Physics Numericals

Q1: Convert 1 mm of Hg into Pascal. (Take density of Hg = (13.6 times 10^3 , text{kg/m}^3) and ( g = 9.8 , text{m/s}^2 )).

Solution:
Given:
Height of mercury column: ( h = 1 , text{mm} = 0.001 , text{m} )
Density of mercury: ( rho = 13.6 times 10^3 , text{kg/m}^3 )
Gravitational acceleration: ( g = 9.8 , text{m/s}^2 )
Formula: ( P = rho cdot g cdot h )
Calculation:
[
P = 13.6 times 10^3 cdot 9.8 cdot 0.001 = 133.28 , text{Pa}
]

Q2: At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in the barometer is replaced by water? (Take density of mercury = (13.6 times 10^3 , text{kg/m}^3)).

Solution:
Given:
Height of mercury column: ( h_{text{Hg}} = 0.70 , text{m} )
Density of mercury: ( rho_{text{Hg}} = 13.6 times 10^3 , text{kg/m}^3 )
Density of water: ( rho_{text{water}} = 10^3 , text{kg/m}^3 )
Formula:
[
rho_{text{Hg}} cdot h_{text{Hg}} = rho_{text{water}} cdot h_{text{water}}
]
Calculation:
[
h_{text{water}} = frac{rho_{text{Hg}} cdot h_{text{Hg}}}{rho_{text{water}}} = frac{13.6 cdot 0.70}{1} = 9.52 , text{m}
]

Q3: At sea level, the atmospheric pressure is 76 cm of Hg. If air pressure falls by 10 mm of Hg per 120 m of ascent, what is the height of a hill where the barometer reads 70 cm Hg? Assume atmospheric pressure falls linearly with ascent.

Solution:
Given:
Atmospheric pressure at sea level: ( P_{text{sea}} = 76 , text{cm Hg} )
Pressure at hilltop: ( P_{text{hill}} = 70 , text{cm Hg} )
Rate of pressure drop: ( 10 , text{mm Hg per 120 m} = 1 , text{cm Hg per 120 m} )
Formula:
Pressure difference: ( Delta P = P_{text{sea}} – P_{text{hill}} )
Height: ( h = Delta P cdot text{Rate of ascent} )
Calculation:
[
Delta P = 76 – 70 = 6 , text{cm Hg}
]
[
h = 6 cdot 120 = 720 , text{m}
]

Q4: At sea level, the atmospheric pressure is (1.04 times 10^5 , text{Pa}). Assuming ( g = 10 , text{m/s}^2 ) and density of air to be uniform and equal to (1.3 , text{kg/m}^3), find the height of the atmosphere.

Solution:
Given:
Atmospheric pressure: ( P = 1.04 times 10^5 , text{Pa} )
Density of air: ( rho = 1.3 , text{kg/m}^3 )
Gravitational acceleration: ( g = 10 , text{m/s}^2 )
Formula: ( h = frac{P}{rho cdot g} )
Calculation:
[
h = frac{1.04 times 10^5}{1.3 cdot 10} = 800 , text{m}
]

Q5: Assuming the density of air to be (1.295 , text{kg/m}^3), find the fall in barometric height in mm of Hg at a height of 107 m above sea level. (Take density of mercury = (13.6 times 10^3 , text{kg/m}^3)).

Solution:
Given:
Height: ( h = 107 , text{m} )
Density of air: ( rho_{text{air}} = 1.295 , text{kg/m}^3 )
Density of mercury: ( rho_{text{Hg}} = 13.6 times 10^3 , text{kg/m}^3 )
Gravitational acceleration: ( g = 10 , text{m/s}^2 )
Formula:
Pressure drop due to air: ( P = rho_{text{air}} cdot g cdot h )
Equivalent height in mercury: ( h_{text{Hg}} = frac{P}{rho_{text{Hg}} cdot g} )
Calculation:
[
P = 1.295 cdot 10 cdot 107 = 1387.15 , text{Pa}
]
[
h_{text{Hg}} = frac{1387.15}{13.6 times 10^3 cdot 10} = 0.010 , text{m} = 10 , text{mm Hg}
]