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 :

Options

A. `2piSYalphaDeltaT`
B. `SYalphaDeltaT`
C. `piSYalphaDeltaT`
D. `2SYalphaDeltaT`

Answer

D

Explanation

`DeltaL=LalphaDeltaT`
Now , `(F)/(S)=(DeltaL)/(L)Y`
`F=alphaDeltaTYS`
So, (for equilibrium of each semi-circular part)
`F=2alphaDeltaTYS`