A manometer reads the pressure of a gas in an encloure as shown in fig. when some of the gas is removed by the pump, the manometer reads as in fig. the liquid used in mano-meters is mercury and the atmospheric pressure is 76 cm of mercury.
(i) Given the absolute and gauge pressure of the gas in the two cases (in units of cm. of mercury).
(ii) How would the levels change in case(b) if `13.6cm` of water are poured into the right limb of manometer? (Ignore the change in volume of the gas).

Explanation

Here, atmospheric pressure, `P=76 cm` of mercury
(i) in fig., pressure head, `h=+20 cm`
`:.` Absolute pressure =`P+h = 76+20 = 96 cm` of mercury
Gauge pressure =`h= 20 cm` of mercury
Pressure head `h=-18 cm`
Absolute pressure = `P+h = 76+(-18) = 58 cm` of mercury
Gauge pressure = `h=-18 cm` of mercury
(ii) Here, `13.6 cm` of water added in right limb is equivalent to `(13.6)/(13/6) = 1cm` of mercury column.
i.e., `h'=1cm` of mercury column
Now pressure at `A` , `P_(A) = P+h'=76+1 = 77cm`
Let the difference in mercury levels in the two lumbs be `h_(1)`, then pressure at `B, P_(B) = 58+h_(1)`
As, `P_(A) =P_(B) :. 77=58+h_(1)` or `h_(1)=77-58 = 19cm ` of mercury column.