A 100 N force stretches the length of a hanging wire by 0.5 mm. The force required to stretch a wire, of the same material and length but having four times the diameter, by 0.5 mm is

To solve this problem, we need to understand the relationship between the force, the material properties, and the dimensions of the wire. The force required to stretch a wire is given by Hooke's Law, which states that the force $F$ is proportional to the extension $\Delta L$ and the cross-sectional area $A$, and inversely proportional to the original length $L$. The formula is:

$F = k \cdot \Delta L$

where $k$ is the stiffness of the wire, which depends on the material and the cross-sectional area. For a cylindrical wire, the cross-sectional area $A$ is given by:

$A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$

Given that the original wire is stretched by a force of 100 N, we can write:

$100 = k \cdot 0.5$

Now, consider the new wire with four times the diameter. The new diameter $d'$ is:

$d' = 4d$

The new cross-sectional area $A'$ is:

$A' = \frac{\pi (4d)^2}{4} = \frac{\pi \cdot 16d^2}{4} = 4\pi d^2$

Since the stiffness $k$ is proportional to the cross-sectional area, the new stiffness $k'$ is:

$k' = 4k$

Now, we need to find the new force $F'$ required to stretch the new wire by 0.5 mm. Using Hooke's Law for the new wire:

$F' = k' \cdot 0.5$

Substitute $k' = 4k$:

$F' = 4k \cdot 0.5 = 4 \cdot 100 = 400$

Therefore, the force required to stretch the new wire by 0.5 mm is:

$\boxed{400 \, \text{N}}$