We would like to prepare a scale whose length does not change wit… - Vidyadip

Question

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say $10cm$. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If $a_{iron} = 1.2 \times 10^{-5}/ K$ and $a_{brass} = 1.8 \times 10^{-5}/ K$, what should we take as length of each strip?

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Answer

According to the problem, $L_1- L_b = 10cm$ where, $L_1$ = length of iron scale $L_b$ = Length of brass scale This condition is possible if change in length both the rods is remain same at all temperatures. Change in length of iron rod,$\Delta\text{L}=\alpha_{\text{I}}\text{L}_{\text{I}}\Delta\text{T}$
Change in length of brass rod,$\Delta\text{L}=\alpha_\text{B}\text{L}_\text{B}\Delta\text{T}$
As the cahange will equal in both the rods, so$\alpha_1\text{L}_1\Delta\text{T}=\alpha_\text{B}\text{L}_\text{B}\Delta\text{T}$
$\Rightarrow\alpha_1\text{L}_1=\alpha_\beta\text{L}_\beta$
$\Rightarrow\frac{\text{L}_1}{\text{L}_\text{B}}=\frac{\alpha_\text{B}}{\alpha_\text{I}}$
Here, $\alpha_\beta=1.8\times10^{-5}\text{K}^{-1},\alpha_\text{I}=1.2\times10^{-5}\text{K}^{-1}$$\therefore\ \frac{\text{L}_\text{I}}{\text{L}_\text{B}}=\frac{1.8\times10^{-5}}{1.2\times10^{-5}}=\frac{3}{2}$
$\text{L}_1=\frac{3}{2}\text{L}_\text{B}$
As, $\text{L}_\text{I}-\text{L}_\text{B}=10\text{cm}$$\therefore\ \frac{3}{2}\text{L}_\text{B}-\text{L}_\text{B}=10$
$\Rightarrow\frac{1}{2}\text{L}_\text{B}=10$
$\Rightarrow \text{L}_\text{B}=20{\text{cm}}$

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