Question

Two Straight metallic strips each of thickness t and length l 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

• A. $t/\left[\right(a_1+a_2\left)\Delta T\right]$
• B. $t/\left[\right(a_2-a_1\left)\Delta T\right]$
• C. $t(a_1+a_2)\Delta T$
• D. $t(a_2+a_1)\Delta T$

Answer 1

The correct option is B $t/\left[\right(a_2-a_1\left)\Delta T\right]$

let $\Delta l$ be the difference between final lengths of two strips and $\Delta r=t$ be the difference in their radii.

Then, since $l=r\theta$

Hence, $\Delta l=\theta \Delta r$ since $\theta$ is same for both arcs.

We know that $l^{\prime }=l(1+\alpha \Delta T)$

Therefore, $\theta =\frac{l}{r}=\frac{l({\alpha }_2-{\alpha }_1)\Delta T}{t}$

$r=\frac{t}{{(\alpha }_2-{\alpha }_1)\Delta T}$