Answer
(a) $0.52\times 10^{-7}m$
(b) aluminum rod
Work Step by Step
(a) For aluminum, the change in length is given as
$\Delta L_{Al}=\frac{FL}{\pi r^2Y_{Al}}$
We plug in the known values to obtain:
$\Delta L_{Al}=\frac{(8400N)(0.55m)}{\pi(0.85m)^2(6.9\times 10^9N/m^2)}$
$\Delta L_{Al}=0.295\times 10^{-7}m$
For brass, the change in length is given as
$\Delta L _{B}=\frac{FL}{\pi r^2Y_B}$
We plug in the known values to obtain:
$\Delta L_B=\frac{(8400N)(0.55m)}{\pi(0.85m)^2(9.0\times 10^{10}N/m^2)}$
$\Delta L _{B}=\Delta L_{Al}+\Delta L_B$
We plug in the known values to obtain:
$\Delta L=0.295\times 10^{-7}m+0.226\times 10^{-7}m=0.52\times 10^{-7}m$
(b) From part (a), the change in length of the aluminum is $\Delta L_{Al}=0.295\times 10^{-7}m$ and the change in length of the brass rod is $\Delta L_{B}=0.226\times 10^{-7}m$. Comparing theses two values, we conclude that the aluminum rod has the greatest change in its length.
