Two metal strips, each of length `l` and thickness `d` at temperature `T_(0)` are riveted together so that their ends coincide. One strip is made of metal `A` having a linear coefficient `alpha_(B)` where `alpha_(A)gtalpha_(B)` . When this bimetallic-strip is heated to a temperature `(T_(0)+DeltaT)` , one strip becomes longer than the other and the bimetallic strip bends into an arc of a circle. What is the radius of curvature `R` of the strip ?

Explanation

Let `A` and `B` two straight strips of equal thickness `d` and length `l_(0)` at temperature `T_(0)`
when temperature is incresed by `DeltaT` , strip `A` expands more `(alpha_(a)gtalpha_(B))` . This cause bimetallic strip, as a whole to bend into an arc.
Let `R_(A)` and `R_(B)` are radii of curature of arc.
At temperature `T_(0)+DeltaT`
`l_(A)=l_(0)(1+alpha_(A)DeltaT)` , `l_(B)=l_(0)(1alpha_(B)DeltaT)`
`l_(A)=R_(A)theta` , `l_(B)=R_(B)theta` , `theta` : angle subtended by two strip in bent position
`l_(A)-l_(B)=l_(0)(alpha_(A)-alpha_(B))DeltaT` , `l_(A)-l_(B)=(R_(A)-R_(B))theta`
`theta=(l_(A)-l_(B))/(R_(A))-(R_(B))=(l_(0)(alpha_(A)-alpha_(B))DeltaT)/(R_(A)-R_(B))=(l_(0)(alpha_(A)-alpha_(B))DeltaT)/(d)` (i)
`l_(A)+l_(B)=2l_(0)+l_(0)(alpha_(A)+alpha_(B))DeltaT`
`theta(R_(A)+R_(B))=2l_(0)+l_(0)(alpha_(A)+alpha_(B))DeltaT`
Mean radius `R=(R_(A)+R_(B))/(2)=(2l_(0)+l_(0)(alpha_(A)+alpha_(B))DeltaT)/(2theta)`
`=(2l_(0)+l_(0)(alpha_(A)+alpha_(B))DeltaT)/[[2l_(0)+l_(0)(alpha_(A)+alpha_(B))DeltaT]//d)`
`([2l_(0)+l_(0)(alpha_(A)+alpha_(B))DeltaT]d)/(2(alpha_(A)-alpha_(B))DeltaT)`
As `(alpha_(A)+alpha_(B))DeltaTlt lt 2(asalpha_(A),alpha_(B)lt lt1)`
`R=(d)/((alpha_(A)-alpha_(B))DeltaT)` .