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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 a metal strip of cross-sectional area S and length L. L is slightly less than 2 $π$ R. To fit the ring on the wheel, it is heated, so that its temperature rises by $∆$ 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 $α$ . and its Young's modulus is Y, the force that one part of the wheel applies on the other part is

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$2 πSYα ∆ T$

$πSYα ∆ T$

$SYα ∆ T$

$2 SYα ∆ T$

answer is D.

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Detailed Solution

Elongation due to change in temperature,

$∆ l = Lα ∆ T$ …(i)

where, $α$ = coefficient of linear expansion,

which is compensated by elastic strain.
When temperature becomes normal, then

$∆ l = \frac{TL}{YS}$ …(ii)

From Eqs. (i) and (ii), we get

$\frac{T L}{Y S} = L α ∆ T$

$\Rightarrow$ $T = Y S α ∆ T$

At equilibrium, force exerted by one-half on other,

$F = 2 T = 2 Y S α ∆ T$

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