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 $30 \mathrm{~cm}$.