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. `2pi SY alpha Delta T`
B. `SY alpha Delta T`
C. `pi SY alpha Delta T`
D. `2 SY alpha Delta T`

Answer

D

Explanation

If `Delta T` is the increase in temperature, then increase in length L of metal ring is
`Delta L =alpha L Delta T` or `(Delta L)/(L)=alpha Delta T`
Let T be the tension developed in the ring, then
`Y=(T)/(S)xx(L)/(Delta L)` or `T=(YS Delta L)/(L) =YS alpha Delta T`
Refer to Fig. 7(CF). 48 shown here, if F is the force that one part of the wheel applies on the other part, then `F=2 T =2 SY alpha Delta T`