Subjective Type

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 10 cm. 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 what should we take as the length of each strip?

Solution

Required scale can be made with the help of iron and brass rod so that their one end are at A coincide and the other ends remain at distance PB = 10 cm at any temperature. Letinitial length of iron and brass rods at any temperature T are and respectively.
$$\begin{aligned}&L_{11}-L_{1 B}=10 \mathrm{cm}\\&\therefore \alpha=\frac{\Delta L}{L_{0} \Delta T} \text { or } \alpha=\frac{L_{2}-L_{1}}{L_{1} \Delta T}\\&L_{2}=L_{1}=L_{1} \alpha \Delta T\\&L_{2}=L_{1}(1+\alpha \Delta T)\end{aligned}$$
If the combination of rods are heated by temperature $$\Delta T$$, then length becomes $$L_{21}$$ and $$L_{2 B}$$
$$L_{2 I}-L_{2 B}=10 \mathrm{cm}$$ (again according to question)
$$L_{11}\left(1+\alpha_{1} \Delta T\right)-L_{1 B}\left(1+\alpha_{B} \Delta T\right)=10$$
$$L_{11}+\alpha 1 L_{11} \Delta T-L_{1 B}-L_{1 B} \alpha_{B} \Delta T=10$$
$$L_{11}-L_{1 B}+\left(\alpha_{1} L_{11}-\alpha_{B} L_{1 B}\right) \Delta T=10$$
$$10+\left(\alpha_{1} L_{11}-\alpha_{B} L_{1 B}\right) \Delta T=10[\text { from }(i)]$$
Or $$\alpha_{1} L_{11}-\alpha_{B} L_{1 B}=0$$
$$\alpha_{1} L_{11}=\alpha_{B} L_{18}$$
$$\frac{L_{11}}{L_{1 B}}=\frac{\alpha_{B}}{\alpha_{1}}=\frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}=\frac{18}{12}=\frac{3}{2}$$
$$\therefore \frac{L_{11}}{L_{1 B}}=\frac{3}{2}$$
Then let $$\quad L_{11}=3 x$$ and $$L_{18}=2 x$$
$$L_{11}-L_{1 B}=10$$
$$3 x-2 x=10$$
$$x=10$$

$$\therefore$$ Length of iron $$\operatorname{rod}=3 \times 10=30 \mathrm{cm}$$
Length of brass $$\operatorname{rod}=2 \times 10=20 \mathrm{cm}$$
The difference between IInd end remain = 10 cm.

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