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A copper rod of length $l_1$ and an iron rod of length $l_2$ are always maintained at the same common temperature $T$. If the difference $(l_2 -l_1)$ is $15\,cm$ and is independent of the value of $T,$ the $l_1$ and $l_2$ have the values (given the linear coefficient of expansion for copper and iron are $2.0 \times 10^{-6}\,C^{-1}$ and $1.0\times10^{-6}\,C^{ -1}$ respectively)
$l_1 = 15\,cm,\,\,l_2 = 30\,cm$
$l_1 = 30\,cm,\,\,l_2 = 15\,cm$
$l_1 = 10\,cm,\,\,l_2 = 25\,cm$
$l_1 = 25\,cm,\,\,l_2 = 10\,cm$
Solution
Suppose the length of the copper rod at $0^{\circ} \mathrm{C}$ is $\ell_{\mathrm{co}}$ and that of the iron rod at $0^{\circ} \mathrm{C}$ is $\ell_{\mathrm{io}}$. The lengths at temperature $\theta$ become $\ell_{c \theta}=\ell_{co}\left(1+\alpha_{c} \theta\right)$ and $\ell_{\mathrm{i \theta}}=\ell_{\mathrm{io}}\left(1+\alpha_{i} \theta\right)$
Subtracting $\ell_{\mathrm{i \theta}}-\ell_{c \theta}=\left(\ell_{\mathrm{io}}-\ell_{\mathrm{co}}\right)+\left(\ell_{\mathrm{io}} \alpha_{i}-\ell_{\mathrm{co}} \alpha_{\mathrm{c}}\right) \theta \ldots(i)$
We now desire that $\left(\ell_{i f}-\ell_{c \theta}\right)=\left(\ell_{\text {io }}-\ell_{c o}\right)=10 \mathrm{cm}$
at all values of $\theta .$ For this, the term in $\theta$ in equation
$(1)$ must vanish identically. (i.e.,) $\left(\ell_{\text {io }} \alpha_{i}-\ell_{\text {co }} \alpha_{c}\right)=0$
or $\frac{\ell_{\text {io }}}{\ell_{\text {co }}}=\frac{\alpha_{c}}{\alpha_{i}}$
$\frac{\ell_{\mathrm{io}}}{\left(\ell_{\mathrm{io}}-\ell_{\mathrm{co}}\right)}=\frac{\alpha_{\mathrm{c}}}{\left(\alpha_{\mathrm{c}}-\alpha_{\mathrm{i}}\right)}=\frac{17 \times 10^{-6}}{6 \times 10^{-6}}=\frac{17}{6}$
Hence $\ell_{\text {io }}=\frac{17}{6}\left(\ell_{\text {io }}-\ell_{co}\right)=\frac{17}{6} \times 10=\frac{170}{6}$
$=28.3 \mathrm{cm}$ and $\ell_{\mathrm{co}}=18.3 \mathrm{cm}$
