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What must be the lengths of steel and copper rods at $0^o C$ for the difference in their lengths to be $10\,cm$ at any common temperature? $(\alpha_{steel}=1.2 \times {10^{-5}} \;^o C^{-1})$ and $(\alpha_{copper} = 1.8 \times 10^{-5} \;^o C^{-1})$
$30\, cm$ for steel and $20\,cm$ for copper
$20\,cm$ for steel and $30\,cm$ for copper
$40\,cm$ for steel and $30\,cm$ for copper
$30\,cm$ for steel and $40\,cm$ for copper
Solution
$L_{f}=L_{i}(1+\alpha \Delta T)$
$L_{S_{f}}=L_{S_{i}}\left[1+\alpha_{S} \Delta T\right]$
$\Rightarrow \Delta L_{\text {steel}}=L_{S_{i}} \alpha_{S} \Delta T$
$L_{C_{f}}=L_{C_{i}}\left[1+\alpha_{C} \Delta T\right]$
$\Rightarrow \Delta L_{\text {copper}}=L_{C_{i}} \alpha_{C} \Delta T$
For $\Delta L_{\text {steel }}=\Delta L_{\text {copper }}$
$\Rightarrow L_{S_{i}} \alpha_{S} \Delta T=L_{C_{i}} \alpha_{C} \Delta T$
$\Rightarrow \frac{L_{S_{i}}}{L_{C_{i}}}=\frac{\alpha_{C}}{\alpha_{S}}=\frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}=\frac{3}{2}$
$\therefore \frac{L_{S_{i}}}{L_{C_{i}}}=\frac{3}{2}$ in $(1)$ only
