Question

What is the difference between the pressure on the bottom of a pool and the pressure on the water surface?
A) $gh$
B) $\dfrac{g}{h}$
C) $0$
D) $none$

Answer 1

As we go down deeper in the pool our body feels more difficult to move inside the water this is due to the pressure exerted by the water on above our body. Additional pressure due to the fluid is given by $P = \rho gh$.
Formula used :- Pressure due to the weight of the liquid is given by $P = \rho gh$.
Where P is the pressure, h is the height of the liquid and $\rho$ is the density of the liquid, and g is the acceleration due to gravity.

Complete step by step answer:
Here is the diagram related to the above condition.

Since there is atmospheric pressure on the top surface of the water so,
$\Rightarrow {P_{top\;surface}} = {P_{atmospheric}}$
${P_{top\;surface}} = {P_{atmospheric}}$
As we go through deep in water the pressure gets added to this atmospheric pressure by the factor $\rho g$times the depth.
So the pressure observed at the bottom of the swimming pool would be
$\Rightarrow {P_{bottom\,\;surface}} = {P_{atmospheric}} + \rho g(depth)$
Since the depth of the pool is $h$, pressure at bottom of the pool would be
$\Rightarrow {P_{bottom\,\;surface}} = {P_{atmospheric}} + \rho g(h)$
We have to find the difference between the pressure at the top surface and bottom surface so
$\Rightarrow {P_{bottom\;surface}} - {P_{top\;surface}} = \left( {{P_{atmospheric}} + \rho g(h)} \right) - {P_{atmospheric}}$
Simplifying the equation
$\Rightarrow {P_{bottom\;surface}} - {P_{top\;surface}} = \rho gh$
Water has a density of 1.
$\Rightarrow \rho = 1$
Putting the value of $\rho$, we get
$\Rightarrow {P_{bottom\;surface}} - {P_{top\;surface}} = gh$
Hence the difference between the pressure at the top surface and bottom surface is $gh$.
Option (A) is the correct option.
Atmospheric pressure is also due to weight of the air above a given height.
The atmospheric pressure at earth’s surface varies a little due to the large-scale flow of the atmosphere induced by earth’s rotation .
1 atmospheric pressure = 1 atm = ${P_{atmospheric}}$
${P_{atmospheric}} = 1.01 \times {10^5}\dfrac{N}{{{m^2}}} = 101kPa$

Note:- Pressure is the weight of the liquid divided by the area supporting this weight.
We notice that air pressure changes on an elevator ride that transports us many stories, but we need to dive a meter only or so below the surface of the swimming pool to feel the pressure increase.
The difference is that the water is much denser than air about 775 times as dense.