A wooden wheel of radius $R$ is made of two semicircular parts. The two parts are held together by a ring made of a metal strip of cross-sectional area $S$ and length $L$, which 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 the 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
Answer: (D) $2SY\alpha\Delta T$
Concept. After cooling, the ring tries to shrink back by $L\alpha\Delta T$ but the wheel stops it. It stays stretched with strain $\alpha\Delta T$, so its tension is $T=SY\alpha\Delta T$. The force between the two halves of the wheel is then found by looking at one half.
Step 1. Tension in the ring. Strain $=\alpha\Delta T$, stress $=Y\alpha\Delta T$, tension $T=SY\alpha\Delta T$.
Step 2. Forces on one half-wheel with its half-ring. Cut along the joint: the ring pulls the half with tension $T$ at each of the two cut ends, both pulling it towards the other half. Total $2T$.
Step 3. Equilibrium. The other half must push back with $2T$.
Answer. $2SY\alpha\Delta T$ (Option D).
Solution by Sreeraj P, M.Sc Physics
