$2\mathrm{.}20$ LAB: Axial stretch a uniform rod in tension

The following equation computes the stretch in a uniform steel rod subject to a tension force using the following relation:

The equation above is correct for all of the quantities in base SI units. So$,$ length l in m$,$ cross$-$sectional area A in m$2,$ force F in N$,$ and elastic modulus E in Pa$.$ The cross$-$sectional area for a circular rod of diameter d is A $=\backslash$pi d $2/4.$ We will consider steel, for which E $=200\backslash$times $109$ Pa$.$

The code below attempts to compute the stretch for a rod when the user enters the diameter, length, and force. The code has one or more Python errors. In addition, there is an incorrect number somewhere. Using hand calculations and proper unit conversions, you find that the correct result for d $=25$ mm$,$ l $=1$ m$,$ and F $=1000$ N is about $\backslash$delta $=0\mathrm{.}010$ mm$.$ You could do the same calculations for different d$,$ l$,$ and F to test the code.

Ex: If the input is:

$25$

$1$

$1000$

then the correct last line of output would be:

The rod will stretch $0\mathrm{.}010$ mm$.$