A wooden wheel of radius R is made of two semicircular part . 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 `2piR`. To fit the ring on the wheel, it is heated so that its temperature rises by `DeltaT`and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircle parts together. If the coefficient of linear expansion of the metal is `alpha`, and it Young's modulus is Y, the force that one part of the wheel applies on the other part is :

Explanation

When the ring is heated and then allowed to cool, its new length `l. = 2piR`
` therefore ` Change in length of ring, `Delta l = l. - l`
` = 2pi R - l`
` therefore ` Strain developed in ring, `(Delta l)/(l) = (2pi R - l)/(l)`
If T is the tension developed in the ring, then
`Y = T/A .(l)/(Delta l)`
` rArr T = (YA Delta l)/(l) = (YA (2pi R - l) )/(l)`
For each semiconductor part,

If one part of the wheel applies force F on the other part, then
` F = 2T = (2Y A (2pi R - l) )/(l)`
` = 2YA ( (2piR - l)/(l) )`