Download Document

Solved Assignment of Unit IST for Class 10
Subject: Science, Mathematics, English, History, Hindi,Urdu:
SCIENCE
CLASS 10TH CH: 1
LIFE PROCESSES
LIFE PROCESSES
Life Processes are those basic functions of living beings which are essential for their
survival. They are the same in all types of living forms whether unicellular or
multicellular, plants or animals.
1. Maintenance:- All living beings are made of protoplasmic structures. They have to
be kept in functional state whether an organism is active or inactive. Functional state
of protoplasmic structures is maintained only if they are kept in dynamic state with
breakdown and build-up processes going on simultaneously.
2. Metabolism:- It is the sum total of all chemical reactions which occur in a living
being due to interaction amongst its molecules. All functions of organisms are due to
metabolism. Metabolism has two components, anabolism and catabolism.
Anabolism orconstructive metabolism consists of build-up reactions where complex
molecules are formed from simpler ones, e.g., formation of glycogen from glucose or
proteins from amino acids. Energy is required for anabolicreactions.
Catabolism or destructive metabolism consists of breakdown reactions where
complex substances are broken down into simpler substances, e.g., respiration (breaks
glucose into carbon dioxide and water).
3. Nutrition:- It is required by all living beings for providing energy and body
building materials. Body building materials are usually carbon based so that food
sources are also carbon based. Plants manufacture their own food in the process of
photosynthesis. Animals obtain food from outside. Food obtained from outside is first
broken down into simpler soluble substances for absorption. Inside the cells, the
simple substances are converted into various complex bio-chemicals to form
components of protoplasm.
4. Respiration:- Every living being requires energy for working of body machinery,
its maintenance, repair, replacement and bio-synthesis. Energy is obtained by break
down of carbon based molecules in the process of respiration. Oxidation-reduction
reactions are common chemical reactions involved in respiratory break down of
molecules. Most organisms use oxygen obtained from outside for the process of
respiration.
5. Growth:- It is irreversible increase in body cells that occurs in young organisms
prior to reaching maturity. Plants have the ability to continuously grow. Growth is
possible if build-up reactions (Anabolism) are more abundant than break down
reactions (Catabolic reactions). For this, the organisms must prepare or obtain food
materials more than their requirement for maintenance.
6. Exchange of materials:- There is a regular exchange of materials between the
living organisms and their environment. Living beings obtain nutrients, water and
oxygen from their environment. They give out undigested materials, carbon dioxide
and waste products. Single-celled organisms have the entire surface in direct contact
with the environment. They do not possess any specific structures for intake and
explosion of materials. Diffusion, facilitated transport and active transport are
involved for movement of substances across the cell membrane.
In multicellular organisms specialized structures have been formed for different
functions e.g., ingestion, egestion, exchange of gases.
7. Transportation:- In multicellular organisms, all the cells are not in direct contact
with the environment. They have specific structures for exchange of gases, ingestion
and digestion of food materials. However, every cell of the body has to be provided
with food, water and oxygen. Similarly, carbon dioxide and wastes have to be taken
away from every cell. Therefore, a mechanism of transportation is found. It is
circulatory system in animals and vascular tissues in plants.
8. Excretion:- A number of waste products are formed as by-products of metabolism.
They are usually toxic and are removed from the body. The process of removal of
waste products from the body is called excretion.
9. Irritability:- Every living organism is aware of its surroundings. It responds to
changes in the environment.
NUTRITION
Nutrition may be defined as a collective phenomenon in which an organism is
capable of capturing food essential not only for maintaining vital life processes, but
also for repair and growth of tissues. The whole process of nutrition involves
introduction of food, its digestion by various juices and enzymes and break down into
smaller simpler and soluble molecules, absorption of the soluble food and lastly
expulsion of undigested matter formed during the process.
MODES OF NUTRITION: - The nutrition of all the organisms may broadly be
grouped into three categories, which are:a) Autotrophic Nutrition Or Holophytic:- It is a mode of nutrition in which
organisms are able to build up their own organic food from inorganic raw materials
with the help of energy. The organisms performing Autotrophic nutrition are called
Autotrophs, Auto meaning self and trophe meaning nourishment. Autotrophic
nutrition is a characteristic feature of all the green coloured pigments called as
chlorophyll. These plants have an ability to synthesise all the essential organic
compounds. From inorganic compounds like CO2 and H2O in presence of sunlight
with the help of a process called Photosynthesis. Some animals, that contain
chlorophyll in their bodies also, show this property of synthesizing their own food
material e.g.; Euglena and some green bacteria like sulphur and methane bacteria.
b) Heterotrophic Nutrition:- It is a mode of nutrition in which organisms obtain
readymade organic food from outside source. The organisms that depend upon
outside sources for obtaining organic nutrients are called heterotrophs, (hetero or
heteros and trophe meaning nourishment). It is a characteristic feature of all animals
and non green plants, that are unable to utilize carbon and synthesis organic
compounds necessary for life, but depends upon organic sources of carbon. They are
thus dependent upon autotrophic organisms (Plants) and are called as heterotrophs. It
is of the following types:1) Saprophytic Nutrition:-In this type of nutrition, an organism lives upon dead
organic sources such as dead plants and dead animals. These usually secrete
dissolving and digesting enzymes and absorb the liquidified molecules so formed e.g.;
yeast, bread moulds and dung moulds etc.
2) Parasitic Nutrition:-In this type of nutrition, an organism lives totally at the
expense of others and derives its food material and shelter from the other .These
organisms which derive food material are called parasites and the organism from
which food is derived is called as host. This type of nutrition is termed as parasitic or
holozoic nutrition .It is also known as parasite-host relationship e.g. Cuscuta, Ascaris
etc.
3) Holozoic Nutrition:- It is a mode of heterotrophic nutrition which involves intake
of solid pieces of food. Since solid food is taken in, Holozoic nutrition is also called
ingestive nutrition. Holozoic nutrition (GK. Holo-Whole, Zoon-Animal) is found in
animals and protozoan protists. The food may consist of another animal, plant or its
parts. Depending upon the source of food, Holozoic organisms are of three typesHerbivores, Carnivores, Omnivores.
Steps in Holozoic Nutrition:There are five steps in holozoic nutrition- ingestion, digestion, absorption,
assimilation and egestion.
1. Ingestion (L.ingestus-taken in.):-It is taking in of solid food with the help of
temporary or permanent mouth. Different animals use different organs for catching,
holding and putting the food into mouth. Cutting and tearing the solid food into small
pieces is common for ingestion.
2. Digestion:-The ingested food consists of complex insoluble organic substances.
The conversion of complex insoluble food ingredients into simple absorbable form is
called digestion. It is a catabolic process which occurs with the help of digestive
enzymes.
3. Absorption:- The digestive food is absorbed from the digestive tract and
transported to all body parts. It is picked up by all the living cells.
4. Assimilation:- Inside the living cells, the absorbed food materials are used in
obtaining energy and formation of new components for repair and growth of cells.
Assimilation is an anabolic process as it takes part in synthesis of proteins,
polysaccharides, fats and other macromolecules.
5.Egestion:- (L. egestus-discharge):- The whole of ingested food is seldom digested.
The undigested components of food are thrown out of the body as faecal matter. The
process is called egestion.
DIFFERENCE BETWEEN AUTOTROPHIC AND HETEROTROPHIC
NUTRITION
Autotrophic Nutrition
1. Food is self manufactured.
Heterotrophic Nutrition
1. Food is obtained readymade from outside.
2. An external source is not required. The
2. An external source of energy is required required energy is present in the food
for synthesis of food.
obtained from outside
3. Inorganic substances are not much
required.
3. Inorganic substances constitute the raw
materials for manufacturing food.
4. An external or internal digestion is
4. Digestion is absent.
required for conversion of complex organic
materials into simpler and soluble ones.
5. Chlorophyll is absent.
5. Chlorophyll is present for trapping light
energy.
6. Organisms performing autotrophic
nutrition function as producers.
Examples : Green plants, Some bacteria and
some protists.
6. Organisms performing heterotrophic
nutrition function as consumers.
Examples: Animals, Many protists and
Monerans.
DIGESTION
Digestion may be defined as hydrolysis or breakdown of complex organic
macromolecules into their diffusible and simpler organic micromolecules by
certain mechanical and biochemical changes with the help of bio-catalytic enzymes
so that they can be absorbed in the gastrointestinal tract for further utilization. It
starts in the mouth and continues into small intestines.
TYPES OF DIGESTION :Digestion is of two types: - Intracellular and Extracellular
1) Intracellular Digestion: - In unicellular organisms like Amoeba the food is taken
through its surface (Endocytosis) and the digests it within the cell. This type of
digestion occurring within the cell is known as Intracellular Digestion.
2) Extracellular or Intercellular Digestion:- In higher multicellular organisms the
food is put into a cavity or canal called alimentary canal or digestive tract or Gut or
GIT where food is digested and then absorbed into the blood to reach the different
parts of the body for proper utilization. The undigested food if any is passed out of the
body through the other end of the alimentary canal. This type of digestion (outside the
cell) is known as Extracellular or Intercellular digestion.
DIGESTION IN MAN:- Digestion of food material ingested by a man does not
take place in a particular region of the digestive tract or alimentary canal. It is
completed in different parts which may be described as under
1. Digestion of food in Mouth (buccal cavity):- In man the process of digestion
starts in the mouth which prepares food material for the digestion.
(i) Mechanical Digestion:- In mouth, food is subjected to grinding action of teeth
called as mastication. It is brought about by movement of lower jaw against upper
jaw. Food is broken into small pieces to increase the surface area and to make
enzymatic action efficient.
(ii) Chemical Digestion :-Three pairs of salivary glands i.e., parotids, sub-maxillary
(sub-mandibular) and sub-lingual are activated by the stimulus of sight, smell and
touch (conditional reflex) and they release saliva. Saliva is a watery fluid with ph 6.8.
It contains large number of enzymes, of which salivary amylase (ptyalin) is of great
importance, causing chemical digestion of starch and maltose into isomaltose.
The masticated and partially digested food is then rounded off in a form of small balls
called as bolus with the help of tongue and is passed through the oesophagus or food
pipe or gullet into the stomach by the peristaltic movements of the oesophagus.
2. Digestion in Stomach:- During digestion, stomach performs following functions:(i) Storage:- Food is stored for some time in the stomach. This is very important
function as it makes continuous feeding unnecessary.
(ii) Chemical Digestion:- As food reaches stomach, a group of cells called G-cells
(gastrin producing cells) in its walls produce hormone gastrin. This hormone
activates gastric glands which release gastric juice. The partially digested food
material is acted upon by gastric juice containing HCl, pepsin, rennin, and gastric
lipase enzymes. The gastric juice is a watery acidic fluid with ph 1-2 is a composite
secretion of at least three different types of cells - parietal cells, chief cells and
mucous cells. The parietal cells secrete HCL, The chief cells supply pepsin and other
enzymes like rennin etc. and the mucous cells secrete mucin in the stomach
 HCl acts as preservative and does not allow putrefaction of food. It changes
inactive pepsinogen into active pepsin. It lowers ph in the stomach and
provides low ph (acidic medium) for the action of pepsin.
 Pepsin breaks down the proteins into smaller molecules known as peptidase.
 Renin causes coagulation of milk by changing soluble caesinogen into
insoluble caesin. Caesin reacts with calcium to form calcium paracaesinate.
 Gastric lipase cause fat digestion in the stomach.
 Mucin protects the lining of stomach against the action of HCl.
(iii) Mechanical Digestion :- The walls of stomach show slight churning movements
which breaks food into fine paste and mix with various juices in stomach turning the
food into soup called chyme. This chyme moves towards pyloric end of stomach
where pyloric sphincter contracts and juts of food pass into the duodenum for further
action.
3. Digestion in Duodenum: - Digestion in duodenum can be explained as follows:(i) Inhibition of secretion of gastric juice:- As food reaches in duodenum, a group of
cells in its walls called mucosal cells secrete a hormone called enterogastrone which
Inhibits secretion of gastric juice.
(ii) Release and functions of Bile :- Another group of cells in the walls of duodenum
called I-cells secrete hormone cholecystokinin which cause gall bladder to contract
and release bile in the duodenum.
Functions of Bile:It causes emulsification of fat i.e., breaks fat globules into small fat droplets.
It neutralizes acidic medium in food and forms an alkaline medium.
It acts as antiseptic and laxative.
The fat soluble vitamins are dissolved in cholesterol of Bile and then they are
absorbed in blood.
(iii) Release of pancreatic juice:- Another group of cells in the walls of duodenum
called S-cells secrete hormone secretin which stimulates pancreas and release
pancreatic juice.
Pancreatic juice is a watery alkaline fluid with ph 7.8. It contains following enzymes:
i. Trypsin :- It occurs in an inactive form called trypsinogen. It is rendered active by
enterokinase (an enzyme secreted by mucosa cells of duodenum). Trypsin changes
proteins into peptides. Trypsin can digest variety of proteins like collagen, fibrinogen
etc. but not keratin.
ii. Amylase :- It changes starch into maltose and isomaltose.
iii. Lipase :- It changes fat into fatty acids and glycerol.
iv. Chymotrypsin:- It causes coagulation of milk and changes soluble caesinogen
into insoluble caesin.
The peristaltic movements of walls of duodenum push food into small
intestine (ileum).
4. Digestion in small intestine (ileum):- The intestinal
glands i.e., Brunner's glands and Crypts of Lieberkuhn
release intestinal juice or succus entericus. Intestinal juice
is a watery yellowish alkaline fluid with ph 8.6. It contains
following enzymes:
i. Enterokinase :- It changes inactive trypsinogen into active trypsin.
ii. Diamino peptidases :- They break peptides into smaller peptides.
iii. Carboxy peptidases :- They break small peptides into amino acids.
iv. Maltase :- It breaks maltose into glucose.
v. Sucrase :- It breaks sucrose into glucose and fructose.
vi. Lactase :- It breaks lactose into glucose and galactose.
vii. Lipase :- It breaks fats into fatty acids and glycerol.
viii. Nucleases :- They include D-nuclease and R-nuclease. They break nucleic acids
into nucleotides.
ix. Nucleotidases :- They break nucleotides into nucleosides and phosphates.
The fully digested food in small intestine contains water, amino acids, glucose,
fructose, galactose, fatty acids, glycerol, phosphates, nucleosides etc. and is called
chyle. This fully digested food is then diffused through the epithelial cells of the
intestinal villi (finger like projections of the intestine) into the blood vessels which
carries it to the cells.
4. Digestion in large intestine:- The process of the digestion does not take place
in this region. It simply secrets mucus, which serves as a lubricant for easy passage
of the undigested matter in the form of faeces. The intestinal walls also absorb water
from the undigested food material passed into it which is ultimately egested through
the anus.
Dental Caries: -It is localized softening and destruction of enamel and dentine of
teeth forming cavities that reach the pulp. Dental caries is also called dental decay. It
is caused by bacterium streptococcus mutans. It feeds on food particles especially
sugars and produces acids. The acids are growing on food particles streptococcus
mutans multiplies rapidly and forms a dental plaque. The plaque covers the teeth. Its
bacteria secrete the acids that cause dental caries. Saliva which normally neutralises
the acid and kills the bacteria is unable to protect the teeth because of the plaque.
Brushing of teeth after meals removes the plaque. Bacteria are unable to multiply and
produce acids. However, if plaque formation is allowed to persist, softening of enamel
and dentine will allow the micro-organisms to reach the pulp of the teeth. This results
in inflammation and infection resulting in acute pain, total decay and falling of teeth.
OPENING AND CLOSING OF STOMATA
It occurs due to turgor pressure of guard cells. When turgor pressure increases in
guard cells, these walls are stretched apart and stomata open. When the turgor
pressure falls in guard cells, these walls come close and stomata closes.
Mechanism of opening and closing of stomata:Starch-Sugar Hypothesis :-
Phosphorylase
This theory has been put forward by Sayre and Scarth to explain the mechanism of
opening and closing of stomata. According to this theory during day time CO2 in
guard cells is released by the process of respiration and is utilized in photosynthesis.
This increases ph of guard cells. Under high ph starch in guard cells changes into
sugars in the presence of enzyme phosphorylase.
Starch
Glucose-6-PO4
Sugar increases osmotic pressure of guard cells. They absorb water from subsidary
cells. This increases turgor pressure of guard cells. Their walls are stretched
outwards and stomata opens.
During night, CO2 in guard cells released in respiration is not utilized in
photosynthesis. It accumulates in guard cells dissolves in water to form carbonic acid.
This reduces ph of guard cells. Under low ph , conversion of starch into sugar does not
occur. This reduces osmotic pressure of guard cells. They loose water by ex-osmosis.
Their turgor pressure decreases and stomata close.
PHOTOSYNTHESIS (Photos-Light, Synthesis-putting together)
Photosynthesis may be defined as an anabolic process in which green plants
manufacture complex organic food substances (carbohydrate) from simple inorganic
compounds like carbon dioxide and water in presence of sSunlight
unlight with the aid of chlorophyll and evolve out oxygen as a byproduct of the
process. Thus photosynthesis is a process in which radiant energy is converted into
chemical energy
6CO2+ 6H2O
Chlorophyll
C6H12O6 + 6O2
Sunlight
In other words photosynthesis is a series of oxidation- reduction reaction in which
CO2 is reduced and H2O is oxidized to produce carbohydrates and oxygen.
Mechanism of Photosynthesis:Sunlight
Photosynthesis is formation of organic food from carbon dioxide and water with the
help of sunlight inside chlorophyll containing cells. Oxygen is produced as byproducts.
6CO2+ 12H2O
Sunlight Glucose
chlorophyll
C6H12O6 + 6H2O + 6O21
Oxygen comes from water. Hydrogen of water is used to reduce carbon dioxide to
form carbohydrate.
light Energy
2H2O
2H2 + O2
Chlorophyll
CO2 + 2H2
Energy
[CH2O] + H2O
Carbohydrate
Actually, photosynthesis occurs in two steps, photochemical and biochemical.
1. Photochemical phase (Light or Hill Reaction):- The reactions of this phase are
driven by light energy. They are of two steps- photolysis of water and formation of
assimilatory power.
a. Photolysis of water: - Light energy splits up water into two components. The step
requires an oxygen evolving complex (formerly called z-complex) having manganese
ions. Calcium and chlorine are also required.
2H2O
OEC
O2+ 4H+ + 4e-
Mn, Ca, Cl
b. Formation of Assimilatory power: - Electrons released by photolysis of water are
picked up by chlorophyll a molecules. On absorption of light energy, each
chlorophyll a molecule throws out an electron with gain of energy. This is primary
reaction of photosynthesis which converts light energy into chemical energy.
Electrons travel along an electron transport system, releasing energy in the process.
The energy is used in the formation of ATP (adenosine triphosphate) from ADP and
inorganic phosphate. Synthesis of ATP and ADP and inorganic
Phosphate (pi) with the help of light energy is called Phototophosphorylation.
ADP + Pi + energy ----------- ►ATP
The electrons ultimately activate NADP (nicotinamide dinucleotide phosphate)
and makes it combine with hydrogen to form NADPH2.
NADP+ + 2e- + 2H+ ------------ ► NADPH + H+ (NADPH2)
Both ATP and NADPH2 together form assimilatory power.
2. Biosynthetic phase (Dark or Blackman's Reaction):- It is actually light
independent reaction which can occur both in light as well as in dark. It requires the
energy and reducing power contained in assimilatory power of light reaction.
Common pathway of biosynthetic phase is Calvin cycle. Carbon dioxide combines
with ribulosebiphosphate in the presence of enzyme ribulosebiphosphate carboxylase
or rubisco. It produces two molecules of phosphoglyceric acid (PGA).
RuBP + CO2
rubisco
2PGA
In the presence of ATP, phosphoglyceric acid is reduced by NADPH2 to
form Glyceraldehydes phosphate (GAP).
PGA + ATP + NADPH2 ------------ ► GAP + NADP + ADP + Pi
A part of glyceraldehydes phosphate is changed into dihydroxyacetone
phosphate. The two condense and form glucose. Ribulosebiphosphate is regenerated
to combine with carbon dioxide again. Glucose undergoes condensation to form
starch.
Raw material of the photosynthesis:- The process of photosynthesis requires various
raw materials essential to synthesize energy complex compounds called
carbohydrates. These include:
(1) Chlorophyll:-The chlorophyll or green pigment of the plant are the most active
and important pigments of the photosynthesis. These are regarded as key pigments of
the photosynthesis because of their remarkable ability of absorbing light energy,
which is then converted into chemical energy during the process of photosynthesis.
All the green plants contain chloroplasts, which give the coloring material and are
accordingly called as photosynthetic- organelles of the plants.
(2) CO2:- All green plants utilize free atmospheric carbon dioxide during the process
of photosynthesis to synthesis energy rich complex organic molecules called
carbohydrates. These photosynthetic plants fix free atmospheric carbon dioxide during
daytime when light energy is available to them. But during the night, the process
stops.
(3) Water:- Water is another raw material for the process of photosynthesis. Plants
absorb the required amount of water by root hairs and pass it on to the leaves through
xylem, where it is utilized during the process of photosynthesis to synthesis energy
organic compounds.
(4) Light:- The sunlight is a natural and prime source of energy for photosynthesis. It
has been verified by experiments that the rate of photosynthesis remains highest in red
light. Ordinary light consists of seven colors (VIBGYOR) and during photosynthesis;
chlorophyll does not use all the seven colors. It absorbs mostly red and violet portions
only. However, the green color reflects back which gives chlorophyll a greenish
appearance.
Factors affecting photosynthesis:1.Temeperature:- It is one of the most important factor which affects rate of
photosynthesis. Photosynthesis starts at a temperature as low as 0 degree centigrade. It
is called minimum temperature. As the temperature increases, rate of photosynthesis
also increases. For every increase in temperature by 10 C, rate of photosynthesis
doubles.
Photosynthesis is maximum at 30C - 35C. It is called optimum temperature. An
increase in temperature above 35C again decreases rate of photosynthesis till at 50 C 55 C, photosynthesis stops completely, it is called maximum temperature. The three
ranges of temperature are called cardinal points.
Temperature has no effect on photochemical reaction of photosynthesis; however, it
has a definite effect on enzyme controlled dark reaction of photosynthesis.
Temperature also controls opening and closing of stomata and thus affects diffusion of
CO2 in the leaves of plants.
2. Light: - Light is another factor which affects rate of photosynthesis. Infact
photochemical reaction of photosynthesis is dependent on light. Both quality as well
as intensity of light affects rate of photosynthesis. Photosynthesis is maximum in red
and blue light, as they have longer wavelength.
An increase in the intensity of light increases rate of photosynthesis. An increase in
light intensity increases temperature and also causes the opening and closing of
stomata. However, a very high increase in intensity of light decreases rate of
photosynthesis as it causes closure of stomata. It is for this reason, that rate of
photosynthesis is maximum during dusk and dawn hours when compensation point is
reached i.e., amount of CO2 released in respiration is equal to the amount of CO2
utilized in photosynthesis.
3.CO2 :-It is one of the raw materials in the process of photosynthesis. Infact dark
reaction utilizes CO2 for synthesis of sugars. Plants obtain CO2 from atmosphere
through their stomata. Hydrophytes obtain CO2 from water in the form of carbonates
and bicarbonates.
Concentration of CO2 in atmosphere is only 0.03% while as plants utilize more
amount of CO2 and increases rate of photosynthesis. If CO2 yyconcentration in
atmosphere is raised to 1%, rate of photosynthesis will increase 20 times. This on
nature CO2 always acts as a limiting factor.
4. Availability of water:- It is the raw material in the process of photosynthesis.
Infect water provides protons and electrons for photochemical reactions. It increases
turgor pressure of guard cells and causes stomata to open. It also hydrates protoplasm
as chemical reaction of photosynthesis can occur only in hydrated state.
Plants absorb water from soil by their roots; only 1% of this absorbed water is
required in the process of photosynthesis. Thus under natural conditions, Water is not
a limiting factor in the process of photosynthesis. However, it becomes a limiting
factor during mid summer day when rate of transpiration is excessive.
5.Content of chlorophyll :- According to some authors, an increase in the content of
chlorophyll increases rate of photosynthesis. However, some authors maintain that
there is a relation between the content of chlorophyll and the rate of photosynthesis.
Shade plants have more content of chlorophyll than sun plants, but still sun plants
have higher rate of photosynthesis than the shade plants.
Activities to demonstrate:-
1) Importance of Chlorophyll:- Take a variegated leaf of a garden plant that has
been exposed to sunlight for few hours. Test it for starch with iodine test. Only green
parts of the plant leaf will turn blue, showing that chlorophyll is necessary for
photosynthesis.
2) Importance of Light:- Take a destarched potted plant, which has been kept in dark
for 3 to 4 days. Cover one of its leaves completely with a carbon paper so that no
light falls on it. Keep the plant in light for 4 to 6 hours. Test the covered leaf and
uncovered leaf for starch with iodine test. The covered leaf will show negligible
amount of starch, while the uncovered leaf will give positive test for starch. The
process clearly shows that light are necessary for photosynthesis.
3) Necessity of Carbon dioxide:- Take two de starched potted plants and cover them
with transparent polythene bags, so that no fresh air enters into them. Keep NaoH
(Soda lime) that would absorb CO2 in one pot and NaHCO (Sodium Bi-Carbonate)
solution that would produce more CO2 in the other pot. Keep both the pots in the
sunlight for 4 to 6 hours and test one leaf from each for starch. The leaf from the first
Pot will show no starch due to the absence of CO2, while the leaf from the second pot
will give positive test for the starch, thereby showing that CO2 is necessary for
photosynthesis.
4) Evolution of oxygen:- Take a beaker filled with water . Add a pinch of baking
soda (NaHC03) to it and put a Hydrilla plant (Aquatic plant) in it. Cover the plant with
a funnel. Invert a test tube containing water over the stem of the funnel. Keep this
apparatus in the bright sunlight. After some time bubbles start emerging out from the
plant, which gets collected in the upper part of the test tube. Remove the test tube and
test the gas with a lighted splinter, it keeps on glowing showing that the gas is a
supporter of combustion. Thus, the experiment clearly shows that O2 is evolved during
photosynthesis.
RESPIRATION
Respiration is a biochemical process of stepwise oxidative breakdown of organic
compounds inside living cells releasing small packets of energy at various steps.
Respiration is an essential physiological activity of all living organisms by which they
obtains energy for carrying out various vital metabolic activities of the body.
However, it is a chemical activity taking place within the protoplasm of a cell, which
results in the liberation of energy. Energy liberated during oxidative breakdown of
respiratory substrate is partly stored in ATP. The rest is dissipated as heat. The
process of respiration involves the following steps.
C6H12O6
Enzymes
6CO2 + 6H2O + 38ATP
1) External Respiration or Breathing:- It refers to those mechanism by which air is
brought into the body from the atmosphere and expulsion of CO2 from the body into
the atmosphere. The exchange of the gases takes place at the respiratory surface such
as gills, tracheae or lungs.
2) Transport of Respiratory gases:- This phase involves transport of oxygen from
respiratory surface to the body tissue and CO2 from the tissues to the respiratory
surface. In higher animals it takes place mainly through blood.
3) Internal or Tissue Respiration:- This phase of respiration involves consumption
of oxygen by the body cells and production of CO2 as a result of oxidative processes
resulting in the liberation of energy necessary for the biological work of the body.
The distinct phases of respiration are represented in the diagram.
AEROBIC AND ANAEROBIC RESPIRATION
ENZYMES
Sachs (1890) discovered that respiration can occur with or without oxygen.
Therefore, there are two types of respiration, aerobic and anaerobic.
1. Aerobic Respiration (Gk. Aer-air, bios-life)
It is a multi step complete oxidative breakdown of respiratory substrate into carbon
dioxide and water with the help of oxygen acting as a terminal oxidant. Aerobic
respiration is the usual mode of respiration in all higher organisms and most of the
lower organisms. The reason is that it yields maximum amount of energy.
C6H12O6
enzymes
6CO2 + 6H2O + 686 kcal or 2870 kj
The energy is stored in some 38 molecules of ATP.
Aerobic respiration occurs in two steps, glycolysis and Krebs cycle.
a.Glycolysis (Gk. Glykys- sugar, lysis-breakdown):- Glycolysis or EMP (Embden,
Meyerhof and Parnas) pathway is the first step of respiration which is common to both
aerobic and anaerobic modes of respiration. It occurs in cytoplasm. Respiratory
substrate is double phosphorylated before it undergoes lysis to produce 3-carbon
compounds glyceraldehydes phosphate. NADH2 and ATP are produced when
glyceraldehyde is changed to pyruvate. The net reaction of glycolysis is:
C6H12O6 + 2ADP + 2Pi + 2NAD+
Enzymes
2C3H4O3 + 2ATP + 2NADH2
In cytoplasm
b. Krebs cycle (Krebs, 1940):- It is also known as citric acid cycle or tricarboxlic
acid cycle (TCA cycle). Pyruvic acid or pyrovate enters mitochondria. It undergoes
oxidative decarboxylation to produce acetyl CoA, carbon dioxide and NADH2. Acetyl
CoA enters Krebs cycle. Here two d ecarboxylation, four dehydrogenations
and
oneenzymesphosphorylation or ATP synthesis occur.
Pyruvate + NAD+ + CoA
oxidative
Acetyl CoA + NADH2 + CO2
decarboxylation
Acetyl CoA + 3NAD++ FAD CoA + 3NADH2 + FADH2 + 2CO2
Krebs cycle
NADH2 and FADH2 liberate electrons and hydrogen ions. They are use in
building up ATP molecules and activating oxygen molecules to combine with
hydrogen for forming water. Water formed in respiration is called metabolic water. As
oxygen is used at the end of Krebs cycle for combining with hydrogen, the process is
called terminal oxidation. The overall equation of aerobic respiration using glucose as
substrate is
Glucose in cytoplasm
Pyruvate
in mitochondria
CO2 +H2O + Energy (38 ATP)
no O required O required
2
2
2. Anaerobic Respiration (Gk. An-without, Aer-Air, BOIS- Life):- It is a multi
step breakdown of respiratory substrate in which atleast one end product is organic
and which does not employ oxygen as an oxidant. Anaerobic respiration occurs in
many lower organisms, e.g. certain bacteria, yeast. In human body it occurs regularly
in red blood cells and during heavy exercise in muscles (striated muscles). Anaerobic
respiration occurs entirely in the cytoplasm. It has two steps. The first step is
glycolysis. Here, respiratory substrate glucose breakdown into two molecules of
pyruvate, ATP and NADH2. Pyruvate is converted into ethyl alchol (C2H50H) in yeast
and certain bacteria. It is changed to lactic acid (CH3CHOH.COOH). In muscle cells
when oxygen utilization is faster than its availability as during vigorous exercise. It
creates an oxygen debt inin cytoplasm
the body. No such change occurs in blood corpuscels.
In Yeast
Glucose in cytoplasm C3H4O3 in cytoplasm
no O2 required
Pyruvate
no O2 requiredz
C2H5OH + CO2 + Energy (2 ATP)
Ethanol
In Muscle cells
Glucose in cytoplasm C3H4O3 in cytoplasm CH3CHOHCOOH + Energy (2 ATP)
no O2 required
Pyruvateno O2 required
Lactic acid
DIFFERENCE BETWEEN AEROBIC AND ANAEROBIC RESPIRATION
Aerobic Respiration
1. It is common method of respiration.
2. It is completed in three steps-glycolysis,
Krebs cycle and terminal oxidation.
3. It requires oxygen.
4. Respiratory substrate is completely
broken down.
5. They are inorganic.
6. End products show little toxicity.
7. It occurs partly in cytoplasm and partly in
mitochondria.
8. An electron transport chain is required.
9. It releases 686 kcal or 2870 kj energy per
mole of glucose.
10. The liberated energy is used in forming
36-38 ATP molecules per mole of glucose.
Anaerobic Respiration
1. It occurs permanently only in few
organisms. In other it may occur as a
temporary measure to Overcome shortage of
oxygen.
2. There are two steps-glycolysis and
anaerobic breakdown of pyruvic acid.
3. Oxygen is not required.
4. Respiratory substrate is incompletely
broken down.
5. Atleast one end product is organic.
Inorganic products may or may not be
present.
6. The organic end product is generally
toxic.
7. It is carried out entirely in cytoplasm.
Mitochondria is not required.
8. Electron transport chain is not required.
9. Energy liberated is 36-50 kcal or 150-210
kj per mole of glucose.
10. The liberated energy is used in synthesis
of 2ATP molecules per mole of glucose.
Mechanism of Respiration:- There are different mechanisms for process of
respiration in different form of organism. The mechanisms of respiration in some of
organisms may be described as under:1. Respiration in Simple organisms:- In the simplest forms of life like Amoeba,
Euglena, Paramecium, Algae and Spirogyra, the respiratory gases may diffused in
and out of the body through general body surface.
2. Respiration in Insects: - In insects, respiration of gases takes place through a
system of internally air filled tubules called trachea. These opens into the exterior
environment by paired apertures called as spiracles these tissues carry air directly into
the tissues of the body and bring out carbon dioxide out from them.
3). Respiration in Aquatic animals: - In majority of higher aquatic animals like
prawns and fish, the process of respiration or gaseous exchange takes place by a
special respiratory organ called as gills. The gills are made up of a large number of
gill plates, which increase surface area of the gills. Each gill is provided with a large
number of membranous gill lamellae the bold of gills absorb dissolved oxygen from
water when it mover over them and carbon dioxide from blood goes out into the water
i.e., from the higher concentration of oxygen and carbon dioxide towards their lower
concentration.
4). Respiration in plants: - In plants the gaseous exchange or respiration takes place
through the stomata of the leaves, lenticels of woody stems and surface of the roots.
The diffused air passes through the stomatal opening into the mesophyll cells of the
leaves. Similarly air is diffused inn through the small microscopic openings in stem
and roots of a plant and carbon dioxide is diffused out into the outer atmosphere or
soil.
MECHANISM OF BREATHING IN MAN
Human beings like other land animals breathe through
their noses with the help of pair of lungs located in an
airtight thoracic cavity. The lungs are spongy, air filled
sac's, which do not have any muscle tissue and thus
cannot expand or contract at their own. The process of
breathing is accomplished through changes in volume
and air pressure of the thoracic cavity. The lungs respond
passively to the pressure changes within a chest cavity
due to contraction and relaxation of muscles of ribs and
movements of diaphragm during inspiration and expiration. In normal breathing, air
enters into the nasal chamber through nose, where it is cleaned and warmed by the
ciliated epithelium. The warmed and cleaned air then passes into the windpipe or
tracheae through larynx and epiglottis. The tracheae at its lower portion bifurcate into
two bronchi, each entering into a lung lobe, where a bronchus divides extensively by
giving out various small branches called bronchioles. Finally the air is deposited in
microscopic air sacs called alveoli. These are lined by a layer of epithelial cells and
surrounded by a network of blood capillaries. The air in the lungs diffuses through the
walls of these blood capillaries into the main blood stream and carbon dioxide in turn
diffuses out into the lungs, where from it is expelled out into the external environment.
This gaseous exchange is completed within a few seconds, while the blood is passed
through the alveoli.
Diffusion: -The process of movement of molecules from one region to another in
accordance to concentration gradient i.e., from the region of higher concentration to
the region of lower concentration is called as diffusion. It takes place in all kinds of
matter i e solid, liquid and gas, but the process is faster in gases than solids or liquids.
The process of diffusion is a main method of transport of material in unicellular
organisms like Amoeba, Euglena, Paramecium, Algae and Chlamydomonas etc. and
some of the multicellular organisms like Sponges, Hydra etc.
Osmosis: - When two liquids of different densities are separated from each other by a
selectively permeable membrane, the water of the liquids flows towards the liquid of
higher density. This process in which water moves from a solution of lower density to
the solution of lower density to the solution of higher density or from a dilute to a
concentrated solution through a selectively permeable membrane is called osmosis.
TRANSPORTATION
Transportation is the movement of materials from one part to another, usually from
the region of their availability to the region of their use, storage or elimination.
Transportation occurs in all organisms, from microscopic ones to large sized trees and
animals.
Transport in plants: - Plants absorb sufficient quantity of water from soil by means
of root hairs through the process of osmosis but they also take in minerals by the
process of diffusion. Some part of this water is used up by the plant during various
processes and the rest evaporates from the stem and roots. The evaporation of this
surplus water from the aerial parts of a plant is known as transpiration. In general,
transpiration may be cuticular, lenticular or stomatal. The cuticular transpiration takes
place through the cuticle found on the surface of the stem and leaves. The lenticular
transpiration takes place through the lenticels found on the stem. The stomatal
transpiration takes place through the stomata situated on the leaves. It is through the
process of transpiration, that the water along with dissolved mineral salts is taken up
and transported up through the xylem. Evaporation of water from the leaves through
stomata causes a drop in the turgor pressure, which makes the xylem cells to act as a
single continues column and cause uptake of water from the soil.
Xylem (Wood) :- It is a complex tissue which transports sap (water and minerals).
Xylem has four types of cell-xylem fibres, xylem parenchyma, tracheids and vessels.
Vessels and trancheids are called tracheary elements because they take part in
transport of sap. Vessels are long multicellular tubes which are formed by end to end
union of several cells. Trancheids are elongated cells with pointed ends. Both the
tracheary elements have lignified walls with pits or other thin unlignified areas for
element to element movement of water. Xylem parenchyma takes part in lateral flow
of water. Trancheids are conducting elements of non-flowering plants. Vessels occur
mostly in angiosperms where they form the main conducting elements. The number of
tracheids is small in angiosperms.
Phloem :- It is complex tissues which takes part in transport of food. Phloem has four
types of cells-sieve tubes, companion cells, and phloem parenchyma and phloem
fibres. Only phloem fibres are dead cells. Others are living cells. Sieve tubes are
conducting channels of phloem. They are eleongated multicellular tubular channels
formed by end to end union of numerous sieve tube elements. The end walls or septa
between adjacent sieve tube elements are bulged out and have pores. They are called
sieve plates. Sieve tube elements do not have a nucleus. Their functioning is
controlled by adjacent nucleated companion cells.
Transport of Water and Minerals:- There is a continuos system of water conducting
channels (vessels and tracheids) from near the root tips to near the shoot tips. In the
roots the surface cells are in contact with soil particles and soil water. Ions and water
are absorbed from the soil. They are pulled and pushed up by various forces to reach
every cell requiring the same. The various steps involved in transport of water and
minerals are as follows:
1. Mineral Absorption:- It occurs in the growing parts of the root. Both the surface
or epiblema cells as well as root hairs take part in mineral absorption. Mineral
absorption is an active process which involves expenditure of energy. Being an active
process, mineral absorption occurs against concentration gradient. It creates a
difference in the concentration of ions between the roots and the soil, with more salts
being present inside the root then in the soil StfMren.
2. Absorption of water:- Root hair zone is the region of water absorption. The inside
of the root has higher osmotic concentration than the soil solution. Root hairs are in
contact with soil interspaces having capillary water. The root hairs pick up water
which is transferred inwardly due to still high osmotic concentration. It reaches the
cells surrounding the xylem channel. Salts accumulated in the basic part of xylem
channel cause osmotic entry of water into xylem and form column of water. It also
creates the positive pressure known as root pressure.
3. Development of negative pressure :- Leaves and other aerial parts of the plant
are continuously losing water in the vapour form in the process of transpiration.
Nearly 99% of the absorbed water is lost during transpiration. Major part of
transpiration is stomatal transpiration. Intercellular spaces of the leaves are in contact
with mesophyll cells as well as outside air through stomata. Outside air is seldom
saturated with water vapours while the intercellular spaces are nearly always saturated
with water due to evaporation from the wet walls of mesophyll cells. Therefore, water
vapours diffuse from intercellular spaces to outside. More water vapours come from
mesophyll cells to replace them. The process continues. Loss of water by mesophyll
cells increases their suction pressure. They withdraw water from the xylem channels.
As there are billions of mesophyll cells withdrawing water from xylem channels,
water column present in the xylem comes under tension or negative pressure.
Transport of food and other substances:- Food materials are translocated from
the region of their manufacture or storage to the region of their utilisation. The region
of supply of food is called source while the area of utilisation is called sink. The
direction of translocation can be downward, upward or both. The food manufactured
by leaves spaces into the storage region and other sinks in the downward direction as
well as towards growing points and developing fruits in the upward direction. The
translocating nutrients consist of soluble carbohydrates (mostly sucrose), amino acids,
organic acids, hormones and other organic solutes. Translocation occurs thorough
phloem. The channels of transport are sieve tubes (sieve cells in non flowering plants).
Sieve tubes are specialised for this purpose. They are devoid of nuclei and internal
membranes. The cytoplasm of one tube cells is continuous with that of adjacent sieve
tube cells through sieve plates. The force required for translocation is produced by
companion cells which live adjacent to sieve tube cells.
Translocation: - Translocation is the movement of dissolved substances from one
part of plant to another through the xylem in accordance to the concentration gradient
of various mineral salts present in the soil.
BLOOD
Blood is described as a connective tissue, which provides one of the means of
communication between the cells of different parts of a body and the external
environment. It is a fluid containing living cells, which are capable of doing metabolic
changes. It performs several vital functions of the body and hence is usually referred
as the seat of the soul. Blood constitutes about 7% of the body weight (about 5.6 Lts
in an 80 kg man). This proportion is les in women and considerable greater in children
and gradually decreasing until the adult level is reached.
Composition of the blood: - Blood of higher animals including man is a viscous
complex fluid tissue of red colour. It is made up of two main components viz.
1) Plasma: - Plasma represents an intercellular substance of straw colour. It
constitutes about 55% of the total volume of the blood. Chemically plasma is
composed of water (90 - 92%) plasma proteins, inorganic and organic in salts and a
liquid called serum. It also contains a coagulative substance called fibrinogen and an
anticoagulant called heparin or hirudin.
2. Formed elements: - The cellular elements constitute about 45% of the total volume
of the blood. These are short and their destruction and replacement goes on constantly
during the life of an animal. These include.
i) Red blood cells of Erythrocytes: - The erythrocytes are produced in the bone
marrow of bones. Each erythrocyte is a round biconvex disc shaped, thinnest at the
centre having no nuclei at maturity. The average diameter of each erythrocyte is 106m. The individual erythrocytes are pale yellow in colour, but in aggregation they
appear to be reddish in colour. Each erythrocyte is bounded by a thin membrane
composed of lecithin and cholesterol, enclosing an elastic substance called as stroma
and an iron containing pigment called haemoglobin. The presence of haemoglobinthe red pigment serves as the carrier of oxygen. The average life span of erythrocytes
is about 120 days in mammals, after this time they are disposed off either by liver or
by spleen.
ii) White blood cells or Leucocytes: - White blood cells do not contain any pigment
and are therefore colourless. They are larger, and fewer in number than the RBC's
(1:600). They are formed in red bone marrow and in the lymph glands. The average
life span of human leucocytes is about 12 to 13 days. The chief function of WBC's is
to provide immunity to the body by producing special proteins called antibodies,
which protects body against the infection of bacteria, viruses and debris etc. The
mature white blood cells are grouped in to two main categories, granulocytes or
granular leucocytes and agranulocytes or agranular leucocytes depending upon the
presence of visible granules in their cytoplasm. The granulocytes are in turn of three
types viz. Eosinophil, Basinophil and neutrophil and the agranulocytes are of two
types viz lymphocytes and monocytes.
iii) Blood platelets: - These are small, flat granular corpuscles or colourless cells,
which are smaller than RBC. These are probably formed in the red bone marrow and
contain a substance called thromboplast in which it acts as one of the enzymes
involved in the series of chemical changes resulting in the clotting of blood at the site
of an injury. The life span of these corpuscles is only 2-3 days. Hence these are
constantly replenished by red bone marrow cells called as Megakaryocytes.
Function of Blood:- Blood has many functions, the most important ones are
summarized as under:1. Transport of oxygen and carbon dioxide: - Blood transports oxygen from the
respiratory surface and thus helps in respiration.
2. Transport of food: - Blood carries soluble food from the intestine to the liver and
body cells, where it is required for cellular activities. The nutritive substances
transported by the blood are glucose, amino acids, fats, minerals vitamins and water
3. Transport of waste products: - Blood transports various waste products, produced
during the cellular activities of the body. These waste products are harmful and
require immediate elimination.
4. Chemical Co-ordination:- Blood distributes various hormones to different parts of
the body. These hormones are produced by the endocrine glands of the body and helps
in the co-ordination of the body.
5. Maintenance of pH: - The plasma proteins of the blood act as buffer system and
prevent any shift in the pH of the blood because of the amphoteric properties of these
proteins.
6. Water balance: - Blood maintains water balance in the blood by bringing about
constant exchange of water between the circulating blood and the tissue fluids.
7. Transport of heat: - Blood allows transfer of heat energy from the deeper tissue to
the surface of the body.
8. Defence against infection: - Blood protects the body from various infections
caused by the micro-organisms like bacteria, viruses etc. with the help of WBC's.
9. Temperature Regulation: - Blood maintains the body temperature by distributing
heat within the body.
10. Blood loss: - Blood prevents excessive loss of blood in an injury with the process
of blood coagulation.
THE TUBES-BLOOD VESSELS
Human blood flows inside tubes called blood vessels. Blood vessels are of three typesarteries, veins and capillaries.
1.Arteries:- They are blood vessels which carry blood coming from heart to various
organs of the body. Blood flows inside the arteries with jerks due to pumping activity
of the heart. As the blood is pumped into an artery, it expands. With the flow of blood
from it, the artery contracts partially. Arteries, generally, carry oxygenated blood.
Only pulmonary arteries transport deoxygenated blood from heart to lungs. The wall
of the arteries is thick and elastic.
2.Veins:- They are blood vessels which carry blood from various parts of the body
towards the heart. Blood flows smoothly and slowly inside veins. Internal valves
prevent back flow. Wall is less thickened and less elastic as compared to that of
arteries. Lumen is wide. Veins carry deoxygenated blood except pulmonary veins that
bring oxygenated blood from lungs to the heart. Veins are generally superficial.
3.Capillaries:- They are very narrow blood vessels (4-10 um) having a single layered
wall (endothelium), which form network inside body organs. Movement of blood is
very slow (1 mm/sec) so as to provide time for exchange of materials. The wall has
very fine pores for exchange of substances between blood and tissue fluid.
DOUBLE CIRCULATION
It is a passage of the same blood twice through the heart first on the right side, then on
the left side in order to complete one cycle. Double circulation has two components,
pulmonary circulation and systemic circulation.
(i) Pulmonary Circulation:- It is movement of blood from heat to the lungs and
back. Deoxygenated blood of the body enters the right auricle, passes into right
ventricle which pumps it into pulmonary arch. With the help of two separate
pulmonary arteries the blood passes into the lungs. Here the arteries break up into
arterioles and then capillaries for oxygenation. Capillaries join to form venules and
then veins. Oxygenated blood comes back to left atrium of heart through four
pulmonary veins, two from each lung.
(ii) Systemic Circulation:- It is the circulation of
blood between heart and different parts of the body
except lungs. Oxygenated blood received by left
atrium passes into left ventricle. The left ventricle
pumps it into aorta for supply to different body parts
including walls of the heart with the help of arteries.
Inside the organs the arteries break up into arterioles
and then capillaries. Capillaries provide oxygen and
nutrients to tissues. They receive carbon dioxide and
wastes from the tissues. Capillaries unite to form venules which join to produce veins.
Veins take the deoxygenated blood which comes back to the heart but now into the
right auricle.
STRUCTURE OF HEART
Position, shape and size of heart:- The
human heart is a cone shaped, muscular organ
situated under the breastbone and between the
lungs inside the thoracic cavity. It is of the
size of human fist, measures about 12cm in
length and 9cm in breadth and lies slightly
towards the left side of the chest cavity. It weighs about 250 gms in an adult female
and 300 gms in an adult male. Externally the heart is covered by pericardium (a
double-layered membranous sac). It protects the heart from mechanical injury. The
space between two pericardial membranes is called the pericardial space, which is
filled with the pericardial fluid. The pericardial fluid keeps the heart moist and reduces
friction between the heart wall and the surrounding tissues when the heart beats.
External Structure of the Heart:
The human heart is a four-chambered organ divided by septa into two halves—the
right half and the left half. Each half consists of two chambers—the upper, small-sized
auricle or atrium and the lower, large- sized ventricle. The ventricles form the larger,
lower part of heart.
Internal Structure of the Heart:
Internally, the heart has the following main components—two atria, two ventricles,
great blood vessels that carry blood to the heart and away from it, various apertures
and valves, and pacemaker tissues.
i. Auricles (atria) the Receiving Chambers:- The auricles or atria are thin-walled
chambers and are separated from each other by an inter-atricular septum. The septum
has an oval, thin area called the fossa ovalis. It marks the position of an opening, the
foramen ovale between the two atria in a foetus.
ii. Ventricles the Discharging Chambers:- The ventricles are thick-walled chambers
and are separated from each other by an obliquely placed inter-ventricular septum.
The wall of the left ventricle is thicker than that of the right ventricle, because the left
ventricle has to pump blood into vessels, which in turn carry the blood to the entire
body. The right ventricle pumps blood to the pulmonary arteries, which carry it to the
lungs.
The walls of the atria are thinner than that of ventricles because they only have to
pump blood into the ventricles.
Great Blood Vessels of Heart :
The blood vessels that enter or leave the heart are called great blood vessels.
Blood Vessels Entering the Heart:
The right atrium receives three blood vessels.
i. Superior (Anterior) Vena Cava or Precaval:- Brings deoxygenated blood from
the head and upper region of the body.
ii. Inferior (Posterior) Vena Cava or Postcaval: - Brings deoxygenated blood from
lower region of the body.
iii. Pulmonary Veins:- The left atrium receives two pairs of pulmonary veins, one
pair from each lung. These bring oxygenated blood from the lungs.
iv. Coronary sinus: Brings deoxygenated blood from the heart’s wall itself. It
consists of two coronary arteries arising from the base of the aorta. These supply
blood to the heart muscles. If coronary arteries get blocked, then it can cause heart
attack.
Blood Vessels Leaving the Heart:
i. Pulmonary Artery:- Arises from the right ventricle and carries deoxygenated blood
to the lungs for purification.
ii. Systemic Aorta:- Arises from the left ventricle and supplies oxygenated blood to
all body parts, except the lungs.
Apertures and Valves of Heart :
There are four valves in the heart which control the flow of blood within the heart and
its passage to various parts of the body through the great blood vessels.
i. The bicuspid valve:- It is also called the mitral valve or left atrio-ventricular valve
guards the opening of the left auricle into the left ventricle.
ii. The tricuspid valve:- It is also called the right atrio- ventricular valve guards the
right atrio-ventricular aperture.
iii. Semilunar or pulmonary valves:- They are present at the base of aortic and
pulmonary arches. These valves check the back flow of blood into the ventricles.
iv. Aortic semilunar valve:- It is present at the point of origin of aorta from the left
ventricle. In all, there are three semilunar valves in the vessels.
Pacemaker tissues of Heart :
Certain tissues in the heart, concerned with the initiation (generation of impulse) and
propagation (conduction) of the heart beat, are called “Pace- Maker” tissue, such as:
1. Sino Atrial Node (S.A. Node):- It is located at the junction of superior vena cava
with right atrium, it initiates and maintains the myocardial activity and its rhythmicity,
(called pace maker of heart).
2. Atrio-Ventricular Node (A.V. Node):- Located posteriorly on right side of the
interatrial septum near coronary sinus, in the destruction of S.A. Node. The function
of pace maker can be taken up by the A.V. Node.
3. Bundle of HIS:- Starts from A.V. Node along interventricular septum at the top.
Impulses travel along bundle of HIS on to ventricles.
4. Purkinje Fibers:Located at the terminal divisions of right and left branch of the bundle of HIS.
Purkinje fibers transmit the impulse at a fast velocity of 4 mts/sec.
Working of Heart (Circulation of blood through heart) :The pumping action of heart starts by the contraction of its muscular
walls. The alternate contraction and relaxation (dilation) continues regularly. The
waves of contraction is initiated by Sino-atrial node (S.A. Node) situated on inner
wall of right atrium. Right atrium is filled with deoxygenated blood, brought through
the right and left superior vena cava from right and left side of the head, neck, chest
and arm. Right and left inferior vena cava brings deoxygenated blood to right atrium
from left and right lower body parts such as abdomen and legs.(Simultaneously left
atrium is filled with oxygenated blood) when both atria are filled with the blood, wave
of contraction starts from S.A. Node and spreads over both the atrai resulting the
contraction of both atria, simultaneously, and blood is pushed into ventricles of their
sides through atrio-ventricular valves. Atrio-ventricular valve (tricuspid valve)
which is a three flap valve present between the right atrium and right ventricle, stops
back flow of blood from ventricles to atrium.
Atrio-ventricular valve which is a two flap (bicuspid valve) valve present between
left atrium and left ventricle and stops back flow of blood from ventricle to atrium.
Just after the filling of ventricles, relaxation starts in the walls of atria due to this,
deoxygenated blood rushes from veins to right atrium and oxygenated blood through
pulmonary vein into left atrium. Now atrio-ventricular node is excited, (Present
near inter auricular septum on the wall of right atrium) wave of contraction of atria,
wave of contraction spreads over to wall of ventricles through bundle of HIS and
Purkinje Fibers. Now both the ventricles contract simultaneously causing pressure of
blood contained in them, blood of right ventricle is forced in pulmonary artery
through semilunar valves, (this valve prevent the backflow of the blood) to the lungs
for gaseous exchange, (oxygenation and carbon dioxide removal and also provide
nutrition to the lungs) after this oxygenated blood comes to left atrium through
pulmonary vein. From left atrium oxygenated blood passes into left ventricle through
left atrio-ventricular valve. Now from left ventricle oxygenated blood is pumped into
aorta through aortic semilunar valve to supply it to all body parts. The heart
contracts about 72 times in the similar fashion in one minute and the total volume of
the blood pumped out by the heart in every minute varies approximately in between 5
- 6 litres. However, the course of blood circulation in man can be represented as
under: Oxygenated blood --------► Left atrium -------- ► Left ventricle ------- ► Aorta ------- ► Arterioles -------- ► Organs ---------- ► Venous capillaries -------►
Venules(Deoxygenated blood) -------- ► Veins ------- ► R. Auricle ------- ► R.
Ventricle------ ► Pulmonary artery ------ ► Lungs (For oxygenation) -------- ►
Pulmonary veins ------- ► Left atrium
CARDIAC OUTPUT
Cardiac output is the output of the heart per minute. In quantitative terms, it is the
amount of blood pumped by each ventricle per minute. The amount of blood pumped
by either ventricle during every beat (called stroke volume) is about 70 ml at rest, and
the heart beats about 70 times per minute at rest.
Hence, cardiac output is expressed as under:
Cardiac output = Stroke volume x Heart rate
= 70 ml x 70/min
= 4900 ml/min
= 5 L/min (approx.)
Cardiac output varies in day-to-day situations. One every day example of such a
situation is exercise, which is associated with an increase in blood flow to the working
muscles of the legs, hands and heart. During exercise, the cardiac output may increase
up to about five-fold even in untrained persons and up to ten-folds in the trained
athlete.
CARDIAC CYCLE
The cardiac cycle is the sequence of events that occur when the heart beats. There are
two phases of cardiac cycle:
(1). Diastole - Ventricles are relaxed.
(2). Systole - Ventricles are contract.\
The heart follows a regular recurring pattern. One contraction (systole) following by
relaxation (diastole) of the heart is known as cardiac cycle.
Suppose the heart rate is 75 per minute,
If 75 beat take 1 minute, then, 1 beat will take 1/75 min = 1 x 60 sec/75 = 0.8 sec
Thus, the duration of the cardiac cycle is 0.8 sec. Out of this, ventricular systole
(generally called "the systole") lasts 0.3 sec and ventricular diastole (generally called
"the diastole") lasts 0.5 sec. When the heart rate increases, the duration of cardiac
cycle decreases. In such a situation, the reduction I'm duration of diastole is greater
than that of systole.
ARTERIAL BLOOD PRESSURE
It is the pressure exerted by the blood on the wall of the blood vessels in which it is
present. It is of two types: a) Systolic Blood Pressure (SBP): - It is the pressure which the blood exerts on the
wall of the blood vessels at the end of systolic contraction of ventricles. In a normal
resting adult, it is about 120 mm Hg.
b) Diastolic Blood Pressure (DBP): - It is pressure, which the blood exerts on the
wall of the arteries when the ventricles are maximally relaxed. In normal resting adult,
it is about 80 mm Hg. blood
Note: - (Blood Pressure in a normal person is equal 120/80mm Hg)
LYMPH
The tissue fluid that bathes the cells is collected in tubes and is then called as lymph. It
is filtered the blood plasma through the capillaries. Although partly reabsorbed into
the capillaries, most of it flows into a system of fine channels, which repeatedly join
together to form a large duct and ultimate the fluid is returned into the blood stream.
This additional system of vessels is called as lymphatic system. It runs parallel to the
veins and forms another medium of circulation in the human body. The lymph is light
yellow in colour and similar in composition to the blood plasma. It is not only found
in the lymphatic vessels and bathing the cells of the body, but also in the various
cavities of the body, such as the Coelomic cavity, Pleural cavity, Pericardial cavity
etc. where it serves as a lubricant.
Functions of lymph :- The chief functions of lymph in the human body are
mentioned as under:
1. It serves as a lubricant for the cells and tissues of the body.
2. It serves to return the interstitial fluid into blood
3. It gives the blood macromolecules of plasma proteins.
4. It carries absorbed fats and lipids from small intestines to the blood.
EXCRETION
It is a process of removal of the various toxic waste products from the body,
produced in the different metabolic processes, undergoing inside the body of an
organism, it eliminates solid, liquid and gaseous waste products produced in the
metabolism and thus maintains the relative constancy of the body’s internal
environment without which life is impossible.
Organs of excretion: - The chief organs of excretion include: 1. Skin: - It excretes out various dissolved salts along with surplus water from the
body. The process takes place through the minute microscopic pores of the skin
mainly in the form of sweat and is referred to as perspiration.
2.Lungs: - These expel out the gaseous wastes like carbon dioxide produced during
the cellular respiration in the body through the process referred to as expiration.
3. Kidneys: - These excrete out nitrogenous wastes produced in the body like
ammonia, urea and uric acid during various metabolic processes. These are excreted
out mainly in the form of urine.
4. Large intestine: - It excretes out the solid wastes like undigested components of
the food material produced during the process of digestion. These excreted or
defecated out through the anus.
5. Excretory system of man: - The excretory system of man consists mainly of two
kidneys, two ureters, a urinary bladder and a urethra as shown under in the diagram.
KIDNEY
Structure of human kidneys: - The human kidneys are reddish brown-paired
structure, which lie along the posterior side of the abdominal wall on either side of the
vertebral column. Each kidney is bean shaped about 10cm long, 6cm wide and 4cm
broad. Each kidney is enclosed in a thin, tough, fibrous, whitish capsule. The outer
surface of each kidney is convex while the inner one is concave. The inner side of
kidney is composed of two main regions, a dark outer region called cortex and a
lighter inner zone called medulla. The cortex contains uriniferous tubules or nephron,
which manufactures the urine; the medulla contains conical projections called renal
pyramids containing tubules, which carry urine from nephron to the pelvis of the
kidney. From this region, the urine is taken to the urinary bladder through a long tuber
called ureters as shown in the diagram.
Structure of Nephron: - The nephron forms the
functional unit of the kidneys. Its one end is
modified into a cup shaped cavity called
Bowman's capsule, which is linked by a small
single layer of squamous epithelial cells. The rest of
the nephron is differentiated into a coiled proximal
convoluted tubule, a U-shaped loop of henle and a
distal convoluted tubule. The distal tubule opens into a branching system of
collecting tubules, which finally opens into a funnel shaped renal pelvis. Each
nephron maintains a close contact with the blood vessels. It filters and removes the
wastes from the blood, which is partially stored in renal pelvis. The urine so formed
is then passed out through a long narrow tube called ureters and stored in a sac called
urinary bladder, where from it is excreted out of the body through urethra.
Process of Mechanism of excretion: - The entire process of excretion involves three
main processes viz.
1.Glomerular filtration: - When the blood enters into the glomerulus through the
afferent arteriole, a part of the water and some dissolved constituent of the blood of
low molecular weight like nitrogenous waste, glucose and mineral salts filter out
through the capillary walls into the surrounding Bowman's capsule by a process
referred to as Glomerular filtration. The filtered fluid or glomerular filtrate
resembles the blood plasma in its chemical composition except for the absence of
large molecules. In man, about 180 liters of fluid is filtered from the blood plasma
through glomerular capillary walls every 24 hours.
2.Tabular Reabsorption: - The glomerular filtrate flows on through the convoluted
tubes, collecting tubule and then into the pelvis of the kidney and down the ureters
into the urinary bladder. As the filtrate flows the proximal convoluted tubule soume
water and physiologically important salts like glucose, amino acids, sodium chloride
and sodium bicarbonate are reabsorbed into the blood through the capillaries around
this portion leaving only the wastes to be excreted out.
3.Active secretion: - As the glomerular filtrate flows through the distal convoluted
tubule, the unwanted substances, which could not be filtered out in the glomerular are
actively secreted out by the tubular walls into the filtrate from the blood. As a result of
this entire process, homeostasis of the blood is maintained and all the waste products
remained in the tubular fluid constitute the urine, which is ready for excretion from
the body.
Haemodialysis: - In patients suffering from renal failure or nephritis, artificial
measures are adopted for removing the accumulated waste products like urea from the
blood. This process is called Haemodialysis and the apparatus used in this process as
artificial kidney. In Haemodialysis, the blood of the patient is taken out from the main
artery and cooled to 30C, then an anticoasulant (heparin) is mixed with it and Dumped
into the apparatus. Inside the apparatus blood flows through tubes bounded by
cellophane membrane, which is permeable to only small molecules like urea, uric
acid, creatanin and mineral ions. The dialyzing fluid used in the apparatus a salt
solution isotonic to blood plasma, so that the blood flowing through the channels or
tubules containing wastes like urea, uric acid, creataninetc diffuses out in the dialyzing
fluid across the cellophane membrane. This process is called dialysis. Finally the
blood coming out of artificial kidney is warmed to body temperature and mixed with
ant- heparin to restore its normal coagulability and then pumped into the body of the
patient through a vein.
MECHANISM OF EXCRETION IN PLANTS
Plants do not have any mechanism to collect, transport and throw out their waste
products. They have adopted varied strategies to protect their living cells from waste
products,
(i) Old leaves :- Waste products are stored in older leaves which soon fall off.
(ii) Old Xylem:- Resins, gums, tannins and other waste products are deposited in the
old xylem which soon becomes nonfunctional, e.g., heart wood.
(iii) Bark:- Bark consists of dead cells which are peeled off periodically. Tannins and
other wastes are deposited in the bark. Incidentally, tannins are raw material for dyes
and inks.
(iv) Central Vacuole:- Most plant waste products are stored in central vacuole of
their cells. They are unable to influence the working of cytoplasm due to presence of
a selectively permeable membrane called tonoplast.
(v) Root Excretion:- Some waste substances are actually excreted by the plant in the
region of their roots.
(vi) Detoxification:- The toxic oxalic acid is detoxified by formation of calcium
oxalate which getscrytallised into needles (raphides), prisms (prismatic crystals), stars
(sphaeraphides) and crystal sand. Excess of calcium is also precipitated as calcium
carbonate crystals, e.g., Cystolith.
(vii) Salt Glands:- They excrete excess salts obtained from the habitat. Hydathodes
also have an excretory function.
Chemical Reactions & Equations:
Introduction: - We are familiar with many chemical changes accessing
around us in our daily life. Some of these are:i)
ii)
iii)
iv)
v)
vi)
A freshly apple exposed to atmosphere becomes brown after some
time.
Iron articles such as pan, or nail hammer etc, gets rusted when left
exposed to humid atmosphere for a long period.
Burning of candle gives CO2.
Food gets digested in our body.
Milk left at room temperature during summer becomes sour and is
converted into a curdy thick mass.
Grapes get fermented.
In all the above cases, the nature and identity of the initial
substance changes. We may also say that during such changes substances are
undergoing permanent changes and new substances are formed. When different
substances react and new substances having different nature and identity are
formed, we say a chemical reaction has occurred. In the permanent unit we shall
learn as to what is actually meant by a chemical reaction. How do we know that
chemical reaction has taken place?
Chemical reaction:- the evaporation of water, desolation of salt in water or
melting of ice are the processes in which the new chemical substances are
formed and hence are termed as physical changes. However the processes in
which the original substances lose their nature and identity and form new
chemical substances with different propositions are called chemical changes.
The process involving a chemical changes is called a chemical reaction. For
example (a) Respiration (b) Burning of candle wax, burning of fuels likes
petrol, diesel, kerosene oil etc.
Reactions & Products:- The substances which take part in a chemical
reaction are called reactants, whereas the substances which are formed as a
result of chemical reactions are called products, e,g. consider the following
chemical reaction:AgNO3 +
Silver Nitrate
NaCl
AgCl
sod. Chloride
+
Silver Chloride
Reactions
Na NO3
Sod. Nitrate
Products
Chemical reactions involve breaking of old chemical bands which exist
between the atoms of reacting substances another making of new chemical
bands between the rearranged atoms of new substances. During a chemical
reaction, atoms of one element do not change into those of another element.
Only
rearrangement of atoms takes place in a chemical reaction.
Again in a chemical reaction, reactants are transformed into products. The
products thus formed have properties which are entirely different from those of
the reactants.
Examples of some Chemical Reactions:1. Burning of magnesium Ribbon in air:- The magnesium metal is
generally available in the laboratory in the form of a ribbon on a wire. It
has a shining surface. However, due to attack of moist air, it is coated
with a white layer of MgO.
Now take a strip of Magnesium ribbon about 2-3cm long clean
the magnesium ribbon by rubbing it with a sand paper. Hold it with a pair
of tongs in the flame using a sprit lamp. We observe that magnesium burn
with a dazzling light and a white powder is obtained in the china dish.
The white powder is found to be that of magnesium oxide. Thus,
magnesium has combined with oxygen of the air to form a new chemical
reaction-MgO. Hence we say that a chemical reaction has taken place in
which magnesium and oxygen are the reactants while as magnesium
oxide is the product.
2Mg + O2
2MgO
2. Reaction between lead nitrate and potassium iodide:- Prepare lead
nitrate solution in one test tube and potassium iodide in another test tube
and then mix the two solutions, a yellow precipitate of lead iodide is
obtained. Another substance formed is potassium nitrate but we cannot
observe it as it remains in the solution. Again to say that a chemical
reaction has taken place in which lead nitrate and potassium iodide are
the reactants while lead iodide and potassium nitrate are the products.
Lead nitrate + Potassium iodide
Lead iodide + Potassium
nitrate
Reactants
Products
3. Reaction Between Zinc & Dilute Sulphuric Acid:- Take a few pieces
of granulated zinc in a small conical flask or a test tube and add dilute
H2SO4 into it. We will observe that the reactions between two reactants
will produce H2 gas. To test the gas evolved, the flask is fitted with a test
tube on bringing a lighted candle near the upper end of the tube, the gas is
found to burn with a popping sound. Thus, the gas evolved is hydrogen.
Another substance formed is zinc sulphate which we cannot see as it
remains in the solution. Again we can say that a chemical reaction has
taken place in which zinc & suphuric acid are the reactants while
hydrogen gas & zinc sulphate are the products.
Zn + H2SO4
ZNSO4 + H2
Dilute
Reactants
Products
Characteristics Of Chemical Reaction:- when a chemical reaction takes
place, some changes are observed=. The easily observable changes that take
place in chemical reaction are called the characteristics of the chemical reaction.
These changes help us to check that a chemical reaction has taken place. Some
important characteristics of chemical reactions are as follows:i)
ii)
iii)
iv)
v)
Some chemical reactions are accompanied by the evolution of a gas.
Some chemical reactions are accompanied by change in state.
Some chemical reactions are accompanied by change in temperature.
Some chemical reactions are accompanied by change in colour.
Some chemical reactions are accompanied by the formation of
precipitate.
Chemical Equations:- A chemical equation is a statement used on
chemistry to represent a chemical reaction. (Or)
The short hand method representing a chemical reaction in terms of symbols
and formulae of the different reactants and products is called a chemical
equation.
There are two ways of representing a chemical reaction:a) The reaction between silver nitrate and sodium chloride can be
represented in terms of word equation as under:-
Silver Nitrate + Sodium Chloride
Silver Chloride+ Sodium
Nitrate
Therefore, the names of the reactants are written on the left hand
side with a (+) sign between them, whereas the names of the products are
written on the right hand side with (+) sign between them. An arrow (→)
is put between the reactants and the products. The arrow sign indicates
that the substances written on the left side are combining to give the
substances written on the right hand side in the equation.
b) In terms of symbols and formulae, the reactions between silver nitrate and
sodium chloride may be represented as under:AgNO3 + NaCl
AgCl + Na NO3
Reactants
Products
This is called a chemical equation. Infect this is the most common
method of representing a chemical reaction.
Steps For Writing A Chemical Equation:For writing a chemical equation, the following point use kept in mind:i)
ii)
iii)
The symbols and formulae of reacting substances-i.e reactants are
written on the left hand side with plus (+) sign between them.
The symbols and formulae of various products are written on the right
hand sign with plus (+) sign between them.
An arrow (
) sign or sign of equality (=) is put between the
reactants and products.
For example, consider a reaction between zinc and dilute sulphuric
acid to produce zinc sulphate iron solution and hydrogen gas.
Zn + H2 SO4
ZnSO4 +H2
Balanced & Unbalanced Chemical Equation:Consider the equation of reaction of magnesium & oxygen to form
magnesium oxide:Mg + O2
MgO
Let us count the number of atoms of each element on the LHS and RHS
of the arrow. It is clear from the above reaction, that the numbers of atoms of
each element are not the same on both sides. Such on equation is called
unbalanced equation. It is also called as skeletal equation.
Again consider the following chemical reaction
Zn + H2 SO4
ZnSO4 +H2
It is clear from the above chemical reaction, that the numbers of atoms of
each element on the two sides of equation are equal. Therefore, such a chemical
equation is called a balanced chemical equation.
Balancing of a Chemical Equation:We know that mass can neither be created nor destroyed in a chemical
reaction. Therefore, the total mass of the elements present the product of a
chemical reaction has to be equal to the total mass of the elements present in the
reactants. In other words, the number of atoms of each element remains the
same, before and after a chemical reaction.
Thus, balancing of a chemical equation means making the number of atoms
of each element equal on both sides of the equation.
The importance of a balanced chemical equation lies in the fact that it
satisfies the law of conservation of mass i.e, in a chemical reaction; total mass
of all the products is equal to the total mass of all the reactants.
Steps involved in the balancing of a Chemical Equation:Step-I:- write down the equation in the word form by writing the names of the
reactants on the left side and those of products on the right side. This step is not
required if the equation is given in terms of symbols and formulae.
Step-II:- write down the symbols and formulae of the various reactants and
products. This gives the skeletal chemical equation.
Step-III:- Enclose the formulae of each reactant and product in a box. This is
done to remember that during balancing of the chemical equation, the formula
of and reactant or product cannot be changed.
Step-IV:- List the number of atoms of different elements on reactant side and
product side.
Step-V:- Now start the process of balancing by choosing the compound which
has maximum number of atoms, irrespective of the fact whether it is a reactant
or a product.
Step-VI:- After selecting the compound with the biggest formula as above, first
balance the element of this compound which has the highest number of atoms.
Then balance other elements one by one, to balance the atoms of an element,
put a small whole number co-efficient before the formula of the compound.
Step-VII:- Atoms of elementary gases such as hydrogen, oxygen nitrogen etc,
are balanced last of all. Sometimes, these are first changed to atomic state. After
balancing, these are changed back to molecular form.
Step-VIII:- Finally check the correctness of the balanced equation by counting
the number of atoms of each element on both sides of the equation. The above
method of balancing of chemical equation is called ‘Hit & Trial Method’ as we
keep on trying to balance the equation using smallest whole number coefficient.
Let us try to balance the following chemical reaction:When steam is passed over heated iron, iron oxide and hydrogen gas is
obtained.
Step-I:- write the equation in word form as:Iron + Steam
Iron Oxide + Hydrogen.
Step-II:- Write the skeletal chemical equation and enclose the formulae in
boxes:Fe
H2O
Fe3O4
H2
Step-III:- count the number of atoms of different elements on both sides of the
equation.
Element
No. of atoms on LHS
No. of atoms on RHS
Fe
1
3
H
2
2
O
1
4
Step-IV:- Start balancing with the compound which contains maximum number
of atoms. It may be a reactant or a product. Now in that compound select the
element which has maximum number of atoms.
According to this rule, select Fe3O4 & the element oxygen in it. There
are 4-0 atoms on the RHS and only 1-0 atoms on the LHS. To balance 0-atoms,
multiply H2O on LHS by 4, we get.
Fe +4 H2O
Fe3 O4 + H2
Step-V: - There are 3-Fe atoms on RHS & 1-Fe atom on LHS. To balance Fe
atoms, multiply Fe on LHS by 3, we get.
3 Fe + 4H2O
Fe3O4 + H2
Step-VI: - There are 8-4 atoms on LHS & 2H – atoms on RHS. To balance Hatoms, multiply H2 by 4, we get.
3 Fe + 4H2O
Fe3O4 + 4H2
Step-VII: - Now check the correctness of the balanced equation:Atom
LHS
RHS
Fe
3
3
H
8
8
O
4
4
Hence, the equation is balanced –i.e. the final balanced equation may be
written as:3 Fe + 4H2O
Fe3O4 + 4H2
Example 2:- Let us illustrate these rules by considering one more example.
When magnesium nitride is treated with water Ammonia gas &
magnesium hydroxide are produced.
Step – I: - Write equation in word form as:Magnesium nitride + water
Ammonia
Magnesium hydroxide +
Step – II: - Write the skeletal chemical equation and enclose the formulae in
boxes.
Mg3N2 + H2O
Mg (OH)2 + NH3
Step – III: - Count the number of atoms of different elements on both side of
the equation.
Element
RHS
No. of atoms LHS
atoms No. of atoms on
Mg
3
1
N
2
1
H
2
5
O
1
2
Step – IV: - Let us seleat Mg3 N2 and balance Mg atoms in it. To balance Mg
atoms on both sides, multiply Mg (OH) 2 + NH3
Mg3N2 + H2O
3 Mg (OH)2 + NH3
Step – V: - Now to balance N-atoms, multiply NH3 on RHS by 2, we get.
Mg3N2 + H2O
3 Mg (OH)2 +2 NH3
Step – VI: - to balance 0-atoms, multiply H2O on LHS by 6, we get.
Mg3N2 + 6H2O
3 Mg (OH)2 + NH3
Step – VII: - Check the correctness of the balanced equation.
LHS
RHS
Mg-atoms
3
3
N-atoms
2
2
H-atoms
12
12
O-atoms
6
6
Making of chemical equation more informative:- The chemical equation
which gives more information about the chemical reaction is known as more
informative or information giving equation. The chemical equations can be
made more informative by following 4 – ways:i. Physical state of the reactants and the products: - The physical state of any
substance is represented by using the symbol(S) for solid, (l) for liquids, (g) for
gaseous and (aq) for aqueous solution. These symbols are written in brackets
after the symbols or formulae of the reactants and products. This will become
more clear from the following examples:a) Zinc metal reacts with dilute sulphuric acid to form zinc sulphate
solution
and hydrogen gas
Zn + H2 SO4
ZnSO4 + H2
Zn(s) + H2 SO4(aq)
ZnSO4 (aq) + H2(g)
or
b) When calcium hydroxide solution reacts with carbon dioxide gas, a
while precipitate of calcium carbonate is formed along with water.
Ca (OH)2 (aq) + CO2 (g)
CaCO3(s) + H2O(l)
(Lime water)
(cal. Carbonate water)
ii) Concentration of the acid: - If acid is present as the one of the reactants, it
may be dilute or concentrated. The symbol (dil) is used for dilute acid. Whereas
symbol 'conc' is used for concentrated acid, e.g
Zn(s) + H2 SO4
Cu(s) + 4HNO3
ZnSO4(aq) + H2(g)
Cu(NO3)2 (aq) + 2NO2(g) + H2
O(l)
(Conc.)
(nitrogen dioxide)
iii) Heat changes accompanying the chemical reaction: There are two types of reactions on the basis of heat charges involvedexothermic and endothermic.
Exothermic Reaction:- These are those reactions in which heat is evolved
during the reaction, e.g when carbon (coke) burns in oxygen to form CO2, a lot
of heat is produced.
C (s) + O2(g)
CO2(g) + Heat
When natural gas mainly methane burns in oxygen of air, it forms CO2 &
H2O. A large amount of heat energy is also produced. (↓ which is).
CH4(g) + 2O2(g)
CO2(g) + 2H2 O(l) + Heat
During respiration, glucose combines with oxygen in the cells of our body
to form CO2 and H2 O along with production of energy.
C6 H12 O6(aq) + 6O2(g)
6H2 O(l) + Energy
The burning of a magnesium wire in air to form magnesium oxide is also an
exothermic reaction because heat & light energy are given out during this
reaction.
2Mg (s) + O2 (g)
2MgO(s) + Heat & light.
Endothermic Reactions: - These are those reactions in which heat is absorbed
during the reaction. An endothermic reaction is usually indicated by writing '+
Heat' or "+ Heat energy" or just "+ Energy" on the reactant side of the equation.
Example:a) The reaction between nitrogen and oxygen to form nitrogen monoxide is
an endothermic reaction because heat is absorbed in this case.
N2 (g) + O2 (g) + Heat
2NO (g)
b) When calcium carbonate is heated it decomposes to form calcium oxide
and carbon dioxide.
CaCO3 (s) + Heat
CaO(s) + CO2
The decomposition of calcium carbonate is an endothermic reaction,
because heat energy is absorbed in this reaction.
c) The electrolysis of water to form hydrogen and oxygen is also an
endothermic reaction. This is because electric energy is absorbed during
this reaction.
2H2O (l) + Heat
2H2 (g) + O2 (g)
d) Photosynthesis is an endothermic reaction. This is because sunlight
energy is absorbed by the green plants during the process of
photosynthesis.
6CO2 (g) + 6H2O (l)
C6H12O6
iv) Conditions under which the reaction takes places: - The conditions
of temperature, pressure and the presence of catalyst, if any may be
represented by writing these conditions above or blew the arrow
drown between the reactants & the products e.g.
500C 200 atm.
N2 (g) + 3H2 (g)
2NH3 (g)
fe
This shows that to get maximum yield of NH3, the most suitable
condition for the above reaction are - a temperature of 500 C, pressure
of 200 atmosphere and presence of iron as catalyst.
Similarly, if a reaction takes place on heating, the sign delta (∆) is
put on the arrow,e.g.
∆
2 KclO3 (s)
2 Kcl(s) + 3O2 (g)
Mn O2
Again, if a reaction takes place in the presence of light, the word "light"
is written on the arrow, e.g.
H2 (g) + Cl2 (g)
Sun light
2Hcl (g)
TYPES
OF
CHEMICAL
REACTION:Depending upon the nature of the reaction - i.e. the type of chemical
change taking place, the various chemical reactions have broadly classified into
the following type:1) Combination reactions (2) Decomposition reactions (3) Displacement
reactions (4) Double Displacement reacts 5) Precipitation reaction (6)
Neutralization reactions (7) Oxidation - Reduction reactions
1) Combination reactions:- The reactions in which two or more substances
combine to form a single substance are called the combination reactions.
Example 1:- Magnesium and oxygen combine when heated to from magnesium
oxide.
2Mg(s) + O2 (g)
2MgO(s)
In the above reaction, two elements, magnesium and oxygen are
combining to form a single compound magnesium oxide.
Therefore, it is a combination reaction.
Example 2:-Hydrogen burns in oxygen to form water is a combination reaction.
2H2 (g) + O2 (g)
2H2O
Example 3:- If we take a small amount of quick lime - i.e calcium oxide in a
beaker and add water to it slowly, they combine vigorously to form slaked lime
- i.e calcium hydroxide.
CaO(s) + H2O (l)
Ca(OH)2 (aq)
The combination reactions are the three types:a) Combination of two elements to form a compound.
b) Combination of an element and a compound to form a new compound.
c) Combination of two compounds to form a new compound.
a)Combination of two elements to form a compound:i)
Carbon or coal burns in air to produce CO2.
C(s) + O2(g)
CO2(g)
ii)
Hydrogen combines with chlorine in the presence of light to give
hydrochloride acid.
H2(g) + Cl2(g)
2HCl(g)
iii)
Sodium combines with chlorine gas to form sodium chloride
2Na(s) + Cl2(g)
2NaCl(s)
b) Combination of an element and a compound to form a new compound:i)
ii)
iii)
Nitric oxide combines with oxygen of the air to form nitrogen
2NO(g) + O2(g)
2NO2(g)
When sulphur dioxide combines with oxygen to form sulphur trioxide
is obtained.
2SO2(g) + O2(g)
2SO3(g)
When carbon monoxide reacts with oxygen, carbon dioxide is
formed.
2CO(g) + O2(g)
2CO2(g)
C) Combination of two compounds to form a new compound:i)
ii)
iii)
Ammonia and hydrogen chloride are colourless gases. They combine
to
form
a
solid
ammonium
chloride.
NH3(g) + HCl(g)
NH4Cl(s)
Calcium oxide reacts with sulphur trioxide to form calcium sulphate.
CaO + SO3
CaSO4
When ammonia reacts with sulphuric acid, ammonium sulphate is
formed.
2NH3(g) + H2SO4(aq)
(NH4)2 SO4 (aq)
Decomposition Reactions:- the reactions in which a compound breaks up into
two or more simpler substances are known as decomposition reactions. Thus,
these reactions are just opposite of combination reactions.
These reactions generally take place when energy is supplied in the
form of heat, light, electricity or catalyst. Thus, there are three types of
decompositions.
A) Decomposition reactions which take place by absorption of heat are
called thermal decomposition reactions.
Experiments:i)
Thermal decomposition of ferrous sulphate:Take a small amount of green coloured
crystals of ferrous sulphate in a dry test
– tube. Hold the test tube in a test tube
holder. Now heat the test tube over a
spirit lamp or burner. We observe that
the green coloured ferrous sulphate
crystals on heating first lose water and
the colour changes to form anhydrous
ferrous sulphate.
FeSO4 + 7H2O(s)
7H2O(g)
∆
FeSO4(s) +
This on further heating gives out characteristic smell o burning
sulphur leaving behind a reddish brown residue of ferric oxide. This is
due to following decomposition reaction:2FeSO4(s)
∆
Fe2O3(s) + SO2(g)↑ + SO3(g)↑
Ferric oxide reddish brown.
ii)
Decomposition of limestone:- When calcium carbonate is heated, it
decomposes to give calcium oxide and carbon dioxide
CaCO3(s)
∆
CaO(s) + CO2(g)
In this reaction on substance calcium carbonate is heated and
so it gets broken down into two simpler substances of calcium oxide
and carbon dioxide, so it is a decomposition reaction.
iii)
Decomposition of Lead Nitrate:- Take a small amount of powdered
lead nitrate in a test tube. Heat it over a flame. Brown fumes of
nitrogen dioxide are found to evolve and a yellow residue of lead
oxide is left behind in the test tube. This is due to the following
decomposition reaction.
2Pb(NO3)2 (s)
∆
2PbO(s) + 4NO2(g) + O2(g)
(B) The decomposition reactions which take place when electric current is
passed through the compound in the molten state or in aqueous solution. These
are called electrolytic decomposition reactions or simply electrolysis.
i) Experiment Electrolysis of Water: - Take a plastic mug. Drill two holes at
its base and fit rubber stoppers in these holes. Now put two carbon electrodes in
these rubber stoppers and seal the holes with some adhesive to prevent leakage
of water. Now connect the electrodes to a 6 volt battery fill the mug with water
such that the electrodes get immersed in water, add a few drops of dil. H 2SO4 to
water. Take two graduated that tubes fill with water and invert them over carbon
electrodes. Switch on the current and leave the apparatus undisturbed for some
time. It is observed that bubbles of gases starts rising from both the electrodes
and the levels of water in the two test – tubes start freeing. Allow it to run till
sufficient amounts of gases have been collected in the two test tubes. Now
switch off the battery we will observe that the test tube covering the cathode,
the amount of gas collected is double than that of the gas collected in the testtube covering the anode.
On testing the gases, the gas in the test tube covering the cathode is
found to be hydrogen – a combustion
gas and in the test-tube covering
the anode, it is found to be oxygen – a supporter of combustion.
The above experiment shows that on supplying electrical energy, water
decomposes into hydrogen and oxygen as shown below:
2H2O(l)
2H2(g) + O2(g)
ii) Electrolytic decomposition of molten sodium chloride:On passing electric current through molten sodium chloride, it
decomposes to give sodium metal and chlorine gas.
2NaCl
2Na + Cl2↑
This reaction is used to obtain sodium metal on a large scale
iii). Electrolytic decomposition of molten Alumina: - On passing electric
current through molten alumina, it decomposes to give aluminum metal and
oxygen gas.
2Al2O3
Electric
Current
4Al + 3O2↑
This reaction is used for the extraction of Aluminum metal.
C) The decomposition reactions which take place on absorption of light. These
are called photo-decomposition reactions or simply photolysis.
Example-I:- Place about 2g of silver chloride in a china dish under sunlight for
some time, we observe that silver chloride is white is colour. Its colour will turn
grey in sunlight.
This is because of decomposition of silver chloride to silver & chlorine
by light.
2AgCl(s)
Sunlight
2Ag(s) + Cl2(g)
II) Photolysis of Hydrogen Iodine: - Hydrogen iodide decomposes in the
presence of ultra violet light into hydrogen and iodine.
2HI
UV-Light
H2 + I2
III) Photolysis decomposition of Hydrogen peroxide: - In the presence of
light, hydrogen peroxide decomposes into water & oxygen.
H2O2
2H2O + O2
Displacement Reactions: - The chemical reactions in which one element takes
the place of another element in a compound are called displacement reactions.
The chemical reactions in which a more active element displaces a less
active element from its compound are called displacement.
In these reactions, an atom or group of atoms in a molecule is replaced
by another atom or group of atoms.
These reactions are very common in chemistry. These reactions occur
mostly in solution form and a more active metal displaces less active metal or a
more active non-metal may displace a less active non-metal form its compound.
Experiment-1:- Take about 10ml of CuSO4 solution in two test tubes A and B.
Also take iron nails and clean them by rubbing with a sand paper. Tie a thread
on one iron nail and suspend it in test-tube A. Wait for about 20-30 minutes.
Now take out the nail and observe the change in the colour of the solution of
test-tube A by comparing with the solution of test-tube B and the change in
colour of the nail that remained suspended in the test-tube A by comparing with
the 2nd nail. The following changes are observed:i)
ii)
Blue colour of copper sulphate solution fades and it changes into light
green colour.
The iron nail that remained suspended has a brownish coating on its
surface.
These changes show that the following reaction has taken
place:Fe(s) + CuSO4(aq)
(Blue solution)
FeSO4 + Cu(s)
(light green solution)
Thus the more active metal, iron displaces less active metal
copper from copper sulphate.
Experiment-2:- Take AgNO3 solution in a beaker. The solution of AgNO3 is
clear and colourless. Place a copper wire in the solution and keep the beaker
undisturbed for some time. It will be observed that the solution becomes blue
and a shiny coating of silver gets deposited on the wire. This suggests that when
copper wire is added to the solution, copper nitrate is formed and therefore. The
solution becomes blue, the reaction may be written as:Cu + 2AgNO3
Cu(NO3)2 + 2Ag
Example-1:- when zinc reacts with dilute H2SO4 or dilutes HCl, hydrogen gas
is evolved. The reaction is as under:Zn(s) + H2SO4
ZnSO4(aq) + H2(g)↑
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2g)↑
This shows that Zn is more reactive than hydrogen
hydrogen from sulphuric acid or hydrochloric acid.
it displaces
Example-2:- when a strip of zinc metal is placed in copper sulphate solution,
then zinc sulphate and copper
obtained.
CuSO4(aq) + Zn(s)
ZnSO4(aq) + Cu(s)
Thus, zinc being more reactive than copper *** copper from CuSO4
solution.
Example-3:- when a piece of Mg metal is placed in CuSO4 solution, the MgSO4
solution and copper metal are formed.
CuSO4(aq) + Mg(s)
MgSO4(aq) + Cu(s)
Double Displacement Reactions: - The reactions in which two compounds
react to form two other compounds by mutual exchange of atoms or group of
atoms are called double displacement reactions. (or)
Those reactions in which two compounds react by an exchange of ions to
form two new compounds are called double displacement reactions.
Experiment: - Take about 3ml of sodium sulphate solution and barium chloride
solution in two test tubes. Both the solutions are clear and colourless. Now add
sodium sulphate solution to barium chloride. We will observe that a white
substance which is insoluble in water is formed. This insoluble substance due to
the formation of insoluble barium sulphate. The reaction may be written as:Bacl2(ag) + Na2SO4
BaSO4(ag) + 2NaCl(ag)
Example 1:- When dilute H2SO4 is added to pieces of iron sulphide, hydrogen
sulphide gas is produced and soluble ferrous sulphide is formed. Hence it is a
double displacement reaction.
Fe S(s) + H2SO4
FeSO4(aq) + H2S(g)↑
Example 2:- When Ag NO3 solution is added to NaCl solution, and then a
white precipitate of AgCl is formed along with sodium nitrate solution.
Ag NO3 + NCl(aq)
AgCl(s)↓ + NaNO3(aq)
(white ppt.)
In this double displacement reaction, two compounds silver nitrate and
sodium chloride react to form two new compounds river chloride and sodium
nitrate.
5. Precipitation Reaction:- Those reactions in which aqueous solutions of two
compounds on mixing react to form an insoluble compound which separates
out as a solid called precipitate or such an insoluble compound is formed when
gas compound is called precipitation reactions.
Example 1:- When barium chloride solution is added to sodium sulphate
solution, then a white precipitate of barium sulphate is formed along with NaCl
solution.
BaCl2(aq) + Na2SO4(ag)
BaSO4(s)↓ + 2NaCl(ag)
(white ppt.)
Example 2:- When Copper sulphate in aqueous form reacts with hydrogen
sulphate gas, a black precipitate of copper sulphide & sulphuric acid is formed.
CuS↓ + H2SO4(ag)
CuSO4(ag) + H2S(g)
(black ppt.)
Example 3:- When Aluminum chloride reacts with an aqueous solution of
ammonium hydroxide, a white precipitate of aluminum hydroxide and
ammonium chloride are formed.
Al(OH)3↓ + 3NH4Cl
AlCl3(ag) + 3NH4OH(ag)
(White ppt.)
6. Neutralization Reactions:- A reaction in which an acid reacts with a base to
form salt & water is called a neutralizatiom reaction - i.e.
Acid + Base
Salt + water
Examples:i)
When a base sodium hydroxide reacts with an acid like hydrochloride,
sodium and water are formed.
Na OH
+ HCl
(Base)
ii)
KOH
(acid)
+
HCl
(Base)
iii)
(acid)
NH4OH +
(Base)
iv)
NaCl + H2O
NaOH
(salt)
(water)
KCl
+ H2O
(salt)
HCl
(water)
NH4Cl + H2O
(acid)
+ CH3COOH
(Base)
(Acid)
CH3COONa + H2O
(Sod. Acetate)
7. Oxidation - Reduction Reactiom:- Oxidation is a reaction which involves
the addition of oxygen or removal of hydrogen e.g,
4Na + O2
2Na2O
2Mg + O2
2Mgo
4H2S + O2
2H2O + 2S
4HCl + MnO2
MnCl + 2H2O + Cl2
On the other hand, reduction is a reaction which involves the addition of
hydrogen or removal of oxygen, e.g
Or
H2 + S
H2S
Cl2 + H2S
2HCl2 + S
ZnO +C
Zn + CO
CuO + H2
Cu + H2O
These oxidation or reduction reactions cannot occur alone. This is
because if one substance loses oxygen, these must be another substance which
gains oxygen. Similarly, if one substance which gains hydrogen. Thus, if one
sustaince iss oxidizea, the othet must be reduced. This means that oxidation and
reduction always go side by side - i.e they occur simultaneously in a reaction.
Thus, for this reason, the oxidation - reduction reactions are also called as redox
reactions.
Example-1:- When copper oxide iss heated with hydrogen, then copper metal
and water are formed.
CuO + H2
Heat
Cu + H2O
Now in this reaction, CuO is changing into Cu - i.e oxygen is being
removed from copper oxide. Now by definition, removal of oxygen from
substance is called reduction, so we can say that copper oxide is being reduced
to copper.
Also in this reaction, Hydrogen is changing into H2O - oxygen is being
added to hydrogen. Now by definition, addition of oxygen to a substance is
called oxidation, so we Can say that hydrogen is being oxidised to water.
Now from the above reaction, we find that hydrogen is being oxidised to
water and at the same time, copper oxide issue being reduced to copper. This
shows that oxidation reduction occur together. The oxidation reduction reaction
between copper oxide & hydrogen Can be shown more clearly as follows:-
CuO + H2
Cu + H2O
Example-2:- when ferric oxide reacts with aluminium, iron & aluminium oxide
are obtained.
Fe2O3 + 2Al
2Fe + Al2O3
Example3:- When hydrogen sulphide reacts with chlorine, then sulphur and
hydrochloride acid are formed.
H2S(g) + Cl2(g)
S(s) + 2HCl(g)
Oxidation
Removal of hydrogen
H2S + Cl2
S + 2HCl
Addition of hydrogen (reduction)
In this reaction:a) His has lost hydrogen to format S- hence H2 S has been oxidised.
b) Cl2 has gained hydrogen to formed HCl hence Cl2 has been reduced.
Effects of Oxidation Reaction is Everyday Life:Oxidation has damaging effects on metals as well as on food. The
damaging effect of oxidation on metals is studied as corrosion and that on food
is studied as rancidity. Thus there are two common effects of oxidation
reactions which we observe in daily life. These are:i) Corrosion ii) Rancidity
i) Corrosion:- The process of slowly eating up of the metals due to attack of the
atmospheric gases such as oxygen, CO 2, H2 S, water vapor etc. on the surface
of the metals so as to convert the metal into oxide carbonate, sulphide etc. is
known as corrosion.
The most common example of corrosion is rusting i. e, corrosion of iron
when an iron article remains exposed to moist air for a long time, its surface is
covered with a brown & non-sticky substance called rust. It is mainly hydrate
ferric oxide (Fe2O3. XH2O). It is formed due to attack of oxygen gas and water
vapour present in the air on the surface of iron.
2Fe(s) + 3/2 O2(g) + xH2O
Fe2O3 . xH2O(s)
Rust
Similarly, copper lose its luster and shine after sometime. The surface of
these objects acquires a green coating of basic copper carbonate. CuCO3.
Cu(OH)2 when exposed to air. This is due to attack of O2, CO2 & water vapour
present in the air on the surface of copper.
2Cu(s) + CO2(g) + H2O(l)
CuCO3 ‘ Cu(OH)2
Likewise silver tarnishes i.e loses lustre and becomes dull on exposure
to air. This is due to the formation of a
ting of black silver sulphide (
)
on its surface by the corrosion of H2S gas present in the air lead or stainless
steel lose their lustre due to corrosion.
The corrosion causes damage to car bodies, bridges, iron railings ships
and many other articles especially made of iron. The corrosion of iron is a
serious problem; every year on enormous amount of money is spent to replace
damaged iron objects.
ii)Rancidity:When food materials containing fats and oil are left from sometime, they
get spoiled and give foul smell and unpleasant taste. This is because when fats
and oils are oxidised they become rancid and their smell as well as taste change.
This change in odour and flavour of oily and fatty foods by oxidation is called
rancidity.
To prevent, this usually substances which prevent oxidation are added to
food containing fats and oils. These substances are called 'antioxidants'.
Keeping food in air tight containers also help to slow down oxidation and this
increases the life if food materials.
In some food-stuffs, the air present around them is replaced by nitrogen in
the packet containing the food-stuffs. Thus, oxidation of the food-stuff is
preventing them from oxidation.
TEXT BOOK QUESTIONS:Q1:-When should a magnesium ribbon be cleaned before burning in air?
Ans: - When magnesium ribbon remains exposed to moist air, a white layer of
MgO is formed on its surface. This hinders the burning of magnesium. Hence
this layer is first removed by rubbing with sand paper before burning.
Q2:- Write the balanced equation for the following chemical reactions.
i) Hydrogen + Chlorine
Hydrogen Chloride
ii) Barium Chloride + Aluminum Sulphate
Barium Sulphate+
Aluminum Chloride
iii) Sodium + Water
Ans: - i)
Sodium hydroxide + Hydrogen.
Hydrogen + Chlorine
Hydrogen Chloride
H2 + Cl2
HCl – (skeletal equation)
H2 + Cl2
2 HCl – (Balanced equation)
ii) Barium Chloride + Aluminum Sulphate
Barium Sulphate+
Aluminum Chloride.
BaCl2 + Al2(SO4)
3BaCl2 + Al2(SO4)3
iii) Sodium + water
Na + H2O
2 Na + 2H2O
BaSO4 AlCl3 – (skeletal equation)
3BaSO4 + 2AlCl3
sodium hydroxide + hydrogen.
NaOH + H2 - [skeletal equation]
2NaOH + H2 – [Balanced equation]
Q3:- Write a balanced chemical equation with state symbols for the
following reactions:a) Solution of Barium Chloride and Sodium sulphate in water react to give
insoluble barium sulphate and solution of sodium Chloride.
b) Sodium hydroxide solution in water reacts with hydrochloric acid solution in
water to produce sodium chloride solution & water.
Ans: - a) The word equation for the given reaction is:Barium Chloride + Sodium sulphate
Chloride
BaCl2(aq) + Na2SO4(aq)
Barium Sulphate + Sodium
BaSO4(s) + NaCl(aq)
[skeletal equation]
BaCl2(aq) + Na2SO4(aq)
BaSO4(s) +2 NaCl(aq)
b) the word equation for the given reaction is:Sodium hydroxide + hydrochloric acid
NaOH(aq) + HCl(aq)
2NaOH(aq) + 2HCl(aq)
sodium chloride + water
NaCl(aq) + H2O(l) [skeletal equation]
2NaCl(aq) + H2O(l) [balanced
equation]
Q4:- A solution of a substance 'X' is used for white washing.
i)
ii)
Name the substance 'X' and write its formula.
Write the reaction of the substance 'X' named above with water.
Ans: - i) The substance 'X' used for white washing is quick lime or calcium
oxide.
ii) When quick lime is mixed with water, the following reaction takes place;CaO(s) + H2O(l)
Ca(OH)2(aq)
Q5:- Why the amount of gas is collected in one of the test tubes in activity
6.7 is doubled of the amount collected in the other? Name this gas.
Ans: - Water decomposes on electrolysis to given hydrogen and oxygen as
under:-
2H2O(l)
2H2(g) + O2(g)
Thus, hydrogen and oxygen are formed in the ratio 2:1 by volume. The
double volume of the gas collected by hydrogen.
Q6:- Why does the colour of CUSO4 solution change when an iron nail is
dipped in it.
Ans: - Iron is more reactive then copper. It displaces copper from CuSO4
solution according to the following reaction:Fe(s) + CuSO4(aq)
FeSO4(aq) + Cu(s)
Thus, as CuSO4 reacts to form
copper sulphate solution fades.
sulphate, the blue colour of
Q7:- Identity the substances oxidized on the substances that are reduced in
the following reactions:a) 4Na(s) + O2(g)
b) CuO(s) + H2(g)
Ans:- i) 4Na(s) +O2(g)
2Na2O(s)
Cu(s) + H2O(l)
2Na2O(s)
Here, Na had gained oxygen to form Na2O.
Therefore, Na has been oxidized to Na2O. Obviously, therefore O2 has been
reduced.
iii)
CuO(s) + H2(g)
Cu(s) + H2O(l)
Here, CuO has not lost oxygen to form Cu. Hence, CuO has been
reduced to Cu. H2 has gained oxygen to form H2O. Hence hydrogen (H2) has
been oxidized to H2O.
EXERCISE
Q1:- What is a balanced chemical equation? Why should chemical
chemucal equation be balanced?
Ans: - Balanced Chemical Equation: - Already defined.
According to low of conservation of mass, the total mass of products
must be equal to the total mass of reactants. This is possible only if the number
of atoms of each element is same on the two sides of the equation.
Q2:- Translate the following statements into chemical equations and then
balance them.
a) Hydrogen gas combines with nitrogen to form ammonia.
b) Hydrogen sulphate gas burns in air to give water and sulphur dioxide.
c) Barium chloride reacts with aluminum sulphate to give aluminum
Chloride and precipitate of barium sulphate.
d) Potassium metal reacts with water to give potassium hydroxide and
hydrogen gas.
Ans: - a) The word equation for the given statement is:Hydrogen + Nitrogen
Ammonia
Now the skeletal chemical equation is:H2(g) + N2(g)
NH3(g)
Therefore, balanced chemical equation is:3H2(g) + N2(g)
2NH3(g)
b) The word equation for the given statement is:Hydrogen Sulphide + Oxygen
Water + Sulphur dioxide
Now the skeletal chemical equation is:H2S(g) + O2(g)
H2O(l) + SO2(g)
Therefore, the balanced chemical equation is:2H2S(g) + 3O2(g)
2 H2O(l) + 2SO2(g)
c) The word equation for the given statement is:Barium Chloride + Aluminium Sulphate
Barium
Aluminium Chloride +
Sulphate
Now the skeletal chemical equation is:
BalCl2(aq) + Al2(SO4)3(aq)
AlCl3(aq) + BaSO4(s)
Therefore, the balanced chemical for the given statement is:
3BalCl2(aq) + Al2(SO4)3(aq)
2 AlCl3(aq) + 3BaSO4(s)
d) The word equation for the given statement is:Potassium + Water
Potassium hydroxide + Hydrogen
Now the skeletal chemical equation is:
K(s) + H2O(l)
KOH(aq) + H2(g)
Therefore, the balanced chemical equation us:2K(s) + 2H2O(l)
2KOH(aq) + H2(g)
Q3:- Balance the following chemical equations:a) HNO3 + Ca(OH)2
Ca (NO3)2 + H2O
The balanced chemical equation is:
2HNO3 + Ca(OH)2
Ca (NO3)2 + 2H2O
b) NaOH + H2SO4
Na2SO4 + H2O
The balanced chemical equation is:2NaOH + H2SO4
c) NaCl + AgNO3
Na2SO4 + 2H2O
AgCl + NaNO3
The above chemical reaction is already balanced.
d) BaCl2 + H2SO4
BaSO4 + HCl
The balanced chemical equation is:BaCl2 + H2SO4
BaSO4 + 2HCl
Q4:- Write the balanced chemical equations for the following reactions:a) Calcium hydroxide + Carbon dioxide
Calcium Carbonate +
Water
Ca(OH)2 + CO2
CaCO3 + H.O
b) Zinc + Silver nitrate
Zinc nitrate + Silver
Zn + AgNO3
Zn(NO3)2 + Ag - [skeletal equation]
Zn + 2AgNO3
Zn(NO3)2 + 2Ag – [Balanced equation]
c) Aluminium + Copper Chloride
Aluminium Chloride + Copper.
Al + CuCl2
AlCl3 + Cu – [skeletal equation]
2AlCl3 + 3Cu – [Balanced equation]
Or 2Al + 3CuCl2
d) Barium chloride + Potassium sulphate
Barium sulphate +
Potassium Chloride
BaSO4 + KCl – [skeletal equation]
BaSO4 + 2KCl – [Balanced equation]
BaCl2 + K2SO4
BaCl2 + K2SO4
Q5:- Write thee balanced chemical equation for the following and identify
the type of reaction in each case.
a) Potassium bromide (aq) + Barium iodide (aq)
KBr(aq) + BaI2(aq)
Potassium iodide
(aq) + Barium Btomide(s)
KI(aq) + BaBr2
This is a skeletal equation.
The balanced equation is as under:2KBr(aq) + BaI2(aq)
2KI(aq) + BaBr2(aq)
In the above reaction since two compounds on reacting from two new
compounds of potassium iodide and barium bromide therefore it is a double
displacement reaction.
b) Zinc Carbinate(s)
ZnCO3(s)
Zinc oxide (s) + Carbon dioxide(g)
ZnO(s) + CO2(g)
The reaction is already balanced. Now as a single compound decomposes in this
case to form two simpler substances, therefore it is a decomposition reaction.
c) Hydrogen(g) + Chlorine (g)
Hydrogen Chloride (g)
H2(g) + Cl2(g)
HCl(g) - [skeletal equation]
H2(g) + Cl2(g)
2HCl(g) – [Balanced equation]
In the above reaction, as two substances combine to form a single
compound, so it is a combination reaction.
d)magnesium + Hydrochloric acid
(g)
Magmesium Chloride(aq) +Hydrogen
Mg(s) + HCl(aq)
MgCl2(aq) + H2(g)
This is a skeletal equation. The balanced Chemical equation is:
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
As Mg displaces hydrogen from HCl, therefore the above reaction is a
displacement reaction.
Q6:- Why respiration is considered as exothermicreaction? Expain.
Ans:- During reapiration we need energy. We get this energy from food we eat.
During digestion food is broken
simpler substances. For example
rice, bread,
contain carbohydrates. These carbohydrates
are broken down to form to glucose. This glucose combines with oxygen in the
cells of our body producing energy.
C6H12O6 + 6O2
6CO2 + 6H2O + Energy
The above reaction is called respirationreaction. Thus, during respiration
energy is released and hence, it is considered as an exothermic reaction.
Q7:- In the refining of silver the recovery of silver from AgNO3 involved
displacement by copper metal. Write down the reaction involved.
Ans: -
Cu(s) + 2Ag NO3(aq)
Cu (NO3)2(aq) +2Ag(s)
Copper(II) nitrate
Q8:- A shiny brown colourd element 'X' on heating in air becomes black in
colour.Name the element 'X' and the black coloured compound formed.
Ans: - The element 'X' must be copper because copper is a shiny brown colourd
element which on heating in air combines with oxygen of the air to form black
copper oxide.
2Cu(s) + O2(g)
2CuO(s)
Thus, black coloured compound is copper (II) Oxide - CuO.
1.Light - Reflection & Refraction
Introduction:-
We see a variety of objective in the world around us However,
we are to unable to see anything in a dark room. On lighting up the room, things
become visible. Also during the day, we are able to see the world around us due to
the light from the sun. Infact sunlight falling on the objects is reflected or scattered
by them in our eyes and we are able to observe them. We may therefore consider
light as a form of energy which produces in us the sensation of sight. Thus, light is
an indispensable tool to explore the colorful beauty of nature such as blue color of
sky, white color of clouds, red sunshine, and sunset, rainbow, beautiful colors of
birds etc. A small source of light casts a sharp shadow of an opaque object become
very small, light has tendency to bend around it. This effect is known as Diffraction of
light.
Moreover, an object reflects light that falls on it. This reflected light, when received
by over eyes, enables us to see things. We are able to see through a transparent
medium, as light is transmitted through it. There are a number of common
wonderful phenomena, associated with light such as image formation of mirrors, the
twinkling of stars, bending of light by a medium, so on. A study of the properties of
light helps us to explore them by observing the common optical phenomena around
us.
Some Important Terms reflected with light:-
1. Source:- A source of light is an object , from which light is given out. Some
sources of light are natural and many others are manmade sources. For us on
earth, Sun is the important source of light. An Electric lamp, an oil lamp, a
candle etc. are some of the manmade sources of light.
The sources of light are of two kinds:-
(a)
Self Luminous: - A luminous source is that which possesses light of its own.
eg, Sun, Stars, Electric lamps, Candle etc.
(b)
Non-Luminous:- A source which does not possess light of its own are
called Non-Luminous source. It receives light from an external source and
scatters it to the surroundings, e.g, the moon, a book, a table, a stone etc.
2. Medium:-A medium is a substance through which light propagates or tries to
do so.
There are three types of media of light:-
a) Transparent:- It is a medium through which
light propagates easily e.g, air, water, glass etc.
b) Translucent:- It is a medium through which light propagates partially e.g,
paper, thick polythene etc.
c) Opaque:- It is a medium through which light cannot propagate. E.g., Wood,
Metals, Wall etc.
3) Ray:- A ray light is the straight line path along which light travels. It is a
represented by an arrow head on a straight line. The arrow head represents
the direction of propagation of light. A number of rays combined together
from a beam of lights.
Reflection of light:- When light traveling in a medium A(say) strikes a boundary
leading to another medium B, a part of the incident light is thrown back into the
original medium A. This phenomenon is called reflection of light, thus, reflection of
light is the phenomenon of bouncing back of light in the same medium on the
striking the surface of any object.
Law of reflection of light:-
First law:- According to this law, the incident ray, the reflected ray and the normal at
a
point
of
incidence all lie in the same
plane.
Here in the figure, light falling on plane mirror MM” along AO is incident ray
and OB is reflected ray. ON is normal to the mirror at O. According to first law OA, Ob
and ON all lie in the same plane. Neither of the two rays or normal goes down into
the paper or cones up out of the paper.
Second Law:- According to this law, the angle of incident(i) is always equal to the
angle of reflection(r)
i.e, <AON=<NOB
or <i= <r
These laws of reflection are applicable to all types of reflecting surface
including spherical surface.
Further when a ray of light is incident normally or perpendicularly on a mirror,
<i=0. According to the second law of reflection ray would go along the normal itself.
Here in the figure the incident ray is along No reflected ray and reflected ray is along
ON. It means that a ray of light incident normally on the mirror retraces its path on
reflection.
Characteristics of Image formed by plane Mirrors:Following are the important characteristics of image formed by plane mirrors:-
i)
ii)
iii)
iv)
v)
The image of a real object is always virtual. Such an image cannot be
taken on a screen.
The image formed in a plane mirror is always erect i.e, upside of the
image is upside of the object and vice-versa.
The size of the image in a plane mirror is always the same as the size of
the object
The image formed in a plane mirror is as far behind the mirror as the
object is in front of the mirror.
The image formed in a plane mirror is laterally inverted i.e, the left side
of the object becomes the right side of the image and vice-versa.
Spherical Mirror:-
The reflecting surface of a spherical mirror is curved inwards or out wards. Infect
a spherical mirror is that mirror whose reflecting surface is a part of a hallow sphere
of glass. One side of mirror is well polished and reflecting and the other side of the
mirror is opaque.
Spherical mirrors are of two types:-
(a)
Concave Mirror:- It is spherical mirror in
reflecting surface is towards the centre of
sphere of which the mirror is a past- i.e
reflection of light.occures at concave
surface or the best in surface. The other
which
the
(b)
surface of the concave spherical mirror is opaque or non- reflecting.
Convex mirror:- It is that spherical mirror in which reflecting surface is way
from the centre of the sphere of which the mirror is apart - i.e, reflection
of light occurs at convex surface or bulging out surface. The surface
(shaded) of the mirror is opaque or non-reflecting.
Some important terms related to Spherical mirror:-
(i)
Center of curvature:- The centre of curvature of a spherical mirror is
the centre of the hallow sphere of the glass of which the spherical
mirror is a part. The centre of curvature is usually represented by the
letter(C)
(ii)
Pole: - The pole of spherical mirror is the centre of a mirror. It is also
called the vertex of the mirror and is usually denoted by the letter ( P)
(iii)
Radius of curvature:- The radius of curvature of a spherical mirror is the
radius of the hallow sphere of glass of which the spherical mirror is the
part. It is usually represented by letter (R)
(iv)
Principal Axis:- The straight line passing through the centre of curvature
(c) and pole ( P) of the spherical mirror produced on both sides is called
principal axis. Here in the figure XX` IS THE PRINCIPAL AXIS. Principal
axis is normal to the mirror at its pole.
(v)
Aperture:- The aperture of a spherical mirror is the diameter of the
reflecting surface of there mirror. It is also called linear.
(vi)
Principal focus :- The principal focus of a mirror is a point on the
principal axis of the mirror at which rays of light incident on a mirror in
a direction parallel to the principal axis actually meet after reflection in
case of concave mirror where as in case of convex mirror appear to
diverge after reflection from the mirror. It is usually denoted by F.
The principal focus (F) of a concave mirror is a real point and
always lies in front of the concave mirror where as in case of convex mirror it is
virtual point and always lays behind thee convex mirror.
vi)
Focal length:- the distance of the principal focus ( F)
Sign convention for mirrors: In the ray diagrams of canvas as well as convex spherical mirror we use
new Cartesian sign conventions for measuring various distances. According to the
new Cartesian sign convention, we have:i)
ii)
iii)
iv)
v)
vi)
vii)
The principle axis of the
mirror is taken along x-axis &
poll P of the mirror is taken on
the left side of the mirror i.e.
light is incident on the mirror
from the left hand side.
The object is taken on the
left side of the side i.e. light is
incident on the mirror from the
left hand side.
All the distance parallel to the principal axis of the spherical mirror
measured from the pole of the mirror.
The distance measured in the direction of incident light is taken as
positive.
The distance measured in the direction opposite to the direction of
incident light is taken as negative.
The heights measured up words –i.e. above the x-axis and
perpendicular to the principal axis of the mirror is taken as positive.
The heights measured downwards- i-e. below the x- axis and
perpendicular to the principal axis of the mirror are taken as negative.
Images formed by spherical mirrors using ray diagrams:-
When an object is placed in front of a concave mirror, its image is formed by
reflection in the mirror. Every point on objects acts like a point source, from which an
infinite n, of rays originate. In order to the image of the point object, an arbitrarily
large number of rays emanating from the point object can be considered .However
for the sake of simplicity, We take any two rays of light, whose paths of simplicity, we
take any two rays of light, Whose paths on reflection from the mirror are known to
us. The point where these two reflected rays actually meet or appear to come from is
the real/natural image of the point object. The fallowing x-rays of light are commonly
used, out of which any two can be chosen concave mirrors. →Spherical.
Rule-1:- a ray of light following
on the mirror parallel to the
principal axis after reflection
will pass through the principal
focus in case of concave mirror
or appear to diverge from the
principal focus of a convex mirror.
Rule2:A ray of light passing
through the principal focus of
a concave mirror or a ray
which is directed towards the
principal focus of a convex
mirror after reflection will emerge parallel to the principal axis. This rule is just the
reverse case of Rule- 1.
Rule -3:- a ray of light passing through the center of curvature of concave mirror or
directed in the direction of center of curvature of a convex mirror after reflection is
reflected back along the same path the rays of light come back along the same path
because the incident rays fall in the mirror along the normal to the reflecting surface.
Rule 4:- A ray of light incident obliquely towards the pole p on the concave mirror or
convex mirror is reflected obliquely. The incident and reflected rays fallow the laws
of reflection at the point of incident making equal angles with the principal axis.
It is important to note that in all the cases the laws of reflection are followed.
At the point of incident the incident ray is affected in such a way that the angles of
reflection equal the angle of incident.
Image formation by cancave mirror:-
The type f image is formed by a concave mirror depends primarily on the
position of the object. In front of the mirror As an object approaches from infinity
towards the spherical mirror there occurs changed in the nature position and sign of
the image in other words if the distance of the object from the mirror is changed, a
corresponding change in the nature of position and size of the mage takes place
different place different cases involved in case of a concave mirror are as under:-
Case-1:- when the objects infinity:- Consider an object AB is situated very for off
from the mirror. Two rays Ad and AP from the top A of the abject are parallel to one
another and incident to the principal axis. These rays are reflected at points D and p
on the mirror and intersect at A` which i8s the real image of the top A of the object.
From A` draw A` B` perpendicular to the principal axis of the mirror Therefore A`B` is
the image of object A at infinity the image is formed at the focus F. It is real inverted
and such smaller in size then the object.
Case 2:- When the object is beyond, center of curvature(C):- In this case
consider on object AB is placed beyond the
center of curvature (C). A ray AD starts from top
A of the object and goes parallel to the principal
axis. After reflection it passes through the
principal focus (F). Another ray AE passing
through the center of curvature straights the
mirror normally and set races its path after reflection. These reflected rays intersect
at A` which is the image of A similarly the image of b is objected at B`
Thus the image of the object is formed between the principal focus (F) and
center of curvature (c) It is real inverted and smellier in size.
Case-3: when the object is at the centre
curvature:When the object is placed at the center of
curvature, the image A`B` of the object AB is also
formed at the center of curvature(C). It is real,
inverted and of same size as that the object.
Case 4: when the object is placed between focus (f) and center of curvature
(C):When the object AB is placed
between the principal focus (F) and the
center of curvature ( C), the image A`B`
of the object is formed beyond the center of curvature ( C). It is real, inverted and
very much magnified. (Diagram at page no 9)
Case 5: When the object is placed at the
focus (F):When the object AB is placed at the
center of principal focus (F) of the mirror, the
image of the object AB is formed at infinity.
However the the image is real, inverted greatly
magnified.
Case 6:- when the object is placed between the le and focus of the mirror:-
When the object is placed between the pole P and the principal focus (F) of the concave
mirror, the reflected rays of the object does not meet anywhere in front at the mirror but when
these rays are reduced back words meet at A` where a virtual image formed the perpendicular AB on
the principal axis gives the position of the image.
Hence the image is formed behind the concave mirror. It is virtual each and magnified in A’
size
Image formation by concave mirror:-
In case of convex mirror the image of an object formed is essentially virtual
erect and diminished in size. The different cases formed in case convex mirror are as
under.
CASE 1 :- When the object is at infinity:- Consider an object lying at infinity the
parallel rays coming from the object lying infinity appear to be diverging from the
principal focus thus the image is formed at the principal focus . However the image
formed is virtual erect and very much diminished in size.
Case 2 When the object is placed between pole and infinity:- Consider an object AB
placed between the pole P of the convex mirror and the infinity the image is formed
between the pole and the principal focus of the mirror. It is virtual erect and smaller
in size then that of the object.
USES OF SPHERICAL MIRRORS:- Some practical applications of sphrcal mirrors are:(A) Concave mirror:i)
ii)
iii)
iv)
v)
vi)
A concave mirror is used as a reflector in torches search lights, head lights
of motor vehicles etc. To get powerful parallel beams of light.
A concave mirror is used as doctors head mirror to focus light o body parts
like eyes, ears, nose, throat etc to be examined.
It is also used as shaving mirror and a makeup mirror as it can form erect
and magnified image of the face.
The dentist use concave mirrors to observe large images of the teeth of
patients.
Large concave mirrors are used to concentrate sunlight to produce heat in
solar cookers
Large concave mirrors are also used in reflecting type telescopes.
(B) Convex mirrors:i)
ii)
It is used as a reflector in street lamps. As a result light from lamp diverges
over a large area.
Convex mirror is used as rear view mirror in automobiles because:a) It produces erect images of object in front it.
b) The size of the image formed is smaller and therefore has wide view of
coverage.
c) All images are formed between pole and focus.
Mirror formula:- In a spherical mirror the distance of the object from its pole is
called the object distance and is denoted by U the distance of the image from the
pole of the mirror is called the image distance denoted by V. The distance of the
principal focus from the pole of the of the mirror is called focal length denoted by F
there is a relationship between these three quantities known as mirror formula and
expressed as .
I
I
_______
+
_______
Object distance
I
I
I
Image distance
=
_______
Focal length
I
___ + ___ = ___
u
v
f
Derivation:Consider an object AB placed perpendicular to principal axis beyond the centre of
curvature (C) in front of concave mirror. The incident ray BD parallel to the principal
axis is reflected from point D such that the ray reflected rays passes through the
focus. Another incident ray BCE is reflected back as ECB. Draw a perpendicular DM
on principal axis.
Now in as ABC=A``C`
∠ACB=∠A`CB-(V.O.∠s)
∠BAC=∠`B`A`C`-(each 90º )
:.
∆ABC=
∆ A`B`C-(AA similarity)
:.
AB
AC
____ = _____
A`B`
AB
_____
A`C`
PA - PC
= __________
A`B`
PC - PA
Also, in ∆s A`B`F` & MDF
∠A`F`B` = ∠MFD −−−− [ V.O. ∠ s ]
∠B’A’F = DMF ------- [ each 90º]
:.
∆A`B`F ~ ∆ MDF ----- [ AA similarity ]
DM
MF
______ =
______
A`B
A`F
AB
MF
Or
______
=
________
[ ·.· DM = AB ]
A`B`
MA` - MF
Now if the aperture of mirror is small, than M will almost coincide P, then
AB
=
A`B`
PF
II
PA` - PF
.·. From I and II , we get .
PA` − PC
PF
______________
=
PC − PA`
Or
− u − ( −2f )
__________
PA` − PF
=
−f
ghghghh
− 2f − (− V)
Or
− u + 2f
− 2f + V
− V − (− f)
=
− f_____
− V+ f
Or − (u − 2f )
_____________
−f
=
___________
− (2f − v)
−(v−f)
Or
(u − 2f ) (v − f ) = f (2f − v)
uv − uf − 2vf + 2f2 = 2f2 − vf
= 2f2 + 2vf + uf − vf − 2f2
uv
Or
uv
= vf + uf
Or
uv
_____
Uvf
=
vf
uf
____ + ____
uvf
------ [†ing b/s by uvf ]
uvf
Or
1___
f
= ___1___ +
u
___1___
v
Magnification:- The linear magnification produced by a concave mirror is defined as
the ratio of height of the image ( h1) to height of the object ( ho). It is represented by
`m`.
:- Linear magnification(m)=
height of image(h2)
height of object (ho)
or m= h1
ho
The magnification (m) of a spherical mirror is also related to the object
distance (u) and the image distance (v) as:It is important to that the height of the object is taken to be +ve as the
object is usually placed above the principal axis. The height of the image should be
taken as +ve for virtual images, but it is to be taken as -ve for real images. A -ve sign
in the value of the magnification indicates that the image is real where as a +ve sign
in the value of magnification indicates the image is virtual.
Relation between radius of curvature 7 focal length of spherical mirrors:- it has
been found that the focal length of a spherical mirror –i.e. either concave or convex
mirror is equal to half of its radius of curvature. If (f) is the length and R is the radius
of curvature, then
F=R/2
Derivation:- Consider a ray AB parallel to principal
axis falling on a concave mirror at B. After reflection
the ray passes through the focus. The dotted line cb is
the normal at point b.there fore ∠Abc is the ∠ of
incidence ‘i’ and ∠fBc is the ∠ of reflection ‘r’ now
according law of reflection.
∠I=∠r
Or ∠ FBC = ∠ FCB
or ∠ ABC = ∠ FBc
{∠ FBC =∠ FCB Att. ∠s}
 ∠ FBC = ∠ Fcb
 .FC=FB -------- I { sides opposite to equal ∠ s }
Now if the aperture of the cancave mirror is small thenthe ray AB will be very close
to P. therefore FB will be approximately equal to Fp.
i.e FP = FB -----II
:- from I and ii we get.
FC =FP
Now Fc= Fp = R, where ‘R’ is the radius of curvature.
FP + FP = R
2FP = R
2f = R
 f = R/2
It is important to mole that focal length of a mirror depends only on the radius of
curvature.
Problem 1:- Find the focal length of a mirror of radius of curvature 1m.
Sol.
Here focal length, f = ?
Radius of curvature, R = 1m
We know that
F = R/2 =1/2 m => F = 0.5m ans
TEXT BOOK
Problem 1.1:- A concave mirror8 used for rear- view on an automobile has radius of
curvature of 3.00m. If a bus is located at 5.oom from this mirror find the position,
nature and size of the image.
Sol;- here radius of curvature R= +3.oo m.
Object distance,
u = - 5.oo m.
image distance
V=?
Height of image
hi = ?
We know that F = R/2 = 3.oo/2 = 1.5m
Also we know that
1/f = 1/u + 1/v
1/1.5 =1/-5 + 1/v => 1/1.5 =- 1/5 + 1/v.
0r 1/1.5 + 1/5 =1/v or 1/15/10 + 1/5 = 1/v.
Or 10/15 + 1/5
= 1/v.
10 =3/15 = 1/v => 13/15 = 1/v => V = 15/ 13 = 1.15 m
:- the image is formed at the back of the mirror.
Again we know that
Magnification, m =h1/ ho = - y/u
 m = -v/u = -1.15/-5.oo = + 0.234
Thus the image is virtual erect and smaller in size.
Problem 1.2: An object 4.0 cm in size is placed at 25.0 cm insert of a concave mirror
of foal length 15.ocm at what distance from the mirror should screen be placed and
size of the image.
Sol:- here size of object, ho= + 4.om.
Distance of object, u = -25.ocm.
Focal length , f = -15.0cm
Distance of the image, v = ?.
Size of the image, hi = ?.
know that 1/f = 1/u + 1/v
1/-15 = 1/-25 + 1/v
Or -1/15 = -1/25 + 1/v
Or 1/v = -1/15 + 1/25
we
Or 1/v = -5+3 = -2/75
75
or -2v = 75 => v = 75/2 = - 37.5 cm Ans.
:- screen should be placed at 37.5 cm from the mirror and the image is real.
Again we know that
Magnification, m = hi/ho – v/u or hi/+4.o = - {-37.5)
- 25.o
 - 25.0hi = 37.5 * 4.
 `hi = 37.5*4
-25
= -6.0cm.
Thus the image is inverted and enlarged.
Problem 2:- The radius of curvature of a spherical mirror is 20cm . what is its focal
length ?
Problem 3:- Find the focal length of a convex mirror, whose radius of curvature is 32
cms.
Extra
Problem 4:- An object 4cm in size is placed at a distance of 25.0cm from a concave
mirror of focal length 15.0cm. Find the position, nature and the size of the image
formed.
Problem 5:- An erect image 3times the size of the object is obtained with a concave
mirror of radius of curvature 36cms. What is the position of the object?
Problem 6:- A 2.5cm candle is placed 12cm away from a concave mirror if a focal
length 30cm.Give the location of the image and the magnification.
REFRACTION OF LIGHT :- When a light travels in an isotropic medium , it propagates
in a straight line , but it changes its path while moving from one medium to another .
In other words we can say that light bonds at the interface (a line dividing the two
mediums) of the two media, when light moves from one medium to another. This
phenomenon is called refraction.
Thus, refraction is the phenomena of bending of light as it passes from one
medium to another .The bending of a light can take place into possible ways.
In first case when light travels from a rare to a denser, medium, it bends
towards the normal at the interface of two media (fig.1)
In the second case when light travels from a denser to a rare medium, it bends
away from normal to the interface of two media (fig. I)
Laws of reflection:- The bending of a light while traveling from one medium to the
another to the other takes place in accordance with certain laws reoffered to
as fallows:First law:- The incident ray refracted ray and normal to the surface at the point of
incidence of all lie in and the same plane.
Second law:- {the incident ray, refracted ray and normal to the to the interface of
two media at the point of incidence all} X the period of a refractive index and
sine of angles at a point in a medium is constant.
i.e.
n sin I = constant
For the two media in contact, we have
n1 sin i1 = n2 sin i2 ______ I
Now since i1 = I & i2 = r
:- from I we get
Ni sin I = n2 sin r
Or
sin i = n2
sin r
n1
= n2
Thus the radio of sine of angle of incidence to the sine of angle of reflection is
constant for the pair of media in contact. This constant is written as 1n2 as it.
Represents refractive index of media 2 with respect to medium 1.
Note: - these relationships were discovered by small experience mentally in 1621 and
do this law in is called as Snell’s law of reflection.
Third law
:- Whenever light goes from one medium to another the frequency of
light does not change . However, the velocity of light and the wave light of light
change.
Refraction through a Rectangular Glass Slab
A ray of light AO traveling in air is incident on a glass slab PQRS at point O. On
entering the glass slab, it gets refracted and bends towards the normal ON`. A second
change of direction takes place when the refracted ray of light OB traveling in glass
emerges i.e, comes out into air at point B. Since the ray of light OB now goes from a
denser medium ‘glass’ into the rarer medium ‘air’, it bends away from the normal
BN’, and goes in the direction BC.
From the figure, the incident ray AO and the emergent ray BC are parallel to each
other. These are parallel to each other because the extent of bending of the ray of
light at points O and B on the opposite. Parallel faces PQ and SR of the rectangular
glass slab are equal and opposite. The incident ray AO bends towards THE NORMAL
AT POINT o WHERE AS REFRACTED RAY ob BENDS AWAY FROM the normal at point
B by an equal amount .Thus, he light emerges from a parallel sides glass slab in a
direction parallel with that in which it enters the glass slab action parallel with that in
which it enters the glass slab through the emergent ray BC is parallel to the incident
ray AO, but the emergent ray has been side displaced or laterally displaced from the
original path of the incident ray by a ┴ distance CD, where as the original path of
incident light is AOD, but the emergent light goes along BC, the displacement
between them being CD. Another point to be noted is that in this case the refraction
of light takes place twice---- first at point) and then at point B. The angle which the
emergent ray makes with the normal is called the angle of emergence.
The case of light falling normally i.e ┴an a glass Slab:- If the incident ray falls
normally to the surface of a glass slab, then these is no bending of the ray of light
and it goes straight. Here in the figure a ray of light AO traveling in air falls on a glass
slab normally or perpendicularly at point O, so it does not bend on entering the glass
slab or as coming out of glass slab. It goes straight in the direction AOBC. Since the
incident ray goes along the normal to the surface, the angle of incidence in this case
is zero. Similarly, if a ray of light falls normally or perpendicularly to the surface of
water, even then, there is no bending of light ray and it goes straight through water.
Refractive Index: - We know that when a ray of light travels obliquely from one
transparent into another changes its direction in the second medium . The extent of
the change in direction that takes place in a given pair of media is expressed in terms
of the refractive index. The refractive index can be linked to a physical quantity, the
relative speed of propagation of light in different media. Light travels the fastest in
vacuum with the highest speed of 3x108m in air, the speed of light is only marginally
less compared to that in vacuum. It reduces considerably in glass or water. The value
of the refractive index for a given water The value of the refractive index of light in
the two media.
Consider a ray of light traveling from medium 1 into medium 2. Let V1 be the speed
of light in medium 1 and V2 be the speed of light in medium 2. The refractive index of
medium 2 with respect to medium 1 is given by the ratio of the
speed of light in medium 1 and the speed of in medium 2. This is
usually represented by the symbol n21 or 1n2.
This can be expressed in an equation FORM AS:-
1
n2 = V1
V2
= Speed of light in medium 1
Speed of light in medium 2
By the same argument, the refractive index of medium 1w.r.1 medium 2 is
represented asm2 or 2n,. It is given by:2n1 = V2 = Speed of light in medium 2
V1
Speed of light in medium 1
Now if medium 1 is vacuum or air, then the refractive index of medium 2 is
considered w.r.t vacuum. This is called the absolute refractive index of the medium.
If ‘c’ is the speed of light in air and V is the speed of light in the medium, then the
refractive index of the is given by:-
nm = Speed of light in air
= C
Speed of light in medium
V
The absolute refractive index of a medium is simply called its refractive
index.
For Example, let us take ‘glass’ as the medium , we know that the speed of light in air
is 3x108m/s at the speed of light in common glass is 2x108m/s.
ng = speed of light in air
= 3x108
Speed of light in glass
2x108 = 3/2
= 1.5
Thus the refractive index of glass w.r.t air is 1.5. This means that the ratio of the
speed of light in air or vacuum to the speed of light in glass is equal to 1.5.
Similarly, refractive index of water is as under:N
water = speed of light in air
speed of light in water
= 3*108 M/S
= 3 =1.33
2.25*108 m/s 2.25
As refractive index is a ratio of two velocities, it has no units. It is a pure
number.
We know that
Refractive index of the medium, nm = c
v
=> v = C
nm
:Larger the value of `nm` smaller is the value of V. It means in a denser medium,
speed of light s smaller and in a rarer medium, the speed of light is larger.
Tx.B
Q5 the refractive index of diamond is 2.42. What is the meaning of this statement.
Ans. He refractive index of diamond is the longest and is = 2.42
:- The optical density of diamond is the largest.
V As n = C
V
=> V = C
n
Therefore, when `n` is largest, v is smaller -i.e. velocity of light in diamond is
minimum.
Refraction by Spherical Lenses:Spherical Lenses: - spherical lens is a piece of a transparent refracting material, which
is bound by two surfaces. Often both the surfaces of a len are spherical. On passing
through a lens, light is spherical lenses are two types:(i)
(ii)
Concave lens or converging lens.
Concave lens or diverging lens
(i) Convex lens:- The lens which is thick in the middle and thin at the edges is called
convex lens. The two surfaces P & Q binding the less are convex – i,e. bulging out
having their own centre of curvature & radius of curvature of the two surfaces may
be equal or unequal. A convex lens is called an CONVERGING LENS, because it
converges the rays of light falling on it.
(ii) Concave lens: - The lens which is thin in the lens middle and thick at the edges is
known as concave lens. The two surfaces P` and Q` binding the lens are concave i.e`
curved in words having their own centre of curvature & radius of curvature. The radii
of curvature of the two surfaces may be equal or unequal. A concave lens is called a
diverging lens, because it diverges the rays of light falling on it.
Some basic terms connected with convex & concave lens :-
(i)
Centre of curvature: - A lens either convex or concave has two spherical
surfaces. Each of these surfaces forms part of a sphere. The Centers of
these spheres are called centre of curvature of the lens. They are
represented usually by c1 and c2.
(ii)
Principal axis:- An imaginary straight line passing through centers of
curvature of the two surfaces of the lens. The line c1c2 represents the
principal axis.
(iii) Aperture: - The aperture of a lens is the diameter of the circular edge of
the lens. Here in the diagram AB represents the aperture.
(iv) Optical Center: - The optical centre of a lens is a point on the principal of
axis of the lens, such that a ray of light passing through It goes un deviated
- i.e. the ray passing through optical centre of the lens suffer no refraction.
Here in the figure the optical centre is represented by the point O.
v)
Principal focus: - A convex lens has two surfaces and hence it has two
principal points.
 First principal focus:- it is the position of a point object on the principal
axis of the lens for with the image formed by the lens is at the infinity. The
first principal focus of convex lens is represented by the point F1. The rays
starting from an object point at f1, on refraction through convex lens
become parallel to principal axis and would form the image of the object
point at infinity.
 Second principal focus: - in the convex lens, it is the position of an image
point on the principal axis of the lens, when the point object is situated at
infinity. It is represented by the point F.
In case of concave lens, the first principal focus is the virtual position
of a point object on the principal axis of the lens for which the image
formed by the concave lens is at infinity. It is also represented by Fi, where
as the second principal focus is the position of image point on the principal
axis of the lens. When the point object is situated at infinity, it is again a
virtual point and represented by Fi.
vi)
Focal length: - In case of convex or concave lens, it is the distance between
the principal focus of the lens from optical centre O of the lens. It is
generally denoted by 'f'.
Formation of Images by a convex lens: - A convex lens forms the image of an
object placed in front of it. Any point on the object acts as a point source from which
an infinite number of rays start. These rays suffer refraction on passing through the
lens. We can locate the position of the image, the size and nature of the image by
considering any Two of the following special rays:-
a) The ray incident on the lens in a direction parallel to the principal focus of the
convex lens, on refraction passes through the principal focus of the located on
the other side of the lens.
b) The ray passing through optical centre of the convex lens passes straight i.e.
undeviated after refraction through the lens.
c) The rays passing through principal focus of convex lens, incident on the lens
becomes parallel to the principal axis of the lens, after refraction through the
lens.
In case of a convex lens, as the object is brought closer to the lens from infinity,
six different cases arise from the consideration of position, nature and size of the
image formed. These 6 different cases are as follows:-
i) Case-I When the object is at infinity:- When
the
object is at a considerable distance, we say the
object is at infinity. Now when the rays of light
coming from the object are parallel to the
principal axis of the convex lens. After
refraction, these rays meet at the principal
focus of the lens on its other side giving rise to the formation of image of the object.
Thus, it is clear from the figure that the image is formed at the focus of the lens.
It is real, inverted and much smaller than the object.
ii)Case-II When the object is beyond 2F1 (object beyond 2f):-
When the object is placed beyond 2F1, the image of the object is formed at A'B'
and it is formed between F2 and 2F2. It is real, inverted and diminished in size.
iii) Case-III When the object is placed at 2F1 (object at 2F):- by placing the object at
2F1, it means that the object is at a distance equal to twice the focal length of the
convex lens. In other words, the object is placed at a distance of 2F i.e. twice the
focal length of the lens. The image of the object is formed at 2F2 and is formed on
the other side of the convex lens. It is real, inverted and of the same size as that of
the object.
iv) Case-IV When the object is placed between 2F1 and F1:- i.e. object between F
and 2F:In this case when the object is placed between 2F1 and F1, the image A'B' of the
object is formed
beyond 2F2 on the other side of the lens. It is real, inverted and very much
magnified.
v) Case-V When the object is placed at focus F1:When the object AB is placed at principal focus F1, the image of the object is
formed at infinity. It is real, inverted and greatly magnified.
vi) Case-VI When the object is placed between optical centre(O) and focus F1:In this case, the object AB is placed between the
principal focus F1 and the optical centre(O), the
rays after passing the lens does not appear to
meet anywhere on the opposite side of the lens.
But when these are produced backwards, the
rays meet at A'. From A' draw A'B' perpendicular
axis of the lens.
Hence the image is formed on the same side of the lens. It is virtual, erect and
greatly magnified in size.
Formation of Images by a concave lens:To locate the position of the image, size and nature of image formed by a concave
lens, we use any two of the following three special rays, whose paths after refraction
through the lens are shown.
a) A ray incident on the concave lens in a direction parallel to the principal axis,
after refraction appears to come from principal focus of concave lens.
b) A ray passing through optical centre of concave lens passes undeviated after
refraction through the lens.
c) A ray of light appearing to meet at the principal focus of a concave lens, after
refraction will emerge parallel to the principal axis of the lens.
The image is formed at a point, where the two refracted rays appear to meet.
For all the positions of the object, the image formed by a concave lens is virtual,
erect and diminished in size. The exact position and size of the image would depend
upon the position of the object. There are only two possible cases in case of concave
lens.
Case l: - When the object lies between optical centre and infinity:As acts as an object and is held
perpendicular to the principal axis of a
concave anywhere between its optical
centre O and infinity.
The image of this object is formed
between optical centre O and principal focus F2 on the same side of the lens as the
object. It is virtual, erect and smaller in size than the object.
Case II: - when the object is at infinity:When the object is at infinity, the image of
object is formed at the second principal
focus on the same side of the lens as the
object. This image is virtual, erect and highly
diminished to almost point size.
this
Sign conventions for lenses:- The new cartisian sign conventions for measuring
various distances or heights in case of lens are similar to the one used for spherical
mirrors. However, all distances in lenses are measured from the optical centre of the
lens.
According to new Cartesian sign conventions:i)
ii)
iii)
iv)
v)
The principal axis of the lens is taken along x-axis of the rectangular Co ordinate system and optical centre O of the lens is taken as the origin.
The object is taken on the left side of the lens -i.e. light is incident on the lens
from left hand side.
All the distances parallel to the principal axis of the lens are measured from
the optical centre of the lens.
The distances measured in the direction of incident light are taken as positive.
The distances measured in the direction opposite to the direction of incident
ray are taken as -ve.
vi)
vii)
The heights measured upwards i.e. above the x-axis and perpendicular to the
principal axis of the lens are taken as +ve.
The heights measured downwards -i.e. below the X axis and perpendicular to
the principal axis of the lens are taken as -ve.
Lens Formula:- It is a relationship between object distance(u) image distance(v) and
focal length(f) of the lens. It is given as:1/v - 1/u = 1/f
The lens formula is general and is valid for any
spherical lens, wherever the object may be
situated.
Derivation:-
Consider an object AB placed perpendicular to the principal axis. BD is the incident
ray parallel to the principal axis which 8 refracted as DB'. Another incident ray
passing through, the optical centre, passes through the lens undeviated. Both the
refracted rays meet at B'. Where the image A'B' is formed.
In triangles ABO and A ' B'S
<AOB = <B'OA' ( V. O. <'S)
<BAO = <B'A 'O (each 90°)
Therefore triangle ABO~ triangle A ' B'S ( AA similarity )
=》 AB/A'B' = OA/OA' ------------(1)
Again in triangles DOF and B'A '
<DFO = <B'FA'
( V.O. <'s)
<DOF = <B'A'F
(each 90°)
Therefore triangle DOF ~ triangle B'A'F (AA similarity )
=》OD/A'B' = OF/A'S
or, AB/A'B' = OF/A'S ---------(II)
Now from I and II, we get
OA/OA'
= OF/A'F
or, OA/OA' = OF/OA'-OF -------(III)
By sign convention, we have
OA =-u, OF= +f, OA'= +v
Therefore, from =n (III) We get,
-u/v = f/v or -uv + uf = vf
or, uf - vf = uv
Dividing b/s by uvf, we get
or, uf/uvf - vf/uvf = 1/f
or, 1/v - 1/u = 1/f
MAGNIFICATION: - The magnification produced by a lens, similar to that for
spherical mirrors, is defined as the ratio of the height of the image and the height of
the object. It is represented by 'm'. If 'h' is the height of the object and h' is the
height of the image given by a lens, then the magnification produced by the lens is
given by:m= height of the image/ height of the object
=h'/h
Magnification produced by a lens is also related to the object distance 'u' and the
image distance 'v':m=h'/h = v/u
Power of a Lens:The power of a Lens is defined as the ability of the lens to converge the rays of
light falling on it. As a convex lens converges the rays of light falling on it, power of
lens is said to +ve. On the other hand, a concave lens diverges the rays of light falling
on it, so the power of a concave lens is said to be -ve.
Further, if the point of convergence of rays of light lies close to the optical centre
of convex lens, its converging ability or power is less. Now If 'P' is the power and 'f' is
the focal length of the lens, then we have
P = 1/f -------(1)
clearly from the above equation, smaller the focal length of the lens, greater is its
power and vice - versa.
The S.I unit of power of a Lens is called diopter which is denoted by D.
Thus, when focal length f = 1
Then we have from (1),
P= 1/1 = 1 diopter.
Therefore, one diopter is the power of a Lens of focal length one metre.
Power of a combination of Lenses: - When a number of thin lenses are placed in
contact with one another, the power of the combination is equal to the algebraic
sum of the power of individual lenses.
If Pl, P2, P3,............ are the powers of individual lenses placed in contact with one
another, the power P of the combination is:P = P1+P2+P3+..........
The individual powers have to be taken with proper sign, +ve for convex lens of
power +5D is placed in contact with a concave lens of power -2D, the power of the
combination is
P= P1 + P2
= 5D + (-2) = 5D - 2D = 3
Thus, this combination would behave as a convex lens of power +3D.
problem:- A thin lens has a focal length of -25cm. What is the power of the lens? is it
convex or concave?
Sol. Here focal length, f = -25.
Power of lens
P=?
We know that
P = 1/f
= 1/-25 ×100
P =-4
-ve sign shows that the lens is concave.
Text Book Questions
Q1:- Name a mirror that can give an erect and enlarged image of an object.
Ans: - A concave mirror gives an erect and enlarged image of an object held between
pole and principal focus of the mirror.
Q2:- A concave mirror produces three times magnified (enlarged ) real image of an
object placed at 10 cm in front of it. Where is the image located?
Sol: - Here linear magnification, m = -3
Object distance, u
= -10 cm
Image distance, v
=?
We know that, m
= -v/u
Therefore, -3 = -v/-10
or, v
= -30am
Therefore, the image is located at 30 cms in front of the mirror.
Q3:- A ray of light travelling in air enters obliquely into water. Does the light ray bend
towards the normal or away from the normal? Why?
Ans: - When a ray of light travels from air into water is optically denser than air. On
entering water, speed of light decreases and the light bends towards normal.
Q4:- you are given kerosene, turpentine and water. In which of these does the light
travel faster? (↓ with refractive index 1.44, 1.47 & 1.33 respectively).
Ans: - We know that
Refractive index =
Speed of light in air .
Speed of light in medium
Or speed of light in medium = speed of light in air
Refractive Index.
Therefore, it is clear from the above relation that the speed of light will be
maximum in that medium or substance, which has the lowest refractive index. Now
out of kerosene, turpentine and water, water has the lowest refractive index of 1.33.
So, the light will have maximum speed in water or light will travel fastest in water.
Q5:- A convex lens forms a seal and inverted image of a needle at a distance of
50cms from it. Where is the needle placed in front of the convex lens if the image is
equal to the size of the object? Also find the power of the lens.
Sol:-
Here image distance, v = 50cm.
Object distance, = ?
Power of lens, P = ?
As the image is of the same size as the object,
:. The object must be at a distance equal to twice the focal length of lens.
:.
We have -u = v = 2f = 50cms.
Or
2f = 50 =› f = 25cms
Now we know that
P = 1/f
= 1/25*100
= 4dioptre
Exercises
Q1:- A spherical mirror and a thin spherical lens have each a focal length of – 15 cm.
the mirror and the lens are likely to be-
a)
b)
c)
d)
Both concave
b) Both convex
the mirror is concave, but the lens is convex
the mirror is convex, but the lens is concave.
Ans:- As per new Cartesian sign conventions, the focal length of a concave mirror and
focal length of a concave lens, both are –ve. Therefore both are concave.
:. (a) Choice is correct.
Q2:- We wish obtain an erect image of an object using a concave mirror of focal
length 15cm. what should be the range of distance of the object from the mirror?
What is the nature of the image? Is the image larger or smaller than the object?
Draw a ray diagram to show the image formation in this case.
Ans: - To obtain an erect image of an object in a concave mirror, the object is placed
between the pole and focus of the mirror. Therefore, the object is to be placed in
front of the mirror art any position such that u ‹ 15cms. The nature of the image is
virtual and erecdt behind the mirror. The image is larger than the object.
Q3:- Name the type of used in the following situations:a) Head lights of a car (b) side rear view mirror of a vehicle.
( c) solar furnace, support your answer with reason.
Ans:- (a) For head lights of a concave use a concave mirror. The light source is held at
the focus of the mirror. On reflection, a strong parallel beam of light emerges.
(b) A convex mirror is used as side rear view mirror, because its field of view is larger
and it forms virtual, erect and diminished images of objects behind.
C) For solar furnace, we use a concave mirror. Light from the sun on reflection from
the mirror is concentrated at the focus of the mirror producing heat.
Q4:- one – half of a convex lens is covered with a black – papers will this lens produce
a complete image of the object? Verify your answer experimentally. Explain your
observations.
Ans: - Yes, it will produce a complete image of
object. This can be verified experimentally by
observing the image of a distance object like
a screen, when lower half of the lens is
covered with a black paper. However the
intensity or brightness of image will reduce.
the
tee on
Q5:- An object 5cm in length is held 25cm away from a converging lens of focal
length 10cm. draw the ray diagram and find the position, size and nature of the
image formed.
Sol: -
Here object size, h1 = 5cm
Object distance, u = -25cm
Also focal length, f = 10cm.
Image distance, v = ?
Size of image, h = ?
We know that
1/v - 1/u = 1/f
Or 1/v
= 1/f + 1/u = 1/10 – 1/25
= 5-2
50
Or 1/v = 3/50
=⟩ 3v = 50
Or v = 50/3 = 16.67cms
As v is +ve, the image formed is real and it forms on the right side of the lens.
Now
m =h2/h1 = v/u
Or
h2/5 = 16.67/-25
Or
h2 = 16.67/-5 = -3.3cm
-ve sign shows that the image is inverted.
Q6:- A concave lens of focal length 15cm forms an image 10cm from the lens. How
far is the object placed from the lens?
Draw the ray diagram.
Sol:-
Here focal length, f = 15cm
Image distance, v = 10cm
Object distance, u = ?
We know that
1/v -1/u =1/f
Or 1/v – 1/f = 1/u
Or 1/10 -1/-15 = 1/u
=›
u= -30cm.
Q7:- An object is placed at a distance of 10cm from a convex mirror of focal length
15cms. Find the position and nature of the image?
Sol: - Here object distance, u = -10cm.
Focal length, f = 15cm
Image distance, v = ?
We know that
1/v + 1/u = 1/f
Or
1/v = 1/f – 1/u
= 1/15 – 1/-10
= 1/15 + 1/10 = 2+3/30 = 5/30 = 1/6
=›
v = 6cm
Now +ve sign of v indicates that the image is at the back of the mirror. It must be
virtual, erect and smaller in size than the object.
Q8:- The magnification produced by a plane mirror is +1. What that these mean?
Ans:- we know that m = h2/h1
Now since m = +1
:. h2/h1 = 1
=›h2 = h1
:. Size of image = size of the object
Also + sign of m indicates that the image is erect and hence virtual.
Q9:- An object 5.0cm in length is placed at a distance of 20cms in front of a convex
mirror of radius of curvature 30cms. Find the position of image, its nature and size.
Sol:- here object size, h1 = 5.0cm
Object distance, u = -20cm.
Image distance, v = ?
Radius of curvature, R = 30cm
MATHEMATICS
Chapters: Number System, Polynomial, Linear equation in one
variables and Lines & Angles
Students are asked to complete these chapters.
History