Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I wanted to find out what the height of some mercury reaching 1m in a tube when inverted and placed in a beaker, again of Mercury, would be?

I know that:

  • Atmospheric Pressure ($P$) $ \approx 10^5 Pa = 100,000 \frac{N}{m^2} = 10 \frac{N}{cm^2}$
  • Density of Hg ($\varrho$) $ \approx 13.6 \frac{g}{cm^3} $
  • Gravity ($g$) $ \approx 9.8 \frac{m}{s^2} = 980 \frac{m}{s^2} $
  • $ P = \varrho h g $

But I get: $$ h = \frac{P}{\varrho \cdot g} = \frac{10 N cm^3 s^2}{13.6 g \cdot 980 cm \cdot cm^2} = \frac{10N s^2}{13.328 g} \approx 1.33 s^2 g^{-1} $$

I wanted a value in cm, what went wrong?

share|improve this question

closed as off topic by J. M., LVK, Chris Eagle, t.b., sdcvvc Aug 27 '12 at 16:37

Questions on Mathematics Stack Exchange are expected to relate to math within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

    
I voted to migrate to physics.SE . –  J. M. May 1 '12 at 15:27
    
You've just been careless. Yatharth. Dropping units, getting your arithmetic wrong... –  TonyK May 1 '12 at 16:19
    
@J.M. I didn't see Physics.SE on SE (Yeah, I actually went to the site and checked; not finding anything like it, I decided to put this Question on Math.SE instead) –  YatharthROCK May 1 '12 at 16:26
    
@TonyK Can you fix it for me, please? I don't really see where I went wrong :( –  YatharthROCK May 1 '12 at 16:27
    
You dropped the newtons in the last approximation. $N-s^2/g$ does come out a length. And $10/13.328 \not \approx 1.33$ –  Ross Millikan May 1 '12 at 16:43
add comment

1 Answer 1

up vote 1 down vote accepted

If you want everything in centimeters, you should convert the Newton (which has units $\text{kg}\cdot\text{m}\over\text{s}^2$) to dynes. Converting $1\,\text N$ to dynes gives $1\,{\text N}= 10^5{\text{g}\cdot \text{cm}\over \text{s}^2}$. Note your units for the gravitational constant are off, you should have $g\approx 980\,{\text{cm}/\text{s}^2}$.

When doing calculations and checking units, I find it easier to separate the "unit part" and the "number part": $$\eqalign{ h={P\over\rho\cdot g} &= {10{\text{N}\over\text{cm}^2}\over 13.6{\text{g}\over\text{cm}^3}\cdot 980{\text{cm}\over s^2} } \cr &= {10\cdot10^5{\text{g}\cdot\text{cm}\over \text{s}^2} \cdot{1\over\text{cm}^2}\over 13.6{\text{g}\over\text{cm}^3}\cdot 980{\text{cm}\over \text{s}^2 } }\cr &= {10\cdot10^5 \over 13.6 \cdot 980 } {\text{g}\cdot\text{cm}\over \text{s}^2}\cdot{1\over\text{cm}^2}\cdot {\rm{s}^2\over\text{cm} }\cdot {\rm{cm}^3\over\text{g}}\cr &={10^6\over13.6\cdot980 }\,\text{cm}\cr &\approx 75\text{cm}. } $$

Note this is slightly off as we used somewhat crude approxiations for $g$ and $\rho$.

share|improve this answer
    
Question resolved, thanks! –  YatharthROCK May 2 '12 at 8:17
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.