91Ó°ÊÓ

1. The temperature rise,$\mathrm{Δ°Õ}={32}^0\mathrm{C}$
2. The fixed distance of the rod,${\mathrm{L}}_0=3.77\text{″¾}$
3. The coefficient of linear expansion,$\mathrm{Î\pm }=25×{10}^{−6}{/}^0\mathrm{C}$

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. Due to the temperature rise, the bar moves upwards and its height changes from the mean position i.e. from the center. Let, that change in height be, x. Also, thermal expansion takes place due to the temperature rise. To calculate the value of x, consider a right-angled triangle as shown in the figure. By using the Pythagoras theorem and thermal expansion coefficient, x can be found.

In this figure, x is the change in height. “±ô’ is the length after the temperature increases, and l₀is the original length i.e.${\mathbf{l}}_{\mathbf{0}}\mathbf{=}{\mathbf{L}}_{\mathbf{0}}\mathbf{/}\mathbf{2}\mathbf{.}$

Formula:

The linear expansion of a body, $\mathrm{Δ³¢}=\mathrm{³¢Î\pm }\mathrm{Δ°Õ}$ …(¾±)

Where$\mathrm{Î\pm }$ is the coefficient of linear expansion of body.

Consider half of the bar. The length of the half bar will be 1.88 m. and its length after increasing the temperature can be given using equation (1)Then, by using Pythagoras theorem in the right angled triangle we get,

$\begin{array}{c}{\mathrm{x}}^2={\mathrm{l}}^2−{{\mathrm{l}}_0}^2\\ ={{\mathrm{l}}_0}^2{\left(1+\mathrm{Î\pm }∆\mathrm{T}\right)}^2−{{\mathrm{l}}_0}^2\text{(fromequation(i))}\end{array}$

Since the change in length is small, therefore${(1+\mathrm{Î\pm Δ°Õ})}^2$can be differentiated and replaced by $(1+2\mathrm{Î\pm Δ°Õ})$.

$\begin{array}{c}{\mathrm{x}}^2={{\mathrm{l}}_0}^2+2{{\mathrm{l}}_0}^2\mathrm{Î\pm Δ°Õ}−{{\mathrm{l}}_0}^2\\ =2{{\mathrm{l}}_0}^2\mathrm{Î\pm Δ°Õ}\end{array}$

i.e.,$\mathrm{x}={\mathrm{l}}_0\sqrt{2\mathrm{Î\pm Δ°Õ}}$

Substituting the given values in the above formula, we get

$\begin{array}{c}\mathrm{x}=(1.88\text{m})\sqrt{2(25×{10}^{−6}{\text{/}}^{\text{0}}\text{C})\left({32}^0\mathrm{C}\right)}\\ =7.5×{10}^{−2}\text{″¾}\end{array}$

Hence, the value of the rise is $7.5×{10}^{−2}m$