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 ${\alpha _{iron}}$ $= 1.2 \times 10^{-5}\,K^{-1}$ and ${\alpha _{brass}}$ $= 1.8 \times 10^{-5}\,K^{-1}$ what should we take as length of each strip ?
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 ${\alpha _{iron}}$ $= 1.2 \times 10^{-5}\,K^{-1}$ and ${\alpha _{brass}}$ $= 1.8 \times 10^{-5}\,K^{-1}$ what should we take as length of each strip ?
According to question, $l_{\text {iron }}-l_{\text {brass }}=10 \mathrm{~cm} \ldots$ $(1)$
Constant at all temperature.
Suppose length at $0^{\circ} \mathrm{C}$ is $l_{0}$ and it becomes $l$ due to increase in temperature $\Delta \mathrm{T}$
$\therefore l_{\text {iron }}=l_{0 \text { iron }}\left(1+\alpha_{\text {iron }} \Delta \mathrm{T}\right) \text { and }$
$l_{\text {brass }}=l_{0 \text { brass }}\left(1+\alpha_{\text {brass }} \Delta \mathrm{T}\right)$
$\therefore$ From equation $(1), l_{0 \text { iron }}\left(1+\alpha_{\text {iron }} \Delta \mathrm{T}\right)-l_{0 \text { brass }}\left(1+\alpha_{\text {brass }} \Delta \mathrm{T}\right)=10 \mathrm{~cm}$ But the difference of their lengths remains constant,
$\therefore l_{\text {0iron }} \alpha_{\text {iron }}=l_{\text {0brass }} \alpha_{\text {brass }}$
$\therefore \frac{l_{0 \text { iron }}}{l_{0 \text { brass }}}=\frac{\alpha_{\text {brass }}}{\alpha_{\text {iron }}}=\frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}=\frac{3}{2}$
$\therefore l_{\text {0iron }}=\frac{3}{2} l_{0 \text { brass }}$
Now, from equation $(1)$,
$\frac{3}{2} l_{0 \mathrm{brass}}-l_{0 \mathrm{brass}}=10$
$\therefore \frac{1}{2} l_{0 \mathrm{brass}}=10$
$\therefore l_{0 \mathrm{brass}}=20 \mathrm{~cm}$
$\therefore$ Original length of brass scale is $10 \mathrm{~cm}$.
Now, again from equation $(1)$,
$l_{\text {0iron }}-l_{0 \text { brass }}=10$
$l_{\text {0iron }}-20=10$
$\therefore$ $l_{\text {0iron }}=30 \mathrm{~cm}$
$\therefore$ Original length of iron scale is $30 \mathrm{~cm}$.
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