The heights of mercury surfaces in the two arms of the manometer … - Vidyadip

Question

The heights of mercury surfaces in the two arms of the manometer shown in figure are $2\ cm$ and $8\ cm$. Atmospheric pressure $= 1. 01 x 10^5N/rn^2$. Find.

Thepressure of the gas in the cylinder and.
The pressure of mercury at the bottom of the $U$ tube.

✓

Answer

Since, the tube is opened from the top, this means, total pressure of the gas will be equal to the pressure due to the mercury column $+$ tmospheric Pressure.

$\therefore$ Pressure of the gas in cylinder $=$ Atmospheric Pressure $+$ Pressure due to the mercury column.
Now, Pressure due to the mercury column $=\rho g\left(h_2-h_1\right)$
where, $\rho$ is the density of the mercury $=13.6 \mathrm{~g} / \mathrm{cm}^3$,
$g$ is the acceleration due to gravity $=9.8 \mathrm{~m} / \mathrm{s}^2=980 \mathrm{~cm} / \mathrm{s}^2, \mathrm{~h}_2$ is $8 \ cm$ and $\mathrm{h}_1$ is $2 \ cm$ .
$\therefore$ Total Pressure of the gas $=1 \mathrm{~atm}+(8-2) \times 13.6 \times 980$
$=1.013 \times 10^5 \mathrm{~Pa}+6 \times 13.6 \times 980 \mathrm{dyne} / \mathrm{cm}^2$
$=101300+7996.8 \mathrm{~Pa}$
$=181268 \mathrm{~Pa}$
$=1.8 \times 10^5 \mathrm{~Pa}$

The pressure of the mercury at the bottom of the $U -$tube.

$=$ The pressure of mercury column on the open side $+$ Atmospheric pressure.
$=13.6 \times 10^6 / 1000 \times 9.8 \times 0.08+1.013 \times 10^5$
$=1.12 \times 10^5 \mathrm{~Pa}$

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