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(B) Let the lengths of the two rods at temperature $T$ be $\ell_1(T)$ and $\ell_2(T)$.At a temperature $T + \Delta T$,the new lengths are $\ell_1' = \

Vedclass · Accepted Answer

$\frac{\ell_1}{\ell_2} = \frac{\alpha_2}{\alpha_1}$

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(B) Let the lengths of the two rods at temperature $T$ be $\ell_1(T)$ and $\ell_2(T)$.At a temperature $T + \Delta T$,the new lengths are $\ell_1' = \ell_1(1 + \alpha_1 \Delta T)$ and $\ell_2' = \ell_2(1 + \alpha_2 \Delta T)$.The difference between the lengths is $\Delta \ell = \ell_1 - \ell_2$.For the difference to be independent of temperature,the change in length of both rods must be equal,i.e.,$\Delta \ell_1 = \Delta \ell_2$.Therefore,$\ell_1 \alpha_1 \Delta T = \ell_2 \alpha_2 \Delta T$.This simplifies to $\ell_1 \alpha_1 = \ell_2 \alpha_2$.Rearranging the terms,we get $\frac{\ell_1}{\ell_2} = \frac{\alpha_2}{\alpha_1}$.

Two rods of different materials have lengths $\ell_1$ and $\ell_2$ whose coefficients of linear expansion are $\alpha_1$ and $\alpha_2$ respectively. If the difference between the two lengths is independent of temperature,then:

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