the pressure exerted by a liquid column of depth 0.5m on the base on the base of its the container is 5000 nm.find the density of the liquid
Answers
Explanation:
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.