MCAT Physics Question — Absolute Pressure
 by
 Apr 17, 2017
 MCAT Question of the Day
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The atmospheric pressure exerted on the surface of the fluid is equal to 101,000 N/m^{2} (1 atm). To calculate the gauge pressure at each point, you must apply the formula, P_{gauge} = ρgh, where P_{gauge }is the pressure due to the water column of height h, ρ is the density of the fluid, g is the gravitational acceleration constant, and h is the depth.
P_{abs (A) }= P_{atm }+ P_{gauge (A)}
P_{abs (A) }= (101,000 N/m^{2})+ ρ_{water}gh_{A}
P_{abs (A) }= (101,000 N/m^{2}) + (1000 kg/m^{3})(10 m/s^{2})(10 m)
P_{abs (A)} = 201,000 N/m^{2}
^{ }
P_{abs (B) }= P_{atm }+ P_{gauge (B)}
P_{abs (B) }= (101,000 N/m^{2})+ ρ_{water}gh_{B}
P_{abs (B) }= (101,000 N/m^{2}) + (1000 kg/m^{3})(10 m/s^{2})(30 m)
P_{abs (B)} = 401,000 N/m^{2}
^{ }
Therefore, the P_{B }= 2P_{A }making answer choice A the correct answer. Furthermore, an increase in depth of 10 m in water equals 1 atmosphere. Therefore, the total pressure at A is 2 atm while the total pressure at B is 4 atm.
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