25. The atmospheric pressure decreases with the
increase in height from the sea level because a).
fluid pressure is directly proportional to the depth
from free surface b). fluid pressure increases with
the decrease in depth c). fluids exert pressure on
boats and aeroplanes only d). fluids do not have
mass.
Answers
Answer:
Variation of Pressure with Depth in a Fluid
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Define pressure in terms of weight.
Explain the variation of pressure with depth in a fluid.
Calculate density given pressure and altitude.
If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense. Consider the container in Figure 1.
A container with fluid filled to a depth h. The fluid’s weight w equal to m times g is shown by an arrow pointing downward. A denotes the area of the fluid at the bottom of the container and as well as on the surface.
Figure 1. The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.
Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid mg divided by the area A supporting it (the area of the bottom of the container):
P
=
m
g
A
.
We can find the mass of the fluid from its volume and density:
m = ρV.
The volume of the fluid V is related to the dimensions of the container. It is
V = Ah,
where A is the cross-sectional area and h is the depth. Combining the last two equations gives
m
=
ρ
A
h
.
If we enter this into the expression for pressure, we obtain
P
=
(
ρ
A
h
)
g
A
.
The area cancels, and rearranging the variables yields
P = hρg.
This value is the pressure due to the weight of a fluid. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation P = hρg represents the pressure due to the weight of any fluid of average density ρ at any depth h below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. Example 2: Calculating Average Density: How Dense Is the Air? illustrates this situation.
EXAMPLE 1. CALCULATING THE AVERAGE PRESSURE AND FORCE EXERTED: WHAT FORCE MUST A DAM WITHSTAND?
In Example 1. Calculating the Mass of a Reservoir from Its Volume, we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 2.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be 1.96 × 1013 N).
A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.
Figure 2. The dam must withstand the force exerted against it by the water it retains. This force is small compared with the weight of the water behind the dam.
Strategy for (a)
The average pressure
¯¯¯¯