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:
\[
P = 1.295 \cdot 10 \cdot 107 = 1387.15 \, \text{Pa}
\]
\[
h_{\text{Hg}} = \frac{1387.15}{13.6 \times 10^3 \cdot 10} = 0.010 \, \text{m} = 10 \, \text{mm Hg}
\]