When composite rod is free, composite rod is free, compositive length increase to `2.002m` from temprature `20^(@)C` to `120^(@)C`. When composite rod is fixed between the support. There is no change in component length. Find `Y` and `alpha` of steel if
`Y_(cu) = 1.5 xx 10^(13 N//m^(2) alpha_(cu) = 1.6 xx 10^(-5//@)C`

Explanation

`Deltal = l_(s)alpha_(s)DeltaT + l_(c)alpha_(c)DeltaT`
`0.002 = [1.5 alpha_(s) + 0.5 xx 1.6 xx 10^(-5)]'100`
`alpha_(s) = (1.2 xx 10^(-5))/(1.5) = 8 xx 10^(-6)^(@)C`
If there is no change in composite length, thermal force of steel and copper rod should be equal
`F_(st) = F_(cu) , Y_(s)Aalpha_(s)Deltat = Y_(cu)A alpha_(cu)Deltat`
`(Y_(s))/(Y_(C)) = (alpha_(C))/(alpha_(S)), Y_(S) = Y_(C) xx (alpha_(c))/(alpha_(s)) = (1.5 xx 10^(13) xx 1.6 xx 10^(-5))/(8 xx 10^(-6))`
`Y_(S) = 3 xx 10^(13 N // m^(2)`