question_answer

A bimetallic strip is formed of two identical strips of copper and brass, (\[{{\alpha }_{c}},\,{{\alpha }_{b}}\] are temperature coefficient of expansion of copper and brass respectively and \[{{\alpha }_{c}}>\,{{\alpha }_{b}}\]). When heated to a temperature t, the radius of curvature of the strip becomes proportional to

A)  \[{{({{\alpha }_{c}}-{{\alpha }_{b}})}^{-1}}\]                       

B)  \[\frac{{{\alpha }_{b}}}{{{\alpha }_{c}}}\]

C)  \[\frac{{{\alpha }_{b}}\times {{\alpha }_{c}}}{{{\alpha }_{b}}+{{\alpha }_{c}}}\]                 

D)  \[({{\alpha }_{c}}-{{\alpha }_{b}})\]

Correct Answer: A

Solution :

\[(R+r)\theta ={{\ell }_{0}}(1+{{\alpha }_{c}}\Delta t)\] \[(R-r)\theta ={{\ell }_{0}}(1+{{\alpha }_{b}}\Delta t)\] \[\frac{R+r}{R-r}=\frac{1+{{\alpha }_{c}}\Delta t}{1+{{\alpha }_{b}}\Delta t}\] Solving for R we get \[R\propto \frac{1}{\Delta t({{\alpha }_{c}}-{{\alpha }_{b}})}\]