At temperature $T_{0}$, two metal strips of length $l_{0}$ and thickness $d$, is bolted so that their ends coincide. The upper strip is made up of metal $A$ and have coefficient of expansion $\alpha_{A}$ and lower strip is made up of metal $B$ with coefficient of expansion $\alpha_{B} \cdot\left(\alpha_{A}>\alpha_{B}\right)$. When temperature of their blastic strip is in-seased from $T_{0}$ to $\left(T_{0}+\Delta T\right)$, on strip become longer than the other and blastic strip is bend in the form of a circle as shown in fig. Calculate the radius of furvature $R$ of the strip:
(a) $R:=\frac{\left[2+\left(\alpha_{A}+\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T}$
(b) $R=\frac{\left[2-\left(\alpha_{A}+\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T}$
(c) $R=\frac{\left[2+\left(\alpha_{A}-\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T_{i}}$
(d) $R=\frac{\left[2-\left(\alpha_{A}-\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta \mathcal{k}}$