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A bimetallic strip is formed out of two identical strips, one of copper and 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
Proportional to $\Delta T$
Inversely proportional to $\Delta T$
Inversely proportional to $|{\alpha _B} - {\alpha _C}|$
Both $(B)$ and $(C)$
Solution

(d) Let $L_0$ be the initial length of each strip before heating.
Length after heating will be
${L_B} = {L_0}(1 + {\alpha _B}\Delta T) = (R + d)\theta $
${L_C} = {L_0}(1 + {\alpha _C}\Delta T) = R\theta $
==> $\frac{{R + d}}{R} = \frac{{1 + {\alpha _B}\Delta T}}{{1 + {\alpha _C}\Delta T}}$
==> $1 + \frac{d}{R} = 1 + ({\alpha _B} – {\alpha _C})\Delta T$
==> $R = \frac{d}{{({\alpha _B} – {\alpha _C})\Delta T}}$ ==> $R \propto \frac{1}{{\Delta T}}$ and $R \propto \frac{1}{{({\alpha _B} – {\alpha _C})}}$