High School

A manometer is filled with water. On the first side, the pressure is standard sea level pressure (P1 = 101325 kPa). The second side is open to the room. The fluid is 50 cm higher on the second side than the first side. What is the pressure in the room? Note: The density of water is 1000 kg/m^3.

A) 101.3 kPa
B) 102.6 kPa
C) 100.2 kPa
D) 103.8 kPa

Answer :

Final answer:

The pressure in the room measured with a water-filled manometer with a 50 cm difference in water level is calculated using the hydrostatic pressure formula. After calculations, the pressure in the room is determined to be 96.4 kPa, which is not one of the provided options, indicating a possible error in the question.

Explanation:

The question asks for the pressure in the room using a water-filled manometer where the first side (P1) is at standard sea level pressure (101325 Pa) and the second side is open to the room, showing a 50 cm difference in water level. To find the pressure in the room (P2), we need to apply the concept of hydrostatic pressure, which is the pressure exerted by a fluid due to gravity. The formula for hydrostatic pressure is p = hpg, where h is the height of the fluid, p is the density of the fluid, and g is the acceleration due to gravity.

Since the water column is higher on the second side by 50 cm, we know that the pressure in the room is less than the standard atmospheric pressure by the amount of hydrostatic pressure of the 50 cm water column. In this case:

  • h = 0.50 m (since 50 cm = 0.50 m)
  • p (density of water) = 1000 kg/m3
  • g = 9.81 m/s2
  • Hydrostatic pressure, p = hpg = 0.50 m * 1000 kg/m3 * 9.81 m/s2 = 4905 Pa (or 4.905 kPa)

Thus, the pressure in the room is P1 - hydrostatic pressure = 101325 Pa - 4905 Pa = 96420 Pa (or 96.4 kPa), which is none of the options provided in the question. It appears there may have been a calculation or conceptual error in the multiple-choice options given.

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