# Measuring the nearest order of magnitude [closed]

The world's largest ball of a string is about $R=2 m$ in radius. To find the nearest order of magnitude, what is the total length $L$ of the string in the ball?

I have tried this in the following way

Total volume occupied by string $v=(cross\ section\ area\ d^2)\times (length\ L)$

So, $$L=\frac{4R^3}{d^2}$$

In question value of $R$ is given to be $2$ but value of $d$ is not given.

So either I am going wrong or question is incorrect.

## closed as off-topic by AccidentalFourierTransform, Jon Custer, user191954, John Rennie, Aaron StevensOct 19 '18 at 13:59

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• Why not make an order-of-magnitude guess for $d$? If you don't trust your intuition make more than one guess and see if it matters. – rob Jun 1 '14 at 19:09
• What rob said. Moreover, that's the entire purpose of questions like these. The question intentionally forgets to tell you how big $d$ is, forcing you resort to a guess and deal with the uncertainty that unavoidably enters into your result. This is the difference between an actual physics problem as opposed to mere applied math problems. – David H Jun 1 '14 at 19:22
• @DavidH So now how should i move move ahead? Should i guess the value of d or keep the answer in form of d? by keeping the answer in form of d it will not match with answer. – Freddy Jun 1 '14 at 19:28
• @Freddy You obviously can't stop at just the formula that gives $L$ as a function of $d$, because the problem asks you to give a number for $L$, not a function, and you don't have a number yet. So it's time to guess a value for $d$. This step makes some students uncomfortable at first because the worry over how to guess "the right value". Take a deep breath and guess anyway. You can always guess again. – David H Jun 1 '14 at 20:12
• @DavidH after 3 try i finally got the answer :) – Freddy Jun 1 '14 at 20:48

Here is an example of how one might be systematic about deciding between appropriate guesses. Start by identifying an obvious upper bound on the order of magnitude of $d$. The ball radius $R$ is on the order of about $1$ meter, and we know $d$ should be small with respect to $R$, so let's start by guessing $$d\approx 10^{-2} \text{ m}=1\text{ cm}.$$

That's about the width of a human finger. Are strings usually thinner than that? Sure they are. So let's try another order of magnitude smaller where $d\approx 1\text{ mm}$. Using the formula for $L$,

$$L=\frac{4R^3}{3d^2}=\frac{32}{3d^2}=10^7\text{ m},$$

or about 10,000 kilometers. This might seem like too large of a result at first since since it's larger than the radius of the Earth, but on the other hand it's only half as long as the Great Wall of China.

Having found a reasonable estimate of $L$, the next step would be to investigate progressively smaller orders of magnitude guesses for $d$ until you're confident anything smaller would be unreasonable. Looking at $d\approx 10^{-4} \text{ m}$ then, you can verify for yourself that the length estimate this results in is about twice the distance from the Earth to the Moon (!!). It would take a ray of light 4 seconds to travel from end of the string to the other. This is just too big. Thus, the previous guess appears to be the Goldilocks guess we desire.

• i have taken $$L=\frac{4R^3}{d^2}$$ because $$V=\frac{4}3\pi r^2$$ $$\frac\pi 3=approximately\ 1$$ – Freddy Jun 1 '14 at 22:05
• after that i have guessed $d=4mm$ so i got final answer $6\times 10^6 m$ Is it ok – Freddy Jun 1 '14 at 22:08

I do have this question in my book. I'll share how they have approached this question.

Let us assume the ball is spherical with $R= 2\ \mathrm m$.

The string in the ball is not closely packed and there are uncountable gaps between adjacent sections of the string. To allow for these gaps, let us somewhat overestimate the cross-sectional area of the string by assuming the cross section is square with an edge length of $d=4\ \mathrm{mm}$. Then, with a cross-sectional area of $d^2$ and a length $L$, the string occupies a total volume of,

$$V= \text{(cross-sectional area)}\times\text{(length)}= d^2 \times L$$

This is approximately equal to the volume of the ball given by $4/3 \pi R^3$ , which is about $4R^3$ because $\pi$ is about $3$. Thus we have

$$d^2L = 4R^3$$

\begin{aligned} L&= 4R^3 / d^2\\ &=4\times(2^3)/(4\times10^{-3})^2\\ &= 2\times10^6\ \mathrm m\\ &= 10^6\ \mathrm m\\ &=10^3\ \mathrm{km}\\ \end{aligned}