NEET Biology Class 12 Microbes In Human Welfare Notes

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.

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.

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.

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:

  1. 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.
  2. 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.

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 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:

  1. Biocontrol: Biocontrol is the use of biological methods for controlling plant diseases and pests.
  2. 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.

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.

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.

NEET Biology Class 12 Strategies For Enhancement In Food Production Notes

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.

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.

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:

  1. Inbreeding and
  2. Out-breeding

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.

  1. Out-crossing
  2. 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.
  3. 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).

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.

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.

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.

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.

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.

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

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

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.

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.

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

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

  1. N2
  2. O2
  3. CH3
  4. He

Answer: 3. CH3

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

  1. 4.4 g CO2
  2. 0.64g SO
  3. 28 g CO
  4. 16g O. [C= 12. O -16. S = 32]

Answer: 1. 4.4 g CO2

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

  1. 6.022 x 1023
  2. 1.806 x 1024
  3. 6.022 x 1022
  4. 6.022 x 1024

Answer: 3. 6.022 x 1023

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

  1. One
  2. Two
  3. Three
  4. Four

Answer: 3. Three

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

Class 10 Physical Science Solution WBBSE

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

  1. x-ray
  2. y-ray
  3. Infrared ray
  4. 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

  1. 75°
  2. 15°
  3. 7-5°
  4. 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?

  1. V∞ I
  2. V∞ I2
  3. V ∞ I-1
  4. V ∞ r-2

Answer: 1. V∞ I

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

  1. Q=WV
  2. Q=V/W
  3. Q=V/W2
  4. Q = W/V

Answer: 4. Q = W/V

Class 10 Physical Science Solution WBBSE

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

  1. Mass number increases
  2. Atomic number increases
  3. Mass number decreases
  4. Atomic number decreases

Answer: 2. Atomic number increases

Question 10. To which group of the long 

  1. Group 1
  2. Group 16 (e)
  3. Group 17
  4. Group 2

Answer:  3. Group 17

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

  1. Sodium chloride
  2. Hydrogen chloride
  3. Naphthalene glucose

Answer: 1. Sodium chloride

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

  1. Pure water
  2. An aqueous solution of Sugar cane
  3. Liquid hydrogen chloride
  4. Aqueous solution of acetic acid

Answer: 4. Aqueous solution of acetic acid

Class 10 Physical Science Solution WBBSE

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

  1. NO
  2. NO2
  3. N2O5
  4. HNO3

Answer: 1. NO

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

  1. Al2O3
  2. Al2O3H2O
  3. Al20O3 2H2O
  4. A1F3 . 3NaF

Answer: 3. Al20O3 2H2O

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

  1. Methyl
  2. Ethyl
  3. Propyl
  4. 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).

Class 10 Physical Science WBBSE

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.

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:

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:

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                                       

Or

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

WB Class 10 Physical Science Question Answer

Question 4. Two resistances r1, and r2, 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: 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 –

  1. 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.
  2. Thus, the two atoms, like their nearest neutral gas, gain a symmetrical electron pattern.
  3. 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:

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

WB Class 10 Physical Science Question Answer

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

  1. Galvanization or the layer of melted zinc on iron prevents rusting.
  2. 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

  1. Acetic Acid – Vinegar, preserves fish and flush, to cook pickle.
  2. 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:

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 Fe2O3 with 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

Or

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

Or

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.

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:

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

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

WBBSE Class 10 Physical Science Question Answer

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:

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

⇒ N1 N2 II, N3 N4

Again N1 N2 II, N3 N4and 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

Or

Answer:

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

  1. Length of the object (h1) = 5cm
  2. 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.

WBBSE Class 10 Physical Science Solutions

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

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:

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

Or

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+

  1. From the nucleus of a radioactive element, when a particle is emitted, the mass number of the new nucleus is 4 units less.
  2. The atomic number is less by 2 units So, so a new element is formed a radioactive element.
  3. 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 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.

WBBSE Class 10 Physical Science Solutions

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

  1. Mendeleev’s periodic law is rectified.
  2. 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 :

  1. Liquid ammonia
  2. Liquid CO2
  3. Balanced chemical equation

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

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

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

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

Or

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

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?

  1. N2O
  2. CH4
  3. CO2
  4. 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

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

  1. CO2
  2. C2H4
  3. C2H
  4. C2H2

Answer: 4. C2H2

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

  1. m
  2. nr!
  3. °C-1
  4. °C

Answer:  3. °C-1

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?

  1. Real and inverted
  2. Virtual and inverted
  3. Rea and erect
  4. Virtual and erect

Answer: 4. Virtual and erect

Class 10 Physical Science Solution WBBSE

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

  1. 180°
  2. 30°
  3. 90°

Answer: 1. 0°

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

  1. Volt
  2. Ampere
  3. Coulomb
  4. Ohm

Answer: 4. Ohm

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

  1. Earth line
  2. Live line
  3. Neutral line
  4. Both live and neutral line

Answer: 2. Live line

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

  1. A stream of electrons
  2. A stream of neutrons
  3. A stream of protons
  4. Electromagnetic wave

Answer: 1. A stream of electrons

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

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

  1. 7
  2. 8
  3. 9
  4. 18

Answer: 4.18

Class 10 Physical Science Solution WBBSE

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

  1. NaCl
  2. LiH
  3. KCl
  4. CaO

Answer: 2. LiH

Question 12. Which of the following can conduct electricity?

  1. Molten NaCl
  2. Liquid HCI
  3. Solid NaCl
  4. 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?

  1. Yellow
  2. Green
  3. Deep blue
  4. Brown

Answer: 3. Deep blue

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

  1. Bell metal
  2. Brass
  3. Bronze
  4. Duralumin.

Answer: 2. Brass

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

  1. C3H6
  2. C2H
  3. C2H2
  4. 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:

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.

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

WBBSE Class 10 Physical Science Question Answer

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 CH2 OH.
Answer: Propanal.

Or

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

⇒ 2P × V/2 = PV = K

Or

According to Charle’s law,

Vt = V0 (1+_t/273)

Or, Vt = V0 (1+2)

Or, Vt= 3V0

Therefore, volume changes 3times the initial volume.

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:

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.

Or

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. Slate Lenz’s law related to electromagnetic induction.
Answer: 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).

WB Class 10 Physical Science Question Answer

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.

Or

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.

Or

2C2H5 OH+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 SO2 Fe= 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:

SO2 512g.

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.

Or

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.

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

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-‘.

Or

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-1 or °C-1. γ = K-1 or, °F-1

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

WBBSE Solutions Guide Class 10

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:

We know,

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

= \(\frac{2}{3}\)

= 0.67

Or

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:

  1. 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).
  2. 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).
  3. 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).

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/Rl= \(\frac{1}{R}\)+ \(\frac{1}{R}\) = \(\frac{2}{R}\)

∴Rl = \(\frac{2}{R}\)

Or

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

∴ Total Power

= 2 (60 + 80)

= 2 x 140

= 280w

WBBSE Solutions Guide Class 10

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

Or

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.

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.

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:

Organic Compound is – CH3COOH

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

Or

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

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?

  1. NO
  2. NO2
  3. CFC
  4. CO2

Answer: 4. CO2

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

  1. 2RT
  2. RT
  3. 0.5RT
  4. 11.2 RT

Answer: 3. 0.5RT

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?

  1. 448L
  2. 224L
  3. 44.8L
  4. 22.4L

Answer: 1. 448L

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

  1. Silver
  2. Diamond
  3. Copper
  4. Aluminium

Answer: 2. Diamond

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?

  1. r = v
  2. r= \(\frac{1}{V}\)
  3. r> v
  4. r < v

Answer: 4. r<v

WBBSE Class 10 Physical Science Question Answer In English

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

  1. 180°
  2. 90°
  3. 360°

Answer: 2. 180°

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

  1. One is the point, One is spherical
  2. Both are spherical
  3. One is the point, One is extended
  4. Both are points

Answer: 4. Both are points.

Question 8. The characteristics of a fuse wire are

  1. High resistance high melting point
  2. Low resistance, low melting point
  3. Low resistance high melting point
  4. High resistance low

Answer: 2. High resistance low melting point.

Question 9. Present in a particle

  1. One proton, one neutron
  2. One proton
  3. Two protons two neutrons
  4. One electron

Answer: 3. Two protons, two neutrons.

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

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

  1. Density
  2. Melting point
  3. Boiling point
  4. Radioactivity

Answer:  4. Radioactivity

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

  1. Hydrogen chloride
  2. Calcium oxide
  3. Methane
  4. Ammonia

Answer: 4. Ammonia.

WBBSE Class 10 Physical Science Question Answer In English

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

  1. The mass of the cathode decreases
  2. The mass of the anode increases
  3. The concentration of CuSO4 in the solution decreases
  4. 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?

  1. Violet
  2. Orange
  3. Deep blue
  4. Green

Answer: 1. Violet.

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

  1. FeO
  2. Fe2O3
  3. Fe3O
  4. FeCO3

Answer: 3. Fe2O3

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

  1. CH3 CH2 OH
  2. CH3CHO
  3. CH3 COCH3
  4. CH3 COOH

Answer: 4. CH3 COOH

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).

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-CH3
Or
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

Or

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:

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.

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

  1. The image is real.
  2. 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

WB Class 10 Physical Science Question Answer

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: CH3 COOH, 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 – CH2OH (Ethyl alcohol)

⇒ CH3O – 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:

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

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: 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

Or

According to the following chemical equation:

  1. 2ZnS +3O2 → 2ZnO +2SO2 from 100 mole of ZnS
  2. How many grams of ZnO, and
  3. 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

  1. 8,150 gm of ZnO, and
  2. 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: 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 Class 10 Physical Science Solutions Question 3. Write down the three factors on which the conduction of heat through a solid substance depends.
Answer:

The factors are:

  1. Area of the width intersection
  2. Change of temperature, and
  3. 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 1.6×108 Ω 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:

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

Or

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:

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

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 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:

  1. γ Ray has the highest penetrating power.
  2. 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:

Or

  1. 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.
  2. S (16) O (8), Te (52), and Se (34) belong to Group 16 of the long periodic table according to increasing order to electronegativity.
  3. 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):

  1. Rb> K > Na > Li
  2. Te< Se <S<0
  3. 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.

WB Class 10 Physical Science Question Answer

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

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

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

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

Question 1. The basis of the modern periodic table is

  1. Atomic volume
  2. Atomic number
  3. Atomic size
  4. Atomic weights

Answer:  2. Atomic volume.

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

  1. F
  2. Cl
  3. Br
  4. I

Answer: 2. Cl.

Question 3. Characteristics of transition elements is incomplete

  1. d-orbitals
  2. f-orbitals
  3. p-orbitals
  4. s-orbitals

Answer: 1. d-orbitals.

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

  1. Boron
  2. Carbon
  3. Nitrogen
  4. Oxygen

Answer: 3. Nitrogen.

Class 10 Physical Science WBBSE

Question 5. The long form of periodic table based on 

  1. Atomic number
  2. Atomic mass
  3. Number of neutrons
  4. None of these

Answer: 1. Atomic number.

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

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

  1. 7
  2. 10
  3. 6
  4. 14

Answer: 1. 7

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

  1. 5%
  2. 0.01%
  3. 1%
  4. 11%

Answer: 3.1%.

Question 8. The shape of the NaCl crystal is 

  1. Tetrahedral
  2. Octahedral
  3. Hexagonal
  4. Icosahedral

Answer: 2. Octahedral.

Question 9. Which has the largest first ionization energy?

  1. 4
  2. na
  3. K
  4. Rb

Answer: 1. 4.

Class 10 Physical Science WBBSE

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

  1. Br-1
  2. I
  3. I

Answer: 3. I

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

  1. N3-
  2. O2
  3. F
  4. Na

Answer: 4. Na

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

  1. B
  2. Na
  3. Al
  4. Mg

Answer: 3. Al

Question 13. 5f subshell is successively filled up in 

  1. Actinoids
  2. Lanthanoids
  3. Tropical metals
  4. Normal metal

Answer: 1. Actinoids.

Question 14. Which of the following form coloured salts?

  1. Non-metals
  2. Metals
  3. p-block elements
  4. Transmission elements

Answer: 2. Metals.

Class 10 Physical Science WBBSE

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

  1. s
  2. p
  3. d
  4. f

Answer:  2. p

Question 16. Lithium shows a diagonal relationship with

  1. Ne
  2. Mg
  3. Be
  4. Ce

Answer: 2. Mg.

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

  1. Mg
  2. Ne
  3. Ar
  4. K

Answer: 3. Ar.

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

  1. NaCl
  2. KCl
  3. MgO
  4. BaO

Answer: 3. MgO.

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

  1. 2
  2. 4
  3. 6
  4. 10

Answer: 3. 6

Class 10 Physical Science WBBSE

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

  1. CH4
  2. H2
  3. KCN
  4. KCI

Answer: 3. KCN

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

  1. 1.47A°
  2. 1.38A°
  3. 1.54A°

Answer: 3. 1.54A°.

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

  1. Diamond
  2. Benzene
  3. Acetylene
  4. Ethane

Answer:  3. Acetylene.

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

  1. CaCl4
  2. CCI2
  3. Sncl4

Answer: 3. Sncl4

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

  1. Sulphur
  2. Helium
  3. Fluorine

Answer: 2. Helium.

WBBSE Class 10 Physical Science Question Answer

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

  1. Non-Metal
  2. Metal
  3. Metalic oxide

Answer: 3. Metal.

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

  1. N – H
  2. C – H
  3. F- F
  4. O- H

Answer: 3. F – F.

Question 27. Which of the following is more covalent?

  1. SnCI4
  2. SnBr4
  3. Snl4

Answer: 1. SnCl4

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

  1. Sigma bond
  2. λ bond
  3. Co-ordinate bond
  4. Hydrogen bond

Answer: 1. Sigma bond.

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

  1. +1
  2. -1
  3. + 3
  4. -3

Answer: 2.- 1.

Question 30. Which agent conducts electricity during electrolysis?

  1. Water
  2. Ion
  3. Atom
  4. Molecule

Answer: 2. Ion.

WBBSE Class 10 Physical Science Question Answer

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

  1. 1:2
  2. 2:3
  3. 2:1
  4. 1:1

Answer: 3. 2 :1.

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

  1. Atoms
  2. Electrons
  3. Ions

Answer: 3. Ions.

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

  1. Resistance
  2. Time
  3. Current
  4. Non of these

Answer: 1. Resistance

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

  1. Na
  2. H2
  3. CO
  4. SO3

Answer: 1. Na.

Question 35. The unit of quantity of electricity is 

  1. Volt
  2. Coulomb
  3. Ampere
  4. Ohm

Answer: 2. Coulomb.

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

  1. Water
  2. Nitric acid
  3. Boric acid
  4. H2SO4

Answer: 3. Boric acid.

WBBSE Class 10 Physical Science Question Answer

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

  1. Increased
  2. Decreased
  3. Unaltered

Answer: 3. Unaltered.

Question 38. Ammonia is :

  1. Acidic
  2. Basic
  3. Neutral
  4. None of these

Answer: 2. Basic.

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

  1. Basic nature
  2. Polar nature
  3. High solubility
  4. Hydrogen bonding.

Answer: 4. Hydrogen bonding.

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

  1. Exothermic
  2. Endo thermic
  3. Neither of the two
  4. Both the two

Answer: 1. Exothermic

Question 41. Nessler’s reagent is :

  1. K2Hgl4
  2. K2 Hgl4+KOH.
  3. K2Hgl4
  4. KHgl4+KOH

Answer: 2. K2 Hgl4+KOH.

Question 42. Which is most explosive?

  1. NCI3
  2. Pcl3
  3. AsCl3
  4. All

Answer: 1. NCI3

WBBSE Class 10 Physical Science Question Answer

Question 43. With an excess of Cl2, ammonia gives 

  1. NCI3
  2. HCI3
  3. NH4CI
  4. NO2

Answer: 1. NCI3

Question 44. Non-combustible hydride is 

  1. NH3
  2. PH3
  3. AsH3
  4. SbH3

Answer: 1. NH3

Question 45. An aqueous solution of ammonia consists of

  1. H+
  2. OH
  3. NH4
  4. NH+OH

Answer: 4. NH+OH

Question 46. The vapour density of ammonia is 

  1. 7
  2. 8.5
  3. 7.5
  4. 17

Answer: 2. 8.5.

Question 47. H2S is 

  1. Smells like fish
  2. Lighter
  3. Sightly heavier than air
  4. None of these

Answer: 1. Smells like fish.

Question 48. The gas which is absorbed by NaOH 

  1. NH
  2. H2S
  3. O2
  4. None of these

Answer: 2.H2S

WBBSE Class 10 Physical Science Solutions

Question 49. H2 S gas is passed through 

  1. P2O5
  2. CaCl2
  3. CaO
  4. None of the above

Answer: 1. P2O5

Question 50. The molecular weight of H2S is 

  1. 32
  2. 34.
  3. 36
  4. 38

Answer: 2. 34.

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

  1. 20
  2. 30
  3. 78
  4. 88

Answer: 78.

Question 52. The solubility of nitrogen in water is 

  1. Low high
  2. Very low
  3. High
  4. Very high

Answer: 4. Very high.

Question 53. Nitrogen-containing organic fertilizer is

  1. Nitro slim
  2. Ammonium sulphate
  3. Urea
  4. Ammonium nitrate

Answer: 3. Urea.

Question 54. Who discovered nitrogen

  1. Daniel Rutherford
  2. Bohr
  3. Dalton
  4. 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 :

  1. Fe2O3
  2. Al (OH)3
  3. Fe (OH)3
  4. Cr (OH)3

Answer: 3. Fe(OH)3

WBBSE Class 10 Physical Science Solutions

Question 56. Oil of vitriol is 

  1. H2SO3
  2. H2SO4
  3. H2S2O7
  4. H2S2O8

Answer: 2. H2SO4

Question 57. Aquafortis is 

  1. HNO3
  2. HNO2
  3. H2NO3
  4. H2N2O2

Answer: 1.HNO3

Question 58. Oleum is

  1. H2SO5
  2. H2S2O7
  3. H2S2O8
  4. H2SO3

Answer: 2. H2S2O7

Question 59. Most of the plants contain 

  1. Iron
  2. Zin C
  3. Sodium
  4. Potassium

Answer: 4. Potassium.

Question 60. The fluorspar is 

  1. CasO4
  2. Baso4
  3. CaCO3
  4. CaF2

Answer: 4.CaF2

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

  1. Gypsum
  2. Limestone
  3. Dolomite
  4. Carnalite

Answer: 4. Carnalite

Question 62. Haematite is 

  1. Fe3O4
  2. Fe2O3
  3. Fe2O4

Answer: 2. Fe2O3

WBBSE Class 10 Physical Science Solutions

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

  1. Mg
  2. Fe
  3. Cu
  4. Zn

Answer: 2. Fe.

Question 64. Pin iron is also called

  1. Wrought iron
  2. Cast iron.
  3. Steel
  4. Stainless steel

Answer: 2. Cast iron.

Question 65. The metal present in insulin is 

  1. Copper
  2. Iron
  3. Zin C

Answer: Zin C.

Question 66. Which of the following contains nitrogen?

  1. Fats
  2. Proteins.
  3. Carbohydrate
  4. None

Answer: 2. Proteins

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

  1. C2H4
  2. C2H2
  3. CH4
  4. C2H2

Answer: 2. C2H2

Question 68. Marsh gas mainly contains

  1. H2S
  2. CO
  3. CH4
  4. C2H2

Answer: 3. CH4

Question 69. The general formula for alkenes is 

  1. CnHzn
  2. CnH2n+2
  3. CnHzn-2
  4. C2nHzn

Answer: 1. CnHzn.

WBBSE Class 10 Physical Science Solutions

Question 70. Acidic hydrogen is present in 

  1. Ethyne
  2. Ethene
  3. 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.

Question 11. What is the relation between N3- and O2-?
Answer:  N3- and O2- 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 C6H12O6 will 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.

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 Brand 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.)

Class 10 Physical Science Solution WBBSE

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

There are two ways of fixation of nitrogen:

  1. By electric discharge
  2. 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.

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 WBBSE

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.

Physics Class 10 WBBSE

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.

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.

Question 58. N2 + __________ = 2NH3
Answer: 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=_______+6H2O
Answer: 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

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. MgCl2 on 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.

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: 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.

Physics Class 10 WBBSE

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:

  • 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:

  1. The magnitude of charge of ions: It increases with an increase in charge on cation-anion or both.
  2. Size of cation: For a common anion, the lattice energy decreases with an increase in size.
  3. 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 :

  1. Ionic or Electrovalent bond
  2. Covalent bond and
  3. 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.

WB Class 10 Physical Science Question Answer

Question 26. What are the main types of physical bonds?
Answer:

The main types of physical bonds are:

  1. Hydrogen bond
  2. Metallic bond
  3. 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 :

  1. Covalent compounds exist in solid, liquid and gaseous state.
  2.  These compounds have low melting and boiling points.
  3. These (except graphite) are bad conductors and react sowly.
  4. These are soluble in non-polar solvents such as benzene, acetone etc. but are insoluble in polar solvents such as water.
  5. 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.

WB Class 10 Physical Science Question Answer

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:

  1. Electroplating
  2. Electro-refining of metals.
  3. Electro-typing
  4. Extraction of metals.

Question 46. What are the conditions for a good deposit?
Answer:

Conditions for good deposit:

  1. High current density
  2. Low temperature
  3. High metal concentration in the electrolyte
  4. 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:

  1. Cathode: the article is to be electroplated.
  2. Anode: Pure Ag plate
  3. 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……… Ni2-+ 4SO2-

2. Electrodes:

  1. Cathode: The article is to be electroplated.
  2. Anode: Pure Ni-plate.

3. Reaction at the cathode: Ni2-+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 Cu2- 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:

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↓.

WB Class 10 Physical Science Question Answer

Question 54. What is a reaction between NaCO, and HCI?
Answer:

Reaction of Na2 CO3 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.

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 Science Solutions

Question 71. State the reaction of H2SO4 with carbonates and bicarbonates.

.Answer: Reaction of H2SO4 with 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, Al2O3 reacts 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.

WBBSE Class 10 Physical Science Solutions

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:

  1. Phosphoproteins
  2. 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.

WBBSE Class 10 Physical Science Solutions

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

Chapter 8 Physical And Chemical Properties Of Matter Broad Answer Type Questions

Question 1. What are groups and sub-groups?
Answer:

Groups Mendeleev’s periodic table shows that the elements with related chemical properties fall in one below the other forming vertical columns called groups. There are nine group from group I to group VIII and O (zero).
Sub-groups Each of the groups from I to VII has been divided into two. Sub-groups A and B.

Question 2. What are halogen elements? In which do they belong?
Answer:

Halogen elements: The four strongly electro-negative non-metals fluorine (F), chlorine (CI), bromine (Br) and iodine (1) form a family of closely allied elements known as the halogens meaning literally sea self-producer as these elements react with most metals to form compounds similar to sea-salt, sodium chloride.

Position in periodic table: Habzen elements placed in group VII B of the periodic table. The electro-negative character of the elements increases from left to right in a period. So, the strongly electro-negative elements are in the most extreme right group i.e. in group VII B of the period table.

Question 3. What are called alkali metals? In which group do they belong? State the similarity of their chemical properties.
Answer:

Alkali metals: Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb) and Caesium (Cs). These five metals are generally known as alkali metals.

Position in the periodic table: The alkali metals are included in the A subgroup of the first group

Similarity:

  1. Alkali metals are strong electropositive.
  2. The oxides of these metals are strongly basic.
  3. Each has valency 1.

WBBSE Class 10 Physical Science Solutions

Question 4. Discuss the position of hydrogen in the periodic table. Answer: Position of hydrogen in the periodic table :

Reasons for placing in group IA :

  • Like the alkali metals, hydrogen is electropositive.
  • The valence of hydrogen is I, alkali metals are also non-valent.
  • During the electrolysis of a fused chloride of an alkali metal, deposits at the cathode; in the electrolysis of a dilute aqueous solution of hydrogen chloride, hydrogen collects at the cathode.
  • Like alkali metals, hydrogen has a reducing property and forms stable oxides like water (H2O).
  • Like hydrogen, each of the alkali metals. Li, Na, K, Rb Cs, Fr possesses one valence-lectron.

Reasons for placing in group VIIB:

  • Like halogens, hydrogen is a non-metatallic element.
  • The valency of hydrogen and that of each halogen is 1.
  • Halogen molecules are diatomic, a hydrogen molecule is also diatomic.
  • The halogens from metallic halides like, NaCl, KBr, Kl etc. Hydrogen also produces salt-like hydrides like L, H, NaH, CaH, etc with the strongly electropositive element and hydrogen here acts as an electro-negative element.
  • Atoms of halogen elements may transform to negative halide ions by taking one electron each.
  • For example, x+eX; X = F, Cl, Br, I and hydrogen atom also may transform to a negative hydride ion by capturing one electron: H+e→ H.
  • Like the halogens hydrogen also can produce covalent compounds reacting with electro-negative elements like S, O etc.

Due to such varied tendencies for occupying a seat in the periodic table, hydrogen is often called a naughty element or rogue element.

Question 5. Where are the inert gas elements placed and why?
Answer:

Position of inert gas elements in the periodic table:

  • He, Ne, Ar, Kr, Xe, and Rn. These are inert elements. They are each gaseous and monoatomic. These are chemically non-reactive. As these elements do not react, they have zero valency.
  • So they are included in the O (zero) group of the periodic table. Besides Helium, all the inert gases have eight (8) electrons in their outermost orbit. There are only 2 electrons in the outermost orbit of Helium.

Question 6. How is copper purified by the electrolysis method?
Answer:

Purification of copper by electrolysis method :

1. Electrolyte:  15% of CuSO4, solution (aqueous) containing (5-10) % sulphuric acid at 50°C is taken in a voltameter.

2. Electrodes:

  • Cathode: Pure thin copper plate.
  • Anode: Thick impure copper plate.

3. Electrolysis:

  • On electrolysis, copper dissolves from the anode and deposits on the cathode. Thus gradually the anode plate wears out and the cathode plate thickens.
  • The copper obtained in the way 99.99% purity.

4. Reaction: \(\mathrm{CuSO}_4 \rightleftharpoons \mathrm{Cu}^{2-}+\mathrm{SO}_4^{2-}\)

  • At cathode:  Cathode : Cu2-+2e →Cu↓
  • At anode: Cu – 2e → Cu2-

WBBSE Class 10 Physical Science Solutions

Question 7. How is aluminium extracted from the electrolysis method?
Answer:

Extraction of aluminium by electrolysis method :

1. Electrolytes:

  • Alumina (Al2O3) 20%
  • Fused Cryolite (AIF3, 2NaF 60%)
  • CaF2, 20%.

2. Electrodes:

  • Cathode: Inner lining of carbon of the steel tank.
  • Anode: Thick carbon rod suspended into a fused electrolyte

Reaction:

At cathode: Al3 + 3e → Al↓

At anode : 3F – 3e→ 3F

⇒ Al2O3 +6F → 2 AIF3+3O

⇒  6O →3O2

Question 8. Draw the diagram of the covalent radius of hydrogen.
Answer:

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

Question 9. Draw the diagram of the showering comparison of Vanderwaal’s radius and covalent radius.
Answer:

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

Question 10. State the contact process for the manufacture of sulphuric acid.
Answer:

Contact process for the manufacture of sulphuric acid :

Principle: 

1. Formation of SO2: SO2 is prepared by burning sulphur or iron pyrites in excess of air.

Equation:  S+ O2 = SO↑ Or,

4 Fes,+ || O2 = 2Fe2O3 + 8SO2
(Iron pyrities)

2. Formation of sulphur trioxide (SO3):  SO3 is prepared by the oxidation of sulphur dioxide with oxygen (from air) in the presence of platinised asbestos or, V2 O5 as catalyst at 450°C.

Equation: \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3+192.5 \mathrm{Kj}\)

3. Formation of oleum:

Now sulphur trioxide thus produced is not allowed to react with water directly sulphur trioxide absorbs concentrated sulphuric acid 98% turning it as fuming sulphuric acid or oleum.

Equation: SO3+ H2SO4= H2S2O7 (Oleum)

4. Dilution of oleum to sulphuric acid: Pure and concentrated sulphuric acid is produced by adding slowly a requisite amount of water in fuming sulphuric acid.

Equation:

H2S2O7 >+ H2O = 2 H2SO4

Question 11.

  1. Write down the electronic arrangement of the element  17X 35   What is the valency of the element?
  2. Will the element form an onion or cation?
  3. What type of valency will be exhibited when the element combines with sodium?

Answer:

The electronic configuration of the element X is: 2, 8, 7,

  • As an atom of the element tends to capture or share one electron to attain stable of the octet, its valency is 1.
  • It will form an anion by capturing electrons.
  • The element will exhibit electro-valency when it combines with sodium since sodium is a metal, an atom which tends to release one electron.

Question 12. The atomic number of element A is 20 and that of another element B is 17. Write down their electronic configuration. Will they produce on electrova- lent compound or a covalent compound? What will be their valencies in that case?
Answer:

  1. The electronic configuration of A is: 2, 8, 8, 2.
  2. The electronic configuration of B is: 2, 8, 7.

They will produce an electrovalent compound.

Explanation:

An atom A will give up two (2) valence electrons. Each of the two atoms of B will capture one electron. In the process atoms of A and B attain a stable octet state and atoms of A will be positive ions and those of B will be negative ions.

  • A and B thus form an electrovalent compound.
  • As an atom of A loses two electrons.
  • So A has valency 2 (two) and since each atom of B captures one electron, The valency of B will be one (1).

WBBSE Class 10 Physical Science Solutions

Question 13. What is the difference between ionic compounds and covalant compounds?
Answer:

Difference between ionic and covalent compounds:

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

Question 14. How ammonia is prepared in the laboratory?
Answer:

Laboratory preparation of Ammonia:

  • Chemicals required: Ammonium chloride (NH4, CI) and quick lime (CaO) or dry slaked lime (Ca (OH2)
  • Condition: Normally ammonia gas is obtained in the laboratory by heating a mixture of ammonium chloride with slaked lime. Instead of using slaked lime quick lime may also be used.
  • Collection: Ammonia is lighter than air, it may be collected by the downward displacement of air.

Ammonia is not collected through the downward displacement of water because it is highly soluble in water.

Equation of the reaction:

2NH4 CI + Ca (OH)2 = 2NH3 ↑ + CaCl2+ 2H2O

2NH4CH+CaO = 2NH3↑ + CaCl2+H2O

  • Precautions: The ingredients, the test tube, the delivery pipes and the gas jar should be absolutely dry. All the connections in the apparatus should be leak-proof.
  • Drying of ammonia: As ammonia is a basic substance, it cannot be dried by acidic drying agents like cones. H2SO, or P2O5,
  • The gas is absorbed by fused CaCl2, with the formation of an addition compound CaCl2, 8NH3. SO, fused CaCl2, cannot be used to dry ammonia. It is best dried with the basic drying agent, quick lime (CaO).

Question 15. Prove by experiment that ammonia is dissolved in water and produces an alkaline solution.
Answer:

Fountain experiment:

Arrangement of the experiment: A flask is filled with dry ammonia gas and its mouth is corked. The flask is kept inverted position and is clamped to a stand.

Through a hole in the cork one end of a glass tube is introduced inside the flask. This end of the tube inside the flask is shaped into a jet. The other end of the tube dips in some red litmus solution taken in a beaker.

  1. Operation: Now ice or ether is poured upon the flask.
  2. Observation: A blue fountain produces inside the flask.

Explanation:

  • Due to the evaporation of ether, the flask is cooled. Ammonia gas inside the flask contracts, as a result, there is a partial vacuum inside the flask.
  • At this stage, if the stop cock is opened the red litmus solution rushes inside the flask and ammonia is dissolved in it.
  • Due to vacuum created inside there is a formation of the fountain and the solution becomes blue.

Conclusion: This experiment proves that ammonia is highly soluble in water and the aqueous solution is alkaline.

WBBSE Class 10 Physical Science Solutions

Question 16. Draw the diagram of the formation of sodium chloride.
Answer:

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

Question 17. Draw the diagram of sodium fluoride :
Answer:

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

Question 18. What are the main points of difference between metallic conductors on electrolytic conductors?
Answer:

Difference between metallic conductors and electrolytic conductors :

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 8 Physical And Chemical Properties Of Matter Metallic Conductor And Electrolytic Conductors

Question 19. What changes are taking place during the electrolysis of an electrolyte?
Answer:

Changes take place during the electrolysis of an electrolyte:

  • When fused or dissolved in water, electrolyte splits up into oppositely charged particles called ions.
  • On passing an electric current, the cations migrate towards the cathode while the anions migrate towards the anode.
  • The cations on reaching the cathode gain electrons from it and from neutral atoms which get deposited on the cathode.
  • The anions on reaching the anode lose electrons and get converted into neutral atoms which may be collected as such or they may undergo some secondary change to from some other products.

Question 20. What are Biodegradable and Non-biodegradable materials?
Answer:

  1. Biodegradable materials: Materials of vegetable and animal origin invariable decay and decompose into CO2, H2O, N2 or NH3 by the combined action of sometimes natural agencies like air, water, subshene etc. These are called biodegradable materials.
  2. Non-biodegradable materials: Synthetic materials like plastics, polythene, Teflon, PVC etc. are not decomposed by these natural agencies even for a long time. These are called non-biodegradable materials.

Question 21. Briefly explains thermite welding of iron and steel.
Answer:

A mixture of 3 parts of ferric oxide and I part of aluminium powder is called a thermite mixture. When the mixture is ignited ferric oxide is reduced to metallic iron by aluminium. Because of the evolution of large amounts of heat, the temperature rises to Magnesium tape

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

2500c The molten iron thus produced is allowed to fall on the red hot broken rails broken machine parts ect can be welded without removing them from their original sites.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 7 Atomic Nucleus

Class 10 Physical Science Solutions WBBSE

Chapter 7 Atomic Nucleus MCQs

Question 1. Which of the following are α-emitter?

  1. Helium-5
  2. Polonium-212
  3. Tritium

Answer: 2. Polonium-212

Question 2. Control rods of nuclear reactors are made up of:

  1. Cd
  2. Diamond
  3. Graphite
  4. Copper

Answer: 1. Cd

Question 3. The moderator used in atomic pile is :

  1. Heavy water
  2. Uranium
  3. Iron

Answer:  1. Heavy water

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

Question 4. Which one of the following is not a synthetic element?

  1. Np
  2. Cm
  3. PU
  4. U

Answer:  4. U

Class 10 Physical Science Solutions WBBSE

Question 5. Which emits b-particle?

  1. 1H3
  2. 6 C14
  3. 19 K40
  4. All of these.

Answer:  4. All of these.

Question 6. Choose the element which is not radioactive :

  1. Cm
  2. NO
  3. Mo
  4. Md

Answer: 3. Mo

Question 7.  1H1 + 1H3 + 1H4 represents:

  1. β-decay
  2. Fusion
  3. Fussion
  4. D-decay

Answer: 2. Fusion.

Question 8. What is the order of reaction of the decay of 92 U235 ?

  1. Zero
  2. First
  3. Second
  4. Third

Answer: 2. First.

Question 9. Which of the following has the maximum ratio?

  1. 16Ne
  2. 16F
  3. 16O
  4. 16N

Answer: 4.16N

Question 10. Which of the following radioactive elements is soluble in water?

  1. Radium
  2. Tritium
  3. Radon

Answer: 3. Radon.

Question 11. The relative penetrating power of a,ß,y and neutron (n) follows the order:

  1. α > β > γ> n
  2. n > γ > β >α
  3. β >α > n > γ
  4. None of these

Answer: 2. n > γ > B > α.

Class 10 Physical Science Solutions WBBSE

Question 12. Which of the following nuclear reactions will generate an isotope?

  1. Newtron particle emission
  2. Positron emission
  3. β – particle emission
  4. α – particle emission

Answer: 1. Newtron particle emission.

Question 13. Nuclear energy is :

  1. Renewable
  2. Not renewable
  3. Sometimes renewable
  4. It is a matter of debate

Answer: 2. Not renewable.

Question 14. Nuclear fission is initiated by:

  1. Slow neutron
  2. By proton
  3. Fast newtron
  4. By α Particle

Answer: 1. Slow neutron.

Question 15. The source of energy of the sun is :

  1. Fission reaction.
  2. Fusion reaction
  3. Spalation reaction
  4. None of these

Answer: 2. Fusion reaction.

Question 16. Near the nuclear reactor, there is:

  1. Large neutron flux
  2. Large proton flux
  3. Large Particles flux
  4. Large β – ray flux

Answer: 1. Large neutron flux

WBBSE Class 10 Physical Science Question Answer In English

Question 17. A decrease in atomic number is not observed during :

  1. α-emission
  2. β-emission
  3. Positron
  4. Electron capture

Answer: ????

Question 18. Identify the nuclear reaction that differs from the rest :

  1. Positron emission
  2. K-capture
  3. γ-decay
  4. β-decay

Answer:  3. γ- decay.

Question 19. Which of the following are β-emitters?

  1. Carbon-14
  2. Cobalt-60
  3. Tritium-3
  4. All of these

Answer: 4. All of these.

Question 20. Which of the following has the highest value for its radioactivity?

  1. 1g of Ra
  2. Ig of Ras 04
  3. 1g of RaBr2
  4. None of these

Answer: 1. Ig of Ra.

Question 21. Of the following which is not used as a moderator in a nuclear reactor?

  1. Heavy water
  2. Sodium
  3. Graphite

Answer: 2. Sodium.

Question 22. Control rods used in the nuclear reactor are made of:

  1. Nickel
  2. Iron
  3. Graphite
  4. Cadmium

Answer: 4. Cadmium.

WBBSE Class 10 Physical Science Question Answer In English

Question 23. Which of the following processes causes the emission of an X-ray?

  1. Alpha emission
  2. Gamma Emission
  3. Electron emission
  4. Positron Emission

Answer: 3. Electron emission.

Question 24. Choose the incorrect one:

  1. 1 curie = 3.7 × 1010 ds-1
  2. 1 rutherford = 106 ds-1
  3. 1 fermi = 103 ds-1
  4. I becquerel = 1 ds-1

Answer: 3. fermi = 103 ds-1

Question 25. Mark the correct relation :

  1. No = Nett
  2. t= 1.44+0.5
  3. N= No(½)x
  4. To.5 = λln2

Answer: 4. To.5 = λln2

Question 26. Choose the natural element among the following:

  1. Uranium
  2. Astatine
  3. Neptunium

Answer: 1. Uranium.

Question 27. α-ray consists of a stream of

  1. H
  2. He-2
  3. Only electrons
  4. Only Neutrons

Answer: 2. He-2

WBBSE Class 10 Physical Science Question Answer In English

Question 28. Which of the following combinations will give the most stable nuclear?

  1. Odd Z and odd N
  2. Odd Z and even N
  3. Even Z and odd N
  4. Even Z and even N

Answer: 4. Even Z and even N.

Question 29. Emission of a ẞ particle by an atom of element results in the formation of its 

  1. Iso tope
  2. Isomer
  3. Isobar
  4. Isomorph

Answer: 3. Isobar.

Question 30. Stable nuclides cannot be obtained for

  1. Z = 43, N = 35
  2. Z = 61, Z = 89
  3. A > 209
  4. All of these

Answer:  4. All of these.

Question 31. Which of the following nuclei are stable?

  1. 28 Ni60
  2. 6 C11
  3. 92 U233
  4. 4 Be8

Answer: 4. 4 Be8

Question 32. Which of the following nuclei is unstable?

  1. 5B10
  2. 7N14
  3. 8O16
  4. 4Be10

Answer: 4. 4 Be10

WBBSE Class 10 Physical Science Solutions

Question 33. A device used for the measurement of radioactivity is 

  1. Mass spectrometer
  2. Cyclotron
  3. G-M-Counter
  4. Nuclear reactor

Answer: 3. G-M-Counter.

Question 34. Which of the following radiations, the one most easily stopped by air is 

  1. X-rays
  2. α-rays
  3. β-rays
  4. γ -rays

Answer: 1. X-rays.

Question 35. y -Ray:

  1. Consist of particles that have mass
  2. Are energy waves.
  3. Have mass
  4. Are deflected by an electric field

Answer: 2. Are energy waves.

Question 36. Loss of B-particle is equivalent to 

  1. Increase of one proton
  2. Decrease of one neutron
  3. Combination of (1) and 2)
  4. None of the above

Answer: 3. Combination of  (1)and (2)

Question 37. Which of the following is the man-made radio-active disintegration series?

  1. Thorium series
  2. Actinium series
  3. Uranium series
  4. Neptunium series

Answer: 4. Neptunium series.

WBBSE Class 10 Physical Science Solutions

38. Question  13A27 is a stable isotope. It is expected to disintegrate by 

  1. α-emission
  2. β – emission
  3. β– emission
  4. Proton emission

Answer: 3. β-emission.

Question  39. Positron has a mass equal to 

  1. Electron
  2. α – particle
  3. Proton
  4. Deuteron

Answer: 1. Electron.

Question  40. Which of the following has a magic number of protons and neutrons?

  1. 8O17
  2. 13Al27
  3. 9Fl17
  4. 20Ca40

Answer: 4. 20Ca40

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus Very Short Answer Type Questions

Question 1. Out of αβ and y rays which one is called hand rays?
Answer: γ rays.

Question 2. What are the magic numbers?
Answer: Magic numbers are 2, 8, 20, 50, 82 and 126.

Question 3. What is the value of the packing fraction for unstable nuclides?
Answer: Positive.

Question 4. What is the value of the packing fraction for stable nuclides?
Answer: Zero or negative.

Question 5. Do nuclear forces obey inverse square law?
Answer: Nuclear forces are not governed by inverse square law

Question 6. Out of a, ẞ and y rays which one causes maximum damage to the body tissues?
Answer: α-rays.

Question 7. What is the density of the nucleus?
Answer: 1014/ Cm3

Question 8. What is the binding energy for a mass defect of 1 amu?
Answer: 931 5 Mev.

Question 9. What is the n/p ratio of lighter nuclides (Zup to 20)?
Answer: 1.

Question 10. What is the n/p ratio of 1ft1 ?
Answer: 0.

Question 11. What is the lifespan of a radioactive element?
Answer: The life span of a radioactive element is infinite.

Question 12. How many a and b particles are emitted from uranium series?
Answer: The uranium series emits 8α and 6β particles.

Question 13. What is artificial series?
Answer: (4n+1) is artificial series.

Question 14. What is the end product of the 4n series?
Answer: The end product of the 4n series is lead.

Question 15. Is the half-life of a radioactive element dependent on its physical and chemical state?
Answer: The half-life of a radioactive element is independent of its physical and chemical state.

Question 16. How mass defects can be converted into energy?
Answer: The mass defect can be converted into energy by using Einstein’s equation ΔE = mc2

WB Class 10 Physical Science Question Answer

Question 17. Write an example of nuclear fission.
Answer:

92 U 235+ 0 n1 35 Br 87 +57 La146 + 3 0 n1

Question 18. Write an example of an isotope of carbon.
Answer: An example of an isotope of carbon C- 14.

Question 19. What is the value of the velocity of ẞ Particle?
Answer: The velocity of β particle is maybe up to \(\frac{9}{10}\) the velocity of light.

Question 20. Define.-Helions.
Answer: α particles are He ions or bare helium nuclei sometimes called helions.

WB Class 10 Physical Science Question Answer Chapter 7 Atomic Nucleus Fill In The Blanks

Question 1. Radioactivity was proposed by_________
Answer: Madam Curie.

Question 2. Measurement of radioactivity is done by _________
Answer: Geiger-Muller counter.

Question 3. The process of transforming one element into another is known as _________
Answer: Alchemy

Question 4. NO2 and CO2, are the examples of _________
Answer: Isosteres

Question 5. The density of the nucleus is of the order of_________
Answer: 1014

Question 6. Magic numbers are 2, 8, 20, 50, 82 and _________
Answer: 126

Question 7. The difference between the total mass of the particles present in the nucleus of a nuclide and its real mass is called _________
Answer: Mass defect

Question 8. Due to the large mass, α particles possess large _________
Answer: Kinetic energy

Question 9. β-particles are _________ charged
Answer: Negatively

Question 10. γ rays are _________
Answer: Neutral

Question 11. γ-rays have the least effect on _________
Answer: Photographic plates.

Question 12. Isotopes of the same element possess the same number of _________
Answer: Protons.

Question 13. 1H2+ 1H32He4  is a  _________ reaction.
Answer: Fusion.

Question 14. The speed of y-rays is _________ speed of light reaction.
Answer: Equal to

Question 15. The atom bomb is based on nuclear _________ reaction.
Answer: Fission.

Question 16. A hydrogen bomb is based on nuclear _________ reaction.
Answer: Fusion.

Question 17. The source of the energy of the Sun is _________
Answer: Fusion.

Question 18. The number of neutrons in a radioactive isotope of hydrogen is _________
Answer: 2.

Question 19. The rate of disintegration of the active nucleus is known as _________
Answer: Activity.

Question 20. 7N14+0n→ a ____________ + 7H14
Answer: 6C14

Question 21. The energy equivalent to 1 amu is _________ me V.
Answer: 931.5.

Question 22. The time taken for the decay of half the initial amount of a radioactive nuclide is called _________.
Answer: Half-life period.

Question 23. The phenomenon of spontaneous emission of invisible radiation is called _________.
Answer: Radioactivity.

Question 24. B-particles are nothing but _________ moving at high speeds.
Answer: Electrons.

Question 25. The phenomenon in which a stable nuclide is converted artificially into a radioactive substance is called _________.
Answer: Artificial radioactivity.

Question 26. Isotones are the nuclides having the same number of_________.
Answer: Neutrons.

Question 27. Atoms of the different elements possessing the same mass number are called _________.
Answer: Isobars.

Question 28. According to Einstein’s equation, the relation between mass and energy can be written as  _________.
Answer: E = mc2

Question 29. _________ is α- a-emitter.

Answer: Polonium-212.

Question 30. Positron has a mass equal to _________.
Answer: Electron.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus Short Answer Type Questions

Question 1. What do you mean by nuclear forces?
Answer:

Nuclear forces: The forces which held the nuclear together within the small nucleus are called nuclear forces.

These forces exist among P-P, P-n and n-n

Question 2. What are nuclear reactions?
Answer:

Nuclear reactions: Nuclear reactions are the reaction in which the nuclear of an atom undergoes a change.

Question 3. What is packing fraction?
Answer:

Packing fraction: Packing fraction was proposed by Aston and defined as the difference between actual isotopic mass and the mass number.

Question 4. What are nuclear isomers?
Answer:

Nuclear isomers: Nuclear species having the same atomic number and same mass number but different radioactive properties are called nuclear isomers.

Examples: 69Zn (7½ 13.8 hours) and 69Zn (7½, = 57 minute)

Question 5. What is Natural transmutation?
Answer:

Natural transmutation: It is a process in which elements such as radium undergo transmutation on their own.

Question 6. What do you mean by ‘Group Displacement Law?
Answer:

Group Displacement Law:

We know that an a-emission decreases the atomic number of the parent by 2 and P-emission increases the atomic number by 1.

Thus ‘In an a-emission, the parent element will be displaced to a group two places to the left and in ẞ-emission, it will be displaced to a group one place to the right.’

Question 7. What is a spallation reaction?
Answer:

Spallation reaction: High-speed projectiles with energies of approximately 40M V may chip fragments from a heavy nucleus, leaving a small nucleus. This type of reaction is called spallation.

Examples: 92 U 235+ 0 n1 74W187 + 20 1H1+ 350 n1

Question 8. What is meant by ‘projectile capture reactions’?
Answer:

Projectile capture reaction: The bombarding particle is absorbed with or without the emission of γ radiations.

Examples: 92 U 238+2H4 92 U 239+ γ

Question 9. What is the disintegration series?
Answer:

Disintegration series: The whole series of elements starting with the parent radioactive element to the stable end product is called a radioactive disintegration series.

These series are- 4n (4n+1), (4n+2) and (4n+3)

Question 10. What do you mean by ‘Alchemy’?
Answer:

Alchemy: The process of transforming one element into another is known as Alchemy and the person involved in such experiments is called an alchemist.

Question 11. What is a disaster?
Answer:

Isoster: Molecules or ions with the same number of atoms and also the same number of electrons are said to form an isosteric group or more simple isosteres.

Examples: NO2 and CO2

Question 12. What is Radioactivity?
Answer:

Radioactivity: It is a process in which the nucleus of certain elements undergo spontaneous disintegration without excitation by any external me

WBBSE Class 10 Physical Science Solutions

Question 13. What do you mean ny ‘isotopes’?
Answer:

Isotopes: The atoms of an element having the same atomic number but different mass numbers are called isotopes.

Examples: 1H1,1H2,1H3

Question 14. What is Nuclear fission?
Answer:

Nuclear fission: The process of artificial transmutation in which a heavy nucleus is broken down into two lighter nuclei of nearly comparable masses with the release of a large amount of energy is termed nuclear fission.

Examples: 92 U 235+ 0 n1 56Ba140+ 36Kr93+ 3 0 n1

Question 15. What is a half-life period?
Answer:

Half-life period: The half-life period of a radioactive isotope is the time required for one-half of the isotope to decay.

⇒ \(t \frac{1}{2}=\frac{0.693}{\lambda}\)

(λ = Disintegration constant)

Question 16. What is nuclear fusion?
Answer:

Nuclear fusion: A nuclear reaction in which two lighter nuclei are fused together to form a heavier nucleus is called nuclear fusion. 2. 3 4

Examples:  1H2 + 1H3 2He4 +17.6MeV

Question 17. What is Artificial radioactivity?
Answer:

Artificial radioactivity: It is the phenomenon in which the artificial transmutation of a stable nucleus leads to the formation of a radioactive nuclide.

Question 18. What do you mean by ‘Rutherford’?
Answer:

Rutherford: If a radioactive substance has 106 disintegrations per second, it is said to have an activity of one Rutherford.

Question 19. What are magic numbers?
Answer:

Magic numbers: Magic numbers are the numbers 2, 8, 20, 50, 82 and 126. Nuclides having a magic number of either protons or neutrons or both are more stable.

Question 20. What is Radioactive equilibrium?
Answer:

Radioactive equilibrium: Radioactive change being an irreversible process shows equilibrium when a daughter element disinter grates at the same rate at which it is formed from the parent element.

Question 21. What is the packing fraction of
Answer: Packing fraction (isotopic mass-mass number) mass number

Packing fraction= \(\frac{(\text { isotopic mass }- \text { mass number })}{\text { mass number }} \times 10^4\)

= \(\frac{55.92066-56}{560} \times 10^4\)

= – 14.167

So, the packing fraction of 26Fe56 is  -14.167

Class 10 Physical Science WBBSE Question 22. How many a and ẞ particles will be emitted by 84 Ra218 in changing to 82 pb206?
Answer:

Let x and y be the number of a and ẞ particles involved in bringing about the change

⇒ 84 Ra218 82 pb206 +x 2He4+y -1e0

Comparing the mass number =  218

= 206 + 4x + 0y

⇒ 4x = 12

⇒ x = \(\frac{12}{4}\)

⇒ x= 3

Comparing the atomic numbers

84 = 82 + 2x + y

2x – y = 2

y = 2(3)- 2

y = 6- 2

y = 4

The number of a and ẞ particles emitted in the given nuclear reaction is 3 and 4.

Question 23. Calculate the number of neutrons in the remaining atom after the emission of a particle from the 92 U 238 atom.
Answer:

On emission of an a-particles atomic mass

If daughter element 238 – 4 = 234 (Mass)

Atomic number of daughter element = 92-2 = 90

Number of Neutrons =Atomic mass – Atomic number 234  – 90=  144.

Question 24. Short Note-y rays.
Answer:

γ rays: γ rays are high-energy electromagnetic waves. This can be shown by diffraction experiments. The penetration power of the y-rays is 10,000. There is no rest mase of γ -rays.

Class 10 Physical Science WBBSE

Question 25. Write four characteristics of γ-rays.
Answer:

  1. The mass of y ray is none
  2. The nature of y-ray is electromagnetic waves
  3. The relative ionisation capacity of γ-rays is 1.
  4. The penetration power of γ-ray is 10,000.

Question 26. Short Note-a Gaiger Miller counter.
Answer:

Geiger Miller counter:

Particles produce scintillations on Zns screens and can easily be detected. One can detect and count the number of β -Particles with the help of the Geiger Muller counter. It consists of a tube fitted with two electrodes at the two ends and contains an inert gas like neon or argon.

Question 27. Short Note β -particles.
Answer:

β -Particles: By measuring the e/m of the B-particles by deflecting them in the an-electric field and a magnetic field it has been conclusively proved that the β – particles are nothing but high-speed electrons. In artificial radioactivity, however, β – particles can be emitted as positrons also.

Question 28. Short Notes a-particle.
Answer:

α-particles = α-particles are in fact. The ions or bare helium nuclei are sometimes called helions. That the a-particles are He can be easily proved in a laboratory, acquire squirt electrons from the surrounding and get neutralised to produce helium atoms. The spectrum of the gas obtained from the a-particles emitted from a radioactive source confirms this point.

As α-particles are neutralised to give helium atoms, uranium mines are always found to be a very good source of helium gas. One can early count the number of particles emitted by a radioactive source, in a given time from the scintillation in an Instrument called Spintharis cope.

It is from the ionisation in a Gelger-Muller counter. One can also measure the amount of positive charge imparted to a metal plate during the same period, one can now determine the number of units of charge each particle carries by dividing the total charge imparted to the metal by the total number of particles emitted during the period. It is found to be two units (taking the amount of electronic charge as a unit). ]

Question 29. Radioactivity is a nuclear phenomenon Explain it.
Answer:

If we take two compounds of a radioactivity element containing the same amount of the element, the measured amounts of radioactivity in the two cases appear to be the same in both cases.

This fact shows that radioactivity is a property of an element, not of a compound. While discussing the cause of radioactivity Rutherford and Soddy commented that radioactivity was an unclear phenomenon.

They presented the following arguments in flavour of Heir’s view. When an atom of radioactive element emits a particle the element loses its atomic mass by 4 units and is converted into an atom written the Uranum series of radioactive elements.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 7 Atomic Nucleus Uranum And Radioactive Elements

Question 30. Write four characteristics of -rays.
Answer:

  1. The nature of β – rays is electrons
  2. The mass of β- the ray is \(\frac{1}{1856}\) AMU.
  3. The relative ionisation Capacity is 100
  4. The penetration power is 100.

Question 31. Short Note-Soddy Fajan’s rule.
Answer:

Soddy Fajan’s Rule:

A problem arose about the placement of the elements of the Uranium series in the periodic table soon after their discovery. It was observed that the element formed by the emission of a particle followed by the emission of two β – particles starting from U, with the atomic mass and atomic number producing an element which had the same chemical properties as that of U.

Question 32. Establish the equation of the rate of radioactive disintegration for a half-life period.
Answer:

We know \(\frac{\mathrm{dN}}{\mathrm{dt}}\) α N

\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = λN ______ (1)

λ = Radioactive disintegration

Integrating the equation  (1)

N= Noe – λt______ (2)

Taking Lagarithm and simplification

λ = \(\lambda\frac{1}{7} \ln \frac{\mathrm{No}}{\mathrm{N}}=\frac{2.303}{7} \log \frac{\mathrm{No}}{\mathrm{N}}\)

λ =  \(\frac{1}{\frac{t_{01}}{2}} \ln _2=\frac{2.203}{\frac{t_1}{2}} \log 2=\frac{0.693}{t_{\frac{1}{2}}}\)

λ =  \(\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}\)

Class 10 Physical Science WBBSE

Question 33. What is the integration form of the rate of radioactive disintegration?
Answer:

We know, \(\frac{\mathrm{dN}}{\mathrm{dt}}\) α N

\(\frac{\mathrm{dN}}{\mathrm{dt}}\)α N

\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = λn

λ = Radioactive disintegration

Integrating the equation, N Noe – λt

No initial number of atoms of the radioactive element.

Question 34. Define isotope. Write an example of the isotope.
Answer:

Isotope Some elements have the same atomic numbers though they have different numbers. They are called isotopes as they occupy the same position in the periodic table.

Example: 92U238

Question 35. Write about the rate of radioactive disintegration.
Answer: If there are N number of atoms of a radioactive element in a sample at a

The particular instant of time, the rate of disintegration at that instant will be proportion to N.

\(\frac{\mathrm{dN}}{\mathrm{dt}}\)α N

\(\frac{\mathrm{dN}}{\mathrm{dt}}\)= – λN

γ = Constant (radioactive disintegration).

Chapter 7 Atomic Nucleus Broad Answer Type Questions

Question Explain the separation of isotopes through the diffusion method.
Answer:

Though the isotopes have identical chemical properties, some of their physical properties may be quite different because of the difference in their mass numbers.

The most important physical properties of this type are the rate of diffusion, boiling point, and radioactive properties isotopes of elements can be separated by physical methods based on their properties. They may be separated almost completely using a special type of instrument called the mass spectrometer.

Diffusion method:

Since the rate of diffusion of a gas varies inversely to the square root of its molecular mass. Gas molecules with lower atomic mass numbers diffuse more quickly than their heavier sisters.

This principle was very effected actively used by Aston for the separation of isotopes of neon. The more is the difference in the atomic mass numbers, the more effective is the separation.

Class 10 Physical Science WBBSE

Question 2. Write the cause of radioactivity.
Answer:

Cause of radioactivity:

The main reason for the radioactivity of an element, as we have noted earlier, is the instability of its nucleus. Such an unstable nucleus tries to attain stability through the expulsion of particles and rays. If there are too many protons in a nucleus they will fall apart and the element will disintegrate. Again, it the nucleus is too heavy with too many neutrons it disintegrates.

It is observed that in order to attain stability the neutron: Proton ratio must be mear about unity for the elements with low atomic masses while for the heavier elements, it must not exceed 1.5.

Thus we see that most of the radioactive elements have very high atomic masses in 92U234 in comparison to their atomic numbers U has a neutron; proton ratio of 1.537 while radium has 1.56. This is why radium is much more radioactive than uranium.

Question 3. Short Note-Radioactive Equilibrium.
Answer:

Radioactive Equilibrium :

The product element formed by the radioactive disintegration of an element may also be radioactive. The process goes an until the chain is terminated by the birth of a non-radioactive stable element at the end of a series. If we keep a sample of a radioactive element or its salt for a sufficient time, a situation arises when the rate of creation and the rate of disintegration of a product element become equal.

This state for a radioactive series is called a radioactive equilibrium. We see that at the equilibrium.

⇒ \(\frac{\mathrm{dN}_1}{\mathrm{dt}}=\frac{\mathrm{dN}_2}{\mathrm{dt}}=\frac{\mathrm{dN}_3}{\mathrm{dt}}\)

∴ λ1 N1 = λ2 N2 = λ3 N3 = …………

Question 4. Short Note Mass spectrometric method to separate the isotope.
Answer:

Mass spectrometric method:

In this method positive particles of the required gas or vapour formed by bombardment with electrons are first passed through electrical plates with a potential difference (X) of about 1000 volts and two collimating slits. The electrical energy Xe, where e is the charge of the positive particles, is equal to the kinetic energy of the particles.

That is Xe =½ mu2 where m and v are the mass and the charge of the positive particles. The positive particles energing through a slit with almost equal kinetic energy pass a magnetic field (H) between two semicircular.

Question 5. What is binding energy? Short Note-Stability of an atomic nucleus packing fraction.
Answer:

The amount of energy that can be obtained by the amount of mass equal to the mass defect is known as the binding energy.

  1. The atomic mass of an isotope of an element is expected to be equal to the Sun of the fundamental particles present in it. In AMU the mass of a proton or a neutron is nearly equal to unity while that of an electron is negligible.
  2. Atomic masses of isotopes of elements are expected to be approximately whole numbers. Called atomic mass numbers.

Packing faction =  \(\frac{\text { Actual mass of a nucleus }- \text { Mass number }}{\text { Mass number }} \times 10^4\)

WB Class 10 Physical Science Question Answer

Question 6. What are isotones? Define-nuclear isomers.
Answer:

The are elements which differ in both their atomic numbers as well as most numbers but they have the same number of neutrons in their nuclei. They are called isotones.

The two nuclei of an element may have the same atomic number and the same mass number yet differ in their internal nuclear energies. These nuclei are called nuclear isomers.

Question 7. Explain the separation of isotopes by distillation under low pressure.
Answer:

Distillation under low pressure: When a liquid like mercury is distilled under low pressure the lighter isotopes come out more quickly and condense on a cooler surface. The condensate is collected, melted and subjected to the same procedure. B repeating the process the condensate can be made richer in the lighter variety.

Question 8.  What do you mean by isobars and isotones?
Answer: Isobars The nuclide of different chemical elements having the same mass number but different atomic numbers are called isobars.

Examples:

  1. 18 Ar40
  2. 19K40
  3. 20Ca40

Isotones: The nuclides of different chemical elements having the same number of neutrons but different atomic numbers are called isotones.’

Examples: 1H3, 2He4

Question 9. What are the parent and end products of different disintegration series?
Answer: Series

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 7 Atomic Nucleus Atomic Nucleus Different Disintegration Series

Question 10. What are the differences between nuclear fission and nuclear fusion?
Answer:

Difference between nuclear fission and nuclear fusion :

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 7 Atomic Nucleus Atomic Nucleus Nuclear Fission And Nuclear Fusion

Question 11. What do you mean by Curie and Becquerel?
Answer:

  • Curie If a radioactive substance disintegrates at the rate of 3.7 × 1010 disintegrations per second, its activity is said to be 1 curie.
  • Becquerel (SI unit): If a radioactive substance has 1 disintegration per second, it is said to have an activity of one Be equal.

WB Class 10 Physical Science Question Answer

Question 12. A radioactive substance decays at such a rate that after 46 days only 0.25 of its original amount is left to calculate its disintegration constant.
Answer:

Let, the original amount be No then Amount after 46 days, Nt = 0.25 x No For a nuclear decay,

λ = \(\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{No}}{\mathrm{Nt}}\)

= \(\frac{2.303}{46}\) log4 day-1

= 0.0301day-1

Question 13. A certain nuclide has a half-life of 60 min. If a sample containing 600 atoms is allowed to decay for 90 min, what will be the remaining atoms?
Answer:

λ = \(\frac{0.693}{60} \mathrm{~min}^{-1}\)

= \(\frac{2.30^3}{90} \log \frac{\mathrm{No}}{\mathrm{N}}\)

⇒ log \(\frac{\mathrm{No}}{\mathrm{N}}\) = 0.4514 Or,

⇒ log \(\frac{\mathrm{No}}{\mathrm{N}}\)= 2.828

∴N = \(\frac{600}{2.823}\)

N= 212

Question 14. The activity of a sample of radioactive element A100 is 6.02 curie. Its decay constant is 3.7 x 10’S’ which calculates the initial mass of the sample.
Answer:

Activity = \(\lambda \times \frac{\mathrm{wt}}{\text { at. wt }} \times 6.023 \times 10^{23}\)

\(6.02 \times 3.7 \times 10^{10}=3.7 \times 10^4 \times \frac{w}{100} \times 6.023 \times 10^{23}\)

 

W= \(\frac{6.02 \times 3.7 \times 10^{10} \times 100}{3.7 \times 1.04 \times 6.023 \times 10^{23}}\)

W= \(10^{-15} \mathrm{~g}\)

Question 15. The half-life period of C is 5760 years. An old piece of wood has a disintegration rate which is 25% of the disintegration rate of an equal weight of a new piece of wood.
Answer: Let the rate of disintegration of the new piece = 100

The rate of migration of old pieces = 25

K = \(\frac{0.693}{\frac{t_1}{2}}=\frac{0.693}{5760}\)

t= \(\frac{2.303}{K} \log \frac{a}{a-x}\)

=\(\frac{2.303}{K} \log \left(\frac{\mathrm{lo}}{\mathrm{It}}\right)\)

∴ \(=\frac{2.30 \times 5760}{0.693} \log \frac{100}{25}\)

t= 11523 years.

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 6 Current Electricity

WBBSE Class 10 Physical Science Question Answer In English

Chapter 6 Current Electricity MCQs

Question 1. How many electrons constitute the current of IA?

  1. 6.25 x 109
  2. 6.25 x 105
  3. 6.25 × 1018
  4. 6.25 x 106

Answer: 3. 6.25 x 1018

Question 2. The material of a wire of a potentiometer is 

  1. Copper
  2. Steel
  3. Magnani
  4. Aluminium

Answer: 3. Magnanin.

Question 3. The vertical component of Earth’s magnetic field is zero at a place where the angle of dip is :

  1. 60°
  2. 45°
  3. 90°

Answer: 1. 0°

Question 4. A magnetic needle suspended freely orients itself: in a definite direction upward

  1. In a definite direction
  2. In no direction
  3. Upward
  4. Downward

Answer:  2. In no direction.

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

Question 5. The unit of specific conductivity is.  

  1. Ohm – cm-1
  2. Ohm – cm2
  3. Ohm-1-cm-1
  4. Ohm-1 cm2

Answer:  3. ohm-1 cm-1

Question 6. The length of a wire is doubled. Its conductance will be

  1. Unchanged
  2. Halved
  3. Doubled
  4. Quadrupled

Answer:  2. Halved.

Question 7. The resistance of the discharge tube is:

  1. Ohmic
  2. Non – Ohmic
  3. Both
  4. None of these

Answer:  2. Non-ohmic.

WBBSE Class 10 Physical Science Solutions

Question 8. A wire of resistance 1 W is stretched to double its length. The resistance will be:

  1. 3V
  2. 9V
  3. 5V
  4. 15V

Answer: 2. 9V.

Question 9. The specific resistance of a wire depends upon 

  1. Length
  2. Cross-sectional area
  3. Mass
  4. None of these.

Answer:  4. None of these

Question 10. The unit of specific resistivity is 

  1. Ohm – cm-1
  2. Ohm – cm2
  3. Ohm-cm
  4. Ohm-1 cm2

Answer:  3. Ohm – cm

Question 11. The SI unit of electric charge is 

  1. Coulomb
  2. Ampere
  3. Ohm

Answer: 1. Coulomb

 

Question 12. The unit of potential difference is 

  1. Joule
  2. Ohm
  3. Volt
  4. Ampere

Answer:  3. Volt.

WBBSE Class 10 Physical Science Solutions

Question 13. The unit of electrical power is:

  1. Volt
  2. Ohm
  3. Watt
  4. Joule

Answer:  3. Watt

Question 14. What material an electric fuse made of?

  1. Copper
  2. Iron
  3. Silver
  4. Tin-lead alloy

Answer:  4. Tin-lead alloy.

Question 15. Unit of resistance is 

  1. Ampere
  2. Ohm
  3. Coulomb
  4. Joule

Answer:  2. Ohm.

Question 16. Specific resistance of a conductor depends on 

  1. Its length
  2. Its Cross-sectional
  3. It’s material

Answer:  3.  It’s material.

Question 17. Which one is not the conductor of electricity

  1. Copper
  2. aluminu
  3. Gold
  4. Plastic

Answer:  4. Plastic

WBBSE Class 10 Physical Science Solutions

Question 18. The electric meter is a house record:

  1. Charge
  2. Current
  3. Energy
  4. Power

Answer: 3. Energy.

19. The main fuse is connected in:

  1. Live wire
  2. Neutral wire
  3. Both the live and earth wires
  4. Both earth and the neutral wire

Answer:  1. Live wire.

Question 20. In parallel combination of resistances 

  1. P.d. is the same across each resistance
  2. Total resistance is increased
  3. Current is same in each resistance
  4. None of the above

Answer:  1. p.d. is the same across each resistance.

Question 21. For which of the following substances, resistance decreases with increase in temperature?

  1. Copper
  2. Mercury
  3. Carbon

Answer:  3. Carbon.

Question 22. Which of the following is an ohmic resistance?

  1. Carbon-arc lamp
  2. Diodevalve
  3. Nichrome
  4. Junction diode

Answer:  3. Nichrome.

WBBSE Class 10 Physical Science Solutions

Question 23. The amount of heat develops due to current can be known from 

  1. Ampere law
  2. Ohm’s law
  3. Joules law

Answer:  3. Joule’s law.

Question 24. An example of the conservation of electrical energy into mechanical energy is :

  1. Electric cell
  2. Electric lamp
  3. Electric motor

Answer: 3. Electric motor.

Question 25. Ammeter measure :

  1. Resistance
  2. Potential difference
  3. Current

Answer:  3. Current.

Question 26. If the cross-section of the conductor is kept unaltered and the length of the conductor is increased then its resistance :

  1. Increases
  2. Decrease
  3. Remains unaltered
  4. None of the above

Answer:  1. Increases.

WBBSE Class 10 Physical Science Solutions

Question 27. Potential difference is measured by 

  1. Volt meter
  2. Voltmeter
  3. Galvano meter

Answer:  2. Volt meter.

Question 28. The action of electric current on magnet was first observed by 

  1. Oersted
  2. Joule
  3. Barlow
  4. Newton

Answer: 1. Oersted

Question 29. Ohm’s law is related to:

  1. The current and potential difference
  2. Volume and pressure
  3. Temperature density.

Answer: 1. current and potential differences.

Question 30. Electric supply company measures the electricity consumed by the consumers in their household in terms of:

  1. Joule unit
  2. Watt-hour
  3. B.O. T units

Answer:  3.B.O. T units.

Class 10 Physical Science WBBSE

Question 31. The dimensional formula of electric density is :

  1. [MLT-2 A-1]
  2. [MLT3 A-1]
  3. [ML2T3-3A-1]
  4. [ML2T3-A-2]

Answer:  2. [MLT3 A-1]

Question 32. The physical quantity measured by kilowatt-hour :

  1. Electric power
  2. Electric energy
  3. Current

Answer:  2. Electric energy.

Question 33. The correct relation is:

  1. Watt= \(\frac{\text { Volt }}{\text { Ampere }}\)
  2. Watt =  \(\frac{\text { Ampere }}{\text { Volt }}\)
  3. Watt=Volt x Ampere

Answer:  3. Watt Volt x Ampere

Question 34. Electric Flux at a point in an electric field is :

  1. Positive
  2. Zero
  3. Negative

Answer:  2. Zero

Question 35. What is the dimensional formula of electric charge?

  1. [M°L° TA]
  2. [M°L°T-1A]
  3. [M°L°TA-1]
  4. [M°L°T-1A-1]

Answer: 1. [M°L° TA]

Question 36. If six identical cells each having emf of 6V are connected in parallel the emf of the combination is :

  1. IV
  2. 1/6V
  3. 6V
  4. 36V

Answer: 3. 6V.

Class 10 Physical Science WBBSE

Question 37. What is the equivalent resistance of the resistances 3 ohm, 5 ohm and 12 ohm combined in series?

  1. 15 ohm
  2. 18 ohm
  3. 20 ohm
  4. 10 ohm

Answer: 3. 20 ohms.

Question 38. A wire of resistance 1 W is stretched to double its length. The resistance will become :

  1. \(\frac{1}{4}\)
  2. 1
  3. 2
  4. 4

Answer: 4. 4

Question 39. A soap bubble is charged to a potential of 16V. Its radius is doubled. The potential of the bubble now will be

  1. 16V
  2. 4V
  3. 8V
  4. 2V

Answer:  3. 8V.

Question 40. The resistance will be least in a wire with dimension :

  1. L|2, 2A
  2. 2L, A
  3. L, A

Answer:  1. L|2, 2A.

Class 10 Physical Science WBBSE Chapter 6 Current Electricity Very Short Answer Type Questions

Question 1. Define the unit of electric unit.
Answer: The unit of current is ampere. It is the one coulomb of charge flowing through a conductor in one second.

Question 2. What do you mean by the conductivity of a material? Give its SI unit.
Answer: It is the reciprocal of resistivity. Its SI unit is Ohom-‘.

Question 3. How does the drift velocity of electrons in a metallic conductor vary with temperature?
Answer: The drift velocity of a metal decreases with an increase in temperature.

Question 4. What is 1B.O.T?
Answer: The total electric energy expended in 1 hour at the rate of 1 kilowatt is known as 1 B.O.T. (Board of Trade unit.)

Question 5. State the cases in which Ohm’s law is not valid.
Answer: Ohm’s law is not valid for current flowing through gases under low pressure, electrolytes, and semi-conductors.

Question 6. How does the resistance of a conductor depend on the cross-section of the conduction?
Answer: Resistance of the conductor decreases with the increase of cross-section.

Question 7. What is the SI unit of resistance?
Answer: The SI unit of resistance of Ohm.

Question 8. How is the heat generated due to electric current through a resistor related to the strength of the current?
Answer: Heat is directly proportional to square of the current.

Question 9. How is the heat generated due to electric current in a resistor related to the resistance of the resistor?
Answer: Heat is directly proportional to resistance.

Question 10. What is the usual color of a live wire?
Answer: The usual color of a live wire is red.

Question 11. What is the usual color of the earthing wise?
Answer: The usual color of the earthing wire is green

Question 12. What is the emf of a cell?
Answer: The potential difference between the electrodes of a cell in open circuit is called emf. (electromotive force)

Question 13. What is the SI unit of current strength?
Answer: The SI unit of current strength is ampere.

Question 14. What do you mean by relaxation time or mean free time?
Answer: It is the average time interval between the two successive collisions between electron and ion in a conductor.

Question 15. What is current density?
Answer: It is the current flowing per unit area of a conductor.

Question 16. Illustrate a condition in which the electric field is not zero but the potential is zero.
Answer: The electric field on the equatorial line of an electrical dipole is not zero but the potential is zero.

Question 17. What is meant by the steady current?
Answer: A current whose magnitude does not change with time.

Question 18. What is meant by varying current?
Answer: A current whose magnitude changes with time.

Question 19. What is the emf of a simple voltaic cell?
Answer: The emf of a simple voltaic cell is 1.08 volt.

Question 20. Which effect of electric current is demonstrated in an electro-magnet?
Answer: Magnetic effect.

WB Class 10 Physical Science Question Answer Chapter 6 Current Electricity Fill In The Blanks :

Question 1. Coulomb =____________ × second.
Answer: Ampere.

Question 2. The equivalent resistance is smaller than the ____________ resistance in a parallel combination of resistances.
Answer: Smallest.

Question 3. Power potential difference x ____________
Answer: Current.

Question 4. I2rt is some electrical
Answer: Work.

Question 5. Resistance of a wire is ____________ proportional to the length of the wire.
Answer: Directly.

Question 6. The watt-hour is the practical unit of ____________energy.
Answer: Electrical.

Question 7. Number of B.O.T. unit =\(\frac{\text { ampere } \times \text { volt }}{1000}\) × ____________
Answer:
Hour

Question 8. The SI unit of current is ____________
Answer: Ampere

Question 9. When electric current does not change the direction it is called ____________
Answer: Direct current

Question 10. Watt= Volt x ____________
Answer: Ampere

Question 11. The electromagnets are made using the current.
Answer: Magnetic

Question 12. Usually the resistance of a conductor ____________ with the rise in temperature
Answer: Increases effect of electricity with the rise in temperature.

Question 13. In Ohm’s law, the constant ____________ and other physical conditions are remaining
Answer: Temperature

Question 14. The temperature remaining constant, the resistance between two opposite forces of a unit cube of a conductor is called its ____________
Answer: Specific resistance

Question 15. Ohm’s law is related to ____________
Answer: Current and potential difference

Question 16. Specific resistance of a conductor depends on____________
Answer: It’s material

Question 17. The unit of electrical power is ____________
Answer: Watt

Question 18. A ____________ is used to measure the potential difference between two points in a section of an electrical circuit.
Answer: Voltmeter

Question 19. An ammeter is used to measure the  ____________  flowing through an electric circuit where the ammeter is connected in series.
Answer: Current

Question 20. A short circuit occurs due to accidental direct contact of the life and the ____________ wires.
Answer: Neutral

Question 21. The total resistance of a number of resistors in series is equal to the ____________  of the resistances of the component resistors.
Answer: Sum

Question 22. Electromotive force is some ____________ not a force.
Answer: Energy.

WBBSE Class 10 Physical Science Question Answer

Question 23. In current electricity, the potential difference between two points is measured by the work done when unit ____________ flows from one point to the other.
Answer: Change

Question 24. If an electrical instrument of power 1 watt works for 1 hour, the electrical energy expended is known as 1 ____________
Answer: Watt-hour

Question 25. Potential difference between two points one ____________ when work is done to carry one coulomb charge between the points is 1 Joule.
Answer: Volt

Question 26. Resistance is the natural property of every material body by virtue of which the body ____________ flow of electric charge through it.
Answer: Opposes

Question 27. The current flowing through a conductor is directly proportional to the potential difference between its ends when ____________  and other physical conditionsof the conductor remain unaltered.
Answer: Temperature

Question 28. Ohm’s law is not valid for current flowing through ____________  under low pressure, electrolytes and semiconductors.
Answer: Gases

Question 29. The unit for specific resistivity is ____________
Answer: Ohm-cm

Question 30. Electric flux at a point in an electric field is ____________
Answer: Zero

Question 31. The material of a wire of a potentiometer is ____________
Answer: Magnanin

Question 32. Ammeter measures  is ____________
Answer: Current

WBBSE Class 10 Physical Science Question Answer Chapter 6 Current Electricity Short Answer Type Questions

Question 1. What is the definition of ohm?
Answer: Ohm: If one-ampere current flowing through a conductor establishes 1-volt potential difference between the two ends of the conductor, the resistance of the conductor is one ohm.

Question 2. What do you mean by potential difference?
Answer: Potential difference: It is the electrical condition of a point in an electric field or on a current-carrying conductor that indicates whether electrons will flow from it or it from another connected point.

Question 3. What is the definition of resistance from Ohm’s law?
Answer: Definition of resistance from Ohm’s law : The resistance of a conductor is a ratio of the potential difference between its ends to the current flowing through it.

Question 4. What is the definition of a coulomb?
Answer: Coulomb It is the quantity of electric charge that passes through silver nitrate solution deposits 0.00 1118g silver at the cathode.

Question 5. What do you mean by a combination of resistance?
Answer: Combination of resistance: In different electrical circuits more than resistance are connected together. This is known as a combination of resistance.

Question 6. What is a voltmeter?
Answer: Voltmeter: A voltmeter is used to measure the potential difference between two points in a section of an electrical circuit; the voltmeter is connected parallel to the section.

Question 7. Why resistivity is also called specific resistance?
Answer:  Explanation: Resistivity is also called specific resistance as the resistivity of a material is the resistance offered by the material of specified dimensions unit length unit cross-sectional area.

Question 8. What is the internal resistance of a cell?
Answer: Internal resistance: The small resistance offered by the electrolyte of a cell to the electric charges flowing through it from the negative to the positive plate is known as to the internal resistance of the cell.

Question 9. State ohm’s law.
Answer:  Ohm’s law (1826): The temperature and other physical conditions remaining constant the current flowing between any two points of a conductor is proportional to the potential difference between them.

Question 10. The specific resistance of copper 20°C is 1.6×10-6 ohm-cm, what do you mean by it?
Answer: Specific resistance: The specific resistance of copper at 20°C is 1.6×10-6 ohm-cm, it means: Resistance across the opposite faces of a copper cube of 1 cm side, at 20°C is 1.6×10 ohm.

Question 11. What is an ammeter? What do you mean by 1-ampere electric charge?
Answer: Ammetre: The current is measured by an instrument called an ammeter. When I coulomb of charge flows through any cross-section of a conductor in 1 second, the electric current flowing through it is said to be 1 ampere.

Question 12. What do you mean by potential difference?
Answer: Potential difference: The potential difference between two points in an electric circuit is defined as the amount of work done moving a unit charge from one point to the other point.

Question 13. What is the resistance of a conductor? What do you mean by the 1 ohm resistance of the conductor’?
Answer: Resistance of conductor: The resistance of a conductor is the ratio of the potential difference across the ends of a conductor to the current flowing through the conductor. The resistance of a conductor is 1 ohm if a potential difference of Ivolt across the end of the conductor makes a current of 1 ampere pass through it.

Question 14. What do you mean by 1-volt potential difference’?
Answer: Potential difference: The potential difference between two points in an electric circuit is defined as the amount of work done moving a unit charge from one point to the other point.

WB Class 10 Physical Science Question Answer

Question 15. Write the properties of electric charges. The properties of electric changes 
Answer: Electric charges:

Properties of Electric changes:

  1. Like charges repel each other and unlike charges attract each other.
  2. Electric charge is always conserved.

Question 16. Define-electric current what is the S.I. unit of potential difference? Answer:

Electric current:

  1. The electric is the flow of electric charges in a conductor such as a metal wine.
  2. The S.I. unit of potential difference is volt.

Question 17 What do you mean by electric charge? Is electric charge a scalar quantity? Write S.I. unit of electric charge.
Answer:

Electric charge:

  1. Electric charge is the physical property of a matter that causes it to experience a force when placed in an electromagnetic field.
  2. Yes, the electric charge is a scalar quantity.
  3. The S.I. unit of electric charge is coulomb

Question 18. The define-terminal voltage of the cell. What is the e.m.f of a cell?
Answer:  Terminal And E.M.f of a cell

  1. Terminal voltage of the cell:  When the current is drawn from a cell, the cell is in a closed circuit, The potential difference between the electrodes of the cell is called the terminal voltage of the cell.
  2. E.m.f of a cell: The e.m.f. of a cell is the energy spent our work done per unit charge is taking a unit positive charge around the complete circuit containing the cell, when the circuit is open.

Question 19. What is meant by “The potential at a point is 1 volt”?
Answer: The potential at a point:  The potential at a point is one volt if one joule of work is done in moving one colomb of charge from infinity to that point in an electric field.

Question 20. Define the electromotive force of the cell electromotive force of cell :
Answer:

The electromotive force of cell:

  1. The work done in carrying a unit positive charge once through a complete circuit is called the electromotive force of the cell.
  2. The mathematical expression of the electromotive force of the cell: E = V + v.

WB Class 10 Physical Science Question Answer

Question 21. What is the electromotive force?
Answer:

Electromotive force:

  1. In a cell, the chemical action creates a difference in the concentration of electrons between the electrodes of the cell, which results in to the potential difference between the electrodes called electromotive force.
  2. When no current is drawn from a cell, the cell is in the open. circuit, the potential difference between the terminals of the cell is the electromotive force of the cell.

Question 22. Define the resistivity of a conductor.
Answer:

The resistivity of a conductor: The resistivity of a conductor is the resistance of the conductor of unit length and unit area of cross-section at a constant pressure.

Question 23. Short Note-Insulator.
Answer:

Insulator: There are substances that are very poor conductors of electricity having very low values of resistivity are called insulators.

Example: Wood, Glass.

Insulators are used to protect us from electric shocks.

Question 24. What is an electromagnet?
Answer: 
Electromagnet: If soft iron is kept in a current-carrying solenoid, then that soft iron behaves like a magnet so long as the current passes. This magnet is known as an electromagnet.

Question 25. What is an electric motor?
Answer:

Electric motor:

  1. The device or machine which converts electrical energy into mechanical energy is known as an electric motor.
  2. An insulated copper coil, wound over a suitable frame rotates in a magnetic field when electric current passes through the coil.

Question 26. Deduce the mathematical form of Ohm’s law.
Answer:

The mathematical form of Ohm’s law:

Let VA and VB be the potentials at the ends A and B of the conductor AB respectively.

So, the potential difference between the points is VA – VB = V (say) Now, if current I  flows through the conductor,

Then following Ohm’s law, να I or,

V = RI (R = constant, the resistance of conductor)

Or, \(\frac{\mathrm{V}}{\mathrm{I}}\) = R

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Ohms Law

Question 27. On what factors does the resistance of a conductor depend? Answer:

Factors upon which the resistance of a conductor depends:

1. Effect of length: Temperature, material, and area of cross-sector remaining constant, the resistance (R) of a conductor is proportional to its length

∴ Rα I (When temperature, material, and cross-section are constant)

2. Effect of cross-section: Temperature, material, and length are constant.)

WB Class 10 Physical Science Question Answer

Question 28. Define equivalent resistance.
Answer:

Equivalent resistance: The single resistance, instead of multiple resistance in a circuit, keeps the voltage and current unchanged, which is called the equivalent resistance of those resistances.

Question 29. What is specific resistance?
Answer:

Specific resistance: We know, R α I when A is constant

R α \(\frac{1}{\mathrm{~A}}\) when 1 am constant

When both the length and the area of the cross-section of a conductor very then from the law of joint variation, we can write,

R α \(\frac{1}{\mathrm{~A}}\)

R= ρ \(\frac{1}{\mathrm{~A}}\) (ρ)(rho) is the constant is proportionality and is known as the specific of resistivity]

Now, if 1 = 1 and A = 1, then R = ρ

Specific resistance Definition:

The specific resistance or resistivity of a material is numerically equal to the resistance of a conductor of the material of length 1 meter and area of cross- section 1 m2.

Question 30. Define 1 kilowatt-hour?
Answer: If a machine of 1 kilo-hour operates for one hour the amount of energy spent is known as kilo-watt hour. This amount of energy is also known as the Board of Trade unit (B.O.T. units).

1 K.W.H = \(\frac{1 \text { B.O.T (watt } \times \text { hour })}{1000}\)

= \(\frac{\text { volt } \times \text { ampere } \times \text { hour }}{1000}\)

Question 31. What is watt-hours? Whose unit is it?
Answer:

Watt-hour: If an electrical machine of power one watt operates for one hour then one watt-hour amount of energy is said to be spent.

1 watt-hour = 1 watt x 1 hour  = 1 watt x 3600 sec. = 3600 J

The unit of electrical energy is watt-hour.

Question 32. What do you mean by the statement? ‘Potential difference between two points in an electric field is 5 volts’?
Answer:

Explanation: ‘Potential difference between two points in an electric field is 5 volts’-this statement means an external agent has to do 5 joules work to carry 1-coulomb positive charge from a point at the lower potential to a point at the higher potential in the electric field.

Class 10 Physical Science Solution WBBSE

Question 33. How can the strength of the motor be increased?
Answer:

The strength of an electric motor can be increased by :

  1. Increasing the current in the armature.
  2. Increasing the strength of the magnetic field.
  3. Increasing the number of turns in the armature.

Question 35. What do you mean by series combination?
Answer:

Series combination: In this combination, resistances are so connected that extreme end of one resistance is joined to the beginning end of the next resistance soon. In this connection same current. flows through all the resistances.

If three resistors r1, r2, and r3, are connected in series, the same current I passes through each then their equivalent resistance R will be.

R = r1+ r2+ r3

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Series Combination Resistance

Question 36. What do you mean by parallel combination?
Answer:

Parallel combination:

A number of resistors are said to be connected in parallel when they are placed side by side and their corresponding ends joined together so that the main current is distributed among them. If the individual resistances is parallel combination are r1+ r2+ r3 then their equivalent resistance R is given by

\(\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\)

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Parallel Combination Of Resistance

So, to create a low resistance out of a few respectively high resistances, those needed to be connected to a parallel combination. Equiva- lent resistance in parallel combination is lesser than the lowest of the individual resistances.

Question 37. State Fleming’s left-hand rule.
Answer:

Fleming’s left-hand rule:

If the thumb, the first finger, and the middle finger of the left hand be held mutually perpendicular to each other in such a way that the first finger points in the direction of the magnetic field and the second finger to that of the current, then the thumb will indicate the direction at motion of the conductor.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Flemings Left Hand Rule Of Magnetic Field

Question 38. State Ampere’s swimming rule.
Answer:

Ampere’s swimming rule: If a man is imagined to be swimming along a current carrying wire in the direction of the current (south or north) with his face turned towards a freely rotating magnetic needle, then the north pole of the needle will be deflected towards his left hand.

Class 10 Physical Science Solution WBBSE

Question 39. Prove that, I =\(\frac{Q}{t}\)
Answer:

If Q amount of charge blows through a cross-section of a conductor in time t, then the current I is given by

I = \(\frac{Q}{t}\)

Question 40. Prove, VA– VB = \(\frac{W}{q}\)
Answer:

If W be the work done in moving a charge q from point B to point A then the potential difference (VA– VB) between the two points (VA– VB) = \(\frac{W}{q}\)
WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Potential Differences

Question 41. Draw the graph between the current (I) VS potential difference.
Answer:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Current And Potential Differences

Question 42. What do you mean by conductors?
Answer:

Conductors: The substances which allow an electric current to flow easily through them are called conductors.

  • The conductors have low resistance and resistivity. Resistances of conductors increase with the increase of temperature.
  • Metals are good conductors of electricity having low volume of resistivity.
  • Example: Silver, Copper

Question  43. Draw the graph of the resistivity of the conductor with an increase in temperature in the case of conductors.
Answer:

Graph of the resistivity of the conductor:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Graph Of Conductor With Increase Of Temperature

P = Resistivity of the conductor.
T = Temperature.

Question 44.

  1. Define-electrical power.
  2. What do you mean by one kilowatt-hour of electrical energy?

Answer:

  1.  Electrical power: Electric power is the amount of electric energy consumed in a circuit per unit of time.
  2. One kilowatt: One-kilowatt hour is the amount of electric energy consumed by a 1000W electric device when it operates for one hour.

Class 10 Physical Science Solution WBBSE

Question 45. What will the resistance of the bulb of 220V-100W?
Answer:

The power rating of (220V-100W) electric bulb means that the resistance of

Its filament while glowing is

R=  \(\frac{\mathrm{V}^2}{\mathrm{P}}\)

R= \(\frac{(220)^2}{100}\)

R= 484 Ω

Question 46. What is meant by earthing?
Answer:

Earthing By earthing we mean that the metal body of the appliance is connected to a thick copper wire, which is buried deep in the earth, and its other end is connected to a copper plate surrounded by a mixture of charcoal and common salt.

Question 47. Electric cell as the source of EMF’-Explain.
Answer:

EMF:

  • A source of electricity without the presence of active machinery parts is called an electric cell.
  • By the transformation of chemical energy to electric energy, an electric cell generates a steady electric current in a current.

Question 48. 1 kilo watt hour = 3.6 x 106 Joule; Prove it.
Answer: 1 kilowatt-hour 1000 watt-hour.

1 watt = 1 Js-1

1 hour (60 × 60) second = 3600 second.

1 kilo watt-hour = 1000 Js1× 3600 second = 3600000 J.

1 kilo watt-hour = 3.6 x 106 .

Question 49.

  1. State Lenz’s law.
  2. Explain Lenz’s law as a consequence of the principle of conservation of energy.

Answer:

Lenz’s law:

  • The direction of induced emf in a circuit is such that it always arrases the very cause for which it is due.
  • Lenz’s law is a consequence of the principle of conversation of energy. Lenz’s law is a form of the how of conservation of energy i.e., Lenz’s law can be derived from the principle of conservation of energy.

Question 50. Draw the graph between resistivity vs increase in temperature in case on semiconductor.
Answer:

The temperature in case of semiconductors:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Resistivity Vs Increase In Temperature Semiconductor

  • Presistivity of the conductor
  • T= Temperature

Question 51. Draw the graph between the resistivity and temperature in the case of a superconductor.
Answer:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Resistivity Vs Increase In Temperature Superconductor

P = Resistivity of the conductor

T = Temperature

Class 10 Physical Science Solution WBBSE Chapter 6 Current Electricity Broad Answer Type Questions

Question 1. State Joule’s laws of heating effect of current.
Answer:

Joule’s Laws (1841):

1.  First Law:

The amount of heat produced in a conductor in a given interval of time of proportion to the square of the current passed. Thus if H be the amount of heat generated in a conductor having resistance R when current 1 passes through it in time t, then

⇒ Hα I2 (When R and t are kept constant)

2. Second law:

The amount of heat produced by a given current in a given time is proportional of the resistance of the conductor.

⇒  H α R (When I and t are kept constant)

3. Third Law:

The amount of heat produced in a given conductor by a given current is proportional to the time for which the current passes.

⇒  Hα t (When I and R are kept constant)

Combining the three laws, we have:

⇒ H α I2 RT (When 1, R and t vary) or,

H = \(\frac{I^2 R T}{J}\)

IRT (J= mechanical equivalent of heat = 4.2 joule (calorie)

If I am in ampere, R in ohm, t in second, and H in calorie, then

H= \(\frac{I^2 R T}{J}\)

H= 0.241 RT calorie.

Question 2. Short Note: 

  1. Ohmic resistance
  2. Non-ohmic resistance

Answer:

1. Ohmic Resistance :

The resistors, which obey Ohm’s Law are said to have ohmic resistances.

Example: All metal or metallic alloy.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Ohmic And Resistance

2. Non-ohmic resistance :

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Non Ohmic Resistance

The resistors which do not obey Ohm’s law are said to have non-ohmic resistance.

Example: Electronic value.

Question 3. Two resistors of 30W and 60W are connected in parallel in an electric circuit. How does the current passing through the two resistors compare?
Answer:

The potential difference across 30W = B potential difference across 60W

i.e I1 R1= I2 R2 Or,

⇒ \(\frac{I_1}{I_2}=\frac{R_2}{R_1}\)

= \(\frac{60 \Omega}{30 \Omega}\)

= 2

Question 4. Short Note-Direct Current.
Answer:

Direct Current:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Direct Current

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Direct Current And Time

If a resistor connects with the two terminals of an electric cell, then through the resistor a steady current of constant magnitude flows in the same direction (from the positive pole of the cell to the negative pole) and this is called a direct current.

Class 10 Physical Science Solution WBBSE

Question 5. A resistance of 6 Ω is connected with a cell of em. f. 1.5 and negligible internal resistance calculate the current flowing through it.
Answer:

V= IR Or, I= \(\frac{V}{R}\)

= \(\frac{1.5}{6}\)

= \(\frac{1}{4}\)

= 0.25 Amp

Question 6. Find the specific resistance of the material of a wire of length 100cm, area of cross-section 0.2 cm,2, and resistance 2 ohms.

Answer:

R = P× \(\frac{1}{\mathrm{~A}}\)

∴ P= \(\frac{\mathrm{RA}}{\mathrm{1}}\)

= \(\frac{20 \times 0.2}{100}\)

= 40 ×10-4  Ohm-cm

So,

R = 20hm

1 = 100 cm

A = 0.2 cm2

P= ?

Question 7. Find effective resistance of the resistors 2 ohm, 4 ohm, 5 ohm connected in

  1. Series
  2. Parallel.

Answer:

We know, for series combination equivalent resistance

R= r1+ r2+r3

= 2+4+5

= 11 ohm.

We also know, for parallel combination equivalent resistance.

⇒ \(\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\)

= \(\frac{1}{2}+\frac{1}{4}+\frac{1}{5}\)

= \(\frac{19}{20}\)

∴ \(\frac{1}{R}\)= \(\frac{19}{20}\) Or,

= \(\frac{20}{19}\)

R = 1.05 Ohm

Question 8. Short Note-Electric Circuit.
Answer:

Electric Circuit: A continuous conducting path between the terminals of a source of electricity, is called an electric circuit.

Open electric → An electric circuit in which the flow of current stops, because of an open switch is called an open electric circuit. Closed electric

Open Electric Circuit:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Open Electric Circuit

Closed Electric Circuit:

Circuit → An electric circuit in which a current flows continuously, because the switch is closed is called a closed electric circuit.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Close Electric Circuit

Question 9. There are two copper wires of equal length. The radius of one is twice the other. Find the ratio of their resistances.

Answer:

We know, R= p \(\frac{1}{A}\)

= \(p \frac{1}{\lambda r^2}\)

(r= radius of the wire)

Since, the length and material are the same.

⇒ \(\mathrm{R} \alpha \frac{1}{\mathrm{r}^2}\)

i.e. \(\frac{R_1}{R_2}=\frac{r_2^2}{r_1^2}=\frac{\left(2 r_1\right)^2}{r_1^2}\)

= 4

∴ The thinner wire has a resistance four times the resistance of the thicker wire.

Question 10. Establish the relation between the emf, terminal voltage and internal resistance.
Answer:

Let, a cell of emf E and internal resistance r is connected to an external resistance R

The total resistance of the circuit = R+r

So, the current drawn from the cells

I= \(\frac{e . m . f \text { of the cell }}{\text { total resistance }}\)

I= \(\frac{E}{R+r}\)

E= I (R+r)

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Terminal Voltage

The terminal voltage of the cell, V = IR voltage (v) drops due to internal resistance -I As the work is done carrying a unit positive charge once through a complete circuit,

E = V + v

v = E-V

Internal resistance r = \(\frac{V}{I}\)

=  \(\frac{E-V}{I}\)

= \(\frac{E-V}{V / R}\)

= \(R \cdot\left[\frac{E}{V}-1\right]\)

Question 11. A current of 0.5 ampere passes through a wire of resistance 2.5 ohm. for 1 hour. Find the heat produced.
Answer: We know,

H  = \(\frac{I^2 R T}{J}\)

= \(\frac{I^2 R T}{4^2}\)

∴ H \(\frac{(0.5)^2 \times 2.5 \times 3600}{4^2}\)

= 535.7 caloric

So,

I = 0.5 ampere

R = 2.5 ohm

t = 1hr.

= 3600 sec.

H =?

Question 12. The resistance of a wire of cross-section area 0.01 cm2 is 10 ohm. What is the length of the wire? The specific resistance of the wire is 50 × 10-6 ohm-cm.
Answer:

We know,

R = p = \(\frac{\ell}{\mathrm{A}}\) Or,

l= \(\frac{\mathrm{RA}}{p}\) Or,

l= \(\frac{10 \times 0.01}{50 \times 10^{-6}}\)

∴ l= 2000 cm

So,

R= 10 Ohm

A= 0.01 cm2

p= 50 × 10-6

Ohmcm

1= ?

Question 13. What is the difference between e.m.f and p.d?
Answer:

Difference between e.m.f. and p.d. : e.m.f (electromotive force):

WBBSESolutions For Class 10 Physical Science And Environment Chapter 6 Current Eletricity Difference Between Electro Motive And Polential

Question 14. A resistance of 6Ω is connected with a cell of e.m.f., 1.5 V, and negligible internal resistance, Calculate the current flowing through it.

Answer: V = IR

I = \(\frac{V}{R}\)

I = \(\frac{1.5}{6}\)

I = \(\frac{1}{4}\)

= 0.25 Amp.

Question 15. A cell of e.m.f. 1.8v is connected to an external resistance of 22 when p.d. recorded at its terminal is 1.6v. Find the internal resistance of cell.

Answer: E = 1.8V, V 1.6 volt, R = 2.2

Internal resistance = \(\frac{R(E-V)}{V}\)

= \( \frac{2(1.8-1.6)}{1.6}\)

= 0.25.Ω

WBBSE Solutions For Class 10 Physical Science And Environment Chapter 5 Light

WBBSE Class 10 Physical Science Question Answer In English

Chapter 5 Light MCQs

Question 1. The formation of the spectrum is due to:

  1. Reflection
  2. Refraction
  3. Dispersion of light

Answer: 3. Dispersion of light.

Question 2. A person uses the spectacle of power + 2D. He is suffering from:

  1. Presbyopia
  2. Astigmatism
  3. Long-sightedness of hypermetropia
  4. Short-sightedness of myopia

Answer: 3. Long-sightedness of hypermetropia.

Question 3. A person cannot see objects clearly beyond 2m. The power of the lens required to correct his vision will be :

  1. + 2D
  2. -1D
  3. + ID
  4. 0.5D

Answer: 4. 0.5D.

Question 4. Myopia is due to:

  1. Older age
  2. Shorting of eyeball
  3. Irregular change in focal length
  4. Elongation of the eyeball.

Answer: 4. elongation of the eyeball.

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

Question 5. Rainbow is formed due to a combination of 

  1. Refraction and absorption
  2. Refraction and scattering
  3. Dispersion and total internal reflection
  4. Dispersion and focussing

Answer: 3. Dispersion and total internal reflection.

Question 6. Which of the prism is used to see the infrared spectrum of light?

  1. Rock salt
  2. Flint
  3. Nicol
  4. Crown

Answer: 1. Rock salt.

WBBSE Class 10 Physical Science Question Answer In English

Question 7. Which of the following colours suffers maximum deviation in a prism?

  1. Yellow
  2. Green
  3. Blue
  4. Orange

Answer: 2. Blue.

Question 8. The splitting of white light into several colours on passing through a glass prism is due to:

  1. Reflection
  2. Diffraction
  3. Interference
  4. Refraction

Answer: 4. Refraction.

Question 9. A convex lens will become less convergent in

  1. Oil
  2. Water
  3. Both
  4. None

Answer: 2. water.

Question 10. The source of ultraviolet light is :

  1. Electric bulb
  2. Carbon arc lamp
  3. Red hot iron bulb
  4. Sodium vapour lamp

Answer: 2. carbon arc lamp.

Question 11. Which of the following materials cannot be used to make a lens?

  1. Water
  2. Plastic
  3. Clay
  4. Glass

Answer: 2. Clay.

WBBSE Class 10 Physical Science Question Answer In English

Question 12. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be :

  1. Plane only
  2. Concave only
  3. Convex only
  4. Either plane or convex

Answer: 4. Either plane or convex.

Question 13. A ray of light suffers refraction through an equilateral prism. The deviation produced by the prism does not depend on the:

  1. Angle of incidence
  2. Size of the prism
  3. colour of light
  4. material of the prism

Answer: 3. size of the prism.

Question 14. A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from:

  1. its second focus
  2. its optical centre
  3. its first focus
  4. the centre of curvature of its second surface

Answer: 1. its second focus

Question 15. What is the value of W1/f1+ W2/f2

  1. 2
  2. 3
  3. 5
  4. 0

Answer: 4. 0

Question 16. The terminal colours of the pure spectrum of white light are :

  1. Blue and violet
  2. Red and orange
  3. Red and violet

Answer: 3. Red and violet.

Question 17. Yellow colour is called :

  1. Primary
  2. Middle colour
  3. Secondary colour
  4. Top colour

Answer: 2. Secondary colour

WBBSE Class 10 Physical Science Question Answer In English

Question 18. The___________ colours can be seen distinctly in pure spectrum.

  1. Six
  2. Eight
  3. Seven
  4. Nine

Answer: 3. Seven.

Question 19. The dispersion of white light was first observed by:

  1. Sir Issac Newton
  2. Boyle
  3. Max Well

Answer: 1. Sir Issac Newton.

Question 20. The arrangement of the seven colours of spectrum is:

  1. VIBGYOR
  2. RAINBOW
  3. VBIGOYR

Answer: 1. VIBGYOR.

Question 21. The object is placed in between ‘f’ and ‘2f’ in convex lens, the size of image will be

  1. Greater in size the object
  2. Equal in size of the object
  3. Greater in size the object.
  4. Smaller in size than the object

Answer: 1. Greater in size the object

Question 22. To for m virtual image on the same side of the object in case of a convex lens, the object should be placed :

  1. In between the lens and the focus
  2. In between ‘f’ and ‘2f’
  3. At ‘2f’ from the lens
  4. At a distance greater than ‘2f’ from the lens

Answer: 1. In between the lens and the focus.

Question 23. Linear magnification is equal to :

  1. (Length of image)/(Length of the object).
  2. (Length of image) × (Length of the object).
  3. (Length of image) + (Length of the object).
  4. (Length of image) – (Length of the object).

Answer: 2. (Length of image)×(Length of the object)

WBBSE Class 10 Physical Science Question Answer In English

Question 24. A mirror produces a magnified erect image of an object. The nature of the mirror is

  1. Convex
  2. Concave
  3. None of these
  4. Plane

Answer: 2. Concave.

Question 25. A plane mirror produces a magnification of:

  1. -1
  2. + 1
  3. Zero
  4. Between 0 and + a

Answer: 2. + 1.

Question 26. Focal length of a convex lens will be maximum for:

  1. Blue light
  2. Red light
  3. Greenlight
  4. Yellow light

Answer: 3. Red light.

Question 27. The number of images observable between two parallel plane mirror is:

  1. 2
  2. 4
  3. 11
  4. Infinite

Answer: 4. Infinite.

Class 10 Physical Science WBBSE

Question 28. When a mirror is rotated through an angle 0, the reflected ray from it turns through an angle of :

  1. θ
  2. θ/2
  3. 0

Answer: 4. 2θ

Question 29. A ray is incident at an angle 38° with a mirror. The angle between normal and deflected ray is :

  1. 90°
  2. 75
  3. 38°
  4. 52°

Answer: 4 52°

Question 30. The focal length f of a speherical mirror of radius of curvature R is :

  1. \(\frac{R}{2}\)
  2. R
  3. \(\frac{3}{2}\) R
  4. 2R

Answer: \(\frac{R}{2}\)

Question 31. A concave mirror of focal length f (in air) is immersed in water (μ = 3). The focal length of the mirror in water will be :

  1. \(\frac{4}{3}\) f
  2. \(\frac{9}{5}\) f
  3. \(\frac{3}{5}\) f
  4.  f

Answer: 4. f

Class 10 Physical Science WBBSE

Question 32. A plane mirror is approaching a person at a speed of 5 cm S-1. At what speed will his image approach him?

  1. 10 cms-1
  2. 20 cm S-1
  3. 5cms-1
  4. 15cm S-1

Answer: 1. 10 cm S-1

Question 33. If an object is 30 cm away from a concave mirror of focal length 15 cm, the image will be :

  1. Erect
  2. Reflection of light
  3. Diminished
  4. Of same size

Answer: 4. Of the same size

Question 34. Mirage is a phenomenon due to:

  1. Reflection of light
  2. Refraction of light
  3. Diffraction of light
  4. Total internal reflection of light

Answer: 4. Total internal reflection of light.

Question 35. The principle behind optical fibres 

  1. Total internal reflection
  2. Total external reflection
  3. Diffraction
  4. Both (1) and (2)

Answer: 1. Total internal deflection

Question 36. A convex lens is dipped in a liquid whose refractive index is euqal to the refractive index of the lens. Then its local length will:

  1. Become zero
  2. Become infinite
  3. Increase
  4. Reduce

Answer: 2. Become infinite.

Class 10 Physical Science WBBSE

Question 37. What is the value of the refractive index of a diamond?

  1. 2.47
  2. 3.47
  3. 1.47
  4. 0.47

Answer: 1/. 2.47.

Question 38. Write an example of an isotropic medium.

  1. Air
  2. Water
  3. Iron
  4. Fire

Answer: 1. Air.

Question 39. Write the value of speed of light.

  1. 3 x 106 m/s
  2. 3 x 108 m/s
  3. 3 x 107 m/s
  4. 3 x 109 m/s

Answer: 3. 3 x 108 m/s.

Question 40. What is the value of refractive index of alcohol?

  1. 1.37
  2. 2.37
  3. 4.37
  4. 0.37

Answer: 1. 1.37.

WB Class 10 Physical Science Question Answer Chapter 5 Light Very Short Answer Type Questions

Question 1. What is the far point of a normal eye?
Answer: Infinity.

Question 2. Which colour of light travels faster in vacuum?
Answer: Light rays of all colour travel with same speed in vacuum.

Question 3. What is the least distance of distinct vision?
Answer: 25 cm.

Question 4. How is deviation produced in a prism related to the angle of the prism and angles of incident and emergency?
Answer: δ= i1+i2 – A

Question 5. Is dispersion of white light in a vacuum possible?
Answer: No.

Question 6. Define the power of a lens.
Answer: The reciprocal of the focal length of the lens is called its power.

Question 7. What will be the focal length and power of a plane glass plate?
Answer:  f = α; Power = 0

Question 8. Why is a night-angled prism a better reflector than a plane mirror?
Answer: As in the prism no part of the light refracted.

Question 9. Does a beam of light give a spectrum on passing through a hollow prism containing air, explain?
Answer: No, because all colours of light pass through vacuum or air same speed and there is no dispersion.

Question 10. What kind of lens one should use to correct myopic eye?
Answer: Concave lens.

Question 11. What kind of lens one should use to correct the astigmatic eye?
Answer: Cylindrical lens.

Question 12. What is monochromatic light?
Answer: The light which consists of only one colour is known as monochromatic light.

Question 13. What is a double convex lens?
Answer: If both the surfaces of a convex lens are convex then the lens is called a double convex lens. It is used in camera, telescopes etc.

Question 14. Will the focal length of a convex lens vary for refraction of light of different colours through it?
Answer: For different colours of light the convex lens will have different focal length.

Question 15. What is the edge of a prism?
Answer: The intersection of the refracting surfaces is called the edge of the prism.

Class 10 Physical Science Solution WBBSE

Question 16. An object is placed at the focus of a concave lens, where will be its image formed?
Answer:

u = f; f = –  ve

So, \(\frac{1}{v}\)– \(\frac{1}{u}\) = \(\frac{1}{f}\) Or,

\(\frac{1}{v}\)=  \(\frac{1}{f}\)+ \(\frac{1}{u}\) Or,

\(\frac{1}{v}\)=  \(\frac{1}{f}\)+ \(\frac{1}{f}\)

= 0

∴ v= α

Question 17. Name the type of mirror which has focal length equal to infinity.
Answer: A plane mirror.

Question 18. What are the types of reflection of light?
Answer:

The types of reflection of light are:

  1. Regular reflection
  2. Irregular reflection

Question 19. Write the expression of Newton’s Equation.
Answer: The expression of Newton’s Equation xy = f2.

Question 20. What is the mathematical expression of linear magnification?
Answer: The mathematical expression of linear magnification is m=\(\frac{v}{u}\)

Question 21. Write two classifications of image.
Answer: The two classifications of image are real and virtual.

Question 22. Write an example of opaque substance.
Answer: An example of opaque substance is iron.

Question 23. What will be the nature of the lens for u-v graph.
Answer: The nature of the graph is rectangular hyperbola.

Class 10 Physical Science Solution WBBSE

Question 24. Write the expression of dispersive power of lens materil.
Answer:

W = \(\frac{δμ}{μ- 1}\)

Question 25. Write the mathematical expression of generalised snell’s law.
Answer: The mathematical expression of generalised snell’s law :

⇒ μ2= sin i1

= μ sin i 2

Question 26. Write a characteristic of the image formed by a plane mirror.
Answer: The image is erect but laterally inverted.

Question 27. What is the value of refractive index of Benzene?
Answer: The value of refractive index of Benzene is 1.501.

Question 28. What are the two types of lens?
Answer: The two types of lens are convex and concave.

Question 29. Write the mathematical expression of lens maker’s formula.
Answer:

\(\frac{1}{f}\)= (μ – 1) = (1/r1 – 1/r2)

Question 30. Write the mathematical expression of the refraction index.
Answer:

⇒ \(\frac{sin r}{sin i}\) =  2μ1

WBBSE Solutions Guide Class 10 Chapter 5 Light Fill In The Blanks

Question 1. The light which consists of only one colour is called_____ light.
Answer: Monochromatic.

Question 2. Yellow colour is a _______colour.
Answer: Secondary.

Question 3. A transparent body absorbs all the colours of white light except the light of colour of its_________
Answer: Own.

Question 4. The elementary colours of white light are arranged in order of _______
Answer: Wavelength.

Question 5. _______First observed the dispersion of white light.
Answer: Sir Issac Newton.

Question 6. Lenses are generally of two types, concave and _______
Answer: Convex.

Question 7. The _________colours can be seen distinctly in pure spectrum.
Answer: Seven.

Question 8. A convex lens is called a _______lens.
Answer: Converging.

Question 9. The middle colour is _______
Answer: Yellow.

Question 10. Red flower appears _______green light.
Answer: Black.

Question 11. Yellow and blue colours are _________colours colours of each other. 
Answer: Complementary.

Question 12. A beam of rays of light parallel to the principal axis of a convex lens, will passe through the______focus after refraction.
Answer: Principal.

Question 13. Magenta colour is made by mixing red and ______colour.
Answer: Blue.

Question 14. During dispersion of light, light is splitting up into________ colours
Answer: Seven.

Question 15. A ________ lens is used as a magnifying glass.
Answer: Convex.

Question 16. The arrangement of seven colours of the spectrum is ________
Answer: VIBGYOR.

Question 17. A plane mirror produces a magnification of ________
Answer: +1.

Question 18. The focal length f of a spherical mirror of radius of curvature R is.
Answer:

⇒ \(\frac{R}{2}\)

Question 19. Mirage is a phenomenon due to ________.
Answer: Total internal reflection of light

Question 20. The band of different colours obtained due to dispersion of white light is called the________.
Answer: Spectrum

Question 21. The lens which is thinner at the centre and wider at the two edges is known as a ________ lens.
Answer: Convex

WB Class 10 Physical Science Question Answer

Question 22. If a ray be incident on a convex lens parallel to its principal axis, then the refracted ray through the lens passes through the ________  of the lens.
Answer: Focus

Question 23. The straight line that joins the centres of the spherical surfaces forming the lens is called the  _______ axis of the lens.
Answer: Principal

Question 24. The distance of the principal focus from the optical centre of a lens is called its principal _______ length.
Answer: Focal

Question 25. An object place perpendicular to the principal axis of a convex lens will have its image _______ to the principal axis.
Answer: Perpendicular

Question 26. The focal length of a convex lens will be maximum for _______
Answer: Red light

Question 27. The number of images observable between two parallel plane mirror is_______
Answer: Infinite

Question 28. The formation of the spectrum is due to _______
Answer: Dispersion

Question 29. The terminal colours of pure spectrum of white light are
Answer: Red and violet

Question 30. _________of the prism is used to see infrared spectrum of light.
Answer: Rocksalt.

WB Class 10 Physical Science Question Answer Chapter 5 Light Short Answer Type Questions

Question 1. What is a centre of curvature of a lens?
Answer:

Centre of curvature: The spherical surface of a lens is a part of sphere. The centre of this sphere is known as centre of curvature of the surface of lens.

Question 2. What is Focal plane?
Answer:

Focal plane:  The plane passing through the principal focus of a lens and perpendicular to the principal axis of the lens is known as the focal plane of the lens.

Question 3. What is Radius of curvature?
Answer:

The radius of curvature: The radius of curvature of a lens is the radius of the glass sphere from which the surfaces of the lens are cut.

Question 4. What is Focal length?
Answer:

Focal length: The distance between the optical centre and the focus is known as focal length.

Question 5. What do you mean by the principal axis of a convex lens? 
Answer:

Principal axis of a convex lens joining the centres of curvature of the two spherical surfaces of a convex lens is called its principal axis.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Principle Axis

Question 6. What is optical centre?
Answer:

Optical centre: If a ray of light strikes the surface of a lens in such a way that the emergent ray from the other surface is parallel to it then the corresponding refracted ray passes through a definite point on the principal axis. This point (A) is the optical centre of the lens.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Optical Centrs

Question 7. What does the term ‘thin’ signify when it is related to a convex lens? Do all rays of light suffer deviation while crossing through a thin convex lens?
Answer:

A thin convex lens is one, the thickness at the middle of which is extremely small compared to the radii of curvature of its surfaces. Rays directed towards the optical centre do not suffer deviation while crossing through a convex lens.

WB Class 10 Physical Science Question Answer

Question 8. What is a spectrum?
Answer:

Spectrum: The band of different colours obtained due to dispersion of white light is known as a spectrum.

Question 9. A concave mirror of a small aperture forms a sharper image why?
Answer:
It is because such a mirror is free from the defect due to spherical aberration.

Question 10. What are luminous and non-luminous object?
Answer:

Bodies which emit light of themselves are luminous sources. Bodies which cannot emit light of themselves are non-luminous sources.

Question 11. State the principle of reversibility of light path.
Answer:

If a ray starting from a point reaches a second point after suffering any number of reflections and refractions and then be reversed in direction, it will retrace its path and reach the first point provided that all other conditions under which reflections and refractions took place remain unaltered.

Question 12. Define real image.
Answer:  If reflected or refracted rays actually converge to a point, the point is called a real image.

Question 13. Define virtual image.
Answer:
If the reflected or refracted rays only appear to diverge from a point, the point is called a virtual image.

Question 14. Write the differences between real and virtual image.
Answer:

  • A real image is formed when a reflected or refracted ray actually converges to it, but in case of a virtual image, the reflected or refracted ray appear to diverge from it.
  • A real image can be seen directly with eye. It can also be cast on a screen. But a virtual image can only be seen with eye but it cannot be cast on a screen.

Question 15. Define image.
Answer:

It the rays diverging from a paint undergo changes in direction by reflection, refraction of by both, ultimately they either converge to a point or appear to diverge from a point.

Question 16. What is incident ray and reflected ray?
Answer:

  • The light ray striking a reflecting surface is called the incident ray.
  • The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.

Question 17. What is plane of incidence and reflection?
Answer:

The plane containing the reflected ray and normal, is called the plane of reflection. The plane containing the reflected ray and the normal, is called the plane of reflection.

Physics Class 10 WBBSE Question 18. Difine Lateral inversion.
Answer: The interchange of the left and right sides in the image of an object in a plane mirror is called the lateral inversion.

19. Write two uses of plane mirror.
Answer:

  • In a kaleidoscope, three plane mirrors inched with each other at 60° are used.
  • In the solar heating devices such as solar cooker, etc plane mirror is used to reflect the light rays incident from sun. The substance to be heated.

Question 20. What is the principle of simple periscope.
Answer: This instrument is used for looking over an obstacle which blocks the path of light from an object to the eyes of an observer.

Question 21. Define

  1. Centre of curvature
  2. Redius of curvature
  3. Principle axis

Answer:

  • The centre of curvature of a mirror is the centre of the sphere of which the mirror is a apart.
  • The radius of the sphere of which the spherical mirror is a part, is called the radius of curvature of the mirror.
  • It is the straight line joining the pole of the mirror to its centre of curvature.

Question 22. Define

  1. Pole
  2. Aperture.

Answer:

  • The geometric centre of the spherical surface of the mirror is called the pole of the mirror.
  • The part of the mirror which is exposed to the incident light is called the aperture of the mirror. Thus it is the surface of the mirror from which reflection occurs.

Physics Class 10 WBBSE Question 23. Define focus and focal length.
Answer:

  • Focus-The focus of a concave mirror is a point on the principle axis through which the light rays incident parallel to the principal axis pass after reflection from the mirror.
  • The focus of a convex mirror is a point on the principal axis at which the light rays incident parallel to the principal axis, appear to meet after reflection from the mirror.
  • Focal length-The distance of focus from the pole of the mirror is called the focal of the mirror.

Question 24. Write the differences between convex mirror and concave mirror.
Answer:

Concave mirror And Convex mirror:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Differences Of Convex And Concave

Question 25. Write two characteristics of the image formed by a plane mirror.
Answer:

  • The dista nces of the object and its image is measured from the mirror are equal.
  • The line joining the object and its image intersects the mirror perpendicularly.

Question 26. Define

  1. First principal focus.
  2. Second principal focus

Answer:

  • Rays either diverging from or directed through the first principal focus of a spherical surface, after refraction at the surface becoms parallel.
  • Rays moving parallel to the principal axis after refraction at a spherical surface either coverge to or appear to diverge from a point on the principal axis, known as second principal focus of the surface.

Question 27. What is Myopia?
Answer:

Myopia: The defect where for point is less than infinity is known as myopia. The defect is also called short Sightedness.

Question 28. How does the ray passes through the prism in the minimum deviation position?
Answer: In the minimum deviation position, the ray passes symmetrically through the prism i.e. the incident ray and emergent ray are equally inclined to the respective faces of prism.

Physics Class 10 WBBSE Question 29. On what factors does the angular dispersion depend?
Answer:

The angular dispersion depends on :

  1. Refraction angle of the prism
  2. Nature of the material of the prism
  3. Wavelength of the incident light

Question 30. Why does a convex lens of glass = 1.5 behave as a diverging lens when immersed in a liquid of “μ = 1.65?
Answer:

The refractive index of glass with respect to the liquid [/latex]\frac{1.5}{1.65}[/latex]) is less than 1.

Question 31. Why does a diamond sparkle with great brilliance?
Answer: Multiple total internal reflections of light occur within the diamond. Diamond cutter uses this fact.

Question 32. What type of mirror would you use as shaving mirror?
Answer: A concave mirror of larger focal length is used for shaving purpose. Because it produces a magnified erect image of the face when placed within the focus of the mirror.

Question 33. The refractive index of diamond is much greater than the ordinary glass. Is this fact of some use to a diamond-cutter?
Answer: A critical angle [sin-1(\(\frac{1}{μ}\))] reflection takes place easily. of a diamond is much smaller and total internal

Question 34. How would you determine whether a mirror is plane, concave or convex?
Answer:

A plane mirror produces an erect image of the same size as that of the object. A concave mirror produces an erect and magnified image of an object when placed within its focus. A convex mirror produces always an erect and diminished image of an object.

Physics Class 10 WBBSE Question 35. What is a Linear magnification of lens.
Answer: Linear magnification of lens: The ratio of the length of the image and that of the object is called the linear magnification.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Linear Magnification

Question 36. What is Focus of a convex lens?
Answer:

Focus of a convex lens: If a beam of parallel rays, travelling parallel to the principal axis of a convex lens are refracted by the lens, the rays become converg- ing and intersect each other at a particular point on the axis. The point is known as the focus of the convex lens.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Fecus Of Convex Lens

Class 10 Physical Science WBBSE Question 37. Which rays should be considered n drawing ray diagrams for images formed by a convex lens?
Answer:

The following points should be considered :

  • A beam of rays parallel to the principal axis after refraction through the lens pass through the principal focus.
  • A beam of rays passing through principal focus emerge parallel to the principal axis, after refraction through the lens.
  • A beam of rays through the optical centre pass out undeviated.

Question 38. What is a pure spectrum?
Answer:

Pure spectrum: The spectrum in which the constituent colours do not overlap on each other and are seperated distinctly into elementary colours is known as a pure spectrum.

39. What is an impure spectrum?
Answer:

Impure spectrum: The spectrum in which the constituent colours partially superpose on each other and are not seperated distinctly into elementary colours is known as an impure spectrum.

40. Is dispersion possible without refraction?
Answer:

Question Explanation: Dispersion without refraction is not possible. In an optical medium different colours travel with different speed and so deviate with different magnitudes. Thus, the constituent colours of a poly chromatic light deviate by different amounts i.e. they are dispersed after refraction.

41. Why blue is used after washing white shirts?
Answer:

Explanation Blue: Whitens yellow or orange colour. So, white shirts after washing may have some yellow or orange stain. To convert such stains to white, blue is used.

Class 10 Physical Science WBBSE Question 42. What is the optical plane?
Answer:

Optical plane: It is an imaginary vertical plane that cuts the principal axis perpendicularly and passes through the optical centre of a lens.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Optical Plane

Question 43. Draw a neat diagram of the formation of image of an object at infinity by a convex lens.
Answer:

Formation of the image of an object at infinity by a convex lens :

  • Distance of object: At infinity.
  • The image formed: On the focal plane.
  • Nature of image: Real and inverted.
  • Size of image: Very much smaller than the object.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Infinity By A Convex Lens

Question 44. The critical angle for a material c is 30°. Find its refractive index.
Answer:

⇒ μ=1/ sinc=1/sin 30% = 2.0

Question 45. What is the focal length of a concave mirror of radius of curvature of 16.0 cm?
Answer:

The radius of curvature = 16cm. Focal length.

= ½ × Radius of curvature

= (\(\frac{1}{2}\) x 6) cm

= 8 cm.

Class 10 Physical Science WBBSE

Question 46. A lens has a power of -2.5D. What is the focal length and the nature of the lens?
Answer: P = – 2.5 D

Focal length =\(\frac{1}{P}\)

= \(\frac{1}{-2.5}\)

= -0.4m

= 40 sin.

Question 47. A lens has a focal length of -25cm. What is the power of the lens and what is the nature?
Answer:

Power = P = \(\frac{1}{f in m}\)

= \(\frac{1}{-0.25}\)

= – 4D.

Question 48. If the refractive index of glass in 1.5 and Zi=55, find the angle of refraction of glass.
Answer:

μ= \(\frac{sini}{sin r}\)

sin r = \(\frac{sin i}{μ}\)

=\(\frac{sin 55°}{1.5}\)

= 33.6°.

Physics Class 10 WBBSE Chapter 5 Light Broad Answer Type Questions

Question 1. Draw a labelled diagram to show refraction of a light ray through a prism.
Or

Path of a ray of light through a prism
Answer:

Path of a ray of light through a prism:

  • Path of a ray of light through a prism Let ABC be the principal section of a prism AB and AC are two refracting surfaces. BC is the base. PQRS is the path of a ray. It is to be noticed that in i a prism.
  • The refracted ray and the emer- gent ray bend towards the base of the prism, when the surrounding medium is optically rarer than the material of the prisms.
  • But, if the medium outside the prism be optically denser than the prism material, the refracted and emergent rays bend oppositely i.e they bend towards the angle of the prism.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light PAth Of A Ray Light Through A Prism

Question 2. What are the conditions of formation of pure spectrum?
Or

Production of pure spectrum
Answer:

Conditions of formatin of pure spectrum:

  • Slit should be very narrow to allow, rays of light to pass as minimum as possible.
  • The slit should be placed at the focus so that the emergent light rays may be Parallel.
  • Prism should be placed at the minimum deviation position of spec- trum i.e. at the mean deviation of yellow spectrum.
  • The second lens (L2 L2) should be placed in such a position so that the screen may be in the focal plane of the lens.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Production Of Pure Spectrum

Question 3. How will you explain twinkling of stars?
Answer:

Twinkling of stars:

The light coming from stars is refracted continuously by different layers of the atmosphere before coming to us. Due to there repeated refractions, The apparent position of a star is different from its actual position. The temperature and density of air in the atmosphere continuously changing and so apparent position of the star also changes continuously. Due to this continuous change in the apparent position of a star, it appears to be twinkling.

Question 4. A ray of light shows no dispersion on emerging from a glass slab. Explain why?
Answer:

A rectangular glass slab cut diagonally into two pieces behaves like a pair of prisms placed side by side in reverse order. So the light dispersed by the first prism, recombines by the second prism. Thus there is no dispersion of light in a rectangular glass slab, bulb it is only laterally displaced.

Question 5. State the laws of reflection.
Answer:

The laws of reflection:

  • The incident ray, the refracted ray and the normal to the refracting Surface at the point of incidence lie in the same plane.
  • 2. For a given pair of media and for a given colour of light, sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction.
  • Thus if i and r be the angles of incidence and of refraction respectively when the ray passes from medium 1 to medium 2.

⇒\(\frac{sini}{sin r}\) = constant

= 1μ2

Physics Class 10 WBBSE Question 6. Deseribe about the action of periscope.
Answer:

The mirror M, forms a virtual image I1, of the object O and this image 1, acts as virtual object-for the mirror M2, which forms an image I2, which is seen by the eye the final image F2, is also virtual. Since light rays from the object are reflected by two mirrors, the lateral inversion caused by the first mirror is reversed by the second mirror. This males the final image to appear with on lateral invession.

Question 7. State the laws of refraction.
Or
What is snell’s law?
Answer: 

  1. Laws of refraction: The incident ray, the reflected ray and the normal to the refracting. Surface at the paint of incidence lie in the same plane.
  2. For a given pair of media and for a given colour of light, the sine of the angle of incidence bears a constant ratio in the sine of the angle of refraction.

Thus, it i and r be the angles of incidence and of refraction respectively when the ray passes from medium 1 to medium 2.

\(\frac{sini}{sinr}\)= Constant

= 1μ2

The second law of refraction was devolped by W, Snell and is know a Snell’s law.

Question 8. Establish the relation between 1μ2 and 2μ1.
Answer:

From Snell’s law:  sin1/sinr = 2μ1

Where 2 u1, is the R. I. of medium 1 w.r.t. medium combining equations we get

2×2μ1= \(\frac{sin1}{sinr}\)×\(\frac{sinr}{sin1}\) = 1

Or

2X = 1/2μ1

Thus R.I. of medium 2 w.r.t. medium 1 is the reciprocal of R.I of medium 1 w.r.t. medum 2

Physics Class 10 WBBSE Question 9. State generalised Snell’s law.
Answer: If i1 and i2 be the angle of incidence and of refraction respectively then we have from Snell’s law

sini1/sini2 =1, μ2

We can write

1, μ2 21

⇒μ21

∴ sini1/sini2= μ21

Or,μ1 sini1 = μ2 sini2 → Generalised Snell’s Law

Question 10. How far from a concave mirror of radius 2m would you place the object to get an image magnified 3 times?
Answer:

Given m = ± 3 \(\frac{v}{u}\) i.e.u= 7.3u

For real image v = -3u, r = -2m and u is negetive.

So, from relation \(\frac{1}{v}\)+ \(\frac{1}{u}\)= \(\frac{2}{r}\)

Or, – \(\frac{1}{3u}\) – \(\frac{1}{u}\) = – \(\frac{2}{2}\)

For virtual image u =+ 3u, r = -2m and u is negative

So =-1 or, u= = 0.67m

= \(\frac{1}{3u}\) – \(\frac{1}{u}\) = – \(\frac{2}{2}\)  Or \(\frac{1-3}{3u}\) = -1

u= \(\frac{2}{3}\)

= 0.67 m

WBBSE Class 10 Physical Science Question Answer In English Question 11. A concave mirror forms an image of 20 cm high object on a screen placed 5m away from the mirror. The height of the image is 50 cm. Find the focal lenght of the mirror and the distance between mirror and the object.
Answer:
Given size of object = 20cm.

Size of the image = 50cm.

u = -5m.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light concave Mirror

= \(\frac{-10}{7}\)

= −1.43m

And distance between the mirror and the object = 2m.

Question 12. Light falls from glass of refractive index 1.5 to air. Find the angle of incidence for which the angle of deviation is 90°.
Answer:

Refractive index of glass is 1.5. So, the critical angle of glass to air medium is given by

⇒ − q= sin-1(\(\frac{1}{u}\)) sin-1(\(\frac{1}{1.5}\))

= 41.8°

So, reflection in the glas medium takes only when the angle of incidence exceeds  41.8°. Now the angle of deviation. Would be 90° when the angle between the incident ray and the reflected ray becomes 90° i.e. the angle of incidence be equal to 45°.

Question 13. Where should an object be placed from a converging lens of focal length 20cm, so as to obtain a real image of magnification 2 ? u =2, or, v = + 2u, f= + 20cm
Answer:

Given m =\(\frac{u}{u}\)= 2 or  , v= +2u, f= +20cm

∴ \(\frac{1}{v}\)– \(\frac{1}{u}\) = \(\frac{1}{f}\) or,

=  \(\frac{1}{2u}\) – \(\frac{1}{u}\)

= \(\frac{1}{20}\) or,

= \(\frac{1-2}{2u}\)

∴ u= \(\frac{− 20×1}{2}\)

= – 10 cm

i.e. object to be placed 10 cm in fornt of the lens.

WBBSE Class 10 Physical Science Question Answer In English

Question 14. Two thin lens of focal lengths 15cm. and 30cm. respectively are kept in contact with each other. What is the power of the combined system?
=
Answer: Given f1= 15cm, f2 = 30cm.

Then focal length F of the equivalent lens

\(\frac{1}{F}\)= 1/f1+1/f2

\(\frac{1}{15}\) + \(\frac{1}{30}\)

= \(\frac{2+1}{30}\)

= \(\frac{1}{10}\)

Power of the combination = \(\frac{100}{F}\)

= \(\frac{100}{10}\)

= +10

Question 15. The dew drops deposited on hairy leaves appear glittering way?
Answer:

The hairs out the leaf and the surface tension of water create an air gap under the dew drop. Hence some of the rays trying to pass from the drop to the side air gap suffer total internal reflection. Such reflected rays may reach our eyes after refraction through the upper surface of the drop and the drop appears glittering.

16. Show that, for a spherical mirror the focal length is half of its radius of curvature.
Answer:

P = Pole

F= Focus of the mirror.

PF = f

C = Centre of curvature.

∠BP’C = ∠P’CF (alternate angles.) and

∠BP’C = ∠CP’F (law of reflections i = r)

Hence ∠’P’CF = ∠CP’F

ΔFP’ C is isosceles. Hence P’F = FC

If the aperture of the mirror is small, the point P i very close to the point P, then P’F = PF

∴ PF=  FC or, PF + PF = PF + FC

or , 2PF = PC or, PF =  ½ PC

or, f = ½ R.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Mirror Lenght

Question 17. Establish relation between the angle of deviation (Dm) and the refractive index of the material of the prism (μ).
Answer: Let, the angle of deviation be D, the refractive index of the material of the prism be μ.

For minimum deviation, we have ¡=i’

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Deviation Angle

Question 18.

  1. Write Cauchy’s equation 
  2. Define-angular dispersion.

Answer:

1. The Cauchy’s equation: = А+B/ λ2

A, B = Cauchy’s constant

2. The angular dispersion is defined as the difference between the angles of deviation of the extreme colours, red and violet of the spectrum.

WBBSE Class 10 Physical Science Question Answer In English

Question 19. 

  1. What is the condition for dispersion with deviation and deviation without dispersion?
  2. What is the pure and impure spectrum?

Answer:

1. Condition for dispersion with deviation

⇒(μv-1)A = (μ’y-1)A’

The condition for devotion without dispersion :

⇒ δyr y-1)A(w-w’)

2. 

  • Pure spectrum: A spectrum in which the colours present do not overlap one another so that different colours occupy distinct and separate positions is called pure spectrum.
  • Impure spectrum: A spectrum in which the colours present overlap one another and hence can not be seen distinctly separated from one another is called an impure spectrum.

Question 20. Light of two colours (say X, Y) are sent through a prism. Suppose X bents more than Y. Which colour travels more slowly in the material of the prism?
Answer:

We know, D, (μx-1)A

D1 = (μy-1)A

⇒ Dy>Dy

⇒ μ x> μy

C/ Vx>C/ Vy

∴ Vy > Vx

The colour X travels more slowly in the material of the prism.

Question 21. Draw a ray diagram for the formation of a concave mirror when the object is at focus F.
Answer:

AB is an object placed at the focus F. The ray AD is incident on the mirror parallel axis and it gets reflected along DF, passing through the focus f. AE appears to be coming from the centre of curvature C, so it gets reflected back along EA. The two reflected rays DF and EA are parallel to each other.

The image is formed at infinity (very far from the mirror) which is real, inverted and highly magnified.

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Concave Mirror Of Focus

When the object is at focus F.

Thus when the object is at the focus F, the image is at infinity. It is

  1.  Real
  2. Inverted, and
  3. Highly magnified.

Question 22. How the angle of minimum deviation of a glass prism changes when it is immersed in a liquid of R.I. greater than I? sin A+D m
Answer:

Minimum deviation of a glass prism:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Minimum Deviation Of Glass Prism

Hence the angle of minimum deviation decreases.

Question 23. Define-Absorption Spectrum.
Answer:

Absorption Spectrum:

WBBSESolutions For Class 10 Physical Science And Environment Chapter 5 Light Absorption Spectrum

The missing wavelengths provide information about the absorbing material. Such a spectrum is called an absorption spectrum. The absorption spectrum may be of two types depending on the absorbing material and the conditions such as temperature.

Question 24 . An object 4 cm. high is placed at a distance of 10cm from a convex lens of focal length 20cm. Find the position, nature and size of the image.
Answer:

O= 4cm, u=-10 cm, f= + 20cm.

\(\frac{1}{V}\) = \(\frac{1}{f}\) +\(\frac{1}{u}\)

= \(\frac{1}{20}\) – \(\frac{1}{10}\)

= – \(\frac{1}{20}\)

V= -20 cm

\(\frac{1}{o}\) – \(\frac{V}{u}\)

I = \(\frac{V}{u}\) × 4

= \(\frac{-20}{-10}\) ×4

= 8cm

Question 25. Two lenses one of focal length 20 cm. and another focal length -15cm, are placed in contact. What is the focal length and power of the combination?
Answer: f, 20cm=0.2m f15cm – 0.15m

Power of the first lens, P1 = \(\frac{1}{0.2}\) = 5D

Power of the second lens, P2 = \(\frac{-1}{0.2}\)

= 6.67 D

Power of combination, P = P1 + P2 = 5 – 6.67 = -1.67D

Focal length, f= \(\frac{1}{1.67}\)  0.599m

The -ve sign indicates that the combination acts as a concave lens.

Question 26. An object 4 cm. high is placed at a distance of 10cm from a convex lens of focal length 20cm. Find the position nature and size of the image.
Answer:

0=  4cm, u = -10cm, f= + 20cm,

\(\frac{1}{V}\) =\(\frac{1}{F}\)+ \(\frac{1}{u}\)

= \(\frac{1}{20}\) – \(\frac{1}{10}\)

= – \(\frac{1}{20}\)

V= – 20 cm

\(\frac{I}{O}\)= \(\frac{V}{u}\) or, \(\frac{V}{u}\)×0

= \(\frac{-20}{-10}\)×4

= 8cm