A steel wire W has a length / and a circular cross-section of radius r - AQA - A-Level Physics - Question 24 - 2018 - Paper 1

Using radius $r_X = 2r$, the cross-sectional area becomes:

$A_X = \pi (2r)^2 = 4\pi r^2$

Plugging this back into our formula gives:

$e_X = \frac{F (2L)}{4\pi r^2 Y}$

To find the extension:

Since we want $e_X = \frac{e}{4}$ and we know:

$e = \frac{F L}{\pi r^2 Y}$

We can manipulate it to find:

$\frac{e_X}{e} = \frac{F (2L)}{4\pi r^2 Y} \cdot \frac{\pi r^2 Y}{F L} = \frac{2}{4} = \frac{1}{2}$

The selection corresponds to the correct length and radius for the new wire X to produce the desired extension of $\frac{e}{4}$.