Class 12 Exam > Class 12 Videos > Density and Pressure - Fluids, Physics
Density and Pressure - Fluids, Physics Video Lecture - Class 12
FAQs on Density and Pressure - Fluids, Physics Video Lecture - Class 12
| 1. What is the formula to calculate density? |  |
Ans. The formula to calculate density is: density = mass/volume. It is a measure of how much mass is contained in a given volume of a substance.
| 2. How does density affect the pressure of a fluid? | |
Ans. Density plays a crucial role in determining the pressure of a fluid. According to the equation of fluid pressure, pressure = density x gravitational acceleration x height, an increase in density will result in an increase in pressure, given that the other variables remain constant.
| 3. What are the units of density and pressure? | |
Ans. The SI unit of density is kilogram per cubic meter (kg/m^3), while the SI unit of pressure is pascal (Pa). However, other commonly used units for density include grams per cubic centimeter (g/cm^3) and kilograms per liter (kg/L), while pressure is often measured in atmospheres (atm) and pounds per square inch (psi).
| 4. How does the density of a fluid affect its buoyancy? | |
Ans. The density of a fluid determines its buoyancy. When an object is placed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by the object. If the density of the object is greater than the density of the fluid, it will sink. Conversely, if the density of the object is less than the density of the fluid, it will float.
| 5. How does pressure change with depth in a fluid? | |
Ans. In a fluid, pressure increases with depth. This is due to the weight of the fluid above pushing down and exerting a force on the fluid below. The relationship between pressure and depth is given by the equation: pressure = density x gravitational acceleration x depth. As the depth increases, the pressure also increases.
Text Transcript from Video
Welcome back.
To just review what I was doing
on the last video before
I ran out of time, I said that
conservation of energy tells
us that the work I've put into
the system or the energy that
I've put into the system--
because they're really the
same thing-- is equal to the
work that I get out of the
system, or the energy that
I get out of the system.
That means that the input work
is equal to the output work,
or that the input force times
the input distance is equal to
the output force times the
output distance-- that's just
the definition of work.
Let me just rewrite this
equation here.
If I could just rewrite this
exact equation, I could say--
the input force, and let me just
divide it by this area.
The input here-- I'm pressing
down this piston that's
pressing down on this
area of water.
So this input force-- times
the input area.
Let's call the input 1, and
call the output 2 for
simplicity.
Let's say I have a piston
on the top here.
Let me do this in a good color--
brown is good color.
I have another piston here, and
there's going to be some
outward force F2.
The general notion is that I'm
pushing on this water, the
water can't be compressed, so
the water's going to push up
on this end.
The input force times the input
distance is going to be
equal to the output force times
the output distance
right-- this is just the law of
conservation of energy and
everything we did with
work, et cetera.
I'm rewriting this equation, so
if I take the input force
and divide by the input area--
let me switch back to green--
then I multiply by the
area, and then I just
multiply times D1.
You see what I did here-- I just
multiplied and divided by
A1, which you can do.
You can multiply and divide by
any number, and these two
cancel out.
It's equal to the same thing
on the other side, which is
F2-- I'm not good at managing
my space on my whiteboard--
over A2 times A2 times D2.
Hopefully that makes sense.
What's this quantity right here,
this F1 divided by A1?
Force divided by area, if you
haven't been familiar with it
already, and if you're just
watching my videos there's no
reason for you to be, is
defined as pressure.
Pressure is force in a given
area, so this is pressure--
we'll call this the
pressure that I'm
inputting into the system.
What's area 1 times
distance 1?
That's the area of the tube
at this point, the
cross-sectional area,
times this distance.
That's equal to this volume
that I calculated in the
previous video-- we could
say that's the
input volume, or V1.
Pressure times V1 is equal to
the output pressure-- force 2
divided by area 2 is the output
pressure that the water
is exerting on this piston.
So that's the output
pressure, P2.
And what's area 2 times D2?
The cross sectional area, times
the height at which how
much the water's being displaced
upward, that is
equal to volume 2.
But what do we know about
these two volumes?
I went over it probably
redundantly in the previous
video-- those two volumes are
equal, V1 is equal to V2, so
we could just divide both
sides by that equation.
You get the pressure input is
equal to the pressure output,
so P1 is equal to P2.
I did all of that just to show
you that this isn't a new
concept: this is just the
conservation of energy.
The only new thing I did is I
divided-- we have this notion
of the cross-sectional area,
and we have this notion of
pressure-- so where
does that help us?
This actually tells us-- and
you can do this example in
multiple situations, but I like
to think of if we didn't
have gravity first, because
gravity tends to confuse
things, but we'll introduce
gravity in a video or two-- is
that when you have any external
pressure onto a
liquid, onto an incompressible
fluid, that pressure is
distributed evenly throughout
the fluid.
That's what we essentially just
proved just using the law
of conservation of energy, and
everything we know about work.
What I just said is called
Pascal's principle: if any
external pressure is applied to
a fluid, that pressure is
distributed throughout
the fluid equally.
Another way to think about it--
we proved it with this
little drawing here-- is, let's
say that I have a tube,
and at the end of the
tube is a balloon.
Let's say I'm doing this
on the Space Shuttle.
It's saying that if I increase--
say I have some
piston here.
This is stable, and
I have water
throughout this whole thing.
Let me see if I can use that
field function again-- oh no,
there must have been a
hole in my drawing.
Let me just draw the water.
I have water throughout this
whole thing, and all Pascal's
principle is telling us that
if I were to apply some
pressure here, that that net
pressure, that extra pressure
I'm applying, is going to
compress this little bit.
That extra compression is
going to be distributed
through the whole balloon.
Let's say that this right here
is rigid-- it's some kind of
middle structure.
The rest of the balloon is going
to expand uniformly, so
that increased pressure I'm
doing is going through the
whole thing.
It's not like the balloon will
get longer, or that the
pressure is just translated
down here, or that just up
here the balloon's going to get
wider and it's just going
to stay the same length there.
Hopefully, that gives you a
little bit of intuition.
Going back to what I had drawn
before, that's actually
interesting, because that's
actually another simple or
maybe not so simple machine
that we've constructed.
I almost defined it as a
simple machine when I
initially drew it.
Let's draw that weird thing
again, where it looks like
this, where I have
water in it.
Let's make sure I fill it, so
that when I do the fill, it
will completely fill, and
doesn't fill other things.
This is cool, because this is
now another simple machine.
We know that the pressure in is
equal to the pressure out.
And pressure is force divided
by area, so the force in,
divided by the area in, is equal
to the force out divided
by the area out.
Let me give you an example:
let's say that I were to apply
with a pressure in equal
to 10 pascals.
That's a new word, and it's
named after Pascal's
principle, for Blaise Pascal.
What is a pascal?
That is just equal to 10 newtons
per meter squared.
That's all a pascal is-- it's
a newton per meter squared,
it's a very natural unit.
Let's say my pressure in is 10
pascals, and let's say that my
input area is 2 square meters.
If I looked the surface of the
water there it would be 2
square meters, and let's say
that my output area is equal
to 4 meters squared.
What I'm saying is that I can
push on a piston here, and
that the water's going to push
up with some piston here.
First of all, I told you what my
input pressure is-- what's
my input force?
Input pressure is equal to input
force divided by input
area, so 10 pascals is equal to
my input force divided by
my area, so I multiply
both sides by 2.
I get input force is equal
to 20 newtons.
My question to you is what
is the output force?
How much force is the
system going to push
upwards at this end?
We know that must if my input
pressure was 10 pascals, my
output pressure would
also be 10 pascals.
So I also have 10 pascals is
equal to my out force over my
out cross-sectional area.
So I'll have a piston here,
and it goes up like that.
That's 4 meters, so I do 4
times 10, and so I get 40
newtons is equal to
my output force.
So what just happened here?
I inputted-- so my input force
is equal to 20 newtons, and my
output force is equal to 40
newtons, so I just doubled my
force, or essentially I had a
mechanical advantage of 2.
This is an example of a
simple machine, and
it's a hydraulic machine.
Anyway, I've just
run out of time.
I'll see you in the next video.
To just review what I was doing
on the last video before
I ran out of time, I said that
conservation of energy tells
us that the work I've put into
the system or the energy that
I've put into the system--
because they're really the
same thing-- is equal to the
work that I get out of the
system, or the energy that
I get out of the system.
That means that the input work
is equal to the output work,
or that the input force times
the input distance is equal to
the output force times the
output distance-- that's just
the definition of work.
Let me just rewrite this
equation here.
If I could just rewrite this
exact equation, I could say--
the input force, and let me just
divide it by this area.
The input here-- I'm pressing
down this piston that's
pressing down on this
area of water.
So this input force-- times
the input area.
Let's call the input 1, and
call the output 2 for
simplicity.
Let's say I have a piston
on the top here.
Let me do this in a good color--
brown is good color.
I have another piston here, and
there's going to be some
outward force F2.
The general notion is that I'm
pushing on this water, the
water can't be compressed, so
the water's going to push up
on this end.
The input force times the input
distance is going to be
equal to the output force times
the output distance
right-- this is just the law of
conservation of energy and
everything we did with
work, et cetera.
I'm rewriting this equation, so
if I take the input force
and divide by the input area--
let me switch back to green--
then I multiply by the
area, and then I just
multiply times D1.
You see what I did here-- I just
multiplied and divided by
A1, which you can do.
You can multiply and divide by
any number, and these two
cancel out.
It's equal to the same thing
on the other side, which is
F2-- I'm not good at managing
my space on my whiteboard--
over A2 times A2 times D2.
Hopefully that makes sense.
What's this quantity right here,
this F1 divided by A1?
Force divided by area, if you
haven't been familiar with it
already, and if you're just
watching my videos there's no
reason for you to be, is
defined as pressure.
Pressure is force in a given
area, so this is pressure--
we'll call this the
pressure that I'm
inputting into the system.
What's area 1 times
distance 1?
That's the area of the tube
at this point, the
cross-sectional area,
times this distance.
That's equal to this volume
that I calculated in the
previous video-- we could
say that's the
input volume, or V1.
Pressure times V1 is equal to
the output pressure-- force 2
divided by area 2 is the output
pressure that the water
is exerting on this piston.
So that's the output
pressure, P2.
And what's area 2 times D2?
The cross sectional area, times
the height at which how
much the water's being displaced
upward, that is
equal to volume 2.
But what do we know about
these two volumes?
I went over it probably
redundantly in the previous
video-- those two volumes are
equal, V1 is equal to V2, so
we could just divide both
sides by that equation.
You get the pressure input is
equal to the pressure output,
so P1 is equal to P2.
I did all of that just to show
you that this isn't a new
concept: this is just the
conservation of energy.
The only new thing I did is I
divided-- we have this notion
of the cross-sectional area,
and we have this notion of
pressure-- so where
does that help us?
This actually tells us-- and
you can do this example in
multiple situations, but I like
to think of if we didn't
have gravity first, because
gravity tends to confuse
things, but we'll introduce
gravity in a video or two-- is
that when you have any external
pressure onto a
liquid, onto an incompressible
fluid, that pressure is
distributed evenly throughout
the fluid.
That's what we essentially just
proved just using the law
of conservation of energy, and
everything we know about work.
What I just said is called
Pascal's principle: if any
external pressure is applied to
a fluid, that pressure is
distributed throughout
the fluid equally.
Another way to think about it--
we proved it with this
little drawing here-- is, let's
say that I have a tube,
and at the end of the
tube is a balloon.
Let's say I'm doing this
on the Space Shuttle.
It's saying that if I increase--
say I have some
piston here.
This is stable, and
I have water
throughout this whole thing.
Let me see if I can use that
field function again-- oh no,
there must have been a
hole in my drawing.
Let me just draw the water.
I have water throughout this
whole thing, and all Pascal's
principle is telling us that
if I were to apply some
pressure here, that that net
pressure, that extra pressure
I'm applying, is going to
compress this little bit.
That extra compression is
going to be distributed
through the whole balloon.
Let's say that this right here
is rigid-- it's some kind of
middle structure.
The rest of the balloon is going
to expand uniformly, so
that increased pressure I'm
doing is going through the
whole thing.
It's not like the balloon will
get longer, or that the
pressure is just translated
down here, or that just up
here the balloon's going to get
wider and it's just going
to stay the same length there.
Hopefully, that gives you a
little bit of intuition.
Going back to what I had drawn
before, that's actually
interesting, because that's
actually another simple or
maybe not so simple machine
that we've constructed.
I almost defined it as a
simple machine when I
initially drew it.
Let's draw that weird thing
again, where it looks like
this, where I have
water in it.
Let's make sure I fill it, so
that when I do the fill, it
will completely fill, and
doesn't fill other things.
This is cool, because this is
now another simple machine.
We know that the pressure in is
equal to the pressure out.
And pressure is force divided
by area, so the force in,
divided by the area in, is equal
to the force out divided
by the area out.
Let me give you an example:
let's say that I were to apply
with a pressure in equal
to 10 pascals.
That's a new word, and it's
named after Pascal's
principle, for Blaise Pascal.
What is a pascal?
That is just equal to 10 newtons
per meter squared.
That's all a pascal is-- it's
a newton per meter squared,
it's a very natural unit.
Let's say my pressure in is 10
pascals, and let's say that my
input area is 2 square meters.
If I looked the surface of the
water there it would be 2
square meters, and let's say
that my output area is equal
to 4 meters squared.
What I'm saying is that I can
push on a piston here, and
that the water's going to push
up with some piston here.
First of all, I told you what my
input pressure is-- what's
my input force?
Input pressure is equal to input
force divided by input
area, so 10 pascals is equal to
my input force divided by
my area, so I multiply
both sides by 2.
I get input force is equal
to 20 newtons.
My question to you is what
is the output force?
How much force is the
system going to push
upwards at this end?
We know that must if my input
pressure was 10 pascals, my
output pressure would
also be 10 pascals.
So I also have 10 pascals is
equal to my out force over my
out cross-sectional area.
So I'll have a piston here,
and it goes up like that.
That's 4 meters, so I do 4
times 10, and so I get 40
newtons is equal to
my output force.
So what just happened here?
I inputted-- so my input force
is equal to 20 newtons, and my
output force is equal to 40
newtons, so I just doubled my
force, or essentially I had a
mechanical advantage of 2.
This is an example of a
simple machine, and
it's a hydraulic machine.
Anyway, I've just
run out of time.
I'll see you in the next video.
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Jan 01, 2025 Last updated |
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