Q.
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 $\mathrm{\Delta}T$ 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
A) $2\pi SY\alpha\mathrm{\Delta}T$
B) $SY\alpha\mathrm{\Delta}T$
C) $\pi SY\alpha\mathrm{\Delta}T$
D) $2SY\alpha\mathrm{\Delta}T$
Explanation
(d) If temperature increases by $\mathrm{\Delta}T$,
Increase in length $L,\mathrm{\Delta}L = L\alpha\mathrm{\Delta}T$
$\therefore\frac{\mathrm{\Delta}L}{L} = \alpha\mathrm{\Delta}T$
Let tension developed in the ring is $T$
[image:9_image18.emf]
$\therefore\frac{T}{S} = Y\frac{\mathrm{\Delta}L}{L} = Y\alpha\mathrm{\Delta}T$
$\therefore T = SY\alpha\mathrm{\Delta}T$
$F = 2T$ (From figure)
Where, $F$ is the force that one part of the wheel applies on the other part
$\therefore F = 2SY\alpha\mathrm{\Delta}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.