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A manometer roads 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 $76 \mathrm{~cm}$ of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of $\mathrm{cm}$ of mercury.
(b) How would the levels change in case (b) if $13.6 \mathrm{~cm}$ of water (immiscible with mercury) are poured into the right limb of the manometer? Ignore the small change in the volume of the gas.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of $\mathrm{cm}$ of mercury.
(b) How would the levels change in case (b) if $13.6 \mathrm{~cm}$ of water (immiscible with mercury) are poured into the right limb of the manometer? Ignore the small change in the volume of the gas.
Solution:
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Verified Answer
The atmospheric pressure, $P=76 \mathrm{~cm}$ of mercury
(a) From figure $(a)$,
Pressure head, $h=20 \mathrm{~cm}$ of mercury
$\therefore$ Absolute pressure $=p+h=76+20=96 \mathrm{~cm}$ of mercury
Also, Gauge pressure $=h=20 \mathrm{~cm}$ of mercury
From figure $(b)$,
pressure head, $h=-18 \mathrm{~cm}$ of mercury
$\therefore$ Abolute pressure $=p+h=76+(-18)$
$=58 \mathrm{~cm}$ of mercury
Also, Gauge pressure $=\mathrm{h}$
$=-18 \mathrm{~cm}$ of mercury.
(b) When $13.6 \mathrm{~cm}$ of water is poured into the right limb of the manometer of figure $(b)$, the, using the relation:
Pressure $=\rho g h=\rho^{\prime} g^{\prime} h^{\prime}$
We get,
$\mathrm{h}^{\prime}=\frac{\rho \mathrm{h}}{\rho^{\prime}}=\frac{1 \times 13.6}{13.6}=1 \mathrm{~cm}$ of mercury
$\left[\mathrm{P}^{\prime}=\right.$ density of mercury $]$
therefore, pressure at the point $B$
$\mathrm{P}_{\mathrm{B}}=\mathrm{P}+\mathrm{h}^{\prime}=76+1=77 \mathrm{~cm}$ of mercury.
If $h^{\prime \prime}$ is the difference in the mercury levels in the two
limbs, then taking $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}$
$\Rightarrow 58+h^{\prime \prime}=77 \Rightarrow h^{\prime \prime}=77-58=19 \mathrm{~cm}$ of mercury.
(a) From figure $(a)$,
Pressure head, $h=20 \mathrm{~cm}$ of mercury
$\therefore$ Absolute pressure $=p+h=76+20=96 \mathrm{~cm}$ of mercury
Also, Gauge pressure $=h=20 \mathrm{~cm}$ of mercury
From figure $(b)$,
pressure head, $h=-18 \mathrm{~cm}$ of mercury
$\therefore$ Abolute pressure $=p+h=76+(-18)$
$=58 \mathrm{~cm}$ of mercury
Also, Gauge pressure $=\mathrm{h}$
$=-18 \mathrm{~cm}$ of mercury.
(b) When $13.6 \mathrm{~cm}$ of water is poured into the right limb of the manometer of figure $(b)$, the, using the relation:
Pressure $=\rho g h=\rho^{\prime} g^{\prime} h^{\prime}$
We get,
$\mathrm{h}^{\prime}=\frac{\rho \mathrm{h}}{\rho^{\prime}}=\frac{1 \times 13.6}{13.6}=1 \mathrm{~cm}$ of mercury
$\left[\mathrm{P}^{\prime}=\right.$ density of mercury $]$
therefore, pressure at the point $B$
$\mathrm{P}_{\mathrm{B}}=\mathrm{P}+\mathrm{h}^{\prime}=76+1=77 \mathrm{~cm}$ of mercury.
If $h^{\prime \prime}$ is the difference in the mercury levels in the two
limbs, then taking $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}$
$\Rightarrow 58+h^{\prime \prime}=77 \Rightarrow h^{\prime \prime}=77-58=19 \mathrm{~cm}$ of mercury.
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