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