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The figure below shows an instrument used to measure pressure.

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c

9.1×104 N/m2\text{9.1} \times 10^4 \text{ N/m}^2

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Here are the solutions to questions 6 and 7.

6. The figure below shows an instrument used to measure pressure.

Part a

Name the instrument.

The instrument is a Mercury barometer.

Part b

State the observation if the tube was tilted assuming the instrument is functioning normally.

If the tube was tilted, the vertical height of the mercury column (measured from the level of mercury in the reservoir to the top of the column) would remain the same. The length of the mercury column along the tilted tube would increase, but its vertical height, which indicates atmospheric pressure, would not change.

Part c

9.1×104 N/m2\text{9.1} \times 10^4 \text{ N/m}^2

State and explain the effect on the level of the mercury in the tube if some air was trapped in A.

If some air was trapped in region A (the space above the mercury column), the level of the mercury in the tube would fall. This is because the trapped air would exert its own pressure, which would add to the pressure of the mercury column, thus reducing the height of the mercury column needed to balance the external atmospheric pressure.

7. A sucker stuck on a flat surface was pulled by means of a spring balance as shown in figure below. When the sucker was just about to be pulled off the surface, the spring balance reading was 40 N. Given that the area of the sucker was 4.4 cm2^2, determine the air pressure in Nm−2^{-2}.

  1. Identify the given values.

    Force F=40 NF = 40 \text{ N}
    Area A=4.4 cm2A = 4.4 \text{ cm}^2

  2. Convert the area to square meters.

    Since 1 cm=10−2 m1 \text{ cm} = 10^{-2} \text{ m}, then 1 cm2=(10−2 m)2=10−4 m21 \text{ cm}^2 = (10^{-2} \text{ m})^2 = 10^{-4} \text{ m}^2.

    A=4.4 cm2×10−4 m21 cm2=4.4×10−4 m2A = 4.4 \text{ cm}^2 \times \frac{10^{-4} \text{ m}^2}{1 \text{ cm}^2} = 4.4 \times 10^{-4} \text{ m}^2

  3. Calculate the pressure using the formula P=FAP = \frac{F}{A}.

    P=40 N4.4×10−4 m2P = \frac{40 \text{ N}}{4.4 \times 10^{-4} \text{ m}^2} P=400.00044 N/m2P = \frac{40}{0.00044} \text{ N/m}^2 P≈90909.09 N/m2P \approx 90909.09 \text{ N/m}^2

  4. Round the answer to an appropriate number of significant figures (2 significant figures, based on the given values).

    P≈9.1×104 N/m2P \approx 9.1 \times 10^4 \text{ N/m}^2

    The air pressure is 9.1×104 N/m2\boxed{\text{9.1} \times 10^4 \text{ N/m}^2}.

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