question_answer

A bimetallic strip is formed out of two identical strips, one of copper and the other of brass. The Coefficients of linear expansions of 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 \[R\] Find \[R\]

A) \[t/\left( {{\alpha }_{B}}+{{\alpha }_{C}} \right)\Delta t\]

B) \[\Delta t/\left( {{\alpha }_{B}}+{{\alpha }_{C}} \right)t\]

C) \[t/\left( {{\alpha }_{B}}-{{\alpha }_{C}} \right)\Delta t\] 

D) \[\frac{\left( {{\alpha }_{B}}-{{\alpha }_{C}} \right)t}{\Delta t}\]

Correct Answer: C

Solution :

Let \[{{\ell }_{0}}\] be the length and \[t\]the thickness of each strip. On heating, length of brass rod \[{{\ell }_{1}}={{\ell }_{2}}\left( 1+{{\alpha }_{B}}\Delta T \right)\]

By the geometry of the figure, we have

\[{{\ell }_{1}}={{R}_{1}}\theta \]

\[=\left( R+\frac{1}{2} \right)\theta \]

\[\therefore \left( R+\frac{1}{2} \right)\theta ={{\ell }_{0}}\left( 1+{{\alpha }_{B}}\Delta T \right)..(i)\]

Similarly for copper strip, \[{{\ell }_{2}}={{\ell }_{0}}\left( 1+{{\alpha }_{C}}t \right)={{R}_{2}}\theta \]

or \[\left( R-\frac{1}{2} \right)\theta ={{\ell }_{0}}\left( 1+{{\alpha }_{C}}T \right)..(ii)\]

Dividing equation (i) by (ii), we get \[\Rightarrow \frac{\left( R+\frac{1}{2} \right)}{\left( R-\frac{1}{2} \right)}=\frac{1+{{\alpha }_{B}}\Delta T}{1+{{\alpha }_{C}}\Delta T}\]

\[\Rightarrow \left( R+\frac{t}{2} \right)\left( 1+{{\alpha }_{C}}\Delta T \right)\]

\[=\left( R-\frac{t}{2} \right)\left( 1+{{\alpha }_{B}}\Delta T \right)\]

\[\Rightarrow R+R{{\alpha }_{C}}\Delta T+\frac{t}{2}+\frac{t}{2}{{\alpha }_{C}}\Delta T\]

\[=R+R{{\alpha }_{B}}\Delta T-\frac{t}{2}-\frac{t}{2}{{\alpha }_{B}}\Delta T\]

If \[\Delta T\]is small, \[\alpha \] is still small, so we can neglect their product. And, therefore, we have \[R{{\alpha }_{B}}\Delta T-R{{\alpha }_{C}}\Delta T=t\]

Or \[R=\frac{t}{\left( {{\alpha }_{B}}-{{\alpha }_{C}} \right)\Delta T.}\]