Question 47 of 126

A wooden wheel of radius $\mathrm{R}$ 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 $\Delta \mathrm{T}$ 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 Youngs' modulus is $Y$, the force that one part of the wheel applies on the other part is :

A$\mathrm{2 \pi S Y \alpha \Delta T}$B$\mathrm{S Y \alpha \Delta T}$C$\mathrm{\pi S Y \alpha \Delta T}$D$\mathrm{2 S Y \alpha \Delta T}$

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Answer: (D) $\mathrm{2 S Y \alpha \Delta T}$

If temperature increases by $\Delta \mathrm{T}$,
Increase in length $\mathrm{L}, \Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T}$
$\therefore \quad \frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha \Delta \mathrm{T}$
Let tension developed in the ring is $\mathrm{T}$.
$\therefore \quad \frac{\mathrm{T}}{\mathrm{S}}=\mathrm{Y} \frac{\Delta \mathrm{L}}{\mathrm{L}}=\mathrm{Y} \alpha \Delta \mathrm{T}$
$\therefore \quad \mathrm{T}=\mathrm{S} \mathrm{Y} \alpha \Delta \mathrm{T}$
From FBD of one part of the wheel,
$\mathrm{F}=2 \mathrm{T}$
Where, $F$ is the force that one part of the wheel applies on the other part.
$\therefore \mathrm{F}=2 \mathrm{SY} \alpha \Delta \mathrm{T}$