Question

The wooden wheel of radius $R$ is made of two semicircular parts (see figure). The two parts are held together by a ring made of 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 just steps over the wheel. As it cools down to the surrounding temperature it presses the semicircular part 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. $2\pi SY\alpha \Delta T$
• B. $SY\alpha \Delta T$
• C. $\pi SY\alpha \Delta T$
• D. $2SY\alpha \Delta T$

Answer 1

The correct option is D $2SY\alpha \Delta T$

$\Delta L=L\alpha \Delta T$
Tension in the ring is T.
$\frac{T}{A}=\frac{\Delta L}{L}Y$
$T=\alpha \Delta TYS$
So, $F=2T$
$F=2\alpha \Delta TYS$