Last time we were discussing some of the fundamental
concepts of surface tension and we will continue
with that. Let us say that we want to see
some application of surface tension as a very
simple case.
To do that we will take an example when there
is a capillary tube which is small in radius,
it is dipped into a larger volume of fluid
and let us say that this was the initial level
of the fluid and now once this capillary is
introduced these fluid is expected to either
rise or fall and we will see that what dictates
that it should rise or fall. To understand
that let us assume that the fluid is rising
over the capillary.
If it rises over the capillary then what dictates
is rising and to what extent it should rise?
Let us try to make an estimate of that. First
question should come that why it should rise
or why it should fall? When there is a fluid
it is a collection of molecules after all
and these molecules are subjected to various
interactions. One type of interaction is the
intermolecular forces of attraction between
the similar types of molecules which are like
so called cohesive forces.
The other one is the intermolecular force
of interaction between the molecules of the
fluid and the molecules at the boundary. So
when you have that type of interaction that
is known as addition. So it is between two
different types of entities. Now, if the addition
wins over the cohesion then that means the
surface at the end has a net attraction towards
the fluid and it likes the fluid.
Earlier we used a terminology called as hydrophilic
material. So such types of substrates which
like water are called as hydrophilic ones.
A better terminology would be waiting because
when we say hydro it means water so to say.
But it may be any other fluid also. So better
we say that those are waiting surfaces. So
the surface is which want to be wet with the
fluid that is there.
Let us say that there is a waiting fluid and
it has formed a meniscus like this. This meniscus
can be of a very complicated shape but let
us just for sake of simplicity assume that
it is a part of a hemisphere or like it is
a hemispherical gap so to say. Now, if you
try to identify that what are the various
forces which are related to the development
of this meniscus. We have discussed about
this earlier.
So, just if we reiterate that if you have
a meniscus like this loosely speaking there
will be a force because of surface tension
which is acting over the periphery. There
is a pressure acting from this side, there
is a pressure acting from the other side,
the bottom side and it is basically filled
with the same fluid which is also there in
the reservoir.
So, if we say that at the top there is some
gas or vapor and at the bottom there is a
liquid then let us say this is P liquid and
let us say this is P gas. So, which one will
be better here P liquid or P gas in this configuration?
Do you expect P gas to be more because it
has to balance the P liquid and the component
of the surface tension in that direction.
So, if you have P gas > P liquid that means
there is a pressure differential across the
meniscus.
So, if you have the pressure in the gas and
the atmospheric pressure obviously the pressure
in the liquid is not atmospheric pressure
across the meniscus and there is a differencing
pressure and that difference how do you relate?
We had derived a formula delta P = sigma * 1/R1
+ 1/R2. So, if we consider it a part of sphere
R1=R2=R. So we are approximating the shape
of the meniscus in that way.
Otherwise it may have different curvature
at like different planes. But this is a simplistic
assumption so if R1=R2=say small r, the small
r is not same as the radius of the capillary.
So this is different because the small r is
the radius of curvature of the meniscus which
is not same as the radius of the capillary.
So, we have this as 2 sigma/small r, obviously
we need to relate small r with capital R.
And to do that we can say have a small geometrical
construction. Let us say that this angle is
theta. So, when you have this angle as theta
this is the angle made by the tangent to the
interface with the vertical. So that should
be same as the angle between the normal to
the interface and the horizon. So, let us
say that that is theta. This normal is in
the direction of the radius of curvature of
the interface.
So, this length is what small r and what is
this length? This is capital R because this
is say the central line of the tube. So this
is capital R. So, how you can relate small
r with capital R? So, small r will be capital
R/cos theta. So we can simplify these 2 sigma/r
as 2 sigma/capital R and then cos theta will
go in the numerator. And what is this delta
P? This delta P is the difference in pressure
between the outside atmosphere and what is
there just inside.
What is the pressure at the outside atmosphere?
The pressure at the outside atmosphere is
same as what is the pressure at this level.
So, if you know what is the pressure at this
level then from that you can find out a relationship
between the pressure difference of these 2
and that pressure difference is nothing but
because of the presence of this much of liquid
color. So, if this is let us say that it has
average height of h.
So, this is nothing but h * rho of the fluid
* g. From here you can find out an expression
for h which is 2 sigma cos theta/rho gR. This
expression is very simple and this expression
is valid only if this meniscus is under a
static equilibrium condition. It is not moving
but still it gives a lot of good insight because
it tells you that if there is a capillary
rise say in these kind of a situation we call
it a capillary rise because as if the fluid
is rising from the reservoir towards the capillary.
So, that capillary rise this dictated by many
things one of the features of course is the
angle theta which is the contact angle and
which depends on the combination of the fluid.
So, if we say that for a glass water combination,
glass water air combination say this theta
is close to 0 may be. So, that means that
it requires 3 different phases so to say,
I mean glass is the solid phase which forms
the surface of the capillary they do have
water here and may be air here.
But if you replace water we say mercury it
may be something different and it may be possible
that instead of a rise it has a fall. That
is dictated by the competition between the
adhesion and the cohesion force and eventually
it is manifested by what is the value of the
contact angle. So, in that case say if the
contact angle say is 140 degree which is like
if it is mercury, glass, air that may be like
one of the possibilities.
So, when you have these one that means you
will get if you have theta in that range obviously
h is negative. That means instead of rise
it is a fall, the meniscus shape also will
be just The other important factors which
is one of the decisive factors is what is
capital R. Because in capital R is large then
no matter what ever is the contact angle this
effect will be small and we will virtually
be that.
At the same time, if R is very small then
this h can be very very large. We get such
examples in nature very nicely. That is if
you have trees, these trees absorb water from
the ground there is no pump which is existing
in the nature as an artificial mean of pumping
the fluid from the root to the topmost leaves
and branches but you will see tall trees also
get nutrient from the ground.
So, when they are transported by the fluid
medium which is the so called ascent of sap
in terms of the biological terminology. It
goes vertically such a large distance, it
covers such a huge height just with the consideration
that this R is small. So those are really
very narrow capillaries and then the capillary
acts like a pump. So, just because of the
surface tension it can attain a large height
of transport.
And that kind of very beautiful mechanism
prevails in nature and that makes the plants
at least the tall trees sustain their lives.
Now, just to have a slight variation from
these let us consider an example when you
do not have just one fluid as liquid and another
fluid as gas but may be both of the fluids
are liquids and they are immiscible ones.
If they are miscible liquids obviously when
you put them one with the other they may mix
with each other.
And the clear meniscus may be not be found.
But if they are not miscible with each other
there may be a clear meniscus that is form.
Let us look into an example with the short
movie and let us try to see that what kind
of meniscus may be form with such an example.
So, in this example what we will try to see
is that if there are 2 different liquids which
are there side by side then when these 2 different
liquids are put in a capillary tube sometimes
magically because of the result in surface
tension transport that is created. The entire
column may move from one end to the other
end and we will see one such example here.
So there will be two different fluids, those
liquids which have already been put and see
that it is just like moving magically. It
is not that there is a pump that is being
put or there is no other driving force that
has been created to induce the motion. Here
we will not first be bothered about the motion
we have first considering the equilibrium.
So, let us consider such a case but not a
moving case but two different liquids which
are keeping meniscus in equilibrium.
So, let us say that you have a capillary tube.
In the capillary tube now you have say two
liquids on the top you have say water and
in the bottom say you have mercury. With the
given tube materials say it is glass, it assumes
this particular meniscus shape. The outside
is say filled up with mercury. So there is
mercury in the outside and water is being
poured from inside. Now, if you see let us
concentrate our focus on the meniscus.
Let us say that the water is being poured
from this height and there has been a so called
depression of the meniscus from the top levels.
Let us say that this is h. Again when you
see such a simple analysis you should keep
in mind that this has many approximations.
When I say that this height is h obviously
it has no sanctity because we are disregarding
the variation in height from one end to the
other when we calculate the pressure difference.
Whenever we calculated the pressure difference
here delta P as h rho g that h, we assume
some average height may be the center one
we took as a reference. But it is not actually
a uniform height. The entire meniscus has
a variation in height and one of the approximations
maybe that not to use this as the height but
use the center line one in state. Just like
what we did in the other example. So maybe
take this as h. What is the error in these?
When you are concentrating this as the h if
you referred to such a figure you are neglecting
the shaded volume. So the shaded volume has
a contribution. Effectively if you see it
is the weight of the volume of the liquid
that has been sustained through the surface
tension and the pressure differential. And
there these shaded volumes also have their
own roles. It is only an approximation by
which we are neglecting it.
There may be cases when this gives rise to
some significant error and fortunately in
most cases it does not give rise to that much
error. So it is fine as a practical approximation.
Now, if you consider this surface and try
to see that what are the different types of
forces which are acting on this surface you
may evaluate the surface tension force just
like what you did in the previous example.
So, when you are considering the contact angle
say we are interested about the angle that
is made by the mercury with the glass. So
that is measured with respect to this one.
So, this is basically the theta. So, it is
important to see what is the sign convention
for the definition. So when we are referring
to the mercury it should be from the solid
with in the mercury domain that is how the
angle is there.
Otherwise with a different notation may be
180 – theta also be taken as an equivalent
representation of the effect of the contact
angle. So, this angle will be 180 degree – theta
and for writing the force balance that angle
will be useful. So, the same formula will
be applicable here. That is if you have delta
p again assuming spherical nature and all
those things you will have that as 2 sigma
cos of 180 degree – theta/R.
So, now on which side the pressure will be
more mercury or water? And how can you calculate
the differential of that pressure? When you
consider the waters height it is basically
h * rho water * g. That is the height of the
column of water. When you consider the mercury
size it is h * rho mercury * g. So the difference
in pressure is like h * difference of rho
mercury – rho water * g that is your delta
p.
So from here you will get an expression for
what is h? And this h is clearly a depression
in this case that has been by this theta which
is roughly 140 degree as an example. So if
you know the surface tension coefficient then
it is possible to put that in this expression
and find out what should be the h under these
conditions. So again this has many approximations
but it gives some kind of idea that what should
be the estimation for capillary rise or capillary
depression.
With surface tension one may also have a dynamic
nature of the meniscus and when you have a
dynamic nature of the meniscus it is not that
you just have to consider this type of equilibrium
at the interface. You may have to consider
overall dynamical nature of one fluid as it
is displacing the other and moving in the
capillary. In the process there are may be
many things. In the static condition we have
a contact angle.
In the dynamic condition this contact angle
may change and the change of this contact
angle may be because of the dynamic nature
of the forces which are acting on the system.
One of such forces is the viscous force. Then
you also have a dynamically evolving may be
force of interaction. So it is possible to
have all the forces of interaction which are
not just like constants but those are evolving
dynamically as the shape of the meniscus is
changing.
Because as the shape of the meniscus is changing
you have different arrangements of the molecules
close to the wall and that may dynamically
give rise to different contact angles. So,
the contact angle that we are referring to
in these cases is commonly known as a static
contact angle. But when the dynamics is evolving
one may have a dynamic contact angle which
depends on the relative interplay of various
forces which are acting.
And since the two important forces dictating
these type of capillary advancement in a dynamic
condition or the surface tension and the viscous
forces. So, there relative interplay has a
strong role to play in determining the contact
angle that evolves dynamically. Surface roughness
also has a strong role to play because that
dictates the proper intermolecular interaction
close to the surface.
So, there is a very rich physics that takes
place close to the interface in a dynamic
condition and this elemental study does not
focus on that. It gives just a broad idea
of if it is a static condition what can be
the consequence of surface tension. But at
least it gives us an idea that surface tension
may be a very important force in our small
scale and as the radius becomes smaller and
smaller its effect becomes more and more prominent.
With this background, now we will move into
more general considerations for equilibrium
of fluid elements. Here, whenever we were
discussing the surface tension force we were
assuming that the pressure is being distributed
in a particular way and we were intuitively
using some concept of high school physics
that if you have a depth of h then what should
be the variation in pressure because of that
depth of h. Now we will look into it more
formally.
So, we will go into the understanding of fluid
statics. We start with an example of a fluid
element which is in static equilibrium and
we take an example of a 2 dimensional fluid
elements just for simplicity. We have taken
such examples earlier what these examples
signify that you have a uniform width in the
other direction perpendicular to the plane
of the board. So, let us say that delta x
and delta y are the dimensions of this fluid
element.
The fluid element is at rest. When this fluid
element is at rest that means we assured that
it is non-deforming because deforming fluid
element is definitely not a fluid element
at rest and when it is non-deforming we are
clear that there is no sheer which is acting
on it. That means there is only normal force
which is acting on all the phases of the fluid
element.
So, we can designate the state of stress on
each phase of the fluid element by pressure.
So, let us try to do that. Let us say that
we are only writing forces along the x direction.
Just for simplicity similar equations will
be valid for the y direction. When you have
say the left phase under consideration just
like these let us say that P is the pressure
on the left phase and the force corresponding
to that is p * delta y * say 1 which is the
width of the fluid element.
When you come to the opposite phase we are
bothered about these phases right now because
we are only identifying the forces acting
along x because we will write equation of
the equilibrium along x not that forces are
not there on the other phases. So, this is
not a complete free-body diagram. It only
just shows the forces along x direction. So,
if the pressure here is p what should be the
pressure here? That is under question.
Will it be p, will it be something different
from p in general. We are not really committed
to what are the other forces which are acting
on it. Then maybe any other body force which
is acting on it along x and y. So, if the
pressure here is p then the question is will
the pressure here be p or something else in
general. Special case of may also be p here.
But we are talking about a more general consideration.
Mathematically speaking what question we are
trying to answer, we have a function here
say p we want to find out the value of the
function at a different location say this
location is x. We are interested to now find
out the value of the function at x + delta
x in terms of what is the value of the function
at x. The function here is p. That means we
want to see that what is p at x + delta x
in terms of what is p at x and that we can
easily do by using a Taylor series expansion.
So that we will do, we will write these as
p at x + the first order partial derivate
of p with respect to x * delta x + and so
on. There are infinite number of terms but
as you take delta x very small maybe you may
neglect the higher order terms in comparison
to the dominating term and the gradient. Keeping
that in mind that we are treating with cases
where delta x, delta y are very small. So,
delta x, delta y all tending to 0.
So, this will become from the expression that
we have here what we can write this will be
p that will be the pressure here. We will
keep this in mind so later on whenever we
encounter any function we will use the Taylor
series expansion to identify what is the change
that is taking place across different cases
of fluid elements because that we will have
to do very commonly many of our analysis.
So, these multiplied by the area on which
it is acting is the force due to pressure
on this phase. Let us say that there is a
body force which is also acting on the fluid
element. So the body force let us say that
bx is the body force mass acting along x.
So, bx is body force mass along x. So, what
will be the total body force which is acting
on this?
Along x first you have to find out what is
the mass of the fluid element, what is that?
It is the density times the volume of the
element that is delta x * delta y. So, this
is the mass of the fluid element that times
the body force with mass gives the total body
force along x. So, these are the forces which
are acting on the fluid element. Now, let
us try to answer another question.
Are these still the force, only forces which
are acting if the fluid element is under rigid
body motion? That is the fluid element is
moving like a rigid body there is no internal
deformation but as a whole it is just like
a solid that is getting displaced. That maybe
displaced linearly or angularly but it is
having a motion. But the motion is a rigid
body motion. If that is the case then are
these the only forces?
See what forces we have identified? Surface
forces we have identified, body forces we
have identified. So, the question boils down
to that are this is the only the surface forces
even if the fluid is under rigid body motion.
The answer is what? See, what is the difference
between a fluid element at rest and fluid
element under rigid body motion? The only
difference that when it is under rigid body
motion it might be having like velocity acceleration
and so on.
But in terms of the surface forces which are
acting if the fluid element is non-deforming
then there is no sheer component of force.
So, for a non-deforming fluid element there
is no difference between the surface forces
which are presented in this diagram and the
surface forces which are there when it is
a moving with acceleration. So, this type
of forcing description is equally valid if
the fluid is under rigid body motion.
So, we identify the situation not just at
fluid under rest but also rigid body motion.
We will see such examples where the rigid
motion of the fluid will be very interesting
like you may have rotation of fluid element
like a rigid body. We will see that what kind
of situation it creates. So, broadly this
is also studied under the category of fluid
statics not because it is a static condition
but in terms of the characteristic of the
fluid the deformable nature is not highlighted
here.
And that is why we may use broadly similar
concepts and we will learn these concepts
together under the same umbrella because they
are very very related, in once case it will,
it may have an acceleration in another case
it may not be. Otherwise it is very very similar.
So, let us say that it is under rigid body
motion and therefore let us say it has some
acceleration along x. Say ax is the acceleration
which is there along x.
So, we can write the Newton’s second law
of motion for the fluid element and when we
do that what do we get the result in force
which is acting along x = the mass of the
fluid element time x times acceleration along
x. So, it is p * delta y – the other term
which is there on the opposite phase + rho
* delta x * delta y * bx = rho * delta x * delta
y that is the mass times the acceleration
along x.
Delta x * delta y we will get cancelled from
both sides these are small but not = 0 these
are tending to 0. So, you can cancel from
both sides at the end what final expression
you will get? So, these will be the expression
which relates the pressure gradient with the
body force that is acting and if there is
any acceleration that acts on that the fluid
element is having. Similar expressions are
valid for the motion along y, so we are not
repeating it again.
With this kind of a general idea, so this
is a very general expression. This general
expression just considers that there is a
body force and the fluid is having some acceleration
in a particular direction subjected to the
body force. But it is a non-deformable fluid
element. With that understanding we will try
to identify that what is the variation of
pressure just due to the effect of gravity
as a body force in a fluid element at rest.
We consider that there is a free surface of
a fluid this is a symbol in fluid mechanics
that we will be using to designate a free
surface. This is a triangle with 2 horizontal
lines very short horizontal dashes or lines
at the bottom. This is a kind of a technical
representation of free surface. We consider
that we are interested about some depth usually
that direction in which depth varies is typically
taken as z direction.
This is just a common notation in most of
the books that the vertical direction across
which the gravity is acting of course the
opposite to the action of gravity because
gravity will be vertically downwards and opposite
to that is considered as a z axis just as
a common notation. Let us try to write this
kind of equation for these fluids which is
at rest, it is of a substantial depth.
So, we are interested to find out what is
the pressure at this point, which is at a
depth h from the free surface. This there
is no acceleration of these fluids it is under
absolute rest. So, the ax or here az term
will be 0. So, you will have – this one.
When you have the z direction you also have
a horizontal direction like x and for x you
can write again similar expression.
So, if you write it for x, what is the body
force which is acting along x? There is no
body force which is acting along x because
only body force to which this is subjected
is the gravity. There is no body force along
x and there is no acceleration that is it
is having along x. So, the second expression
is even more simple but it gives us a very
important in sight.
What is that that if you are not having a
body force along a particular direction and
the fluid is under rest then pressure does
not vary with the same fluid along that direction.
That means for a horizontal, along a horizontal
line you are not having any pressure variation
in a continuous fluid system and this is one
of the basic principles that we use for measurement
of pressure differentials, as you have seen
in examples of manometers earlier.
So, this is something which is of very consequence
but it is an obvious conclusion. What we get
from the first equation? So we, let us try
to replace the bz, what is bz? See this is
the z direction and this is acting in the
opposite direction. So, this is – g. So,
you have from this and since pressure is not
varying with x, so you can write this as dp
dz in place of partial derivative because
it is now just a function of a single variable.
So, you can write this as – dp dz = rho
g. So, dp is – rho g dz. So, if you want
to find out what is the pressure difference
between say 2 points A and B. So you have
to integrate it with respect to the z variation
from say A to B, when it is a say we take
our reference such that the origin is located
here that means at Az = 0 at B it is – h.
So, you can write PB – PA = - integral of.
Now it is important to see that what is the
lengths scale that we are considering over
which this variation is taking place. If this
h is quite large there may be a significant
variation in density over it just like consider
the atmosphere which is above the surface
of that. As you go more and more above the
surface of that you expect the density to
change because the temperature changes and
so on.
And therefore the density in many cases may
not be treated as a constant. So, if it is
treated as a constant then it can only come
out of the integral. Similarly, you also are
probably working on length scale over with
G is not changing. If you are taking a large
height like what people who are dealing with
atmospheric sciences, for them the length
scales are large length scales over which
you may have even a change in acceleration
due to gravity.
But if you consider that such a situation
is not there just for simplicity, so if rho
* g is a constant that is the best way to
say because a very tough mathematician will
say that I do not care whether rho is varying
whether g is varying I am happy in bringing
this term out of the integrals so long as
rho * g is a constant.
So, may be mathematicians way of looking into
it is that rho varies in a particular way
g varies in a particular way but those variation
effects get cancelled out somehow so that
rho * g is a constant maybe very hypothetical
but for our case to bring it out of the integral
the product being a constant that will solve
our purpose. So, then what you have? Then
you have PB – PA = rho gh that means PB
= PA + rho gh, which is your very well know
expression.
Now, important thing is that see we are not
expressing the pressure at B just in an absolute
sense we are expressing it relative to the
pressure at A. Many times this pressure at
A say this is atmosphere, so that may be taken
as a reference. So, if this is taken as a
reference as p = 0 as an example. So, whatever
is the atmospheric pressure say we call it
0. That means, any other pressure we are expressing
relative to the atmospheric pressure.
So, then in that case PB is the pressure relative
to P atmosphere, if p = 0 is the atmosphere.
It is not definitely = 0 but if you have p
= 0 we, I mean that is just the choice of
your reference. So that it express any other
pressure in terms of that as a reference.
So, any pressure which is expressed in terms
of the atmospheric pressure that is a relative
way of expressing the pressure.
It is not that always you have to express
relative to atmospheric pressure but atmospheric
pressure being a well know standard under
a given temperature. So, reference with respect
to atmospheric pressure is something which
is a very standard reference that we many
times used. So, reference pressure relative
to atmosphere, so we are talking about a reference
where the reference is the atmosphere then
whatever pressure is there at any other point
we call that as a gauge pressure.
This is just a terminology. So, gauge pressure
means that any pressure relative to atmospheric
pressure. So, that means it is nothing but
P absolute – P atmosphere. There is a difference
between the absolute pressure and the atmospheric
pressure that is as good as taking the atmospheric
pressure as 0 reference and mentioning the
pressure relative to that. So, this is a very
simple exercise but from this we learn something.
What we learn something?
So, whenever we have an expression we should
keep in mind what are the assumptions under
which it is valid. So, we will develop the
habit of not using any formula like a magic
formula. This is very very important. Formula
based education is very bad. So, whenever
you have a formula and you want to use it
try to be assured that it is valid for the
condition in which you are applying if not
exactly but at least approximately.
So, when you are using pressure = h rho g,
what are the assumptions under which it is
valid? So, obviously rho * g is a constant
and there is no other body force which is
acting on it and fluid is at rest. That is
these are the assumptions that are there with
such a simple expression. Now, with this kind
of concept one may utilize device, one may
utilize this type of concept in making devices
for measuring atmospheric pressures just like
you have barometers.
Whenever we will be learning a concept we
will try to give examples of measurement devices
which try to utilize those concepts as all
of you know a barometer maybe utilize to measure
the atmospheric pressure. So, how it is there?
You have say inverted tube and this inverted
tube is say put in a bath of some fluids say
mercury and let us say that it is there up
to this much height.
Now, this much of height is there, there are
various forces which are acting on this. So,
one is you have, we are of course neglecting
the local surface tension effect and the capillary
formation. You must keep in mind that as this
radius become smaller and smaller the effect
of the curvature maybe more and more important
because surface tension effect will be more
and more important and there may be significant
errors in reading because of them.
Now, if we just neglect that effect from the
time being then you have atmospheric pressure
acting from this side. If you assume that
there is a vacuum here, there is a big question
mark whether there will be vacuum or not.
But let us for the sake of simplicity assume
that there is a vacuum. Then whatever pressure
is there which is acting from the side that
is balance by the height of the liquid column
which is there on the top.
So, from that you can get an estimation for
what is the pressure here, let us say that
P is the atmospheric pressure. So, p * the
surface area on which it is acting is the
force that is being sustained by the weight
of the liquid column. So, that is nothing
but that. So, it is like h rho g that * the
area and area gets cancelled from both sides.
So, you get this P if it is vacuum as the
h rho g.
But if it not a vacuum, let us say that there
is some pressure here which is the vapor pressure
of the fluid which is occupying this and it
is common that such vapor pressure will be
there. Why? Because if it is a saturated liquid
it is likely to have its own vapor on the
top of that and that will always exert some
pressure. So, it is never a vacuum in an ideal
sense. So, we can say that P – P vapor is
actually what is being balance by this way,
so that is the h rho g.
So, if there is a vapor pressure, you cannot
just use h rho g for the estimation of the
atmospheric pressure but you have to make
a correction because of the presence of the
vapor and that is the function of the temperature
because vapor pressure varies with temperature.
Very commonly the mercury is one of the fluids
that is being used for this purpose and why
mercury is being used?
Obviously because it is quite dense it will
not occupy a very large height for representing
the atmospheric pressure. If you use any other
fluid it may occupy a great height, so it
may be an unmanageable device, unmanageably
long device. Also the vapor pressure of mercury
is quite small in most of the temperature
ranges and therefore this connection is not
that severe. These two are the important results.
There are many other reasons which are always
into the picture when you select a fluid for
measurement of a pressure, so like in a barometer.
A barometer is a very interesting device we
have discuss something quite seriously but
I would just like to, I mean share a kind
of a story associated with barometer, a very
well know story and I am sure that many of
you have heard about it.
A long time back in a high school examination
there was a question. It was a physics examination
and the question was that how can you measure
the height of a building using a barometer?
Now, all though all of you or most of you
have heard about the story you will try to
get a morale out of that story and you will
try to keep that in mind whenever we are going
to learn something.
So, what the student replied in answer, the
answer was that you just have a thread you
connect the barometer with the thread go to
the roof of the house just drop that barometer
with the thread and then like the total height
that total height of the thread that is required
to bring the barometer to the ground maybe
+ whatever is the portion of the barometer
we will give the height of the building.
Now as seen most of the examinations by this
student was given a big 0 and the expectation
was it was justified that why he was given
a big 0 because it was expected that the answer
should reveal some basic concept in physics.
It was a physics examination but this does
not reveal any basic concept in physics, so
he was given a 0 but then the student went
for an arbitration. It was a democratic system
even in that time.
So, the student said that no like, I mean
let my answer be reviewed. So, there was a
panel. The panel said that okay may be you
were not aware that what kind of expectation
that we are having from your answer. So, we
give you another chance. So, you think about
a solution for this question which reflects
your understanding in physics and we will
evaluate you from that. So, student said that
let me be given sometime.
So, he was given 5 minutes to think about
that. So, he was thinking for 5 minutes and
when he was thinking for 5 minutes and he
was still not coming up with an answer then
like the evaluators where very happy that
he might be failed again. So, they said that
no you could not come up with an answer, so
we are sorry. Then he said that no actually
there are lots of answers have come to my
mind.
So, I am not sure that what should I say and
that is why I was not giving a response and
then he was asking permission that I mean,
am I going to be allowed to speak of that
remaining answers. Then they said that fine,
I mean whatever you have thought you just
tell then he said that I will, what I will
do is I will drop the barometer from the top
of the building and I will measure the time
that it takes to reach the ground and h = half
g t square.
So, from the time I can measure the height.
It reflects some understating of physics but
it is a bit distracting because the barometer
like it may be damaged and like and so on.
Then he said that no then if you want a different
answer maybe what I will do, I will try to
make a pendulum out of the barometer swing
it one in the bottom ground level another
on the top level and we will measure the time
period.
And this time period difference will give
the difference in g between the 2 heights
and since g is the function of height it will
tell us that what is the height difference
between the 2 and I mean, still the examiners
were not happy and they were still they were
ready to pass him because these were like
the reflecting some of the basic concepts
in physics.
Then he said that even if I am given different
opportunity what I will do is I will climb
across the staircase and in the side of the
staircase you can just put the barometer one
after the other till you travels the entire
height. The number of times you put it you
multiply with the length it is the basic length
measurement principle. So, from that you get
that and he said that there are many answers
which are coming to my mind.
But given me an entire freedom what I will
do I will not really put this much of effort,
I will go to house master and tell that see
now I have a beautiful new barometer for you
I am giving it you please tell me what is
the height of the house and the housemaster
will obviously tell that because it is like
a gift, free gift that the house master is
having.
And then he say that perhaps you are not expecting
all these answers from me you are expecting
to, we to give a very structured answer that
the barometer measures the atmospheric pressure.
So, from the difference in the level of atmospheric
pressure in the 2 heights we can easily say
that what is the difference in height between
the 2 and since this is the most structured
answer I hope that you will be happy with
that.
And then of course the evaluators passed him
and he was quite successful and the name of
the student is Niels Bohr, who later on, I
mean like discovered many many beautiful phenomena
in physics. So, the whole idea is that I would
always encourage you may be, I mean none of
us like Niels Bohr, I mean we are not born
with those special abilities but at least
whenever you are having an opportunity to
think of solutions do not always go for a
structured solution.
Try to think about different possibilities
whenever you are thinking about designing
a measurement principle. Whenever you are
thinking about solving a particular problem
just try to think about various possibilities
some of the possibilities may not be very
very encouraging very very welcome but at
least these possibilities will give us some
kind of clue that what could be alternatives.
Some of the alternatives may be discarded.
They may not be very smart but they will at
least give us a scope of thinking lateral
and that is how one may improve in science,
technology and research. And such a simple
example like barometer, I mean one is always
reminded of that kind of a story and I feel
that it is not something to bad to share with
you. So, what we will do is we will not go
further ahead today we will stop here.
In the next class we will just make a plan
of what we will do in the next class. Now
we have found out a particular way in which
you have an estimation of variation in pressure
because of a body force which is acting and
for fluid element which may be at rest or
subjected to acceleration. So, we will utilize
this principle to calculate 2 important things.
One is if there is a plain surface which is
emerged in a fluid what is the total force
which is there on the plain surface because
of the pressure distribution.
Now we have realized that pressure is like
a distributed force because it varies with
the depth so at different depths you have
different pressures. Therefore, it will be
like a simple statics problem where you have
a distributed force on surface to find out
what is the total force which is acting. If
you have a curved surface we will see that
the technique may be a bit different.
But broadly we can utilize some of the concepts
of pressure distribution on a plain surface
even to calculate force on curve surface.
So, in the next class we will see what are
the forces on plain and curve surfaces which
are there in a fluid at rest and then to see
that what are the consequences and we will
work out some problems related to that. So,
we stop here today. Thank you.
