An equilateral triangle ABC is formed by two copper rods AB and BC and one is aluminium rod which heated in such a way that temperature of each rod increases by `DeltaT.` Find change in the angle `angleABC.` (Coefficient of linear expansion for copper is `alpha _(1).` and for aluminium is `alpha2`.).

Explanation

Suppose, `AB = l _(1), AC = _(2) and BC = l _(3)`
`therefore cos theta = ( l _(3) ^(2) + l _(1) ^(2) - l _(2) ^(2))/( 2 l _(3) l _(1))`where ` angleABC = theta `
`therefore 2l _(3) l _(1) cos theta = l _(3) ^(2) + l_(1) ^(2) - l _(2) ^(2)`

Integrating on both side,
`2 ( l _(3) dl _(1) + l _(1) dl _(3)) cos theta - 2 l3l _(1) sin theta d theta =2l _(3) dl _(3) +2l _(1) dl _(1) -2l _(2) dl _(2)`
Dividing by 2,
`(l_(3) dl _(1) + l _(1) xx dl _(3)) cos theta -l _(3) l _(1) sin theta d theta =l_(3) dl _(3) + l _(1) dl _(1) - l _(2) dl _(2)`
Now `dl _(1) = l _(1) alpha _(1) Delta T, dl _(2) = l _(1) alpha _(2) Delta T, dl _(3) = l _(3) alpha _(3) Delta T ` then,
`(l _(3) xx l _(1) alpha _(1) Delta T + l _(1) xx l _(3) alpha _(3) Delta T) cos theta - l _(3) l _(1) sin theta d theta =`
`l _(3) xx l _(3) alpha _(3) Delta T + l_(1) xx l _(1) alpha _(1) Delta T - l _(2) xx l _(2) alpha _(2) Delta T`
Now let `l _(1) = l _(2) = l _(3) = l and alpha _(3) = alpha _(1)`
`therefore (l ^(2) alpha _(1) Delta T + l ^(2) alpha _(1) Delta T) cos theta - l ^(2) sin theta d theta = l ^(2) alpha _(1) Delta T + ^(2) alpha _(1) Delta T - l ^(2) alpha _(2) Delta T`
`cos theta = cos 60^(@) = 1/2 ` (Equilateral triangle)
`therefore 2l ^(2) alpha _(1) Delta T xx 1/2 - l ^(2) sin theta d theta = 2l alpha _(1) Delta T - l ^(2) alpha _(2) Delta T`
`therefore l alpha _(1) Delta T - l ^(2) sin theta d theta = 2l ^(2) alpha _(1) Delta T - l ^(2) alpha _(2) Delta T ` Dividing by `l ^(2),`
`alpha _(1) Delta T - sin theta d theta = 2 alpha _(1) Delta T - alpha _(2) Delta T`
`- alpha _(1) Delta T + alpha _(2) Delta T = sin theta d theta `
`therefore d theta = ((alpha _(2) - alpha _(1))Delta T )/( sin theta )`
`therefore d theta =(2 (alpha _(2) - alpha _(1)) Delta T)/( sqrt3) [ because sin theta = sin 60^(@) = (sqrt3)/(2) ]`