Marksparks .arkansas, Author at WBBSE Solutions

NEET Biology Class 12 Organisms And Populations Notes

February 14, 2025August 30, 2023 by Marksparks .arkansas

Organisms And Populations

Ecology is the study of interactions among organisms and between the organism and its physical (abiotic) environment. Ecology is concerned with 4 levels of biological organization: Organisms, Populations, Communities, and biomes.

Organism And Its Environment

Physiological ecology:

(Ecology at the organismic level) is the study of the adaptation of an organism to environments in terms of survival and reproduction.

The rotation of the earth and the tilt of its axis cause annual variations in temperature and seasons. Major biomes (desert, rainforest, tundra, etc.) are formed due to these variations and precipitation (rain and snow).

Biology Class 12 Notes For Neet

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NEET Biology Class 12 Organisms And Populations Mean Annual Temperature And Mean Annual Precipitation

NEET Biology Class 12 Organisms and Populations Notes

Regional and local variations within a biome lead to the formation of different habitats.
Life exists even in extreme and harsh habitats.
For example, the Rajasthan desert, rain-soaked Meghalaya forests, deep ocean trenches, torrential streams, Polar Regions, high mountain tops, thermal springs, and compost pits.
Our intestine is a habitat for many microbes.
The physicochemical (abiotic) components (water, light, temperature, soil, etc.) and biotic components (pathogens, parasites, predators, competitors, etc.) lead to variation in different habitats.

“class 12 organisms and population notes “

Abiotic Factors

Biology Class 12 Notes For Neet

1. Temperature:

The most ecologically relevant environmental factor.
The temperature on land varies seasonally. It gradually decreases from the equator towards the poles and from plains to mountain tops. It ranges from subzero levels (in polar areas and high altitudes) to >500C (in tropical deserts).
The average temperature in thermal springs and deep-sea hydrothermal vents is above 1000 C.
Mango trees cannot grow in temperate countries (Canada, Germany, etc.). There is no Snow leopard in Kerala forests.
Tuna fishes are rare beyond tropical latitudes in the ocean.
Temperature affects the kinetics of enzymes, basal metabolism, and other physiological functions of the organism.

NEET Biology Class 12 Organisms And Populations Notes

Based on a range of thermal tolerance, organisms are of 2 types:

Eurythermal: They can tolerate a wide range of temperatures.
Stenothermal: They can tolerate only a narrow range of temperatures

2. Water:

It is the second most important factor.
Desert organisms have special adaptations to limited water.
The productivity & distribution of plants is dependent on water.
For aquatic organisms, water quality (pH, chemical composition) is important. The salt concentration (salinity in parts per thousand) is less than 5 in inland waters, 30-35 in the sea, and > 100 in some hypersaline lagoons.

“organism and population neet pyq “

Based on the tolerance to salinity, organisms are 2 types:

Euryhaline: Tolerates a wide range of salinities.
Stenohaline: Tolerates only a narrow range of salinity. Many freshwater animals cannot live for long in seawater and vice versa because of osmotic problems.

Biology Class 12 Notes For Neet

3. Light:

Plants need sunlight for photosynthesis.
Small forest plants (herbs and shrubs) are adapted to photosynthesize optimally under very low light because they are overshadowed by tall, canopied trees.
Many plants depend on sunlight for photoperiodism (For example, flowering). Many animals use the diurnal and seasonal variations in light intensity and photoperiod for timing their foraging, reproductive, and migratory activities. The sun is the ultimate source of light and temperature on land.
Deep (>500m) in the oceans, the environment is dark and there is no energy available from the Sun. The spectral quality of solar radiation is also important for life. The UV spectrum is harmful to many organisms.
Not all the color components of the visible spectrum are available for marine plants.

Organisms and Populations NEET Notes

4. Soil:

Nature & properties of soil differ due to climate, weathering process, sedimentation, method of soil development, etc.
Soil composition, grain size, and aggregation determine the percolation and water-holding capacity of the soils. These characteristics and parameters like pH, mineral composition, and topography determine the vegetation and animals in an area. In the aquatic environment, the sediment s determine the type of benthic animals.

Responses to Abiotic Factors

Organisms maintain a constant internal environment (homeostasis) despite varying external environmental conditions. This is possible by following processes.

1. Regulate:

It is the maintenance of homeostasis by physiological and behavioral means. It ensures constant body temperature (thermoregulation), constant osmotic concentration (osmoregulation), etc. For example, All birds and mammals, have very few lower vertebrates and invertebrates.

Thermoregulation in mammals:

The success of mammals is mainly due to their ability to maintain a constant body temperature. In summer, when the outside temperature is higher than the body temperature (370C), sweating occurs.

This results in evaporative cooling and brings down body temperature. In winter, when the temperature is below 370C, shivering occurs. It produces heat and raises the body temperature.

2. Conform:

99% of animals and nearly all plants cannot maintain a constant internal environment. Their body temperature or osmotic concentration changes with the surrounding conditions. They are called conformers.
In aquatic animals, the osmotic concentration of body fluids changes with that of the ambient osmotic concentration.
Thermoregulation is energetically expensive, especially for small animals (shrews, hummingbirds, etc.). They have a larger surface area relative to their volume. So they lose body heat very fast when it is cold outside.
Then they have to expend much energy to generate body heat. Therefore very small animals are rare in polar regions.

“organisms and population class 12 ppt “

NEET Biology Class 12 Organisms And Populations Diagrammatic Representation Of Organismic Response

NEET Biology Organisms and Populations Important Notes

3. Migrate:

Many animals like birds move away temporarily from stressful habitats to a more hospitable area and return when the stressful period is over.
For example, During winter, Keolado National Park (Bharatpur,
Rajasthan) hosts migratory birds coming from Siberia and other extremely cold northern regions.

4. Suspend:

In bacteria, fungi, and lower plants, thick-walled spores help to survive unfavorable conditions. Under suitable conditions, they germinate.
In higher plants, seeds and some vegetative reproductive structures serve to tide over periods of stress by reducing their metabolic activity. They germinate under favorable moisture and temperature.

In animals: Examples are

Hibernation of bears during winter.
Aestivation of some snails and fishes during summer.
Diapause (a stage of suspended development) of many zooplankton in lakes & ponds.

Adaptations

Biology Class 12 Notes For Neet

Adaptation is the morphological, physiological & behavioral attribute that enables an organism to survive and reproduce in its habitat.
Many adaptations have evolved over a long evolutionary time and are genetically fixed.

Adaptations of kangaroo rat in North American deserts:

Internal fat oxidation gives water as a byproduct if there is no external source of water.
Ability to concentrate urine so that a minimal volume of water is used to remove excretory products.

Adaptations of desert plants:

Presence of thick cuticles on leaf surfaces. Their stomata are arranged in deep pits to minimize water loss through transpiration.
A special photosynthetic pathway (CAM) that enables their stomata to remain closed during day time.
Desert plants like Opuntia have no leaves (they are reduced to spines). Photosynthesis is done by stems.

“organisms and population class 12 one shot “

Adaptations of mammals:

Mammals from colder climates have shorter ears and limbs to reduce heat loss. (This is called Allen’s Rule).
Aquatic mammals like seals have a thick layer of fat (blubber) below their skin that acts as an insulator and reduces loss of body heat.

Physiological and biochemical adaptations:

Archaebacteria are found in hot springs and deep-sea hydrothermal vents where the temperature is >1000C. Many fish thrive in Antarctic waters (temperature is below 00C).
Many marine invertebrates and fishes live at great depths in the ocean where the pressure is >100 times the normal atmospheric pressure.
At a high-altitude place (>3,500 m) we feel altitude sickness. Its symptoms are nausea, heart palpitations, and fatigue. This is due to low atmospheric pressure. So the body does not get enough O₂.
Gradually, we acclimatize to the situation and the body compensates for low O₂ availability by increasing RBC and breathing rate and decreasing the binding capacity of hemoglobin.

Behavioural adaptations:

Desert lizards bask in the sun and absorb heat when their body temperature is low, but move into shade when the ambient temperature starts increasing. Some species burrow into the soil to hide and escape from the above-ground heat

Organisms and Populations Class 12 NEET Key Concepts and Summary

Populations

A population is a group of individuals of the same species that live in a given geographical area, share or compete for similar resources, and potentially reproduce.
For example, All the cormorants in a wetland, rats in an abandoned dwelling, teakwood trees in a forest tract, bacteria in a culture plate lotus plants in a pond, etc.
Population ecology is an important area of ecology as it links ecology to population genetics & evolution.

Population Attributes

1. Birth rates: Refer to per capita births. For example, In a pond, there were 20 lotus plants last year, and through reproduction 8 new plants were added.

Hence, the current population = 28
The birth rate = 8/20 = 0.4 offspring per lotus per year.

2. Death rates: Refer to per capita deaths.

For example,4 individuals in a laboratory population of 40 fruit flies died during a week. Hence, the death rate = 4/40 = 0.1 individuals per fruit fly per week.

3. Sex ratio: A population has a sex ratio. For example,60% of the population is females and 40% males.

Class 12 Biology Notes For Neet

4. Age pyramid:

It is the structure obtained when the age distribution (% of individuals of a given age or age group) is plotted for the population.
For the human population, age pyramids generally show the age distribution of males and females in a combined diagram.

NEET Biology Class 12 Organisms And Populations Age Pyramid

“organism and population class 12 ppt “

Population size or population density (N):

It is the number of individuals of a species per unit area or volume. For example, the population density of Siberian cranes at Bharatpur wetlands in any year is <10. It is millions for Chlamydomonas in a pond.
Population size is also measured in % cover or biomass. For example, In an area, 200 Parthenium plants and a single huge banyan tree are seen. In such cases, measuring % cover or biomass is meaningful to show the importance of the banyan tree.
The total number is a difficult measure for a huge population.
In such cases, relative population density (without knowing absolute population density) is used. For example, the number of fish caught per trap indicates the total population density in the lake.
In some cases, indirect estimation of population sizes is performed. For example, Tiger census in national parks & tiger reserves based on pug marks & fecal pellets.

Class 12 Biology Notes For Neet

Population Growth

The population size changes depending on factors like food availability, predation pressure & weather.
Changes in population density give some idea about the population – whether it is flourishing or declining.

4 basic processes that fluctuate the population density:

Natality (B): It is the number of births in a population during a given period.
Mortality (D): It is the number of deaths in a population during a given period.
Immigration (I): It is the number of individuals of the same species that have come into the habitat from elsewhere during a given period.
Emigration (E): It is the number of individuals in the population who left the habitat and went elsewhere during a given period.

Natality and immigration increase the population density and mortality and emigration decreases the population density

NEET Biology Class 12 Organisms And Populations Population Density

NEET Biology Class 12 Chapter Organisms and Populations Detailed Notes

If N is the population density at time t, then its density at time t +1 is

N_t+1= N_t+ [(B + I) – (D + E)]

Population density increases if B+I is more than D+E.
Otherwise, it will decrease.

Under normal conditions, births & deaths are important factors influencing population density. The other 2 factors have importance only under special conditions. For example, for a new colonizing habitat, immigration may be more significant to population growth than birth rates.

Growth Models

Class 12 Biology Notes For Neet

1. Exponential growth:

Resources (food and space) are essential for unimpeded population growth.
If resources are unlimited, each species shows its full innate potential to grow in number. Then the population grows exponentially or geometrically.
If population size = N, birth rates (per capita births) = b, and death rates (per capita deaths) = d, then the increase or decrease in N during a unit period t (dN/dt) will be

dN/dt = (b – d) × N

Let (b–d) = r, then

dN/dt = rN

The r (‘intrinsic rate of natural increase) is an important parameter for assessing the impacts of any biotic or abiotic factor on population growth.

r value for the Norway rat = 0.015

r value for the flour beetle = 0.12

r value for human population in India (1981) = 0.0205

The integral form of the exponential growth equation is

N = N₀e_rt

“organism and population /class 12 ppt “

Where,

N_t = Population density after time t

N₀ = Population density at time zero

r = Intrinsic rate of natural increase

e = The base of natural logarithms (2.71828)

NEET Biology Class 12 Organisms And Populations Population Growth Curves

NEET Study Material for Organisms and Populations Chapter

a = Exponential growth (J-shaped curve)

b = Logistic growth (Sigmoid curve)

Class 12 Biology Notes For Neet

2. Logistic growth:

No population in nature has unlimited resources for exponential growth. This leads to competition among individuals for limited resources.
Eventually, the ‘fittest’ individuals survive and reproduce.
In nature, a given habitat has enough resources to support the maximum possible number, beyond which no further growth is possible. It is called carrying capacity (K).
A population with limited resources shows initially a lag phase, phases of acceleration & deceleration, and finally an asymptote. This type of population growth is called Verhulst-Pearl Logistic Growth.

It is described by the following equation:

⇒ $\mathrm{dN} / \mathrm{dt}=\mathrm{rN}\left(\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right)$

Where,

N = Population density at time t

r = Intrinsic rate of natural increase

K = Carrying capacity

Since resources for growth for most animal populations are finite the logistic growth model is a more realistic one.

Life History Variation

Populations evolve to maximize their reproductive fitness or Darwinian fitness (high r-value). Under a particular set of selection pressures, organisms evolve toward the most efficient reproductive strategy.
Some organisms breed only once in their lifetime (Pacific salmon fish, bamboo) while others breed many times (most birds and mammals).
Some produce a large number of small-sized offspring (Oysters, pelagic fishes) while others produce a small number of large-sized offspring (birds, mammals).
These facts indicate that life history traits of organisms have evolved due to limited abiotic and biotic components of the habitat.

Population Interactions

Class 12 Biology Notes For Neet

Organisms interact in various ways to form a biological community.
Interaction between two species is called Interspecific interactions. They include

Name of the Interaction And Species:

NEET Biology Class 12 Organisms And Populations Name Of Interaction And Species

Organisms and Populations Class 12 NCERT Notes for NEET

In predation, parasitism, and commensalisms, the interacting species live closely together.
In a broad ecological context, all carnivores, herbivores, etc. are predators. About 25 % of insects are phytophagous.
If a predator overexploits its prey, then the prey might become extinct. It results in the extinction of predators.
Therefore, predators in nature are ‘prudent’.

Importance of predators:

Predators control prey populations:

When certain exotic species are introduced into a geographical area, they spread fast due to the absence its natural predators in the invaded land.
For example, the prickly pear cactus introduced into Australia in the early 1920s caused havoc by spreading.
Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) was introduced into the country.

Biological control:

Biological control methods are based on the ability of the predator to regulate the prey population.
Predators maintain species diversity in a community, by reducing the intensity of competition among competing prey species.
For example, the starfish Pisaster is a predator in the rocky intertidal communities of the American Pacific
Coast. In an experiment, when all the starfishes were removed from an enclosed intertidal area, more than 10 species of invertebrates became extinct within a year, due to interspecific competition.

Defenses of prey species to lessen the impact of predation:

Camouflage (cryptic coloration) of some insects and frogs.
Some are poisonous and so avoided by predators.
The monarch butterfly is highly distasteful to its predator bird. It is due to a special chemical in its body. It is acquired during its caterpillar stage by feeding on a poisonous weed.
Thorns (Acacia, Cactus, etc.) are the most common morphological means of defense of plants. Many plants produce chemicals that make the herbivore sick, inhibit feeding or digestion, disrupt its reproduction, or kill it.
For example, Calotropis produces highly poisonous cardiac glycosides. Therefore cattle or goats do not eat it.
Nicotine, caffeine, quinine, strychnine, opium, etc. are defenses against grazers and browsers.

2. Competition:

It is a process in which the fitness of one species (‘r’ value) is significantly lower in the presence of another species. Interspecific competition is a potent force in organic evolution.
Competition occurs when closely related species compete for the same limited resources.
Unrelated species can also compete for resources.
For example, Flamingoes and fishes in some shallow South American lakes compete for zooplankton.
Competition occurs in abundant resources also. For example, In interference competition, the feeding efficiency of one species is reduced due to the interfering and inhibitory presence of other species, even if resources are abundant

Class 12 Biology Notes For Neet

Evidence for competition:

The Abingdon tortoise in the Galapagos Islands became extinct within a decade after goats were introduced on the island, due to the greater browsing efficiency of the goats.
‘Competitive release’: A species, restricted to a small geographical area (due to the presence of competitively superior species), expands its distributional range when the competing species is experimentally removed.
Connell’s field experiments: Connell’s field showed that on the rocky sea coasts of Scotland, the larger & competitively superior barnacle Balanus dominates the intertidal area, and excludes the smaller barnacle Chathamalus from that zone.

Gause’s ‘Competitive Exclusion Principle:

It states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually.
This may be true in limited resources, but not otherwise.
Species facing competition may evolve mechanisms that promote co-existence rather than exclusion. Example, ‘resource partitioning’.

Resource partitioning:

If two species compete for the same resource, they could avoid competition by choosing different times for feeding or different foraging patterns.
For example, MacArthur showed that five closely related species of warblers living on the same tree were able to avoid competition and co-exist due to behavioral differences in their foraging activities.

3. Parasitism:

Many parasites are host-specific (they can parasitize only a single host species). They tend to co-evolve. i.e., if the host evolves special mechanisms against the parasite, the parasite has to evolve mechanisms to counteract them, to be successful with the same host species.

Adaptations of parasites: Loss of sense organs, presence of adhesive organs or suckers to cling on to the host, loss of the digestive system, high reproductive capacity, etc.

The life cycles of parasites are often complex. Example,
Human liver fluke depends on 2 intermediate hosts (a snail and a fish) to complete its life cycle.
The malarial parasite needs mosquitoes to spread to other hosts.
Parasites harm the host. They may reduce the survival, population density, growth, and reproduction of the host.
They may make the host physically weak and more vulnerable to predation.

Types of parasites:

1. Ectoparasites:

Parasites that feed on the external surface of the host. Example,
- Lice on humans.
- Ticks on dogs.
- Ectoparasitic Copepods on many marine fishes.
- Cuscuta plant on hedge plants.
Cuscuta has no chlorophyll and leaves. It derives its nutrition from the host plant.
Female mosquito is not considered a parasite, because it needs our blood only for reproduction, not as food.

2. Endoparasites:

Parasites that live inside the host body at different sites (liver, kidney, lungs, RBC, etc).
The life cycles of endoparasites are more complex.
They have simple morphological and anatomical features and high reproductive potential.

Brood parasitism in birds:

Here, the parasitic birds lay eggs in the nest of their host and let the host incubate them.
During evolution, the eggs of the parasitic bird have evolved to resemble the host’s egg in size and color to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.
For example, Brood parasitism between cuckoo and crow.

4. Commensalism:

Examples:

Orchid (+) growing as an epiphyte on a mango branch (0).
Barnacles (+) growing on the back of a whale (0).
Cattle egret (+) and grazing cattle (0). The egrets forage close to where the cattle are grazing. As the cattle move, the vegetation insects come out. Otherwise, it is difficult for the egrets to find and catch the insects.
Sea anemone (0) and clown fish (+). Stinging tentacles of sea anemones give protection to fish from predators.

Population Growth, Adaptation, and Interactions NEET Notes

5. Mutualism:

Examples:

Lichen: It is a mutualistic relationship between a fungus and photosynthesizing algae or cyanobacteria.
Mycorrhizae: Associations between fungi and the root of higher plants. The fungi help the plant in the absorption of essential nutrients from the soil while the plant provides the fungi with carbohydrates.
Mutualism between plants and animals through pollination and seed dispersion: Examples:
- Fig trees and wasps: The fig species is pollinated only by its ‘partner’ wasp species and no other species. The female wasp pollinates the fig inflorescence while searching for suitable egg-laying sites in fruits. The fig offers the wasp some developing seeds, as food for the wasp larvae.
- Orchids show a diversity of floral patterns: They can attract the right pollinator insects (bees and bumblebees) to ensure pollination. Not all orchids offer rewards.
- ‘Sexual deceit’ of Ophrys (Mediterranean orchid): One petal of its flower resembles a female bee in size, color and markings. So male bees ‘pseudocopulates’ with the flower and is dusted with pollen. When this bee ‘pseudocopulation’ with another flower, it transfers pollen to it.

If the female bee’s color patterns change slightly during evolution, pollination success will be reduced unless the orchid flower co-evolves to maintain the resemblance of its petal to the female bee

NEET Biology Class 12 Biotechnology And Its Applications Notes

February 14, 2025August 30, 2023 by Marksparks .arkansas

Biotechnology And Its Applications

Biotechnology has a wide range applications such as biopharmaceuticals, therapeutics, diagnostics, genetically modified crops for agriculture, processed food, bioremediation, waste treatment, and energy production.

Biology Class 12 Notes For Neet

Biotechnology has 3 critical research areas:

Providing the best catalyst in the form of an improved organism usually a microbe or pure enzyme.
Creating optimal conditions through engineering for a catalyst to act.
Downstream processing technologies to purify the protein/organic compound

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Applications In Agriculture

Applications In Agriculture

Options for increasing food production:

Agro-chemical based agriculture.
Organic agriculture.
Genetically engineered crop-based agriculture.

Genetically Modified Organisms (GMO) or transgenic organisms are the plants, bacteria, fungi and animals whose genes are altered by manipulation.

Biology Class 12 Notes For Neet

Advantages of genetic modification in plants:

It makes crops more tolerant to abiotic stresses (cold, drought, salt, heat, etc.).
Pest-resistant crops reduce the use of chemical pesticides.
It helps to reduce post-harvest losses.
It increases the efficiency of mineral usage by plants (it prevents early exhaustion of fertility of soil).
It enhances the nutritional value of food. example, Vitamin A-enriched rice.
To create tailor-made plants to supply alternative resources (starches, fuels, pharmaceuticals, etc.) to industries.

“short notes on biotechnology “

Pest Resistant Plants:

They act as bio-pesticides.
It reduces the need for insecticides.
For example, Bt cotton, Bt corn, rice, tomato, potato, soybean, etc.

Bt Cotton:

Some strains of Bacillus thuringiensis have proteins that kill insects like coleopterans (beetles) lepidopterans (tobacco budworm, armyworm), and dipterans (flies, mosquitoes).

B. thuringiensis forms a toxic insecticidal protein (Bt toxin) crystal during a particular phase of their growth. It does not kill the Bacillus as it exists as inactive protoxins.
When an insect ingests the toxin, it becomes active due to the alkaline pH of the gut which solubilizes the crystals.
The toxin binds to the surface of mid-gut epithelial cells and creates pores. It causes cell swelling lysis and death of the insect.
Bt toxin genes were isolated from B. thuringiensis and incorporated into crop plants such as cotton.
Most Bt toxins are insect-group-specific. The toxin is
Coded by a gene named cry. for example, proteins encoded by the genes cryIAc and cryIIAb control the cotton bollworms of cryIAb control corn borer.

Biology Class 12 Notes For Neet

Nematode resistance in tobacco plants:

A nematode Meloidogyne incognitia infects the roots of tobacco plants causing a reduction in yield.
It can be prevented by RNA interference (RNAi) strategy.
RNAi is a method of cellular defense in all eukaryotic organisms. It prevents the translation of a specific mRNA (silencing) due to a complementary dsRNA molecule.
The source of this complementary RNA is from infection by RNA viruses or mobile genetic elements (transposons) that replicate via an RNA intermediate.
Using Agrobacterium vectors, nematode-specific genes (DNA) are introduced into the host plant. It produces both sense and anti-sense RNA in host cells.
These RNAs are complementary. So they form double-stranded (ds) RNA.
It initiates RNAi and silences the specific mRNA of the nematode. Thus the parasite cannot survive in a transgenic host expressing specific interfering RNA.

NEET Biology Class 12 Biotechnology And Its Applications Notes

NEET Biology Class 12 Biotechnology and Its Applications Notes

Applications In Medicine

Medicine:

The recombinant DNA technology helps for the mass production of safe and more effective therapeutic drugs.
The products from non-human sources induce unwanted immunological responses. However recombinant therapeutics do not have such problems.
At present, about 30 recombinant therapeutics have been approved. Of these, 12 are being marketed in India.

“importance of biotechnology “

1. Genetically Engineered Insulin:

Insulin is used to manage adult-onset diabetes.
Insulin from the pancreas of animals (cattle and pigs) causes allergies or other types of reactions to the foreign protein.
Now, it is possible to produce human insulin using bacteria.
Insulin consists of two short polypeptide chains (chain A and chain B) that are linked by disulfide bridges.
In mammals, insulin is synthesized as a pro-hormone (pro-insulin). It needs processing to become a mature and functional hormone.
The pro-hormone contains an extra stretch called the C peptide. This is removed during maturation into insulin.
In 1983, Eli Lilly (an American company) prepared two
DNA sequences corresponding to A and B chains of human insulin and introduced in plasmids of E. coli to produce insulin chains.
Chains A and B were combined by creating disulfide bonds to form human insulin

NEET Biology Class 12 Biotechnology And Its Application Pro Hormone Or Proinsulin

Biotechnology and Its Applications NEET Notes

2. Gene Therapy:

It is a method to correct a gene defect in a child/embryo.
Here, genes are inserted into a person’s cells and tissues to treat a hereditary disease. It compensates for the nonfunctional gene.
The first clinical gene therapy (1990) was given to a 4-year-old girl with adenosine deaminase (ADA) deficiency.
This disorder is caused due to the deletion of the gene for adenosine deaminase (an enzyme crucial for the immune system to function).
This can be cured by bone marrow transplantation or by enzyme replacement therapy (injection of ADA). But these are not completely curative.
In gene therapy, lymphocytes from the patient’s blood are grown in a culture.
Then, a functional ADA cDNA (using a retroviral vector) is introduced into lymphocytes.
Then, they are returned to the patient. This should be periodically repeated as these cells are not immortal.
If the ADA gene (from marrow cells) is introduced into cells at early embryonic stages, it could be a permanent cure.

Biology Class 12 Notes For Neet

3. Molecular Diagnosis:

Early diagnosis of diseases is not possible using conventional methods (serum and urine analysis).
It is possible by techniques such as Recombinant DNA technology,

“application of biotechnology “

PCR & ELISA. PCR (Polymerase Chain Reaction):

The presence of a pathogen is normally suspected only based on symptoms.
By this time, the concentration of pathogens is already very high in the body.
However, very low concentrations of bacteria or viruses can be detected by amplification of their nucleic acid by PCR.

Uses of PCR:

To detect HIV in suspected AIDS patients.
To detect gene mutations in suspected cancer patients.
To identify many other genetic disorders.
A single-stranded DNA or RNA, tagged with a radioactive molecule (probe) is hybridized to its complementary DNA in a clone of cells followed by detection using autoradiography.
The clone having a mutated gene will not appear on the photographic film, because the probe will not have complementarity with the mutated gene.

NEET Biology Biotechnology and Its Applications Important Notes

ELISA (Enzyme-Linked Immuno-Sorbent Assay):

It is based on the principle of antigen-antibody interaction.
Infection by pathogens can be detected by the presence of antigens (proteins, glycoproteins, etc.) or by detecting the antibodies synthesized against the pathogen.

Biology Class 12 Notes For Neet

Transgenic Animals

These are the animals whose genome has been altered by the introduction of an extra (foreign) gene by manipulation.

Examples are transgenic rats, rabbits, pigs, sheep, cows, and fish.
Over 95% of the transgenic animals are mice.

Benefits of transgenic animals:

To study the regulation of genes and their action on normal Physiology & development:

For example, Study of complex factors such as insulin-like growth factor. Genes (from other species) that alter the formation of this factor are introduced and the biological effects are studied.
This gives information about the biological role of the factor in the body.

“biotechnology notes “

To Study the contribution of genes in the development of a disease and thereby new treatments:

For example, transgenic models for many human diseases such as cancer, cystic fibrosis, rheumatoid arthritis & Alzheimer’s.

Class 12 Biology Notes For Neet

Biological products:

Some medicines contain expensive biological products. Transgenic animals are used to produce useful biological products by introducing genes that code for a particular product.
For example, human protein (α-1-antitrypsin) is used to treat emphysema, products for the treatment of phenylketonuria (PKU) and cystic fibrosis, etc.
In 1997, Rosie (the first transgenic cow) produced human protein-enriched milk (2.4 gm per liter).
It contains the human α-lactalbumin and is a nutritionally more balanced product for human babies than natural cow milk.

Best Short Notes for Class 12 Biology Biotechnology and Its Applications

Vaccine safety testing:

Transgenic mice are used to test the safety of the polio vaccine.
If it is reliable, they can replace the use of monkeys to test the safety of vaccines.

Chemical safety testing (toxicity testing):

Some transgenic animals carry genes that make them more sensitive to toxic substances than non-transgenic animals.
They are exposed to the toxic substances and the effects studied. It gives immediate results.
The problem of unpredictable results: Genetic modification may cause unpredictable results.

“biotechnology applications “

Ethical Issues

The problem of unpredictable results: Genetic modification may cause unpredictable results.

Indian Government has set up organizations like GEAC (Genetic
Engineering Approval Committee) to make decisions about the validity of GM research and the safety of GM organisms for public services.

Problems of patent:

Certain companies have patents for products and technologies that make use of genetic materials, plants, etc.
That has been identified, developed, and used by farmers and indigenous people of a country. Examples, are Basmati rice, and herbal medicines (turmeric, neem, etc).

Class 12 Biology Notes For Neet

Basmati rice:

Basmati rice has a unique aroma & flavor. India has 27 varieties of Basmati.
In 1997, an American company got patent rights on Basmati rice through the US Patent and
Trademark Office. This allowed the company to sell a ‘new’ variety of Basmati. This was derived from
Indian farmer’s varieties. Indian Basmati was crossed with semi-dwarf varieties and claimed as a novelty.
Other people selling Basmati rice could be restricted by patent.

Biotechnology and Its Applications Class 12 NEET Key Concepts and Summary

Biopiracy:

It is the use of bio-resources by multinational companies and other organizations without proper authorization from the countries and people concerned.

Most of the industrialized nations are poor in biodiversity and traditional knowledge.
The developing and underdeveloped world has rich biodiversity and traditional knowledge related to bio-resources.
It has to develop laws to prevent unauthorized exploitation of bio-resources and traditional knowledge.
Indian Parliament has cleared the second amendment of the
Indian Patents Bill that has considered patent terms emergency provisions and research and development initiatives.

NEET Biology Class 12 Biotechnology Principles And Process Notes

February 14, 2025August 30, 2023 by Marksparks .arkansas

Biotechnology Principles And Process

Biotechnology is the technique of using live organisms or their enzymes for products & processes useful to humans.
The European Federation of Biotechnology (EFB)defines
Biotechnology as ‘the integration of natural science and organisms, cells, parts thereof, and molecular analogs for products and services’.

Biotechnology deals with:

Microbe-mediated processes (making curd, bread, wine etc).
In vitro fertilization (test-tube baby program).
Synthesis and use of a gene.
Preparation of DNA vaccine.
Correcting a defective gene.

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Principles Of Biotechnology

Core techniques of modern biotechnology:

Genetic engineering: The technique in which genetic material (DNA & RNA) is chemically altered and introduced into host organisms to change the phenotype.
Maintenance of sterile ambiance: It is necessary in chemical engineering processes for growing desired microbe/eukaryotic cell for the manufacture of antibiotics, vaccines, enzymes, etc.

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Basic steps in genetically modifying an organism:

Identification of DNA with desirable genes: Traditional hybridization techniques lead to the inclusion and multiplication of undesirable genes along with desired genes. Genetic engineering helps to isolate and introduce only desirable genes into the target organism.
Introduction of the identified DNA into the host: A vector DNA such as a plasmid is used to deliver an alien piece of DNA into the host organism.
Maintenance of introduced DNA in the host and transfer of the DNA to its progeny: A piece of alien DNA has no the sequence called Origin of replication (ori) needed for starting replication. So, it cannot multiply itself in the progeny cells of the organism. Hence alien DNA is integrated into the recipient genome (it has ori). It multiplies & inherits along with host DNA.

The first recombinant DNA (rDNA) was produced by Stanley Cohen & Herbert Boyer (1972). They isolated an antibiotic resistance gene by cutting out a DNA piece from a plasmid. This gene was linked with a native plasmid of Salmonella typhimurium.

Biology Class 12 Notes For Neet

Tools Of Recombinant DNA Technology

1. Restriction Enzymes (‘molecular scissors’):

These are the enzymes that cut DNA at specific sites into fragments.

They belong to a class of enzymes called nucleases. In 1963, two enzymes responsible for restricting the growth of bacteriophage in E. coli were isolated.
One enzyme added methyl groups to DNA. The other (restriction endonuclease) cut DNA.
More than 900 restriction enzymes have been isolated from over 230 strains of bacteria.

Naming of the restriction enzymes:

The first letter indicates genus and the second two letters indicate species of the prokaryotic cell from which they were isolated. for example, EcoRI comes from E. coli RY 13 (R = the strain.
Roman numbers = The order in which the enzymes were isolated from that strain of bacteria).

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Types of Restriction Enzymes:

Exonucleases: They remove nucleotides from the ends of the DNA.

Endonucleases: They cut at specific positions within the DNA.

They bind to specific recognition sequences of the DNA and cut the two strands at specific points.
The first restriction endonuclease is Hind II. It cuts DNA molecules by recognizing a specific sequence of 6 base pairs. This is called the recognition sequence for Hind II.
Restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA.
It is a sequence of base pairs that read the same on the two strands in the 5′ → 3′ direction and in the 3′ → 5′ direction.

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For example:

5′ —— GAATTC —— 3′

3′ —— CTTAAG —— 5′

NEET Biology Class 12 Biotechnology Principles And Processes Restriction Endonuclease Enzyme

Biotechnology Principles and Processes NEET Notes

Steps in the formation of recombinant DNA by action of restriction endonuclease enzyme EcoRI

Restriction enzymes cut the strand a little away from the center of the palindrome sites but between the same two bases on the opposite strands.
This leaves single-stranded overhanging stretches at the ends. They are called sticky ends. They form H-bonds with their complementary cut counterparts. This stickiness facilitates the action of the enzyme DNA ligase.
When cut by the same restriction enzyme, the resultant DNA fragments have the same kind of sticky-ends and these are joined together by DNA ligases.

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NEET Biology Class 12 Biotechnology Principles DNA Ligases

Separation and isolation of DNA fragments

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DNA fragments are separated by a technique called gel electrophoresis.

NEET Biology Class 12 Biotechnology Principles Gel Electrophoresis

DNA fragments can be seen as bright orange colored bands when they are stained with ethidium bromide and exposed to UV radiation.
DNA bands are cut out from agarose gel. This is called elution. These purified DNA are used to construct recombinant DNA by joining them with cloning vectors.

2. Cloning Vector

It is a DNA molecule that can carry a foreign DNA segment and replicate inside the host cells.

Plasmids, bacteriophages, etc.
Plasmids are autonomously replicating the circular extrachromosomal DNA of bacteria. Some plasmids have only
1-2 copies per cell. Others have 15-100 copies per cell.
Bacteriophages (high number per cell) have very high copy numbers of their genome within the bacterial cells.
When the cloning vectors are multiplied in the host, the
Linked piece of DNA is also multiplied to the numbers equal to the copy number of the vectors.

Features required for cloning into a vector:

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1. Origin of replication (ori):

This is a sequence where replication starts.
A piece of DNA linked to or can replicate within the host cells. This also controls the copy number of linked DNA.
So, to get many copies of the target DNA, it should be cloned in a vector whose origin support high copy number

NEET Biology Class 12 Biotechnology Principles And Process Notes

NEET Biology Biotechnology Principles and Processes Important Notes

2. Selectable marker (marker gene):

It is a gene that helps to select the transformants and eliminate the non-transformants.
Transformation is a procedure through which a piece of
DNA is introduced in a host bacterium. Such a bacterium is called a transformant. If transformation does not take place, it is non-transformant.
Selectable markers of E. coli include the genes encoding resistance to antibiotics like ampicillin, chloramphenicol, tetracycline, kanamycin etc. Normal E. coli cells have no resistance against these antibiotics.

3. Cloning sites:

To link the alien DNA, the vector needs a single or very few recognition sites for restriction enzymes.

More than one recognition site generates several fragments. It complicates gene cloning.
Ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic-resistant genes. ligation of foreign DNA at the Bam HI site of the tetracycline resistance gene in vector pBRr322.

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As a result, the recombinant plasmid is formed. If ligation does not occur, it is called a non-recombinant plasmid.

NEET Biology Class 12 Biotechnology Principles Non Recombinant Plasmid

Restriction sites: HindIII, EcoR I, BamH I, Sal I, Pvu II, Pst I, Cla I, ori
Antibiotic resistance genes: ampR andtetR.
Rop: codes for the proteins involved in the replication of plasmid.
The recombinant plasmids lose tetracycline resistance due to the insertion of foreign DNA.
When the plasmids are introduced into E. coli cells, 3 types of cells are obtained:
1. Non-transformants: They have no plasmid. So they are not resistant to either tetracycline or ampicillin.
2. Transformants with non-recombinant plasmid: They are resistant to both tetracycline & ampicillin.
3. Transformants with recombinant plasmid: They are resistant only to ampicillin.
Recombinant plasmids can be selected from nonrecombinant ones by plating transformants on an ampicillin medium. Then the transformants are transferred on a tetracycline medium.
The recombinants grow in an ampicillin medium but not in a tetracycline medium. However, non-recombinants grow on the medium containing both antibiotics.
Thus, one antibiotic resistance gene helps to select the transformants. The inactivated antibiotic resistance gene helps to select recombinants.
Selection of recombinants due to the inactivation of antibiotics requires simultaneous plating on 2 plates having different antibiotics.
Therefore, alternative selectable markers have developed to differentiate recombinants from nonrecombinants based on their ability to produce color in the presence of a chromogenic substrate.
In this, a recombinant DNA is inserted within the coding sequence of an enzyme, β-galactosidase. So, the enzyme is inactivated. It is called insertional inactivation.
Such colonies do not produce any color. These are identified as recombinant colonies.
If the plasmid in bacteria has no insert, it gives blue-colored colonies in the presence of a chromogenic substrate.

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4. Vectors for cloning genes in plants & animals:

Genetic tools of some pathogens can be transformed into useful vectors for delivering genes to plants & animals. For example,
Agrobacterium tumefacient (a pathogen of many dicot plants) can deliver a piece of DNA (T-DNA) to transform normal plant cells into a tumor. These tumor cells produce the chemicals required by the pathogen.
The tumor-inducing (Ti) plasmid of A. tumefacient is modified into a cloning vector that is not pathogenic to the plants but is able to use the mechanisms to deliver genes of interest to plants.
Retroviruses in animals can transform normal cells into cancerous cells. So, they are used to deliver desirable genes into animal cells.

3. Competent Host (For Transformation with Recombinant DNA)

DNA is a hydrophilic molecule. So, it cannot pass through cell membranes.
To avoid this problem, bacterial cells are treated with a specific concentration of a divalent cation (e.g. calcium).
So, DNA enters the bacterium through pores in the cell wall.
Such cells are incubated with recombinant DNA on ice.
Then they are placed briefly at 420C (heat shock) and put them back on ice.
This enables the bacteria to take up recombinant DNA

Other methods to introduce alien DNA into host cells:

Micro-injection: In this, recombinant DNA is directly injected into the nucleus of an animal cell.
Biolistics (gene gun): In this, cells are bombarded with high-velocity micro-particles of gold or tungsten coated with DNA. This method is suitable for plants.
‘Disarmed pathogen’ vectors: They infect the cell and transfer the recombinant DNA into the host.

Class 12 Biology Notes For Neet

Processes Of Recombinant DNA Technology

1. Isolation of the Genetic Material (DNA):

The bacterial cells/plant or animal tissue are treated with enzymes like lysozyme (bacteria), cellulase (plants), chitinase (fungus) etc.
The cell is broken releasing DNA & other macromolecules (RNA, proteins, polysaccharides and lipids).
RNA is removed by treating it with ribonuclease. Proteins are removed by treatment with protease. Other molecules are removed by appropriate treatments.
When chilled ethanol is added, purified DNA precipitates out as a collection of fine threads in the suspension.

2. Cutting of DNA at Specific Locations:

Purified DNA is incubated with the restriction enzyme at optimal conditions. As a result, DNA digests.
Agarose gel electrophoresis is employed to check the progression of a restriction enzyme digestion.
DNA is negatively charged. So it moves towards the anode. The DNA fragments separate according to their size through the sieving effect of the agarose gel (a polymer extracted from seaweeds).
The smaller fragment moves farther. The process is repeated with the vector DNA. After cutting the source DNA and vector DNA, the cut-out gene of interest from the source.
DNA and cut vector are mixed and ligase is added. It creates recombinant DNA.

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3. Amplification of Gene of Interest using PCR:

Polymerase Chain Reaction (PCR) is the synthesis of multiple copies of the gene of interest in vitro using 2 sets of primers & the enzyme DNA polymerase.
Primers are small chemically synthesized oligonucleotides that are complementary to the regions of DNA.

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Steps of PCR:

Denaturation: It is the heating of target DNA (gene of interest) at a high temperature (940 C) to separate the strands.
Each strand acts as a template for DNA synthesis.
Annealing: It is the joining of the two primers (at 520 C) at the 3’ end of the DNA templates.
Extension: It is the addition of nucleotides to the primer with the help of a thermostable DNA polymerase called
Taq polymerase:
- It is isolated from a bacterium, Thermus aquaticus. It remains active in high temperatures during the denaturation of double-stranded
- DNA. Through continuous replication, the DNA segment is amplified up to 1 billion copies.

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The amplified fragment can be used to ligate with a vector for further cloning.

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Biotechnology Principles and Processes Class 12 NEET Key Concepts and Summary

4. Insertion of Recombinant DNA into Host Cell:

Using any method, the ligated DNA is introduced into recipient cells. They take up DNA from its surroundings.
If a recombinant DNA-bearing ampicillin resistant gene is transferred into E. coli cells, the host cells become ampicillin-resistant cells.
If the transformed cells are spread on agar plates containing ampicillin, only transformants will grow.
Untransformed recipient cells will die.

5. Obtaining the Foreign Gene Product:

The aim of recombinant DNA technology is to produce a desirable protein.
If a protein-encoding a foreign gene is expressed in a heterologous host, it is called a recombinant protein.
The cells with foreign genes can be grown in the laboratory.
The cultures are used to extract the desired protein and purify it by using separation techniques. The cells can also be multiplied in a continuous culture system.
Here, the used medium is drained out from one side while the fresh medium is added from the other.
It maintains the cells more physiologically active and so produces a larger biomass. It yields more desired protein.

“biotechnology processes “

Bioreactors:

These are the vessels in which raw materials are biologically converted to specific products, enzymes, etc., using microbial plant, animal, or human cells.
Bioreactors are used to produce large quantities of products. They can process 100-1000 liters of culture.
A bioreactor provides the optimal growth conditions (pH, temperature, substrate, salts, vitamins, oxygen) for achieving the desired product.
The most commonly used bioreactors are of stirring type (stirred-tank reactor).

NEET Biology Class 12 Biotechnology Principles and Processes Notes

NEET Biology Class 12 Biotechnology Principles Stirred Tank Reactor

NEET Biology Class 12 Chapter Biotechnology Principles and Processes Detailed Notes

It is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents. The stirrer facilitates even mixing and oxygen availability. Alternatively, air can be bubbled through the reactor.

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The bioreactor has:

An agitator system
An oxygen delivery system
A foam control system
A temperature control system
pH control system
Sampling ports (for periodic withdrawal of the culture).

6. Downstream Processing:

It is a series of processes such as separation and purification of products after the biosynthetic stage.
The product is formulated with suitable preservatives.
Such formulation undergoes thorough clinical trials and strict quality control testing.

NEET Biology Class 12 Human Health And Diseases Notes

February 14, 2025August 30, 2023 by Marksparks .arkansas

Human Health And Diseases

Health:

Health is a state of complete physical, mental and social well-being. Health is affected by genetic disorders, infections, change in life style (food, water, rest, exercise, habits etc). Mind influences the immune system (through neural and endocrine systems).

When the functioning of organs or systems of the body is adversely affected, it is called a disease.
Diseases may be infectious (transmitted from one person to another) or non-infectious.
Disease-causing organisms are called Pathogens. Parasites are pathogens as they harm the host

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Common Infectious Diseases In Man

1. Bacterial Diseases:

Typhoid: Pathogen is Salmonella typhi.
- Mode of transmission: It enters small intestine through food and water and migrates to other organs via blood.
- Symptoms: Sustained high fever (39o-40o C), weakness, stomach pain, constipation, headache & loss of appetite. Intestinal perforation and death may occur. Widal test is used for confirmation of the disease.
Pneumonia: Pathogen is Streptococcus pneumoniae and
- Haemophilus influenzae: It infects lung alveoli. The alveoli get filled with fluid leading to respiratory problems.
- Mode of transmission: Inhaling the droplets/aerosols released by an infected person. Sharing glasses and utensils with an infected person.
- Symptoms: Respiratory problems, fever, chills, cough, headache. In severe cases, lips and fingernails turn grey to bluish colour.

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2. Viral Diseases:

Common cold: The Pathogen is Rhinoviruses. It infects the nose and respiratory passage but not the lungs.
- Mode of transmission: Inhaling droplets resulting from cough or sneezes. Through contaminated objects.
- Symptoms: Nasal congestion and discharge, sore throat, cough, hoarseness, headache, tiredness etc. Last for 3-7 days.

3. Protozoan Diseases:

Malaria: Pathogen is Plasmodium sp. (P. vivax, P.malariae and P. falciparum). Most serious (malignant) malaria is caused by P. falciparum.
- Mode of transmission: By female Anopheles mosquito.
- Symptoms: Haemozoin (a toxin released by Plasmodium) causes chill and high fever recurring every 3-4 days
Amoebiasis (Amoebic dysentery): Pathogen is Entamoeba histolytica
- Mode of transmission: Houseflies (mechanical carriers) transmit parasites from faeces to food & water.
- Symptoms: Constipation, abdominal pain and cramps, stools with excess mucous and blood clots.

Human Health And Disease Class 12 Disease Chart

Life Cycle Of Plasmodium:

NEET Biology Class 12 Human Health And Diseases Life Cycle Of Plasmodium

Class 12 Biology Notes For Neet

4. Helminth Diseases:

Ascariasis: Pathogen is Ascaris (Intestinal parasite).
- Mode of transmission: Soil, water, vegetables, fruits etc.
- contaminated with faeces containing eggs of parasites.
- Symptoms: Internal bleeding, muscular pain, fever, anaemia and blockage of intestinal passage.
Filariasis (Elephantiasis): Pathogen is Filarial worms or Wuchereria (W. bancrofti & W. malayi).
Mode of transmission: Bite of female Culex mosquito.
Symptoms: Filarial worms live in lymphatic vessels (usually of lower limbs). It causes chronic inflammation of the organs in which they live for many years. Limbs and genital organs may be deformed.

NEET Biology Class 12 Human Health and Diseases Notes

5. Fungal Diseases:

Ringworms: Pathogens are Microsporum, Trichophyton and epidermophyton. They are seen in the groin, b/w toes etc.
- Mode of transmission: From soil or by using towels, cloths, comb etc. Heat and moisture help fungi to grow.
- Symptoms: Dry, scaly lesions on skin, nails, scalp etc. Intense itching.

Human Health and Diseases NEET Notes

Prevention And Control Of Diseases

Personal hygiene: Keep the body clean. Use clean drinking water, food etc.

Public hygiene: Proper disposal of wastes and excreta.

Periodic cleaning and disinfection of water reservoirs, pools, cesspools and tanks.
Avoid contact with infected persons or their belongings (to control air-borne diseases).
Standard practices of hygiene in public catering.
Control and eliminate the vectors (For example, Mosquitoes).
- Avoid stagnation of water.
- Regular cleaning of household coolers.
- Use of mosquito nets.
- Introduce larvivorous fishes like Gambusia in ponds.
- Spraying insecticides in ditches, drainage and swamps.
- Provide doors and windows with wire mesh.

These precautions can avoid vector-borne diseases like Malaria, Filariasis, Dengue and Chikungunya.

Class 12 Biology Notes For Neet

Immune System

It is the system that gives immunity to the body.
It plays a role in allergic reactions, auto-immune disease and organ transplantation.
It includes lymphoid organs, tissues, cells & antibodies.

“human health and disease class 12 short notes “

Lymphoid Organs

These are the organs where the origin, maturation & proliferation of lymphocytes occur. It is 2 types

Primary and
Secondary.

1. Primary lymphoid organs:

Class 12 Biology Human Health And Disease Notes

Here, immature lymphocytes differentiate into antigensensitive lymphocytes. E.g. Bone marrow & thymus.
Bone marrow is the site of the formation of blood cells.
The thymus is large during birth but gradually reduces in size and becomes very small size in puberty.

2. Secondary lymphoid organs:

The organs, to which matured lymphocytes migrate, interact with antigens and then proliferate to become effector cells.
Examples are the spleen, lymph nodes, tonsils, Peyer’s patches, Mucosal associated lymphoid tissue (MALT) and appendix.
Spleen: Bean-shaped organ. Contains lymphocytes and phagocytes. It removes worn-out RBCs & microorganisms from blood. It is a reservoir of erythrocytes in the foetus.
Lymph nodes: Found in the lymphatic system. They trap microorganisms or other antigens. Trapped antigens activate lymphocytes and cause an immune response.
MALT: Located within the lining of respiratory, digestive & urinogenital tracts. It constitutes 50% of lymphoid tissue.

Class 12 Biology Notes For Neet

Immunity

It is the ability of the immune system to fight the pathogens. It is 2 types

Innate and
Acquired.

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1. Innate (inborn) immunity: It is the non-specific immunity present at the time of birth.

It includes 4 types of Barriers:

Physical barriers: Example, Skin (Prevent entry of foreign bodies), Mucus coating of the respiratory, gastrointestinal and urogenital tracts to trap microbes.
Physiological barriers: For example, gastric HCl, saliva, tears etc.
Cellular barriers: Phagocytes like WBC [For example, Neutrophils or Polymorphonuclear leukocytes (PMNL), monocytes and natural killer lymphocytes], macrophages etc.
Cytokine barriers: Virus-infected cells secrete proteins called interferon which protect non-infected cells from further viral infection.

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2. Acquired immunity: It is pathogen-specific immunity developed during a lifetime.

It is characterized by memory, i.e. during the first encounter with a pathogen, the body produces a primary response in low intensity.
A second encounter of the same pathogen causes a secondary (anamnestic) response in high intensity.
Primary and secondary immune responses are carried out with B-lymphocytes (B-cells) and T-lymphocytes (T-cells).
- B-lymphocytes: Produce antibodies.
- T-lymphocytes: Help B-cells to produce antibodies

Human Health And Disease Class 12 Disease Chart

NEET Biology Class 12 Human Health And Diseases Structure Of An Antibody Molecule

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Structure of an antibody molecule: An antibody has 4 polypeptide chains: 2 light chains and 2 heavy chains (H₂L₂).
Types of antibodies: IgG, IgA, IgM, IgE and IgD.

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Types of Acquired Immune Response

Humoral immune response/Antibody-mediated immunity (AMI): It is the immune response mediated by antibodies. Antibodies are found in blood plasma. So called a Humoral immune response.

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Cell-mediated response / cell-mediated immunity (CMI):

It is the immune response mediated by T-lymphocytes (T cells).
The body can differentiate ‘self’ and ‘non-self’ and the CMI causes Graft rejection.
Tissue matching & blood group matching are essential before undertaking any graft/ transplant.
After this, the patient should take immuno-suppressants all his life.

Types of Acquired Immunity: Acquired immunity is 2 types

Active and
Passive.

1. Active immunity:

It is the immunity in which antibodies are produced in a host body when the host is exposed to antigens (for example, living or dead microbes or other proteins).

Biology Class 12 Notes For Neet

It is a slow process. It is produced in 2 ways:

Natural Active Immunity: It is developed during natural infection by microbes.
Artificial Active Immunity: It is developed by injecting the microbes deliberately during immunization.

2. Passive immunity: Here, readymade antibodies are directly given to the body.

It is 2 types:

Natural Passive Immunity: For example
- Antibodies (IgG) from mother → Placenta → Foetus
- Antibodies (IgA) in colostrum → infants
Artificial Passive Immunity: For example, Anti-tetanus serum (ATS)

Class 12 Biology Human Health And Disease Notes

Immunization

This is based on the ‘memory’ of the immune system. 2 types:

1. Active Immunization (Vaccination):

In this, the preparation of vaccine (antigenic proteins of pathogen or inactivated pathogen) is introduced into the body. It results in the development of antibodies.

During actual infection, the antibodies neutralize antigens.
The vaccines also generate memory B and T-cells. They recognize the pathogen quickly.
Example, Polio vaccine, Hepatitis B vaccine, DPT vaccine etc.
Vaccines are produced using DNA recombinant technology (For example, Hepatitis B vaccine produced from Yeast).

2. Passive Immunization:

It is the direct injection of pre-formed antibodies or antitoxin. It requires for quick immune response.
For example, Immunization against Tetanus, snake venom etc

Class 12 Biology Human Health And Disease Notes

Allergies

It is the exaggerated response of the immune system to certain antigens present in the environment.
Allergens: Substances causing allergy. E.g. mites in dust, pollens, animal dander, fur etc.
Antibodies produced against the allergens are of IgE type.
Allergy is due to the release of chemicals like histamine and serotonin from the mast cells.
Symptoms: Sneezing, watery eyes, running nose, difficulty
in breathing, skin rashes etc.
Determination of cause of allergy: The patient is exposed to or injected with very small doses of possible allergens, and the reactions are studied.
Treatment: Drugs like anti-histamines, adrenaline and steroids quickly reduce the symptoms of allergy.
Asthma: Asthma is a respiratory disease due to allergy.
Modern-day lifestyle results lowering of immunity and more sensitivity to allergens.

Autoimmunity

It is the condition in which the body attacks self-cells due to genetic and other unknown reasons.
It leads to auto-immune disease. For example, Rheumatoid arthritis.

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AIDS (Acquired Immuno Deficiency Syndrome):

It is a deficiency of the immune system.
It is caused by HIV (Human Immunodeficiency Virus), a
retrovirus having RNA genome.
AIDS was first reported in America (1981).

Class 12 Biology Human Health And Disease Notes

1. Transmission:

Sexual contact with an infected person.
Transfusion of contaminated blood & blood products.
Sharing of infected needles.
From the infected mother to her child through the placenta.

NEET Biology Class 12 Chapter Human Health and Diseases Detailed Notes

2. High-risk people of getting HIV:

Individuals with multiple sexual partners.
Drug addicts who take drugs intravenously.
Individuals who require repeated blood transfusion.
Children born to an HIV-infected mother.

3. HIV does not spread by touch or physical contact. It spreads only through body fluids.

4. There is a time lag (from a few months to 5-10 years) between the infection and the appearance of symptoms.

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5. Replication of retrovirus:

NEET Biology Class 12 Human Health And Diseases Replication Of Retrovirus

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6. Life cycle:

HIV enters the body → To macrophages (acts as HIV factory) → RNA genome replicates in the presence of Reverse transcriptase to form viral DNA → Viral DNA incorporates into host DNA → Infected cells produce virus particles → HIV enters into helper T-cells (TH) →Replicates & produce progeny viruses → Attack other helper T-cells → T-cells decrease → Weaken immunity.

7. HIV infected person may be infected with Mycobacterium, viruses, fungi and parasites like Toxoplasma.

8. Diagnosis: ELISA test (Enzyme-linked immuno-sorbent Assay).

9. Treatment: Anti-viral drugs are partially effective. They can only prolong the life of the patient.

Human Health And Disease Class 12 Notes Neet

10. Prevention of AIDS:

Educate people about AIDS.
Making blood (from blood banks) safe from HIV.
Use of disposable needles and syringes.
Advocating safe sex and free distribution of condoms.
Controlling drug abuse.
Regular check-ups for HIV in susceptible populations.

NEET Study Material for Human Health and Diseases Chapter

Cancer

Cancer is an abnormal and uncontrolled multiplication of cells resulting in the formation of tumours (masses of cells).
Normal cells show contact inhibition (contact with the other cells inhibits their uncontrolled growth). Cancer cells do not have this property.

Types of Tumours:

Benign tumours: Confined to the place of their origin. They do not spread to other parts. Cause little damage.
Malignant tumours: Mass of proliferating cells (neoplastic or tumour cells) that grow rapidly, invade and damage the surrounding normal tissues. Due to active division and growth, they starve normal cells by competing for nutrients.

Cells sloughed from tumours reach other sites via blood where they form a new tumour. This is called metastasis.

Class 12 Biology Human Health And Disease Notes

Cancer Causes (Carcinogens):

Physical agents: Example, Ionizing radiations like X-rays and gamma rays and non-ionizing radiations like UV.
Chemical agents: Tobacco smoke (a major cause of lung cancer), vinyl chloride, caffeine, nicotine, mustard gas etc.
Biological agents: For example, oncogenic viruses, c-onc (cellular oncogenes or proto-oncogenes) etc. When C-onc in normal cells is activated, the cells become oncogenic.

Cancer detection and diagnosis:

Human Health And Disease Class 12 Notes Neet

Biopsy: A thin piece of the suspected tissue is stained and examined under a microscope (histopathological studies).
In case of leukaemia: Biopsy & histopathological studies.
Blood & bone marrow tests for increased cell counts.
Radiography (use of X-rays): CT (Computerized tomography) scan and MRI (Magnetic Resonance Imaging).
Use of Antibodies against cancer-specific antigens.
Molecular biology technique: To detect cancer-related genes. Such individuals should avoid carcinogens (for example, tobacco smoke).

Cancer Treatment:

Radiotherapy: Tumour cells are irradiated lethally, without damaging surrounding normal tissues.
Chemotherapy: Use of chemotherapeutic drugs. Many drugs have side effects like hair loss, anaemia etc.
Immunotherapy: The patients are given biological response modifiers (for example, αinterferon) which activate their immune system and help in destroying the tumour.
Surgery.
Most cancers are treated by a combination of surgery, radiotherapy and chemotherapy

Human Health And Disease Class 12 Notes Neet

Drugs Smoking And Alcohol Abuse

Drugs (opioids, cannabinoids & coca alkaloids):

Opioids

They bind to specific opioid receptors in the, for example, CNS and gastrointestinal tract. morphine, heroin, brown sugar.
Morphine: Morphine is extracted from the latex of the poppy plant, Papaver somniferum. It is a sedative and painkiller, and useful for surgery.

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NEET Biology Class 12 Human Health And Diseases Chemical Structure Of Morphine

Human Health and Diseases Class 12 NCERT Notes for NEET

Heroin: Heroin(smack or diacetylmorphine) is a white, odourless, bitter crystalline compound. It is obtained by acetylation of morphine. It is taken by snorting and injection. Heroin is a depressant and slows down body functions.

Human Health And Disease Notes Class 12th

Cannabinoids

They interact with cannabinoid receptors in the brain.
Generally taken by inhalation and oral ingestion.
Natural cannabinoids are obtained from inflorescences of Cannabis sativa (Hemp plant). Its flower tops, leaves & resin are used to make marijuana, hashish, charas & ganja.
They affect the cardiovascular system.
Cannabinoids are abused by some sportspersons.

NEET Biology Class 12 Human Health And Diseases Skeletal Structure Of Cannabinoid Molecule

Coca alkaloid or cocaine (coke or crack)

It is obtained from the coca plant Erythroxylum coca.
It interferes with the transport of the neurotransmitter dopamine.
Cocaine is usually snorted.
It stimulates CNS producing euphoria & increased energy.
Excessive dosage of cocaine causes hallucinations.
Atropa belladonna & Datura are also hallucinogenic plants.

Human Health And Disease Notes Class 12th

Drugs like barbiturates, amphetamines, benzodiazepines, etc. are used as medicines to treat mental illnesses like depression and insomnia. But their abuse results in impairment of physical, physiological or psychological functions.

Smoking

Tobacco is smoked, chewed or used as a snuff.
Tobacco contains nicotine (an alkaloid). It stimulates the adrenal gland to release adrenaline and nor-adrenaline, causing high BP and heart rate.
Smoking causes cancers of the lungs, urinary bladder and throat, bronchitis, emphysema, coronary heart disease, gastric ulcer etc. Tobacco chewing causes oral cancer.
Smoking increases CO₂ content in blood and reduces oxyhaemoglobin. This causes O₂ deficiency in the body.

Adolescence

Adolescence is a period’ and ‘a process’ during which a child becomes mature in terms of his/her attitudes and beliefs for effective participation in society.
Adolescence is a bridge linking childhood and adulthood (period of 12-18 years of age). It is very vulnerable phase of mental and psychological development.

Causes of drug/alcohol use in Adolescence:

Curiosity and Experimentation.
Need for adventure and excitement.
To escape facing problems.
Stress from pressure to excel in academics or examinations.
Television, movies, newspapers, internet etc.
Unstable or unsupportive family structures & peer pressure.

Human Health And Disease Notes Class 12th

Addiction and Dependence

Addiction:

It is a psychological attachment (euphoria and a temporary feeling of well-being) with drugs and alcohol.
With repeated use of drugs, the tolerance level of the receptors increases. Thus the receptors respond only to higher doses leading to greater intake and addiction.

Immunity, Vaccination, and Common Diseases NEET Notes

Dependence:

It is the tendency of the body to manifest a characteristic and unpleasant withdrawal syndrome if the regular dose of drugs/alcohol is abruptly discontinued.
This results in anxiety, shakiness, nausea and sweating.
Dependence leads to social adjustment problems.

“human health and welfare “

Effects of Drug/alcohol abuse:

Reckless behaviour, vandalism and violence.
Coma and death due to respiratory failure, heart failure or cerebral haemorrhage.
Drugs in combination with alcohol may lead to death.
Damage of the nervous system and liver cirrhosis.
Mental and social distress to family and friends.
Social problems like stealing and the spread of infectious diseases (for example, AIDS, hepatitis B).
The use of drugs and alcohol by pregnant women affects the foetus (Foetal alcohol syndrome or FAS).
Loss of sexual drive and necrospermia.
Misuse of drugs by athletes (for example, Narcotic analgesics, anabolic steroids, diuretics and certain hormones to increase muscle strength and bulk and to promote aggressiveness).

Warning signs of drug/alcohol abuse in the Adolescence period:

Drop in academic performance and absence from school.
Lack of interest in personal hygiene.
Withdrawal and isolation.
Depression, fatigue, aggressive and rebellious behaviour.
Change in sleeping and eating habits.
Fluctuations in weight, appetite etc.
Loss of interest in hobbies.
Deteriorating relationships with family and friends.

Side effects of anabolic steroid abuse:

Human Health And Disease Notes Class 12th

1. In males:

Acne.
Mood swings & depression.
Increased aggressiveness.
Reduced testicles.
Decreased sperm.
Kidney & liver dysfunction.
Breast enlargement.
Premature baldness
Enlargement of prostate gland.

2. In females:

Masculinisation
Mood swings & depression
Increased aggressiveness
Excessive hair growth
Abnormal menstrual cycle
Deepening of voice
Enlargement of clitoris

“human health and welfare “

In adolescent males & females: Severe facial and body acne, premature closure of the growth centres of the long bones resulting in stunted growth.

Prevention and control:

Avoid undue peer pressure.
Education and counselling.
Seeking help from parents and peers.
Looking for danger signs.
Seeking professional and medical help.
- Psychologists and psychiatrists.
- De-addiction and rehabilitation programs.

NEET Biology Class 12 Strategies For Enhancement In Food Production Notes

February 14, 2025August 30, 2023 by Marksparks .arkansas

Strategies For Enhancement In Food Production

Animal Husbandry

It is the scientific agricultural practice of breeding and raising livestock.
It deals with the care & breeding of livestock (buffaloes,
cows, pigs, horses, cattle, sheep, camels, goats etc.)
poultry farming and fisheries.
More than 70% of the world’s livestock population is in
India & China. However, the contribution to the world farm produce is only 25%, i.e., the productivity per unit is very low.
Hence new technologies should be applied to achieve improvement in quality and productivity.

Read And Learn More: NEET Biology Class 12 Notes

Management of Farms And Farm Animals

1. Dairy Farm Management (Dairying):

It is the management of animals to increase the yield and quality of milk and its products.
Milk yield depends on the quality of breeds on the farm.
It is important to select good breeds that have high-yielding potential and resistance to diseases.

Ways for the yield potential:

Look after the cattle (housing well, giving adequate water, and maintaining disease-free).
Feeding of cattle in a scientific manner – emphasis on the quality and quantity of fodder.
Stringent cleanliness and hygiene of cattle & handlers while milking, storage, and transport of the milk.
Nowadays, these processes have been mechanized. It reduces the chance of direct contact of the produce with the handler.
Tensure these stringent measures there should be
Regular inspections identify and rectify problems.
Regular visits by a veterinary doctor.

NEET Biology Class 12 Strategies For Enhancement In Food Production Notes

Class 12 Biology Notes For Neet

2. Poultry Farm Management:

Poultry is a domesticated bird used for food or eggs. Examples are chicken, ducks, turkey, and geese.

Components of poultry farm management:

Selection of disease-free and suitable breeds.
Proper and safe farm conditions.
Proper feed and water.
Hygiene and health care.

class 12 biology chapter 9 notes

Animal Breeding:

A breed is a group of animals related by descent and similar general appearance, features, size, etc.
Breeding is the modification of the genotype of an organism to make that organism more useful to humans. Example, Jersy
(improved cattle breed), and Leghorn (improved chicken breed).
Animal breeding aims at increasing the yield of animals and improving the desirable qualities of the produce.

Class 12 Biology Notes For Neet

Breeding is 2 types:

Inbreeding and
Out-breeding

NEET Biology Class 12 Strategies for Enhancement in Food Production Notes

1. Inbreeding:

It is the mating of more closely related individuals within the
same breed for 4-6 generations. This strategy is as follows:
Identify and mate superior males & females of the same breed
Evaluate the progeny obtained and identify superior males and females among them for further mating.
In cattle, a superior female produces more milk per lactation.
A superior male (bull) gives rise to superior progeny.

Inbreeding Advantages:

It increases homozygosity to a pure-line animal. It exposes harmful recessive genes that are eliminated by selection.
It helps in the accumulation of superior genes and the elimination of less desirable genes. This increases the productivity of the inbred population.
Continued inbreeding, especially close inbreeding, may reduce fertility and productivity. This is called inbreeding depression.
To solve this problem, selected animals should be mated with unrelated superior animals of the same breed.

Class 12 Biology Notes For Neet

2. Out-breeding:

It is the breeding of the unrelated animals. It includes outcrossing, cross-breeding, and inter-specific hybridization.

Out-crossing
Cross – Breeding:
- This is the mating of animals within the same breed, but having common ancestors on either side of their pedigree up to generations.
- The offspring of such a mating is known as out-cross.
- It is the best method for animals having low productivity in milk production, growth rate in beef cattle, etc.
- It helps overcome inbreeding depression.
- In this method, superior males of one breed are mated with superior females of another breed.
- The desirable qualities of 2 different breeds are combined.
- The progeny hybrid animals may be used for commercial production or may be subjected to inbreeding and selection to develop new stable superior breeds.
- For example, Hisardale (sheep) developed in Punjab by crossing
- Bikaneri ewes and Marinrams.

Strategies for Enhancement in Food Production NEET Notes

Interspecific hybridization:
- It is the mating of males and females of two different species.
- In some cases, the progeny may combine desirable features of both parents and may be of considerable economic value. For example, Mule (male ass X female horse).

“class 12 biology ch 9 notes “

Controlled Breeding Experiments

1. Artificial insemination:

The semen collected from a male parent is injected into the reproductive tract of the selected female by the breeder.
Semen is used immediately or is frozen and used later.
Frozen semen can also be transported.
The success rate of crossing mature male & female animals is low even though artificial insemination is carried out.

Class 12 Biology Notes For Neet

2. Multiple Ovulation EmbryTransfer Technology (MOET)”

It is a program for herd improvement. It improves 47 chances of successful production of hybrids.

In this, a cow is administered hormones, with FSH-like activity, to induce follicular maturation and superovulation (production of 6-8 eggs per cycle instead of one egg).
The animal is either mated with an elite bull or artificially inseminated. Fertilized eggs at 8–32 cell stages are recovered and transferred to surrogate mothers.
MOET has been demonstrated for cattle, sheep, rabbits, buffaloes, mares, etc.
High milk-yielding breeds of females and high-quality (lean meat with less lipid) meat-yielding bulls have been bred successfully to increase herd size in a short time.

Class 12 Biology Notes For Neet

Bee-Keeping (Apiculture)

It is the maintenance of hives of honeybees to produce honey and beeswax.
The most common species that can be reared is Apis indica.
Honey is a food of high nutritive and medicinal value.
Beeswax is used in the preparation of cosmetics, polishes, etc.
Apiculture can be practiced in an area having bee pastures of some wild shrubs, fruit orchards, and cultivated crops.
Important points for successful bee-keeping:
- Knowledge of the nature and habits of bees.
- Selection of suitable location for keeping beehives.
- Catching and hiving of swarms (group of bees).
- Management of beehives during different seasons
- Handling and collection of honey and of beeswax.
- Bees are the pollinators of crop species such as sunflower,
- Brassica, apple, and pear.
Keeping beehives in crop fields during the flowering period increases pollination. It improves crop and honey yield.

Strategies For Enhancement In Food Production Notes

Fisheries

The fishery is an industry of catching, processing, or selling fish, shellfish, or other aquatic animals (prawn, crab, lobster, edible oyster, etc.).

Freshwater fishes: Catla, Rohu, common carp, etc.
Marine fishes: Hilsa, Sardines, Mackerel, Pomfrets etc.
Fisheries provide income and employment to millions of fishermen and farmers.
Aquaculture: Aquaculture (farming of aquatic organisms) & pisciculture (farming of fishes) are the techniques to increase the production of aquatic plants and animals. Blue
Revolution: The development and flourishing of the fishery industry.

“what is breed in biology “

Plant Breeding

It is the manipulation of plant species to create desired plant types suitable for better cultivation, better yields, and disease resistance.

Green Revolution: The development and flourishing of agriculture. It was dependent on plant breeding.

Class 12 Strategies For Enhancement In Food Production

Classical plant breeding:

Classical plant breeding involves the hybridization of pure lines and artificial selection to produce desirable traits.
Now molecular genetic tools are used for plant breeding.

Desirable traits that breeders have tried to incorporate:

Increased crop yield.
Improved quality.
Increased tolerance of to environmental stresses (salinity, extreme temperatures & drought), and resistance to pathogens.
Increased tolerance of insect pests.

NEET Biology Strategies for Enhancement in Food Production Important Notes

Steps of Plant Breeding

1. Collection of genetic variability:

In wild relatives of many crops, pre-existing genetic
variability is available.
Collection and preservation of wild varieties, species, and relatives of the cultivated species is a prerequisite for effective exploitation of natural genes.
The entire collection of plants/seeds having all the alleles for all genes in a given crop is called germplasm collection.

2. Evaluation and selection of parents:

The germplasm is evaluated for identifying plants with desirable characteristics.
Selected plants are multiplied and used for hybridization.
Pure lines are created wherever desirable and possible.

Class 12 Strategies For Enhancement In Food Production

3. Cross-hybridization of the selected parents:

It is the process in which desired characters are genetically combined from 2 different parents to produce a hybrid plant
For example, the high protein quality of one parent is combined with disease resistance from another parent.

Cross-hybridization Limitations:

Very time-consuming and tedious process.
Hybrids may not combine the desirable characteristics.
Usually, only hundreds ta thousand crosses show the desirable combination.

“what is totipotency class 12 “

4. Selection & testing of superior recombinants

It is crucial to the success of the breeding objective and requires careful scientific evaluation of the progeny.
It yields plants that are superior to both parents.
These are self-pollinated for several generations till they reach a state of uniformity (homozygosity), that the characters will not segregate in the progeny.

5. Testing, release & commercialization:

The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance, etc.
This is done by growing them in the research fields and recording their performance under ideal fertilizer application irrigation and other crop management practices.
The evaluation is followed by testing the materials in farmers’ fields, for at least 3 growing seasons at several
Locations in the country, representing all the agro-climatic zones. The material is evaluated in comparison to the best available local crop cultivar (a check or reference cultivar).

Class 12 Strategies For Enhancement In Food Production

Wheat and Rice:

In India, food production has increased by the development of high-yielding varieties of wheat and rice in the mid-
1960s (Green Revolution).
From 1960 to 2000, wheat production increased from 11 million tons to t75 million tons. The rice production increased from 35 million tons to t89.5 million tons.
Nobel laureate Norman E. Borlaug (International Centre for Wheat & Maize Improvement, Mexico) developed semi-dwarf wheat.
In 1963, high-yielding and disease-resistant wheat varieties like Sonalika & Kalyan Sona were introduced in India.
Semi-dwarf rice varieties were derived from IR-8, (developed at International Rice Research Institute (IRRI),
Philippines) and Taichung Native-1 (from Taiwan). Later better-yielding semi dwarf varieties Jaya and Ratna were developed in India.

Sugar cane:

Saccharum barberi (grown in north India, but with poor sugar content & yield) was crossed with Saccharum officinarum (tropical canes in south India, thicker stems and higher sugar content but not grow well in north India) and got a hybrid sugar cane having desirable qualities like high yield, thick stems, high sugar, and ability grow in north India.

Strategies for Enhancement in Food Production Class 12 NEET Key Concepts and Summary

Millets:

Class 12 Strategies For Enhancement In Food Production

Hybrid maize, jowar & bajra developed in India. It includes high-yielding varieties resistant to water stress.

Plant Breeding for Disease Resistance

It enhances food production and helps to reduce the use of fungicides and bactericides.
Resistance of the host plant is the genetic ability to prevent the pathogens from disease.

Some plant diseases:

Fungal: Rusts. for example, brown rust of wheat, red rot of sugarcane and late blight of potato.
Bacterial: Black rot of crucifers.
Viral: Tobaccmosaic, turnip mosaic, etc.

“what is totipotency class 12 “

Methods of breeding for disease resistance

1. Conventional breeding: The steps are:

Screening germplasm for resistance sources.
Hybridisation of selected parents.
Selection and evaluation of the hybrids.
Testing and release of new varieties.

Some crop varieties bred by Conventional method:

NEET Biology Class 12 Strategies For Enchancement In Food Production Of Some Crop Varieties Bred By Conventional Method

NEET Biology Class 12 Chapter Strategies for Enhancement in Food Production Detailed Notes

Conventional breeding is constrained by the availability of a limited number of disease-resistance genes.

Strategies For Enhancement In Food Production Class 12

2. Mutation breeding:

Mutation (sudden genetic change) can create new desirable characteristics not found in the parental type.
Mutation breeding is the breeding by mutation using chemicals or radiations (e.g. gamma rays) to produce plants with desirable characteristics. Such plants are selected and multiplied directly or used as a source in breeding.
For example, In mung bean, resistance to yellow mosaic virus and
powdery mildew was induced by mutations
Resistant genes from wild species have been introduced in the high-yielding cultivated varieties. example, In Bhindi (Abelmoschus esculentus), resistance to yellow mosaic virus was transferred from a wild species. It resulted in a new variety of A. esculentus called Parbhani kranti.
Transfer of resistance genes is achieved by sexual hybridization between the target and the source plant.

Plant Breeding for Developing Resistance to Insect Pests

Insect resistance in host crop plants may be due to morphological, biochemical, or physiological characteristics.

Hairy leaves: Eaxmple, resistance tjassids in cotton, and cereal leaf beetle in wheat.
Solid stems in wheat lead to non-preference by the stem sawfly.
Smooth-leaved and Nectar-less cotton varieties do not attract bollworms.
High aspartic acid, low nitrogen, and sugar content in maize lead to resistance to maize stem borers.

Sources of resistance genes for breeding are cultivated varieties, germplasm collections of crops, or wild relatives.

Strategies For Enhancement In Food Production Class 12

Some crop varieties bred for insect pest resistance:

NEET Biology Class 12 Strategies For Enchancement In Food Production Of Some Crop Varieties For Insect Resistance

“what is totipotency class 12 “

Plant Breeding for Improved Food Quality

More than 840 million people in the world do not have adequate food. 3 billion people suffer from micronutrients, protein, and vitamin deficiencies (‘hidden hunger’).
Breeding crops with higher levels of nutrients is called
Biofortification. It helps to improve public health.

Strategies For Enhancement In Food Production Class 12

Objectives of breeding for improved nutritional quality:

To improve Protein content and quality.
To improve Oil content and quality.
To improve Vitamin content.
To improve Micronutrient and mineral content.

Examples of hybrids with improved nutritional quality:

Maize hybrids have twice the amount of amino acids, lysine & tryptophan compared to existing maize hybrids.
The wheat variety, Atlas 66, has a high protein content.
Iron-fortified rice varieties contain over five times as much iron as in common varieties.

Vegetable crops rich in vitamins & minerals:

Released by Indian Agricultural Research Institute, New Delhi.

Vitamin A enriched carrots, spinach, pumpkin.
Vitamin C-enriched bitter gourd, bathua, mustard, and tomato.
Iron & calcium enriched spinach & bath.
Protein-enriched beans (broad, lablab, French & garden peas).

Plant Breeding Class 12 Notes

Single Cell Protein (SCP)

It is an alternate source of proteins for animal and human nutrition. example, microbes like Spirulina.

Spirulina is rich in protein, minerals, fats, carbohydrates & vitamins. It is grown on materials like wastewater from postprocessing plants, straw, molasses, animal manure & sewage.
This also reduces environmental pollution.
A 250 Kg cow produces only 200 g of protein/day.
But 250g of a micro-organism like Methylophilus methylotrophus produces 25 tonnes of protein/day.

NEET Study Material for Strategies for Enhancement in Food Production Chapter

Tissue Culture

A technique of growing plant cells/tissues/organs in a sterile culture medium under controlled aseptic conditions.

The ability to generate a whole plant from any cell/explant is called totipotency. An explant is any part of a plant that is grown in a test tube under sterile nutrient media.
The nutrient medium must provide a carbon source (such as sucrose), inorganic salts, vitamins, amino acids, and growth regulators like auxins, cytokinins, etc.
The method of producing thousands of plants in a very short time through tissue culture is called micropropagation. These plants will be genetically identical original plant, i.e., they are somaclones.
Tomato, bananas, apples, etc. are produced by this method.
Tissue culture is used for recovering healthy plants from diseased plants. The meristem (it will be free of virus) from the infected plant is removed and grown in vitro to obtain virus-free plants. Scientists have cultured meristems of banana, sugarcane, potato, etc.

Plant Breeding Class 12 Notes

Somatic hybridization:

It is the fusion of protoplasts from different varieties of plants (with desirable characteristics) to get hybrid protoplasts. It can be grown from a new plant called somatic hybrid.
Protoplasts can be isolated after digesting the cell walls of single cells of plants.
A protoplast of tomatoes has been fused with that of potatoes, forming new hybrid plants called pomatwith characteristics of tomato and potato.
But it has all the desired characteristics for its commercial utilization.

NEET Biology Class 12 Microbes In Human Welfare Notes

February 14, 2025August 30, 2023 by Marksparks .arkansas

Microbes In Human Welfare

Several microbes such as bacteria, viruses, fungi, etc. are useful to man in many ways. Some of them are given below:

Microbes In Household Products

Lactobacillus or Lactic acid bacteria (LAB):

It converts milk curd by producing acids that coagulate and partially digest the milk proteins.
Fresh milk can be converted to curd by adding some curd containing LAB. It also increases vitamin B₁₂ in curd.
In the stomach, LAB helps to check pathogens.

Biology Class 12 Notes For Neet

Bacterial fermentation (anaerobic respiration):

Read And Learn More: NEET Biology Class 12 Notes

In dough is used to make foods such as dosa, idle, etc.
The puffed-up appearance of dough is due to the production of CO₂.

“microbes in human welfare “

Baker’s Yeast (Saccharomyces Cervidae):

It is used to make bread by fermenting dough.
Toddy is made by fermenting sap from palms.
Microbes are used to ferment fish, soya beans & bamboo shoots and to produce cheeses.
Swiss cheese has large holes due to the production of CO₂ by
Propionibacterium sharmanii (a bacterium).

Roquefort cheese is ripened by growing a fungus on it.

Biology Class 12 Notes For Neet

Microbes In Industrial Products

Microbes In Industrial Products: Production of beverages, antibiotics, etc. on an industrial scale, requires growing microbes in very large vessels (fermenters).

Fermented beverages:

Saccharomyces cerevisiae (Brewer’s yeast) is used in the production of beverages by fermenting malted cereals and fruit juices to produce ethanol.
Wine and beer are produced without distillation.
Whisky, Brandy, Rum, Gin, Arrack, etc. are produced by distillation of fermented broth.

“chapter 10 biology class 12 notes “

Antibiotics:

Chemical substances produced by some microbes can kill or retard the growth of pathogens.
They are used to treating plague, whooping cough, diphtheria, leprosy, etc.
Penicillin: The first antibiotic discovered by Alexander
Fleming: He observed that Staphylococci could not grow around a mold (Penicillium nota tum) growing in unwashed culture plates. He extracted penicillin from it.
Earnest Chain and Howard Florey established its full potential as an effective antibiotic.
Fleming, Chain & Florey were awarded the Nobel Prize (1945).

Chemicals, enzymes & other bioactive molecules:

Biology Class 12 Notes For Neet

1. Organic acids: Acid-producer microbes include

Aspergillus Niger (a fungus): Citric acid
Acetobacter aceti (a bacterium): Acetic acid
Clostridium bretylium (a bacterium): Butyric acid
Lactobacillus (a bacterium): Lactic acid

2. Alcohol: Yeast (S. Cervidae) is used to produce ethanol.

NEET Biology Class 12 Microbes In Human Welfare Notes

NEET Biology Class 12 Microbes in Human Welfare Notes

3. Enzymes:

Lipases: Used in detergent formulations. Help to remove oily stains from the laundry.
Pectinases & Proteases: To clarify bottled juices.
Streptokinase: Produced by Streptococcus. Used as a ‘clot buster’ to remove clots from the blood vessels of patients who have myocardial infarction.

4. Cyclosporine A: Produced by Trichoderma polysporum (fungus). Used as an immunosuppressive agent in organ transplant patients.

5. Statins: Produced by Monascus purpure us (a yeast). Used as blood-cholesterol-lowering agents. It inhibits the enzymes responsible for the synthesis of cholesterol.

Class 12 Biology Notes For Neet

Microbes In Sewage Treatment

Microbes In Sewage Treatment: Sewage (municipal wastewater) contains large amounts of organic matter and microbes. Sewage is treated in Sewage Treatment Plants (STPs) to make it less polluting.

It includes 2 stages:

Primary treatment:
- It is the physical removal of particles. It includes
- Removal of floating debris by sequential filtration.
- Removal of the grit (soil & pebbles) by sedimentation.
- The settled solids form the primary sludge and the supernatant forms the primary effluent.
Secondary treatment (Biological treatment):
- Primary effluent is passed into large aeration tanks and constantly agitated. This allows vigorous growth of useful aerobic microbes into a flock (bacteria associated with fungal filaments form mesh-like structures).
- These microbes consume the organic matter in the effluent. This reduces the BOD (Biochemical Oxygen Demand) of the effluent.

“ch 10 bio class 12 notes “

BOD:

The amount of O₂ consumed by bacteria to oxidize all organic matter in one liter of water. It is a measure of organic matter present in the water. The greater the BOD more is its polluting potential. The effluent is then passed into a settling tank where the bacterial ‘flocs’ are sediment. This sediment is called ‘activated sludge’.

Class 12 Biology Notes For Neet

A small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum.

The remaining sludge is pumped into large tanks called anaerobic sludge digesters.
Here, some anaerobic bacteria digest the bacteria and fungi in the sludge by producing gases like CH₄, H₂S, and CO₂. These gases form the biogas.
The effluent is released in natural water bodies like rivers and streams.
The Ministry of Environment & Forests has initiated the Ganga Action Plan & Yamuna Action Plan to save water

Microbes in Human Welfare NEET Notes

Microbes In The Production Of Biogas

Microbes In The Production Of Biogas:

Biogas: Biogas is a mixture of gases (mainly CH₄) produced by microbial activity. It is used for cooking & lighting.
Methanogens: Methanogens grow anaerobically on cellulosic material and produce CH₄, for example, Methanobacterium.
Methanobacterium: Methanobacterium is found in the anaerobic sludge and rumen of cattle (for cellulose digestion).
The dung of cattle (gobar) is rich in these bacteria. Dung can be used for the generation of biogas (Go bar gas).
The Biogas plant consists of A concrete tank (10–15 feet deep) that collects bio-waste and a slurry of dung.
A floating cover is placed over the slurry, which keeps on rising as the biogas is produced.
An outlet that is connected to a pipe supplies biogas.
An outlet to remove spent slurry (used as fertilizer).

Indian Agricultural Research Institute (IARI) and Khadi and Village Industries Commission (KVIC): Developed technology for biogas production in India.

Microbes In Human Welfare Class 12 Notes

Microbes As Biocontrol Agents

Microbes Biocontrol Agents:

Biocontrol: Biocontrol is the use of biological methods for controlling plant diseases and pests.
Chemical pesticides: Chemical pesticides and insecticides kill both useful and harmful organisms and cause pollution.

Microbial biocontrol agents:

Bacillus do huringiensis (Bt): To control butterfly caterpillar.

The dried spores of Bt (available in sachets) are mixed with water and sprayed on vulnerable plants such as brassicas and fruit trees. These are eaten by the caterpillar. In their gut, the toxin is released, and the larvae get killed.
Scientists have introduced B. the huringiensis toxin genes in implants. For example, Bt cotton.

“ch 10 biology class 12 notes “

Trichoderma sp (fungus): These are free living present in the root ecosystems. They control several plant pathogens.

Microbes In Human Welfare Class 12 Notes

Baculoviruses (Especially genus Nucleopolyhedrovirus):

Attacks insects and other arthropods.
These are suitable for species-specific, narrow-spectrum
insecticidal applications. This is desirable in IPM (Integrated
Pest Management) program to conserve beneficial insects.

Microbes As Biofertilizers

Microbes Biofertilizers:

Class 12 Microbes In Human Welfare Notes

Biofertilizers: Biofertilizers are organisms that enrich the nutrient quality of the soil for example, Bacteria, fungi, cyanobacteria, etc.

Rhizobium: Rhizobium (symbiotic bacteria in root nodules of leguminous plants) fix atmospheric N₂.

Free-living bacteria in the soil (for example, Spirillum and
Acetobacter) enrich the nitrogen content of the soil.
Mycorrhiza: Symbiotic association of fungi (For example, genus of Glomus) with plants.
The fungus gets food from plants.
The fungal symbiont performs the following:
Absorb phosphorous from the soil and pass it to the plant.
Give resistance to root-borne pathogens and tolerance to salinity and drought.
Give overall increase in plant growth and development.

“microbes in human welfare images “

Cyanobacteria (Blue-green algae): Autotrophic microbes.

They fix atmospheric nitrogen. For example, Anabela, Nostoc,
Oscillatoria etc. In paddy fields, Cyanobacteria serve as important biofertilizers.
It adds organic matter to the soil and increases its fertility.

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 2

April 21, 2025August 24, 2023 by Marksparks .arkansas

WBBSE Class 10 Physical Science Question Answer In English

Multiple Choice Questions Physical Science And Environment

Question 1. Which among the following gases absorb long wavelength infrared radiation emitted from the earth’s surface

N₂
O₂
CH₃
He

Answer: 3. CH₃

Question 2. At STP, 2-24 L is occupied by

4.4 g CO₂
0.64g SO
28 g CO
16g O. [C= 12. O -16. S = 32]

Answer: 1. 4.4 g CO₂

Read And Learn More: WBBSE Solutions For Class 10 Physical Science And Environment

Question 3. How many molecules of CO₂ will be produced when I mole C reacts completely with 1 mole O₂?

6.022 x 1023
1.806 x 1024
6.022 x 1022
6.022 x 1024

Answer: 3. 6.022 x 1023

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 2”

Question 4. For a solid, how many types of thermal expansion coefficients are there?

One
Two
Three
Four

Answer: 3. Three

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 2

Question 5. Which one of the following has the highest wavelength?

x-ray
y-ray
Infrared ray
Ultraviolet ray

Answer: 3. Infrared ray

Question 6. In the case of refraction if the angle of incidence and the angle refraction are 45° and 30° respectively, then the angle of deviation is

75°
15°
7-5°
37.50

Answer: 2. 15°

Question 7. Temperature remains unchanged if the potential difference between the two ends of a conductor is V and the current through the conductor is I, which of the following is true?

V∞ I
V∞ I²
V ∞ I^-1
V ∞ r^-2

Answer: 1. V∞ I

Question 8 . The relation among electromotive force (V), work (W) and charge (Q) is

Q=WV
Q=V/W
Q=V/W2
Q = W/V

Answer: 4. Q = W/V

Question 9. For the atom produced by ẞ-particle emission from a radioactive atom

Mass number increases
Atomic number increases
Mass number decreases
Atomic number decreases

Answer: 2. Atomic number increases

Question 10. To which group of the long

Group 1
Group 16 (e)
Group 17
Group 2

Answer: 3. Group 17

Question 11. Solid state of which of the following compounds is composed of ions?

Sodium chloride
Hydrogen chloride
Naphthalene glucose

Answer: 1. Sodium chloride

Class 10 Maths	Class 10 Social Science
Class 10 English	Class 10 Maths
Class 10 Geography	Class 10 Geography MCQs
Class 10 History	Class 10 History MCQs
Class 10 Life Science	Class 10 Science VSAQS
Class 10 Physical Science	Class 10 Science SAQs

Question 12. Which of the following has the highest ability to conduct electricity?

Pure water
An aqueous solution of Sugar cane
Liquid hydrogen chloride
Aqueous solution of acetic acid

Answer: 4. Aqueous solution of acetic acid

“Class 10 WBBSE Model Question Paper 2023, Physical Science and Environment Set 2 study material”

Question 13. In the first step of fixation of nitrogen which of the following compounds as a result of lightning

NO
NO₂
N₂O₅
HNO₃

Answer: 1. NO

Question 14. Which of the following is the formula of bauxite, ore of aluminum?

Al₂O₃
Al₂O₃H₂O
Al₂₀O₃2H₂O
A1F₃ . 3NaF

Answer: 3. Al₂₀O₃2H₂O

Question 15. Which of the following is the alkyl group containing two carbon atoms

Methyl
Ethyl
Propyl
Isopropyl

Answer: 2. Ethyl

Class 10 Physical Science WBBSE Physical Science And Environment Answer The Following Questions

Question 1. Write down the unit of calorific value of the fuel.
Or
Does the temperature increase of decrease with an increase in altitude in the strato- sphere?
Answer:

Kilo Joule/Kg OR The temperature increases.

Question 2. Which radiation, coming from the sun is prevented by the ozone layer from falling on the earth’s surface?
Answer: Ultra Violet Ray (UV ray).

Question 3. State whether the following. the statement is true or false: The volume of gas molecules is taken into consideration in Avogradro’s raw.
Answer: False.

Question 4. The product of volume and pressure of how many grams of Ni gas is 244 liter atmosphere at STP? [N= 14].
Answer: 280g.

Question 5. State whether the following statement is True or False: The constituent particles of a material change position during the conduction of heat through it.
Or, The width and the cross-section of a conductor remain unchanged, what is the relation between the thermal resistance and thermal conductivity of that conductor?
Answer:

False or $R_T \alpha \frac{1}{K}$ [d and A constant]

Question 6. What will lie the angle of incidence when a ray of light passes through the centre of curvature of a concave mirror?
Answer: 0°.

Question 7. How many rectangular surfaces are there in a prism?
Answer: 3.

Question 8. Give an example of a semiconductor.
Answer: Silicon (Si).

Question 9. A thin wire and a thick wire of the same conducting material have the same length. Which one of them will carry more current connected to the same potential difference?
Answer: Thick wire.

Question 10. Mention one misuse of the nuclear fission reaction.
Or
Which law explains the release of huge amounts of energy in nuclear fusion?
Answer:

Atom Bomb Or E = mc [m = Decreasing mass, C = Vacuum, E = Produced energy).

Question 11. Match the right column with the left column:

Left column- Right column

Oxide layers protect from attack by water vapor- AI

Group 1 cement of the logn periodic table having the least reducing property accelerates the rusting of iron – LI

When the metal remains exposed to air metal slowly develops green patches on its surface – Cu

Group 2 cement of the long periodic table having the least atomic radius – Bc

Question 12. Draw the Lewis dot structure of N? molecules, (atomic number of N is 7)
Answer:

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Structure Of N Molecule

Question 13. Which kind of electricity is used in electrolysis?
Answer: Direct Current (dc)
Or
Write down the cathode reaction in the electrolysis of acidulated water using platinum electrodes.
Answer: H + e → H, H+H → H₂↑

Question 14. In electroplating gold on brass, what is the electrolyte used?
Answer: Solution of Potassium Aurocyanide [K[Au(CN)2].

Question 15.What colour is formed in the reaction of ammonia with Nessler’s reagent?
Answer: Copper-brown color.

“WBBSE 2023 Madhyamika Physical Science and Environment, Set 2 Model Question Paper download”

Question 16. Write down the formula of the precipitate formed when H’S gas is passed through an aqueous solution of silver nitrate.
Or
Write the name of the compound which is formed by the reaction of nitrogen with magnesium metal nl a high temperature.
Answer: Ag₂S or Magnesium Nitride (Mg₂ N₂).

Class 10 Physical Science WBBSE

Question 17. What is the value of the H-C-H bond angle in methane? 1 Or, Write the IUPAC name of CH CH₂COOH- 1.
Answer: 109° 28′ OR Propanoic Acid.

Question 18. What is the industrial source of CGN7?
Answer: The gaseous substances contained in the petroleum mine or coal mine.

WB Class 10 Physical Science Question Answer Physical Science And Environment Answer The Following Questions

Question 1. What is the concept of sustainable development?
Answer:

The concept of sustainable development

Sustainable development means obtaining and utilizing natural resources discriminately so that they do not get exhausted completely, keeping the future generation in mind.

Question 2 . 1 g of a gas at 7°C and 2-atmosphere pressure occupies a volume of 410 mL. Determine the molar mass of the gas. (R=0-082 litre atmosphere mole’ 11C1}
Or
A fixed mass of gas occupies a volume of 273 cm³ at STP. At what pressure the above gas will occupy a volume of 300 cm3 at 27″C?2
Answer:

Given

1 g of a gas at 7°C and 2-atmosphere pressure occupies a volume of 410 mL.

Mass of gas (W) = 1 g.

Temperature (T) = 7° c=(273+7) k=280k.

Pressure(P0 = 2 atm and volume (v) = 410 ml =0.41L

Let the molar mass of gas = M

From PV = ($\left(\frac{W}{M}\right)$)

M= $\frac{W R T}{P V}$ Or,

M= $\frac{1 \times 0.082 \times 280}{2 \times 0.41}$

= 28

∴ The molar mass of gas = 28 g. mol^-1

Answer: Primary Pressure of the gas (P₁) = 76 cm mercury pressure, Temperature (T₁)=273k and volume (V₁)=273 cm and final temperature (T₂)=(27+273) K = 300k, volume (V₂)= 300 cm. Let, final pressure = P₂.

From the combined law of Charles and Boyle’s

⇒ $\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$ Or

⇒ $P_2=\frac{P_1 V_1 T_2}{T_1 V_2}$

⇒ $P_2=\frac{76 \times 273 \times 300}{273 \times 300}$

P₂= 76

∴ The final pressure of the gas = 76 cm mercury pressure

Question 3. What is the refractive index of a medium?
Or
Which type of detection of vision is rectified a convex lens?
Answer:

The refractive index of a medium is $\frac{\sin i}{\sin r}$ = 1μ₂

That is the refraction of the second medium in respect of the first medium

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Second In medium In Respect Of The First Medium

Suppose, near a person with a long type of defect of vision, there is point N and near his eyes, there is point N. The vocal length of the convex lens

will be such that after refraction through the lens the irregular image of point N¹ is formed at point N¹. So, the person sees point N¹ at point N¹

“WBBSE Madhyamika 2023 Physical Science and Environment, Set 2 Model Question Paper with solutions”

Question 4. Two resistances r₁, and r_², when connected separately to the same potential difference, it was seen that the current flowing through r₁, was six times the current flowing through r₁. Determine the ratio of and r₂
Answer:

Given

Two resistances r₁, and r_², when connected separately to the same potential difference, it was seen that the current flowing through r₁, was six times the current flowing through r₁.

Let, Potential difference = V and through resistances r¹ , and r², the current following is I₁ and I₂ respectively.

I = 6I₂

In case of r₁ V= I ______________(1)

In case of r₂V = r____________ (2)

By comparing 1 and 2 we get

I₁ r₁= I₂r₂

Or $\frac{r_1}{r_2}=\frac{I_1}{I_2}=\frac{I_2}{6 I_1}=\frac{1}{6}$

∴ r₁ : r₂= 1:6

Question 5. How did Kossel explain the formation of ionic bonds?
Or
Liquid hydrogen chloride cannot conduct electricity, but molten sodium chloride can conduct electricity. Explain.
Answer:

Koss el’s explanation –

At the time of the chemical bond between two different elements, an electro-positive atom of the element gives out one or more electrons from its orbit and an other electro-negative atom of another element receives the refused electron in its outermost orbit.
Thus, the two atoms, like their nearest neutral gas, gain a symmetrical electron pattern.
Then, the Cation and the Anion with the coulombian attraction force, form an ionic bond.

Or
Hydrogen chloride is a compound with the same valency; so, there is no ion even when it is liquid. So, liquid hydrogen chloride is not able to conduct electricity.

Again NaCl is a ionic compound even in the solid state, it has Na* and Cl- ions in it.

Question 6 . Distinguish between Sodium chloride and naphthalene by two physical properties.
Answer:

Difference between Sodium chloride and naphthalene by two physical properties

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Physicla Properties

Question 7. Between two aqueous solutions, one is Ferric chloride and the other is Aluminium
chloride. How would you identify the Ferric chloride solution using an aqueous solution of ammonia? Answer with a balanced chemical equation.
Answer:

In two solutions, by adding the solution of ammonia separately, brown residual is formed. That solution can be identified as Ferric Chloride. Moreover, the other solution with white gum-like residual is the solution of aluminum chloride.

Equation:

FeCI₃ +3NH₄OH → Fe(OH)₃↓+3NH₄ CI

AICI₃ +3NH₄OH → Al(OH)₃ +3NH₄ CI

Question 8. Why zinc blend can be called both mineral and ore of zinc?
Or
Mention two ways of preventing rusting of iron.

Answer: Zinc blend (ZnS) is naturally hard and metallic are found in mines. Hence it is called the ore of zinc.
Or

Galvanization or the layer of melted zinc on iron prevents rusting.
Tar Clour etc. prevent rusting.

Question 9. Write with a balanced chemical equation what happens when methane is burnt in oxygen.
Or
Mention one use of each acetic acid and ethyl alcohol.
Answer: CH₄ (methane) +2O₂– CO₂+H₂O
Or

Acetic Acid – Vinegar, preserves fish and flush, to cook pickle.
Ethyl alcohol – Resin, scent artificial rubber and fiber, medicine when produced takes ethyl alcohol as a soluble agent.

WBBSE Class 10 Physical Science Question Answer Physical Science And Environment Answer The Following Questions

Question .1 What is meant by the moler volume of a gas? Mention two reasons for the deviation of real gases from the behavior of ideal gases.
Answer:

Moler volume of a gas:

In a certain temperature and pressure the volume of 1-mole quantity of a gaseous element is called the molar volume of the gas.

Ideal gas molecules are similar to mass point, but real gas molecules, no matter how small. They are, and cannot be ignored. Molecules of ideal gases have no force of attraction and distraction among them. Molecules of real gases show attractive force.

Question 2. How many grams of ammonium sulfate is required to prepare 558 g of by the reduction of Fe₂O₃with Al at high temperature? How many moles of Fe₂O₃required in the reaction? [Fe=55 8. Al = 27.0 = 16]
Or
By heating 32.1 g ammonium chloride with calcium hydroxide 10.2g NH₃ 33.3 g CaCI, and 10.8 g H₂O are obtained. How many grams of calcium hydroxide take part in the reaction? How many mole of NH₃, and how many liters of NH₃, at STP are formed in the reaction? (N = 14. H=1)
Answer

Fe₂O₃+2 Al→ 2Fe + Al₂O₃

2×278→ 2×55.8g

2×55.8g Fe needs 2x27g Al.

558g Fe needs = $\frac{2 \times 27 \times 558}{2 \times 55.8}$

= 270g Al.

Again,

2x27g Al reacts with 1 mol Fe₂O₃.

270g Al reacts with mol Fe₂O₃

Answer: As per the law of the regularity of mass, the mass of NH₄ Cl+ mass of Ca(OH)₂=

NH₃ mass + CaCl₂ mass + H₂O mass

32.1g+ mass of Ca(OH)₂

= 10.2g+33.3g + 10.8g

Mass of Ca(OH)₂= (10.2 +33.3 + 10.8)g-32.1g = 54.3g-32.1g = 22.2g

We know, 17g NH₃ = 1 mol NH₃

10.2g NH₃ =$=\frac{1}{17} \times 10.2 \mathrm{~mol}$x 10.2 mol

= 0.6mol

In STP, 1 mol. NH, volume = 22.4litre

∴ 0.6 mol. NH₃ volume = 22.4×0.6 litre = 13.44 litre.

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 2 PDF”

Question 3. Which quantities remain fixed in the definition of the volume expansion coefficient of a gas? Name a non-metal which is a good conductor of heat.
Or
What is meant by ‘the linear expansion coefficient of copper is 17 x10^-6°C? Why does the value remain the same even in Kelvin scale?
Answer:

The quantities remaining fixed are
The pressure of the gas, and
Mass of the gas.

Diamond is a nonmetal which is a good conductor of heat.
Or
The linear expansion coefficient of copper is 17×10^-6means if 1°C temperature is increased of any copper rod, the length of the rod will increase by 17×10^-6portion of the primary length of the rod.

Since the change of 1°C and the change of 1 K are the same, the value remains the same in Kelvin Scale also.

Question 4. What type of mirror is used by dentists? Why a ray of light does not deviate as a result of refraction through a glass slab?
Answer:

The dentists use a concave mirror

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 The Dentist Use Concave Mirror

Here PQRS is a rectangular glass slab. ABCD is the direction of a ray of light. N₁N₂ and N₃ N₄ are two perpendiculars on B and C points where the reflection takes place. Al point B, because of refraction of light, ABN₁ = (angle of incidence) and CBN₂ =r, (refraction angle). At point C, because of refraction, BCN₃ =r₂ (angle of incidence) DCN₄ = i₂ (refraction angle).

Question 5. When an object is placed 20 cm away from a convex lens, no image is obtained on either side of the lens. What is the focal length of the lens? If the refractive index of glass with respect to air is 1.5, what is the refractive index of air with respect to glass?
Or
The length of an object is 5 cm. An image of length 10 cm is obtained when it is placed at.a distance of 2 cm in front of a convex lens. What is the linear magnification and image distance?
Answer:

Given

The length of an object is 5 cm. An image of length 10 cm is obtained when it is placed at.a distance of 2 cm in front of a convex lens.

The slab of glass is rectangular. The two opposite sides are parallel. Hence,

⇒ N₁ N₂ II, N₃N₄

Again N₁ N₂ II, N₃N₄and transversal is BC.

<N₂ C= <BCN₃ (Alternative Angle)

r₁ = r₂, Again, refraction angle i₂ = i₁ AB || CD i₁ = i₂.

∴ The ray does not deviate

The focal length of the lens = 20cm

At a distance of 20 cm from the convex lens, an object is placed. Then the refracted rays of light will be parallel.

∴ In respect of air, the refraction index of glass will be

⇒ $g^\mu \text { air }=\frac{1}{\text { air } \mu g}=\frac{1}{1.5}$

= 0.67

Air μg = 1.5

Answer:

In respect of glass, a refractive index of air will be –

Length of the object (h₁) = 5cm
Length of the image (h₂ ) = 10cm

$WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Glass Refractine Index$

∴ Linear magnification = m

= $\frac{h_2}{h_1}$

= $\frac{10}{5}$

∴ m = 2

∴ Image distance (V) = mu = (2×2) m = 4cm.

“Class 10 WBBSE Madhyamika Model Question Paper 2023, Physical Science Set 2 practice questions”

Question 6. Write in brief the basic principle of hydroelectric power generation.
Answer:

The basic principle of hydroelectric power generation

The principle of hydropower is that the potential energy of water that is stored at great heights in the dam is changed to kinetic energy by allowing the water to flow at high speed. The kinetic energy of flowing water is utilized to produce electricity.

Question 7. A current of 1A flow when an electric bulb is connected to 220 V mains. What would be the current when the same bulb is connected to 110V mains?
Or
Find the ratio of resistances for two bulbs of 220V-60W and 220V-60W.
Answer:

A current of 1A flow when an electric bulb is connected to 220 V mains.

Potential difference (V₂) = 220V and current flow in the bulb (I₁) = 1A.

Again Potential difference (V₂) = 110 V

Let, the current flow in the bulb for the second time = I,

Resistance of the bulb = R

∴ V₁ =I₁R ___________(1)

V₂ =I₂R______________(2)

∴ $\frac{V_1}{V_2}=\frac{I_1}{I_2}$ Or,

⇒ $I_2\frac{V_2 I_2}{V_1}$

⇒ $I_2\frac{110 \times 1}{220}$

I₂= 0.5A

Answer:

R= $\frac{V_2}{P}$______________(1)

$=\frac{220 \times 220}{60} \Omega$____________(2)

$\frac{110 \times 110}{60} \Omega$____________(3)

⇒ $2 \div 3=\frac{R_1}{R_2}=\frac{220 \times 220}{60} \times \frac{60}{110 \times 110}$

= $\frac{R_1}{R_2}$

∴ R₁: R₂

Question 8. Explain why a new element is formed by a particle emission but no new element is formed by y-ray emission from a radioactive element.
Answer:

The nature of a particle is like He²⁺

From the nucleus of a radioactive element, when a particle is emitted, the mass number of the new nucleus is 4 units less.
The atomic number is less by 2 units So, so a new element is formed a radioactive element.
Again, the Y-ray has a small wavelength. It is electromagnetic. In the case of y-rays, the atomic number is unchanged So, no new element is formed.

Question 9. Write down Dobereiner’s law of triads. Arrange Cl, Br, I, F in increasing ‘order of their oxidizing power.
Or
What is the important conclusion of Moscley’s experiment? What is me impor- tance of this conclusion in regard to the periodic table?
Answer:

Dobereiner’s law of triads.

Dobereiner’s law of triads states that the atomic mass of the middle element of a triad is he arithmetic mean of the atomic masses of the two other elements. Increasing order of oxidizing power-1<Br<el<F.
Or
Answer: Mosley concluded that there were three unknown elements between aluminum and gold. There were only 92 elements up to and including uranium and 14 rare-earth elements.

The importance of this conclusion in the case of the periodic table:

Mendeleev’s periodic law is rectified.
The elements are arranged in the periodic table by atomic number instead of atomic masses.

Question 10. Write two differences between the conduction of electricity through a metallic wire and an electrolyte during electrolysis. In the electrolytic refining of copper metal, impure copper rod is used as which electrode?’
Answer:

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Conduction Through Metallic Wire And Conduction Through Electrolyte

Question 11. Write the names of the chemicals used and the balanced chemical equation in the industrial production of urea.
Answer:

The chemicals are :

Liquid ammonia
Liquid CO₂
Balanced chemical equation

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Balanced Chemical Euaqtion

“WBBSE Class 10 Model Question Paper 2023, Physical Science and Environment Set 2 exam pattern”

Question 12. (A) and (B) are two unsaturated hydrocarbons, each containing 2 carbon atoms. On reaction with bromine per molecule and (B) adds two molecules of bromine per molecule. Write the structural formula of (A) and (B). Write a balanced chemical equation of the reaction of (B) with bromine.
Answer:
Or
Write a balanced chemical equation of the reaction of sodium hydroxide with acid. Which one between just and polyethylene is environment-friendly for packaging and why?

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 Hydrocarbons

(A) and (B) are two unsaturated hydrocarbons, each containing 2 carbon atoms. On reaction with bromine per molecule and (B) adds two molecules of bromine per molecule.

The balanced chemical equation of the reaction of (B) with bromine

HC ≡ CH+2Br₂—CHBr₂–CHBr₂( 1, 1, 2, 2-Tetrabromithane)

Answer:

The equation is –

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 2 The equation Of Acetic Acid

Jute is environment-friendly. It is formed with cellulose organic polymer. So, it is decomposed by bacteria and fungi, etc. easily. Hence, it is environmentally friendly.

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3

April 21, 2025August 24, 2023 by Marksparks .arkansas

WBBSE Class 10 Physical Science Question Answer In English

Multiple Choice Questions Physical Science And Environment

Question 1. Which of the following greenhouse gases has the maximum contribution towards global warming?

N₂O
CH₄
CO₂
H₂O

Answer: 3. CO₂

Question 2. According to Boyle’s law what is the PV-P graph?

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Boyles Law Of PVP Graph

Answer: 2 PV-P Graph

Read And Learn More: WBBSE Solutions For Class 10 Physical Science And Environment

Question 3. If the vapour density of a carbon-containing gaseous substance.is 13. Which of the following can be its molecular formula?

CO₂
C₂H₄
C₂H
C₂H₂

Answer: 4. C₂H₂

Question 4. The unit of coefficient of linear expansion of a solid is

m
nr!
°C-1
°C

Answer: 3. °C-1

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 3”

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3

Question 5. An object is placed in between the optical centre and the focus of a thin convex lens. What is the nature of the image of the object?

Real and inverted
Virtual and inverted
Rea and erect
Virtual and erect

Answer: 4. Virtual and erect

Question 6. When a ray of light is incident perpendicularly on a transparent glass slab, what will be its angle of deviation?

0°
180°
30°
90°

Answer: 1. 0°

Question 7. Which of the units given below is the SI unit of resistance?

Volt
Ampere
Coulomb
Ohm

Answer: 4. Ohm

“Class 10 WBBSE Model Question Paper 2023, Physical Science and Environment Set 3 study material”

Question 8. In a domestic electric circuit, the fuse wire is connected to which of the following? live line

Earth line
Live line
Neutral line
Both live and neutral line

Answer: 2. Live line WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3

Question 9. β-ray emitted from a radioactive clement is an electromagnetic wave

A stream of electrons
A stream of neutrons
A stream of protons
Electromagnetic wave

Answer: 1. A stream of electrons

Question 10. How many groups are there in the long periodic table?

Answer: 4.18

Class 10 Maths	Class 10 Social Science
Class 10 English	Class 10 Maths
Class 10 Geography	Class 10 Geography MCQs
Class 10 History	Class 10 History MCQs
Class 10 Life Science	Class 10 Science VSAQS
Class 10 Physical Science	Class 10 Science SAQs

Question 11. Information of which of the following compounds’ octet rule is not obeyed?

NaCl
LiH
KCl
CaO

Answer: 2. LiH

Question 12. Which of the following can conduct electricity?

Molten NaCl
Liquid HCI
Solid NaCl
Aqueous solution of glucose

Answer: 1. Molten NaCl

Question 13. What will be the colour of (the resulting solution when excess aqueous ammonia is added to an aqueous solution of copper sulphate?

Yellow
Green
Deep blue
Brown

Answer: 3. Deep blue

Question 14. In which of the following alloys zinc is present?

Bell metal
Brass
Bronze
Duralumin.

Answer: 2. Brass

Question 15. Which of the following is a saturated hydrocarbon?

C₃H₆
C₂H
C₂H₂
C₂H₆

Answer: 4. C₂H₆

WBBSE Class 10 Physical Science Question Answer Physical Science And Environment Answer The Following Questions

Question 1. Mention one use of biogas.
Or
What is the role of NO in the decomposition of ozone in the ozone layer?
Answer:

Use of biogas

Biogas-1st is used as a fuel.
Or
Role of NO – No reacts with ozone(O₃) and decomposes NO₂ and O₂.

Question 2. Among charcoal, petrol and ethanol which one is a fossil fuel?
Answer: Petrol is a fossil fuel.

“WBBSE 2023 Madhyamika Physical Science and Environment, Set 3 Model Question Paper download”

Question 3. Under constant pressure, at what temperature in degrees Celsius, the volume of an ideal gas will be zero according to Charles’ law?
Answer: The volume of an ideal gas will be zero at a temp. of -273° C.

Question 4. What is the unit of M in the equation PV –RT? (Symbols have usual meaning)
Answer: Unit of M-g/mol.

Question 5. Whether the following statement is ‘true’ or ‘false’? The real expansion of any liquid depends on the expansion of the vessel in which it is kept.
Or
Among iron, invar and copper which one has the least coefficient of linear ex- pansion?
Answer: True.
Or
Answer: Invar has the least coefficient of linear expansion.

Question 6. Between the angle of incidence and the angle of refraction which one is greater when light travels from a rarer to a denser medium?
Answer: The angle of incidence is greater than the angle of refraction.

Question 7. What type of mirror is used in the viewfinder of a motor car?
Answer: Concave mirror.

Question 8. How does the resistance of a semiconductor change with the increase of temperature?
Answer: The resistance of a semiconductor decreases with an increase in temperature.

WBBSE Class 10 Physical Science Question Answer

Question 9. Which type of energy is transformed into electrical energy in a dynamo?
Answer: Mechanical energy.

Question 10. Arrange [5 and 7-rays in ascending order of their penetrating power.
Or
Which kind of nuclear reaction is the source of the sun’s energy?
Answer: α<β<γ.
Or
Nuclear fusion.

Question 11. Match the right Column with the left Column:

Left Column – Right Column

An alkali metal K – The most electronegative element F

An element whose anion Fe accelerates the rusting of iron – Extracted from haematite – Fe

Extracted from haematite K – An alkali metal – K

Most electronegative element Cl – An element whose anion accelerates the rusting of iron Cl

Question 12. What type of chemical bond is present in CaO?
Answer: CaO has an ironic bond between them.

Question 13. What is used as a cathode to electroplate silver over a copper spoon?
Or
Give an example of a compound whose aqueous solution is a weak electrolyte.
Answer:

A copper spoon is used as a cathode.
Or
Weak electrolyte → NH₃

Question 14. During electrolysis which electrode is called a cathode?
Answer: The electrode connected to the negative terminal of a battery is called the cathode.

Question 15. State one use of liquid ammonia.
Or
Write the formula of the precipitate formed when an aqueous ammonia solution is added to an aqueous solution of aluminium chloride.
Answer:

Use of liquid Ammonia-used as a refrigerant in ice making.
Or
NH₄Cl precipitate is formed.

Question 16. In the laboratory preparation of nitrogen, an aqueous solution of which compound is mixed will aqueous solution of ammonium chloride and healed?
Answer: A concentrated solution of aqueous sodium nitrate is used.

Question 17. Write the IUPAC name of CH₃, CH₂CHO.
Or
Wine the structural formula of positional CH₃ CH₂ CH₂OH.
Answer: Propanal.

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Position Of Isomer

Question 18. Mention one use of polytetrafluoroethylene.
Answer: Use coating articles and cookware to make them non-sticky.

WB Class 10 Physical Science Question Answer Physical Science And Environment Answer The Following Questions

Question 1. What is methane hydrate?
Answer:

Methane hydrate

Methane hydrate is a crystalline solid consisting of a methane molecule surrounded by a cage of interlocking water molecules.

Question 2. The pressure of a fixed mass of a gas at a temperature of 0°C. is doubled while the volume is halved. What will be the final temperature of the gas?
Or
Under constant pressure fixed mass of a gas is heated from 0°C to 546°C. What is the ratio of the final volume of the gas to its initial volume?
Answer:

Given

The pressure of a fixed mass of a gas at a temperature of 0°C. is doubled while the volume is halved.

No, the final temperature will not change As, PV = K according to Boyle’s law.

⇒ 2P × V/2 = PV = K

According to Charle’s law,

V_t = V₀ (1+_t/273)

Or, V_t = V₀ (1+2)

Or, V_t= 3V₀

Therefore, volume changes 3times the initial volume.

“WBBSE Madhyamika 2023 Physical Science and Environment, Set 3 Model Question Paper with solutions“

Question 3. What is meant by the optical centre of a convex lens?
Or
Why does the earth’s sky appear blue during daytime?
Answer:

Optical centre of a convex lens

If a ray of light strikes one surface of a lens that emergent ray from the other surface is parallel to it, and the corresponding refracted ray passes through a definite point on the principal axis. The point is the optical centre of the lens.

According to Rayleigh’s law,

Scattering a 1/ γ⁴ [y wavelength of light]

The lesser the wavelength of light, more will be the scattering of light, since we know, the wavelength of blue is much less and is sensitive to our eyes. So sky appears blue during day time.

Question 4. State Lenz’s law related to electromagnetic induction.
Answer:

Lenz’s law related to electromagnetic induction

According to Lenz’s law, the induced current will be such that it will oppose the cause producing it.

Question 5. Write with an example how according to Lewis’s concept a covalent bond is formed.
Or
Why the bond in sodium chloride cannot he expressed as Na-CI?
Answer:

Each atom achieves a Lewis octet by forming a double bond. A covalent bond takes two electrons it occurs between two atoms and the electrons are shared euqally. Example water (H₂O).
Or
The bond in Sodium chloride cannot be expressed as Na-Cl because the solution contains only Na+ and Cl- ions and water, but not the metal Na(S).

Question 6. Give one example each of a liquid and a solid covalent compound.
Answer:

Liquid covalent bond → H₂O.

Solid covalent bond → CH₄

Question 7. Write with a balanced chemical equation what happens when H₂S gas is passed through an aqueous copper sulphate solution.
Answer: CuSO₄ +H₂S→ CUS+H₂SO₄

Question 8. Write down the cathode reaction when an aqueous solution of MSO₂ (M = metal) is electrolysed. Write with reason whether the reaction is oxidation or reduction.
Or
Give one use of each copper and aluminium.
Answer:

Cathode reaction –

M²⁺+ 2e → M.

Reductions placed at the cathode.

Coppered used to make household utensils.

Aluminium is used to make the external parts of aeroplanes.

Question 9. What is the condition of the substitution reaction of methane with chlorine? Write tile balanced chemical equation of the first step of the reaction.
Or
Write with a balanced chemical equation what happens when ethanol reacts with metallic sodium.
Answer:

Condition → displacement of one atom occurs.

→ Occurs in unsaturated hydrocarbons.

2C₂H₅OH+2Na →2C₂H₅ONa+H₂

Class 10 Physical Science WBBSE Answer The Following Questions

Question 1. Establish an ideal gas equation on the basis of Boyle’s law, Charles’s law and Avogadro’s law.
Answer: Equation according to Boyle’s law.

⇒ PV=K (P = Pressure of gas, V= its volume, K= proportionality constant.) Charles law – VIT K = Constant.

⇒ Avogadro’s law – PV=nRT [ R= molar gas constant]

Question 2. SO₂ required for the industrial production of sulphuric acid is produced by burning iron pyrites in excess air current.

The chemical equation of the reaction is given below:

4FeS₂+11O₂→ 2Fe₂O₃ +8SO₂. How many grams of FeS₂ is required for the production of 512g of SO₂Fe= 56, S= 32, O=16).
Or
By heating 200g of a metal carbonate 112g of metal oxide and a gaseous compound are produced. Vapour density of the gaseous compound is 22. How many moles of the gaseous compound is produced in the reaction?
Answer:

Given

By heating 200g of a metal carbonate 112g of metal oxide and a gaseous compound are produced. Vapour density of the gaseous compound is 22.

SO₂512g.

Molecular mass of SO₂ →32+32 = 64

Moles of SO₂ = 512/64 = 8mol.

For 8moles of SO₂

4moles of FeS₂ is required

Molecular mass of FeS₂ → 56 +64 = 120g.

Therefore, 4moles of FeS₂ → 120 x 4 = 480g.

480g of FeS₂ is required.

200g → 112g

MCO₃ MO

Vapour dencity = 22. Mass of the gaseous component = 200-112= 8g.

Molecular weight = 2 x V.D

= 2 x 2.2

= 4.4

No of moles= 8/4.4

= 1.81 moles.

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 3 PDF”

Question 3. What is thermal conductivity? What is its SI unit?
Or
Define the coefficient of surface expansion, Write its SI unit.
Answer:

Thermal conductivity

The coefficient of thermal conductivity at the material of a substance is numerically equal to the quantity of heat that conducts in one second normally through a slab of unit length and unit area, that difference of temperature between its end faces, being one degree. Its SI Unit is Jm’ Or, Wm’ K-‘.

It is defined as the fractional change in the surface area of the solid per degree rise in temperature.

β = $\frac{S_2-S_1}{S_1\left(t_2-t_1\right)}$ S₁ and S₂ are the surface areas of a solid at t₁ and t₂ respectively. Y= coefficient of surface expansion unit

α = K^-1or °C^-1. γ = K^-1 or, °F^-1

β = K^-1 Or °C^-1 Or °F γ = K^-1 Or °C^-1Or °F^-1

Question 4. How can an erect and magnified image be formed with the help of a convex lens? With the help of which type of lens long-sightedness can be rectified?
Answer: An erect and magnified image can be formed by keeping the object between the focus and the optical centre of a convex lens.

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Optical Centre Of Convex Lens

Long-sightedness can be rectified using a concave lens.

Question 5. If the velocity of light in a medium is 2x10s m/s, what will be the refractive index of that medium?
Or
The refractive index of a medium with respect to air is V₂. If the angle of incidence of a ray of light in air is 45° determine the angle of deviation for that ray in case of refraction.
Answer:

If the velocity of light in a medium is 2x10s m/s

We know,

⇒ $\mu\frac{v}{c}=\mu=\frac{2 \times 10^8}{3 \times 10^8}$

= $\frac{2}{3}$

= 0.67

Given,

⇒ μ = $\sqrt{2}$. We know

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Deviation Ray

= r= $\sin ^{-1}\left(\frac{1}{2}\right)$

= r= 30°

∴ i-r

= S= 45 – 30 = 15°

∴ Angle of Direction is 15°

Question 6. Write ionic’s laws related to the heating effect of current.
Answer:

Joule’s Laws:

The amount of heat produced in a conductor in a given interval of time is proportional to the square of the current passed. H∞l (when R and t are constant).
The amount of heat produced by a given current in a given time is proportional to the resistance of the conductor. H∞ R (when I and t are constant).
The amount of heat produced in a given conductor by a given current is proportional to the time for which the current passes. H∞t (when I and R are constant).

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 3 sample paper

Question 7. Calculate the equivalent resistance when a wire of resistance 10 ohm is divided in two equal parts and connected in parallel combination.
Or
There are two 60-watt lamps and two 80-watt fans in a house. The lamps and fans run 5 hours daily. Find out the expense in a month if a unit of electricity costs Rs. 4/-. (assume one month 30 days).
Answer:

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Electricity

Equivalent resistance

→ 1/R^l= $\frac{1}{R}$+ $\frac{1}{R}$ = $\frac{2}{R}$

∴R^l = $\frac{2}{R}$

Given, 2 60W bulb & 2 80W bulb.

∴ Total Power

= 2 (60 + 80)

= 2 x 140

= 280w

Question 8. Compare the charge and ionising power of the a and y rays. Mention one use of radioactivity.
Answer:

Pow in BOT unit

⇒ $\frac{280 \times 5}{1000}$

= 1. 4 B.O.T

Total expense = 1.4 x 4 x 30 = Rs. 168

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Ionising Power Of Rays

One use of radioactivity: To estimate age: Dating technique.

Question 9. What is meant by the ionisation energy of an atom of an element? Arrange Li, Rb, K and Na in the increasing order of their ionisation energy.
Or
Mention the similarity of properties of hydrogen with one property of Group 1 elements and two properties of Group 17 elements.
Answer:

Ionisation energy of an atom of an elemen

It is the minimum energy required to remove a valence from a neutral gaseous atom Rb < K < Na < Li.

Like all other Group 1 elements it has one electron in its outermost shell. Like all other group 17 elements it acquires noble gas configuration on gaining one electron.

Question 10. What is present along with pure alumina in the molten mixture which is electrolysed for the extraction of aluminium by electrolysis? What are used as cathode and anode in this electrolysis?
Answer:

Fluospar and Cryolite is used in the molten mixture.

Cathode – Graphite

Anode – Graphite rods.

“WBBSE Class 10 Model Question Paper 2023, Physical Science and Environment Set 3 exam pattern”

Question 11. Write the conditions and balanced chemical equation for the industrial production of ammonia by Haber’s process.
Answer:

Hober’s Process.

$\mathrm{N}_2+3 \mathrm{H}_2 \underset{550^{\circ} \mathrm{C}}{\stackrel{200 \mathrm{atn}}{\longrightarrow}} 2 \mathrm{NH}_3$

Condition.

A pressure of 200 atm is required. Higher than this pressure may damage the plant.

A temperature of 550° C is maintained below this temp the process will be very slow.

A catalyst like finely divided iron is used, so that the maximum surface area of the catalyst can be used & molybdenum is used to increase its efficiency.

“Class 10 WBBSE Madhyamika Model Question Paper 2023, Physical Science Set 3 practice questions”

Question 12. The molecular formula of an organic compound is C₂H₄O₂. The compound is soluble in water and on the addition of NaHCO₃. to the aqueous solution of the compound CO₂ evolved. Identify the organic compound. Write, with conditions and a balanced chemical equation, the reaction of the compound with ethanol.
Or
Compare three properties of organic and inorganic compounds..
Answer:

The molecular formula of an organic compound is C₂H₄O₂. The compound is soluble in water and on the addition of NaHCO₃. to the aqueous solution of the compound CO₂ evolved.

Organic Compound is – CH₃COOH

CH₃COOH + C₂H₅OH H₂SO₄ → CH₃ -C-O-C₂H₅+H₂O

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 3 Organic Compounds And Inorganic Compounds

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 1

April 21, 2025August 23, 2023 by Marksparks .arkansas

WBBSE Class 10 Physical Science Question Answer In English

Multiple Choice Question And Answers Physical Science And Environment

Question 1. Which among the following gases does not help in the depletion of ozone in the ozone layer?

NO
NO₂
CFC
CO₂

Answer: 4. CO₂

Question 2. What is the value of PV for 11.2 litres of an ideal gas at STP?

2RT
RT
0.5RT
11.2 RT

Answer: 3. 0.5RT

Read And Learn More: WBBSE Solutions For Class 10 Physical Science And Environment

Question 3. According to the following chemical equation CH₄+2O₂→ CO₂ +2H₂O What volume of 0, will be required to burn 10 moles of CH₄ at STP?

448L
224L
44.8L
22.4L

Answer: 1. 448L

Question 4. Which among the following substances has the highest heat conductivity?

Silver
Diamond
Copper
Aluminium

Answer: 2. Diamond

WBBSE Madhyamika Model Question Paper 2023 Physical Science And Environment Set 1

Question 5. If a beam of red light and a beam of violet light are incident at the same angle on the inclined surface of a prism from an air medium and produce angles of refraction r and v respectively, which of the following is correct?

r = v
r= $\frac{1}{V}$
r> v
r < v

Answer: 4. r<v

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 1”

Question 6. A point source of light is placed at the centre of curvature of a concave mirror. The angle of deviation of the rays incident on the mirror from this source and reflected from it is

0°
180°
90°
360°

Answer: 2. 180°

Question 7. Coulomb’s law related to electric charges is applicable when the two charges

One is the point, One is spherical
Both are spherical
One is the point, One is extended
Both are points

Answer: 4. Both are points.

Question 8. The characteristics of a fuse wire are

High resistance high melting point
Low resistance, low melting point
Low resistance high melting point
High resistance low

Answer: 2. High resistance low melting point.

Question 9. Present in a particle

One proton, one neutron
One proton
Two protons two neutrons
One electron

Answer: 3. Two protons, two neutrons.

“Class 10 WBBSE Model Question Paper 2023, Physical Science and Environment Set 1 study material”

Question 10. Which of the following is not a periodic property of elements?

Density
Melting point
Boiling point
Radioactivity

Answer: 4. Radioactivity

Question 11. In which of the following compounds there is no existence of molecules?

Hydrogen chloride
Calcium oxide
Methane
Ammonia

Answer: 4. Ammonia.

Question 12. Which of the following statements is correct in the case of electrolysis of CuSO₄ solution using Cu electrodes?

The mass of the cathode decreases
The mass of the anode increases
The concentration of CuSO₄ in the solution decreases
The concentration of CuSO₄ in the solution remains unchanged

Answer: 4. The concentration of CuSO₄ in the solution remains unchanged.

Question 13. What colour is produced when H₂S gas is passed through an alkaline aqueous solution of sodium nitroprusside?

Violet
Orange
Deep blue
Green

Answer: 1. Violet.

Class 10 Maths	Class 10 Social Science
Class 10 English	Class 10 Maths
Class 10 Geography	Class 10 Geography MCQs
Class 10 History	Class 10 History MCQs
Class 10 Life Science	Class 10 Science VSAQS
Class 10 Physical Science	Class 10 Science SAQs

Question 14. The formula of red haematite, an area of iron, is

FeO
Fe₂O₃
Fe₃O
FeCO₃

Answer: 3. Fe₂O₃

Question 15. By the reaction of aqueous NaHCO₃ with which of the following compounds CO₂ is produced?

CH₃ CH₂ OH
CH₃CHO
CH₃COCH₃
CH₃COOH

Answer: 4. CH₃COOH

Class 10 Physical Science WBBSE Physical Science And Environment Answer The Following Questions

Question 1. Which fuel gas is harvested from the coal bed?
Answer: Methane
Or
Name a gas present in the air, the increase in the amount of which causes global warming.
Answer: CO₂

Question 2. Name an energy source which can be used for sustainable development.
Answer: Solar energy.

Question 3. Write whether the following statement is true or false. The speed of the gas molecules contained in a closed vessel at fixed temperature and pressure is the same.
Answer: True.

Question 4. What is the nature of the V versus T graph according to Charles’s law?
Answer: If the straight line increase backwards, it will go through the main point.

Question 5. Write whether the following statement is true or false. Among copper, invar and iron the linear expansion coefficient of iron is the lowest.
Answer: True.
Or
What is the unit of volume expansion coefficient?
Answer: K^(-1).

“WBBSE 2023 Madhyamika Physical Science and Environment, Set 1 Model Question Paper download”

Question 6. What is meant by the pole of a spherical mirror?
Answer: It refers to the middle point of the spherical mirror.

Question 7. Write down one use of X-Ray.
Answer: Medical science.

Question 8. Name a machine where electrical energy is converted to mechanical energy.
Answer: Electric motor.

Question 9. Apart from the live wire, what are the two other wires in the household circuit?
Answer: Neutral wire and earth wire.

Question 10. Which kind of nuclear reaction produces energy in a nuclear reactor?
Answer: Nuclear division.
Or
Give an example of a natural radioactive element.
Answer: Uranium / Thorium.

Question 11. Match the Right column with the left column.
Answer:

Left column – Right column

Krypton – A noble element

Neptunium – Transuranic element

Copper – Prepared by carbon reduction of the oxide of the metal

Zinc – In the alloy brass, the metal whose percentage amount is higher than that of the other metal.

Question 12. Between chloroform and sodium chloride which is not soluble in water?
Answer: Chloroform.

Question 13. Name e metal which is extracted by the process of electrolysis.
Answer: Aluminium (Al)
Or
Which is the anode in the electroplating of silver or brass spoon?
Answer: Silver.

Class 10 Physical Science WBBSE

Question 14. Which energy causes chemical reactions during electrolysis?
Answer: From electrical energy to chemical energy.

Question 15. Show with the help of an appropriate litmus paper that the aqueous solution of ammonia is alkaline in nature.
Answer: Red litmus paper will become blue.

Question 16. NaOH+H₂S →______+H₂O
Answer: Na₂S.

Question 17. Write one use of Urea.
Answer: Fertiliser.

Question 18. Write the structural formula of propanone.
Answer: CH₃-CO-CH₃Or
Wohler first prepared an organic compound from an inorganic compound in the laboratory. What is the organic compound?
Answer: Urea.

Question 19. Give an example of a biodegradable natural polymer.
Answer: Cellulose of Protein.

Class 10 Physical Science WBBSE Physical Science And Environment Answer The Following Questions

Question 1. Write with reason in which layer among the layers of the atmosphere the pressure is the highest.
Answer:

The pressure is the highest in Troposphere, because among all the layers of the atmosphere, the temperature is the lowest ture Moreover, when temperature decreases, pressure increases.

Question 2. Find out the ratio of the Volumes occupied by 32g O₂ and 44g CO₂ gases at 27°C temperature and 700 mm Hg pressure (C= 12, O= 16).
Answer:

Molar mass of O₂ = (16+16) 32 g = 1 mol

Molar mass of CO₂ = (12+32) 44 g = 1 mol

So, Volume of both is = 22.4 litre.

Ratio of volume = 1:1

A fixed mass of a gas occupies a volume of 520 cm3 at -13°C temperature. Keeping the pressure unchanged, when the gas is heated the volume of the gas increases to 700 cm. What is the final temperature of the gas in degrees Celsius?
Answer:

Given

A fixed mass of a gas occupies a volume of 520 cm3 at -13°C temperature. Keeping the pressure unchanged, when the gas is heated the volume of the gas increases to 700 cm.

Mass of the gas const. (m)

According to the problem

T₁ =-13°C= 273° + (-13°)K

= 260K.

V₁= 520 cm³

P₁ = 76 cm [Mercury pressure]

T₂ = ?

V₂ = 700 cm³

P₂ = 76cm.

⇒ $\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

⇒ $\frac{76 \times 520}{260}$

∴ $=\frac{76 \times 700}{T_2}$

T₂ = 350

= (350-273)

= 77°C.

Final temperature = 77° C./350K.

“WBBSE Madhyamika 2023 Physical Science and Environment, Set 1 Model Question Paper with solutions”

Question 3. Mention two features of the image formed by a simple camera.
Answer:

The image is real.
The image is inverted.

Or
Where in front of a concave mirror image of an extended object placed at infinity will be formed by the mirror? Mention one feature of the image.
Answer: The object should be placed within the focus of the mirror. The image will be real and inverted.

Question 4. Mention one similarity and one dissimilarity between electromotive force and potential difference?
Answer:

Similarity – Both are related to energy.

Dissimilarity:

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 1 Electromotion Force And Potential Difference

Question 5. By giving an example of an iconic compound show that its ions do not obey the octet rule.
Answer:

PCI₂ (The electron number in the outermost orbit is 10. P is the central atom.
Or
Explain why the melting point of sodium chloride is much greater than that of glucose.
Answer: Salts have high intermolecular forces. Hence, they have a higher melting point than covalent compounds like glucose.

Question 6. Show that F forms an ionic bond with Na but it forms a covalent bond with H (The atomic number of H F and Na are 1, 9 and 11 respectively).
Answer:

H and F both are non-metals; so, they share their electrons and make covalent bonds. But as Na is Metal. It gives e to F and Na+, and F^– are formed and they form ionic bonds.

Question 7. Write with the balanced chemical equation, what happens when Nitrogen gas is passed over calcium carbide heated at 1100°C.
Answer:

⇒ $\mathrm{CaC}_2+\mathrm{N}_2 \stackrel{1100^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{CaCN}_2+\mathrm{C}$

Question 8. Write the balanced chemical equation of the reaction for the formation of metallic iron from ferric oxide by thermit process. Write an application of the process.
Answer:

⇒ $\mathrm{Fe}_2 \mathrm{O}_3+2 \mathrm{Al} \stackrel{\text { Heat }}{\longrightarrow} 2 \mathrm{Fe}+\mathrm{Al}_2 \mathrm{O}_3$

Or
Write the balanced chemical equation of the reaction that occurs when a piece of metallic CuSO What information is obtained from this reaction about the relative position of Cu and Fe in the activity series of metals?
Answer:

Fe + CuSO₄– FeSO₄ + Cu↓

On the submerged portion of Fe, the red metallic copper layer is formed

Question 9. Select the members of a homologous series from the following compound and arrange them in increasing order of their molecular weights: CH₃COOH, CH₃CH₂OH, CH₃OCH₃, CH₃OH, C₂ H₄, C₂H₆, CH₃CH₂OH, C₃H₄
Answer:

The members are CH₃CH₂OH, CH₃OH. Arranged in increasing order – CH₃OH, CH₃C H₂OH.
Or
Write with an example what is meant by functional group.
Answer:

Function groups are specific substituents within melecules that are respon- sible for the characteristic chemical reaction of those melecules. Example-

⇒ CH₃ – CH₂– OH (Ethyl alcohol)

⇒ CH₃– O – CH₃ (Diethyl ether)

Class 10 Physical Science Solution WBBSE Physical Science And Environment Answer The Following Questions

Question 1. State Avogadero’s law.

At a certain fixed temperature and pressure, the molar volumes (V/n) of the real gases are nearly equal and at STP the limit is 22.4 L met. How Avogadro’s law can be arrived at from this information obtained from experiments?
Answer:

Avogadero’s law

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Given

At a certain fixed temperature and pressure, the molar volumes (V/n) of the real gases are nearly equal and at STP the limit is 22.4 L met.

1 mol of gas contains 6.022×10²³ numbers of atoms.

6.022×10²³ numbers of atomic volumes are very merged in respect of the volume of gas.

Question 2. A and B react to produce C according to the following chemical equation. 2A+ B →2C A, B and C are the formulas for three gaseous substances. The vapour densities of A and B are 32 and 16 respectively. Find out the vapour density of C.

Answer:

Given

A and B react to produce C according to the following chemical equation. 2A+ B →2C A, B and C are the formulas for three gaseous substances. The vapour densities of A and B are 32 and 16 respectively.

2A+B-2C

Vapour density of A = 32

Vapour density of B = 16

We know,

Vapour density = 1/2 x molecular mass.

∴ Molecular mass of A = (2 x 32) = 64 g

∴ Total Molecular mass reactant (64×2) + 32 = 160 g

B(2×16)= 32 g

‍ ∴ Mass of 2C = 160g

Mass of C= $\frac{160}{2}$g

C= 80g

∴ Vapour density of C = 1/2 x molecular mass

= $\frac{1}{2}$ x 80=40 g

According to the following chemical equation:

2ZnS +3O₂ → 2ZnO +2SO₂from 100 mole of ZnS
How many grams of ZnO, and
How many mol of SO, will be produced? (Zn65.5, S = 32, O= 16)

Answer:

2ZnS +3O₂ → 2ZnO +2SO₂

⇒ 2 x (65.5 + 32) → 2 x (65.5+ 16)

⇒ 195→ 163

1= 163/195

ZnO = 8,150gm

195→ (32+32) = 128

195→128

1 → 128/195

9750→ $\frac{128 \times 9750}{195}$

= 6400gm

= $\frac{6400}{64}$

= 100 mole SO₂

8,150 gm of ZnO, and
100 mole of SO, will be produced.

Or
Give an example of the volume expansion of a liquid on heating. The area of a solid substance at a temperature of T₁ K is A sqm and that at a temperature of T₁K is A, sqm. Write down the mathematical expression of or the coefficient of area expansion with unit of that solid substance.
Answer:

Given

The area of a solid substance at a temperature of T₁ K is A sqm and that at a temperature of T₁K is A, sqm.

The volume of alcohol in an alcohol thermometer increases at the time of increase in temperature.

At temperature T₁ the area is A₁ of the solid

At T₂ the area is A₂

∴ The mathematical expression of the coefficient of area expansion is $\beta=\frac{A_2-A_1}{T_2-T_1}$

Unit cm/°c in CGS

mr^-1 in SI.

“WBBSE Madhyamika Model Question Paper 2023, Physical Science and Environment Set 1 PDF”

Question 3. Write down the three factors on which the conduction of heat through a solid substance depends.
Answer:

The factors are:

Area of the width intersection
Change of temperature, and
Length of the solid substance.

Question 4. What is the dispersion of light? Will there be a dispersion of white light within a glass slab after retraction when white light is incident on the glass slab at an angle of 45°?
Answer: The phenomenon of splitting white light into its constituent colours is known as the dispersion of light.
Answer: Yes.

Question 5. The principal section of the prism is an equilateral triangle If a ray of light is incident at an angle of 30° on one of the refracting surfaces and emerges at an angle of 45° from the other refracting surface, what is the angle of deviation?

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 1Equilateral Triangle

δ =i₁+i₂– A

= 30°-45-60° (As it is an equilateral triangle)

= 15°

Question 6. Two metallic conductors A and B of the same length have resistivities of 1.6×10⁸Ω and 3.2×10^-8 m respectively. These two conductors are separately connected to the same potential difference what would be the ratio of their cross sections in order to have the same current flowing through each of them?
Answer:

Given

Two metallic conductors A and B of the same length have resistivities of 1.6×10⁸Ω and 3.2×10^-8 m respectively. These two conductors are separately connected to the same potential difference

R= $\frac{P l}{A}$ and

V=IR

R= $\frac{V}{l}$

R = $\frac{V}{l}$

= $\frac{P l}{A}$

As potential difference(v), current (I)and Length(l) are

∴ $\frac{P}{A}$ Constant

∴ $\frac{P_1}{A_1}=\frac{P_2}{A_2}$

⇒ $\frac{1.6 \times 10^{-8}}{A_1}=\frac{3.2 \times 10^{-8}}{A_2}$

⇒ $\frac{A_2}{A_1}=\frac{2}{1}$

⇒ A₁: A₂ = 1:2

The series combination of two 10Ω-ohm resistances is connected in parallel combination with a 20-ohm resistance. Determine the equivalent resistance of the final combination.
Answer:

Given

The series combination of two 10Ω-ohm resistances is connected in parallel combination with a 20-ohm resistance.

Series combination:

∴ Equivalent resistance = Rp.

∴ Rp = R₁+R₂

= 10Ω+10Ω = 20 Ω

∴ Rp =20 Ω

WBBSE Madhyamik Model Question Paper 2023 Physical Science And Environment Set 1 Physical Science And Environment Equivalent Resistance

Parallel combination

Equivalent resistance = $\frac{1}{R p}$

⇒ $\frac{1}{R p}=\frac{1}{R_1}+\frac{1}{R_2}$

= $\frac{1}{20}+\frac{1}{20}$

= $\frac{1+1}{20}$

= $\frac{2}{20}$

= $\frac{1}{10} \Omega$

Rp=10 Ω

“Class 10 WBBSE Madhyamika Model Question Paper 2023, Physical Science Set 1 practice questions”

Question 7. What is meant by electrical power? The dating of a bulb is written as 220V- 100W. What is its meaning?
Answer:

Electrical power

Electrical power is the rate of consumption of electrical energy in respect of the time of some electric machine. We mean that the bulb is to be used at 220 volts potential difference to glow fully and the electrical energy will be spent at the rate of 100 joule/second or a power of 100w will be required.

WB Class 10 Physical Science Question Answer Question 8. From which part of the atom is the radioactive rays emitted? Which of the radioactive rays has the highest penetrating power and which has the highest ionising power?
Answer:

From the nucleus:

γ Ray has the highest penetrating power.
α Ray has the highest ionising power.

Question 9. Mention the dissimilarity of properties of hydrogen with one property of group 1 elements and two properties of group 17 elements.
Answer:

Dissimilarity with group 1 elements Hydrogen is non-metal and diatomic gas. Group I elements or bases are monoatomic hard substances.

Hydrogen is an electropositive element; halogens are electronegative.
Hydrogen is a reductant, but halogens are oxidants.

Arrange as directed:

Na(11), K (19), Li (3), and Rb (37) belong to Group 1 of the long periodic table according to decreasing order of atomic radius.
S (16) O (8), Te (52), and Se (34) belong to Group 16 of the long periodic table according to increasing order to electronegativity.
Ca(20), Be (4), Sr (38), and Mg (12) belong to group 2 of the long periodic table according to decreasing order of reducing power.

(The atomic numbers have been given within the first brackets after the symbols of the elements):

Rb> K > Na > Li
Te< Se <S<0
Sr>Ca> Mg> Be [Reducing powers increases from up to down]

Question 10. On what basis electrolytes have been classified as strong and weak system slectrolytes? Give an example of a strong electrolyte.
Answer:

The electrolytes are classified as strong and weak on the basis of how much they are ionised. Strong electrons are ionised almost completely the weak electro- lytes are ionised a little.

NaOH is a strong electrolyte.

“WBBSE Class 10 Model Question Paper 2023, Physical Science and Environment Set 1 exam pattern”

Question 11. Write mentioning the name of the catalyst and condition, how nitric oxide is manufactured by oxidising ammonia with the help of aerial oxygen. Write also the balanced chemical equation of the reaction.
Answer: Catalyst- Pt or Pt Rh. condition 5-7 atm, 700° – 800° c, 0.0014 second.

⇒ $4 \mathrm{NH}_3+5 \mathrm{O}_2 \frac{\text { Pt or Pt Rh (Catalyst })}{5-7 \mathrm{~atm}, 700-800^{\circ} \mathrm{C}, 0.0014 \mathrm{sec}}=4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O}$

Question 12. Two different organic compounds A and B have the same molecular formula of C₂H₆O, A reacts with metallic sodium to produce hydrogen gas but B does not react with metallic sodium. Write structural formulas of the compounds A and B. Write the balanced chemical equation of the reaction of A with metallic sodium. A = CH₃CH₂OH B = CH₃OCH₃

Two different organic compounds A and B have the same molecular formula of C₂H₆O, A reacts with metallic sodium to produce hydrogen gas but B does not react with metallic sodium.

⇒ $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \mathrm{Na} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{ONA}+\frac{1}{2} \mathrm{H}_2$

Or
Write the condition for the reaction of the addition of hydrogen to ethylene. Write the balanced chemical equation of the reaction. Mention one use of CNG.

Answer: In the presence of Pt, Pd or raney nickel, the reaction of the addition of hydrogen is dove

⇒ $\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \frac{\mathrm{Pt}, \mathrm{pd}}{\text { or, RaneyNi}} \mathrm{CH}_3–\mathrm{CH}_3$

CNG is used as fuel.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter

April 21, 2025August 22, 2023 by Marksparks .arkansas

Class 10 Physical Science WBBSE Chapter 8 Physical And Chemical Properties Of Matter MCQs

Question 1. The basis of the modern periodic table is

Atomic volume
Atomic number
Atomic size
Atomic weights

Answer: 2. Atomic volume.

Question 2. Which of the following has the maximum value of electron affinity?

Answer: 2. Cl.

Read And Learn More: WBBSE Solutions For Class 10 Physical Science And Environment

Question 3. Characteristics of transition elements is incomplete

d-orbitals
f-orbitals
p-orbitals
s-orbitals

Answer: 1. d-orbitals.

Question 4. The element with the highest first ionization potential is

Boron
Carbon
Nitrogen
Oxygen

Answer: 3. Nitrogen.

Class 10 Physical Science WBBSE

Question 5. The long form of periodic table based on

Atomic number
Atomic mass
Number of neutrons
None of these

Answer: 1. Atomic number.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter

Question 6. How many periods are there in the Mendeleev’s periodic table

Answer: 1. 7

“WBBSE Class 10 Physical Science and Environment Chapter 8 solutions, Physical and Chemical Properties of Matter”

Question 7. The percentage of noble gas present in the air is

5%
0.01%
1%
11%

Answer: 3.1%.

Question 8. The shape of the NaCl crystal is

Tetrahedral
Octahedral
Hexagonal
Icosahedral

Answer: 2. Octahedral.

Question 9. Which has the largest first ionization energy?

Answer: 1. 4.

Class 10 Physical Science WBBSE

Question 10. Which one of the following has the largest size?

Br^-1
I
I^–

Answer: 3. I^–

Question 11. Which one of the following is the smallest in size?

N^3-
O²^–
F^–
Na^–

Answer: 4. Na^–

Question 12. The element in the third group and third of the periodic table is

Answer: 3. Al

Question 13. 5f subshell is successively filled up in

Actinoids
Lanthanoids
Tropical metals
Normal metal

Answer: 1. Actinoids.

Class 10 Maths	Class 10 Social Science
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Class 10 Geography	Class 10 Geography MCQs
Class 10 History	Class 10 History MCQs
Class 10 Life Science	Class 10 Science VSAQS
Class 10 Physical Science	Class 10 Science SAQs

Question 14. Which of the following form coloured salts?

Non-metals
Metals
p-block elements
Transmission elements

Answer: 2. Metals.

Question 15. In the long-form periodic table, the block containing non-metals is

Answer: 2. p

Question 16. Lithium shows a diagonal relationship with

Answer: 2. Mg.

Question 17. The iso-electronic atom of Ca²⁺ of Ca²⁺is

Answer: 3. Ar.

Question 18. Which of the following substances has the highest melting point?

NaCl
KCl
MgO
BaO

Answer: 3. MgO.

Question 19. The total number of electrons that take part in forming bonds in N₂ is

Answer: 3. 6

Question 20. The compound which contains both ionic and covalent bonds is

CH₄
H₂
KCN
KCI

Answer: 3. KCN

Question 21. C-C bond length in the saturated compound is

1.47A°
1.38A°
1.54A°

Answer: 3. 1.54A°.

Question 22. The shortest carbon bond distance is found in the?

Diamond
Benzene
Acetylene
Ethane

Answer: 3. Acetylene.

Question 23. Which of the following is the most polar?

CaCl₄
CCI₂
Sncl₄

Answer: 3. Sncl₄

Question 24. Which of the following exists as monoatomic?
Answer:

Sulphur
Helium
Fluorine

Answer: 2. Helium.

“Class 10 WBBSE Physical Science Chapter 8 solutions, Physical and Chemical Properties of Matter study material”

Question 25. Which class of substance is a common component of all electrovalent compounds?

Non-Metal
Metal
Metalic oxide

Answer: 3. Metal.

Question 26. Which of the following bonds will be non-polar

N – H
C – H
F- F
O- H

Answer: 3. F – F.

Question 27. Which of the following is more covalent?

SnCI₄
SnBr₄
Snl₄

Answer: 1. SnCl₄

Question 28. Which among the following bonds shows the maximum bond strength?

Sigma bond
λ bond
Co-ordinate bond
Hydrogen bond

Answer: 1. Sigma bond.

Question 29. The element with the atomic number 9 car exhibits an oxidation state of

+1
-1
+ 3
-3

Answer: 2.- 1.

Question 30. Which agent conducts electricity during electrolysis?

Water
Ion
Atom
Molecule

Answer: 2. Ion.

Question 31. What is the volume ratio of produced hydrogen and oxygen from the electrolysis of water?

Answer: 3. 2 :1.

Question 32. The flow of current in an electrolyte is due to

Atoms
Electrons
Ions

Answer: 3. Ions.

Question 33. The amount of an ion discharged during electrolysis is not directly proportional to

Resistance
Time
Current
Non of these

Answer: 1. Resistance

Question 34. On the electrolysis of a q. solution of sodium sulphate, on the cathode, we get

Na
H₂
CO
SO₃

Answer: 1. Na.

Question 35. The unit of quantity of electricity is

Volt
Coulomb
Ampere
Ohm

Answer: 2. Coulomb.

Question 36. For the electroplating of nickel, nickel sulphate is mixed with

Water
Nitric acid
Boric acid
H₂SO₄

Answer: 3. Boric acid.

Question 37. During electrolysis of copper using copper electrodes of copper in solution is

Increased
Decreased
Unaltered

Answer: 3. Unaltered.

Question 38. Ammonia is :

Acidic
Basic
Neutral
None of these

Answer: 2. Basic.

Question 39. The high heat of the vaporization of ammonia is due to its

Basic nature
Polar nature
High solubility
Hydrogen bonding.

Answer: 4. Hydrogen bonding.

Question 40. The reaction N₂+3H₂, = 2NH₃, is

Exothermic
Endo thermic
Neither of the two
Both the two

Answer: 1. Exothermic

Question 41. Nessler’s reagent is :

K₂Hgl₄
K₂ Hgl₄+KOH.
K₂Hgl₄
KHgl₄+KOH

Answer: 2. K₂ Hgl₄+KOH.

Question 42. Which is most explosive?

NCI₃
Pcl₃
AsCl₃
All

Answer: 1. NCI₃

Question 43. With an excess of Cl_2, ammonia gives

NCI₃
HCI₃
NH₄CI
NO₂

Answer: 1. NCI₃

Question 44. Non-combustible hydride is

NH₃
PH₃
AsH₃
SbH₃

Answer: 1. NH₃

Question 45. An aqueous solution of ammonia consists of

H⁺
OH^–
NH₄
NH⁺OH^–

Answer: 4. NH⁺OH^–

Question 46. The vapour density of ammonia is

Answer: 2. 8.5.

Question 47. H₂S is

Smells like fish
Lighter
Sightly heavier than air
None of these

Answer: 1. Smells like fish.

Question 48. The gas which is absorbed by NaOH

NH₃
H₂S
O₂
None of these

Answer: 2.H₂S

Question 49. H₂ S gas is passed through

P₂O₅
CaCl₂
CaO
None of the above

Answer: 1. P₂O₅

Question 50. The molecular weight of H₂S is

Answer: 2. 34.

Question 51. The percentage of nitrogen in the air by volume is

Answer: 78.

Question 52. The solubility of nitrogen in water is

Low high
Very low
High
Very high

Answer: 4. Very high.

Question 53. Nitrogen-containing organic fertilizer is

Nitro slim
Ammonium sulphate
Urea
Ammonium nitrate

Answer: 3. Urea.

Question 54. Who discovered nitrogen

Daniel Rutherford
Bohr
Dalton
Lavoisier

Answer: 3. Daniel Rutherford.

Question 55. In the contract process for the production of H₂SO₄, on an industrial scale, the impurities of arsenic are removed by :

Fe₂O₃
Al (OH)₃
Fe (OH)₃
Cr (OH)₃

Answer: 3. Fe(OH)₃

WBBSE Class 10 Physical Science Solutions

Question 56. Oil of vitriol is

H₂SO₃
H₂SO₄
H₂S₂O₇
H₂S₂O₈

Answer: 2. H₂SO₄

Question 57. Aquafortis is

HNO₃
HNO₂
H₂NO₃
H₂N₂O₂

Answer: 1.HNO₃

“WBBSE Class 10 Physical Science Chapter 8, Physical and Chemical Properties of Matter solved examples”

Question 58. Oleum is

H₂SO₅
H₂S₂O₇
H₂S₂O₈
H₂SO₃

Answer: 2. H₂S₂O₇

Question 59. Most of the plants contain

Iron
Zin C
Sodium
Potassium

Answer: 4. Potassium.

Question 60. The fluorspar is

CasO₄
Baso₄
CaCO₃
CaF₂

Answer: 4.CaF₂

Question 61. Which of the following is not an area of calcium?

Gypsum
Limestone
Dolomite
Carnalite

Answer: 4. Carnalite

Question 62. Haematite is

Fe₃O₄
Fe₂O₃
Fe₂O₄

Answer: 2. Fe₂O₃

Question 63. Which of the elements is present in haemoglobin?

Answer: 2. Fe.

Question 64. Pin iron is also called

Wrought iron
Cast iron.
Steel
Stainless steel

Answer: 2. Cast iron.

Question 65. The metal present in insulin is

Copper
Iron
Zin C

Answer: Zin C.

Question 66. Which of the following contains nitrogen?

Fats
Proteins.
Carbohydrate
None

Answer: 2. Proteins

Question 67. Which of the following gases is used in welding?

C₂H₄
C₂H₂
CH₄
C₂H₂

Answer: 2. C₂H₂

Question 68. Marsh gas mainly contains

H₂S
CO
CH₄
C₂H₂

Answer: 3. CH₄

Question 69. The general formula for alkenes is

CnHzn
CnH2n+2
CnHzn-2
C2nHzn

Answer: 1. CnHzn.

“WBBSE Class 10 Physical and Chemical Properties of Matter solutions, Physical Science and Environment Chapter 8”

Question 70. Acidic hydrogen is present in

Ethyne
Ethene
Benzene

Answer: 1. Ethyne.

Class 10 Physical Science Solution WBBSE Chapter 8 Physical And Chemical Properties Of Matter Very Short Answer Type Questions

Question 1. Periodic changes of what occurs in a periodic table?
Answer: Physical and chemical properties of sets of elements arranged in some manner change periodically.

Question 2. What is the name ‘Oxygen’ group?
Answer: Chalcagens, group.

Question 3. What are ‘d’-block elements called?
Answer: ‘d’-block elements are called transition elements.

Question 4. Is radioactivity a periodic property?
Answer: Radioactivity is not a periodic property.

Question 5. Between Na and Nat which one has greater in size.
Answer: Na has greater in size than Na.

Question 6. Are vander whal’s radii larger than covalent radii?
Answer: Van der Waal’s radii are larger than covalent radii.

Question 7. Have ionisation energies a positive value?
Answer: Ionisation energies have a positive value.

Question 8. What are the elements of the second period called?
Answer: Elements of the second period are called bridge elements.

Question 9. Is the second I.P is always greater than the first I.P. of a given species?
Answer: The second I.P. is always greater than the first I.P. of a given species.

Question 10. What is the valency of an alkali metal?
Answer: The valency of an alkali metal is 1.

“Class 10 WBBSE Physical Science Chapter 8, Physical and Chemical Properties of Matter easy explanation”

Question 11. What is the relation between N^3- and O^2-?
Answer: N^3- and O^2-are iso-electronic ions.

Question 12. Between Br and I which one has a metallic character?
Answer: Between Br and I, iodine has a metallic character.

Question 13. What is the valency present in NaH?
Answer: Electrovalency is present in NaH.

Question 14. Which one of NaCl and C₆H₁₂O₆will have a higher melting point?
Answer: NaCl.

Question 15. Between MgCl₂, and CHCI₃, which one is electro-valent in nature?
Answer: MgCl₂

Question 16. What do you mean by valence shell?
Answer: The outermost shell of an atom is known as the valence shell.

Question 17. What is the nearest inert element of a chlorine atom?
Answer: The nearest inert element of the chlorine atom is argon (Ar).

Question 18. How many long pair (s) is/are present in XeOF₄?
Answer: 1 (one).

Question 19. What is the type of bonding in ferric chloride?
Answer: The type of bonding in ferric chloride is covalent.

Question 20. What is the valency of carbon in CH₄, compound?
Answer: The valency of carbon is 4 in the CH₄ compound.

Question 21. What is called cryolite?
Answer: AIF₃, 3NaF is called cryolite.

Question 22. Is mercury an electrolyte?
Answer: Mercury is a good conductor but not an electrolyte.

Class 10 Physical Science Solution WBBSE

Question 23. Which agent conducts electricity through an electrolyte?
Answer: Jons conduct electricity through an electrolyte.

“WBBSE Class 10 Physical Science Chapter 8 solutions, Physical and Chemical Properties of Matter PDF”

Question 24. What is the function of acid or alkali mixed in water during electrolysis of water?
Answer: Acid or alkali increases the number of ions that help electrolysis.

Question 25. Name two non-electrolytes.
Answer: Sugar solution and glycerine are two non-electrolytes.

Question 26. Name a metal and a non-metal which can be used as electrodes.
Answer: Metal Platinum

Question 27. Between Br^–and OH^– which ion will be discharged first at the anode?
Answer: Between Br^– and OH^–,OH^– ion will be discharged first at the anode as the position of OH^– ion is lower than that of Br^– in the electrochemical series.

Question 28. What is called fluorspar?
Answer: CaF, is called fluorspar.

Question 29. What is Nessler’s reagent?
Answer: Nessler’s reagent Nessler’s reagent is an alkaline solution (KOH) of potassium mercuric iodide [K₂ Hgl₄]

Question 30. Which Particular substance is used to dry ammonia gas?
Answer: Ammonia is dried with calcium oxide (CaO).

Question 31. What is the amount of ammonia present in liquor ammonia?
Answer: 35% (by weight)

Question 32. What is the density of H₂S at NTP?
Answer: The density of H₂S gas is 1.53 g/L at NTP.

Question 33. Which acid is used to prepare H₂S gas in the laboratory?
Answer: Dilute sulphuric acid.

Question 34. Mention one identifying test for H₂S.
Answer: head acetate paper turns black when it is held in HS gas.

Question 35. Name the chemicals used in the laboratory method of preparation of nitrogen.
Answer:

Chemicals required: Ammonium chlorides (NH₄CI) and sodium nitrite (NaNO₂.)

Question 36. What are the ways of fixation of nitrogen?
Answer:

There are two ways of fixation of nitrogen:

By electric discharge
The biochemical reaction through bacteria

Question 37. What is the molecular weight of nitric acid?
Answer: The molecular weight of nitric acid is 63.

Question 38. What is the specific gravity of sulphuric acid?
Answer: 1.84.

Question 39. Which Acid is called the ‘King of chemicals”?
Answer: Sulphuric acid (H₂SO₄) is called the ‘king of chemicals’.

Question 40. What is laughing gas?
Answer: Nitrous oxide (N₂O) is called laughing gas.

Question 41. Name a reducing acid.
Answer: Hydrochloric acid (HCI) is an example of a reducing acid.

Question 42. What is royal water?
Answer: Aqua regia is also known as royal water.

Question 43. What is the basic component of chuni (ruby), Panna (emerald) etc.?
Answer: The basic component of chuni and panna is Al₂O₃

Question 44. What is called Mohr’s salt?
Answer: FeSO₄, (NH₄)₂, SO₄, 6H₂O is called Mohr’s salt.

Question 45. What is Kipp’s base?
Answer: A mixture of FeS+H₂SO₄, is called Kipp’s base.

“WBBSE Class 10 Physical Science Chapter 8, Physical and Chemical Properties of Matter important questions“

Question 46. What is called Fool’s gold?
Answer: CuFeS₂ is called Fool’s god.

Question 47. What is Hydrolith?
Answer: Calcium hydride is known as hydrolith.

Question 48. What is ketose?
Answer: The carbohydrate-containing keto (C = O) group is called ketose.

Question 49. What is amylum?
Answer: Starch is called amylum.

Question 50. What is TEL?
Answer: TEL is commonly used antiknock compound tetraethyl lead, [(C₂H₅)₄, Pb]

Class 10 Physical Science Solution WBBSE Chapter 8 Physical And Chemical Properties Of Matter Fill In The Blanks

Question 1. The energy released when an electron is added to a neutral gaseous atom is called ________ the atom.
Answer: Electron affinity

Question 2. There are ________ elements in the fourth period of the periodic table.
Answer: 18

Question 3. Hydrogen belongs to the________ period.
Answer: IA (one)

Question 4. Helium belongs to the ________ group.
Answer: 0 (zero)

Question 5. Na⁺ ion is ________ in size than no-atom.
Answer: Smaller

Question 6. On Pauling’s electronegativity scale, the element next to F is ________
Answer: Oxygen.

Question 7. Elements of group IA are called ________ metals.
Answer: Alkali

Question 8. The term ‘periodic’ means ________ of anything at regular intervals.
Answer: Recurrence

Question 9. The elements belonging to zero group are chemically________
Answer: Insert

Question 10. 10 strong electro-positive elements are ________ good
Answer: Reducing agent

Question 11. Covalent compounds do not-produce ________ in solution or in fused states, so these are non-electrolytes.
Answer: Ions

Question 12. Almost all the participants in a covalent compound are________
Answer: Non-metals.

Question 13. In an ammonia molecule, each hydrogen atom is related to one pair of shared ________ so the covalency of hydrogen is 1.
Answer: Electrons.

Question 14. Valency of an element in a covalent compound is measured by its relation to the number of shared ________
Answer: Electron-pairs

Question 15. The valency of an element in an electrovalent compound is the number of ________ an atom of it gains or loses in forming the compound.
Answer: Electrons

Question 16. A chemical bond is a strong force of attraction that holds; together ________ in a molecule or crystal.
Answer: Atoms

Question 17. Electrovalency is the type of chemical bonding established by the actual transference of one or more ________ electron (s).
Answer: Valence.

Question 18. In covalency, two or more atoms of the same or different non-metallic elements combine chemically by the process of ________ one or more pairs of valence electrons.
Answer: Sharing.

Question 19. Electrovalent compounds are formed by strong attractive electrostatics between ions of opposite charge.
Answer: Forces.

Question 20. In the formation of all electrovalent compounds, a________ is involved.
Answer: Metal.

Question 21. Formation of NaCl involves ________In energy.
Answer: Decrease.

Question 22. Helium atom contains electron.
Answer: Two (2).

Question 23. Chemical reactions of covalent compounds usually ________
Answer: Slow.

Question 24. Solid NaCl is a ________ conductor of electricity.
Answer: Bad.

“Class 10 Physical Science and Environment Physical and Chemical Properties of Matter solutions, WBBSE syllabus”

Question 25. Covalent compounds are generally soluble in ________ solvents
Answer: Non-polar.

Question 26. ________ Is a vessel in which electrolysis is carried out.is involved.
Answer: Voltameter.

Question 27. In electrolytes, electrical conduction occurs by the migration of________
Answer: Ions.

Question 28. Rubber is________ of electricity.
Answer: Non-conductor.

Question 29. Mercury is a good conductor of electricity but is not an________
answer: Electrolyte.

Question 30. The aqueous solution of sugar is ________
answer: Non-electrolyte.

Question 31. In electrolysis________ conduct electricity.
Answer: Ions.

Question 32. Between H⁺and Al³⁺ _______ion is discharged first at the anode.
Answer: H^–

Question 33. 2H⁺ 2e → ________
Answer: 2H

Question 34. In electroplating, the plating metal is used as________
Answer: Cathode.

Question 35. Aluminium is extracted by the electrolysis of a molten mixture of bauxite and ________
Answer: Cryolite.

Question 36. Ammonia has a typical ________ smell
Answer: Pungent.

Question 37.________ reagent is used for the identification of ammonia.
Answer: Nessler’s.

Question 38. Ammonia is dried with ________
Answer: Calcium oxide.

Question 39. Ammonia is ________ for eyes
Answer: Harmful.

Question 40. 2NH₄Cl+Ca(OH)₂ = __________ +CaCl₂+ 2H₂O
Answer: 2NH₃

Question 41. Ammonia gas is collected by __________ the Displacement of air.
Answer: Downward.

Question 42. An aqueous solution of ammonia is __________
Answer: Alkaline.

Question 43. When ammonia l Eakes from the factory we should wash our eyes with __________
Answer: Water.

Question 44. In ammonia, we have nitrogen and __________
Answer: Hydrogen.

Question 45. Ammonia is dried by passing through __________
Answer: Quick line.

Question 46. Ammonia is __________ soluble in water.
Answer: Highly.

“WBBSE Class 10 Chapter 8 Physical Science, Physical and Chemical Properties of Matter step-by-step solutions

Question 47. H₂S gas is absorbed in __________ Solution
Answer: Sodium hydroxide.

Question 48. __________ Is used in order to dry H₂S
Answer: P₂O₅

Question 49. H₂S is a weak __________ acid.
Answer: Dibasic.

Question 50. Molecular formula of sulphuretted hydrogen is__________
Answer: H₂S

Question 51. H₂S +H₂ SO₄ (cons.) = 5 + __________+ 2 H₂O
Answer: SO₂

Question 53. H₂S gas is 1.5 times heavier than __________
Answer: Air.

Question 54. H₂S gas turns lead acetate __________
Answer: Black.

Question 55. Solubility of nitrogen in water very __________
Answers: Low.

Question 56. Nitrogen is a __________ gas.
Answer: Colour less.

Question 57. Liquid nitrogen is used as __________
Answer: Condenser.

“WBBSE Class 10 Physical and Chemical Properties of Matter, Physical Science Chapter 8 key concepts”

Question 58. N₂ + __________ = 2NH₃Answer: 3H₂

Physics Class 10 WBBSE

Question 59. CaC₂+ N₂ = __________ +C
Answer: CaCN₂

Question 60. Natural source of nitrogen is __________
Answer: Air.

Question 61. Nitrogen gas is collected by the __________ displacement of water.
Answer: Downward.

Question 62. The solubility of nitrogen in water is about __________ m/L at NTP.
Answer: 23.5.

Question 63. Nitrolim reacts with steam to form calcium carbonate and
Answer: Ammonia.

Question 64. Nitrogen is a relatively __________ element
Answer: Non-reactive.

Question 65. MnO₂+4HCI = MnCI₂ +_____ 2H₂O
Answer: Cl₂

Question 66. 4NH₃+ 5O₂=_______+6H₂OAnswer: 4NO

Question 67.____________ acid is known as aquafortis.
Answer: Nitric.

Question 68. The molecular formula of oleum is __________
Answer: H₂S₂O₇.

Question 69. AgNO₃ + HCl = __________ +HNO₃
Answer: AgCl.

Question 70. For dehydration
Answer: Sulphuric acid.

Question 71. The boiling point of sulphuric acid is
Answer: 338°C.

Question 72.__________ is also known as chile saltpetre.
Answer: NaNO₃

Question 73. NaCl + H₂SO₄ = ___________HCl (150°C – 200°C)
Answer: NaHSO₄

“WBBSE Class 10 Physical Science Chapter 8, Physical and Chemical Properties of Matter summary”

Question 74. At high temperatures, HNO₃ decomposes as:
Answer: 4HNO₃ = 4NO₂ +_________ + 2H₂O

Question 75. C₁₂H₂₂O₁₁(sugar) + cone. H₂SO₄ = __________(11HO+H₂SO₄ )
Answer: 12C.

Question 76. The colour of fuming nitric acid is __________
Answer: Yellowish.

Question 77. Concentrated HNO₃ is not used to prepare HCl since the former is highly __________
Answer: Volatile.

Question 78. Concentrated sulphuric acid has a great affinity to __________
Answer: Water.

Question 79. 3 HCI + HNO₃ (cone) = NOCI + __________ +2H₂O (Nitrosyl Chloric)
Answer: 2 (CI).

Physics Class 10 WBBSE

Question 80. Gold is alloyed with copper or silver to make it __________
Answer: Hrder.

Question 81. Zinc oxide is an __________ oxide.
Answer: Amphoteric.

Question 82. The chief ore of aluminium is __________
Answer: Bauxite (AI₂O₃, 2H₂O).

Question 83. The formula of malachite is __________
Answer: CuCO₃, Cu (OH)₂

Question 84. Brass is an alloy of __________
Answer: Copper and zinc.

Question 85. FeSO₄ , 7H₂O is known as __________
Answer: Green vitriol.

Question 86. MgCl₂on electrolysis gives Mg.
Answer: Molten.

Question 87. Bell metal is an alloy of __________
Answer: Copper and tin.

Question 88. Stainless steel is an alloy of __________
Answer: Chromium, nickel and iron.

Question 89. Generally, the density of iron is __________
Answer: 7.85 g/ml.

Question 90. AlC₃ evolves __________ when treated with water.
Answer: Methane.

Question 91. Wurtz reaction is used for the preparation of __________
Answer: Alkanes.

Question 92. _______________ is the abbreviation of polyvinyl chloride.
Answer: PVC.

Question 93. The monomer of____________ is tetrafluoro ethene.
Answer: Teflon.

Question 94. Vitamin C is soluble in____________
Answer: Water.

Question 95. The carbon chains in alkalines are____________
Answer: Zig-Zag.

Question 96. The function group of organic acid is____________
Answer: – COOH.

Question 97. Proteins contain various kinds of acids____________
Answer: Amino.

Question 98._______ and proteins are the polymerisation products of amino____________ acids.
Answer: Polypeptides.

Question 99. Starch is____________saccharide.
Answer: Poly.

Physics Class 10 WBBSE Chapter 8 Physical And Chemical Properties Of Matter Short Answer Type Questions

Question 1. What is atomic volume?
Answer:

Atomic volume: Atomic volume of an element is the volume in cm3 occupied by one gram atom of the element in the solid state hence it is also called gram- atomic volume.

Question 2. What is Mendeleev’s periodic law?
Answer:

Mendeleev’s periodic law (1869): The physical and chemical properties of elements are the periodic function of their atomic numbers.

“WBBSE Class 10 Physical Science Chapter 8 Physical and Chemical Properties of Matter, definitions and examples”

Question 3. What is modern periodic law?
Answer:

Modern periodic law (Moseley): It states that the physical and chemical properties of elements are the periodic function of their atomic numbers.

Question 4. What is ionic radius?
Answer:

Ionic radius: it is defined as the effective distance from the nucleus of an ion upto which it has an influence in the ionic bond.

Question 5. What is the metallic radius?
Answer:

Metallic radius

It is defined as one half of inter-nuclear distance between two nearest metal atoms in a metalic lattice is called metallic radius.

r mettalic $=\frac{\text { Inter-nuclear distance between two nearest metal atoms in metallic lattice }}{2}$

Question 6. What is Ionisation energy (IE) or Ionisation potential (IP) ?
Answer:

Ionisation energy or Ionisation potential: First ionisation energy is defined as the amount of energy required to remove one valence electron from an isolated neutral gaseous atom resulting in the formation of a monovalent positive ion.

Question 7. What is electron affinity?
Answer:

Electron affinity (EA): First electron affinity or simply electron affinity is the amount of energy released when one electron is added to a neutral gaseous atom to form a monovalent negative ion.

Physics Class 10 WBBSE

Question 8. What is the necessity of arranging elements in the periodic table?
Answer:

Necessity of arranging elements in the periodic table: It was observed from the early days that there were few groups of elements each of which had almost identical chemical and physical properties. So, a systematic arrangement of elements is essential since it is difficult to remember the individual properties of elements.

Question 9. What is periodicity?
Answer:

Periodicity: It is the recurrence of elements with similar propertes after certain regular intervals when these are arranged in the increasing order of their atomic numbers.

Question 10. What is the definition of electronegativity?
Answer:

Electronegativity (Pauling’s definition): It is the attractive force which an atom, bonded by a covalent bond exerts on the bond pair of electrons responsible for the covalent bond. The electro-negativity of the elements increases from left to night of the periods in a periodic table.

Question 11. What are s-block elements?
Answer:

S-block elements: These are the elements of IA or 1 group (alkali metals; configuration ns’) and IIA or 2 group (alkaline earth metals; configuration ns3). These are so named because the last electron in them enters S-oribitals.

Question 12. What are p-block elements?
Answer:

P-block elements: These are the elements in which the last electron enters p- orbital of valence shell. The elements with configurations ns2 np’ to ns2 np constitute this block. Thus p-block consists of elements of group IIIA (13), IVA (14), VA (15), VIA (16), VIIA (17) and zero groups (18).

Question 13. What is covalent radius?
Answer:

Covalent radius: It is defined as one-half of the distance between the centres of nuclei of two similar atoms held together by a purely covalent single bond.

Question 14. Explain how the atomic size of the electron change in periodic table?
Answer:

Atomic size: The distance of the outermost orbit from the nucleus of a spherically shaped atom is called the atomic size. The atomic size gradually decreases from left to right of the period upto the group VIIB, again increasing in the end element.

Question 15. What is Vander waal’s radius?
Answer:

Vander Waal’s radius: It is defined as one-half of the internuclear distance between two similar, adjacent atoms belonging to two neighbouring molecules of the same substance in the solid state.

r vander waals= $\frac{\text { (Internuclear distance between two non-bonded nearest neighbouring atoms) }}{2}$

Question 16. How many periods and groups are pressnt in Mendeleev’s periodic table?
Answer: Mendelev’s periodic table contains seven periods and nine groups.

Question 17. What are the conditions necessary for the formation of Ionic bond?
Answer:

The condition necessary for the formation of Ionic bond:

Formation of cation from a neutral atom having low ionisation energy.
Formation of an anion from a neutral atom with high value of electron affinity.
Formation of crystal lattice from oppositely charged ions involving large release of energy.

Question 18. What are the causes of chemical combination?
Answer:

Cause of chemical combination :

Tendency to acquire nobe gas configuration.
Tendency to acquire minimum energy.

Question 19. What are the characleristics of ionic compound’s?
Answer:

The characleristics of ionic compound’s

All ionic compounds are usually crystalline solids and are composed of ions even in the solid state.
lonic compounds have low valatility high density and high stability.
Ionic solid have high melting points and boiling points due to the presence of strong attractive forces between the oppositely charged ions.
lonic compounds are highly soluble in polar solvents (such as water) having high dielectric constant (80) but insoluble in organic solvents (such as benzene, alcohol, ether etc.)

Question 20. What is Ionic bonding?
Answer.

Ionic bonding: The electrostatic force of attraction which holds the oppositely charged ions together is called ionic bond or electrovalent bond and the compounds which are formed by the transference of electrons from one atom to another are known as ionic or electrovalent compounds. The number of electrons which an atom loses or gains while forming an ionic bond is known as electrovalency.

Question 21. What are the factors on which Lattice energy depends?
Answer:

Lattice energy depends upon the following factors:

The magnitude of charge of ions: It increases with an increase in charge on cation-anion or both.
Size of cation: For a common anion, the lattice energy decreases with an increase in size.
Size of anion: For a common cation, the lattice energy decreases, with the increased size of anion.

Question 22. What are the common types of bonds?
Answer: Depending upon the mode of acquiring the nearest noble gas configuration.

There are three common types of bonds :

Ionic or Electrovalent bond
Covalent bond and
Coordinate or Dative bond.

Question 23. State the common reason behind the phenomena of electron release or electron capture by atoms to form ions.
Answer:

Explanation: An element goes into chemical bondage with another element since it tends to attain the stable state of its nearest inert element in the periodic table. To fulfil this, an atom either gains or loses electrons 70 possess 8 electrons or 2 electrons (for hydrogen atoms) in its outermost orbit. As a result of losing electrons, atoms are transformed to positive ions and on gaining electrons atoms transform to negative ions.

Question 24. What is the electronic theory of bonding?
Answer:

Electronic theory of bonding: Atoms combine by transfer of electrons (ionic bonding) or by sharing of electrons (covalent bonding).

Question 25. What is Lattice energy?
Answer:

Lattice energy: The lattice energy of an ionic solid is the amount of energy released when the required number of cations and anions combine to form one mole of an ionic solid.

Question 26. What are the main types of physical bonds?
Answer:

The main types of physical bonds are:

Hydrogen bond
Metallic bond
Vander Waal’s interactions.

Question 27. What is a covalent bond?
Answer:

Covalent bond (G.N. Lewis, 1916): A covalent bond is formed by the mutual sharing of electrons between the atoms, both of which are short electrons. The compound so formed is called covalent compound. The member of electrons contributed by an atom for sharing is known as its covalency.

Depending upon the number of electrons shared between two atoms being one, two or three, we have single covalent bond (: or =), double covalent bond (: : or=) and triple covalent bond (::: or≡).

Question 28. What are the exceptions of the octet rule?
Answer:

Exceptions of octet rule: There are many atoms which do not obey the octet rule and may contain six (as in BF₃), or ten (as in PF₃) electrons.

Question 29. What are the characteristics of covalent compounds?
Answer:

Characteristics of covalent compounds :

Covalent compounds exist in solid, liquid and gaseous state.
These compounds have low melting and boiling points.
These (except graphite) are bad conductors and react sowly.
These are soluble in non-polar solvents such as benzene, acetone etc. but are insoluble in polar solvents such as water.
Due to the directional nature of covalent bonds. These compounds show stereoisomerism.

Question 30. What is octet rule?
Answer:

Octet rule: It states that atoms react because they have a tendency to complete their octet (or duplet) i.e. to have eight electrons in the valence shell (or two electrons if only one shell is present) They can do so by losing, gaining or sharing electrons.

Question 31. Write down the electronic configuration of a chlorine atom, a chlorine ion.
Answer:

Electronic configuration of chlorine atom: 2, 8, 7 electronic configuration of a chlorie ion is: 2, 8

WB Class 10 Physical Science Question Answer

Question 32. Water is generally a good solvent for ionic compounds why?
Answer:

Explanation:

The dielectric constant of water is high (80): The electrostatic force of attraction between oppositely charged ions gets reduced to ($\frac{1}{8}$) of the original force in air. Hence, ions of an electrovalent compound get separated which is then solvated by water.

Question 33. Why does the tendency of sharing electrons grow in many nonmetallic atoms during the formation of covalent compounds?
Answer:

Explanation: During the formation of covalent compounds of non-metallic atoms, each participant involved in the process tends to attain a duplet or octet stable state. To attain such stable state they share electrons.

Question 34. SnCl is a poor conductor of electricity. Why?
Answer:

Explanation: SnCl₄, being a covalent compound does not ionise and hence, is a poor conductor of electricity.

Question 35. What are conductors?
Answer:

Conductors: A substance which allows the electric current to flow through it is called a conductor, for example, Cu, Ag, Al etc.

Question 36. What is Non-conductors?
Answer:

Non-conductors: A substance which does not-conductor. for example, wood, gas, rubber etc.

Question 37. What are electrolytic conductors or electrolytes?
Answer:

Electrolytic conductors or electrolytes: These are compounds which in the fused state on in solution in a suitable. solvent (particulary water) conduct an electric current and undergo distinct chemical Decomposition during the process of conducting current.

For example, Aqueous solution of H₂SO₄, NaOH, NaCl etc.

Question 38. Why is mercury considered as non-electrolyte?
Answer:

Explanation: Mercury is a liquid metal. It conducts electricity but during the passage of electric current, it is not decomposed. So, it is considered as non- electrocyte.

“WBBSE Class 10 Physical and Chemical Properties of Matter, Physical Science Chapter 8 revision notes”

Question 39. What is Electrolysis?
Answer:

Electrolysis: The process of chemical decomposition of an electrolyte in solution or in the state by the passage of electric current is called electrolysis.

Question 40. What is electrolytic conduction?
Answer:

Question Electrolytic conduction: The movement of ions towards oppositely charged electrodes is known as electrolytic conduction.

Question 41. What is called Electrodes?
Answer:

Electrodes: The two metallic or graphite strips or rods, placed in a voltameter to pass electrolyte during electrolysis are known as electrodes.

Question 42. What is called Electrolytic dissociation?
Answer:

Electrolytic dissociation: The splitting up of an electrolyte into ions is known as electrolytic dissociation. This is a reversible process. This means that the electrolyte molecules break up partly into ions and the ions in solution constantly reunite to form the undissociated molecules.

Question 43. What are Metallic conductors?
Answer:

Metallic conductors: These are the substances (metals) which allow the current to pass through them but do not undergo any change in themselves. for example,Cu, Ag, An etc.

Question 44. What are Non-metallic conductors?
Answer:

Non-metallic conductors: These are the substances (non-metals) which allow the current to pass through them out and do not undergo any change in themselves, for example, Graphite, gas-carbon etc.

Question 45. What are the applications of electrolysis :
Answer:

Application of electrolysis:

Electroplating
Electro-refining of metals.
Electro-typing
Extraction of metals.

Question 46. What are the conditions for a good deposit?
Answer:

Conditions for good deposit:

High current density
Low temperature
High metal concentration in the electrolyte
The electrolyte must be complex salt of the metal to be deposited.

Question 47. Explain the electroplating of Ag.
Answer:

Electroplating of Ag:

Electrolyte: Potassium argento cyanide solution K [Ag (CN)₂]

⇒ K [Ag (CN)₂]→ K ̄+Ag‍ ̄+2CN ̄

Electrodes:

Cathode: the article is to be electroplated.
Anode: Pure Ag plate
Reaction at cathode: Ag + e → Ag↓

WB Class 10 Physical Science Question Answer

Question 48. Explain the electroplating of Ni.
Answer:

1. Electroplating of Ni: Electrolyte Nickel sulphate (NiSO) with Boric acid (H₃BO₃)

⇒ NiSO₄……… Ni^2-+ ₄SO^2-

2. Electrodes:

Cathode: The article is to be electroplated.
Anode: Pure Ni-plate.

3. Reaction at the cathode: Ni^2-+2e→ Ni↓

Question 49. What is the cathodic reaction? Give one example.
Answer:

Cathodic reduction: The reduction of ions taking place at a cathode of a cell is known as cathode reduction.

Example: Reduction of Cu^2- ions to cu at the cathode of Daniell cell is an example of cathodic reduction.

Question 50. Anhydrous HCI is a bad conductor of electricity but aqueous HCI is a good conductor. Why?
Answer:

Explanation: Being covalent in nature, anhydrous HCI is a bad conductor. However, in aqueous solution, it ionsies to give H and Cl- ions which conduct the electricity.

Question 51. State two physical properties of HCI :
Answer:

Physical properties of HCI :

It is a colourless gas with a choking smell and strongly fuming in moist air, hydrogen chloride is 1.27 times as heavy as air; it neither burns nor supports burning.
It is highly soluble in water. At 0°C, 450 ml of hydrogen chloride gas is dissolved in 1 ml of water.

Question 52. How is pure hydrogen chloride prepared?
Answer: Preparation of pure hydrogen chloride Pure hydrogen is obtained by the action of water upon silicon tetrachloride.

Equation: SiCI₄ + 4H₂O= Si (OH)₄+ 4 HCI

Question 53. State the reaction of HCI with Na.
Answer:

Reaction of HCI with N

Hydrogen chloride neither burns in the air nor supports combustion. However, burning sodium continues to burn in the gas with a bright yellow flame producing hydrogen and anhydrous sodium chloride.

Equation: 2Na + 2HCl = 2NaCl + H₂↓.

Question 54. What is a reaction between NaCO, and HCI?
Answer:

Reaction of Na₂CO₃ with HCI :

HCl reacts with Na₂CO₃ to liberate carbon dioxide.

Na₂CO₃ + 2HCI = 2NaCl + CO₂ ↑+ H₂O.

Question 55. Why hydrogen chloride is not collected over displacement of water?
Answer:

Reason: Hydrogen chloride is highly soluble in water. It has been found that at 0°C, 450 ml of hydrogen chloride is dissolved in 1 ml of water. So, it is not collected over displacement of water.

Question 56. How copper and silver react with hydrochloric acid?
Answer: Generally, copper and silver are not attacked by hydrochloric acid. In the presence of air, copper and silver react very slowly producing corresponding chlorides and water.

Equations:

2Cu + 4 HCI + O₂= 2CuCl₂ + 2H₂O

4Ag + 4HCI + O₂ 4AgCl + 2H₂O

Question 57. How does ammonia react with hydrogen chloride?
Answer: Reaction of ammonia with hydrogen chloride: Ammonia in contact with hydrogen chloride gives dense white fumes of solid ammonium chloride. This reaction is an example of the formation of a solid product by the interaction of two gases.

Equation: NH₃ (g) + HCI (g) NH₄CI (S)

Question 58. Show the presence of chloride ions in hydrochloric acid.
Answer:

Presence of chloride ion in hydrochloric acid: When hydrochloric acid is added to a solution of silver nitrate, a white precipitate of the chloride of silver is obtained.

Silver chloride is insoluble in nitric acid but soluble in ammonium hydroxide.

Equation: AgNO₃ + HCI + AgCl ↓ + HNO₃

Question 59. State the identification of hydrogen chloride gas.
Answer:

Identification of hydrogen chloride gas: Dense white fumes are formed when a glass rod, moistened with strong ammonium a solution, is held in the hydrogen chloride gas.

Equation: NH₃ (g) + HCI (g) NH₄CI (s)

Question 60. Why nitric acid is prepared at a lower temperature (200°C) in the laboratory?
Answer:

Nitric acid is prepared at a lower temperature (200°C) in the laboratory because.

At high temperatures, nitric acid decomposes.

4 HNO₃= 4NO₂ +O₂ + 2H₂O.

Nitric acid vapour attacks the glass surface of the retort.
Sodium sulphate (Na₂SO₄) formed at higher temperatures sticks to glass and is difficult to remove from the glass retort.

Question 61. What are the physical properties of nitric acid?
Answer:

Physical properties of nitric acid :

Pure nitric acid is a colourless, fuming liquid of specific gravity 1.52 at 15°C
It boils at 86°C and freezes at -42°C into a transparent crystalline substance.
It is highly soluble in water.

WB Class 10 Physical Science Question Answer

Question 62. State the reaction of nitric acid with alkalis.
Answer:

Reaction of nitric acid with alkalis: As nitric acid is a strong acid, it rapidly reacts with and neutralises alkalis to form salt and water.

Equations:

1. NaOH + HNO₃= NaNO₃ + H₂O

2. NH₄ OH + HNO₃= NH₄NO₃+ H₂O

Question 63. What is aqua regia? What is its use?

Answer:

Aqua regia: A mixture of conc. HNO, (1 vol.) and conc. hydrochloric acid (3 vols.) is known as aqua regia.

Au + 4HCI + HNO₃= HAUCI₄+NO+ 2H₂O

(Soluble chloro auric acid)

Use of aqua regia: It dissolves gold and platinum.

“Class 10 WBBSE Physical Science Chapter 8 Physical and Chemical Properties of Matter, multiple-choice questions”

Question 64. What is fuming sulphuric acid or oleum?
Answer:

Fumming sulphuric acid or oleum: Fumming sulphuric acid or oleum is obtained when SO₃ is passed over 98% (approximately) sulphuric acid.

Equation: H₂SO₄+SO₃= H₂S₂O₇ (oleum)

Question 65. What is passive iron?
Answer:

Passive iron: Cold and concentrated nitric acid or fuming nitric acid or fuming nitric acid when comes in contact with iron produces passive iron is chemically inactive.
Cause of passivity: The nitric acid when comes in contact with iron, an insoluble coating of iron oxide (Fe₃O₄) forms on iron at initial stage. The coating makes iron chemically inactive.

Question 66. What is acid rain?
Answer:

Acid rain:

The oxides NO₂ and SO₂ reacting with moisture and oxygen of air corres ponding produce HNO₃ and H₂SO₄. The acids dissolve in rainwater. The rain coming down on earth carrying these acids is known as acid rain.
The function of acid rain The acids present in the rainwater damage buildings monuments, and statues by corrosion. Soil also becomes acidic which causes degradation of soil that in turn causes decline in forest area and agricultural productivity.

Question 67. State the effect of SO₂ pollution.
Answer:

Effect of SO₂ pollution:

SO₂ creates problems in eyes and also in the lungs,
Many diseases like asthma, bronchitis etc. affect if SO₂ is inhaled.

Question 68. What is stone cancer?
Answer:

Stone cancer: In the atmosphere, sulphur dioxide (SO) gas reacts with oxygen and water vapour producing sulphuric acid. The acid being dissolved in rainwater comes down on marble walls. The marble walls thus corrode, the corrosion of the marble is called stone cancer.

Question 69. State physical properties of H₂SO₄.
Answer:

Physical properties of :

It is a colourless, odourless heavy oily liquid.
Its specific gravity is 1.84 and its b.p. (boiling point) is 338°C.
It is soluble in water and it is a corrosive acid.

Question 70. Water is not added on concentrated sulphuric acid to make it dilute why?
Answer:

Explanation: Huge amount of heat is produced if water is added to concentrated sulphuric acid. As a result, water all on a sudden being volatilised spreads all around and Creates a Problem. So conc H₂SO₄ acid is slowly added to with constant stirring.

“WBBSE Class 10 Physical and Chemical Properties of Matter solutions, Chapter 8 worksheet

Question 71. State the reaction of H₂SO₄ with carbonates and bicarbonates.

.Answer: Reaction of H₂SO₄with carbonates and bicarbonates: At ordinary temperature, it liberates carbon dioxide from carbonates and bicarbonates.

Equations:

Na CO₃ + H₂SO₄ = NaSO₄+ CO₂↑ +H₂O

NaHCO₃ + H₂SO₄= NaH₂SO₄ + CO₂↑ +H₂O.

Question 72. Explain the following: Concentrated nitric acid turns yellow in sun light. Answer: Explanation:

Nitric acid turns yellow because of its decomposition forming nitrogen dioxide (NO) gas.

Equation: 4HNO₃ = 4NO₂+ 2H₂O +O₂

Question 73. Concentrate Nitric acid can be kept is a vessel made of aluminium-Why?
Answer:

Reason: Concentrate Nitric acid does not react with aluminium, so it can be kept in the aluminium vessel.

Question 74. What is thermite? What is its use?
Answer:

Thermite: Thermite is a mixture of aluminium powder and ferric oxide.

Use: It is used for welding purposes.

Question 75. State what happens when? Zn is added to the caustic soda solution.
Answer:

Explanation: Zinc reacting with caustic soda solution produces sodium zincate and hydrogen gas.

Equation: Zn + 2NaOH = Na₂ ZnO₂ + H₂↑

Question 76. What is stainless steel? State one use of it.
Answer:

Stainless steel: It is an alloy which composes iron (Fe) = 73%, Chromium (Cr) = 18% Nickel (Ni) = 8% and carbon (C) Uses of stainless steel: It is broadly used to make instruments.

Question 77. Why alumina (Al₂O₃) cannot be reduced by carbon?
Answer:

Explanation: At high temperatures, Al₂O₃reacts with carbon to form aluminium carbide.

Equation: 2Al₂O₃ + 9C → AI₄C₃+6CO

Question 78. Why aluminium cannot be obtained by the electrolysis of bused AICI,?
Answer:

Explanation: Aluminium cannot be obtained by the electrolysis of fused AICI₃, since it is covalent solid and sublimes at 453K.

79. Although less conducting than copper, aluminium is used for power transmission. Why?
Answer:

Explantion: Aluminium being lighter and cheaper than copper is used for power transmission.

Question 80. Why did aluminium act as a good reducing agent?
Answer:

Explanation: Reducing the character of a substance depends upon its affinity for oxygen. Aluminium due to its high affinity for oxygen is a good reducing agent. It reduces a large number of oxides of other metals.

For example: 2A1+ Fe₂O₃ → Al₂O₃ + 2Fe + heat.

Question 81. Explain the following: Anode mud in copper refining contains silver and gold.
Answer:

Explanation: Silver and gold being less electro-positive cables because it is a good conductors. Moreover, being a noble metal, it is not affected by the atmosphere.

Question 82. What is organic chemistry?
Answer:

Organic chemistry: All carbon-containing compounds except oxides of carbon, metal carbonate, bicarbonate, hydrogen cyanide and metallic cyanides are organic compounds and the chemistry of organic compounds is called organic chemistry.

Question 83. What are Monosaccharides?
Answer:

Monosaccharides: These are the compounds such as glucose and fructose which do not break into simpler compounds on hydrolysis.

Question 84. What is Moltose?
Answer:

Moltose: It is obtained by the partial hydrolysis of starch by enzyme diastase present in malt.

Question 85. What is Lactose?
Answer:

Lactose: It occurs in the milk of all animals cow’s milk contains about 5% Lactose while human milk contains about 7% lactose.

Question 86. What is Cellulose?
Answer:

Cellulose: It is the main structural material of wood and other plants. Cotton is about 80% cellulose.

Question 87. What are Amino acids?
Answer:

Amino acids: These are organic compounds containing both amino and carboxylic group in their molecules. They are represented by the general formula.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter Amino Acids

Question 88. What are essential amino acids?
Answer:

Essential amino acids: Human body can synthesise 10 out of 20 a-amino acids found in proteins. The remaining 10 must be present in our diet and are called essential amino acids.

Question 89. What is Zwitter ion?
Answer:

Zwitter ion: In aqueous solution, the acidic carboxyl group donates a proton to the basis amino group to form an internal salt called a dipolar ion or zwitter ion. Although it is neutral overall, it contains both a positive and a negative charge.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter Zwitter Ion

Question 90. What are peptides
Answer:

Peptides: These are condensation products of self-amides formed by the reaction of two or more amino acid molecules.

WBBSE Class 10 Physical Science Solutions

Question 91. What is peptide linkage?
Answer:

Peptide linkage:

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter Peptide Linkage

The bond Between the Carboxyl group of an acid molecule and nitrogen of the other amino acid molecule is known as the peptide bond or peptide linkage.

Question 92. What is Denaturation?
Answer:

Denaturation: After coagulation, proteins lose their physiological activity and certain other properties. This phenomenon is known as denaturation.

Question 93. What are Enzymes?
Answer:

Enzymes: These are also proteins which acts as catalyst in many biochemical reactions. Enzymes are specific in action due to their specific structural arrangement.

Question 94. What is the biuret test?
Answer:

Biuret test: To alkaline solution of proteins add a dilute solution of CuSO Formation of violet colour confirms proteins.

Question 95. What are vitamins?
Answer:

Vitamins: These are the biomolecules needed in small quantities, that regulates many biochemical function and prevent the development of many deficiency diseases.

Question 96. What are fats?
Answer:

Fats: These are organic compounds composed of carbon, hydrogen and oxygen. They are made of glycerol and fatty (organic) acids. Fats may be of animal or vegetable origin.

Question 97. State examples of conjugated proteins and derived proteins.
Answer:

Conjugated proteins:

Phosphoproteins
Glycoproteins.

Derived proteins: This type of protein is obtained after the partial hydrolysis of protein of very high molecular weight by acid, base or enzyme to simplier proteins.

Question 98. What are the types of simple proteins?
Answer:

Types of simple proteins :

Albumins → Example, milk, serum etc.

Globumins → Example, egg, yolk, tissues etc.

Glutemins→ Example, wheat, rice etc.

Prolomins → Example, barley, wheat etc.

Scleroproteins →Example, Keratin, fibroin etc.

Question 99. What are Alkynes?
Answer:

Alkynes: Those unsaturated hydrocarbons containing triple bonds (S) between two adjacent carbon atoms in their molecules are called as alkynes.

For example, Acetylene

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter Alkynes

Question 100. What is Isomerism?
Answer:

Isomerism: In organic chemistry when the same molecular formula represents two or more compounds which differ in their physical and chemical properties, then such compounds are called isomers and the phenomenon is called isomerism.

Question 101. What is catenation?
Answer:

Catenation: The property by virtue of which carbon forms covalent linkage chains is called catenation.

Question 102. What are the types of structural isomers?
Answer:

Types of structural isom:

Chain isomerism; For example,

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter Chain Isomerism