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Two straight metallic strips each of thickness $t$ and length $\ell$ are rivetted together. Their coefficients of linear expansions are $\alpha_1$ and $\alpha_2$. If they are heated through temperature $\Delta T$, the bimetallic strip will bend to form an arc of radius
$\left.t /\left\{\alpha_1+\alpha_2\right) \Delta T \right\}$
$t /\left\{\left(\alpha_2-\alpha_1\right) \Delta T\right\}$
$t\left(\alpha_1-\alpha_2\right) \Delta T$
$t\left(\alpha_2-\alpha_1\right) \Delta T$
Solution
(b)
Let the angle subtended by the arc formed be $\theta$. Then
$\theta=\frac{\ell}{ r }$ or $\theta=\frac{\Delta \ell}{\Delta r }=\frac{\ell_2-\ell_1}{ r _1- r _2}$
$\therefore \quad \theta=\frac{\ell\left(\alpha_2-\alpha_1\right) \Delta T }{ t }$
or $\frac{\ell}{ r }=\frac{\ell\left(\alpha_2-\alpha_1\right) \Delta T }{ t }$
So, $r=\frac{t}{\left(\alpha_2-\alpha_1\right) \Delta T}$
