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 × 10^-5/ K and a_brass = 1.8 × 10^-5/ K, what should we take as length of each strip?

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Answer

According to the problem, L₁- L_b = 10cm where,

L₁ = 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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