Samacheer Class 11 Physics - Properties of Matter

Stress: The force per unit area is called as stress.
Stress, σ = \(\frac { Force }{ Area }\) = \(\frac { F }{ A }\)
Strain: Strain is defined as the ratio of change in size to the original size of an object. It measures the degree of deformation.

Elastic behaviour of solid. In a solid, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to the neighbouring molecules. When deforming force is applied on a body so that its length increases, then the molecules of the body go far apart.

Steel is more elastic than rubber. If equal stress is applied to both steel and rubber, the steel produces less strain. So Young’s modulus is higher for steel than rubber. Hence steel is more elastic than rubber.

When a spring balance has been used for a long time, it develops elastic fatigue, the spring of such a balance takes a longer time to recover its original configuration and therefore it does not give correct measurement.

When a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.
Upthrust or buoyant force = weight of liquid displaced.

The law of floatation states that a body will float in a liquid if the weight of the liquid displaced by the immersed part of the body equals the weight of the body.

The coefficient of viscosity of a liquid is defined as the viscous force acting tangentially per unit area of a liquid layer having a unit velocity gradient in a direction perpendicular to the direction of flow of the liquid.

Streamlined flow:
* When a liquid flows such that each particle of the liquid passing a point moves along the same path and has the same velocity as its predecessor then flow is said to be streamlined flow.
* The velocity of the particles is constant.
* The path taken by the particle in this flow is a curve.
Turbulent flow:
* When the speed of the moving liquid, exceeds the critical speed v c, the motion becomes turbulent.
* The velocity changes both in magnitude and direction from particle to particle.
* The path taken by the particle in this flow becomes erratic and whirlpool. Like circles.

Reynold’s number R c is a critical variable, which decides whether the flow of a fluid through a cylindrical pipe is streamlined or turbulent.
R c = \(\frac { ρVD }{ η }\)

According to Bernoulli’s theorem, the sum of’ pressure energy, kinetic energy, and potential energy per unit mass of an incompressible. Non-viscous fluid in a streamlined flow remains a constant.

A liquid in a steady flow can possess three kinds of energy. They are
(i) Kinetic energy, KE = \(\frac { 1 }{ 2 }\)mv²
(ii) Potential energy, PE = mgh
(iii) Pressure energy, respectively E p = PV

Surface tension is defined as the force acting on a unit length of an imaginary line drawn on the free surface of the liquid, the direction of the force being perpendicular to the line so drawn and acting parallel to the surface. The SI unit and dimensions of surface tension are Nm -1 and MT -2, respectively.

The angle between tangents drawn at the point of contact to the liquid surface and solid surface inside the liquid is called the angle of contact for a pair of solid and liquid. It is denoted by θ.

Cohesive – The force between the like molecules which holds the liquid together is called ‘cohesive force’.
Adhesive – When the liquid is in contact with a solid, the molecules of these solid and liquid will experience an attractive force which is called ‘adhesive force’.

In a liquid whose angle of contact with solid is less than 90°, suffers capillary rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

When a drop of water is placed on oil, the cohesive force of water molecules dominates the adhesive force between water and oil molecules. Hence drop of water contracts to a spherical shape.

This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem.
III. Long Answer Questions:

Hooke’s law states that for a small deformation when the stress and strain are proportional to each other.
It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A spring is suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire as shown in Figure.
Using a vernier scale arrangement the extension produced on the wire is measured. From the experiment, it is known that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ΔL.
It is directly proportional to the original length L and inversely proportional to the area of cross-section A. A graph is plotted using F on the X-axis and AL on the Y-axis. This graph is a straight line passing through the origin as shown in Figure.
Variation of ∆L with F
∴ ∆L = (slope)F
Multiplying and dividing by volume,
V = AL,
F(slope) = \(\frac { AL }{ AL }\) ∆L
Rearranging, we get
i.e., the stress is proportional to the strain in the elastic limit.

There are three types of modulus of elasticity. They are:
* Young’s modulus
* Bulk modulus
* Rigidity modulus
Young’s modulus: When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young’s modulus.
SI unit of Young’s modulus is Nm -2 or pascal.
Bulk modulus: Bulk modulus is defined as the ratio of volume stress to the volume strain.
Bulk modulus,
The negative sign in equation (1) means that when pressure is applied to the body, its volume decreases.
The rigidity modulus or shear modulus:
It is defined as rigidity modulus or Shear modulus,

When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy. Let us consider a wire whose un-stretch length is L and the area of cross-section is A. Let a force produce an extension l and it is assumed that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire. The work is done in stretching the wire by dl,
dW = F dl
The total work done in stretching the wire from 0 to l is
W = \(\int_{0}^{l} \mathrm{~F} d l\) … (1)
From Young’s modulus of elasticity,
Y = \(\frac { F }{ A }\) x \(\frac { L }{ l }\) ⇒ F = \(\frac { YAl }{ L }\) … (2)
Substituting equation (2) in equation (1), we get
W = \(\int_{0}^{l} \frac{\mathrm{YAl}}{\mathrm{L}} d l\)
Since l is the dummy variable in the integration, we can change l to l’ (not in limits), therefore

Let us consider a water sample of a cross-sectional area in the form of a cylinder. Let h 1 and h 2 be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively as shown in Figure (a). Let F 1 be the force acting downwards on level 1 and F 2 be the force acting upwards on level 2, such that, F 1 = P 1 A and F 2 = P 2 A Let the mass of the sample to be m and under equilibrium condition, the total upward force (F 2 ) is balanced by the total downward force (F 1 + mg), otherwise, the gravitational force will act downward which is being exactly balanced by the difference between the force F 2 – F 1
F 2 – F 1 = mg = F G … (1)
Where m is the mass of the water available in the sample element. Let p be the density of the water then, the mass of water available in the sample element is
m = ρV = ρA(h 2 – h 1 ])
V= A (h 2 – h 1 )
A sample of water with base area A in a static fluid with its forces in equilibrium
Hence, gravitational force,
F G = ρA(h 2 – h 1 )g
On substituting the value of W in equation (1)
F 2 = F 1 + mg
⇒ P 2 A = P 1 A + ρA(h 2 – h 1 )g
Cancelling out A on both sides,
P 2 = P 1 + ρ (h 2 – h 1 )g … (2)
If we choose level 1 at the surface of the liquid (i.e., air-water interface) and level 2 at a depth ‘h ’ below the surface (as shown in Figure), then the value of h 1 becomes zero (h 1 = 0) when P 1 assumes the value of atmospheric pressure (say P a ). In addition, the pressure (P 2 ) at a depth becomes P. Substituting these values in the equation,
p 2 = p 1 + ρ(h 2 – h 1 )g
we get
p = p a + ρgh
This means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where P a is the atmospheric pressure = 1.013 x 10 5 P a. If the atmospheric pressure is neglected then
p = ρ gh
For a given liquid, p is fixed and g is also constant, then the pressure due to the fluid column is directly proportional to vertical distance or height of the fluid column.

Pascal’s law states that if the effect of gravity can be neglected then the pressure in a fluid in equilibrium is the same everywhere. Let us consider any two points A and B inside the fluid imagined. A cylinder is such that points A and B lie at the centre of the circular surface at the top and bottom of the cylinder.
Let the fluid inside this cylinder be in equilibrium under the action of forces from outside the fluid. The forces acting on the circular, top, and bottom surfaces are perpendicular to the forces acting on the cylindrical surface. Therefore the forces acting on the faces at A and B are equal and opposite and hence add to zero.
As the areas of these two faces are equal, the pressure at A = pressure at B. This is the proof of Pascal’s law when the effect of gravity is not taken into account.

Archimedes principle states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it, and its upthrust acts through the centre of gravity of the liquid displaced.
Proof:
Consider a body of height h lying inside a liquid of density p, at a depth x below the free surface of the liquid. The area of a cross-section of the body is a. The forces on the sides of the body cancel out.
Pressure at the upper face of the body,
P 1 = xρg
Pressure at the lower face of the body,
P 2 = (x + h)pρg
Thrust acting on the upper face of the body is
F 1 = P 1 a = xρga
acting vertically downwards.
Thrust acting on the lower face of the body is
F 2 = P 2 a = (x+h)ρga
acting vertically upwards.
The resultant force (F 2 – F 1 ) is acting on the body in the upward direction and is called Upthrust (U).
∴ U = F 2 – F 1 = (x + h)ρg – xρga = ahρg
But, ah = V, the volume of the body = volume of liquid displaced.
U = Vρg = Mg
[∴ M = Vρ = mass of liquid displaced] i.e., upthrust or buoyant force.
= Weight of liquid displaced.
This proves the Archimedes principle.

Let us consider a sphere of radius r which falls freely through a highly viscous liquid of coefficient of viscosity η. Let the density of the material of the sphere be p and the density of the fluid be c.
Gravitational force acting on the sphere,
F G = mg = \(\frac { 4 }{ 3 }\)πr³pg (downward force)
Upthrust, U= \(\frac { 4 }{ 3 }\)πr³σg (upward force)
viscous force F = 6πηrv t
At terminal velocity v t.
downward force = upward force.

Consider a liquid flowing steadily through a horizontal capillary tube.
Let v = (\(\frac { V }{ t }\)) be the volume of the liquid flowing out per second through a capillary tube.
It depends on
(i) coefficient of viscosity (η) of the liquid
(ii) radius of the tube (r).
(iii) the pressure gradient (yj Then,
So, equating the powers of M, L, and T on both sides, we get
a + c = 0, – a + b – 2c = 3, and – a – 2c = 1 On solving three equations, we get
a = – 1, b = 4, and c = 1
Therefore, equation (1) becomes,
v = kη -1 r(\(\frac { p }{ l }\)) 1
Experimentally, the value of k is shown to be \(\frac { π }{ 8 }\), we have
V = \(\frac{\pi r^{4} \mathrm{P}}{8 \eta l}\)
The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube.

(i) Liquid drop: Let us consider a liquid drop of radius R and the surface tension of the liquid is T.
The various forces acting on the liquid drop are,
(a) Force due to surface tension F T = 2πRT towards the right
(b) Force due to outside pressure, F\(\mathrm{F}_{\mathrm{P}_{1}}\) towards right
(c) Force due to inside pressure,
\(\mathrm{F}_{\mathrm{P}_{2}}\) = P 2 πR² towards left
As the drop is in equilibrium,
\(\mathrm{F}_{\mathrm{P}_{2}}\) = F T + \(\mathrm{F}_{\mathrm{P}_{1}}\)
P 2 2πR² = 2πRT + P 1 πR²
⇒ (P 2 – P 1 )πR² = 2πRT
Excess pressure is ∆P = P 2 – P 1 = \(\frac { 2T }{ R }\)
(ii) Liquid bubble: A soap bubble of radius R and the surface tension of the soap bubble be T is as shown in Figure. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble. Hence, the force on the soap bubble due to surface tension is 2 x 2πRT. The various forces acting on the soap bubble are,
(a) Force due to surface tension
F T = 4πRT towards right
(b) Force due to outside pressure,
F P = P 1 πR² towards right
(c) Force due to inside pressure,
\(\mathrm{F}_{\mathrm{P}_{2}}\) = P 2 πR² towards left
As the bubble is in equilibrium,
\(\mathrm{F}_{\mathrm{P}_{2}}\) = F T + \(\mathrm{F}_{\mathrm{P}_{1}}\)
P 2 πR² = 4πRT + P 1 πR²
⇒ (P 2 – P 1 )πR² = 4πRT
Excess pressure is ∆P = P 2 – P 1 = \(\frac { 4T }{ R }\)
(iii) Air bubble: Let us consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. Let P 1 and P 2 be the pressures outside and inside the air bubble, respectively. Now, the excess pressure inside the air bubble is ∆P = P 1 – P 2.
In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,
(a) The force due to surface tension acting towards right around the rim of length 2πR is F T = 2πRT
(b) The force due to outside pressure P 1 is to the right acting across a cross-sectional area of πR² is \(\mathrm{F}_{\mathrm{P}_{1}}\) = P 1 πR²
(c) The force due to pressure P 2 inside the bubble, acting to the left is \(\mathrm{F}_{\mathrm{P}_{2}}\) = P 2 πR²
As the air bubble is in equilibrium under the action of these forces, \(\mathrm{F}_{\mathrm{P}_{2}}\) = F T + \(\mathrm{F}_{\mathrm{P}_{1}}\).

The rise or fall of a liquid in a narrow tube is called capillarity. Let us consider a capillary tube which is held vertically in a beaker containing water; the water rises in the capillary’ tube to a height h due to surface tension.
The surface tension force F T, acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T is resolved into two components.
(i) Horizontal component T sinθ and
(ii) Vertical component T cosθ acting upwards, all along the whole circumference of the meniscus.
Total upward force = (T cosθ) (2πr)
= 2πrT. cosθ
Where θ is the angle of contact, r is the radius of the tube. Let ρ be the density of water and h be the height to which the liquid rises inside the tube. Then,
The upward force supports the weight of the liquid column above the free surface, therefore,
If the capillary is a very fine tube of the radius (i.e., the radius is very small) then \(\frac { r }{ 3 }\) can be neglected when it is compared to the height h. Therefore,
T = \(\frac{r \rho g h}{2 \cos \theta}\)

Let us consider a pipe AB of varying cross-sectional areas a 1 and a 2 such that a 1 > a 2. A non – viscous and incompressible liquid flows steadily through the pipe, with velocities v 1 and v 2 in areas a 1 and a 2, respectively as shown in Figure.
Let m 1 be the mass of fluid flowing through section A in time ∆t, m 1 = (a 1 v 1 ∆t)ρ
Let m 2 be the mass of fluid flowing through section B in time ∆t, m 2 = (a 2 v 2 ∆t)ρ
For an incompressible liquid, mass is conserved m 1 = m 2
a 1 v 1 ∆tρ = a 2 v 2 ∆tρ
a 1 v 1 = a 2 v 2 ⇒ av = constant
Which is called the equation of continuity. It is based on the conservation of mass in the flow of fluids. In general, av – constant.

Construction: It consists of two wider tubes A and A’ (with cross-sectional area A) connected by a narrow tube B (with the cross-sectional area a). A manometer in the form of a U-tube is also attached between the wide and narrow tubes as shown in Figure. The manometer contains a liquid of density ‘ρ m ‘.
Theory:
Let P 1 be the pressure of the fluid at the wider region of tube A. Let us assume that the fluid of density ‘ρ’ flows from the pipe with speed ‘v 1 ’ and into the narrow region, its speed increases to ‘v 1 ‘.
According to Bernoulli’s equation, this increase in speed is accompanied by a decrease in the fluid pressure P 2 at the narrow region of the tube B. Hence, the pressure difference between tubes A and B is noted by measuring the height difference (∆P = P 1 – P 2 ) between the surfaces of the manometer liquid.
From the equation of continuity, we can say that Av 1 = av 2 which means that
v 2 = \(\frac { A }{ a }\)v 1
Using Bernoulli’s equation,
IV. Numerical Problems:

Let diameter of capillary tube = d mm
Let the radius of capillary tube r 1 = r mm
Capillary rise h 1 = 30 mm
Let the radius of another capillary tube r 2 = \(\frac { 2 }{ 3 }\) r
Let the capillary rise of another capillary tube be h 2
We know that

Length of a cylinder l = 1.5 m
Diameter of a cylinder d = 4 x 10 -2 m
Tangential force F t = 4 x 10 5 N
Rigidity modulus η R = 6 x 10 10 Nm -2
Twist produced θ = ?

Excess of pressure inside the liquid due to surface tension,
∆P = \(\frac { 2T }{ R }\)
Where T – surface tension
In the case of soap bubbles, the excess pressure inside the soap bubble,
∆P b = \(\frac { 4T }{ R }\)
Excess of pressure of air inside the bigger bubble
∆P bigger = \(\frac { 4T }{ 4 }\) = T
Excess of pressure of air inside the smaller bubble
∆P smaller = \(\frac { 4T }{ 2 }\) = 2T
Air pressure different between the smaller bubble and the atmosphere will be equal to the sum of excess pressure inside the bigger and smaller bubbles.
Excess pressure inside a single soap bubble
= \(\frac { 4T }{ R }\) = 4T = T
∴ Pressure different of single soap bubble less than the radius of both T < 3T.

Let the mass of Ag block be x kg,
Tension in the string T = 37.12 N.
Relative density of liquid R liq = 0.72
Relative density of silver R Ag = 10
According to the principle of flotation
V pg = mg
∴ Vp=m … (1)
Weight of Ag block = mg = Vρg = R Ag V g
Weight of Ag block W = 10 Vρg [ ∵ m = Vρ]
Force of buoyancy F B = R liq (Vρ)g = 0.72 V ρg
Apparent weight W app = 10 V ρg – 0.72 Vρg
= 9.78 Vρg
= 9.78 mg
Tension in the string = Apparent weight
37.12= 9.78 mg
Mass m = \(\frac { 37.12 }{ 9.28 }\) = 4 kg

Initial pressure P 2 = 5 x 10 5 Nm -2
Final pressure P 1 = 4. 5 x 10 5 Nm -2
Initial velocity V 1 = 0
Final velocity v 2 = v 2
Density of water ρ = 10³ kg/m³
Using Bernoulli’s theorem, we can write,
V. Conceptual Questions:

When oil comes out of one hole with high velocity, the pressure in the tin decreases. To have a continuous flow of oil proper pressure is to be maintained inside the tin. To achieve this, atmospheric air has to be entered inside the tin. For this purpose, only another hole is made out. Hence two holes are made to empty an oil tin.

The area of a sharp edge is much less than the area of a blunt edge. For the same total force, the effective force per unit area is more for the sharp edge than the blunt edge. Hence, a sharp knife cuts easily than a blunt knife.

While going up in an aeroplane the atmospheric pressure decreases with height. When aeroplane is going up, the ink in the pen tends to ooze out to equalise the pressure. This may spoil the clothes of the passengers. So they are advised to remove ink from the pen.

When we suck through the straw, the pressure inside the straw becomes less than the atmospheric pressure. Due to the pressure difference, the soft drink rises in the straw and we are able to take the soft drink easily.