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A simple barometer is shown. The vertical tube is fixed. a) Determine the atmospheric pressure in SI units. b) Name the part labelled A. c) The tube is accidentally knocked at the top, so that a small

Physics

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Here's the solution for question 12:

12. The figure below shows a simple barometer. The vertical tube is fixed in position.

a) Determine the atmospheric pressure in SI units.

  1. Determine the height of the liquid column.

    The total height of the liquid in the tube from the reservoir surface is 75 cm75 \text{ cm}. The space above the liquid is 25 cm25 \text{ cm}.
    The actual height of the liquid column supported by atmospheric pressure is the difference:

    h=75 cm−25 cm=50 cmh = 75 \text{ cm} - 25 \text{ cm} = 50 \text{ cm}

    Convert this height to meters:

    h=0.50 mh = 0.50 \text{ m}
  2. Identify the liquid and its density.

    A simple barometer typically uses mercury. The density of mercury (ρHg\rho_{Hg}) is 13600 kg/m313600 \text{ kg/m}^3.

  3. Use the formula for pressure due to a liquid column.

    Assume the acceleration due to gravity g=10 m/s2g = 10 \text{ m/s}^2.
    The atmospheric pressure (PatmP_{atm}) is given by:

    Patm=ρHgghP_{atm} = \rho_{Hg} g h

    Substitute the values:

    Patm=13600 kg/m3×10 m/s2×0.50 mP_{atm} = 13600 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.50 \text{ m} Patm=68000 PaP_{atm} = \boxed{\text{68000 Pa}}

    b) Name the part labelled A.

    Part A is the space above the liquid column in the sealed tube, which is a vacuum.

    Torricellian vacuum\boxed{\text{Torricellian vacuum}}

    c) The tube is accidentally knocked at the top, so that a small crack is caused at the top of the tube. Air leaks in through the crack. State and explain the effect, if any, this has on the level of the liquid in the tube.

    Effect: The level of the liquid in the tube will fall.

    Explanation: When air leaks into the space above the liquid (the Torricellian vacuum), the trapped air exerts pressure on the surface of the liquid inside the tube. This air pressure adds to the pressure exerted by the liquid column, meaning the external atmospheric pressure now has to support both the air pressure and the liquid column. To maintain equilibrium, the height of the liquid column must decrease.

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