An equilateral triangle ABC if formed by two Cu rods AB and BC and one Al rod

it is heated in such a way that temperature of each rod increases by `Delta T.` Find change in the angle ABC. [Coeff. Of linear expansion for Cu. is `alpha_1,` Coeff. of linear expansion for Al is `alpha_2]`

Explanation

Consider the diagram shown
Let `" "l_(1)=AB,l_(2)=AC,l_(3)=BC`
`therefore" "cos theta = (l_(3)^(2)+l_(1)^(2)-l_(2)^(2))/(2l_(3)l_(1))" "("assume"angleABC = theta)`
`implies " "2l_(3)l_(1)costheta=l_(3)^(2)+l_(1)^(2)-l_(2)^(2)`
Differentiating 2 `(l_(3)dl_(1)+l_(1)dl_(3))costheta - 2l_(1)l_(3)sinthetad theta`
`" "=2l_(3)dl_(1)+2l_(1)dl_(1)-2l_(2)dl_(2)`
Now, `" "dl_(1)=l_(1)alpha_(1)Deltat" "("where, " Deltat= "change in temperature")`
`" "dl_(2)=l_(2)alpha_(1)Deltat implies dl_(3)=l_(3)alpha_(2)Deltat`
and `" "l_(1)= l_(2)=l_(3)=l`
`(l^(2)alpha_(1)Deltat + l^(2) alpha_(1)Deltat)cos theta + l^(2) sin theta d theta = l^(2)alpha_(1)Deltat + l^(2)alpha_(1)Deltat-l^(2)alpha_(2)Deltat`

`" "sin theta d theta = 2alpha_(1) Deltat (1-cos theta)-alpha_(2)Deltat`
Putting `" "theta = 60^(@)" (for equilateral triangle)"`
`" "d thetaxx sin 60^(@) = 2 alpha_(1)Deltat (1- cos 60^(@))-alpha_(2)Deltat`
`" "= 2 alpha_(1)Delta t xx(1)/(2)-alpha_(2)Deltat = (alpha_(1)-alpha_(2))Deltat`
`implies " "d theta= "change in the angle "angleABC`
`" "=((alpha_(1)-alpha_(2))DeltaT)/(sin60^(@))=(2(alpha_(1)-alpha_(2))DeltaT)/(sqrt(3))" "(because Deltat=DeltaT" given")`