What must be the lengths of steel and copper | vedclass

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(A) Let $L_S$ and $L_C$ be the lengths of steel and copper rods at $0^o C$.At any temperature $T$,the lengths are $L_S(1 + \alpha_S T)$ and $L_C(1 + \

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$30\, cm$ for steel and $20\,cm$ for copper

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(A) Let $L_S$ and $L_C$ be the lengths of steel and copper rods at $0^o C$.At any temperature $T$,the lengths are $L_S(1 + \alpha_S T)$ and $L_C(1 + \alpha_C T)$.The difference in lengths is given as $10\,cm$ at any temperature $T$,so:$L_S(1 + \alpha_S T) - L_C(1 + \alpha_C T) = 10$$(L_S - L_C) + (L_S \alpha_S - L_C \alpha_C)T = 10$For this to be independent of $T$,the coefficient of $T$ must be zero:$L_S \alpha_S - L_C \alpha_C = 0 \Rightarrow L_S \alpha_S = L_C \alpha_C$$L_S (1.2 	imes 10^{-5}) = L_C (1.8 	imes 10^{-5})$$L_S / L_C = 1.8 / 1.2 = 3 / 2$Also,from the first equation,$L_S - L_C = 10$.Substituting $L_S = 1.5 L_C$ into $L_S - L_C = 10$:$1.5 L_C - L_C = 10 \Rightarrow 0.5 L_C = 10 \Rightarrow L_C = 20\,cm$.Then $L_S = 1.5 	imes 20 = 30\,cm$.

$S$.No.	$l (m)$	$\Delta T (^{\circ}C)$	$\Delta l (m)$
$(1)$	$2$	$10$	$4 \times 10^{-4}$
$(2)$	$1$	$10$	$4 \times 10^{-4}$
$(3)$	$2$	$20$	$2 \times 10^{-4}$
$(4)$	$3$	$10$	$6 \times 10^{-4}$

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 $(\alpha_{copper} = 1.8 \times 10^{-5} \;^o C^{-1})$

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