## College Physics (4th Edition)

(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.
(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.