$A$ manometer reads the pressure of a gas in an enclosure as shown in Figure $(a)$. When a pump removes some of the gas,the manometer reads as in Figure $(b)$. The liquid used in the manometers is mercury and the atmospheric pressure is $76 \; cm$ of mercury.
$(a)$ Give the absolute and gauge pressure of the gas in the enclosure for cases $(a)$ and $(b)$,in units of $cm$ of mercury.
$(b)$ How would the levels change in case $(b)$ if $13.6 \; cm$ 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) For figure $(a)$:
Atmospheric pressure,$P_{0} = 76 \; cm$ of $Hg$.
The difference between the levels of mercury in the two limbs gives the gauge pressure. Since the level in the right limb is higher,the gauge pressure is $20 \; cm$ of $Hg$.
Absolute pressure $= P_{0} + \text{Gauge pressure} = 76 + 20 = 96 \; cm$ of $Hg$.
For figure $(b)$:
The level in the right limb is lower than the left limb,so the gauge pressure is $-18 \; cm$ of $Hg$.
Absolute pressure $= P_{0} + \text{Gauge pressure} = 76 - 18 = 58 \; cm$ of $Hg$.
$(b)$ When $13.6 \; cm$ of water is poured into the right limb,it exerts an additional pressure. Since the relative density of mercury is $13.6$,a column of $13.6 \; cm$ of water is equivalent to $1 \; cm$ of $Hg$.
Let $h$ be the new difference between the levels of mercury in the two limbs.
The pressure in the right limb at the level of the mercury surface in the left limb is $P_{R} = P_{0} + 1 \; cm$ of $Hg = 76 + 1 = 77 \; cm$ of $Hg$.
The pressure in the left limb is $P_{L} = P_{\text{gas}} + h = 58 + h$.
Equating the pressures at the same horizontal level: $58 + h = 77 \implies h = 19 \; cm$.
Thus,the new difference between the levels of mercury in the two limbs will be $19 \; cm$.