A 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 pi R`. To fit the ring on the whell, 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 semicirular parts together. If te doefficient of linear expansion of the metal is `alpha`, and its Youngs' modulus is dY, the force that on 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`