Answer
(a) The oil should rise by $5.6~cm$ on the side that is open to the atmosphere.
(b) The mercury should rise by $0.37~cm$ on the side that is open to the atmosphere.
Work Step by Step
(a) We can find the height $h_o$ of oil that would have the same gauge pressure as 0.74 cm Hg:
$\rho_o~g~h_o = \rho_m~g~h_m$
$h_o = \frac{\rho_m~h_m}{\rho_o}$
$h_o = \frac{(13.6~g/cm^3)~(0.74~cm)}{0.90~g/cm^3}$
$h_o = 11.2~cm$
The oil should rise by $5.6~cm$ on the side that is open to the atmosphere, while the oil falls by $5.6~cm$ on the side with the air tank.
(b) If the manometer used mercury instead, the mercury should rise by $0.37~cm$ on the side that is open to the atmosphere, while the mercury falls by $0.37~cm$ on the side with the air tank.
