A rod of original length L₁ changes its temperature from T₁ toT₂. Let $α$ is temperature coefficient of linear expansion and L₂ is length of temperature T₂. Choose the correct options.

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$L_{2} = L_{1} [1 + α (T_{2} - T_{1})], where T_{2} - T_{1} and α are not large$

Formula in option (1) is valid in all conditions

$L_{2} = L_{1} e^{\int_{T_{1}}^{T_{2}} αdT}, where α is a function of temperature$

$L_{2} = L_{1} e^{α (T_{2} - T_{1})}, where α is constant$

answer is A, B, C.

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Detailed Solution

When $α$ is a constant, $dL = αLΔT$

$\begin{array}{l} \Rightarrow \frac{dL}{L} = αΔT \Rightarrow [\log L]_{L_{1}}^{L_{2}} = [αT]_{T_{1}}^{T_{2}} \\ \Rightarrow \log \frac{L_{2}}{L_{1}} = α (T_{2} - T_{1}) \Rightarrow L_{2} = L_{1} e^{α (T_{2} - T_{1})} \end{array}$

When $α$ is a function of temperature,

$\begin{array}{l} \int_{L_{1}}^{L_{2}} dL = \int_{T_{1}}^{T_{2}} αdT \Rightarrow \log \frac{L_{2}}{L_{1}} = \int_{1}^{2} αdT \\ \Rightarrow L_{2} = L_{1} e^{\int_{T_{1}}^{T_{2}} αdT} \end{array}$