Question

At what depth below the surface of water will the pressure be equal to 1 atmosphere. The atmospheric pressure is ${10}^{5}Pa$. Take density of water as ${10}^{3}kg{m}^{-3}$and $g=9.8m{s}^{-2}$.

Answer 1

Step 1: Given data

$$\begin{gathered} Atmosphericpressure\left(P\right)={10}^{5}Pa \\ Densityofwater\left(\rho \right)={10}^{3}kg{m}^{-3} \\ g=9.8m{s}^{-2} \end{gathered}$$

Let us assume the depth below the surface of the water at which pressure will be equal to 1 atm as$h$.

Step 2: Calculating the depth

$$\begin{gathered} P=\rho \times g\times h \\ {10}^{5}Pa={10}^{3}kg{m}^{-3}\times 9.8m{s}^{-2}\times h \\ \Rightarrow h=\frac{{10}^5}{{10}^3\times 9.8} \\ =10.2m \end{gathered}$$

Hence, at a depth of $10.2m$ below the surface of the water, the pressure will be equal to atmospheric pressure.