Answer
(a) The diameter of the wire is $1.3~mm$
The tensile stress is $8.4\times 10^7~N/m^2$
(b) The maximum weight that can be hung from the wire is $571~N$
Work Step by Step
(a) $Y = \frac{F/A}{\Delta L/L}$
$Y$ is Young's modulus
$F$ is the force
$A$ is the cross-sectional area
$\Delta L$ is the change in length
$L$ is the original length
The Young's modulus of copper is $120\times 10^9~Pa$
We can find the radius of the wire:
$A = \frac{F~L}{Y~\Delta L}$
$\pi~r^2 = \frac{F~L}{Y~\Delta L}$
$r = \sqrt{\frac{F~L}{\pi~Y~\Delta L}}$
$r = \sqrt{\frac{(120~N)(3.0~m)}{(\pi)(120\times 10^9~Pa)(2.1\times 10^{-3}~m)}}$
$r = 0.674\times 10^{-3}~m$
$r = 0.674~mm$
Since the diameter of the wire is twice the radius, the diameter is $1.3~mm$.
We can find the tensile stress:
$\frac{F}{A} = \frac{F}{\pi~r^2}$
$\frac{F}{A} = \frac{120~N}{(\pi)~(6.74\times 10^{-4}~m)^2}$
$\frac{F}{A} = 8.4\times 10^7~N/m^2$
The tensile stress is $8.4\times 10^7~N/m^2$
(b) We can find the maximum weight:
$\frac{F}{A} = 4.0\times 10^8~N/m^2$
$F = (4.0\times 10^8~N/m^2)~(\pi~r^2)$
$F = (4.0\times 10^8~N/m^2)~(\pi)~(6.74\times 10^{-4}~m)^2$
$F = 571~N$
The maximum weight that can be hung from the wire is $571~N$
