Length of mystery object

What everyday object is ${n}^{-1/4}$ metres long, where $n$ is an integer?

Hint: n=

128

Note there may be some cultural dependence here.

• Oh, please don't rewrite the question and pull away the rug under the already given answers. If the question was solved quickly, then that's how it is. There isn't anything sensible you can do about it. – M Oehm Oct 18 '17 at 10:04
• @M Oehm: it's not pulling a rug from under your solution. It remains great. Leave it as it is, or put it in spoiler. But I am thinking most people won't have read this yet, and I want them to have as much fun as possible. – Laska Oct 18 '17 at 10:08
• Does the modified question gurantee a single/unique answer? If so how? – Mea Culpa Nay Oct 18 '17 at 10:40
• The new question is much broader and doesn't feel like it has a single answer... Looking at the current answer there are a multitude of variations of that which are all equally valid... – Chris Oct 18 '17 at 10:40
• @MOehm: In my opinion the first question wasn't great. It wasn't much of a puzzle, it was just do a calculation and recognise the value given - a trivia question at best (I don't think trivia questions are on topic here but may be wrong). – Chris Oct 18 '17 at 10:42

With

n = 128

We have

$128^{(-1/4)} = 2^{(-7/4)} \approx 0.297m$ or $29.7cm$

An everyday object of that length is

An A4 sheet of paper: 21cm wide, and 29.7cm long

For details, and an explanation why theoretically the length is exactly $n^{(-1/4)}$ see M Oehm's answer

• Please explain why this is so. – Laska Oct 18 '17 at 8:50
• Doesn't it imply that a standard printer's print tray(where papers are loaded) is also an eligible candidate for this, on a lighter vein? – Mea Culpa Nay Oct 18 '17 at 10:22

[I wanted this to be a comment to sousben's answer, but the formulas got a bit unwieldy.]

The papers of the A series of ISO 216 have a edge-length ratio of $\sqrt2:1$, and the long side of $\text A(n-1)$ is the short side of $\text A(n)$. $\text A0$ is defined to have a surface of $1\text m^2$, so

\begin{align} l\cdot \frac{l}{\sqrt2}&=1\text m^2\\ l^2&=1\text m^2\cdot\sqrt2\\ l&=1\text m\cdot2^\frac{1}{4} \end{align}

The lengths of the smaller sizes are:

$l_1=l_0/\sqrt{2}^1=1\text m\cdot2^\frac{1}{4}\cdot2^{-\frac{1}{2}}$
$l_2=l_0/\sqrt{2}^2=1\text m\cdot2^\frac{1}{4}\cdot2^{-\frac{2}{2}}$
$l_3=l_0/\sqrt{2}^3=1\text m\cdot2^\frac{1}{4}\cdot2^{-\frac{3}{2}}$
$l_4=l_0/\sqrt{2}^4=1\text m\cdot2^\frac{1}{4}\cdot2^{-\frac{4}{2}}$

The rule

$x^n\cdot x^m=x^{n+m}$

can be used to simplify $l_4$ to

$l_4=1\text m\cdot2^{\frac{1}{4}-2}=2^{-\frac{7}{4}}\text m$

• This is an international standard. So, despite the OP's remark about cultural dependence, no such remark is needed. On the contrary, such a remark would be more appropriate for local alternatives. – Rosie F Oct 18 '17 at 18:26