} \sqrt{\left(\Delta L^2+2 L \Delta L\right)}
\end{aligned}
$$
As increase in length $\Delta L$ is very small as compared to original length.
i.e. $\Delta L^2 \ll L$, neglecting $(\Delta L)^2$, we get
$$
x=\frac{1}{2} \times \sqrt{2 L \Delta L}
$$
We know that, $\Delta L=L \alpha \Delta T$
$$
\text { So, } \begin{aligned}
x &=\frac{1}{2} \sqrt{2 L \times L \alpha \Delta t}=\frac{1}{2} L \sqrt{2 \alpha \Delta t} \\
&=\frac{10}{2} \times \sqrt{2 \times 1.2 \times 10^{-5} \times 20}=5 \times \sqrt{4 \times 1.2 \times 10^{-4}} \\
&=5 \times 2 \times 1.1 \times 10^{-2}=0.11 \mathrm{~m}=11 \mathrm{~cm}
\end{aligned}
$$
$$
L=10 \mathrm{~m}, \Delta t=20^{\circ} \mathrm{C}
$$
Let us consider the diagram.
By Pythagoras theorem in right angled triangle in figure.
$$
\begin{aligned}
&\left(\frac{L+\Delta L}{2}\right)^2=\left(\frac{L}{2}\right)^2+x^2 \Rightarrow x=\sqrt{\left(\frac{L+\Delta L}{2}\right)^2-\left(\frac{L}{2}\right)^2} \\
&=\frac{1}{2} \sqrt{(L+\Delta L)^2-L^2}=\frac{1}{2} \sqrt{\left(L^2+\Delta L^2+2 L \Delta L\right)-L^2} \\
&=\frac{1}{2} \sqrt{\left(\Delta L^2+2 L \Delta L\right)}
\end{aligned}
$$
As increase in length $\Delta L$ is very small as compared to original length.
i.e. $\Delta L^2 \ll L$, neglecting $(\Delta L)^2$, we get
$$
x=\frac{1}{2} \times \sqrt{2 L \Delta L}
$$
We know that, $\Delta L=L \alpha \Delta T$
$$
\text { So, } \begin{aligned}
x &=\frac{1}{2} \sqrt{2 L \times L \alpha \Delta t}=\frac{1}{2} L \sqrt{2 \alpha \Delta t} \\
&=\frac{10}{2} \times \sqrt{2 \times 1.2 \times 10^{-5} \times 20}=5 \times \sqrt{4 \times 1.2 \times 10^{-4}} \\
&=5 \times 2 \times 1.1 \times 10^{-2}=0.11 \mathrm{~m}=11 \mathrm{~cm}
\end{aligned}
$$
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