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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?

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

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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

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 $\rho$.

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Question resolved, thanks! –  YatharthROCK May 2 '12 at 8:17

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