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(B) Let the length of the strips be $\ell$ and their thickness be $t$. When heated by $\Delta T$,the new lengths are $\ell_1 = \ell(1 + \alpha_1 \Delt

Vedclass · Accepted Answer

$t / \{(\alpha_2 - \alpha_1) \Delta T\}$

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(B) Let the length of the strips be $\ell$ and their thickness be $t$. When heated by $\Delta T$,the new lengths are $\ell_1 = \ell(1 + \alpha_1 \Delta T)$ and $\ell_2 = \ell(1 + \alpha_2 \Delta T)$.Assuming $\alpha_2 > \alpha_1$,the strip with $\alpha_2$ forms the outer arc.Let $r$ be the radius of the neutral axis (the center of the bimetallic strip). The thickness of each strip is $t/2$ from the neutral axis,so the radii of the two strips are $r_1 = r - t/2$ and $r_2 = r + t/2$.The angle $	heta$ subtended by the arc is the same for both: $	heta = \ell_1 / r_1 = \ell_2 / r_2$.Substituting the values: $\frac{\ell(1 + \alpha_1 \Delta T)}{r - t/2} = \frac{\ell(1 + \alpha_2 \Delta T)}{r + t/2}$.Since $\alpha \Delta T$ is very small,we approximate $1 + \alpha \Delta T \approx 1$ for the denominator,but keep the difference in the numerator: $\frac{\ell_2}{\ell_1} = \frac{r + t/2}{r - t/2}$.Using the component-dividendo rule: $\frac{\ell_2 - \ell_1}{\ell_2 + \ell_1} = \frac{(r + t/2) - (r - t/2)}{(r + t/2) + (r - t/2)} = \frac{t}{2r}$.Since $\ell_2 - \ell_1 = \ell(\alpha_2 - \alpha_1) \Delta T$ and $\ell_2 + \ell_1 \approx 2\ell$,we get $\frac{\ell(\alpha_2 - \alpha_1) \Delta T}{2\ell} = \frac{t}{2r}$.Solving for $r$,we get $r = \frac{t}{(\alpha_2 - \alpha_1) \Delta T}$.

Two straight metallic strips each of thickness $t$ and length $\ell$ are riveted together. Their coefficients of linear expansion are $\alpha_1$ and $\alpha_2$. If they are heated through a temperature change $\Delta T$,the bimetallic strip will bend to form an arc of radius:

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