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(B) Let the lengths of the rods at temperature $T$ be $L_1$ and $L_2$. At a temperature $T + \Delta T$,the new lengths are $L_1' = L_1(1 + \alpha_1 \D

Vedclass · Accepted Answer

$\frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1}$

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(B) Let the lengths of the rods at temperature $T$ be $L_1$ and $L_2$. At a temperature $T + \Delta T$,the new lengths are $L_1' = L_1(1 + \alpha_1 \Delta T)$ and $L_2' = L_2(1 + \alpha_2 \Delta T)$. The difference between the lengths is independent of temperature,so $L_1' - L_2' = L_1 - L_2$. Substituting the expressions: $L_1(1 + \alpha_1 \Delta T) - L_2(1 + \alpha_2 \Delta T) = L_1 - L_2$. Expanding this,we get $L_1 + L_1 \alpha_1 \Delta T - L_2 - L_2 \alpha_2 \Delta T = L_1 - L_2$. Subtracting $(L_1 - L_2)$ from both sides,we get $L_1 \alpha_1 \Delta T = L_2 \alpha_2 \Delta T$. Since $\Delta T 
eq 0$,we have $L_1 \alpha_1 = L_2 \alpha_2$. Therefore,$\frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1}$.

The rods of length $L_1$ and $L_2$ are made of materials 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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