The density of air near earth’s surface is $1.3{\text{\hspace{0.17em}kg\hspace{0.17em}m}}^{-3}$and the atmospheric pressure is$\text{1.0}\times {\text{10}}^5{\text{\hspace{0.17em}Nm}}^{-2}$. If the atmosphere had uniform density, same as that observed at the surface of the earth, the height of the atmosphere to exert the same pressure is

Given:

• Density of air, ρ = 1.3 kg/m³
• Atmospheric pressure, P = 1.0 × 10⁵ N/m²
• Acceleration due to gravity, g = 9.8 m/s²

P = ρ g h

Where:

• P = Atmospheric pressure
• ρ = Density of air
• g = Acceleration due to gravity
• h = Height of the atmosphere

h = P / (ρ.g) h = (1.0 × 10⁵) / (1.3 × 9.8) h ≈ 7850 m

Final Answer

The height of the atmosphere, assuming uniform density, is approximately 7850 meters.