A bimetallic strip is formed out of two ident | vedclass

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(B) Let $L_0$ be the initial length of each strip before heating. Let $d$ be the total thickness of the bimetallic strip.After heating,the length of t

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$\frac{1}{\Delta T}$

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(B) Let $L_0$ be the initial length of each strip before heating. Let $d$ be the total thickness of the bimetallic strip.After heating,the length of the brass strip is $L_B = L_0(1 + \alpha_B \Delta T) = (R + d) 	heta$.The length of the copper strip is $L_C = L_0(1 + \alpha_C \Delta T) = R 	heta$.Dividing the two equations,we get $\frac{R + d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$.This simplifies to $1 + \frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$.$\frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T} - 1 = \frac{1 + \alpha_B \Delta T - 1 - \alpha_C \Delta T}{1 + \alpha_C \Delta T} = \frac{(\alpha_B - \alpha_C) \Delta T}{1 + \alpha_C \Delta T}$.Since $\alpha \Delta T \ll 1$,we can approximate $1 + \alpha_C \Delta T \approx 1$.Thus,$R \approx \frac{d}{(\alpha_B - \alpha_C) \Delta T}$.Therefore,$R \propto \frac{1}{\Delta T}$.

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