Question

Calculate the absolute pressure at the bottom of a 20 m deep pool of water when -- (atmospheric pressure = 1.013 x 10⁵ Pa. , Density of water = 1000 kg/m³)

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{atmospheric}\mathrm{pressure},{\mathrm{P}}_0=1.013\times {10}^5\mathrm{Pa} \\ \mathrm{density}\mathrm{of}\mathrm{water},\mathrm{\rho }=1000\mathrm{kg}/{\mathrm{m}}^3 \\ \mathrm{depth},\mathrm{h}=20\mathrm{m} \\ \mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Absolute}\mathrm{pressure}\mathrm{at}\mathrm{the}\mathrm{bottom},\mathrm{P}={\mathrm{P}}_0+\mathrm{\rho gh} \\ \mathrm{P}=\left(1.013\times {10}^5\right)+\left(1000\times 9.8\times 20\right) \\ \mathrm{P}=\left(1.013\times {10}^5\right)+\left(196000\right) \\ \mathrm{P}=\left(101300\right)+\left(196000\right) \\ \mathrm{P}=297300\mathrm{Pa} \end{gathered}$$