calculate pressure in CGS units using following data : $$
\mbox{Specific gravity of mercury},\gamma_{Hg}=13.6\\
\mbox{Density of water}, \rho=10^3{\rm kg/m^3}\\
\mbox{Gravity}, g=9.8{\rm m/s^2}\\
\mbox{height}, h=75{\rm cm}$$
I know, $P=h\rho g$
i have also converted all the data into CGS units $\gamma_{hg}=13.6,\rho=10^6{\rm g/m^3},\ g=980{\rm m/s^2},\\ h=75{\rm cm}$
After that I thought it is easy, i just have to substitute the value in the equation.
But then i saw specific gravity of mercury (frankly i saw this for first time) i thought what is this. Now i don't know what to do.
I think it it has some relation with $g$ as there is word specific gravity.
I have no idea what to do.
Best Answer
Since this has been open for a while now, here's the solution. The thing with unit conversions is to remember to be systematic and to double-check for typos and power-of-ten slipups. Even after fifteen years of unit conversion problems I find stupid little mistakes when I write the things out in full like I have below; if I try to "save time" by not writing the conversions out all the way, I make the mistakes but I don't find them.
The density of water is \begin{align} \rho_\mathrm{H_2O} &= 10^3\, \mathrm{\frac{kg}{m^3}} \mathrm{ \times \frac{10^3\,g}{1\,kg} \times\left(\frac{1\,m}{10^2\,cm}\right)^3 } \\&= 1\,\mathrm{\frac{g}{cm^3}}, \end{align} so the density of mercury is $\rho_\mathrm{Hg} = \gamma_\mathrm{Hg}\rho_\mathrm{H_2O} = 13.6\,\mathrm{g/cm^3}$. (This is one of the plusses of CGS units, that the density of water is unity and specific gravities and densities have the same values.)
The acceleration due to gravity is \begin{align} g &= \mathrm{ 9.8\,\frac {m}{s^2} \times \frac{100\,cm}{1\,m} }\\&= 980\,\mathrm{\frac{cm}{s^2} } \end{align}
So the pressure under a 75 cm column of mercury is \begin{align} P = \rho g h &= \mathrm{ 13.6\,\frac{g}{cm^3} \times 980\,\frac {cm}{s^2} \times 75\,cm }\\ &=\mathrm{ 0.9996\times10^6 \,\frac{dyne}{cm^2} \approx 1\,megabarye } \end{align}