Question

A bar of length $L_0=3.77\text{m}$ is placed on two rigid supports as shown in the figure. As a result of rise in temperature by $32{}^{\circ }\text{C}$ , bar cracks at its centre and buckles upward as shown in the figure. The fixed distance between the supports is also $L_0=3.77\text{m}$ and coefficient of linear expansion of the bar is $25\times {10}^{-6}/{}^{\circ }\text{C}$. Find the rise $x$ of centre of the bar.

• A. $0.075\text{m}$
• B. $0.065\text{m}$
• C. $0.055\text{m}$
• D. $0.045\text{m}$

Answer 1

The correct option is A $0.075\text{m}$

Given:
Change in temperature, $\Delta T=32{}^{\circ }\text{C}$
Length of bar, $L_0=3.77\text{m}$
Coefficient of linear expansion $\alpha =25\times {10}^{-6}/{}^{\circ }\text{C}$
Let half - length of the bar $\left(\frac{L_o}{2}\right)$is expanded by $\Delta x$ then,
$\Delta x=\frac{L_0}{2}\times \alpha \Delta T$
$\Rightarrow \Delta x=\frac{3.77}{2}\times 25\times {10}^{-6}\times 32=1.5\times {10}^{-3}\text{m}$

Now, from figure, we can say that ,
$x=\sqrt{{(\frac{L_0}{2}+\Delta x)}^2-{\left(\frac{L_0}{2}\right)}^2}$
$\Rightarrow x=\sqrt{{(\frac{3.77}{2}+1.5\times {10}^{-3})}^2-{\left(\frac{3.77}{2}\right)}^2}$
$\Rightarrow x=0.075\text{m}$
Hence, option $\left(a\right)$ is correct.