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Students can Download Chapter 13 Kinetic Theory Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 13 Kinetic Theory

Summary
Introduction
The kinetic theory was developed in the nineteenth century by Maxwell, Boltzmann and other. It gives a molecular interpretation of pressure and temperature of a gas. It also explains gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It help us to find molecular sizes and masses.

Molecular Nature Of Matter
The scientific atomic theory is credited to John Dalton. Atomic theory is not the end of quest, but the beginning. Atoms consist of a nucleons and electrons. The nucleous itself is made up of protons and neutrons. The protons and neutrons are again made up of quarks. Even quarks may not be the end of the story.

There may be string-like elementary entities. In this chapter we shall limit ourselves to understanding the behavior of gases.

Behavior Of Gases
Gas Laws
1. Boyles law
The law states that at a given temperature, the volume of a given mass of gas varies inversely as its pressure.
It can be written as
\(P a \frac{1}{V}\) (at constant T)
PV = constant
PV = μ RT
μ → No. of moles, R → universal gas constant, T → temperature
Boyles law is not obeyed by gasses at all temperatures and pressure. Usually, Boyles, law is obeyed by gases at high temperature and low pressure (Graph is given below).

A real gas which obey this law is called ideal or perfect gas.

Variation of ‘R’ \(\left(=\frac{\mathrm{PV}}{\mathrm{T}}\right)\) with Pressure (for different temperatures)

Variation of R \(\left(=\frac{\mathrm{PV}}{\mathrm{T}}\right)\) with Pressure for temperature T₁ > T₂ > T₃ is shown in the above graph. The above graph shows that, all real gases approach ideal gas behavior at low pressure and high temperature.

At low pressure or high temperature, the molecules are for apart and molecular interactions are negligible. Without interactions the gas behaves like an ideal gas.

Note:
R = 8.324 J mol^-1 k^-1.
Variation of V with P for different temperature:

The above graph shows experimental PV curves for steam at three temperature The dotted line are the theoretical curves. (According to Boyles law). The theoretical value and experimental value comes in agreement at high temperatures and low pressures.

2. Charles law
Charles law states that the volume of a given mass of gas is proportional to its temperature when its pressure is kept constant.
V a T (P is constant)
i.e.,

The graph between V and T:

The above graph shows experimental T-V curves (solid lines) for Co₂ at three pressures compared with Charles law (dotted lines). T is in unit of 300 k and V in units of 0.13 litres.

Question 1.
Why theoretical value does not agree with experimental value?
Answer:
According to Charles law the graph between T and V is straight line. It means that when temperature decreases, the volume of gas decreases and finally becomes zero.

Practically volume will not be zero. Because the molecules require some finite space to exist. This implies that we cannot reduce its temperature to zero value. A zero kelvin temperature is only an idealized concept.

Dalton’s law of Partial Pressures:
It states that, the total pressure of a mixture of ideal gases is the sum of partial pressures.

Proof:
Consider a mixture of non-interacting ideal gases. μ₁, moles of gas 1, μ₂ moles of gas 2 etc. in a vessel of volume V at temperature T and pressure P. Using Boyles law, we can write
PV = (μ₁ + μ₂ +…………..)RT
P = \(\frac{\mu_{1} \mathrm{RT}}{\mathrm{V}}+\frac{\mu_{2} R T}{V}+\ldots \ldots \ldots\)
P = P₁ + P₂ +……………………

Kinetic Theory Of An Ideal Gas
The kinetic theory of gases has been developed by Clausius, Maxwell, Boltzmann and others. The theory is based on the following postulates.

• The gas is a collection of large number of molecules. The molecules are perfectly elastic hard spheres.
• The size of a molecule is negligible compared with the distance between the molecules.
• The molecules are always in random motion
• During their motion, the molecules collide with each other and with the walls of the containing vessel.
• The collisions are elastic and hence the total K.E energy and the total momentum of the colliding molecules before and after collisions are the same.
• The kinetic energy of a molecule is proportional to the absolute temperature of the gas.
• There is no force of attraction or repulsion between molecules.

The pressure of an ideal gas:

Consider molecules of gas in a container. The molecules are moving in random directions with velocity V. This is the velocity of a molecule in any direction. The velocity V can be resolved along x, y and z directions as V_x, V_y, and V_z respectively.

If we assume a molecule hits the area A of the container with velocity V_x and rebounds back with -V_x. (The velocities V_x and V_y do not change because this collision is perfectly an elastic one).

Therefore, the change in momentum imparted to the area A by the molecule
= mv_x^– – mV_x
= 2mV_x

To find the total number of collisions taking place in a time t, consider the motion of the molecules towards the wall. The molecules covers a distance V_xt along the x direction in a time t. All the molecules within the volume AV_xt will collide with the area in a time t.

If ‘n’ is the number of molecules per unit volume, the total number of molecules hitting the area A,
N = AV_xt n.

But on an average, only half of those molecules will be hitting the area, and the remaining molecules will be moving away from the area. Hence the momentum imported to the area in a time t
Q = 2mv_x × \(\frac{1}{2}\) AV_xtn.
= nmV_x²At
The rate of change of momentum,
\(\frac{Q}{t}\) = nmV_x²A
But rate of change of momentum is called force, ie. force F = nmV_x²A
∴ Pressure P = nmV_x² (P = \(\frac{F}{A}\))
Different molecules move with different velocities. Therefore, the average value V_x² is to be taken. If \(\overline{\mathbf{v}}_{\mathbf{x}}^{2}\) is the average value then the pressure.

\(p=n m \bar{v}_{x}^{2}\) ……………….. (1)

\(\overline{\mathbf{v}}_{\mathbf{x}}^{2}\) is known as the mean square velocity. Since the gas is isotropic (having the same properties in all directions), we can write

Kinetic Interpretation of gas laws:

Question 2.
Derive the ideal gas equation from P = \(\frac{1}{3} \mathrm{nm} \overline{\mathrm{v}}^{2}\)
Answer:
The average kinetic energy of the molecule is
KE = \(\frac{1}{2} m \bar{v}^{2}\) …………………….(3)
The eq (2) can be modified as

The average Kinetic energy of a molecule remains constant when the temperature is constant. That is when the temperature varies, \(\overline{\mathrm{KE}}\) also varies accordingly. The kinetic energy of a molecule is related to its absolute temperature by an equation
\(\overline{\mathrm{KE}}\) = \(\frac{3}{2}\)K_BT
Substitute equation (5) in equation (6); we get

N = μ N₀
P V = μ N₀K_BT
P V = μ R T…………………… (8)
(N_BK_B = R)
This is the ideal gas equation

Deduction of Boyles law:
If the temperature is constant for gas, the eq (8) can be written as
PV = Constant
This is called Boyles law

Deduction of Charles law:
If pressure of a gas is constant, the eq (8) can be written as
V a T

This is called Charles law.

Deduction of Avogadro’s Hypothesis:
If P₁, T₁, and V constant, N₀ will be constant, ie. equal Volumes of all gases, under the same conditions of pressure and temperature will contain the same number of molecules. This is known as Avogadro’s hypothesis.

Law Of Equipartion Of Energy
Degrees of freedom:
Degrees of freedom is number of independent ways by which a molecule can possess kinetic energy of translation, rotation and vibration.

Law of equipartition energy:
The total kinetic energy of a molecule is equally divided among the different degrees of freedom.

K.E. per degree of freedom:
The average energy per degree of freedom

Where K_B is Boltzman constant.

Degrees of freedom and energy of monoatomic gas:
A monoatomic atom has 3 degrees of freedom, ie; it can move in x, y and z-direction. The average energy of single monoatomic gas in the x-direction,

Note:
1. If a molecule is restricted to move in plane. It has 2 degrees of freedom.
2. If a molecule is restricted to move in a line, it has only 1 degrees of freedom.

Degrees of freedom and energy of single diatomic molecule (rigid rotator)

Consider a diatomic molecule as a rigid rotator (Rigid rotator means that the molecule does not vibrate). A rigid diatomic molecule has 3 translation degrees of freedom and 2 rotational degrees of freedom.
(Rotational degrees of freedom is shown in the above figure).
∴ Total average energy of diatomic rigid rotator,

Note:
A diatomic molecule has 3 rotational degrees of freedom. But we consider only 2 degrees of freedom. We neglect rotation along the line joining the atoms. Because it has very small moment of inertia.

Degrees of freedom and energy of single diatomic molecule of nonrigid rotator:
Molecules like ‘co’ even at moderate temperatures have a mode of vibration. The vibration energy of a diatomic molecule.

Where K is the force constant of the oscillator and y the vibrational coordinate. The vibration energy mode contain two terms (1) Kinetic energy (2) Potential energy. Hence a single mode of vibration of molecule is considered as 2 degrees of freedom (potential energy and kinetic energy)
∴ The total vibrational energy of a single-mode
= 2 × \(\frac{1}{2}\)K_BT
= K_BT
∴ The total energy of diatomic nonrigid rotator

Degrees of freedom and energy of polyatomic molecule, (non- rigid rotator):
If a polyatomic molecule has ‘f’ modes of vibration, total number of degrees freedom = 3 vibration + 3 rotator+f vibration.
∴ Total average energy of single polyatomic molecule,

Specific Heat Capacity
Monoatomic Gases (Molar specific heat capacity):
The energy of a single monoatomic gas = 3 × \(\frac{1}{2}\) K_BT
The energy of one mole monoatomic gas = 3 × \(\frac{1}{2}\) K_BT × N_A
[one-mole atom contain Avogadro number (N_A) of atoms]

A Diatomic Gas (Molar specific heat capacity):
A rigid diatomic molecule has 5 degrees of freedom : 3 translational and 2 rotational.
∴ energy of single diatomic (rigid) molecule = 5 × \(\frac{1}{2}\) K_BT
for one mole of diatomic molecule, energy U = 5 × \(\frac{1}{2}\) K_BT × N_A

Nonrigid diatomic molecule having a vibrational mode (Molar specific heat capacity):
If the diatomic molecule is not rigid but has a vibration mode. The energy of one mole,

Polyatomic Gas (Molar specific heat capacity):
In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (t) of vibrational modes.

The energy of one mole polyatomic gas,

Note: The experimental value of C_P and C_V of polyatomic gases are greater than the predicted values. The theoretical value and experimental value will be equal when we include vibrational modes of motion in the calculation.

Specific heat capacity of solids:
Consider a solid of N atoms. Each atom is vibrating about its mean position. Each vibration mode has two degrees of freedom (corresponding to potential energy and kinetic energy). Hence an oscillation in one dimension has average energy of

2 × \(\frac{1}{2}\) K_BT = K_BT
∴ Total energy in three dimension
= 3 × K_BT
= 3K_BT.

For one mole of solid, total energy,
U = 3K_BT × N_A
[N_A = Avagadro number]
= 3N_AK_BT
U = 3RT …………………. (1) [∴ R = K_BN_A].
We know specific heat capacity,

Note:
In solids, we do not consider the translational and rotational degrees of freedom. We consider only vibrational degrees of freedom.

Specific heat capacity of water:
We treat water like a solid. A water molecule has 3 atoms. Each atom in the molecule is vibrating about its mean position. A single vibration mode has 2 degrees of freedom (1) potential energy (2) kinetic energy.

ie; The energy of one atom in one dimensional vibration mode =
2 × \(\frac{1}{2}\) K_BT = K_BT

The energy of one atom having 3 dimensional vibration mode = 3 × K_BT
The energy of one H₂0 molecule having 3 dimension vibration mode
U = 3 × 3K_BT × N_A
U = 9RT [∵ R = K_B N_A]
∴ Specific heat capacity,
\(\mathrm{c}=\frac{\mathrm{dU}}{\mathrm{dT}}=\frac{\mathrm{d}}{\mathrm{dT}}(\mathrm{RT})\)
C = 9R.

Mean Free Path
Molecules in a gas have large speeds. Yet a gas leaking from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room. Why?

The molecules in a gas have a finite size. So they collide with other molecules during their motion. As a result, they cannot move straight like path. Their paths are continuously deflected.

Mean free path:
The mean free path is the average distance covered by a molecule between two successive collisions.

Expression for mean free path:

Suppose the molecules of a gas are spheres with diameterd. Let 〈v〉 be the average velocity of the molecule.
The volume covered by a molecule during its motion, in a time Δt = πd² 〈v〉 Δt.
If ‘n’ is the number of molecules per unit volume, the total number of molecules in the above volume
= πd² 〈v〉 Δt n.
The number collisions in a time Dt,
= πd² 〈v〉 Δt n.
Number of collisions in one second,
= n π d² 〈v〉
∴ The time between two successive collisions,
\(\tau=\frac{1}{n \pi d^{2}\langle v\rangle}\).
The average distance between two successive collisions,

In this derivation, we imagined the other molecules to be rest. But actually all other molecules are moving. Hence we must take relative velocity 〈v_r〉 instead of 〈v〉. A more exact treatment gives
\(\ell=\frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^{2}} \dots(1)\)
The mean free path given by the above equation depends inversely on the number density and the size of the molecule.

Students can Download Chapter 5 Digestion and Absorption Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 5 Digestion and Absorption

What is digestion?
This process of conversion of complex food substances to simple absorbable forms is called digestion.

DIGESTIVE SYSTEM:
It includes

• Alimentary canal
• Associated glands.

Alimentary canal:

The human digestive system
1. The alimentary canal begins with the mouth, and it leadsto the buccal cavity or oral cavity.

2. The oral cavity has a number of teeth and a musculartongue. Each tooth is embedded in a socket of jaw bone. This type of attachment is called thecodont.

3. Human being forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Dental formula of adult human
An adult human has 32 permanent teeth which are of four different types (Heterodont dentition).

• incisors (I)
• canine (C)
• premolars (PM)
• molars (M).

Arrangement of teeth in each half of the upper and lower jaw in the order I, C, PM, M is represented by a dental formula which in human is
\(\frac{2123}{2123}\)

• The hardest part of teeth is made up of enamel, helps in the mastication oWood.
• The tongue is attached to the floor of the oral cavity by the frenulum.
• The upper surface of the tongue has small projections called papillae, some of which bear taste buds.
• The oral cavity leads pharynx which serves as a common passage for food and air.
• The oesophagus and the trachea (wind pipe) open into the pharynx.
• A cartilaginous flap called epiglottis prevents the entry of food into the glottis – opening of the wind pipe – during swallowing.
• The oesophagus is a thin, long tube which passing through the neck, thorax and diaphragm and leads to a ‘J’ shaped bag like structure called stomach.
• A muscular sphincter (gastro-oesophageal) regulates the opening of oesophagus into the stomach.

The stomach, located in the upper left portion of The abdominal cavity, has three major parts:

1. A cardiac portion into which the oesophagus opens
2. A fundic region and
3. A pyloric portion which opens into the first part of small intestine

Small intestine is distinguishable into three regions:

1. A ‘U’ shaped duodenum
2. A long coiled middle portion jejunum and
3. A highly coiled ileum

The opening of the stomach into the duodenum is guarded by the pyloric A sphincter. Ileum opens into the large intestine.

Large intestine:
It consists of

• caecum
• colon
• rectum.

Caecum:
It is a small blind sac consists of some symbiotic micro-organisms. A narrow finger-like tubular projection, the vermiform appendix which is a vestigial organ, arises from the caecum. The caecum opens into the colon.

colon:
It is divided into three parts-an ascending, a transverse and a descending part. The descending part opens into the Rectum.

Rectum: It opens out through the anus.
Wall layers of alimentary canal:
It consists of four layers namely

1. Serosa
2. muscularis
3. sub-mucosa &
4. mucosa.

1. Serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues.
2. Muscularis is formed by smooth muscles
3. The submucosal layer is formed of loose connective tissues containing nerves, blood and lymph vessels.
4. The innermost layer lining of the alimentary canal is the mucosa.
   • In duodenum, glands are also present in sub-mucosa.
   • Mucosa forms irregular folds (rugae) in the stomach and small finger-like foldings called villi in the small intestine
   • The cells lining the villi produce numerous projections called microvilli giving a brush border appearance.
   • These modifications increase the surface area.
   • Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal.
   • Mucosal epithelium has goblet cells which secrete mucus that help in lubrication.
   • Mucosa also forms glands in the stomach (gastric glands) and crypts in between the bases of villi in the intestine (crypts of Lieberkuhn).

Digestive Glands:
It includes

1. salivary glands
2. liver
3. pancreas.

1. Salivary glands:
Saliva is mainly produced by three pairs of salivary glands

• parotids (cheek)
• sub-maxillary/sub-mandibular (lower jaw)
• sublinguals (belowthe tongue).

The duct systems of liver, gall bladder and pancreas These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.

2. Liver:

• It is the largest gland of the body weighing about 1.2 to 1.5 kg in an adult human.
• It is situated in the abdominal cavity, just below the diaphragm and has two lobes.
• The hepatic lobules are the structural andfunctional units of liver containing hepatic cells arranged in the form of cords.
• Each lobule is covered by a thin connective tissue sheath called the Glisson’s capsule.
• The bile secreted by the hepatic cells passes through the hepatic ducts and is stored in gall bladder.
• The duct of gall bladder (cystic duct) along with the hepatic duct from the liver forms the common bile. duct
• The bile duct and the pancreatic duct open together into the duodenum which is guarded by sphincter of Oddi.

3. Pancreas:
It is a heterocrine (both exocrine and endocrine) elongated organ situated between the limbs of the ‘U’ shaped duodenum.

Secretions of exocrine and endocrine:

• The exocrine portion secretes an alkaline pancreatic juice containing enzymes
• The endocrine portion secretes hormones, insulin and glucagons.

DIGESTION OF FOOD:

1. The process of digestion is accomplished by mechanical and chemical processes.
2. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. *Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus.

Contents of saliva:
It contains electrolytes (Na⁺, K+⁺, Cl^–, HCO^–“) and enzymes, salivary amylase and lysozyme.
The chemical process of digestion is initiated in the oral cavity by the carbohydrate splitting enzyme, the salivary amylase( pH 6.8).

Digestion of starch:
About 30 percent of starch is hydrolysed here by salivary amylase (optimum pH 6.8) into a disaccharide -maltose.

Enzyme for preventing infections:
Lysozyme present in saliva acts as an antibacterial agent that prevents infections.

• The bolus is then passed into the pharynx and then into the oesophagus by swallowing or deglutition.
• The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis.
• The gastro-oesophageal sphincter controls the passage of food into the stomach.

Mucosa of stomach and gastric glands:
Gastric glands have three major types of cells namely

Digestion in stomach:
1. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme.

2. The proenzyme pepsinogen, in the presence hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach.

3. Pepsin converts proteins into proteoses and peptones (peptides).

4. The mucus and bicarbonates play an important role in lubrication and protection of the mucosal epithelium from concentrated hydrochloric acid- pH (pH 1.8).

Special type of proteolytic enzyme in infants:
Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins. The bile, pancreatic juice and the intestinal juice are the secretions released into the, small intestine.

Pancreatic juice:
It contains inactive enzymes

Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into aqtive trypsin.

Contents of Bile and functions:
The bile released into the duodenum contains bile pigments (bilirubin and bili-verdin), bile salts, cholesterol and phospholipids but no enzymes Bile helps in emulsification of fats, i.e., breaking down of the fats into very smalfmicelles.

Bile also activates lipases. The intestinal mucosal epithelium has goblet cells which secrete mucus. The secretions of the brush border cells of the mucosa alongwith the mucus constitute the intestinal juice or succu sentericus.

Intestinal juice/succus entericus:
It contains enzymes like

1. disaccharidases (eg: maltase)
2. dipeptidases
3. lipases,
4. nucleosidases etc.

How does the intestine protect itself from digestion?
1. The mucus alongwith the bicarbonates from the pancreas protects the intestinal mucosa from acid as well as provide an alkaline medium (pH 7.8) for enzymatic activities.

2. Sub-mucosal glands (Brunner’s glands) also help in this.

The breakdown of biomacromolecules occurs in the duodenum region of the small intestine. The simple substances thus formed are absorbed in the jejunum and ileum regions of the small intestine. The undigested and unabsorbed substances are passed on to the large intestine.

Functions of large intestine:

1. Absorption of some water, minerals and certain drugs
2. Secretion of mucus which helps in adhering the waste (undigested) particles together and lubricating it for an easy passage.

The undigested, unabsorbed substances called faeces enters into the caecum of the large intestine through ileo-caecal valve, which prevents the back flow of the faecal matter. It is temporarily stored in the rectum till defaecation.

• The sight, smell and the presence of food in the oral cavity can stimulate the secretion of saliva.
• Gastric and intestinal secretions are also stimulated by neural signals.
• The muscular activities of different parts of the alimentary canal controlled by neural mechanisms, both local and through CNS.

ABSORPTION OF DIGESTED PRODUCTS:

1. The end products of digestion passthrough the intestinal mucosa into the blood or lymph.
2. It is carried out by passive, active or facilitated transport mechanisms.
3. Small amounts of monosacharides like glucose, amino acids and some of electrolytes like chloride ions are generally absorbed by simple diffusion.
4. Fructose and some amino acids are absorbed with the help of the carrier ions like Na⁺. This mechanism is called the facilitated transport.
5. Transport of water depends upon the osmotic gradient.
6. Various nutrients like amino acids, monosacharides like glucose,electrolytes like Na⁺ are absorbed into the blood by active transport and hence requires energy.

How does fat absorption occur?
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. Then, the fat globules are coated with small protein called as chylomicrons which are transported into the lymph vessels (lacteals) in the villi.

These lymph vessels ultimately release the absorbed substances into the blood stream. The maximum absorption occurs in the small intestine. The absorbed substances finally reach the tissues which utilise them for their activities. This process is called assimilation.

The digestive wastes, solidified into faeces in the rectum initiate a neural reflex causing an urge for its removal. The egestion of faeces to the outside through the anal opening (defaecation) is a voluntary process and is carried out by a mass peristaltic movement.

DISORDERS OF DIGESTIVE SYSTEM:
Jaundice:
The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.

Vomiting:
It is the ejection of stomach contents through the mouth. This reflex action is controlled by the vomit centre in the medulla. „

Diarrhoea:
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of food.

Constipation:
In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly. ‘

Indigestion:
In this condition, the food is not properly digested leading to a feeling of fullness. The causes are inadequate enzyme secretion, anxiety, food poisoning, overeating, and spicy food.

NCERT SUPPLEMENTARY SYLLABUS
Calorific value of carbohydrate, protein and fat:
Carbohydrates, proteins and fats are chief sources of energy in humans. These are oxidized and released energy stored in ATP, it is used for many activities of the cell.

calorific value kcal = 4.184kJ):
It is defined as the amount of heat produced in calories (cal) or in joules (J) from complete combustion of 1 gram food in a bomb calorimeter (a closed metal chamber filled with 02). One kilocalorie is the amount of heat energy needed to raise the temperature of one kilogram of water through 100C(1.80F).

The calorific values of carbohydrates, proteins and fats are 4.1 kcal /g, 5.65 kcal /g and 9.45 kcal/g, respectively. The actual amounts of energy liberated in the body by these nutrients referred to as the physiologic value of the food, and are 4.0 kcal /g, 4.0 kcal Ig and 9.0 kcal /g respectively.

DEFICIENCY DISEASES:
The low amount of nutrients (Vitamin A, iron and iodine) in the diet cause deficiency disorders The important among them is protein energy malnutrition (PEM). lt is major health and nutritional problems in India.

Young children (0-6 years) require more protein for each kilogram of body weight than adults. So they are more vulnerable to malnutrition. Malnutrition leads to permanent impairment of physical and mental growth and childhood mortality and morbidity. The details of the disorders are given below.

The child suffering from PEM can recover if adequate quantities of protein and carbohydrate rich food are given.

Students can Download Chapter 7 Body Fluids and Circulation Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 7 Body Fluids and Circulation

Blood
Blood is a special connective tissue contains

• Fluid matrix
• Plasma
• Formed elements.

Plasma:
It constitute nearly 55 per cent of the blood. 90 – 92 percent of plasma is water and proteins (Fibrinogen, globulins and albumins) contribute 6 – 8 percent of it. Fibrinogens are needed for clotting or coagulation of blood.

Defence mechanism and osmotic balance:
Globulins primarly are involved in defense mechanisms of the body and the albumins help in osmotic balance. Plasma also contains small amounts of minerals like Na⁺, Ca⁺⁺, Mg⁺⁺, HCO₃^–, CI^–, etc. Factors for coagulation or clotting of blood are also present in the plasma in an inactive form.

What is serum?
Plasma without the clotting factors is called serum.

Formed Elements

1. Erythrocytes
2. Leucocytes
3. Platelets

They constitute nearly 45 per cent of the blood.
1. Erythrocytes or red blood cells (RBC):
They are the most abundant and on an average, 5 millions to 5.5 millions of RBCs mm-3 of blood. RBCs are formed in the red bone marrow in the adults. RBCs are devoid of nucleus in most of the mammals and are biconcave in shape. They have a red coloured, iron containing complex protein called haemoglobin.

A healthy individual has 12 – 16 gms of haemoglobin in every 100 ml of blood. These molecules play a significant role in transport of respiratory gases. RBCs have an average life span of 120 days after which they are destroyed in the spleen (graveyard of RBCs).

2. Leucocytes:
They are also known as white blood cells, colourless, nucleated and are relatively lesser in number which averages 6000 – 8000 mm-3 of blood. Leucocytes are generally short lived. The two main categories of WBCs granulocytes and agranulocytes.

• Neutrophils are the most abundant cells (60 – 65 percent) of the total WBCs and basophils are the least (0.5 – 1 percent) among them.
• Neutrophils and monocytes (6 – 8 per cent) are phagocytic cells which destroy foreign organisms entering the body.
• Basophils secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions.
• Eosinophils (2 – 3 per cent) resist infections and are also associated with allergic reactions.

Lymphocytes (20-25 percent) are of two major types – ‘B’ and ‘T’ forms. Both B and T lymphocytes are responsible for immune responses of the body.

3. Platelets:
They are also called thrombocytes, are cell fragments produced from megakaryocytes (special cells in the bone marrow). Blood normally contains 1,500,00 – 3,500,00 platelets mm-3. Platelets are involved in the coagulation or clotting of blood. A reduction in their number can lead to clotting disorders which will lead to excessive loss of blood from the body.

Blood Groups
The ABO and Rh- are widely used all over the world.

ABO grouping
It is based on the presence or absence of two surface antigens on the RBCs namely A and B. The plasma of different individuals contain two natural antibodies (proteins produced in response to antigens). There are four groups of blood, A, B, AB and O.

Who are called as universal donor and receipient?
The group ‘O’ blood can be donated to persons with any other blood groupand hence ‘O’group individuals are called ‘universal donors’. Persons with ‘AB’ group can accept blood from persons with AB as well as the other grbups of blood, such persons are called ‘universal recipients’.

Rh grouping
Rh antigen is observed on the surface of RBCs of majority (nearly 80 percent) of humans. Such individuals are called Rh positive (Rh+ve) and without antigen are called Rh negative (Rh-ve). Rh group should also be matched before transfusions.

What is Rh incompatibility?
It is the mismatching of blood between the Rh-ve blood of a pregnant mother with Rh+ve blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh-ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta.

But during the delivery of the first child, there is a possibility of mixing of the maternal blood to small amounts of the Rh+ve blood from the foetus. In such cases, the mother starts preparing antibodies against Rh in her blood. In subsequent pregnancies, the

Solving of Rh incompatibility:
This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Coagulation of Blood
Blood coagulates or clots in response to an injury or trauma to prevent excessive loss of blood from the body. The network of threads called fibrins in which dead and damaged formed elements of blood are trapped.

Fibrins are formed by the conversion of inactive fibrinogens in the plasma by the enzyme thrombin. Thrombins are formed from another inactive substance present in the plasma called prothrombin.

An enzyme complex, thrombokinase, is required forthe above reaction. An injury stimulates the platelets in the blood to release certain factors which activate the mechanism of coagulation .Calcium ions play a very important role in clotting.

Lymph (tissue fluid)
As the blood passes through the capillaries in tissues, fluid released out is called the interstitial fluid or tissue fluid. It has the same mineral distribution as that in plasma. An elaborate network of vessels called the lymphatic system collects this fluid and drains it back to the major veins.

The fluid present in the lymphatic system is called the lymph. Lymph is a colourless fluid which are responsible forthe immune responses of the body. Lymph is also an important carrier for nutrients, hormones, etc. Fats are absorbed through lymph in the lacteals present in the intestinal villi.

Circulatory Pathways
The circulatory patterns are of two types – open or closed.

Open circulatory system:
It is present in arthropods and molluscs in which blood pumped by the heart passes through large vessels into open spaces or body cavities called sinuses.

Closed circulatory system:
It is present in Annelids and chordates in which the blood is pumped by the heart circulated through a closed network of blood vessels.
All vertebrates possess a muscular chambered heart.
1. Fishes have a 2-chambered heart with an atrium and a ventricle.

2. Amphibians and the reptiles (except crocodiles) have a 3-chambered heart with two atria and a single ventricle.

3. Crocodiles, birds and mammals possess a 4-chambered heart with two atria and two ventricles. In fishes the heart pumps out deoxygenated blood which is oxygenated by the gills and supplied to the body parts from where deoxygenated blood is returned to the heart (single circulation).

In amphibians and reptiles, the left atrium receives oxygenated blood from the gills/lungs/skin and the right atrium gets the deoxygenated blood from other body parts. However, they get mixed up in the single ventricle which pumps out mixed blood (incomplete double circulation).

In birds and mammals, oxygenated and deoxygenated blood received by the left and right atria respectively passes on to the ventricles of the same sides. The two separate circulatory pathways are present in these organisms, hence the animals have double circulation.

Human Circulatory System
Heart:
It is situated in the thoracic cavity, in between the two lungs. It is protected by a double walled membranous bag,pericardium, enclosing the pericardial fluid. It has four chambers, two small upper chambers called atria and two larger lower chambers called ventricles.

A thin, muscular wall called the inter atrial septum separates the right and the left atria, whereas a thick- walled, the inter-ventricular septum, separates the left and the right ventricles. The atrium and the ventricle of the same side are separated by a thick fibrous tissue called the atrioventricular septum.

Tricuspid and bicuspid valve:
The opening between the right atrium and the right ventricle is guarded by a valve formed of three muscularflaps or cusps, the tricuspid valve. Bicuspid or mitral valve guards the opening between the left atrium and the left ventricle.

Semilunar valve:
The openings of the right and the left ventricles into the pulmonary artery and the aorta respectively are provided with the semilunar valves.

Function of semilunar valve:
It allows the flow of blood only in one direction, i.e., from the atria to the ventricles and from the ventricles to the pulmonary artery or aorta. These valves prevent any backward flow.

SAN & AVN:
A specialised cardiac musculature called the nodal tissue is also distributed in the right upper corner of the right atrium called the sino-atrial node (SAN). Another mass of this tissue is seen in the lower left comer of the right atrium close to the atrio-ventricular septum called the atrio-ventricular node (AVN).

A bundle of nodal fibres, atrioventricular bundle (AV bundle) continues from the AVN which passes through the atrio-ventricular septa to emerge on the top of the interventrical sepyum and divides into a right and left bundle. These branches give rise to minute fibres throughout the ventricular musculature, they are called purkinje fibres.

These fibres alongwith right and left bundles are known as bundle of HIS. The nodal musculature has the ability to generate action potentials without any external stimuli.

What is the pacemaker of heart?
The SAN can generate the maximum number of action potentials, i.e., 70 – 75 min^-1, and is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Hence it is called the pacemaker. Our heart normally beats 70 – 75 times in a minute (average 72 beats min^-1).

Cardiac Cycle
As the tricuspid and bicuspid valves are open, blood from the pulmonary veins and vena cava flows into the left and the right ventricle respectively. The semilunar valves are closed at this stage. The action potential is conducted to the ventricular side by the AVN and AV bundle from where the bundle of HIS transmits it through the entire ventricular musculature.

This causes the ventricular muscles to contract, (ventricular systole), the atria undergoes relaxation (diastole). As the ventricular pressure increases the semilunar valves open, allowing the blood in the ventricles to flow through these vessels into the circulatory pathways.

The ventricles relax (ventricular diastole) and the ventricular pressure falls causing the closure of semilunar valves which prevents the backflow of blood into the ventricles.

This sequential event in the heart which is cyclically repeated is called the cardiac cycle and it consists of systole and diastole of both the atria and ventricles. The heart beats 72 times per minute,i.e., that many cardiac cycles are performed per minute.

Cardiac output is the volume of blood pumped out by each ventricle per minute and averages 5000 mL or 5 litres in a healthy individual.

Sound produced in heart:
For example, the cardiac output of an athlete will be much higherthanthat of an ordinary man. During each cardiac cycle sounds are produced, the first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance.

Electrocardiograph (ECG)
ECG is a graphical representation of the electrical activity of the heart during a cardiac cycle. Each peak in the ECG is identified with a letter from P to T that corresponds to a specific electrical activity of the heart.
1. The P-wave represents the electrical excitation Diagrammatic presentation of a standard ECG (or depolarisation) of the atria, which leads to the
contraction of both the atria.

2. The QRS complex represents the depolarisation of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.

3. The T-wave represents the return of the ventricles from excited to normal state (repolarisation). The end of the T-wave marks the end of systole.

Since the ECGs obtained from different individuals have the same shape any deviation from this shape indicates abnormality or disease.

Doube Circulation
It involves two types of circulation,
1. Pulmonary circulation:
The deoxygenated blood pumped into the pulmonary artery is passed on to the lungs from where the oxygenated blood is carried by the pulmonary veins into the left atrium. This pathway constitutes the pulmonary circulation.

2. Systemic circulation:
The oxygenated blood entering the aorta is carried by a network of arteries, arterioles and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins and vena cava and emptied into the right atrium. This is the systemic circulation.

The systemic circulation provides nutrients, O₂ and other essential substances to the tissues and takes CO₂ and other harmful substances away for elimination. A vascular connection between the digestive tract and liver called hepatic portal system.

The hepatic portal vein carries blood from intestine to the liver before it is delivered to the systemic circulation. In Coronary system of blood vessels is present in our body exclusively for the circulation of blood to and from the cardiac musculature.

Regulation Of Cardiac Activity
Normal activities of the heart are auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. Medulla oblangata control the cardiac function through autonomic nervous system (ANS).

Neural signals through the sympathetic nerves (part of ANS) increase the rate of heart beat. On the other hand, parasympathetic neural signals decrease the rate of heart beat. Adrenal medullary hormones can also increase the cardiac output.

Disorders Of Circulatory System
High Blood Pressure (Hypertension):
Hypertension is the term for blood pressure that is higherthan normal (120/80). In this measurement 120 mm Hg (millimetres of mercury pressure) is the systolic, or pumping, pressure and 80 mm Hg is the diastolic, or resting, pressure. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.

Coronary Artery Disease (CAD):
Coronary Artery Disease, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused by deposits of calcium, fat, cholesterol and fibrous tissues, which makes the lumen of arteries narrower.

Angina:
It is also called ‘angina pectoris’. A symptom of acute chest pain appears when no enough oxygen is reaching the heart muscle. It occurs due to conditions that affect the blood flow.

Heart Failure:
It is the state of heart when it is not pumping blood effectively enough to meet the needs of the body. Heart failure is not the same as cardiac arrest (when the heart stops beating) or a heart attack (when the heart muscle is suddenly damaged by an inadequate blood supply).

Students can Download Chapter 15 Waves Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 15 Waves

Summary
The wave is the propagation of disturbance that carries energy from one point to another point, without translatory motion of particles in the medium. There are three types of wave.
1. Mechanical Wave: Requires medium for propagation.
Eg: Sound wave, Matter wave, Seismic wave, etc.

2. Electromagnetic Wave: No medium for propagation.
Eg: Light, X-rays, UV ray, etc.

3. Matter Wave: Wave associated with moving particles (microscopic particle).
Eg: Wave of moving electron, proton, etc.

Generation of longitudinal waves by tuning for k
When prongs of tuning fork moves outward, it compresses the surrounding air and a region of increased pressure is formed. This region is called condensation. When the prongs move inward a region of low pressure called rarefraction is formed. Thus condensations and rarefractions are alternately produced.

Expression for progressive wave (Displacement relation)
A plane progressive wave propagating along positive direction of ‘x’ is given by

The wave propagating along negative ‘x’ direction is given by y (x, t) = a sin (kx + ωt + Φ).
y(x, t) gives the transverse displacement of element at position x at time t.

Crust and Trough:
Crust is the point of maximum positive displacement of wave. Trough is the point of maximum negative displacement of wave.

Transverse and Longitudinal Wave
Based on the direction of propogation and vibration wave can be of two types.

Parameters of wave
Amplitude:
The magnitude of maximum displacement of element of the wave from initial position is called amplitude (a).
Phase and initial phase:
The value (kx + ωt + Φ) is called phase and f is the initial phase. The phase gives the state of motion of wave at position ‘x’ and at time V. Initial phase gives initial state of wave.
Wave length (λ):
The linear distance travelled by the wave in one complete oscillation(or vibration). Or it can be defined as distance between two conT secutive crusts or troughs. It is the distance travelled during time period T.
Wave number/Angular wave number/propagation constant:
Wave number ‘k’ is defined as

Its unit is radian/m
Time period (T):
Time for one complete oscillation/vibration is called time period.
Frequency (ν):
The number of oscillations/vibrations in one second is called frequency. Its unit is S^-1 or Hz (Hetz).
ν = \(\frac{1}{T}\)
Angular velocity or angular frequency (ω):
Angular displacement per unit time is called angular velocity or angular frequency.

The speed of travelling wave [Relation connecting V, ν and λ]

The wave propagating along x direction is represented by y = A sin (kx – ωt + Φ). As wave moves, each point on the wave from (like A) retains its displacement. This is possible only if (kx – ωt) is constant. As the wave moves both x and t are changing to keep (kx – ωt) as constant (x increase with t).

The velocity depends on wavelength and frequency. The wavelength and frequency of wave depends on the properties of medium, i.e. velocity of wave in a medium is determined by

• linear mass density
• Elastic properties.

Speed of wave in stretched string (Transverse wave)
The velocity of wave in stretched string depends on

• linear mass density (µ)
• The tension (T)

Speed of sound wave (Longitudinal wave)
The speed of sound in medium depends on

• density of medium (ρ)
• Modulus of elasticity

Case: 1( In solid )
If solid has Young’s modulus ‘Y’

Case: 2( In liquid )
If liquid has Bulk modulus ‘B’

Case: 3 In gas
The speed of sound waves in gas was determined by Newton. According to Newton, condensations and rare fractions are isothermal processes. Hence modulus of elasticity is equal to pressure.

This is called Newton’s formula.
Correction in Newton’s formula.

Question 1.
Find velocity of sound in air using Newton’s formula (P = 1.013 × 10⁵ ρ =1.239 kg m^-3)
Answer:

Note: The velocity of sound at STP is found to be 332 ms^-1
Laplace’s Formula:
Laplace corrected Newton’s formula taking condensation and rare fraction as adiabatic process. The
modulus of elasticity is now ‘γP‘ where γ = \(\frac{C_{p}}{C_{v}}\). C_p is specific heat capacity at constant pressure and C_v is specific heat capacity at constant volume.

The principle of superposition of waves
It states that when two or more waves pass through a media the net displacement of particle at any time is the algebraic sum of displacements due to each wave.

(or)

The overlapping waves algebraically add to produce a resultant wave.

Reflection of waves
Reflection from rigid boundary:
When a travelling wave is reflected by a rigid boundary, phase reversal (phase difference of π or 180°) will take place.

Reflection from open boundary:
When a travelling wave is reflected by an open boundary, no phase change will happen. The incident and reflected wave superimpose to give maximum displacement at boundary.

Standing waves

When two waves of same amplitude and frequency travelling in opposite direction superimpose the resulting wave pattern does not move to either sides. This pattern is called standing wave.
The wave travelling in positive direction of x axis y₁(x, t) = a sin(kx – ωt)
The wave travelling in negative direction of x axis y₂(x, t) = a sin(kx + ωt)
According to superposition Principle, the combined wave is
y(x, t) = y₁(x, t) + y₂(x, t)
y(x, t) = a sin(kx – ωt) + a sin(kx + ωt)
But sin A + sin B = 2 sin \(\frac{(A+B)}{2} \cos \frac{(A-B)}{2}\)
Hence we get,combined wave as

This wave has an amplitude of ‘2asinkx’, and it is not a moving wave.
Nodes & Antinodes:
The position of maximum amplitude in a standing wave is termed as anti node and position of minimum amplitude (zero) is termed as node.
Node: The amplitude of standing wave is ‘2 a sin kx’. It is zero when kx = 0, π, 2π…. etc.
ie kx = nπ.

Antinode:
The amplitude has maximum value 2a when (2a sin kx = 2a)
sinkx = 1;
ie; kx = π/2, 3π/2, 5π/2 …… etc

But k = \(\frac{2 \pi}{\lambda}\)
Hence we get

n = 0, 1, 2, 3 etc.

1. Standing Waves In Stretched String & Modes Of Vibration Of String:
A string of length L is fixed at two ends. The position of one end is chosen as x = 0, then the position of other end will be x = L. At x = 0, there will be node. To occur node at x = L, it must satisfy

The frequency of vibrations of stretched string of length L is

n = 1,2, 3…etc.
This set of frequencies at which the string can vibrate are called natural frequencies or modes of vibration or harmonics. The above equation shows that the modes of vibration (natural frequencies) of string are integral multiple of lowest frequency
n = \(\frac{V}{2 L}\) (for n = 1)
Fundamental mode(or) First harmonics:
For n = 1
ν₁ = \(\frac{V}{2 L}\)
This is the lowest frequency with which string vibrates. This is called fundamental mode or first harmonic of vibration.
Relation between I and L for first harmonics:

The string vibrate in a single segment as shown in figure.

Second harmonic:

For n = 2

Relation between I and L for second harmonics

Third harmonic:

For n = 3, there is 3^rd harmonic. Thus collection of all possible mode is called harmonic series and n is called harmonic number.

2. Vibrational modes of an air column:
(a) In closed tube:
In closed tube one end is closed and other end is open. Air column in a glass tube partially filled with water is an example of closed system. The air column in tube can be set into vibrations with the help of air excited by tuning fork.

The longitudinal waves thus generated is reflected at the closed end and a node is formed there [reflected and incident wave are out of phase and at the closed end, they superimpose to give minimum displacement]. At the open end, the displacement is maximum and antinode is formed. If L is the length of air column, anti node occurs at x = L.
We know the condition for antinode x = (n + 1/2) λ/2
Therefore L = (n + 1/2) λ/2
for n = 0, 1, 2, …….etc.
The wavelength, λ = \(\frac{2 \mathrm{L}}{(\mathrm{n}+1 / 2)}\) _____(1)
for n = 0, 1, 2, …. etc.
The frequency ν = (n + 1/2) \(\frac{V}{2 L}\) ____(2)
for n = 0, 1, 2, …….etc.
From this equation, it is clear that the air column can vibrate with different modes of frequencies (normal modes or harmonics)
Fundamental Mode(or) First harmonic:
We get fundamental mode when n=0 Substitute this in eq(2), we get
ν₁ = \(\frac{V}{4 L}\)
This is the fundamental frequency. The higher frequencies are odd harmonics of fundamental frequency ie;

(b) Open tube:
In open tube, both ends are open. At both ends, antinodes are formed.
The condition to get antinode x = n λ/2

The frequency (fundamental frequency):
We will get fundamental frequency in open tube,when n = 1. Substitute this in eq(4)
ν₁ = \(\frac{V}{2 L}\)
In open pipe all harmonies are generated whereas in closed pipe only odd harmonies are generated.
For n = 2, ν₂ = 2\(\frac{V}{2 L}\). This is second harmonic.
For n = 3, ν₃ = 3\(\frac{V}{2 L}\). This is 3rd harmonic.
ν₂ = 2 ν₁ & ν₃ = 3 .ν₁. Thus both odd and even harmonics are generated.

Beats:
When two sound waves of nearly same frequency and amplitude travelling in same direction super imposed and periodic variation of sound intensity (wavering of sound or waxing and waning of sound) is produced. This is called beats.
Explanation:
If two tuning forks of slightly different frequencies are sounded together, a regular rise and fall of sound can be heard. The sound travels in the form of condensation and rarefactions.

When two condensations due to two notes reach our ear at the same time, they superimpose to get maximum intensity (waxing of sound). If two rarefactions are reached simultaneously, they superimpose to get minimum intensity (waning of sound).

Analytical treatment of beats
Beats frequency
Suppose two sound waves of u₁ and u₂ propagating in the same direction through a medium. For simplicity let the listener be situated at x = 0 and the amplitudes of waves to be equal ie. a₁ = a₂ = a.
The displacements y₁ and y₂ due to each wave are given by
y₁ = a sin2pu₁t and y₁ = a sin2pu₂t
According to super position principle, the resultant displacement at the same time t is
y = y₁ + y₂
= a sin2pu₁t + a sin2pu₂t
y = a [sin2pu₁t + sin2pu₂t]

It is clear that, the amplitude of resultant wave (A) changes with time. It shows maxim and minima.
The resultant amplitude will be maximum, if,

Hence the amplitude of the resultant wave will be maximum at times

Time interval between successive maxima = \(\frac{1}{v_{1}-v_{2}}\)
resultant amplitude will be minimum if

Time internval between two consecutive minima = \(\frac{1}{v_{1}-v_{2}}\)
Frequency of minima = ν₁ – ν₂
Graphical representation of beats

Doppler Effect
The apparent change in the frequency of sound wave due to the relative motion of source or listener or both is called Doppler effect. It was proposed by John Christian Doppler and it was experimentally tested by Buys Ballot.

Considers source is producing sound of frequency n. Let V be the velocity of sound in the medium and I the wavelength of sound when the source and the listener are at rest. The frequency of sound heard by the listener is
ν = \(\frac{v}{\lambda}\)
Let the source and listener be moving with velocities v_s and v_l in the direction of propogation of sound from source to listener. (The direction S to L is taken as positive)
The relative velocity of sound wave with respect to the source = V – V_s.
Apparent wavelength of sound,
λ¹ = \(\frac{V-V_{s}}{v}\) ____(1)
Since the listener is moving with velocity v f, the rela¬tive velocity of sound with respect to the listener,
V¹ = V – V_l _____(2)

Question 2.
A string fixed one end is suddenly brought in to up and down motion.

1. What is the nature of the wane produced in the string and name the wave?
2. A brass wire 1 m long has a mass 6 × 10^-3 kg. If it is kept at a tension 60N, What is the speed of the wave on the wire?

Answer:
1. Transverse wave

2.

Students can Download Chapter 6 Breathing and Exchange of Gases Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 6 Breathing and Exchange of Gases

What is respiration?
The process of exchange of O₂ from the atmosphere with CO₂ produced by the cells is called breathing, commonly known as respiration.

RESPIRATORY ORGANS:

Human Respiratory System:
The nostrils leads to a nasal chamber through the nasal passage. The nasal chamber opens into nasopharynx the common passage for food and air. Nasopharynx opens through glottis of the larynx region into the trachea.

Sound box in respiratory system:
Larynx is a cartilaginous box which helps in sound production called as the sound box. Glottis is covered by a thin elastic cartilaginous flap called epiglottis to prevent the entry of food into the larynx. Trachea is a straight tube which divides into a right and left primary bronchi.

Each bronchi undergoes repeated divisions to form the secondary and tertiary bronchi and bronchioles ending up in very thin terminal bronchioles.
Each terminal bronchiole gives rise to vascularised bag-like structures called alveoli. The branching network of bronchi, bronchioles and alveoli comprise the lungs.

The two lungs which are covered by a double layered pleura, with pleural fluid between them. It reduces friction on the lung surface. The conducting part transports the atmospheric air to the alveoli. Exchange part is the site of diffusion of O₂ and CO₂ between blood and atmospheric air.

Where is lung fitted in human body?
The lungs are situated in the thoracic chamber which is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs and on the lower side by the dome-shaped diaphragm. The anatomical setup of lungs in thorax is the arrangement essential for breathing, directly alterthe pulmonary volume.

Respiration involves the following steps:

• Breathing or pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out.
• Diffusion of gases (O₂ and CO₂) across alveolar membrane.
• Transport of gases by the blood.
• Diffusion of O₂ and.CO₂ between blood and tissues.
• Utilisation of O₂ by the cells for catabolic reactions and resultant release of CO₂

MECHANISM OF BREATHING:
Breathing involves two stages:

1. Inspiration during which atmospheric air is drawn in and
2. Expiration by which the alveolar air is released out.

How does increase or decrease of pulmonary volume occur?
Inspiration is initiated by the contraction of diaphragm which increases the volume of thoracic chamber .The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber. It increase the pulmonary volume.

An increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure which forces the airfrom outside to move into the lungs,i.e., inspiration Relaxation of the diaphragm and the inter-costal muscles returns the diaphragm and sternum to their normal positions and reduce the thoracic volume.

It reduce the pulmonary volume. This leads to an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causing the expulsion of airfrom the lungs, i.e., expiration.

Instrument used for measuring breathing movements:
A healthy human breathes 12 – 16 times/minute. The volume of air involved in breathing movements can be estimated by using a spirometer.

Respiratory Volumes and Capacities:

EXCHANGE OF GASES:
Alveoli are the sites of exchange of gases. O₂ and CO₂ are exchanged in these sites by simple diffusion.
Rate of diffusion:
Solubility of the gases and the thickness of the membranes are the important factors that affect the rate of diffusion.

Partial pressure of gases:
Pressure contributed by an individual gas in a mixture of gases is called partial pressure and is represented as pO₂ for oxygen and pCO₂ for carbon dioxide. As the solubility of CO₂ is 20 – 25 times higher than that of O₂, the amount of CO₂ that can diffuse through the diffusion membrane.

As the solubility of CO₂ is 20 – 25 times higher than that of O₂ the amount of CO₂ that can diffuse through the diffusion membrane.

The diffusion membrane is made up of three major layers such as

Its total thickness is much less than a millimetre.

TRANSPORT OF GASES:
Blood is the medium of transport for 02 and C02.

1. About 97 per cent of O₂ is transported by RBCs in the blood.
2. The remaining 3 per cent of O₂ is carried in a dissolved state through the plasma.
3. Nearly 20 – 25 per cent of CO₂ is transported by RBCs whereas 70 per cent of it is carried as bicarbonate.
4. About 7 per cent of CO₂ is carried in a dissolved state through plasma.

Transport of Oxygen:
O₂ bind with haemoglobin to form oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of O₂. Binding of oxygen with haemoglobin is related to partial pressure of O₂.

Oxygen dissociation curve is useful in studying the effect of factors like pCO₂, H⁺ concentration, etc., on binding of O₂ with haemoglobin.

Therefore O₂ gets bound to haemoglobin in the lung surface and gets dissociated at the tissues.

Transport of Carbon dioxide:
CO₂ is carried by haemoglobin as carbamino-haemoglobin (about 20 – 25 per cent). This binding is related to the partial pressure of CO₂. When pCO₂ is high and pO₂ is low as in the tissues, more binding of carbon dioxide occurs whereas, when the PCO₂ is low and pCO₂ is high as in the alveoli, dissociation of CO₂ from carbamino-haemoglobin takes place. RBCs contain a very high concentration of the enzyme, carbonic anhydrase which facilitates the reaction in both directions.

At the tissue site where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into blood (RBCs and plasma) and forms HCO₂ and H⁺. At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation 0f CO₂ and H₂O. Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO₂ to the alveoli.

Students can Download Chapter 16 Probability Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 16 Probability

Plus One Maths Probability Three Mark Questions and Answers

Question 1.
Two dice are thrown. The events A, B and C are as follows:

1. A: getting an even number on the first die.
2. B: getting an odd number on the first die.
3. C: getting sum of the numbers on the dice ≤ 5.

Describe the events.
Answer:

1. A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
2. B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
3. C = {(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (3, 1), (2, 3), (3, 2), (1, 4), (4, 1)}

Question 2.
A bag contains 6 red and 12 green balls. Two balls are drawn. What is the probability that one is red and other is green?
Answer:
Here total number of balls = 6 + 12 = 18 Two balls from 18 can be drawn in
¹⁸C₂ = \(\frac{18 \times 17}{1 \times 2}\) = 153
One red ball out of 6 red can be drawn in ⁶C₁ = 6 ways. One green balls from 12 green may be done in ¹²C₁ = 12 ways.
Therefore, number of favorable cases
= 6 × 12 = 72
The probability that one is red and other is green \(=\frac{72}{153}=\frac{8}{17}\).

Question 3.
In class XI of a school 40% of the students study Mathematics and 30% study Biology, 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer:
Let M- Mathematics and B – Biology be the events.
P(M) = \(\frac{40}{100}\) = \(\frac{2}{5}\); P(B) = \(\frac{3}{10}\);
P(M ∩ B) = \(\frac{1}{10}\)
P(M ∪ B) = P(M) + P(B) – P(M ∩ B)

Plus One Maths Probability Four Mark Questions and Answers

Question 1.
Two die are rolled, A is the event that the sum of the numbers shown on the two die is 7 and B is the event that at least one of the die shows up 2. Are the two events A and B

1. Mutually exclusive
2. Exhaustive.

Answer:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5),(2,6), (3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = {(6, 1), (1, 6), (4, 3), (3, 4), (2, 5), (5, 2)}
B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
1. Now; A ∩ B = {(2, 5), (5, 2)} ≠ Φ
Therefore not mutually exclusive.

2. A ∪ B = {(6, 1), (1, 6), (4, 3), (3, 4), (2, 5), (5, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (1, 2), (3, 2), (4, 2), (6, 2)} ≠ S
Therefore not mutually exhaustive.

Question 2.
A letters of the word ASSASSINATION are randomly chosen. Find the probability that letter is

1. a vowels. (2)
2. a consonant (2)

Answer:
There 13 letter in the word, with 6 vowels and 7 consonants.
One letter is selected out of 13 in ¹³C₁ = 13 ways.
1. One vowel is selected out of 6 in ⁶C₁ = 6 ways.
Thus the probability of a vowel = \(\frac{6}{13}\).

2. One consonant is selected out of 7 in ⁷C₁ = 7 ways.
Thus the probability of a consonant = \(\frac{7}{13}\).

Question 3.
If E and F are events such that P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\) and P(E and F) = \(\frac{1}{8}\). Find

1. P(E or F)
2. P(not E and not F)

Answer:
P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\); P(E ∩ F) = \(\frac{1}{8}\)
1. P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

2. P(not E and not F) = P(E’ ∩ F’) = P(E ∪ F)’
= 1 – P(E ∪ F) = 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\)

Plus One Maths Probability Practice Problems Questions and Answers

Question 1.
A die is thrown. Describe the following events: (1 score each)

1. A: a number less than 7.
2. B: a number greater than 7.
3. C: a multiple of 3:
4. D: a number less than 4.
5. E: An even number greater than 4.
6. F: a number not less than 3.

Also find
A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E F’, E ∩ F’.
Answer:

1. A = {1, 2, 3, 4, 5, 6}
2. B = Φ
3. C = {3, 6}
4. D = {1, 2, 3}
5. E = {6}
6. F = {3, 4, 5, 6}

Now; A ∪ B = {1, 2, 3, 4, 5, 6}
A ∩ B = Φ; B ∪ C = {3, 6}, E ∩ F = {6}
F’ = {1, 2}, E ∩ F’ = Φ.

Question 2.
Describe the sample space for the following events: (2 score each)

1. A coin is tossed and a die is thrown.
2. A coin is tossed and then a die is rolled.
3. 2 boys and 2 girls are in a Room X and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
4. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its upper most face is noted.
5. An experiment consists of tossing a coin and then throwing it second time if a head occur. If a tail occurs on the first toss, then a die is rolled once.
6. A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up shows up an head, a die is thrown. If the die shows up an even number, the die is thrown again.

Answer:
1. A coin is tossed then S₁ = {H, T}
A die is rolled then S₂ = {1, 2, 3, 4, 5, 6}
Hence sample space
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

2. A coin is tossed then S₁ = {H, T}
A die is rolled then S₂ = {1, 2, 3, 4, 5, 6}
Hence sample space S = {H1, H2, H3,H4, H5,H6, T}

3. Let B₁, B₂ denote the boys and G₁, G₂ girls in room X, B₁ denote the boys and G₃, G₄, G₅ girls in room Y. Hence sample space
S = {XB₁, XB₂, XG₁, XG₂, YB₁, YG₃, XG₄, XG₅}

4. Three die are R,W, and B, then S₁ = {R, W, B)
A die is rolled then S₂ = {1, 2, 3, 4, 5, 6}
Hence sample space S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

5. A coin is tossed then S₁ = {H, T}
A coin tossed again then S₂ = {H, T}
A die is rolled then S₃ = {1, 2, 3, 4, 5, 6}
Hence sample space
S = {HH, HT, T1, T2, T3, T4, T5, T6}

6. A coin is tossed then S₁ = {H, T}
When the coin shows T, then there is no action. When the coin shows H, a die is thrown
S₂ = {1, 2, 3, 4, 5, 6}
When the die shows {1, 3, 5}, then there is no action.
When the die shows {2, 4, 6}, then a die is thrown again.
Hence;
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46,H61, H62, H63, H64, H65, H66}.

Question 3.
A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?
Answer:
When die is rolled the sample space is S = {1, 2, 3, 4, 5, 6}.
E: die shows 4 = {4}
F: die shows even number = {2, 4, 6}
Now E ∩ F = {4} ≠ Φ
Thus E and F are not mutually exclusive.

Question 4.
Two die are thrown simultaneously. Find the probability of getting 4 as the product.
Answer:
n(S) = 36
A = {(1, 4), (4, 1), (2, 2)}
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\).

Question 5.
A coin is tossed, twice, what is the probability that atleast one tail occurs?
Answer:
Here, S = {HH, HT, TH, TT}; n(S) = 4
A = {TH, HT, TT}; n(A) = 3
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{4}\).

Question 6.
A die is rolled, find the probability of following events: (2 score each)

1. A prime number will appear.
2. A number greater than or equal to 3 will appear.
3. A number less than or equal to one will appear.
4. A number more than 6 will appear.
5. A number less than 6 will appear.

Answer:
Here sample space is S = {1, 2, 3, 4, 5, 6}
1. A: A prime number will appear A = {2, 3, 5};
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

2. B: A number greater than or equal to 3 will appear. B = {3, 4, 5, 6};
P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

3. C: A number less than or equal to one will appear. C = {1};
P(C) = \(\frac{n(C)}{n(S)}=\frac{1}{6}\)

4. D: A number more than 6 will appear.
D = Φ; P(D) = \(\frac{n(D)}{n(S)}=\frac{0}{6}=0\)

5. E: A number less than 6 will appear. E = {1, 2, 3, 4, 5, 6};
P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{6}=1\).

Question 7.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that (2 score each)

1. The card drawn is black.
2. The card drawn is a king.
3. The card drawn is black and a king.
4. The card drawn is either black ora king.

Answer:
1. A pack of 52 cards has 26 black cards. So one black card may be drawn in 26 ways. Therefore; Probability of card drawn is black
= \(\frac{26}{52}=\frac{1}{2}\)

2. A pack of 52 cards has 4 kings. So one king card may be drawn in 4 ways. Therefore; Probability of drawing a king
= \(\frac{4}{52}=\frac{1}{13}\)

3. A pack of 52 cards has 2 black king cards. So one black king card may be drawn in 2 ways. Therefore; Probability of drawing a king
= \(\frac{2}{52}=\frac{1}{26}\)

4. A pack of 52 cards has 26 black cards which include 2 black king and 2 red king cards. So number of cards which are black or king cards
= 26 + 2 = 28.
Therefore; Probability of drawing either a black or king card = = \(\frac{28}{52}=\frac{7}{13}\).

Question 8.
GivenP(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\). Find P(A or B), if A and B are mutually exclusive events.
Answer:
Here; P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\)
Since A and B mutually exclusive events
P(A ∪ B) = P(A) + P(B)