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A bimetallic strip is formed out of two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are $\alpha_C$ and $\alpha_B$. On heating, the temperature of the strip increases by $\Delta T$ and the strip bends to form an arc of radius $R$. Then $R$ is proportional to
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The correct answer is:
$\frac{1}{\Delta T}$
Let $L_0$ be the initial length of each strip before heating. Length after heating will be
$L_B=L_0\left(1+\alpha_B \Delta T\right)$
$=(R+d) \theta$
$L_C=L_0\left(1+\alpha_C \Delta T\right)=R \theta$
$\Rightarrow \quad \frac{R+d}{R}=\frac{1+\alpha_B \Delta T}{1+\alpha_C \Delta T}$
$1+\frac{d}{R}=1+\left(\alpha_B-\alpha_C\right) \Delta T$
$R=\frac{d}{\left(\alpha_B-\alpha_C\right) \Delta T}$
$R \propto \frac{1}{\Delta T}$
$L_B=L_0\left(1+\alpha_B \Delta T\right)$
$=(R+d) \theta$
$L_C=L_0\left(1+\alpha_C \Delta T\right)=R \theta$
$\Rightarrow \quad \frac{R+d}{R}=\frac{1+\alpha_B \Delta T}{1+\alpha_C \Delta T}$
$1+\frac{d}{R}=1+\left(\alpha_B-\alpha_C\right) \Delta T$
$R=\frac{d}{\left(\alpha_B-\alpha_C\right) \Delta T}$
$R \propto \frac{1}{\Delta T}$
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