91Ó°ÊÓ

1. Length of composite bar, L = 52.4cm or 0.524m
2. Coefficient of linear expansion for composite bar, $\mathrm{Î\pm }=1.3×{10}^{\text{- 5}}{\text{/}}^{\text{0}}\text{C}$
3. Coefficient of linear expansion of steel, ${\text{α}}_{\text{1}}=\text{11}×{\text{10}}^{\text{-6}}{\text{/}}^{\text{0}}\text{C}$
4. Coefficient of linear expansion of brass, ${\mathrm{Î\pm }}_2=19×{10}^{-6}{\text{/}}^{\text{0}}\text{C}$.

The propensity of matter to alter its form, area, volume, and density in reaction to a change in temperature is known as thermal expansion. As we see in the given problem, there are two different metals whose coefficients are known to us, and from the given length of the bar, we can find the lengths of the two metals in the strip. Also, we can derive the formula for the coefficient of linear expansion of composite bar.

Formula:

Linear expansion of the body due to thermal expansion, $\mathrm{Δ}L=\mathrm{Î\pm }L\mathrm{Δ}T$…(¾±)

Where $\mathrm{Î\pm }$is coefficient of thermal expansion, L is the length and is the change in temperature.

The bar is made of two metals of length L₁ & L₂ and their expansions are $\mathrm{Δ}{L}_1\&\mathrm{Δ}{L}_2$respectively.

So, the total expansion of the length using equation (i) is given as follows. Further, solving for the coefficient of the linear expansion, we have

$\begin{array}{c}\mathrm{Δ}L=\mathrm{Δ}{L}_1+\mathrm{Δ}{L}_2\\ \mathrm{Î\pm }L\mathrm{Δ}T={\mathrm{Î\pm }}_1{L}_1\mathrm{Δ}T+{\mathrm{Î\pm }}_2{L}_2\mathrm{Δ}T\\ \mathrm{Î\pm }=\frac{{\mathrm{Î\pm }}_1{L}_1+{\mathrm{Î\pm }}_2{L}_2}{L}\mathrm{.................................}\text{(ii})\end{array}$

Hence, the value of the coefficient of linear expansion is $\frac{{\mathrm{Î\pm }}_1{L}_1+{\mathrm{Î\pm }}_2{L}_2}{L}$.

From the values of coefficient of linear expansion of steel and brass, using equation (ii), we can get the relation of lengths as

$$\begin{gathered} 13×{10}^{-6}{\text{/}}^{\text{0}}\text{C}=\frac{\left(11×{10}^{-6}{\text{/}}^{\text{0}}\text{C}\right)L_1+\left(19×{10}^{-6}{\text{/}}^{\text{0}}\text{C}\right)L_2}{L_1+L_2} \\ 13L_1+13L_2=11L_1+19L_2 \\ 2L_1=6L_2 \\ {L}_1=3L_2..........................\left(\text{iii}\right) \end{gathered}$$

Since we know L₁+L₂=52.4cm,we can write the value of the length, L₂ as

role="math" localid="1663070205639" $\begin{array}{c}{L}_2=52.4−L_1\\ L_2=52.4−3L_2\text{(fromequation(iii))}\\ 4L_2=52.4\\ L_2=13.1\text{ c³¾}\end{array}$

Substituting the value of length L₂ in equation (iii), we can get the length L₁ as

$$\begin{gathered} L_1=(3×13.1) \text{cm} \\ =39.3 \text{cm} \end{gathered}$$

Hence, the value of the required length is 39.3cm.

From the calculations of part (b), we get the value of the length L₂ as 13.1cm.