A copper rod of length l_1 and an iron rod of | vedclass

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Question

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_{\m

Vedclass · Accepted Answer

$l_1 = 15\,cm,\,\,l_2 = 30\,cm$

Vedclass · Answer

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 $	heta$ become $\ell_{c 	heta}=\ell_{co}\left(1+\alpha_{c} 	hetaight)$ and $\ell_{\mathrm{i 	heta}}=\ell_{\mathrm{io}}\left(1+\alpha_{i} 	hetaight)$

Subtracting $\ell_{\mathrm{i 	heta}}-\ell_{c 	heta}=\left(\ell_{\mathrm{io}}-\ell_{\mathrm{co}}ight)+\left(\ell_{\mathrm{io}} \alpha_{i}-\ell_{\mathrm{co}} \alpha_{\mathrm{c}}ight) 	heta \ldots(i)$

We now desire that $\left(\ell_{i f}-\ell_{c 	heta}ight)=\left(\ell_{	ext {io }}-\ell_{c o}ight)=10 \mathrm{cm}$

at all values of $	heta .$ For this, the term in $	heta$ in equation

$(1)$ must vanish identically. (i.e.,) $\left(\ell_{	ext {io }} \alpha_{i}-\ell_{	ext {co }} \alpha_{c}ight)=0$

or $\frac{\ell_{	ext {io }}}{\ell_{	ext {co }}}=\frac{\alpha_{c}}{\alpha_{i}}$

$\frac{\ell_{\mathrm{io}}}{\left(\ell_{\mathrm{io}}-\ell_{\mathrm{co}}ight)}=\frac{\alpha_{\mathrm{c}}}{\left(\alpha_{\mathrm{c}}-\alpha_{\mathrm{i}}ight)}=\frac{17 	imes 10^{-6}}{6 	imes 10^{-6}}=\frac{17}{6}$

Hence $\ell_{	ext {io }}=\frac{17}{6}\left(\ell_{	ext {io }}-\ell_{co}ight)=\frac{17}{6} 	imes 10=\frac{170}{6}$

$=28.3 \mathrm{cm}$ and $\ell_{\mathrm{co}}=18.3 \mathrm{cm}$

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