Question

A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modulus of this wire?

Answer 1

The Young modulus is given by:
$E= \frac{F /A}{\Delta L / L_0}$
where
F is the force applied
$L_0$ is the initial length of the wire
$A$ is the cross-sectional area of the wire
$\Delta L$ is the stretch of the wire

The wire in the problem stretches by $0.5%$ of its length, this means 
$\frac{\Delta L}{L_0} = 0.005$

We can also calculate the area of the wire; its radius is in fact half the diameter:
$r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m$
and so the area is
$A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2$

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

$E= \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2$