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. After heating, the lengths of the brass and copper strips are given by:$L_B = L_0(1

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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. After heating, the lengths of the brass and copper strips are given by:$L_B = L_0(1 + \alpha_B \Delta T) = (R + d)	heta$$L_C = L_0(1 + \alpha_C \Delta T) = R	heta$Dividing the two equations, we get:$\frac{R + d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$$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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