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 goes up by  $\Delta T$ and the strip bends to form an arc of radius of curvature R. Then R is.

Let $L_0$  be the original length of the strip.
Co-efficient of linear expansion of brass is greater than that of copper i.e., ${\alpha }_B>{\alpha }_C$ .
$L_B=L_0\left(1+{\alpha }_B\Delta T\right)\left(R+d\right)\theta$ 
$=L_0\left(1+{\alpha }_B\Delta T\right)$ 
Again, $L_C=L_0\left(1+{\alpha }_C\Delta T\right)=R\theta$ 
$\therefore \frac{\left(R+d\right)\theta }{R\theta }=\frac{1+{\alpha }_B\Delta T}{1+{\alpha }_C\Delta T}$ 
Or, $\frac{R+d}{R}=\left(1+{\alpha }_B\Delta T\right)\left(1+{\alpha }_C\Delta T\right),$   
    [By binomial expansion]
Or, $1+\frac{d}{R}=1+\left({\alpha }_B-{\alpha }_C\right)\Delta T-{\alpha }_B{\alpha }_C{\left(\Delta T\right)}^2$   
Or, $\frac{d}{R}=\left({\alpha }_B-{\alpha }_C\right)\Delta TorR=\frac{d}{\left({\alpha }_B-{\alpha }_C\right)\Delta T}$ 

$\therefore R\alpha \frac{1}{\Delta T}andR\alpha \frac{1}{\left|{\alpha }_B-{\beta }_C\right|}$