Two rods, one made of aluminium and the other | vedclass

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Question

$\mathrm{L}_{\mathrm{t}}=\mathrm{L}_{0}(1+\alpha \mathrm{t})$

$\Delta \mathrm{L}=\mathrm{L}_{\mathrm{t}}-\mathrm{L}_{0}=\mathrm{L}_{0} \alpha \math

Vedclass · Accepted Answer

$\frac{\alpha_2}{(\alpha_1+\alpha_2)}$

Vedclass · Answer

$\mathrm{L}_{\mathrm{t}}=\mathrm{L}_{0}(1+\alpha \mathrm{t})$

$\Delta \mathrm{L}=\mathrm{L}_{\mathrm{t}}-\mathrm{L}_{0}=\mathrm{L}_{0} \alpha \mathrm{t}$

If the length of each rod increases by same amount

on raising the temperature to $t,$ then

$l_{1} \alpha_{1} t=l_{2} \alpha_{2} t$

or $\quad l_{1} \alpha_{1}=l_{2} \alpha_{2}$

or $\quad \frac{l_{2}}{l_{1}}=\frac{\alpha_{1}}{\alpha_{2}}$

$1+\frac{l_{2}}{l_{1}}=1+\frac{\alpha_{1}}{\alpha_{2}}$

$\frac{l_{1}+l_{2}}{l_{1}}=\frac{\alpha_{1}+\alpha_{2}}{\alpha_{2}}$ or $\frac{l_{1}}{l_{1}+l_{2}}=\frac{\alpha_{2}}{\alpha_{1}+\alpha_{2}}$

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