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A wooden wheel of radius R is made of two semicircular parts (as in 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 co-efficient of linear expansion of the metal is $α$ and its Young's modulus y then, the force that one part of the wheel applies on the other part is

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2 $π$ SY $α$ $∆$ T

SY $α$ $∆$ T

$π$ SY $α$ $∆$ T

2SY $α$ $∆$ T

answer is B.

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Detailed Solution

Given,

length =L and cross-sectional area =S
Temperature change = △T
Y is Young's Modulus
Now, the force that one wheel component exerts on the other component is,

$Y = \frac{stress}{strain} = \frac{F l}{A ∆ l}______(1)$

$strain = \frac{∆ l}{l} = α ∆ T$

substituting the values,

$Y = \frac{F}{Sα ∆ T} \Rightarrow F = SYα ∆ T$

Hence the correct answer is $SYα ∆ T .$

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