Question

A wire is stretched by 8 mm when a load of 60 N is applied.
What will be the extension of a wire of the same material having four times the cross-sectional
area and twice the original length, when the same load is applied?

Answer 1

Hooke's law is
\[F = kx\]
If you know F = 60 N and x = 8 mm, then you can calculate the spring constant k.

The spring constant is related to Young's modulus E by
\[k = \frac{AE}L\]
where A is cross sectional area and L is length.
The value of E is inherent to a material and does not depend on the geometry of the wire/spring/etc.
Hence
\[k = \frac{F}x \propto \frac{A}L\]