Equating the pressure in mm of Hg on both arms above the line XX ),
we get

pabs. + pwater = pHg + patm.
Now, pwater = \(\cfrac{34}{136}\) = 2.5 mm of Hg.
∴ pabs + 2.5 = 97.5 + 760
pabs = 97.5 + 760 – 2.5
= 855 mm of Hg.
= 855 × pHg × g × 10–5 bar
= \(\cfrac{855}{1000}\) (m) x (13.6 × 1000) (kg/m3) × 9.81 × 10–5
= 1.1407 bar.