Find the Youngs modulus of the rod:
A composite rod is made by joining a copper rod end to end with a second rod of different material but of same cross-section. At 25oC, the composite rod is 1 m in length of which the length of copper rod is 30 cm. At 125oC, the length of the composite rod increases by 1.91 mm. While the composite rod is not permitted to expand through holding it among two rigid walls, it is found that the length of constituents does not change with rise in temperature. Find the Young's modulus and coefficient of linear expansion of the second rod. For copper, α = 1.7 × 10^{- 5} ^{o}C^{-1}, and Y = 1.3 × 10^{11} Nm^{- 2}.
Solution
For copper rod α_{1} = 1.7 × 10^{- 5} ^{o}C- 1, and Y_{1} = 1.3 × 10^{11} Nm^{-2}, l1 = 30 cm = 0.3 m. Let l_{2} be initial length of the rod of other material at 25^{o}C, and α_{2} and Y_{2} be coefficient of linear expansion and Young's modulus for the material of this rod. Then, at initial temperature θ1 = 25^{o}C, l_{1} + l_{2} = 1 m,
l_{2} = 1 - l_{1} = 1 - 0.3 = 0.7 m
Let the final temperature θ_{2} = 125^{o}C, the increases in the length of cooper rod and that of other material be Δl_{1} and Δl_{2}, respectively.
Then,
Δl_{1} + Δl_{2} = 1.91 mm = 1.91 × 10^{-3} m . . . (A)
Now,
Δl_{1} = l_{1} α_{1} (θ_{2} - θ_{1}) = 0.3 × 1.7 × 10^{- 5} × (125 - 25) = 0.51 × 10^{-3} m
Also,
Δl_{2} = l_{2} α_{2} (θ_{2} - θ_{1}) = 0.7 × α_{2} × (125 - 25) = 70 α_{2}
Substituting for Δl_{1} and Δl_{2} in Eq. (A), we get
0.51 × 10^{- 3} + 70 α_{2} = 1.91 × 10^{-3}
or, α_{2} = 2.0 × 10- 5 ^{o}C ^{-1}
Now,
Y = (F/A)/ (Δl/C) = FL/ A Δl ⇒ F = Y A Δl/l
Let A be the cross-section of each of the two rods, thus, tension produced in copper rod F_{1} = Y_{1}AΔl_{1} /l_{1}, and tension generates in the rod of other material, F_{2} = Y_{2} A Δl_{2}/ l_{2}. The tension developed in the two rods will be same, i.e. F_{1}= F_{2}.
Therefore,
(Y_{1} A Δl_{1})/l_{1} = (Y_{2} A Δl_{2})/l_{2}
or Y_{2} = Y_{1} l_{2}Δl_{1}/l_{1}Δl_{2} ⇒ Y_{2} = 1.3 × 10^{11} × 0.7×0.51×10^{-3}/0.3 × 1.40 × 10^{-3}
or Y_{2} = 1.105 × 10^{11} Nm^{-2}