Rods of length L₁ and L₂ are made of materials whose coefficients of linear expansion are α ₁

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10-1.Thermometry, Thermal Expansion and Calorimetry

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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 length is independent of temperatures, then

$\frac{{{L_1}}}{{{L_2}}} = \frac{{{\alpha _1}}}{{{\alpha _2}}}$

$\frac{{{L_1}}}{{{L_2}}} = \frac{{{\alpha _2}}}{{{\alpha _1}}}$

$L_1^2{\alpha _1} = L_2^2{\alpha _2}$

$\alpha _1^2{L_1} = d_2^2{L_2}$

Solution

$\mathrm{L}_{1}-\mathrm{L}_{2}=\mathrm{L}_{1}-\mathrm{L}_{2}$

$\Rightarrow \mathrm{L}_{1}\left(1+\alpha_{1} \Delta \mathrm{T}\right)-\mathrm{L}_{2}\left(1+\alpha_{2} \Delta \mathrm{T}\right)=\mathrm{L}_{1}-\mathrm{L}_{2}$

$\Rightarrow \mathrm{L}_{1} \alpha_{1}=\mathrm{L}_{2} \alpha_{2}$

$\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}=\frac{\alpha_{2}}{\alpha_{1}}$

Standard 11

Physics

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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 length is independent of temperatures, then

Solution

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