A submarine is designed to withstand an absolute pressure of 100 ~atm$100 \mathrm{~atm}$. How deep can it go below the water surface? (Consider the density of water = 1000 ~kg m⁻³$1000 \mathrm{~kg\ m^{-3}}$, 1 ~atm = 1 × 10⁵ ~Pa$1 \mathrm{~atm} = 1 \times 10^5 \mathrm{~Pa}$ and gravitational acceleration g = 10 ~m/s²$\mathrm{g} = 10 \mathrm{~m/s}^2$)

Related Formula

Core Logic

The absolute pressure P$P$ at a depth h$h$ is the sum of the atmospheric pressure at the surface P₀$P_0$ and the hydrostatic gauge pressure ρ g h$\rho g h$. Given: Absolute maximum pressure P = 100 ~atm = 100 × 10⁵ ~Pa = 10⁷ ~Pa$P = 100 \mathrm{~atm} = 100 \times 10^5 \mathrm{~Pa} = 10^7 \mathrm{~Pa}$ Atmospheric pressure P₀ = 1 ~atm = 10⁵ ~Pa$P_0 = 1 \mathrm{~atm} = 10^5 \mathrm{~Pa}$ Density ρ = 1000 ~kg/m³ = 10³ ~kg/m³$\rho = 1000 \mathrm{~kg/m}^3 = 10^3 \mathrm{~kg/m}^3$ Gravity g = 10 ~m/s²$g = 10 \mathrm{~m/s}^2$

Step 1: Solve for Depth

Pattern Recognition

Every 10 ~m$10 \mathrm{~m}$ of water depth adds approximately 1 ~atm$1 \mathrm{~atm}$ of gauge pressure. For 100 ~atm$100 \mathrm{~atm}$ absolute, subtract 1 ~atm$1 \mathrm{~atm}$ for the surface, leaving 99 ~atm$99 \mathrm{~atm}$ gauge pressure. 99 × 10 ~m = 990 ~m$99 \times 10 \mathrm{~m} = 990 \mathrm{~m}$.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids