Question

A wooden wheel of radius R is made of two semicircular parts(see figure). The two parts are field 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 U, the force that one part of the wheel applies on the other part is?

• A. $SY\alpha \Delta T$
• B. $\pi SY\alpha \Delta T$
• C. $2SY\alpha \Delta T$
• D. $2\pi SY\alpha \Delta T$

Answer 1

The correct option is A $SY\alpha \Delta T$

Radius $=R$

Cross-sectional area $=S$

Length $=L$

Temperature change $=△T$

$Y$ is Young's Modulus

Coefficient of Linear expansion $=\alpha$

Now, for the force that one part of the wheel applies on the other part is,

$Y=\frac{Stress}{Strain}=\frac{Fl}{A△l}\left(1\right)strain=\frac{△l}{l}=\alpha △T$

Putting values in equation $\left(1\right)$

$Y=\frac{F}{S\alpha △T}F=SY\alpha △T$

From here, force is given.