question_answer

A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is slightly less than \[2\pi R\]. To fit the ring on the wheel, it is heated so that its temperature rises by \[\Delta T\] and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is \[\alpha \], and its Young's modulus is Y, the force that one part of the wheel applies on the other part is:   AIEEE  Solved  Paper-2012

A) \[2\pi SY\alpha \Delta T\]                               

B) \[SY\alpha \,\Delta T\]

C) \[\pi SY\alpha \,\Delta T\]                              

D) \[2SY\alpha \,\Delta T\]

Correct Answer: D

Solution :

             \[L\Rightarrow S\] \[\Delta L=L\propto \Delta T\] \[\frac{F}{A}=\frac{\Delta L}{L}Y\] \[F=\alpha \,\Delta T\,\,YS\] So, \[T=2\,F\] \[T=2\alpha \,\Delta t\,YS\]