Chapter 14: Q23P (page 440)

A backyard swimming pool with a circular base of diameter is filled to depth $1.5 m$
(a) Find the absolute pressure at the bottom of the pool.
(b) Two persons with combined mass $150 k g$ enter the pool and float quietly there. No water overflows. Find the pressure increase at the bottom of the pool after they enter the pool and float

Short Answer

Expert verified

(a) The absolute pressure at the bottom of the pool: $P = 15 p a$

(b) The pressure increase at the bottom of the pool after they enter the pool and float $P = 523.6 p a$

Step by step solution

Step 1:

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B = ρ_{F l u i d} g V_{d i s p}$

Where $V_{d i s p}$ weight of fluid is displaced at and role="math" localid="1668141302142" $ρ_{F l u i d}$ is the density of the fluid.

Step 2:

Part(a):

Given:Initially,

Depth $h = 1.5 m$

Densityofwater $ρ_{w} = 1000 k g / m^{3}$

The absolute pressure at the bottom of the pool:

$P = ρ_{w} g h P = 1000 \times 10 \times 1.5 P = 15 p a$

Step 3:

Part(b):

Circularbaseofdiameter $6 m$

Radius: $r = 3 m$

Depth $h = 1.5 m$

Let masses of two person is $m_{1} & m_{2}$ then their combine mass $m = m_{1} + m_{2} = 150 k g$

Now when the two persons enter the pool, water level of pool rises by h,

By Archimedes’s principle,

$\begin{array}{rcl} B & = & ρ_{F l u i d} g V_{d i s p} \\ ρ_{w} V g & = & m g \\ V & = & \frac{m}{ρ_{w}} \\ V & = & \frac{150}{1000} \\ V & = & 0.15 m^{3} \end{array}$

Now area of the pool

$A = π \times r^{2} A = π \times 3^{2} A = 9 π$

So, pressure increase $(p)$ is:

$\begin{array}{rcl} P & = & ρ_{w} g h \\ P & = & ρ_{w} g \frac{V}{A} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (∵ V = A \times h) \\ P & = & \frac{1000 \times 10 \times 0.15}{9 π} \\ P & = & 523.6 p a \end{array}$