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Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Fluids2 Marks
Question
A manometer reads the pressure of a gas in an enclosure as shown in Fig.$(a)$ When a pump removes some of the gas, the manometer reads as in Fig. $(b)$ The liquid used in the manometers is mercury and the atmospheric pressure is $76\ cm$ of mercury. Give the absolute and gauge pressure of the gas in the enclosure for cases $(i)$ and $(ii)$ in units of cm of mercury. How would the levels change in case $(i)$ if $13.6\ cm$ of water $($immiscible with mercury$)$ are poured into the right limb of the manometer? $($Ignore the small change in volume of the gas.$)$
✓

Answer

  1. Given : Atmospheric pressure,
$P_0=6 \mathrm{~cm}$ of $\mathrm{Hg}$
In figure $(i)$ pressure head,
$\mathrm{h}_1=+20 \mathrm{~cm}$ of $\mathrm{Hg} .$
Absolute pressure $( P )$ of the gas is greater than the $\mathrm{P}_0$ i.e.,
$P=P_0+h_1 p g$
$=76 \mathrm{~cm}$ of $\mathrm{Hg}+20 \mathrm{~cm}$ of $\mathrm{Hg}$
$=96 \mathrm{~cm}$ of $\mathrm{Hg}$
Gauge pressure is the difference between the absolute pressure and the atmospheric pressure. $\frac{1}{2}$ It means,
Gauge pressure $=P-P_0$
$=96 \mathrm{~cm}$ of $\mathrm{Hg}-76 \mathrm{~cm} \text { of } \mathrm{Hg}$
$=20 \mathrm{~cm}$ of $\mathrm{Hg} .$
In figure $(ii),$ pressure head,
$h_2=-18 \mathrm{~cm}$ of $\mathrm{Hg} \text {. }$
$\therefore$ The absolute pressure of the gas is lesser than the atmospheric pressure is given by
$P=P_0+h_2 p g$
$=76 \mathrm{~cm}$ of $\mathrm{Hg}+(-18 \mathrm{~cm}) \text { of } \mathrm{Hg}$
$=58 \mathrm{~cm}$ of $\mathrm{Hg}$
Gauge pressure $=$ Absolute pressure $-$ Atmospheric pressure
$=58 \mathrm{~cm}$ of $\mathrm{Hg}-76 \mathrm{~cm} \text { of } \mathrm{Hg}$
$=-18 \mathrm{~cm}$ of $\mathrm{Hg}$
It means, Gauge pressure is simply equal to $h \ cm$ of $Hg .$
  1. Given : $13.6\ cm$ of water added in the right limb is equivalent to $\frac{13.6}{13.6}=1\text{cm}$ of $Hg$ column i.e., $h = 1\ cm$ of $Hg$ column, which can be calculated as follows
$\text{h}_{\omega}=13.6\text{Cm}$ of water
Suppose $h_m =$ height of $Hg$ column equivalent to $13.6\ cm$ of water, thus equilibrium.
$\text{h}_{\text{m}}\rho_{\text{m}}\text{g}=\text{h}_{\omega}\rho_{\omega}\text{g}.$
$\text{h}_{\text{m}}=\text{h}_{\omega}\frac{\rho_{\omega}}{\rho_{\text{m}}}=\frac{\text{h}_{\omega}}{\Big(\frac{\rho_{\text{m}}}{\rho_{\omega}}\Big)}$
$=\frac{13.6}{13.6}=1\text{cm}$ of $hg$
The mercury will rise in the left limb such that the difference in the height of $Hg$ column in the two limbs.
$= 20\ cm - 1m$
$= 19 \ cm$ of $Hg$ column.

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