## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 9 - Problems - Page 361: 26

#### 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.

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