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Laws of Motion Test - 2

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Laws of Motion Test - 2
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  • Question 1
    1 / -0

    The height of the mercury column in a mercury barometer has a height of about _______ at sea level equivalent to one atmosphere.

    Solution

    In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of P; given by ρgh. A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of 1 mm is called a torr (after Torricelli)

  • Question 2
    1 / -0

    The density of the atmosphere at sea level is 1.29 kg/m3. Assume that it does not change with altitude. The height up to which the atmosphere extends,

    Solution

    According to the expression to obtain the pressure at a depth h,

    ρgh = 1.29 kg m-3 × 9.8 m s2 × (h) m = 1.01 × 105 Pa

    ∴ h = 7989 m ≈ 8 km

  • Question 3
    1 / -0

    At a depth of 1000 m in an ocean the gauge pressure and the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is 1.03 × 103 kg m-3, g = 10m s-2.)

    Solution

    Here h = 1000 m and ρ = 1.03 × 103 kg m­-3.

    Gauge pressure is P – Pa = ρgh = Pg = 1.03 × 103 kg m­-3 × 10 ms-2 × 1000 m = 103 × 105 Pa ≈ 103 atm.

    Now the force exerted on the window of the submarine

    The pressure outside the submarine is P = Pa + ρgh and the pressure inside it is Pa. Hence, the net pressure acting on the window is gauge pressure, Pg = ρgh. Since the area of the window is A = 0.04 m2, the force acting on it is F = PgA = 103 × 105 Pa × 0.04 m2 = 4.12 × 105 N.

  • Question 4
    1 / -0

    According to the Pascal’s law for transmission of fluid pressure;

    Solution

    Whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is the Pascal’s law for transmission of fluid pressure and has many applications in daily life. A number of devices such as hydraulic lift and hydraulic brakes are based on the Pascal’s law.

  • Question 5
    1 / -0

    What is the pressure on a swimmer 10 m below the surface of a lake?

    Solution

    Here the depth of the lake from the water surface is, h = 10 m and ρ = 1000 kg m-3.

    Take g = 10 m s-2

    As we know the pressure at the bottom is given as,

    P = Pa + ρgh = 1.01 × 105 + 1000 × 10 × 10 = 2.01 × 105 Pa ≈ 2 atm

  • Question 6
    1 / -0

    The excess of pressure, P – Pa, at depth h is called a ____________ at that point.

    Solution

    The pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess of pressure, P −Pa, at depth h is called a gauge pressure at that point.

  • Question 7
    1 / -0

    The value of atmospheric pressure is equal to,

    Solution

    Atmospheric pressure, sometimes also called barometric pressure (after the sensor), is the pressure within the atmosphere of Earth. The standard atmosphere (symbol: atm) is a unit of pressure defined as 1,013.25 mbar (101,325 Pa), equivalent to 760 mm Hg or 760 torr

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