Two roda of equal cross - section one of copper and the other of steel, are joined to from a composite rod of length 2.0 m. At `20^(@)C` the length of the copper is 0.5 m . When the temperatue is raised to `120^(@)C` the length of ther comosite rod incrases to 2.002. m. If the composite rod is fixed betweeen two rigid walls and is, thus, not allowed to expand, it is found that the lenghts of the two component rods also do not change with the increases a temperature. Calculate the Young's modulus and the coefficient of linear expansion of steel. Given `Y_(ca) = 1.3 xx 10^(13)N//m^(2) and alpha_(ca) = 1.6 xx 10^(-5) //""^(@)C`.

Explanation

(a) with increases in temperature due to thermnal expansion the leght of a rod changes,
i.e., L. = L `(1 + alpha Delta theta)`
So, for composite rod
`L_(s). + L_(c ). = L_(c ) (1 + alpha_(c ) Delta theta) + L_(s ) (1 + alpha_(s) Delta theta)`
or `L_(s). + L_(c ). = (L_(s) + L_(c )) + (L_(s) alpha_(s) + L_(c ) alpha_(c)) alpha theta`
According to given problem,
`2.002 = 2 + [0.5 xx 1.6 xx 10^(-5) + 1.5 alpha_(s)] xx 100`
or `1.5alpha_(s) = 2 xx 10^(-5) - 0.8 xx 10^(-5)` i.e., `alpha_(s) = 0.8 xx 10^(-5) // ""^(@)C`
(b) When the rods are fixed between the walls and its lenght remains unchanged
`(DeltaL_(s))_(H)=(DeltaL_(s))_(c)and(DeltaL_(c))_(H)=(DeltaL_(c))_(c)`
Now as `(DeltaL)_(H) = L alpha Delta theta and (Delta L)_(c )=(FL//AY)`
So, `L_(s)alpha_(s)Deltatheta=(FL_(s)//AY_(s))andL_(c)alpha_(c)Deltatheta=(FL-(c)//AY_(c))`
Dividing one by the other `((alpha_(c))/(alpha_(s)))=((Y_(s))/(Y_9c))`
So, `Y_(s)=(alpha_(c))/(alpha_(s))xxY_(c)=(1.6xx10^(-5))/(0.8xx10^(-5))xx1.3xx10^(13)=2.6xx10^(13)N//m^(2)`