The design of a physical instrument requires that there be a constant difference in length of $10$ cm between an iron rod and a copper cylinder laid side by side at all temperatures. If ${\alpha }_{Fe}=11\times {10}^{-6}{{}^{\circ }\mathrm{C}}^{-1}$ and  ${\alpha }_{cu}=17\times {10}^{-6}{{}^{\circ }\mathrm{C}}^{-1}$ , their lengths are:

Since a constant difference in length of $10 \mathrm{cm}$ between an iron rod and a copper cylinder is required therefore

$L_{Fe}-L_{\mathrm{Cu}}=10 \mathrm{cm}$    …(i)

$\Delta L_{Fe}-\Delta L_{Cu}=O$

$\therefore \Delta L_{Fe}=\Delta L_{Cu}$

Linear expansion of iron rod = Linear expansion of copper cylinder

$\Rightarrow L_{Fe}\times {\alpha }_{Fe}\times \Delta T=L_{Cu}\times {\alpha }_{Cu}\times \Delta T$

$\Rightarrow \frac{L_{Fe}}{L_{\mathrm{Cu}}}=\frac{{\alpha }_{\mathrm{Cu}}}{{\alpha }_{Fe}}=\frac{17}{11}$

$\therefore \frac{L_{Fe}}{L_{Cu}}=\frac{17}{11}$    …(ii)

From (i) and (ii)

$L_{Fe}=28.3 \mathrm{cm},L_{\mathrm{Cu}}=18.3 \mathrm{cm}$