Suppose we need to convert $18\text{ km/h}$ speed into $\text{m/s}$. The answer is $5\text{ m/s}$. But $5\text{ m/s}$ has only one significant figure only whereas the $18\text{ km/h}$ has two.

My question is, should not the number of significant figures remain same when we convert units?

And then suppose we were to covert $5\text{ m/s}$ into $\text{km/h}$. Should our answer be $20\text{ km/h}$? I thought of rounding off $18\text{ km/h}$ to $20\text{ km/h}$ because $20\text{ km/h}$ has only one significant figure.


Significant figures are a very crude method of roughly gauging the propagation of uncertainty. They are convenient and easy, and generally get you a more reasonable estimate of your uncertainty than you get by ignoring them. However, occasionally significant figures are not particularly good either, such as this case.

You are absolutely correct that by the standard rules of significant figures a speed of $18 \text{ km/h}$ should convert to $5.0 \text{ m/s}$. Let's look at what this means.

When using significant figures the quantity $18 \text{ km/h}$ means that we have a speed that is somewhere between $17.5 \text{ km/h}$ and $18.5 \text{ km/h}$ which is more commonly written $(18.0 \pm 0.5) \text{ km/h}$. If we convert that to $\text{m/s}$ we get a true representation of our uncertainty as $(5.00 \pm 0.14)\text{ m/s}$.

Now, let's look at the answer given and the significant figures value. The answer would be more explicitly written as $(5.0 \pm 0.5)\text{ m/s}$ which overstates our uncertainty by a factor of about 3.6. The significant figures approach would be more explicitly written as $(5.00 \pm 0.05)\text{ m/s}$ which understates our uncertainty by a factor of 2.8 or so. Both are pretty bad representations of our actual uncertainty of $(5.00 \pm 0.14)\text{ m/s}$, but the significant figures approach is slightly less bad than the answer given.

Now, look what happens if you go the other way. $5\text{ m/s}$ means more explicitly $(5.0 \pm 0.5)\text{ m/s}$. Converting this to $\text{km/h}$ gives $(18.0 \pm 1.8) \text{ km/h}$. The official significant figures approach, as you said, gives $20\text{ km/h}$ which is not even in the actual range of our uncertainty. Furthermore, the implied precision $(20 \pm 5) \text{ km/h}$ overstates the uncertainty by a factor of 2.8. If you instead incorrectly reported the result as $18 \text{ km/h}$ then the implied precision of $(18.0 \pm 0.5)\text{ km/h}$ would understate your uncertainty by a factor of 3.6. Either way you do not get a good representation of both the quantity and the uncertainty, but the significant figures is probably still less bad.

  • $\begingroup$ thanks a ton Dale, I have one more question, suppose we were to covert 5 m/s into km/h. Should our answer be 20 Km/h ( as question says to take care of significant figures). Because 20 km/h has only one significant figure, so I thought of rounding off 18 km/h to 20 km/h. Am I correct? $\endgroup$
    – Arun Arora
    Sep 27 '20 at 15:49
  • $\begingroup$ @ArunArora I updated your question and my answer $\endgroup$
    – Dale
    Sep 27 '20 at 17:07

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