Question

Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is $27.5m$. Assume the density of the water is $1.00\times {10}^{3}Kg/m^3$and that the air above is at a pressure of $101.3kpa$. (b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of$35.0cm$ ?

Short answer

(a) The absolute pressure at the bottom of a freshwater lake is$P=3.71\times {10}^{5}Pa$ .

(b) The force is exerted by the water on the window of an underwater vehicle is$F=3.57\times {10}^{4}N$ .

Step-by-step solution

Pressure:

The pressure in a fluid at rest varies with depth h in the fluid according to the expression

$P=\left(P_0+\rho gh\right)$

Where, $P_0$ is the pressure at $h=0$and p is the density of the fluid, assumed uniform.

The pressure P in a fluid is the force per unit area exerted by the fluid on a surface:

$P=\frac{F}{A}$

Where, P is the pressure, A is area, and F is the force exerted on the fluid.

(a) Calculate the absolute pressure at the bottom of a freshwater lake:

At a depth of 27.5m , the absolute pressure is,

$$\begin{gathered} P=\left(P_0+\rho gh\right) \\ =101.3kpa+\left(1.00\times {10}^{3}Kg/m^3\right)\left(9.8m/s^2\right)\left(27.5m\right) \\ =3.71\times {10}^{5}Pa \end{gathered}$$

Hence, the absolute pressure at the bottom of a freshwater lake is $3.71\times {10}^{5}Pa$.

(b) Force is exerted by the water on the window of an underwater vehicle at this depth:

The inward force the water will exert on the window is:

$$\begin{gathered} F=PA \\ =P\left(\pi r^2\right) \\ =P\pi {\left(\frac{D}{2}\right)}^2 \end{gathered}$$

$$\begin{gathered} F=\left(3.71\times {10}^{5}Pa\right)\left(3.14\times {\left(\frac{35.0\times {10}^{-2}m}{2}\right)}^2\right) \\ =3.57\times {10}^{4}N \end{gathered}$$

Hence, the force exerted by the water on the window of an underwater vehicle is $3.57\times {10}^{4}N$.