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

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(A) Let the lengths of the copper and iron rods at $0^{\circ}C$ be $L_{c0}$ and $L_{i0}$ respectively.The lengths at temperature $T$ are given by $l_1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the lengths of the copper and iron rods at $0^{\circ}C$ be $L_{c0}$ and $L_{i0}$ respectively.The lengths at temperature $T$ are given by $l_1 = L_{c0}(1 + \alpha_c T)$ and $l_2 = L_{i0}(1 + \alpha_i T)$.The difference is $l_2 - l_1 = L_{i0} - L_{c0} + (L_{i0}\alpha_i - L_{c0}\alpha_c)T$.For the difference to be independent of $T$,the coefficient of $T$ must be zero:$L_{i0}\alpha_i - L_{c0}\alpha_c = 0 \implies L_{i0}\alpha_i = L_{c0}\alpha_c$.Given $\alpha_c = 2.0 	imes 10^{-6}\,^{\circ}C^{-1}$ and $\alpha_i = 1.0 	imes 10^{-6}\,^{\circ}C^{-1}$,we have $L_{i0}(1.0 	imes 10^{-6}) = L_{c0}(2.0 	imes 10^{-6})$,so $L_{i0} = 2L_{c0}$.We are given $l_2 - l_1 = 15\,cm$,which implies $L_{i0} - L_{c0} = 15\,cm$.Substituting $L_{i0} = 2L_{c0}$,we get $2L_{c0} - L_{c0} = 15\,cm$,so $L_{c0} = 15\,cm$ and $L_{i0} = 30\,cm$.Since the difference is constant,$l_1 = L_{c0} = 15\,cm$ and $l_2 = L_{i0} = 30\,cm$.

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