PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Punjab State Board PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Long Answer Type Questions:

Question 1.
How to meet the needs of ever-increasing population of our country? List two main steps. Give in a tabulated form some of the crops grown in India.
Answer:
The present population of 1.21 billion people will reach about 1.343 billion people by the year 2020. This population will need about 241 million tonnes of grain production per year. Following two measures will help up to meet the demand.
1. Increase food production of both plants and animals.
2. Sustainable agriculture where by we should minimize using chemicals as fertilizers and insecticides. These can be replaced by biological resources.
Some crop plants grown in India:

Type Some Examples
Cereals or grain crops

Fibre crops

Pulses

Oil seeds

Fodder crops

Root crops

Tuber crops

Sugar crops

Plantation crops

Products from animals

Rice, Wheat, Barley, Ragi, Maize, Jowar, Bajra.

Jute, Cotton, Hemp, Coir.

Grams, Peas, Beans, Masoor, Mung.

Mustard, Groundnut, Sunflower, Coconut, Taramira. Barseem, Oat, Sudan grass.

Sweet potato, Carrot, Radish, Beet.

Potato, Tapioca.

Sugarcane, Beetroot.

Coffee, Tea, Rubber, Coconut.

Fish, Egg, Milk, and Meat.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 2.
How is manure prepared?
Answer:
Method of preparing compost manure:
Manure is a natural organic substance obtained by decomposition of animal wastes and plant residue through the action of microbes. It is of three types
1. Farmyard manure
2. Compost
3. Green manure.

  1. A trench having the desirable size of 4-5 m long, 1.5 to 1.8m broad with a depth of 1.0 to 1.8 m is made.
  2. A layer of about 30 cm in thickness containing well mixed refuse is spread in the trench.
  3. This layer is sprayed with water containing slurry of cow dung.
  4. Another layer of mixed refuse in trench up to the height of 45-60 cm (height of fish layer included).
  5. Top of these two layers is covered by thin layer ot earth.
  6. After a gap of about three months, material is taken out of trench, moistened with water and covered with earth.
  7. Compost is ready for use after gap of 1-2 months.

Question 3.
Differentiate between manures and fertilizers.
Answer:
Differences between manures and fertilizers:

Manure Fertilizers
1. Manures are partially decayed wastes and animal residues by microbes.

2. Organic substances.

3. Voluminous, bulky, difficult to store and transport.

4. Not very rich in minerals like N, P and K.

5. Contain all nutrients, although in small amount.

6. Slow absorption, being less soluble in water.

7. Plenty of humus is added to soil and improves the texture of soil.

1. Fertilizer is a salt or organic compound containing essential plant nutrients.

2. Inorganic salts or organic compounds.

3. Compact, can be easily stored and transported.

4. Rich m minerals like N, P, K.

5. Specific. Every fertilizer contains one or more nutrients.

6. Rapid absorption due to easy solubility in water.

7. Humus is not added to soil.

Question 4.
What is mixed cropping? Discuss the advantages of mixed cropping.
Answer:
Mixed cropping is growing of two or more crops simultaneously on the same piece of land.
Objectives of mixed cropping:

  1. To minimise risk and insure against crop failure.
  2. To reduce cultivation expenses.
  3. To provide balanced nutrition to farmer and his family.

Advantages:

  1. It acts as insurance against possible total crop failure in poor rainfall areas.
  2. It saves time and labour of the farmer.
  3. It provides different types of food materials.
  4. Thus, farmer and his family can get balanced nutrition.

Some of the prominent mixed cropping practices:

  1. Maize + urdbean
  2. Cotton + mungbean
  3. Groundnut + sunflower
  4. Wheat + chickpea
  5. Wheat + mustard

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 5.
Define livestock. Classify the cattle on the basis of their utility and give one example of each.
Answer:
Livestock includes domesticated animals like cows, buffaloes, sheep, goats, pigs, horses, etc. Cattle and buffaloes are most important livestock animals. These are used in agricultural operations and transportation though also provide milk, meat, hides (for leather goods), dung manure and fuel (in biogas plants). There are 30 breeds of cows and 10 breeds of buffaloes in India.

  1. Milch breeds: Milk-yielding varieties of cows but their males are not useful as working animals e.g. Gir, Sahiwal, etc.
  2. Draught breeds: Their males are used as beast of burden and good work animals but their cows are poor milk-yielding e.g. Malvi, Hallikar, etc.
  3. Dual-purpose breeds: Their cows are good milk-yielding while their males are good Work animals and help in agricultural operations e.g. Haryana, Tharparkar, etc.

Indian breeds of cows and buffaloes

1. Cows:

(a) Milch breeds:

  1. Gir
  2. Sahiwal
  3. Red Sindhi

(b) Draught breeds:

  1. Malvi
  2. Nageri
  3. Hallikar
  4. Kangayam

2. Buffaloes:

  1. Murrah
  2. Mehsana
  3. Surti
  4. Nili Ravi

Buffalo milk is richer in fat, tocopherol, proteins, calcium, phosphorus and contains low sodium, potassium, cholesterol. Buffalo milk is ideal for making milk products like khoa, rabri, dahi and ghee.

Question 6.
What is the need of proper shelter to cattle? List the characters of a good animal shelter.
Answer:
Shelter to Cattle: A good animal shelter not only increases the milk-production but also improves the health of animals.
A good animal shelter has following characteristics:
(a) It should provide protection to the animals from unfavourable environmental factors and predators.
(b) It should be clear, dry, airy, spacious and well ventilated (proper sunlight).
(c) It should have arrangement for the hygienic disposal of animal excreta.
(d) It should have arrangement for clear drinking water for animals.
(e) It should have hygienic conditions to protect the animals from various diseases.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 7.
What are additive feeds for cattle? How can one detect that an animal is sick?
Answer:
The cattle feed consists of two main components i.e. roughage and concentrate. However high yielding cows require more feed. Dairy animals require minerals and antibiotics and they are commonly called additive. Additive feed performs following functions:
1. Protects the animals from diseases.
2. They enhance milk yield.
3. Promote growth of the animals.

Symptoms of detection of sick animals:

  1. Laziness, tired and prefer to stay alone.
  2. Stop feeding or take a very little feed.
  3. Walks very slowly.
  4. Fall in milk or egg production.
  5. Waste passed out is dilute.
  6. Rise in temperature with shivering and sneezing.
  7. Secretion of excess of saliva.

Question 8.
Write a short note on variety improvement of poultry farming.
Answer:
Variety Improvement of Poultry Farming. It involves cross-breeding of indigenous varieties with exotic breeds. The improved varieties are developed for the following desirable traits:

  1. Quality and quantity of chicks.
  2. Dwarf broiler parent for commercial chick production.
  3. Summer adaptation capacity/tolerance to high temperature.
  4. Low maintenance requirements.
  5. Reduction in the size of the egg-laying bird with ability to utilise more fibrous cheaper diets formulated using agricultural by-products.

Question 9.
What steps should be taken to improve production of food from animal sources in our country?
Answer:

  1. Introduction of high-yielding varieties.
  2. The output of research work carried out at various centres such as NDRI Kamal, Central Institute of Fresh Water Aquaculture (CIFA) Bhubaneshwar, should be made available to the public for their projects.
  3. Protection of animals from diseases.
  4. Providing proper shelter to animals.

Question 10.
What are the practices adopted to improve crop production?
Answer:
The practices adopted to improve crop production are as follows:

  1. Fertilizers: These are the chemical compounds which are added to the soil to increase the fertility. They make up for the deficiency of the required nutrients and help in increasing the crop production.
  2. Selective Breeding: Disease-resistant seeds are produced by selective breeding. Regular use of high yield variety results in better crop production.
  3. Weed Control: The unwanted plants or weeds are controlled by using certain chemicals called weedicides.
  4. Control of Plant Diseases: Crops should be protected from insects, fungi, animals and other diseases.

It is very useful for increasing crop production. Insects are very harmful to crops. So insecticides should be used to kill insects.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 11.
Discuss the preventive and control measures to check losses of grain during storage.
Answer:
Preventive and control measures
These are used before storing and the grains are stored for future use.
1. This can be done by proper drying of the produce in sun followed by drying in shade.

2. Maintenance of Hygiene. Godowns and stores should be properly cleaned. All sort of dust, dirt, rubbish, webbing or refuse of the previous grain should be swept away. Cracks and holes in the wall, floor or ceiling should be sealed. If old gunny bags are being used, clean them properly, turn inside out and expose to sun or fumigate. Earthen pots should also be cleaned and properly exposed to sun before using them for grain storage.

3. Fumigation. Chemicals which can exist in gaseous state in sufficient concentration to be lethal against the pest is known as fumigants. Aluminium phosphide tablets commonly known as black poison (3 g each) can be used at the rate of 2 tablets per ton grains.

4. Plant Products. The practice of adding small quantity of vegetable oil or mineral oil to grains or legumes to protect them from insect pests and mixing of neem kernel (seed) powder, crushed dried fruit of black pepper or cloves is also effective in controlling insects.

Question 12.
Write a brief account of procedure of Inland Fishery.
Answer:
Inland fishery involves the rearing of fishes in the specially designed breeding ponds near the rivers or other fresh-water natural sources.

It involves the following steps :
Breeding of good male and female culturable fishes in the breeding ponds either by natural breeding or induced breeding. In induced breeding, the male and female breeder fishes are injected with pituitary extract containing FSH and LH hormones (called Hypophysiation) which induce them to spawn within 24 hours. Fertilization occurs in the water so is external.

  1. The fish seed (of fertilized eggs) are collected with the help of shooting net or benchi jal.
  2. The fertilized eggs are kept in hatching pits, called hapas, and young ones are called hatchlings.
  3. Hatchlings are allowed to grow in hapas for about 3-14 days to form fry.
  4. Fries are allowed to feed and grow in the nursery ponds to form the fingerlings.
  5. Fingerlings are allowed to grow in rearing ponds for about 3 months.
  6. Fingerlings are allowed to attain full size in the still larger stocking ponds.
  7. Harvesting or fishing involves the capturing of fully-grown fishes.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 13.
Briefly explain marine fisheries.
Answer:
Marine Fisheries: India’s marine fishery resources include 7500 km of coastline and the. deep seas beyond it. Popular marine fish varieties include pomphret, mackerel, tuna, sardines and Bombay duck. Marine fish are caught using many kinds of fishing nets from fishing boats.

The modern technologies for catching more fish include echosounders and use of satellite. Some marine fish of high economic value are also farmed in sea water. This includes finned fishes like mullets, bhetki and pearl spots, shellfish such as prawns, mussels and oysters.

Short Answer Type Questions:

Question 1.
How human beings depend upon plants and animals for nutrition?
Answer:

  1. Man is omnivorous and takes plants and animals as food.
  2. The plant sources of food are cereals (wheat, rice, maize), pulses, millets, fruits and vegetables.
  3. The animal sources of food are meat, milk, fish, egg, milk products and liver oil.

Question 2.
Define nutrients. Give examples of macronutrients and micronutrients.
Answer:

  • Nutrients: The elements needed for growth^of plants and animals are called nutrients. ^
  • Macronutrients: The mineral elements needed by the plants in large amounts (more than 1 ppm) are called macronutrients.
  • Examples: Nitrogen, Phosphorus, Carbon, Hydrogen, Oxygen, Potassium, Calcium, Sulphur.
  • Micronutrients: The mineral elements needed by plants in small amount (or traces) are called micronutrients.
  • Examples: Iron, Manganese, Copper, Zinc, Boron, Molybdenum, Chlorine.

Question 3.
What are Fertilizers?
Answer:
Fertilizers:

  1. Fertilizers are commercially produced plant nutrients by using different chemicals.
  2. Fertilizers supply Nitrogen, Phosphorus, Potassium (NPK), basically this is used for good vegetative growth (leaves, branches and flowers), giving rise to healthy plants.
  3. Fertilizers are one of the major components for obtaining higher yields specially in high cost farming practices.
  4. They are easy to store and transport.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 4.
What is the need of water to plants? Explain different types of irrigation system.
Answer:
Periodic Irrigation of Crops:
Crops are irrigated periodically due to the following reasons:
1. Plants absorb water from the soil by roots. The amount of water in the soil is not constant throughout the year. It is constantly lost by evaporation and percolation to lower depths of the ground.
2. The water is also lost by the aerial parts of the plants by transpiration.

Irrigation Systems:

1. Canal System: Here reservoirs or rivers supply water to canals. Canal is divided into sub-canals and distributaries. Further field channels may be made. Rotation system is usually followed to provide irrigation water to all the fields at the time of scarcity of water supply.

2. Tanks: Here water is stored, which is available due to run off. Small dams can be made at the base of higher elevation of catchment region. Outflow of water in tanks is kept in control. If it is not done, it may lead to:
(a) Uneven distribution of water.
(b) Shortage of water at tail end and excessive use at the top. This leads to uneven distribution of water.

3. Wells. Wells are made to exploit groundwater. They are of two types i.e. dug wells and tubewells. In dug well, water accumulates due to available groundwater table. From deeper areas of earth, tube wells can tap water. Here water is lifted by diesel or electricity-operated water pumps. Continuous supply can be ensured by this system.

4. River valley system: In Western Ghats of Karnataka and Kerala, many steep and narrow riverine valley are present. Rainfall in these areas is present only for shorter period i.e. 3-4 months. Extra water shifts in river in these months. In Rabi season, no rainfall is there. On slopes, plants like coconuts, arecanuts, coffee, rubber and tapioca are grown. These all are perennial plants. In basement areas, single crop like rice is grown.

5. River lift system: In this system, water is directly drawn from rivers for irrigation purpose. This all is done in areas nearby the rivers and canal flow is insufficient.

Question 5.
How do insect pests attack plants? Give examples.
Answer:
1. Cut root, stem and leaf – weevil attack wheat crops.
2. Suck cell sap from various parts of plants – aphids feed on mustard plants.
3. Bore into stem and fruits – top borer and shoot borer, larvae and caterpillars which bore into stem and fruits. The plant pathogens are transmitted to plants through water, soil, air and seeds.

Diseases caused by these pathogens include

  1. Blast in paddy (rice).
  2. Rust in wheat.
  3. Stem rot in pigeon pea (mung).
  4. Wilt in chickpea (gram).

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 6.
What are weeds? What are the two types of weeds?
Answer:
Weeds: They are the small-sized unwanted plants which grow along with a cultivated crop in a field. Weeds are economically very important as they can severely reduce crop yields by competing for light, water and nutrients.

Based on the morphology of plants, weeds are classified into two types:

  1. Narrow-leaf weeds, e.g., Wild sorghum, Wild oat.
  2. Broadleaf weeds, e.g., Amaranthus viridis, Trianthima.

Question 7.
Discuss the three major methods of weed control.
Answer:
Methods of weed control:

  1. Mechanical Methods: Uprooting, weeding with khurpi, hand hoeing, interculture, ploughing, burning and flooding.
  2. Culture Methods: Proper seedbed preparation, timely sowing of crops, intercropping and crop rotation.
  3. Chemical Methods: Spraying of some chemicals as herbicides or weedicides, is also done in case of heavy infestation.
  4. Biological Control: Use of insects or some organisms which consume and destroy the weed plants.
  5. Examples: Prickly-pear cactus (Opuntia) is controlled by cochineal insect and aquatic weeds are controlled by fish grass carp.

Question 8.
Briefly explain three types of manures on the basis of biological material used.
Answer:
Types of manure:

  1. Farm Yard Manure (FYM): Livestock farm waste i.e. cattle excreta (cow dung and urine) is stored in a pit for decomposition. After 1-2 months this is used as FYM in farming practices.
  2. Compost: The process in which waste material like vegetable waste, animals refuse, domestic waste, sewage waste, straw, eradicated weeds, etc. is decomposed in pits is known as composting. The compost is rich in organic matter and nutrients.
  3. Green Manure: In cultivation field prior to the sowing of the seeds, some crops like sun hemp, guar, etc. are grown. After some time these plants are mulched by ploughing. These green plants thus turn into green manures which help in enriching the soil by N and P.

Question 9.
What are weeds? How does it affect the yield of crop?
Answer:
Weeds. They are unwanted plants which grow on their own along with crop plants. They are a harm to the crops.
Weeds damage the crops: They compete with the crop for nutrients and water in the field. They occupy space meant for crop plants. This leads to poor yield and quality of produce.

Methods of weed control:

  1. Removal by hands: The weed can be uprooted and removed by hand.
  2. Removal by instruments: The weeds can be removed by using a trowel (khurpa).
  3. Removal by using chemicals: Weeds can also be destroyed by spraying special chemicals called weedicides.
  4. These chemicals easily kill the broad-leaved weeds without affecting the crop.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 10.
List any five chemical fertilizers rich in nitrogen.
Answer:
Chemical fertilizers rich in nitrogen:

  1. Urea
  2. Ammonium sulphate
  3. Ammonium nitrate
  4. Sodium nitrate
  5. Calcium ammonium nitrate

Out of these urea is organic while others are inorganic.

Question 11.
How has the excess use of pesticides and fertilizers proved harmful?
Answer:
If pesticides are not used with care, they lead to the disappearance of not only the undesirable insects but even the helpful ones. There is a danger that the pests may become resistant to pesticides and make them ineffective. Indiscriminate use of pesticides and fertilizers can lead to environmental degradation. Excess fertilizers is washed away into surrounding water bodies.

The resulting high concentration of nitrates and phosphates in ground and surface waters make them toxic and unfit for human and animal consumption.

Question 12.
What are the two main crop seasons? List the crops of respective season.
Answer:
1. Kharif Season. (June to October)
Crops grown: Paddy, Soyabean, Arhar, Maize, Cotton, Urad and Moong.
2. Rabi season. (November to April)
Crops grown: Wheat, Gram, Peas, Mustard, Linseed.

Question 13.
Write a note on organic farming.
Answer:
Organic Farming: It is a farming system with minimum or no use of chemicals as fertilisers, herbicides, pesticides, etc. and with maximum input of organic manures, recycled farm wastes, use of bio-agents such as culture of blue-green algae in preparation of biofertilisers, neem leaves or turmeric specifically in grain storage as bio-pesticides, with healthy cropping systems.

Question 14.
Define crop rotation. What is the role of leguminous plants in it?
Answer:
Crop Rotation: The process in which different types of crops are grown alternately in the same field is called crop rotation.

Leguminous crops save the nitrogenous fertilizers because such plants grown during crop rotation fix nitrogen from the air and enrich the soil with nitrate and nitrites. These nitrogen-containing compounds are used by plants.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 15.
What is the role of soil in better yield?
Answer:
Soil provides all the nutrients such as nitrogen, phosphorus, potassium etc. to the crop plants. Soil also serves as a source of water. If soil is deficient in one or more nutrients, the yield of crop will reduce.

Question 16.
What is mixed farming? Define it with suitable examples.
Answer:
Mixed farming can be defined as a system of farming on a particular farm to sustain and satisfy the essential needs of the farmers.
Examples: In mixed farming, crop production is combined with the rearing of livestock, poultry, fish and bees etc.

Question 17.
How can mixed farming sustain agricultural production? Answer With suitable examples.
Answer:
When earnings from one enterprise on small farms is not sufficient to sustain the family, farmer considers about the different possible combinations of enterprises.

In mixed farming from livestock, farmyard manure is made available to be used again in agricultural farms. With exact combination of mixed farming, a better money/ income is available. It provides the farmer work throughout the year. It provides the farmer with all the food needs of the family.

Question 18.
What are the uses of mixed farming?
Answer:

  1.  From livestock, farmyard manure is available to be used again in agricultural farms.
  2. Number of animals can be increased as per food available to them (as per crop availability) to provide milk and milk products.
  3. By this method, straw, husks and chaffs of grains, refuse of household kitchen, shed grains in the field are converted into human food through the agency of cattle, sheep, poultry, pigs etc. as per choice of farmer.
  4. With exact combination in mixed farming, a better income is available.
  5. It provides work to all the members of a family throughout the year. Thus it provides subsidiary occupation to all the members of a household, without the need of employing special labour.

Question 19.
Write a note on crop protection management.
Answer:
Crop Protection Management: In fields, crops have to be protected from weeds, insect-pests and disease-causing organisms like fungi.

All these cause damage to crop plants so much so that most of the crop is lost. Thus, crops can be protected by the following methods:

  1. Use of pesticides.
  2. Use of resistant varieties.
  3. Crop rotation and cropping system.
  4. Summer ploughing.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 20.
Discuss the requirements for a good crop rotation.
Answer:
Requirements for a good Crop Rotation

  1. The area of each crop should be nearly the same year after year.
  2. The rotation should provide roughage and pastures for domestic animals.
  3. Most profitable cash crops with wisdom should be selected for rotation.
  4. Rotated crops can be cared for.
  5. The rotation should include one tilled crop for elimination of weeds.
  6. Organic matter should increase in soil due to rotation and feeding system. By growth of leguminous plant in rotation, nitrogen contents of soil increase.

Question 21.
List group of plants for crop rotation on the basis of one year, two year and three-year rotation.
Answer:

Duration Rotation of Crops
1. One-year rotation

2. Two-year rotation

3. Three-year rotation

1. Maize-mustard
2. Rice-wheat1. Maize-mustard-sugarcane – fenugreek (Methi)
2. Maize-potato-sugarcane-peas1. Rice-wheat-moong-mustard – sugarcane-berseem
2. Cotton-sugarcane-peas-maize-wheat

Question 22.
List five groups of plants according to their soil needs for crop rotation.
Answer:
For crop rotation, plants are put into five groups according to their soil:

Roots Legumes Brassica Others Permanent
Potatoes

Carrots

Beet root

Peas

Runner beans
Broad beans

French beans

Cabbages

Cauliflowers

Broccoli

Turnips

Radishes

Lettuces

Onions

Cucumber

Spinach

tomatoes

Asparagus

Herbs

Fruits

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 23.
Describe biological method of pest control.
Answer:
Biological pest control methods
They can be further classified into three categories:
(a) Breeding disease resistant varieties. It is the development of varieties having resistance to pathogenic infection.
(b) Hyperparasitism. It involves the control of one pathogenic organism with the help of another organism, which parasitizes the pathogen.
(c) Trap crops. Many plants secrete some substances after infestation by some pathogens. These substances are toxic to pathogens. Such hosts are called trap or antagonistic plants.

Question 24.
What is inter-cropping? How does it differ from mixed cropping?
Answer:
Inter-cropping is the growing of two or more crops simultaneously in the same field in definite rows.
Differences between mixed cropping and inter-cropping

Characters Mixed cropping Intercropping
1. Aim

2. Pattern

3. Mixing of seeds

4. Application of fertilizers

5. Harvesting and threshing

6. Application of pesticides

To reduce the chances of crop failure.

No definite pattern of rows.

Seeds are mixed up before sowing.

Fertilizers cannot be applied easily to different crops. Cannot be done separately for crops.

Spraying of pesticides for separate crops not possible.

To enhance the production of crops per unit area.

Grown in definite pattern of rows like 1 : 1, 1 : 2, 1 : 3.

Seeds are not mixed before sowing.

Can be applied as per need of individual crops.

Crops can be harvested and threshed separately.

Can be done easily.

Question 25.
Make a table showing the nutritional values of animal products.
Answer:
Nutritional values of animal products:

Animal products

Percent (%) Nutrients

Fat Protein Sugar Minerals Water
Milk (Cow) 3.60 4.00 4.50 0.70 87.20
Egg 12.00 13.00 * 1.00 74.00
Meat 3.60 21.10 * 1.10 74.20
Fish 2.50 19.00 * 1.30 77.20

Present in very little amount

Question 26.
What does cattle feed consist of?
Answer:
Feed of Cattle: Supply of uncontaminated and balanced diet containing sufficient quantities of required nutrients is an essential need of animal husbandry. Cattle feed is formed of two main components: roughage and concentrate.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 27.
Differentiate roughage and concentrate.
Answer:
Roughage contains large amount of fibre but have low nutrients and include hay, fodder, silage and legumes like barseem, lucerene and cowpea. It also includes Common fodder grasses like Napier grass, Guinea grass and Elephant grass.

The concentrate is a mixture of cereals like maize, oat, barley, jowar, broken grams, rice polish, cotton seeds, gram barn and oilseed cake etc. moistened in water. These are rich in proteins and other nutrients, highly palatable and easily digestible.

Question 28.
Discuss the average daily feed of a cow.
Answer:

  1. Green fodder = 15 – 20 kg
  2. Grain mixture = 4 – 5 kg
  3. Water = 30 – 35 litres
  4. Additive feeds like antibiotics, hormones, minerals, etc.

Additive feed increases the milk yield and also protects the animals from diseases.

Question 29.
Name four animals which provide us food.
Answer:

  1. Cows (milk and meat)
  2. Buffaloes (milk)
  3. Goats (milk and meat)
  4. Pigs (provide pork)
  5. Poultry birds (eggs and meat)
  6. fishes (meat)

Question 30.
Mention the names of animal products which are used as food.
Answer:
Milk, beef (cow’s meat), pork (pig meat), mutton (sheep and goat meat), eggs (poultry birds) and fish meat and by-products of fishery such as fish meal, fish-protein concentrate, oil etc.

Question 31.
Name any two Indian breeds of:
1. Cows
2. Buffaloes
Answer:
1. Cow-breeds : Sahiwal and Gir.
2. Buffalo breeds : Murrah and Mehsana.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 32.
Name two exotic breeds of cows.
Answer:
1. Jersey of USA.
2. Brown-Swiss of Switzerland.

Question 33.
Mention the improved crossbreeds of cows.
Answer:

  1. Karan-Swiss
  2. Karan-Fries
  3. Frieswal

Question 34.
List some Indian breeds of cows and buffaloes.
Answer:
1. Cows:

(a) Milch breeds

  1. Gir
  2. Sahiwal
  3. Red Sindhi

(b) Draught breeds

  1. Malvi
  2. Nageri
  3. Flallikar
  4. Kangayam

(c) General utility breeds

  1. Haryana
  2. Ongole
  3. Tharparkar

2. Buffaloes:

  1. Murrah
  2. Bhadawari
  3. Jaffarabadi
  4. Surti
  5. Nagpuri
  6. Nili Ravi
  7. Mehsana

Question 35.
Briefly explain contribution of Dr. V. Kurien.
Answer:
Contribution of Dr. V. Kurien:
Dr. V. Kurein born on 26th November 1921, is the founder Chairman of the National Dairy Development Board (NDDB), which designed and implemented the world’s largest dairy development programme – the “Operation Flood”. Dr. Kurien is called the architect of India’s modem dairy industry and the father of white revolution.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 36.
Write a note on foot and mouth disease of cattle.
Answer:
It is a viral disease of cows and buffaloes. It is characterized by excessive salivation and reddish granules on their feet and mouth. It can be prevented by vaccinating the cows and by keeping their shelter clean and hygienic.

Question 37.
Write two infectious diseases of each of cows, poultry and fishes.
Answer:

Animal Diseases
1. Cows

2. Poultry

3. Fishes

Anthrax (bacterial) and Foot and mouth disease (viral).

Ranikhet (viral) and Salmonellosis (bacterial).

Viral Haemorrhagic Septicemia (VHS) and Infectious Pancreatic Necrosis (IPN).

Question 38.
Mention a few measures for prevention of diseases in the animals.
Answer:
Preventive measures of diseases in animal

  1. Compulsory vaccination of animals.
  2. Proper disposal of dead animals and animal wastes.
  3. Hygienic handling of all animal products and by-products.
  4. Periodical screening of animals for diseases.
  5. Providing a clean, dry, airy and well-ventilated good animal shelter with hygienic conditions.

Question 39.
List the main reasons for low milk yield of cattle in India.
Answer:
Preventive measures of diseases in animals:

  1. The poor quality of feed.
  2. The shortage of feed.
  3. Low milk-yielding varieties of cattle.

Question 40.
Discuss the importance of poultry as a source of food.
Answer:
Poultry birds supply eggs and meat both being good sources of food. Whole egg contains 36% yolk, 64% proteins and vitamins like A and D. Poultry meat contains proteins (like myosin, globulins, actinomyosin, etc.), fats, vitamins, minerals, etc.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 41.
Name some indigenous and exotic breeds of fowls.
Answer:
1. Indigenous breeds: e.g. Aseel, Ghagus, Basara, Brahma and Chittagong,
2. Exotic breeds:
(a) American breeds e.g. Plymouth rock and Rhode Island Red.
(b) English breeds e.g. Sussex and Australorp.
(c) Mediterranean breeds e.g. Minorcha and White Leghorn.

Question 42.
Make a list of diseases of poultry.
Answer:

  1. Poultry Diseases: The poultry birds suffer from various diseases caused by:
  2. Virus – Fowl pox
  3. Bacteria – Tuberculosis, Cholera, Diarrhoea
  4. Fungi – Aspergillosis
  5. Parasites – Worms, mites, lice, etc.

They also suffer from nutritional deficiency diseases. The most common disease of poultry is ‘Bird Flu’. This disease is damaging the poultry on vast scale.

Question 43.
Write the names of two exotic breeds of poultry.
Answer:

  1. American breeds like Plymouth rock and Rhode Island Red.
  2. English breeds like Australorp and Sussex.
  3. Mediterranean breeds like White Leghorn and Minorcha.

Question 44.
What are important points to remember in poultry farming?
Answer:
Important points in poultry farming:

  1. Maintenance of temperature and hygienic conditions in housing.
  2. Proper poultry feed.
  3. Prevention and control of diseases and pests.
  4. Isolation of diseased birds.
  5. Proper vaccination of birds.
  6. Spraying of disinfectants at regular intervals.

Question 45.
List various measures of prevention of poultry diseases.
Answer:
Prevention of Poultry Diseases. To prevent the poultry from disease the following measures should be taken:

  1. The poultry birds should be kept in good spacious, airy and ventilated shelter.
  2. The shelter should be cleaned properly and regularly. Quick and hygienic disposal of excreta should be ensured.
  3. External parasites should be controlled by applying insecticide solution.
  4. Disinfectant should be sprayed to kill mosquitoes and other external parasites,
  5. Every animal should be vaccinated at regular interval to minimise it against common infections and diseases.

Question 46.
Name three common fresh water and three marine food fishes.
Answer:
Fresh-water food fishes

  1. Labeo rohita – Rahu
  2. Catla catla – Katla or Theila
  3. Wallago attu – Mullee

Marine food fishes
1. Harpodon – Bombay duck
2. Hilsa – Hilsa
3. Sardinella – Salmon

  • Total fish production in India – 7th position in the world.
  • Marine fish production – 10th position in the world.
  • Aquaculture production – 2nd in south East Asian countries.
  • Fish industry contribution – Rs. 400 crores annually as foreign exchange.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 47.
Why is fish meat considered advantageous than meat of other animals?
Answer:

  1. Fish meat contains more proteins (13 – 20%) but less fats.
  2. It has good amount of vitamins A and D and is rich in iodine (for thyroxine formation).
  3. It is more easily digestible than other proteins.
  4. So, fish meat is considered next to the mother’s milk as baby food.

Question 48.
Why are the major carps considered best culturable fishes?
Answer:

  1. These survive even at high temperature and at low oxygen.
  2. These have fast growth rate.
  3. These have easily digestible and nutritive flesh.

Question 49.
Give the economy of fishes.
Answer:

  1. Fish meat is rich in proteins (13 – 20%), vitamins and iodine but has less fats.
  2. Their liver oil is rich source of vitamins A and D.
  3. Fish meal is very rich source of proteins (55 – 70%), so is good food for domesticated animals.
  4. Fish wastes can be used as manure for coffee, tea and tobacco plants.
  5. Fish skin of sharks is used to form hand bags, shoes, tobacco pouches, etc.

Question 50.
Give the functions of followings in fishery:
1. Hapas
2. Nursery ponds
3. Traps
Answer:
1. Hapas: These are hatching pits. These are formed of cloth and supported by bamboo sticks. In these pits, hatching occurs and young ones called hatchlings, emerge and grow in hapas to form fries.
2. Nursery ponds: These are small-sized ponds located near the hapas. In these ponds, fries feed upon the planktons and grow into young ones called fingerlings.
3. Traps: These are used to harvest the fish from the stocking ponds.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 51.
Differentiate capture fishery and culture fishery.
Answer:
Capture fishery involves the catching of fish by various methods while culture fishery involves the rearing of fish in artificial freshwater bodies.

Question 52.
List certain common diseases of fishes.
Answer:
1. Main infectious diseases of fishes are:
(a) Viral Haemorrhagic Septicemia (VHS) and
(b) Infectious Pancreatic Necrosis (IPN).

2. Water-pollution caused diseases of fishes are:
(a) Gill rot (blackening of gills).
(b) Fin rot (cutting down of fins).
(c) Dropsy (swollen belly).

3. Fish-ectoparasites – e.g. fish lice – Argulus.

Question 53.
List some measures to control fish diseases.
Answer:

  1. Pollution of fish farm should be avoided.
  2. Regular monitoring of the level of oxygen, carbon dioxide and pH of the water of fish farm.
  3. Argulus – fish lice can be controlled by adding 2.5 ml/litre of malathion in the pond water.

Question 54.
List the steps of fish seed production by induced breeding techniques.
Answer:
Fish Seed Production by induced breeding technique consists of the following steps:

  1. Use of inducing agents.
  2. Selection of healthy brooders.
  3. Administration of hormones by injection.
  4. Releasing sets of brooders in breeding pool.
  5. Spawning.
  6. Collection of Eggs.
  7. Hatching.
  8. Post care of fish seeds in nursery and rearing ponds.
  9. Transfer of fingerlings in stocking ponds.

Question 55.
What are the uses of honey?
Answer:
Utility of honey:

  1. Honey has great importance for its medicinal value specially in disorders related to digestion, dysentery, vomiting and stomach and liver ailments.
  2. It also helps in growth of our body as it contains iron and calcium.
  3. It is also used as a source of sugar in confectionary items.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 56.
Write the components of honey.
Answer:
Honey is a dense sweet liquid, containing sugars 20-40%, moisture 60-80%, minerals 0.22-0.3%, vitamins 0.2-0.5%, enzymes and pollen.

Question 57.
What are the advantages of beekeeping?
Answer:
Advantages of beekeeping:

  1. Beekeeping is undertaken on commercial basis as an enterprise. Besides honey, other products obtained from beekeeping include wax, royal jelly and bee venom.
  2. Beekeeping requires low investments, therefore, farmers along with agriculture also do beekeeping as an additional income-generating activity.
  3. It also helps in cross-pollination as pollens are transferred from one flower to another by bees while collection of nectar.

Question 58.
List the varieties of honeybee used for beekeeping.
Answer:
Honeybee varieties used for beekeeping
The indigenous varieties of bee used for commercial production of honey are Apis cema indica F., commonly known as Indian bee, Apis dorsata, the rock bee and A-florae little bee. An exotic variety from Italy has been domesticated in India to increase yield of honey. This variety is called Italian bee i.e. Apis mellifera, commonly known as Italian bee.

Question 59.
How are honeybees affected by different factors?
Answer:
1. Honeybees generally get bacterial and viral diseases.
2. Common pests of bees are wasps, wax moths and mites. Wasps are controlled manually, by exposing bees in bee hive to sun or by increasing temperature. King crow and greenbee eater prey upon bees. They can be scared away by some devices.

Question 60.
Explain compound fish culture. List the factors.
Answer:
Composite Fish Culture:
Combination of 6 species is used in the culture system. This combination is highly
advantageous because these fishes do not compete for food among them having different types of food habits. Another advantage is that food available in all the parts/zones of the pond is utilized due to their food habits.

The food habits of six species are catla, a surface feeder, rohu feed in middle zone of the pond i.e. column feeder and mrigal and common carp feed at the bottom, where as grass carp feed on aquatic weeds in the pond. Amongst them three are foreign or exotic, i.e. transplanted from China and 3 species are of Indian origin.

Factors: Important factors to be taken into consideration for fish culture include:

  1. Topography or location of pond.
  2. Water resources and quality.
  3. Soil quality i.e. composition particle size as well as nutrients.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 61.
What are the uses of chemical fertilizers?
Answer:
Uses of fertilizers:

  1. They cover up the mineral deficiency of soil due to excessive and repeated cropping in a field.
  2. They are required in smaller bulk.
  3. They are easy to transport.
  4. They are quickly available to their plant food constituents.
  5. The chemical fertilizers are available both in soil form as well as in solution form.
  6. Chemical fertilizers are soluble in water, hence are easily absorbed by plants.

Very Short Answer Type Questions:

Question 1.
What are the major sources of food for us?
Answer:
Plants and animals.

Question 2.
List the nutrients supplied by food.
Answer:
Proteins, carbohydrates, fats, vitamins and minerals.

Question 3.
Coin the terms for extensive production of :
1. fish
2. milk
3. oil
Answer:
1. Blue revolution
2. White revolution
3. Yellow revolution

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 4.
What term is used for extensive production of pulses?
Answer:
Golden revolution.

Question 5.
List cereal crops which provide carbohydrates for energy.
Answer:
Wheat, rice, maize, minor millets, and sorghum.

Question 6.
Name a few pulses which provide proteins.
Answer:
Gram, Pea, Black gram, green gram, pigeon pea, lentil.

Question 7.
List any six oil seed crops which provide fatty acids.
Answer:
Soyabean, groundnut, seasame, castor, mustard, linseed, niger and sunflower.

Question 8.
From where we get minerals and vitamins?
Answer:
Vegetables, spices and fruit crops.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 9.
Name three fodder crops.
Answer:
Berseem, oat and Sudan grass.

Question 10.
Name the crops grown in rabi season (November to April).
Answer:
Wheat, gram, peas, mustard, linseed.

Question 11.
Name kharif season crops.
Answer:
Paddy, soyabean, arhar, maize, cotton, urad and moong.

Question 12.
How many nutrients are required by plants?
Answer:
16 nutrients.

Question 13.
Name four macronutrients.
Answer:
Nitrogen, phosphorus, potassium and sulphur.

Question 14.
Name nine micronutrients.
Answer:
Iron, Manganese, Boron, Zinc, Copper, Molybdenum, Chlorine, Calcium, Magnesium.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 15.
What is farmyard manure (FYM)?
Answer:
FYM is the decomposed mixture of cattle excreta (dung) and urine along with litter (i.e. bedding material used in night under cattle) and left over organic matter such as roughage or fodder.

Question 16.
What is composting?
Answer:
Composting is a biological process in which both aerobic and anaerobic micro-organisms decompose the organic matter.

Question 17.
What is vermicomposting?
Answer:
The degradation of organic wastes through the consumption by the earthworms is called vermicomposting.

Question 18.
What are nitrogenous fertilizers? Give one example.
Answer:
These fertilizers supply the macronutrient nitrogen. Example. Urea.

Question 19.
What are complex fertilizers?
Answer:
When a fertilizer contains atleast two or more nutrients (N, P2O5 and K2O), it is called complex fertilizer.

Question 20.
Define irrigation.
Answer:
The process of supplying water to crop plants by means of canals, reservoirs, wells etc. is called irrigation.

Question 21.
Name two potassium fertilizers.
Answer:
Potassium sulphate and potassium chloride.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 22.
When even excessive application of manure does not cause pollution?
Answer:
Manures are biodegradable, so they do not cause any damage.

Question 23.
List three important nitrogen-containing fertilizers.
Answer:
Urea, Ammonium Sulphate and Ammonium nitrate.

Question 24.
List three phosphorus-containing fertilizers.
Answer:
Ammonium phosphate, Ammonium hydrogen phosphate, Calcium Superphosphate.

Question 25.
Name two potassium-containing fertilizers.
Answer:
Potassium sulphate, Potassium chloride.

Question 26.
What is being traditionally used as manure in our country?
Answer:
Cow-dung.

Question 27.
Define herbicide.
Answer:
Herbicide is the chemical agent that destroys or inhibits plant growth ; used to destroy weeds in the cultivated patch of land.

Question 28.
List any three manures.
Answer:
Farmyard manure, composted manure and green manure.

Question 29.
What is monoculture?
Answer:
Growing the same crop in a field year after year is called monoculture.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 30.
A fanner grows gram crop between two cereal crops. Which agricultural practice is being followed by him?
Answer:
Crop rotation.

Question 31.
Suppose you are incharge of a grain store. How will you find out the presence of pests? Mention any two indications.
Answer:
1. By presence of living or dead insects.
2. By noticing white powdery materials on the bags or on the floor.

Question 32.
What are weeds?
Answer:
Unwanted plants growing alongwith main crops are called weeds.

Question 33.
Define fungicide.
Answer:
The pesticides that kill fungi are called fungicides.

Question 34.
What is the basic objective of mixed cropping?
Answer:
Minimize risk and insurance against crop failure due to abnormal weather conditions.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 35.
List any two prominent mixed cropping practices.
Answer:
1. Maize and urd bean.
2. Cotton and mung bean.

Question 36.
Which crops can be grown along with wheat during mixed cropping practices?
Answer:
Chickpea and mustard.

Question 37.
What is intercropping?
Answer:
Intercropping is growing of two or more crops simultaneously in the same field in definite rows.

Question 38.
Expand HYV.
Answer:
High Yielding Varieties.

Question 39.
Name any two subsidiary occupations which are part of mixed farming.
Answer:
Daily farming and poultry farming.

Question 40.
What is row inter-cropping?
Answer:
Growing two more crops at the same time with atleast one crop planted in rows.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 41.
What is inter-cropping?
Answer:
Growing two or more crops together in strips wide enough to permit separate crop production using machines but close enough for the crops to interact.

Question 42.
Name any two improved varieties of wheat.
Answer:
Sonara, PPW 154.

Question 43.
Name any two improved varieties of rice.
Answer:
Kasturi, PNR-591-18.

Question 44.
What is green revolution?
Answer:
The tremendous increase in food grains (especially wheat) during the last three decades due to the use of HYV, high dose of fertilizers and irrigation is known as green revolution.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 45.
Define selection.
Answer:
Selection is the sorting out of individual plants or groups of plants from mixed population.

Question 46.
Name the oldest method of crop improvement.
Answer:
Introduction.

Question 47.
Give one example of crop combination used in mixed cropping.
Answer:
Soyabean and pigeon pea.

Question 48.
Define plant breeding.
Answer:
Plant breeding means production of new varieties or strains by a programme of artificial selection spanning several generations of the organism concerned.

Question 49.
Name the branch which deals with feeding, caring and breeding of domestic animals.
Answer:
Animal husbandry.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 50.
What are milch animals?
Answer:
Animals providing milk are called milch animals.

Question 51.
Define the livestock.
Answer:
Domesticated animals reared to provide milk, meat, etc. e.g. cows, buffaloes, sheep, goats, etc.

Question 52.
Give two uses of cattle.
Answer:
1. Cattles provide hide to prepare leather goods.
2. They are useful in agricultural operations like ploughing, harrowing, levelling, etc.

Question 53.
Give two examples of each of indigenous breeds and exotic milch breeds of cows.
Answer:
1. Indigenous breeds. Sahiwal and Red Sindhi.
2. Exotic breeds. Jersey and Brown Swiss.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 54.
List three categories of cattle on the basis of their utility. Give one example of each.
Answer:

  1. Milch breed e.g. Sahiwal.
  2. Draught breed e.g. Hallikar.
  3. General utility breed e.g. Haryana.

Question 55.
Name two high milk-yielding crossbreeds of cows.
Answer:
Karan-Swiss and Karan-Fries.

Question 56.
Name two exotic breeds of cows.
Answer:
Jersey (USA) and Brown-Swiss (Switzerland).

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 57.
Name two breeds of buffaloes.
Answer:
Murrah and Mehsana.

Question 58.
What are the two components of cattle feed?
Answer:
Roughage and Concentrate.

Question 59.
How roughage and concentrate differ from each other?
Answer:
Roughage contains fibres but less nutrients e.g. fodder, while concentrate is rich in proteins e.g. cereal grains.

Question 60.
What is ration of cow?
Answer:
Ration is the amount of food, which is given to animal during 24 hours period. For cow it is about 15 to 20 kg of green fodder and 4-5 kg of grain mixture.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 61.
What are additives?
Answer:
Dairy animals require certain additive feeds, which contain antibiotics, minerals and hormones, they promote growth of animals, facilitate good yield of milk and protect them from diseases.

Question 62.
Name the progeny of cross breed of Brown Swiss and Sahiwal.
Answer:
Karan Swiss.

Question 63.
What is artificial insemination?
Answer:
Introduction of semen of a high quality pedigree bull into the vagina of a healthy female cow by artificial means is called artificial insemination.

Question 64.
Give the term for the process by which a female cow of good breed is stimulated by the hormones to release more ova from its ovaries.
Answer:
Superovulation.

Question 65.
Give the significance of superovulation.
Answer:
It increases the chances of transmission of good characters to progeny.

Question 66.
Who is called “Father of White Revolution”?
Answer:
Dr. V. Kurien.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 67.
What is NDRI? Where is it located?
Answer:
National Dairy Research Institute. It is located at Kamal (Haryana).

Question 68.
Name two bacterial diseases of cattle.
Answer:
Anthrax and Brucellosis.

Question 69.
Name two viral diseases of cattle.
Answer:
Foot and mouth disease and cowpox.

Question 70.
Name two animal products in which carbohydrate is totally absent.
Answer:
Eggs and meat.

Question 71.
Name two non-leguminous dry fodders.
Answer:
Pounded straw of wheat and dry grass.

Question 72.
Give one term for the science dealing with rearing of birds.
Answer:
Poultry.

Question 73.
Name two indigenous breeds of fowls.
Answer:
Aseel and Brahma.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 74.
Name two exotic breeds of fowls.
Answer:
Rhode Island Red and White Leghorn.

Question 75.
Name two high-yielding crossbreeds of fowls.
Answer:
“B-77” and “HH-260”.

Question 76.
What are broilers?
Answer:
Meat-providing birds are called broilers.

Question 77.
What are layers?
Answer:
Egg-laying hens are called layers.

Question 78.
Name one viral and one bacterial disease of fowls.
Answer:
1. Ranikhet (Viral disease)
2. Salmonellosis or Pullorum (Bacterial disease)

Question 79.
What are vegetarian eggs?
Answer:
Infertile eggs are called vegetarian eggs.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 80.
Define pisciculture.
Answer:
Pisciculture is rearing and management of fishes.

Question 81.
Give one term for composite fish-farming.
Answer:
Polyculture.

Question 82.
What is economic importance of fish as food?
Answer:
Fish meat contains more proteins (13 – 22%), less fats, vitamins (A and D) and iodine.

Question 83.
Name two by-products of fishery.
Answer:
Liver oil and fish meal.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 84.
Which two vitamins are present in liver oil of certain fishes?
Answer:
Vitamins A and D.

Question 85.
Name two indigenous breeds of carps used as food fishes.
Answer:
1. Catla (Theila)
2. Labeo (Rahu)

Question 86.
Name two exotic breeds of carps used as food for fishes.
Answer:
Silver carp and Grass carp.

Question 87.
What is fish meal? Give its significance.
Answer:
It is prepared from non-oil type fishes and is rich in proteins (55 – 70%).

Question 88.
Define inland culture fishery.
Answer:
Rearing of fishes in the artificially prepared fresh water ponds.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 89.
Give two examples of marine food fishes.
Answer:
Sardinella (Salmon) and Harpodon (Bbmbay duck).

Question 90.
What is hypophysation?
Answer:
Process by which female and male fishes are injected the hormones of pituitary extract to induce spawning.

Question 91.
Give the term for the newly-hatched young ones of fishes.
Answer:
Hatchlings.

Question 92.
Define fingerlings.
Answer:
Fingerlings are about 4″ – 6″ sized fishes formed from fries in nursery ponds and rearing ponds.

Question 93.
What is harvesting?
Answer:
Capturing of fully grown fishes is called harvesting or fishing.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 94.
Name two methods of fishing.
Answer:
Angling and trapping.

Question 95.
What are hapas?
Answer:
These are rectangular boxes of mosquito net cloth in which hatching of fish eggs occur.

Question 96.
Name two infectious diseases of fishes.
Answer:
1. Viral Haemorrhagic Septicemia (VHS)
2. Infectious Pancreatic Necrosis (IPN)

Question 97.
Name two varieties of Indian fishes.
Answer:
Carps (e.g. Catla) and Cat-fishes (e.g. Mystus).

Question 98.
Mention seafood items other than fishes.
Answer:
Lobsters, prawns, oysters, mussels, etc.

Question 99.
Name the different types of ponds.
Answer:
Nursery ponds, rearing ponds, stocking ponds and broodstock ponds.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 100.
List the products obtained from beekeeping.
Answer:
Honey, wax, royal jelly and bee venom.

Question 101.
“Honey is a dense sweet liquid.” Write its composition.
Answer:
Sugar 20-40%, moisture 60-80%, minerals 0.22 to 0.3%, vitamins 0.2 – 5%, enzymes, pollen etc.

Question 102.
What is the utility of pollen for bees?
Answer:
Pollens serve as protein food for bees.

Question 103.
What is a beehive?
Answer:
A beehive is a wooden box of size 46 × 23 cm made up of wooden chambers for egg-laying, honey collection, as honey reserve.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Punjab State Board PSEB 9th Class Science Important Questions Chapter 14 Natural Resources Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Long Answer Type Questions:

Question 1.
Write a note on formation of soil.
Answer:
Formation of Soil: The formation of soil depends on the parent rock material, the climate and topography of the area, the organisms present in the soil and the time over which the soil has been developing. Over long periods of time, thousands and millions of years, the rocks near the surface of the Earth are broken down by various physical, chemical and some biological processes. The end product of this breaking down is the fine particles of soil.

Processes for soil formation:
1. The Sun: The sun heats up rocks during the day so that they expand. At night, the rocks cool down and contract. The unequal expansion and contraction in different parts of the rock results in the formation of cracks and ultimately rocks break up into smaller pieces.

2. Water: Water helps in the formation of soil in two ways:

  1. Water could get into the cracks in the rocks formed due to uneven heating by the sun. If this water freezes, it will widen the cracks.
  2. Fast flowing water carries big and small particles of rock downstream, causing breakdown of rock particles into smaller, finer particles through their abrasive effects.

3. Wind: Strong winds also break down rocks. They also carry sand from one place to the other like the water does.

4. Living organisms: They also influence the formation of soil. While lichens grow on surface of rocks, they release certain substances that cause the rock surface to powder down and form a thin layer of soil. Likewise, small plants like moss and roots of big trees also break the rocks.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 2.
What are the effects of use of fertilizers and pesticides for long period on soil fertility? What are the causes of soil erosion?
Answer:
Effects of excessive use of fertilizers and pesticides:

  1. Use of these substances over a long period of time can destroy the soil structure by killing the soil micro-organisms that recycle nutrients in the soil.
  2. They also kill the earthworms which are instrumental in making the rich humus.
  3. Fertile soil can quickly be turned barren if sustainable practices are not followed.
  4. Major cause of soil pollution is removal of useful components from the soil and addition of other substances, which adversely affects the fertility of the soil and kills the diversity of organisms living in it, is called soil pollution.

Causes of Soil Erosion:

  1. Wind causes soil erosion by carrying away the top loose soil particles.
  2. Rain causes soil erosion on unprotected topsoil by washing it down.
  3. mproper farming or tilling and leaving the field fallow for long time causes soil erosion.
  4. Frequent flooding of rivers causes soil erosion by removing the fertile top soil of the fields near the river banks.
  5. Deforestation also leads to soil erosion.

Question 3.
Why is soil as resource important for mankind? Mention the constituents of soil.
Answer:
Soil is a rich source of minerals and humus. It is important for growing crops. Soil water is used by plants for various functions.
Soil provides support to crops, grassland and forests thus it is an important natural resource.

Components of Soil. Soil is a mixture, it contains:

  1. Small particles of rocks.
  2. Bits of decayed living organisms which is called humus.
  3. Soil also contains various forms of microscopic life.
  4. It contains nutrients and availability of which depends on the rocks from which it was formed.
  5. Soil water – 25% – 35%
  6. Soil air – 15-25 %

Question 4.
Define soil fertility. How can it be maintained?
Answer:
Soil fertility. It is the ability of soil to provide minerals, water and other nutrients to the plants.
Conservation of Soil fertility:

  1. Adding of manure to the soil.
  2. Rotation of crops.
  3. Keeping the land as such without growing any crop.
  4. Addition of fertilizers.

Artificial methods to maintain soil fertility:

  1. Nitrogenous and other fertilizers are added.
  2. For natural restoration of nutrients, soil is kept uncultivated for certain period.

Role of humus:

  1. Humus increases the soil fertility.
  2. Humus has high retaining capacity for water.
  3. It makes the soil porous and allows water and air to penetrate deep.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 5.
What is air pollution? Write the main sources and preventive measures.
Answer:
Air Pollution. Air pollution refers to the release into the atmosphere of materials that are harmful to man, other animals, plants and buildings or other objects.

Sources of Air pollution:
The major sources of air pollution are fossil fuels (coal and petroleum) and industries. Human Sources. Many activities done by man are the main sources of air pollution. These activities can be divided into following categories.

  1. Combustion activities.
  2. Industrial activities.
  3. Agricultural works.
  4. Use of solvents.
  5. Activities concerned with atomic energy.

Preventive measures for air pollution:
To prevent and control air pollution two types of measures can be adopted.
1. Instead of releasing poisonous gases containing various pollutants into the atmosphere they could be destroyed or used by some other measures.
2. Converting harmful pollutants to harmless products and then releasing them into
the atmosphere.

Control measures for minimizing air pollution:
1. Simple combustible solid wastes should be burnt in incinerators.
2. Automobiles must be either made to eliminate the use of gasoline and diesel oil or complete combustion is obtained in the engine so that harmful products are omitted.

Question 6.
Write a few properties of water. Sketch water Cycle.
Answer:
Water is a liquid at room temperature. It is densiest (heaviest) at about 4°C. The dense water sinks and the lighter frozen water (ice) floats, ice also insulates the water below. This enables the aquatic life to survive under the ice in cold weather.
PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 1

Question 7.
What is nitrogen cycle? Make a simple line sketch to show nitrogen cycle in the biosphere.
Answer:
The nitrogen cycle in the biosphere is regarded as a perfect cycle because the cycling process keeps the overall amount of nitrogen constant in the atmosphere and water bodies. The use of chemical nitrogeneous fertilizers like NPK and urea also help in the maintenance of soil nutrients and nitrogen cycle.

However, some of the nitrogen compounds present in soil get trapped within sedimentary rocks and therefore, they are not available to nitrogen cycle for circulation in the biosphere. However, this loss is compensated by volcanic eruptions and erosions and sedimentary rocks. Both these processes release nitrogen.

Micro-organisms involved in Nitrogen Cycle:

As already learnt, micro-organisms play a very important role in nitrogen cycle in nature. Different organisms are involved in different processes of nitrogen cycle. The main micro-organisms involved in nitrogen cycle are listed below:

Activity:

  1. Nitrogen fixation
  2. Ammonification
  3. Ammonia to nitrites
  4. Nitrification (Nitrites to nitrates)
  5. Denitrification (Nitrates to free Nitrogen)

Organisms:

  1. Rhizobium, blue-green algae
  2. Decay bacteria, fungi
  3. Nitrosomonas
  4. Nitrobacter
  5. Pseudomonas

The nitrogen cycle in the biosphere involves the following steps:

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 2

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 8.
Briefly explain oxygen cycle.
Answer:
Oxygen cycle: Oxygen is very abundant element on earth. It is found in the elemental form in the atmosphere to the extent of 21%. It also occurs extensively in the combined form in the earth’s crust as well as also in the air in the form of carbon dioxide. In the crust, it is found as the oxides of most metals and silicon, and also as carbonate, sulphate, nitrate and other minerals. It is also an essential component of most biological molecules like carbohydrates, proteins, nucleic acids and fats (or lipids).

But when we talk of the oxygen cycle, we are mainly referring to the cycle that maintains the levels of oxygen in the atmosphere. Oxygen from the atmosphere is used up in three processes, namely combustion, respiration and in the formation of oxides of nitrogen. Oxygen is returned to the atmosphere in only one major process, that is, photosynthesis. And this forms the broad outline of the oxygen cycle in nature.
PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 3

Question 9.
Briefly explain carbon cycle in nature.
Answer:
The Carbon Cycle. Carbon is found in various forms on the earth. It occurs in the elemental form as diamond and graphite. In abiotic environment, carbon is present in the following forms:

  1. as carbon dioxide in the atmosphere.
  2. as carbonate and hydrogen-carbonate salts in various minerals.
  3. as dissolved carbonic acid and bicarbonates in water.
  4. as fossil fuels like coal, petroleum and natural gas.
  5. Plants utilise the atmospheric carbon dioxide in photosynthesis to produce carbohydrates, which are taken by herbivores and then pass through small and large carnivores.

Forms of Carbon: Carbon is found in various forms on the earth.

  1. It occurs in the elemental form as diamonds and graphite.
  2. In the combined state, it is found as carbon dioxide in the atmosphere, as carbonate and hydrogen-carbonate salts in various minerals.
  3. All life forms are based on carbon-containing molecules like proteins, carbohydrates, fats, nucleic acids and vitamins.
  4. Both plants and animals release carbon dioxide to the atmosphere as a product of respiration.
  5. By decomposition of organic wastes and dead bodies by decomposers.
  6. By burning of fossil fuels, like coal, and petroleum.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 4

Question 10.
What are the causes of ozone depletion?
Answer:
Ozone depletion: Recently it was discovered that ozone layer was getting depleted. Various man-made compounds like CFCs (carbon compounds having both fluorine and chlorine which are very stable and not degraded by any biological process) were found to persist in the atmosphere.

Once they reached the ozone layer, they would react with the ozone molecules. This resulted in a reduction of the ozone layer and recently they have discovered a hole in the ozone layer above the Antarctica. It is difficult to imagine the consequences for life on earth if the ozone layer dwindles further, but many people think that it would be better not to take chances. Measures should be taken towards stopping all further damage to the ozone layer.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 11.
Write a note on freshwater resources.
Answer:
Fresh Water Resources:
Fresh water resources range from ponds to lakes and large rivers. It has the following characteristics :
(a) Freshwater is exhaustible, however, it is being made available again by oceans through hydrological cycle.

(b) Out of this three per cent, 77.2 per cent is stored in glaciers and ice caps. And 22.4 per cent is groundwater and soil moisture. Remaining 0.36 per cent is found in lakes, rivers, streams and swamps etc. Out of the total water evaporated from oceans 90 per cent falls on the oceans and remaining 10 per cent falls on the land. This water is utilised by various terrestrial ecosystems.

(c) Freshwater is essential for life on earth as well as for survival of human race.

(d) The total water in hydrosphere is 1.4 billion cubic kilometres (Km3). Total ocean
water is 97%. The ocean water cannot be consumed by human beings. Remaining three percent (freshwater) is available for human consumption. The water resources in India have an average run off in river system of 1,869 km2 and 432 km3 groundwater.

Short Answer Type Questions:

Question 1.
What is atmosphere? Name its different layers.
Answer:
Atmosphere: Gaseous envelope surrounding the earth is called atmosphere. Several concentric layers can be identified in vertical profile of atmosphere. Density, temperature and composition differ in these layers. Near the earth’s surface, density is highest and with increase in latitude density decreases. Starting from earth’s surface five concentric layers can be distinguished:

  1. Troposphere
  2. Stratosphere
  3. Mesosphere
  4. Thermosphere
  5. Ionosphere.

Exosphere which forms the outer fringe of atmosphere is highly rarefied and gradually get mixed with other space.

Composition of dry atmosphere:

Components Volume
Nitrogen (N2) 78%
Oxygen (O2) 21%
Carbon dioxide (CO2) 0.03%
Argon 0.93%
Helium, Neon, Ozone, ammonia 0.04%

Helium, Neon and Argon are noble gases.

Question 2.
What is the role of atmosphere in climate control?
Answer:
Role of atmosphere in climate control. Atmosphere covers the earth like a blanket. Air is a bad conductor of heat. The atmosphere keeps the average temperature of the earth fairly steady during the day and even during the course of the whole year. The atmosphere prevents the sudden increase in temperature during the daylight hours.

During the night, it slows down the escape of heat into outer space. Moon, which is about the same distance from the sun that the earth is. Despite that, on the surface of the moon, with no atmosphere, the temperature ranges from -190° C to 110° C.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 3.
What is ozone layer? Write its importance.
Answer:
Ozone layer: It is the protective layer. Ozone in stratosphere is responsible for protecting the earth from high energy ultraviolet radiation. It forms a life-saving screen as it checks the entry of lethal UV- rays. Ozone found in troposphere has warming effect. Ozone is one gas which is harmful as well as useful for human beings.

16th September, 1996 was celebrated as “International Day for the Preservation of Ozone layer”. It was aimed to generate awareness about the dangers of ozone depletion in the stratosphere and it was drawn up by UNEP.

Question 4.
What is meant by ozone shield? How the CFC’s and ozone-depleting substances effect ozone shield?
Answer:
Ozone shield: An equilibrium is established between generation and destruction of O3, leading to a steady-state concentration of ozone layer in the stratosphere between 20 and 26 km above the sea level. The thickness of the vertical column of stratospheric O3 layer, condensed to standard temperature and pressure, average 0.29 cm above the equator and may exceed 0.40 cm above the poles at the end of the winter season. This layer acts as the ozone shield protecting the earth biota from harmful effects of strong UV radiations.

CFC’s produce active chlorine (Cl with CIO radical) in the presence of UV radiations. These radicals catalytically destroy ozone converting it into oxygen. CH4 and N2O also cause ozone destruction.

Question 5.
Write a note on the air pollution caused due to combustion.
Answer:
The mobile combustion sources are the main sources of air pollution especially in the cities. They include the locomotives, automobiles and aircrafts.
The air pollutants from these are:
1. (i) Carbon monoxide (ii) oxides of nitrogen and (in) a mixture of hydrocarbons.
2. The petroleum used as fuel in these sources contains lead as an impurity in the form of tetraethyl lead Pb (C2H5)4, and tetramethyl lead Pb (CH3)4.

Question 6.
Discuss harmful effects of air pollution.
Answer:
Harmful Effects of Air Pollution
1. Air pollution affects the respiratory system causing breathing difficulties and diseases such as bronchitis, asthma, lung cancer, tuberculosis and pneumonia.

2. Burning of fossil fuels like coal and petroleum releases oxides of nitrogen and sulphur. Inhalation of these gases is dangerous. These gases also dissolve in rain to give rise to acid rain.

3. The combustion of fossil fuel also increases the amount of suspended particles in air. These suspended particles could be unbumt carbon particles or substances called hydrocarbons. The presence of high levels of all these pollutants, reduce visibility in cold weather where water also condenses out of air forming smog. Smog is an indication of air pollution.

4. Regular breathing in the polluted air increases allergies, cancer and heart diseases.

Question 7.
What is the role of biotic components in living organisms?
Answer:
The living or biotic components are plants and animals including us and non¬living or physical components are air, water, soil, light and temperature. All these components interact and effect each other, resulting in the establishment of a complex and complete balance in the environment. The environmental components like mountains, rivers, ponds, forests, minerals, coal and even petrol and other natural resources are of great importance to us.

CO2 is fixed in two ways:
1. Green plants convert CO2 into glucose in the presence of sunlight and chlorophyll pigments.
2. Many marine animals use carbonates dissolved in sea water to make their shells. Oxygen is required by eukaryotic and many prokaryotic cells. All the cells need oxygen to break down glucose molecules in order to release energy required to carry out this vital functions of life.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 8.
What are aerosols?
Answer:
Aerosols. Aerosols are certain chemicals released in the air with force in the form of mist or vapour. The important source of aerosols is the jet aeroplane emissions in the outer atmosphere. The aerosols contain fluorocarbons which deplete the ozone layer in the atmosphere.

Question 9.
What are acid rains?
Answer:
Acid Rain. It is the rain which contains small amount of acid in it that is formed from the gases like sulphur dioxide and nitrogen oxides present in polluted air. It causes damage to living and non-living things.

Question 10.
What is smog?
Answer:
Smog. Smoke and fog when combined together forms smog in the presence of sunlight. Various unbumt hydrocarbons produced from the automobile combustion react with oxides of nitrogen to form ozone, peroxyacyl nitrates and aldehydes. They are called photochemical oxidants. Together with smoke and fog they constitute smog which has a harmful effect on humans repiratory and nervous system; it also harm the plants and rubber goods.

Question 11.
Explain the role of sun in soil formation.
Answer:
Role of sun in soil formation:
The sun heats up rocks dtiring the day as a result that they expand. At night, these rocks cool down and contract. Since all parts of the rock do not expand and contract at the same rate, this results in the formation of cracks and ultimately the huge rocks break up into smaller pieces and small pieces further break up into still smaller pieces and fine particles.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 12.
How does water take part in soil formation?
Answer:
Role of water in soil formation:
1. Water enters into cracks formed as a result of uneven heating by sun. As this water freezes, it causes the cracks in the rocks to widen.
2. Flowing water wears away hard rocks over long period of time. The fast flowing water often carries big and small particles of rock downstream. These pieces of rock rub against other rock pieces and the abrasion causes the rocks to wear down into smaller and still smaller pieces.

Question 13.
Discuss the role of wind and living organisms in soil formation.
Answer:
Role of wind in the soil formation. Strong wind rubs against rocks and wear them down. The wind also carries soil particles from one place to another.

Role of living organisms in soil formation:

Living organisms also influence the formation of soil. The lichen that we read about earlier, also grows on the surface of rocks. While growing, they release certain substances that cause the rock surface to powder down and form a thin layer of soil.

Other small plants like moss, are able to grow on this surface now and they cause the rock to break up further. The roots of big trees sometimes go into cracks in the rocks and as the roots grow bigger, the crack further becomes bigger causing the rocks to break down to form soil.

Question 14.
What is the role of atmosphere in movement of air which causes winds?
Answer:
The movement of air causes winds:
1. The atmosphere gets heated from the radiation that is reflected back or re-radiated by the land or water bodies. As a result of heating convection currents are set up in the air. Since land gets heated faster than water, the air over land gets heated faster than the air above water bodies.

2. In coastal regions, during the day, the air above the land gets heated faster and starts rising. So a region of low pressure is created and air over sea moves into this area of low pressure. The movement of air from one region to the other region causes winds.

3. During the day, the direction of wind would be from the sea to the land and at night, both land and sea starts to cool. Since water cools down slower than the land, the air above water would be warmer than the air above land, thus the direction of wind would be from the land to the sea.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 15.
How are rainfall patterns decided?
Answer:

  1. Rainfall patterns are decided by the prevailing wind patterns.
  2. In large parts of India, rains are mostly brought by the southwest or northeast monsoons.
  3. Depressions in the Bay of Bengal have caused rains in some areas, is the common comment during weather ireports.

Question16.
What is the role of atmosphere in causing rain?
Answer:
Role of atmosphere in causing rain:

  1. When water bodies are heated during the day, a large amount of water evaporates and goes into the air.
  2. The wind carries the water vapour to various places.
  3. The air gets heated and rises up carrying the water vapour with it.
  4. As this air rises, it expands and cools causing the water vapour in the air to condense in the form of tiny droplets.
  5. Once the water droplets are formed, they grow bigger by the ‘condensation’ of these water droplets.
  6. When the drops grow big and heavy, they fall down in the form of rain.

Question 17.
Comment “Water is one of the major resources which determine life on land”. List a few other factors also.
Answer:
The availability of water decides not only the number of individuals of each species that are able to survive in a particular area, but it also decides the diversity of life there. Of course, the availability of water is not the only factor that decides the sustainability of life in a region. Other factors like the temperature and nature of soil also matters. But water is one of the major resources which determine life on land.

Question 18.
What is water pollution?
Answer:
Water pollution: Addition of harmful materials to water is termed water pollution. The sources of inland water pollution are community wastewater (sewage) and wastes from industries and agricultural practices. Water pollutants include organic matter, pathogens, chemicals and minerals, solid particles, radioactive wastes and heat.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 19.
What are the sources of water pollution?
Answer:
Sources of Water Pollution:

  1. Agriculture substances such as fertilisers and pesticides are used to increase crop yield, and some amount of these chemicals is washed into the water bodies that pollutes the water.
  2. Sewage from homes and wastes from factories are dumped into rivers or lakes also cause water pollution.
  3. Hot and cold water discharged from industries make a change in temperature, which is harmful for aquatic organisms.
  4. All these affects the balance among various organisms that are found in water bodies.

Question 20.
What are the effects of water pollution?
Answer:
Effects of Water Pollution:

  1. Water pollutants reach the sea directly from the coastal cities and ships, and indirectly with river water from distant places. Oil spilled in tanker accidents is a major threat to ocean life.
  2. The substances like fertilisers and pesticides used in farming, mercury salts used by paper industries could be poisonous. There could also be disease-causing organisms, like the bacteria which causes cholera.
  3. Industrial or household waste reduces the dissolved oxygen in water bodies, thereby affecting the aquatic life.
  4. Aquatic organisms can stay alive in a certain range of temperature. Sudden change in temperature of water bodies is dangerous for aquatic organisms and affects their breeding.

Question 21.
Make a list of various diseases caused by polluted water.
Answer:
Diseases caused by polluted water

  1. Bacterial diseases. Cholera, Typhoid, Diarrhoea, Dysentery.
  2. Viral diseases. Jaundice, Polio etc.
  3. Protozonal diseases. Diseases associated with stomach and intestines eg. Amoebic dysentery, Giardiasis etc.
  4. Helminthic diseases. Infection of some intestinal parasites like Ascaris lumbricodies is through drinking water only.
  5. Guinea worm diseases is through Cyclops present in the drinking water. Through contaminated water they reach to another host i.e. man.

Question 22.
What is greenhouse effect? Show the % age of gases that cause greenhouse effect.
Answer:
Greenhouse effect. Earth’s temperature is maintained by reradiated infra-red radiations by CO2, CH4, O3, NO and NO2 and slightly by water vapours in atmosphere. These gases prevent heat from escaping to outer space, so are functionally comparable to glass panels of a greenhouse and are called greenhouse gases (GHGs) and the process is called greenhouse effect. The CO2 is added to atmosphere mainly by burning fossil fuels, volcanic activities and respiration.
PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 5

Question 23.
List Various Measures for Soil Conservation
Answer:
Various Measures for Soil Conservation:

  1. Stopping clear-cutting of forests and overgrazing checks soil erosion by streams and rivers.
  2. Intensive cropping helps in checking soil erosion. A field always under a crop is protected against erosion.
  3. Bunds around the fields contain rain water and check soil erosion besides washing away of minerals.
  4. Irrigation channels in the fields should be so designed as to carry water at a slow speed.
  5. Drainage canals to carry flood water will protect the fields against erosion.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 24.
How will you determine composition of soil?
Answer:
PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 6
Determination of Soil Composition:

  1. Materials required: 150 cc of sifted soil, a measuring glass cylinder of 1 litre capacity, water, glass rod.
  2. Procedure: Take 150 cc of sifted soil in a glass cylinder. Pour about 750 cc of water over it. Stir the soil well with the help of a glass rod. Take out the rod. Allow the particles to settle. Observe after 30 minutes.
  3. Results: The bottom of the cylinder has a layer of coarse sand. A layer of fine sand lies above it. Then there is a layer of silt. Clay lies above silt.

Turbid water occurs above the clay. It contains clay as well as mineral salts. Humus or organic matter floats over the top of turbid water.

Question 25.
Make an outline sketch of nitrogen cycle.
Answer:
Nitrogen Cycle
PSEB 9th Class Science Important Questions Chapter 14 Natural Resources 7

Very Short Answer Type Questions:

Question 1.
Life exists on which planet?
Answer:
Earth.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 2.
What are the materials required for life?
Answer:
Environment, Heat, Light, Water and food.

Question 3.
What is environment?
Answer:
Environment. The earth and everything which affects the living organisms constitute its environment.

Question 4.
Basic requirment of life are obtained from which sources.
Answer:
Energy from sun and resources present on earth.

Question 5.
What is atmosphere?
Answer:
Atmosphere: It is the multilayered gaseous envelope of air that covers the whole of the planet earth like a blanket.

Question 6.
How much surface of earth is covered by water?
Answer:
About 75 percent.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 7.
How atmosphere covers the earth?
Answer:
Atmosphere covers the earth as a blanket.

Question 8.
What is biosphere?
Answer:
It is the life-supporting zone of earth where the atmosphere, the hydrosphere and the lithosphere interact and make life possible.

Question 9.
What are biotic components of biosphere?
Answer:
All living organisms.

Question 10.
List the abiotic components of bioshphere?
Answer:
Air, water and soil.

Question 11.
Name the gaseous components of atmosphere.
Answer:
Nitrogen, Oxygen, CO2 and Water vapour.

Question 12.
Name the gases present on Venus and Mars planet.
Answer:
95 to 97% CO2.

Question 13.
What is respiration?
Answer:
A process in which O2 is used and CO2 is liberated.

Question 14.
How is CO2 used so that balance is maintained?
Answer:
1. CO2 is used during photosynthesis and carbohydrates are formed.
2. Used as carbonates by marine molluscs.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 16.
How is temperature regulated on earth?
Answer:
Atmosphere regulates the temperature on earth.

Question 17.
What is the minimum and maximum temperature on moon?
Answer:
Minimum temperature = – 190°C
Maximum temperature = 110°C.

Question 18.
What causes wind?
Answer:
Movement of air caused by uneven heating of the atmosphere in different regions of earth.

Question 19.
How are clouds formed?
Answer:
Clouds are formed by condensation of water droplets in the air.

Question 20.
Which wind gets hot : Water to earth surface or from surface of earth to upward.
Answer:
Land (surface of earth) to upward.

Question 21.
During night what is the direction of movement of air?
Answer:
From surface of earth (land) to the sea.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 22.
What is the cause of movement of wind?
Answer:
Interaction of atmospheric components.

Question 23.
List two factors which affect wind.
Answer:
1. Rotation of earth
2. Presence of mountain heights.

Question 24.
What is deforestation?
Answer:
Cutting of trees on large scale is called deforestation.

Question 25.
What is the effect of deforestation?
Answer:
Deterioration of atmosphere.

Question 26.
Is air a good or bad conductor of heat?
Answer:
Air is a bad conductor of heat.

Question 27.
What is the cause of rain on Indian Land.
Answer:
Rain in India is due to monsoon from south-west or east-west direction.

Question 28.
What is smog?
Answer:
Smoke mixed with fog is called smog.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 29.
What does the smog indicate?
Answer:
It indicates pollution of air.

Question 30.
Where is the purest form of water available.
Answer:
Snow/Ice caps.

Question 31.
Write one importance of water for living organisms.
Answer:
All cellular processes take place in water medium in living organisms.

Question 32.
Write one cause of pollution of water in town.
Answer:
Sewage.

Question 33.
What is soil?
Answer:
The top weathered part of earth’s surface is called soil.

Question 34.
What is the role of sun in the formation of soil?
Answer:
Heating of rocks causes cracking and ultimately breaking up into smaller pieces.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 35.
What is the role of wind in soil formation?
Answer:
Wind causes erosion of rocks.

Question 36.
What is soil erosion?
Answer:
Removal of useful components from the soil which affect the fertility of soil is called soil erosion.

Question 37.
What is the importance of soil?
Answer:
Soil supports terrestrial plants and animals and it decides the diversity of life in an area.

Question 38.
How is soil formed?
Answer:
Soil is formed by weathering of rocks.

Question 39.
What are three kinds of water sources?
Answer:
Rainwater, Groundwater and Subsoil water.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 40.
Write one advantage of seawater.
Answer:
It is a house of table salt (common salt).

Question 41.
What are the two types of water resources?
Answer:
1. Freshwater resources.
2. Saltwater (sea) resources.

Question 42.
What are the sources of freshwater?
Answer:

  1. Rainwater
  2. Surface water
  3. Groundwater
  4. Polar ice caps
  5. Ponds and Pools

Question 43.
What do you understand by aquatic habitat?
Answer:
Organisms which are found in water possess aquatic habitat.

Question 44.
How much percent of nitrogen is present in atmosphere?
Answer:
78%.

Question 45.
How is nitrogen used in living organisms.
Answer:
Nitrogen is a component of proteins, nucleic acid (DNA and RNA).

Question 46.
Name two plants which are capable of fixing atmospheric nitrogen.
Answer:
Green pea and other leguminous plants.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 47.
Write two uses of mirco-organisms.
Answer:
Micro-organisms act as biofertilizers. They also produce antibiotics.

Question 48.
Name any two natural cycles operating in nature.
Answer:
1. Carbon cycle
2. Nitrogen cycle

Question 49.
Name the gaseous components of biosphere.
Answer:
CO2, O2 and Nitrogen.

Question 50.
Name the source of energy for the process of photosynthesis.
Answer:
Solar energy.

Question 51.
Define biomass.
Answer:
The total weight of a living organism.

Question 52.
Name two nitrifying bacteria.
Answer:
Nitrosomonas and Nitrobacter.

Question 53.
Define pollution.
Answer:
Pollution is an undesirable change in physical, chemical or biological characteristics of air, water or land caused by pollutants.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 54.
What are pollutants?
Answer:
The substances causing pollution are termed pollutants.

Question 55.
What are the three major types of pollution?
Answer:
Water, air and soil pollution.

Question 56.
What is soil pollution?
Answer:
Soil pollution is removal of useful components from soil and addition of other substances which adversely affect the soil is termed soil pollution.

Question 57.
Which part of solar radiation is absorbed by ozone layer?
Answer:
UV rays.

Question 58.
Name the major surface water pollutant from farm run off and bathroom water.
Answer:
Phosphorus.

Question 59.
Give the source of pathogens in the water.
Answer:
Domestic sewage.

Question 60.
What is the source of aerosols?
Answer:
Aeroplanes.

Question 61.
Which term is used for pollutants that are degraded by natural means?
Answer:
Biodegradable.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Question 62.
How can SO2 pollution of air be checked?
Answer:
By using sulphur-free fuel in automobiles.

Question 63.
Mention the regions where rainfall is highest and lowest in India.
Answer:

  • Minimum rainfall: Arid region having rainfall of 20 to 50 cm.
  • Maximum rainfall: Wet region having rainfall of more than 200 cm.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Punjab State Board PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Long Answer Type Questions:

Question 1.
Discuss the modes of transmission of diseases.
Answer:
Transmission of Communicable Diseases: Diseases are transmitted from the reservoir of susceptible persons in tire following ways:
1. Direct transmission:
(a) Direct contact between infected and healthy person, e.g. chickenpox, leprosy.
(b) Droplet infection from sneezing, coughing, spitting and talking. e.g. T.B, Whooping cough.
(c) Contact with soil which contains saprophytic disease-causing agents.
(d) Bite of an animal, e.g. Rabies, by bite of rabid dog.
(e) Transplacental transmission (from mother to foetus).

2. Indirect transmission:
1. Air-borne diseases such as common cold, pneumonia and tuberculosis. Such disease causing microbes are spread through the air. This occurs through little droplets thrown out by an infected person who sneezes or coughs. Someone standing closeby can breathe in these droplets, and the microbes get a chance to start a new infection.

2. Water-borne diseases such as cholera, diarrhoea, jaundice. Infectious diseases can be spread through water. This occurs if the stool or other wastes from person suffering from an infectious intestinal disease, gets mixed with the drinking water used by people living nearby. The cholera-causing microbes will enter new hosts through the water they drink and cause disease in them.

3. Sexually-transmitted diseases such as Syphilis and AIDS. Both of these pathogens are transmitted by sexual contact from one partner to the other. Such sexually transmitted
diseases are not spread by casual physical contact. Casual physical contacts include handshakes or hugs or sports as wrestling or by any of the other ways in which we touch each other socially.

4. Spread of disease through vectors. Many animals which live with us may carry diseases. These animals carry the infecting agents from a sick person to another potential host. These animals act as intermediate host and are called vectors. Mosquitoes (Anopheles) are vector of a disease called malaria and dengue fever.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 2.
Show the common modes of transmission of diseases.
Answer:
Modes of transmission of diseases
PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill 1

Question 3.
What are general ways of preventing infectious diseases?
Answer:
General ways of preventing infectious diseases:
The general ways of preventing infections mostly relate to preventing exposure. Public hygiene is one basic key to the prevention of infectious diseases.

The following practices are adopted in this method of prevention of diseases:

  1. For air-borne microbes, we can prevent exposure by providing living conditions that are not overcrowded.
  2. For water-borne microbes, we can prevent exposure by providing safe drinking water.
  3. For vector-borne infections, we can provide clean environment. Such a clean environment would not allow mosquito breeding.

Question 4.
Explain acute and chronic diseases.
Answer:
Acute and Chronic diseases: The manifestations of disease will be different depending on a number of factors. One of the most obvious factors that determine how we perceive the disease is its duration. Some diseases last for only very short periods of time, and these are called acute diseases.

The common cold lasts only a few days. Other ailments can last for a long time, even as much as a lifetime and are called chronic diseases. An example is the infection causing elephantiasis, which is very common in some parts of India.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 5.
Make a table showing organ specific and tissue-specific manifestation of diseases.
Answer:
Organ-specific and tissue-specific manifestations depend on the target organ which the microbes target after their entry. They are as follows:

Target organ Specific manifestation
1. Lungs Cough, breathlessness, chest pain and may be bloody sputum as in TB and lung cancer.
2. Liver Inflammation of liver cells leading to jaundice characterized by yellowness of skin and eyes as in

Hepatitis.

3. Intestine Inflammation of intestinal mucosa leading to acute diarrhoea and dehydration as in cholera.
4. Nasal chambers Inflammation of nasal mucosa leading to sneezing, bronchitis, coughing, fever, etc. as in influenza.
5. Brain Headaches, vomiting, fits or unconsciousness.

Question 6.
Expand AIDS. Explain causes/modes of transmission, effects, incubation period, diagnosis, symptoms and preventive measures of AIDS. What is the significance of 1st December?
Answer:
AIDS (Acquired Immuno-Deficiency Syndrome)
(a) Cause. AIDS is caused by a retro-virus-HIV (Human immuno-deficiency virus).
In India, it was first reported in the prostitutes of Chennai in 1986. It is a pandemic disease.

(b) Epidemiology (Transmission). Human infection occurs through:

  1. Unprotected sexual intercourse,
  2. Use of contaminated syringes,
  3. Blood transfusion,
  4. Organ transplantation,
  5. Common razor of the barbers, etc.

Effects: It causes damage, decrease in number of platelets, swollen lymph nodes, ritght sweats, loss of memory, etc. It is a 100% fatal disease.

(c) Incubation period is of about 28 months.

(d) Diagnosis by ELISA test and Western Blot test.

(e) Symptoms: HIV kills the Helper T-lytnphocytes. It is characterized by following symptoms :

  1. Prolonged fever.
  2. Swollen lymph nodes.
  3. Weight loss and loss of appetite.
  4. Unexplained bleeding.
  5. Loss of memory and mental ability.
  6. Patient becomes susceptible to other infectious diseases.
  7. Night Sweats.

(f) Preventive measures: Involves educating the high risk groups use of disposable syringes; screening tests of blood, organs, semen, etc; monogamous relationship avoiding prostitution, polygamy and heterosexuality, using condoms, avoiding the use of common razor; etc.

(g) Treatment: Drugs like AZT (Azidothymidine), TIAS injection and Protease inhibitors, etc. are known to suppress AIDS virus. Efforts are on for a vaccine against the virus. Ist December is observed as World AIDS Day.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 7.
Make a list of common communicable diseases caused by bacteria, viruses, fungi, protozoans and helminthes.
Answer:
Common communicable diseases. The following table shows some important diseases caused by bacteria, fungi and protozoa.

Name of causative agent Diseases
1. Bacteria Pneumonia, Tetanus, Tuberculosis, Cholera, Food Poi­soning, Sexually transmitted diseases.
2. Viruses Chickenpox, poliomyelitis, Influenza, AIDS.
3. Fungi Skin diseases, Food poisoning.
4. Protozoans Malaria, Kala-Azar, Amoebic dysentery, and African sleeping sickness.
5. Helminthes Taeniasis, Cysticercosis, Ascariasis, Elephantiasis.

Short Answer Type Questions:

Question 1.
Define disease and health?
Answer:

  • Disease: It is defined as a condition of the body or a part of it in which functions are disturbed. The word disease means lack of ease and comfort.
  • Health: It is a state of complete physical, mental and social well-being. Health is also linked with social environment and cultural background.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 2.
What precautions could you take in your school to reduce the incidence of infectious diseases?
Answer:

  1. Providing safe and clean drinking water.
  2. Providing clean environment to prevent vector-borne infections.
  3. Adopting public hygienic measures.

Question 3.
Discuss the principle of prevention of communicable disease.
Answer:
Principles of Prevention: Following three limitations are normally confronted while treating an infectious disease:

  • If a person has a disease, his/her body functions are damaged and may never recover completely.
  • Treatment will take time, which means that person suffering from a disease is likely to be bedridden for sometime even if given proper treatment.
  • The person suffering from an infectious disease can serve as the source from where the infection may spread to other people. It is because of such reasons that prevention of diseases is better than their cure.

Question 4.
Differentiate between infectious and non-infectious diseases.
Answer:
Differences between infectious and non-infectious diseases

Characters Infectious diseases Non-infectious diseases
1. Transmission Can be transmitted from an infected person to a healthy person. Cannot be transmitted from an infected person to a healthy person.
2. Causative agents Microorganisms called pathogens. Deficiency of nutrient or hormone: or degeneration of tissue or hypersen­sitivity of body or tumour formation.
3. Nature They are brought about by extrinsic factor. They are brought about by intrinsic factors.
4. Examples…….. Typhoid, Cholera, T.B., AIDS, Malaria etc. Diabetes, Kwashiorkor, Marasmus, Goitre, Cancer, Allergy etc.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 5.
What are congenital diseases? Name the three types. Give examples.
Answer:
Congenital diseases: These are inborn diseases present since birth caused by gene or chromosomal mutation.
Types of congenital diseases:

  1. Diseases caused by gene mutation, e.g. Haemophilia, colorblindness, alcaptonuria and sickle cell anaemia.
  2. Diseases caused by chromosomal mutation, e.g. Down’s syndrome, Klinefelter’s syndrome, Turner’s syndrome.
  3. Diseases caused by environmental factors such as radiations or pollutants. They are non-inheritable,

Question 6.
List various modes of direct transmission of diseases, giving one example of each type.
Answer:

  1. By direct contact with an infected person e.g. Leprosy, chickenpox etc.
  2. By droplet infection e.g. Diphtheria, tuberculosis, etc.
  3. By contact with soil e.g. bacterial cysts of tetanus.
  4. By animal bite e.g. rabies viruse.
  5. Transplacental transmission e.g. viruses of German measles and AIDS.

Question 7.
What are sources of diseases? Name various sources of diseases, giving one example of each.
Answer:
1. Sources of diseases are those sites which are occupied by the pathogens before entering inside the human body. These are also called reservoirs of infection.
2. Types of sources of diseases:
(a) Carriers or vectors e.g. Plasmodium (female Anopheles).
(b) Soil e.g. bacterial cysts of Clostridium tetani.
(c) Air e.g. bacterial cysts of TB.
(d) Food and water e.g. bacterial cysts of Cholera.

Question 8.
Differentiate symptoms and signs.
Answer:
Differences between symptoms and signs:

Symptoms Signs
1. They indicate presence of disease.

2. They are collective manifestations of a number of diseases of a particular part or organ.

1. They provide information about the presence of particular diseases.

2. They are distinct for different diseases.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 9.
Differentiate acute diseases and chronic diseases.
Answer:
Differences between acute diseases and chronic diseases:

Acute diseases Chronic diseases
1. They occur very rapidly but for only short period.

2. Do not cause major effects on general health.

3. Examples: Common cold. Cough.

1. These diseases last for a long time and could be dangerous.

2. They have prolonged and major effects on general health.

3. Examples: T.B., Cancer, Diabetes, Arthritis.

Question 10.
How do antibiotics function?
Answer:
Action of antibiotics: Antibiotics commonly block biochemical pathways important for bacteria. Many bacteria, for example, make a cell-wall to protect themselves. The best illustration is action of penicillin, ft blocks the bacterial processes that build the cell-wall. As a result, the growing bacteria become unable to make cell-walls, and die easily.

Human cells don’t make a cell-wall anyway, so penicillin cannot have such an effect on us. Penicillin will have this effect on any bacteria that use such processes for making cell-walls. Similarly, many antibiotics work against many species of bacteria rather than simply working against one.

Question 11.
Why do antibiotics not work against viral infection?
Answer:
Viruses do not have their own metabolic pathways at all, and that is the reason why antibiotics do not work against viral infections. In case of common cold, taking antibiotics does not reduce the severity or the duration of the disease.

Question 12.
What is hydrophobia? How does it occur in man? Why is it called a neurotrophic disease?
Answer:

  1. Hydrophobia is another name of a viral disease called Rabies.
  2. It is caused by a RNA-virus, Rabies vires, which is injected in the human being along with saliva of rabid animals like dogs, cats, monkeys etc.
  3. Because the virus damages the motor neurons of brain and spinal cord, therefore causes paralysis and death.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 13.
What are vaccines?
Answer:
Vaccines are produced by deliberate infection of animals, recombinant DNA techniques coupled with hybridomas have opened up the way for custom-made monoclonal antibodies for preventive and therapeutic use. The vaccination prepares the body to fight against the attack.

Question 14.
Give a few examples of vector-borne diseases.
Answer:
Vector-borne diseases:

Vector Disease
Tse Tse fly (Glossina)

Sandfly (Phlebotomus)

Female mosquito (Anopheles)

Rat flea (Xenopsilla)

Aedes mosquito

African sleeping sickness

Kala-azar and Oriental sore

Malaria

Bubonic plague

Yellow fever, Dengue.

Question 15.
Write the name of causal organism of the following diseases:

  1. Malaria
  2. Rabies
  3. Influenza
  4. Tuberculosis
  5. Typhoid

Answer:

Disease Casual Organism
1. Malaria

2. Rabies

3. Influenza

4. Tuberculosis

5. Typhoid

Plasmodium vivax

Rabies-virus

Myxovirus influenzae

Mycobacterium tuberculosis

Salmonella typhosa

Question 16.
Draw simple diagrams to show the structure of Staphylococcus, Heliobacter, SARS, Leishmania and Trypanosoma.
Answer:
Structure of disease-causing agents
PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill 2
PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill 3

Question 17.
Name the infectious disease that leads to immunodeficiency and wasting of body parts. Give the scientific name of the pathogen causing the disease and mention the body organs it primarily affects.
Answer:

  1. AIDS is characterized by immunodeficiency and wasting of body parts.
  2. It is caused by Human Immunodeficiency Virus (HIV).
  3. HIV attacks helper T-lymphocytes, so causing cell-mediated immunodeficiency, so making the body more prone to various infections.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 18.
Which disease is called epidemic jaundice? List its main symptoms? How can it be prevented?
Answer:

  1. Epidemic jaundice is commonly called Hepatitis-A.
  2. It is characterized by yellowing of skin, urine and stool due to damage of liver cells and overproduction of bilirubin.
  3. It can be prevented by proper sanitation, use of boiled or ozonised water and intramuscular injection of human immunoglobulins.

Question 19.
Explain the differences between active and passive immunization.
Answer:
Differences between active immunization and passive immunization:

Active Immunization Passive Immunization
1. Antigens are introduced from outside which trigger off the formation of antibodies in the body. 1. Ready-made antibodies are introduced into the body.
2. It does not provide immediate relief. 2. It provides immediate relief.
3. Immunity thus achieved is long-lasting. 3. It is not long-lasting.

Question 20.
How is health at risk in a cyclone?
Answer:
Health is at risk in case of cyclone because:

  1. Social environment is disturbed as it is an important factor in case of individual health.
  2. Garbage collected in places is source of multiplication of microbes and breeding place for various vectors.
  3. Stagnant water will provide breeding surface for mosquitoes and other such disease spreading agents.

Question 21.
Show by simple diagram how airborne diseases are easier to catch the person who is near the infected person.
Answer:
PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill 4
Fig. Air-transmitted diseases are easier to catch the closer we are to the infected person. However, in closed areas, the droplet nuclei recirculate and pose a risk to everybody. Overcrowded and poorly ventilated housing is therefore a major factor in the spread of airborne diseases.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 22.
Draw a diagram showing structure of HIV Virus.
Answer:
Structure of HIV Virus
PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill 5

Question 23.
What is meant by hydrophobia (rabies)? Write its four symptoms. Suggest four preventive measures to check this disease.
Answer:
Rabies: It is caused by bite of rabid or mad dog and other rabid animals.
Causative agent: Rabies virus present in saliva of dog.
Symptoms:

  1. High fever.
  2. Severe headache.
  3. Painful contraction of muscles of throat and chest.
  4. Fear of water.

Preventive measures:

  1. Wound should be cleaned.
  2. Immunise dogs and cats.
  3. Kill highly rabid dogs.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 24.
A person is suffering from chest pain, breathlessness, loss of body weight, persistent cough and produces blood stained sputum.

  1. Name the disease and its causative agent.
  2. Mention two means of its transmission.
  3. Name the vaccine used to prevent this disease.
  4. Who discovered this disease?

Answer:

  1. Person is suffering from pulmonary tuberculosis: Causative agent Bacterium namely Mycobacterium tuberculosis.
  2. Modes of transmission: It is a communicable disease. Droplet infection during sneezing or otherwise.
  3. BCG vaccine can prevent TB.
  4. Robert Koch (1882).

Very Short Answer Type Questions:

Question 1.
Health and diseases are complex problems in which group of organisms?
Answer:
Human.

Question 2.
List four special activities which occur in human body?
Answer:

  1. Heartbeat
  2. Breathing with lungs
  3. Working of brain
  4. Excretion in kidneys.

Question 3.
What will happen due to malfunctioning of kidney?
Answer:
Toxic substances will accumulate in the body.

Question 4.
Why is food necessary?
Answer:
Food is necessary for cell, tissue functions and maintenance.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 5.
What is health?
Answer:
Health is a state of being well enough to function well physically, mentally and socially.

Question 6.
List two other factors which affect health.
Answer:
Personal and community issues both matters for health.

Question 7.
Define disease.
Answer:
Any condition which interferes with the normal functions of the body and impairs the health. It literally means being uncomfortable.

Question 8.
What do you mean by symptoms of a disease? Give example.
Answer:
Symptoms of a disease are things we feel as being wrong, e.g. headache, cough, loose motions etc.

Question 9.
What are the signs of disease?
Answer:
Signs indicate a little more definite indications of presence of disease.

Question 10.
What are two types of diseases on the basis of duration?
Answer:

  • Acute diseases
  • Chronic diseases

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 11.
What do you understand by acute diseases?
Answer:
The diseases which last only for a short period of time.

Question 12.
Define chronic diseases.
Answer:
Diseases which last for long time, even life time are called chronic diseases.

Question 13.
Give two examples of acute diseases.
Answer:
1. Cough and cold
2. Flu

Question 14.
Write example of chronic disease.
Answer:
Elephantiasis.

Question 15.
Which kind of diseases are more harmful to the body?
Answer:
Chronic diseases.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 16.
What is the cause of dysentary?
Answer:
Contaminated food and water.

Question 17.
Name two types of diseases on the basis of their occurrence.
Answer:

  1. Congenital diseases
  2. Acquired diseases

Question 18.
Why are communicable diseases called infectious diseases?
Answer:
Because these are caused by the infection and multiplication of some kind of micro-organisms like bacteria, viruses etc.

Question 19.
What are congenital diseases?
Answer:
Diseases present in the body from the birth. They are mostly hereditary disorders.

Question 20.
Name a disease which is no longer chronic disease.
Answer:
Peptic ulcer.

Question 21.
Name the bacterium responsible for peptic ulcer.
Answer:
Helicobacter pylori.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 22.
Who discovered that Helicobacter pylori causes peptic ulcer and were awarded Nobel prize?
Answer:
Robin Warren and Barry Marshall.

Question 23.
Name a few disease-causing microbes.
Answer:
Viruses, Bacteria, Fungi and some Protozoans.

Question 24.
Name any four diseases caused by bacteria.
Answer:
Typhoid, Cholera, Tuberculosis (TB), Anthrax.

Question 25.
List a few common diseases caused by viruses.
Answer:
Common cold, Influenza, Dengue fever and AIDS.

Question 26.
Write examples of protozoanal diseases.
Answer:
Malaria, Kala-azar, Amoebic dysentary.

Question 27.
Name a cutaneous disease caused by fungi.
Answer:
Ringworm.

Question 28.
Name two 100% fatal diseases.
Answer:
Rabies and AIDS.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 29.
What is the common name of influenza? Give its causative agent.
Answer:
Influenza is commonly called flu. It is caused by Myxovirus influenza virus.

Question 30.
How does Penicillin acts as a useful antibiotic?
Answer:
Penicillin blocks the pathway that build the cell wall as a result growing bacteria are unable to form a cell wall.

Question 31.
Why antibiotic do not affect cough and cold?
Answer:
Cough and cold are mostly caused by viruses and antibiotics fail to act on viruses.

Question 32.
Name three sexually transmitted diseases.
Answer:
AIDS, Syphilis, Gonorrhea.

Question 33.
Sexually transmitted diseases are not spread by which factors?
Answer:
Handshake, embracing, wrestling.

Question 34.
How do bacteria causing T.B. reach lungs?
Answer:
Through nose during breathing.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 35.
How do bacteria causing typhoid enter body.
Answer:
During intake of contaminated food and water.

Question 36.
What is the cause of jaundice?
Answer:
Hepatitis virus.

Question 37.
Who proposed the name malaria from bad air?
Answer:
Macculoch (1827).

Question 38.
Why is rabies called a neurotrophic disease?
Answer:
Because the toxins of Rabies-virus damage the motor neurons of the brain.

Question 39.
Expand the term AIDS.
Answer:
Acquired Immuno Deficiency Syndrome.

Question 40.
Give the full form of HIV.
Answer:
Human-Immuno Deficiency Virus.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 41.
What is the significance of December 1?
Answer:
World AIDS Day falls on December 1.

Question 42.
Which disease is characterized by yellowing of skin?
Answer:
Hepatitis (epidemic jaundice).

Question 43.
Name three modes of transmission of AIDS.
Answer:

  1. Sexual intercourse with infected partner
  2. Use of contaminated syringes and
  3. Contaminated blood transfusion.

Question 44.
Who prepared the first vaccine?
Answer:
Edward Jenner.

Question 45.
Why is it difficult to make antiviral substances?
Answer:
Viruses do not have their own biochemical pathways, instead they utilize the machinery of cells they attack, therefore, it is difficult to make antiviral substances.

Question 46.
Define vectors.
Answer:
Vectors: They are organisms which spread the disease-causing agents from infected person to a healthy person.

Question 47.
Encephalitis attack which organ of body?
Answer:
Brain.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 48.
List a few symptoms of encephalitis.
Answer:
Headache, Vomitting, Unconsciousness.

Question 49.
Name a disease which does not occur after one attack?
Answer:
Small-pox.

Question 50.
Which disese have been eliminated from the world.
Answer:
Small pox.

Question 51.
List some common modes of spread of disease.
Answer:

  1. Direct contact
  2. Air
  3. Indirect contact
  4. Insect bites
  5. Contaminated food and water
  6. Rabid animal.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Question 52.
Name the two ways of preventing diseases.
Answer:

  • General way is preventing exposure.
  • Strong immune system.

Question 53.
Name a few diseases for which vaccines are available.
Answer:
Whooping cough, Diphtheria, Measeles, Polio and Tuberculosis.

Question 54.
What is immune system?
Answer:
The system of animal body, which protects it from various infectious agents and cancer is termed immune system.

Question 55.
List two features for individual health.
Answer:
Good economic conditions of jobs and stress-free life are needed for individual health.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Punjab State Board PSEB 9th Class Science Important Questions Chapter 12 Sound Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Long Answer Type Questions:

Question 1.
(A) Name two different types of waves.
(B) Give an experiment to explain the formation of transverse waves.
(C) Define transverse waves.
(D) What should be the conditions for the production of transverse waves?
(E) Give examples of transverse waves.
(F) Define crest and trough.
Answer:
(A) Types of Waves: Waves are classified according to the direction of vibration of particles of the medium. It can be either in the direction parallel to the direction of propagation of wave or in a direction perpendicular to the direction of propagation. In this way waves are of two types:

  • Transverse waves
  • Longitudinal waves.

(B) Formation of Transverse waves. To understand the formation of transverse waves attach one end of a long string to the hook fixed in the wall as shown in the figure, Hold the other end of the string in your hand. The coloured threads of length 10 cm each with the string at equal distance as shown in the figure.
image
Now give jerk to the slinky. An upward maund or hump is formed in the string which travels along the string towards the fixed end. Such sudden disturbance that lasts for short duration is called pulse i.e. the particles of string move along with the disturbance in the perpendicular direction.
PSEB 9th Class Science Important Questions Chapter 12 Sound 1
As shown in fig. (c) if we continuously give up and down jerks to the free end of the string a number of waves begin to travel along the string forming a wave train. Each part of the string vibrates up and down while the waves travel along the string. So the waves in the string are transverse in nature.

The points (c, c, c …………… ) of maximum displacement in the upward direction are called crests and the points (T, T, T…………) of maximum displacement in the downward direction are called troughs.

(C) Transverse Wave: These are the waves in which particles of the medium vibrate (up and down) in a direction perpendicular to the direction of propagation of wave.

(D) Conditions necessary for formation of transverse waves:

  1. The medium should have property of inertia.
  2. The medium should have property of elasticity so that the particles can come back to their original positions after being disturbed.
  3. The medium should have minimum frictional force between its particles so that the particles may keep vibrating for a long.
  4. Vibrations of plucked stretched string of a violin.

(E) Examples of Transverse Waves.

  1. In loose string or spring: If a string is held in hand and the other end is tied to a fixed support and it is continuously moved up and down then transverse waves are produced.
  2. Wave on the surface of still water: If a pebble is dropped gently on the surface of still water of a pond then ripples are produced on the water surface.
  3. A cork floating on the surface of water would then begin to vibrate up and down then transverse waves are produced. The cork is not displaced with the waves but keeps tossing up and down,

(F) Crests and Troughs:

  • Crests: In transverse waves, the particles of the medium which have maximum displacement in the upward positive direction along Y-axis are called crests.
  • Troughs: In transverse waves, the particles of the medium which have maximum displacement in the downward negative direction along Y-axis are called troughs.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 2.
(A) Define longitudinal wave.
(B) Arrange an experiment to demonstrate the formation of longitudinal wave.
(C) In reference to longitudinal wave, define compressions and rarefactions.
Answer:
(A) Longitudinal wave. Those waves in which the particles of the medium vibrate about their mean position in the direction of propagation of disturbance are called longitudinal waves and the wave motion is called the longitudinal wave motion. Sound in air gets propagated in the form of longitudinal wave motion consisting of regions of compressions and rarefactions.
PSEB 9th Class Science Important Questions Chapter 12 Sound 2
(B) Experiment: Consider, a tuning fork struck gently with a rubber pad so that its prong begins to vibrate [Fig. (a)]. As prong moves towards right, it compresses the layer of air in contact with it. As air has elasticity, the compressed air tends to relieve itself of its strain and moves towards the right to compress the next layer and so on.

Thus, a wave of compression moves towards the right. At the point of compression, there is an increase of pressure and is shown in form of crest C. At the point of rarefaction of concentration of particles is least and has been shown as trough R. When the prong moves towards left, a region of reduced pressure or rarefaction is produced towards right [Fig. (b)].
Examples:

  • Hearing in man,
  • Vibrating tuning fork,
  • Beating diaphragm of drum.

(C) Compressions and Rarefactions:
Compression: The region of high pressure in the longitudinal wave so that the particles of the medium are closer to each other than the normal distance between them. The higher the pressure, the higher is the number of particles per unit volume i.e. higher is the density, is called compression. In fig it is denoted by ‘c’.

Rarefaction: The region of low pressure in the longitudinal wave so that the particles of the medium are far away from each other than the normal distance between them, is called Rarefaction. In fig it is denoted by ‘R’.

Question 3.
Establish the relation between wave velocity, wavelength and frequency of a wave.
Solution:
Suppose
υ = Wave velocity
ν = Frequency of the wave (i.e. frequency of vibrating particles of the medium)
λ = Wavelength of wave.
T = Time period of a vibrating particle (i.e. time taken by particle of the medium to complete 1 vibration)
Distance travelled by the wave during T seconds = λ
∴ Distance covered by the wave in unit time (1 s) = \(\frac{\lambda}{\mathrm{T}}\)
But distance covered in unit time is wave velocity
∴ υ = \(\frac{\lambda}{\mathrm{T}}\)
or υ = \(\frac {1}{T}\) × λ
∴ υ = ν × λ [∵ \(\frac {1}{T}\) = ν]
i. e. Wave velocity = Frequency × Wavelength
This relation is true both for longitudinal and transverse waves.

Question 4.
Distinguish between sound waves and light waves.
Answer:
Difference between sound waves and light waves

Sound Waves Light Waves
1. Sound waves are mechanical waves. Light waves are electromagnetic waves.
2. Sound waves are longitudinal waves in which the direction of vibration of particles of the medium is same as that of propagation of wave. Light waves are transverse waves in which the vibration of the particles of the medium is in a direction perpendicular to the direction of propagation of wave.
3. Sound waves can not travel through vacuum. These require some material medium. Light waves can travel through vacuum.
4. Speed of sound waves in air is 340 m s-1. Speed of light waves in air is very large. It is 3 × 108 m s-1.
5. Sound waves are produced due to vibrations of particles of related medium. Light waves depend upon the change in electromagnetic fields.
6. Sound waves have low frequency and large wavelength. Light waves have high frequency and small wavelength.
7. Sound waves cannot be polarised. Light waves can be polarised.
8. Sound waves produce effect on our ears. Light waves produce effect on our eyes.
9. The velocity of sound waves is independent of the wavelength of wave. The velocity of lightwave depends upon the wavelength of the wave.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 5.
Explain the classification of sound waves on the basis of frequency range.
Answer:
1. Audible waves: Those sound waves to which human ears can respond (i.e. those sounds waves which can be heard by human beings). In human beings the audible range is from 20 Hz to 20,000 Hz. As the human being advances in age his ears become less sensitive to sounds of high frequencies. These waves are produced by vibrations in air column, tuning fork and violin.

2. Ultrasonic waves: The sound waves having frequency above 20 KHz (i.e. 20,000 Hz) are called ultrasonic waves or ultrasonics. Insects of some species can hear such sound waves. Bats, dolphins, etc. produce such sound waves.

3. Infrasonic waves: Those sound waves which have frequency less than 20 Hz are called infrasonic waves or infrasonics. Whale and elephants produce infrasonic waves. These waves are produced before the main high frequency waves of earthquake occur on hearing these waves the animals become terrorised and become impatient.

Question 6.
What are the laws of reflection of sound? How will you prove these laws experimentally?
Answer:
Like light, sound also obeys the laws of reflection these are:
1. The angle made by incident sound and reflected sound with the normal to the reflecting surface at the point of incidence are always equal.
i.e. \(\angle i=\angle r\)
2. The incident sound, the normal at the point of incidence and the reflected sound all lie in the same plane.
PSEB 9th Class Science Important Questions Chapter 12 Sound 3
Experimental Verification:
Take two cardboard tubes A and B about 1 m long and 5 cm in diameter. Mount the tubes as shown in figure facing metal plate as shown in fig. Place a watch at the mouth of the tube A and try to hear the sound by applying ear close to the end of the tube B.

Place a screen S made of cardboard or of some other absorbing material in between the two tubes to prevent sound from reaching our ear directly. It will be observed that the sound is maximum when angles made by tubes A and B with normal are equal i.e., \(\angle i=\angle r\)

Question 7.
List the three characteristics of sound waves. State the factors on which each of these characteristics depends.
Answer:
Characteristics of sound: The three characteristics of sound are

  1. Loudness
  2. Pitch and
  3. Quality or timbre.

1. Loudness: It is the response differently i.e., one sound louder than the other of ear to the intensity of sound. It distinguishes between a loud sound and soft (low) sound. Even two sounds of equal intensity, may hear. Loudness depends on two main factors.
Factors on which sound depends:
(a) Intensity of sound
(b) Sensitivity of ear.
Graphs given below show the wave shape of a loud sound and a soft sound having the same frequency.
PSEB 9th Class Science Important Questions Chapter 12 Sound 4

2. Pitch: Pitch is the sensation which helps a listener to distinguish between a high and a low note. Pitch depends on frequency. The faster the vibration of the source of sound, the higher is the frequency and higher is the pitch,
PSEB 9th Class Science Important Questions Chapter 12 Sound 5
The voice of a child or a lady is shriller than that of a man i.e., the pitch of a lady’s sound is higher than that of a man. The mosquito’s sound is of high frequency and hence high pitch.

3. Quality or timbre. The quality or timber of sound is that characteristic which helps us, to distinguish one sound from another having the same pitch and loudness. It is due to the quality of sound that one can recognise the voice of friend without seeing him.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Short Answer Type Questions:

Question 1.
What is periodic motion? Give three examples.
Answer:
Periodic motion: The motion of a body that repeats itself regularly after a fixed interval of time is called periodic motion. Such type of motion is vibratory motion.
Examples of periodic motion.

  1. Motion of the earth around the sun.
  2. The motion of a swing which moves to and fro (left and right) about its mean position.
  3. The motion of a simple pendulum.
  4. The motion of the hands of a clock.

Question 2.
Define oscillatory motion. Give examples.
Answer:
Oscillatory or vibratory motion. If a body moves to and fro repeatedly about a fixed position (called mean position), its motion is said to be oscillatory or vibratory motion.
Examples of oscillatory motion:

  1. Motion of the pendulum of a wall clock.
  2. Motion of a swing.

Question 3.
Differentiate between transverse waves and longitudinal waves.
Answer:
Differences between transverse and longitudinal waves:

Transverse waves Longitudinal waves.
1. In transverse waves, the particles of the medium vibrate perpendicular to the direction of wave motion. In longitudinal waves, the particles of the medium vibrate along the direction of wave motion.
2. These waves travel in the form of alternate crest and troughs. These waves travel in the form of alternate compressions and rarefactions.
3. These waves can be transmitted through solids or liquid surfaces. These waves can be transmitted through all the three media, viz (i) solids, (ii) liquids and (iii) gases.
4. They do not cause pressure changes in the medium through which they pass. Example. Waves formed over water surface. They cause changes in the pressure of the different parts of the medium through which they pass. Example. Sound waves in air,

Question 4.
How is sound propagated? Can it be propagated through vacuum? out of solid, liquid and gas in which medium speed of sound is maximum and in which it is least?
Answer:
The propagation of sound is in the form of transverse waves therefore sound waves consist of compressions and rarefactions. The source of sound is always in vibrating state. The sound emitted by a vibrating source can always propagates through a medium.

Sound can not travel through vacuum. Transverse waves can travel through all the three mediums solid, liquid and gas. This happens due to elasticity of the medium. Solids are more elastic than liquids and gases. It has been proved experimentally that the speed of sound is maximum in solids the least in gases.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 5.
What are applications of ultrasound?
Answer:
Applications of ultrasound:

  1. Glaton whistle. It is used by hunters. When hunter and hound (hunting dog) get separated an hunter wants to call back dog to help him catch the prey, he blows the Glaton whistle which produces only ultrasonic waves.
  2. These waves can be heard by dog but not by other animals and birds of the forest.
  3. Bats judge the distance of prey or the coming obstacle by sending these waves. By observing the time taken by waves to travel back, they can find the distance of the obstacle/ prey.
  4. Ultrasound waves are used to find the depth of the sea.
  5. Ultrasonic waves are used by doctors for scanning different parts of the body.
  6. These waves are used by dentists to compress the silver filled in the cavity of teeth.
  7. These are used to clean parts located in hard-to-reach places. Objects are cleaned in cleaning solution and ultrasonics are passed in the solution.
  8. Particles of dust, grease get detached due to high-frequency vibrations and get into cleaning solution.

Question 6.
Define the terms wave and wave motion.
Answer:
Wave and wave motion. A wave is a pattern of disturbance which travels through a medium due to repeated vibrations of the particles of the medium, the disturbance being handed over from one particle to the next. The motion of the disturbance is called wave motion.

Question 7.
Distinguish between a wave pulse and a periodic wave.
Answer:

Pulse Periodic wave
1. A pulse is a wave produced by a sudden disturbance of short duration. A periodic wave produced by continuous and regular vibrations of the particles of the medium.
2. Due to pulse, the medium oscillates for a short time and then returns to its undisturbed position. Due to periodic wave, medium vibrates for a long time after being disturbed.
3. It is not repetitive. It repeats itself after a fixed interval.
4. It is formed in a small portion of the medium. It spreads over the entire length of the medium.

Question 8.
What are mechanical or elastic waves? Give examples.
Answer:
Mechanical waves: The waves which require a material medium for their propagation are called mechanical waves. They are also called elastic waves because their propagation depends on the elastic properties of the medium.

Examples of mechanical waves:

  1. Sound waves in air.
  2. Waves over water surface
  3. Waves produced during an earthquake. These are known as seismic waves.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 9.
State two factors on which the speed of sound depends.
Answer:
The speed of sound through a medium depends on following two factors:

  • Nature of the medium.
  • Temperature of the medium.

Question 10.
Explain in brief the dependence of speed of sound on nature of material medium and temperature.
Answer:
Speed of sound in different media. Sound travels fastest through solids and slowest through gases. This is because elasticity of solids is much greater than that of liquids which in turn, is greater than that of gases.

Effect of Temperature. The speed of sound increases with the increase in temperature of the medium through which sound travels.

Question 11.
Define the terms time period and frequency of an oscillating body. Give their units and write the relation between them.
Answer:
Time period: The time taken by an oscillating body to complete one oscillation is called its time period. It is denoted by T. Its SI unit is second(s).
Frequency: The number of oscillations or vibrations completed by an oscillating body in one second is called its frequency. It is denoted by ν (Greek letter nu).
SI unit of frequency = per second (s-1) = cycles per second (cps)
= hertz (Hz).
Relation between time period and frequency.
Let T = time period of an oscillating body. Then,
number of oscillations completed in “T second = 1
number of oscillations completed in 1 second = \(\frac {1}{T}\)
But number of oscillations completed in 1 second = frequency (ν)
∴ ν = \(\frac {1}{T}\)
Hence frequency is equal to the reciprocal of time period.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves:
1. Reflecting boards. In large halls or auditorium large wooden boards are fixed behind the speaker in the form of a concave cylinder. The sound waves on striking the reflecting boards get reflected parallel to the principal axis reaches everyone in the auditorium so that everyone can hear clearly. The working of reflective board is based on reflection.
PSEB 9th Class Science Important Questions Chapter 12 Sound 6
2. Ear trumpet. The sound energy received by the wide end of the trumpet is reflected into much smaller area thereby increasing the intensity of sound. Hence, a person who is hard of heamg, can hear the sound distinctly.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 13.
Distinguish between the terms music and noise.
Answer:

  • Music: The sound which has a pleasing sensation to the ears is called music. It is produced by regular and periodic vibrations, without any sudden change in loudness.
  • Example: The sound produced by plucking the string of a sitar, sound from a tabla, etc.
  • Noise: The sound which has a displeasing on the ears is called noise. It is produced by vibrations at irregular intervals and with sudden change in loudness.
  • Example: The sound produced by an explosion.

Question 14.
How can bats ascertain distances, directions, nature and size of the obstacles without eyes?
Answer:
Bats can produce and receive ultrasonic waves. During flight, a bat emits ultrasonic waves. The bat receives back these waves after being reflected by the obstacle in its path. From the time interval between transmission and reception of ultrasonic waves, the bat gets information about the distance, nature of obstacle and its direction of location. Hence bats can move about freely even in total darkness.

Question 15.
It is observed that some animals get disturbed before earthquake. How?
Answer:
Earthquakes produce low-frequency infrasound before the mainshock waves begin. These infrasonic waves are out of our audible range. But animals are able to detect these waves and hence some animals get disturbed before earthquakes.

Important Formulae:

  1. Wave Velocity (υ) = ν × λ
  2. Frequency (ν) = \(\frac {1}{T}\)
  3. Wave length (λ) = v × T
  4. Total Distance = Velocity × Time

Numerical Problems (Solved):

Question 1.
What will be the frequency of Mohan’s heart when it beats 75 times in 1 minute?
Solution:
Time taken for 75 heart beats = 1 min
= 60 s
and time taken for 1 heartbeat = \(\frac {60}{75}\)
∴ Time period (T) = 0.80 s
We know, frequency (ν) = \(\frac{\text { 1 }}{\text { Time Period(T) }}\)
= \(\frac {1}{0.80}\)
= 1.25 Hertz (Hz)

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 2.
A boat strikes waves of ocean having crest 100 m away. The wave velocity of crest is 20 m s-1. What is the frequency of waves striking the boat?
Solution:
Wave length (λ) = 100 m
Wave velocity (υ) = 20 m/s
Frequency (ν) = ?
We know, υ = ν × λ
20 = v × 100
or ν = \(\frac {20}{100}\)
= 0.2 Hertz (Hz)

Question 3.
A source of wave produces 40 crests in 0.4 s. Find the frequency of wave.
Solution:
A wave cycle consists of one crest and one trough.
Since the given source produces 40 crests, it produces 40 wave cycles in 0.4 s.
No. of wave cycles produced in 0.4 s = 40
No. of wave cycles produced in 1 s = \(\frac {40}{4}\)
= 100
∴ Frequency of wave = 100 Hz

Question 4.
A source produces a sound of wavelength. 1.7 x 10-2 m. If its velocity is 343.4 m s -1, then find frequency of sound,
Solution:
Velocity, υ = 343.4 ms-1
Wave length, λ = 1.7 × 10-2 m;
Frequency, ν = ?
We know, υ = ν × λ
ν = \(\frac {υ}{λ}\)
= \(\frac{343.4}{1.7 \times 10^{-2}}\)
∴ Frequency, ν = 2.02 × 104 Hz

Question 5.
What will be the frequency of the wave, if its time period is 0.05 s?
Solution:
Here Time Period T = 0.05 s
Frequency, ν = ?
But Frequency, ν = \(\frac{\text { 1 }}{\text { Time Period(T) }}\)
= \(\frac {1}{0.05}\)
= \(\frac {100}{5}\)
= 20Hz

Question 6.
A distance displacement of a periodic wave is shown in a graph, if velocity of the wave is 320 m s-1 , then find (a) wavelength (b) frequency.
Solution:
(a) Wavelength = Distance between two consecutive crests
= 50 – 10
= 40 cm
= \(\frac {40}{100}\)m/s
= 0.4 m

(b) Velocity of wave, υ = 320m/s
Wavelength, λ = 0.4 cm
Frequency, ν = ?
We know, υ = ν × λ
or ν = \(\frac {υ}{λ}\)
= \(\frac {320}{0.4}\)
Frequency, ν = 800 Hz

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 7.
Longitudinal waves is produced on a spring. This wave travels with a velocity of 30 cm/s and its frequency is 20 Hz. What is the minimum distance between two consecutive compressions?
Solution:
Here, wave velocity, υ = 30 cm/s
= 0.30 m/s
Frequency, ν = 20 Hz
Wave length, λ = Distance between two consecutive compressions = ?
υ = ν × λ
or λ = \(\frac {υ}{ν}\)
= \(\frac {0.30}{20}\) = 0.15 m

Question 8.
A message was transmitted from boat which returned to the sender after reflection from the bottom of the sea in 0.8 s. If the velocity of sound in water is 1500 ms-1 then find the depth of sea.
Solution:
Velocity of sound, υ = 1500 ms-1
Time taken, t = 0.8 s
Distance travelled by sound = Velocity of wave × Time
Total distance travelled by sound (2d) = 1500 × 0.8 = 1200 m
∴ Depth of sea = \(\frac {1200}{2}\)
= 600 m

Question 9.
The frequency of a tuning fork is 600 Hertz. What will be its time period?
Solution:
Time period of tuning fork = \(\frac{\text { 1 }}{\text { Frequency of Tuning fork }}\)
= \(\frac {1}{400}\)
= 0.0025 s

Question 10.
A stone is dropped in a 44.1 m deep well. If the sound produced by striking of stone with the water surface is heard after 3.13 s then find the velocity of wave in air.
Solution:
Given : Depth of well(h) = 44.1 m
Acceleration due to gravity (g) = 9.8 m s-2
Time taken by stone go to the surface (t)= 313 s of water and sound to return
Suppose stone takes t1 time to reach the water surface, then from
PSEB 9th Class Science Important Questions Chapter 12 Sound 7

Question 11.
A man claps near a cliff and echo is heard after 5 s. If the velocity of sound is 346 m s-1, then what will be the distance between the man and the cliff?
Solution:
Velocity of sound (v) = 346 m s-1
Time taken for echo to be heard (t) = 5 s
Distance travelled by sound (S) = υ × t
= 346 m s-1 × 5 s
= 1730 m
Sound took 5 s to travel twice the distance between man and cliff
Distance between man and cliff = \(\frac {1730}{2}\)
= 865 m

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 12.
A ship produces ultrasonic sound which is collected in 3.42 s after reflection from the surface of sea. If the velocity of ultrasonics is 1531 m s-1, then what is the distance of sea surface from sea?
Solution:
Time taken from transmission to collection of sound (t) = 3.42 s
Velocity of ultrasonics in sea water (υ) = 1531 m s-1
Distance travelled by transmitted sound= 2d where d is the depth of sea
2d = velocity of sound × time
2d = 1531 m s-1 × 3.42 s
2d = 5236 m
d = \(\frac {5236}{2}\)
= 2618 m
∴ Distance of ship from sea surface (d) = 618 m

Very Short Answer Type Questions:

Question 1.
What is sound?
Answer:
Sound: It is a kind of energy which produces in us the sensation of hearing.

Question 2.
In which medium the velocity of sound is more – Solids or Gases?
Answer:
In solids, velocity of sound is more i.e. sound travels faster in solids than in gases.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 3.
What frequency of sound is audible to human ear?
Answer:
Human ear can hear sound of frequency from 20 Hz to 20,000 Hz.

Question 4.
What is the nature of sound Longitudinal wave or Transverse wave?
Answer:
Longitudinal wave.

Question 5.
What should be the properties of the medium for producing sound waves?
Answer:
The medium should have the property of (i) inertia and (ii) elasticity.

Question 6.
What is the relation between frequency, wavelength and wave velocity?
Answer:
Wave velocity = Frequency × Wavelength.

Question 7.
What is the unit of frequency?
Answer:
Hertz (Hz).

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 8.
What is the relation between frequency (ν) and time period (T)?
Answer:
ν = \(\frac {1}{T}\)

Question 9.
On dropping a pebble in still water, what type of waves are produced on the surface of water?
Answer:
Transverse waves.

Question 10.
What kind of sound waves are produced in air?
Answer:
Longitudinal waves.

Question 11.
What is the full form of SONAR?
Answer:
The full form of SONAR is Sound Navigation and Ranging.

Question 12.
What is seismograph for?
Answer:
Seismograph is a device used to measure intensity of earthquake.

Question 13.
Which scale measures the intensity of earthquake measured?
Answer:
Richter scale.

Question 14.
Earthquake of what intensity is considered safe on Richter Scale.
Answer:
Earthquake of intensity upto 5 on Richter Scale is considered safe.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 15.
What is the cause for production of sound?
Answer:
Vibrations.

Question 16.
What is the time for persistence of hearing?
Answer:
It is \(\frac {1}{10}\) = 0.1 s.

Question 17.
What is the velocity of sound on moon?
Answer:
Sound cannot travel on moon because moon has no atmosphere.

Question 18.
Which animal can hear infrasonics?
Answer:
Elephant.

Question 19.
What is audible range for human beings?
Answer:
20 Hz to 20,000 Hz.

Question 20.
What is the minimum distance of the obstacle from the source of sound for hearing distinct echo?
Answer:
17.2 m.

Question 21.
Which has a higher pitch, whistle or a drum?
Answer:
Whistle has higher pitch.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Question 22.
A violin and a sitar may have the same frequency, yet we can distinguish between their notes. Why?
Answer:
This is on account of the difference in quality (timbre) of sound produced by them.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Punjab State Board PSEB 9th Class Science Important Questions Chapter 11 Work and Energy Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Long Answer Type Questions:

Question 1.
What is potential energy? Deduce an expression for the gravitational potential energy of a body of mass’m’ placed at height ‘h’ above the ground. Give some practical examples.
Answer:
Potential Energy: It is the energy possessed by an object when it is raised above or below the surface of earth.
Mathematical Expression: Consider a body of mass ‘m’ raised to a height ‘h’ above the surface of earth to a position ‘B’. In order to lift the body to this height we have to apply minimum force equal to mg, the weight of the body in the upward direction from position ‘A’.
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 1
∴ Work done on the body against the force of gravity (W) = Force × Displacement
W = F × S
But F = mg and S = h
W = mg × h
= mgh

The work done (W = mgh) on the body is against the force of gravity
∴ Ep = W = mgh
Since this work done is against the force of gravity, so this energy is called gravitational potential energy. This potential energy does not depend on the path along which the body is moved.

Practical Examples:
1. In olden days there were no digital watches and the watches used to be wound up. When its spring is open then the watch does not work. When its spring is wound up then it starts working.
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 2
As is shown in Fig. on winding the spring closes and potential energy is stored up. This potential energy due to change in size and shape of the spring helps to move the hands of watch.
2. Try to open a spring with your both hands. Work is, therefore, being done by your hands. By doing this the spring extends in its length and as such potential energy is stored in it.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 2.
What is kinetic energy? Derive a mathematical expression for kinetic energy.
Answer:
Kinetic Energy: It is defined as the energy possessed by a body due to velocity, if body is not in motion then it does not possess kinetic energy.
Mathematical Expression: Suppose a football of mass ‘m’ is at rest and force ‘F’ is applied. Due to the force acting on it, the football covers a distance ‘S’ in’t’ seconds and attains a velocity υ. Thus, acceleration produced in the football is ‘a’
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 3
Work done on the football W = F × S ………..(1)
According to Newton’s second law of motion,
F = m × a …………. (2)
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 4
Equation (8) proves that the work done on the football is stored in it as kinetic energy.
∴ kinetic energy stored = Work done = \(\frac {1}{2}\)mv2

Question 3.
What is law of conservation of energy? Explain the law with the help of an example and prove its reality.
Answer:
Law of conservation of energy: According to this law total energy always remain constant or total energy always remains the same. Though one form of energy can be transformed to some other form of energy but total energy still remains constant.
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 5
Explanation with example: Throw a ball vertically upwards from the surface of earth. You do some work on the ball. This work gets stored in the form of potential energy. This is called gravitational potential energy. As the ball moves up and up its velocity continue to decrease and its kinetic energy continue to decrease and potential energy goes on increasing. When ball reaches the maximum height its kinetic energy becomes minimum i.e. zero but potential energy becomes maximum. We can prove mathematically that at any position in the path of motion of the ball the sum of potential energy and kinetic energy remains same.
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 6
Mathematical Proof for law of conservation of energy:
Consider a ball of mass 10 kg situated at a point 30 m above the surface of earth i.e. point A. Let it be dropped from point A or shown in fig.
At point A
potential energy of ball (P.E.) = m × g × h
= 10 × 10 × 30
= 3000 J
Since, ball is at rest initially thus, at point A it kinetic energy (K.E.) = 0
∴ Total mechanical energy of ball = P.E. + K.E.
= 3000 + 0 = 3000 J ………….(i)

At point B
Ball is above the surface of earth by 20 m.
potential energy (P.E.) = m × g × h
= 10 × 10 × 20
= 2000 J
Using υ2 – u2 = 2gh
υ2 = u2 + 2gh
υ2 = 0 + 2 × 10 × 10
∴ υ2 = 200
∴ Kinetic energy at point B (K.E.) = \(\frac {1}{2}\)mυ2
= \(\frac {1}{2}\) × 10 × (200)
= 1000 J
∴ Total mechanical energy of ball at point B = potential energy + kinetic energy
= 2000 + 1000
= 3000 J. ……………… (ii)
Similarly, at point C and D total mechanical energy will be 3000 J.
These equation prove that total energy is always conserved.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 4.
If the force acting on the object is not in the direction of motion then how will you consider the work done? Explain giving example and also tell when will the work done be minimum and when it will be maximum?
Answer:
When the force acting on the object is not in the direction of motion. A gardener pushes the lawn mower in the forward direction. He applies force F on the handle HH’ facing the ground.
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 7
As is clear from the figure, the gardener is not applying force in the horizontal direction but applies force in a direction making angle θ with the direction of force. In such situation the force acting on the machine drives the roller in the horizontal direction from A to B.
Here the effective force is F cos θ and not F and the vertical component F sin θ balances the weight of the lawn mower.
∴ Work done by the lawnmower (W) = Component of Force × Displacement
= F cos θ × S
1. When the force acts along the direction of displacement then θ = 0° and cos θ = cos 0° = 1
∴ W = F cos θ × S
= F × 1 × S
W = F × S
This time the work done will be maximum.

2. When the force acting on the object is in a direction perpendicular to the direction of motion then θ = 90° and cos θ = cos 90° = 0
∴ W = F cos θ x S
= F cos 90° x S
= F x 0 x S
⇒ W = 0
This time the work done is minimum.

Short Answer Type Questions:

Question 1.
What is work? How can you calculate it? Also give the unit of work.
Answer:
Work: Work done by a force or by an object in the product of applied force and the displacement taking place in the direction of applied force.
If F = force acting on the body
S = displacement of the body in the direction of force
Work done by force, W = F × S ……(i)
we know, when a force acts on a body acceleration is produced in the body.
Thus, if m = mass of body
a = acceleration produced in the body then from Newton’s second law of motion
F = m × a ….(ii)
Using (i) and (ii),
W = m × a × S ………….(iii)

Unit of work: When force is measured in Newton (N) and displacement in metre (m), the
work = Newton × Metre
W = N × m
W = Joule (J)
∴ S.I. unit of work is Joule (J) and C.G.S. unit is erg.
1 Joule = 107 erg.

Question 2.
Show by giving an example that if force acting in the body does not produce any displacement then the work done will be zero?
Answer:
This statement can be understood by the following example.
If a child tries to push a car by applying his, maximum force but the car does not displace even through 1 centimeter, then in the language of physics we can say that the child has done no work.
In this situation suppose the child applies force F and displacement S = 0 then
∴ Work done by the child, W = F × S
= F × 0
⇒ W = 0

Question 3.
A stone tied to one end of string is moved in a circle. How much work is done by the centripetal force in this circular motion?
Answer:
A stone is made to move in a circle as shown the fig. then the finger with which you hold the string will experience some force. The force acting on the moving stone is known as centripetal force.
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 8
This force acts radially inwards towards the centre of the circle. If in this situation the stone gets detached then it will fly off in the direction tangential along AT or BT as shown in fig. Now, the centripetal force acts perpendicular to the direction of motion, thus no work is done by the centripetal force.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 4.
What is power? Write its SI unit also.
Answer:
Power. Harish and Karan climbed up a tree to pluck 60 mangoes each. They started climbing at the same moment of time. Harish got his 60 mangoes in 30 minutes whereas Karan got his 60 mangoes in 60 minutes. That means Karan took more time to do the same work. Both did the same work, both had same energy but their power was not same i.e. Power of Harish is more than that of Karan.

Power: Power of an object or a machine is its time rate of doing work i.e. Power is defined as rate of doing work with time.
∴ Power = \(\frac{\text { Work done }}{\text { time taken }}\)
or P = \(\frac{\text { w }}{\text { t }}\)
If work is measured in joule and time in second then unit of power is ‘watt’.
∴ 1 watt = \(\frac{\text { 1 Joule }}{\text { 1 second }}\)
commercial unit of power is kilowatt.
1 kilowatt = 1000 watt = 1000 joule/second

Question 5.
Prove by an experiment that mechanical energy can be transformed into heat energy.
Or
When we hammer a nail into a wooden block the nail gets heated up? Why?
Answer:
Place a nail on a wooden block and hammer it, some part of the nail goes into the wooden block, when nail is completely into the block due to hammering, when nail in further hammered then nail, hammer and block all gets heated up. When hammer is lifted up for hammering, its potential energy increases due to its position.

When it falls on the nail then its whole energy is transformed into kinetic energy of the nail and the nail goes into the wooden block. When nail is completely embedded into the block then mechanical energy of the hammer converts into heat energy of the nail, block and hammer meaning thereby that mechanical energy gets converted or transformed into heat energy.

Question 6.
Differentiate between Potential Energy and Kinetic Energy.
Answer:
Difference between Potential energy (P.E.) and Kinetic energy (K.E.)

Potential Energy Kinetic Energy
1. The potential energy of an object depends upon its position and size. The kinetic energy of an object is due to its motion or velocity.
2. Potential energy (P.E.) = m × g × h Kinetic energy (K.E.) = \(\frac {1}{2}\)mυ2
3. Potential energy of an object depends upon its height above the ground or its depth below the ground surface. The kinetic energy of an object depends upon its velocity.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 7.
A horse and a dog are running with the same velocity. If the mass of horse is ten times the mass of the dog then what will be ratio of their kinetic energy?
Solution:
Suppose the mass of dog = m
∴ Mass of the horse = 10m
Velocity of horse = Velocity of dog = υ (say)
C:\ch 11.1\PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 9.png

Question 8.
Two masses m and 2m are dropped from height h and 2h. On reaching the ground, which will have a greater kinetic energy and why?
Answer:
K.E. of mass m = P.E. lost by mass m = mgh
K.E. of mass 2m = P.E. lost by mass 2m
= 2m × g × 2h = 4mgh
Hence, mass 2 m will have a greater kinetic energy on reaching the ground.

Question 9.
Two objects having same mass’m’ are moving with velocities υ and 2υ. Find ratio of their kinetic energies.
Answer:
Let bodies A and B each have same mass m and their respective velocities are υ and 2υ
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 10

Question 10.
In what conditions the physical capacity of a man to do work decreases?
Answer:
After sickness and in old age the physical capacity of a man to do work decreases because the energy of his body muscles becomes less.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 11.
Explain clearly the difference between energy and power with the help of an example.
Answer:

  • Energy: Energy is the ability to do work and is equal to the magnitude of total work which could be done by the energy. It has no relation with the time.
  • Power: It is the rate of doing work. It has no relation with the magnitude of total work.
  • Example: A worker taken one hour to complete a piece of work, whereas the other worker takes 2 hours to complete the same piece of work. In this case, both the workers are doing the same work i.e. both are consuming same amount of energy.

But first worker did the same work in half the time as compared to other worker thus. First worker has doubled the power than the other worker.

Important Formulae:

  1. Kinetic Energy = \(\frac {1}{2}\) mυ2
  2. Potential Energy = mgh
    Here m = Mass of the object; g = Acceleration due to gravity, h = Height
  3. Work (W) = Force (F) × Displacement (S)
  4. Power (P) = \(\frac{\text { Work Done (W) }}{\text { Time taken (t) }}\)
    Or Power (P) = \(\frac{\text { F × S }}{\text { t }}\)
    = F × \(\frac{\text { S }}{\text { t }}\) (∵ υ = \(\frac{\text { S }}{\text { t }}\))
    = F × υ
  5. 1 Joule = 1 Newton × 1 metre
  6. 1 watt = \(\frac{\text { 1 Joule }}{\text { 1 Second }}\)
  7. Horse Power = 746 watt
  8. 1 Kilowatt hour = 36,00,000 Joule = 3.6 × 106 Joule
  9. 1 watt hour = 3600 Joule

Numerical Problems (Solved):

Question 1.
J of energy is applied to lift a box of mass 0.5 kg. How much high it would be raised?
Solution:
Potential energy (P.E.) = 1 J
Mass of the box (m) = 0.5 kg
Acceleration due to gravity, (g) = 10 m/s2
Height to which box is raised, (h) = ?
We know, potential energy (P.E.) = m × g × h
1 = 0.5 × 10 × h
1 = 5 × h
or h = \(\frac {1}{5}\) = 0.2
∴ Height to which box would be raised, (h) = 0.2 m

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 2.
A woman pulls water-filled bucket weighing 5 kg from a well 10 m deep in 10 s. What is her power?
Solution:
Mass of water filled bucket, (m) = 5 kg
Depth of well(h) = 10 m
Acceleration due to gravity, (g) = 10 m s-2
Work done by woman(W) = P.E
= m × g × h
= 5 × 10 × 10
= 500 J
Time (t) = 10 s
Now, power = \(\frac{\text { Total work done(W) }}{\text { Total time taken(t) }}\)
= \(\frac {500J}{10s}\)
= 50 J/s
= 50 watt

Question 3.
A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate the amount of work done by him. How much potential energy does he gain? Take r = 9.8 m/s2?
Solution:
Here, m = 50kg, g = 9.8 m/s2, h = 100m
Work done by the body = mgh
= 50 × 9.8 × 100
= 49000 J
= 4.9 × 104 J
Gain in P.E. = Work done = 4.9 × 104 J

Question 4.
A man drops a 10 kg rock from the top of a 20 m ladder. What will be its kinetic energy when it reaches the ground? What will be its speed just before it hits the ground? Does the speed depend on the mass of the rock? (Take g = 10 m s-2)
Solution:
Here u = 0, m = 10 kg, h = 20 m, g = 10 m s-2
We know, υ2 – u2 = 2gh
∴ υ-2 – 0-2 = 2 × 10 × 20
or υ = \( \sqrt{{400}} \)
= 20 m s-1
K.E. = \(\frac {1}{2}\)mυ2
= \(\frac {1}{2}\) × 10 × (20)2
= 2000 J
Speed does not depends on the mass of the rock because the acceleration due to gravity under which the rock falls does not depend on mass.

Question 5.
A rocket of 3 × 106 kg mass takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. Calculate the potential energy, and the kinetic energy. (Take g = 10 m s-2).
Solution:
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 11

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 6.
An electric heater of 1000 W is used for 2 hours a day. What is the cost of using it for a month of 28 days, if 1 unit costs ₹ 3.00?
Solution:
Here, P = 1000 W
= 1 kW
Total time, t = 2 × 28 hours
= 56 hours
Total energy consumed = P × t
= 1kW × 56h
= 56kWh
∴ Cost of 1 kWh = ₹ 3.00
∴ Cost of using electricity for Feb = 3 × 56
= ₹ 168.

Question 7.
The power of a motor pump is 5 kW. How much water per minute the pump an raise to height of 20 m? Take g = 10 ms-2.
Solution:
Here, P = 5 kW = 5000 W, t = 1 min = 60s, h = 20 m, g = 10 ms-2
PSEB 9th Class Science Important Questions Chapter 11 Work and Energy 12

Question 8.
Calculate the electricity bill amount for the month of November of a family if 4 tube lights of 40 W each for seven hours, a TV of 150 W for three hours and two bulbs of 60 W each for four hours are used per day. The cost per unit is ₹ 3.50.
Solution:
Power of each tube light = 40 W
∴ Total power of 4 tube lights = 4 × 40 W
= 160 W
Energy consumed by 4 tube lights each day
= 160 W × 7 h
= 1120 Wh
= \(\frac {1120}{1000}\) = 1.112kWh
Energy consumed by T.V. per day = 150W × 3h
= 450 Wh
= \(\frac {450}{1000}\) = 0.45 kWh
Energy consumed by 2 bulbs per day = 60W × 4h × 2
= 480 Wh
= \(\frac {480}{1000}\) = 0.48 kWh
Total energy consumed per day by all appliances = 1.12 + 0.45 + 0.48 = 2.05 kWh
Total energy consumed in 30 days = 2.05 kWh × 30
= 61.50 kWh
Cost of 1 kWh = ₹ 3.50
Cost of 61.50 kWh = 3.50 × 61.50
= ₹ 215.25

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 9.
A person carrying 10 bricks each of man 2.5 kg. on his head moves to a height 20 metres in 50 seconds. Calculate the power spent in carrying bricks of the person. (g = 10 ms-2)
Solution:
Total mass (m) = 10 × 2.5 kg
= 25 kg.
Time (t) = 50 s
Displacement (S) = 20m
g = 10 ms-1
Force exerted by person (F) = mg
= 25kg × 10 ms-2
= 250 N
Work done = Force × Displacement
= F × S
= 250 N × 20 m
= 5000 J
Power = \(\frac{\text { Work done }}{\text { Time }}\)
= \(\frac {5000J}{50s}\)
= 100 w.

Question 10.
A car of 1000 kg moving with a velocity of 30 m/s stops with uniform acceleration after covering a distance of 50 m on application of brakes. Find the force applied by the brakes on the car and also work done.
Solution:
u = 30 ms-1
υ = 0
S = 50m
Now υ2 – u2 = 2aS
(0)2 – (30)2 = 2 × a × 50
– (30 × 30) = 100 × a
a = –\(\frac {900}{100}\)
∴ a = – 9m/s2
Force, F = m × a
= 1000 × 9
= 9000 N
work done by the brakes, W = F × S
= 9000 × 50
= 450000 N-m
= 4.5 x 105 J

Question 11.
A freely falling hammer of mass 1 kg 7 falls on a nail fixed in a block of wood. If the hammer falls from a height of 1 m then what will be the kinetic energy just before striking the nail? (Take g = 10 m s-2)
Solution:
Mass of the hammer, (m) = 1kg
Height of the hammer (h) = 1 m
Acceleration due to gravity (g) = 10ms-2
using υ2 – u2 = 2gS
υ2 – (0)2 = 2 × 10 × 1
υ2 = 20 ………..(i)
Now kinetic energy of hammer (K.E.) = \(\frac {1}{2}\) × m × υ2
= \(\frac {1}{2}\) × 1 × 20 [From (i)]
= 10 J

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 12.
A car is moving with a speed of 54 km/h. What will be the kinetic energy of the boy of mass 40 kg sitting in the car?
Solution:
Speed of the boy = speed of the car
= 54 km/h
= 54 × \(\frac {5}{18}\)m s-1
= 3 × 5 m s-1
= 15 m s-1
Mass of the boy (m) = 40kg
∴ kinetic energy (K.E.) of the boy = \(\frac {1}{2}\)mυ2
= \(\frac {1}{2}\) × 40 × (15)2
= 20 × 15 × 15
= 4500 J

Question 13.
1 Joule of energy is used for one heart beat. Calculate the power of the heart if it throbs 72 times in one minute.
Solution:
Work done in 1 heart beat = 1 J
∴ Total work done by heart in 72 beats = 72 × 1 J
= 72 J
Time (t) = 1 min
= 1 × 60 s
= 60 s
Power = \(\frac{\text { Work done }}{\text { Time taken }}\)
= \(\frac {72 J}{60s}\)
= 1.2 J/s
= 1.2 Watt

Very Short Answer Type Questions:

Question 1.
State the relation between commercial unit of energy and joules.
Answer:
1 commercial unit of energy (or 1 kWh) = 3.6 × 106 Joule.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 2.
How much work is done on a body of mass 1 kg whirling on a circular path of radius 5m?
Answer:
Work done is zero.

Question 3.
What is the SI unit of power?
Answer:
The SI unit of power is Watt.

Question 4.
A ball is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy when it reaches the maximum height?
Answer:
The kinetic energy of the body changes into its potential energy.

Question 5.
If the heart works 60 joules in one minute, what is its power?
Answer:
Power = \(\frac{\text { Energy }}{\text { Time }}\)
= \(\frac {60J}{60s}\)
= 1 W

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 6.
Name the term used for the sum of kinetic energy and potential energy of a body.
Answer:
Mechanical energy.

Question 7.
How many joules make one-kilowatt hour?
Answer:
1 kilowatt-hour = 3.6 × 106 J.

Question 8.
What should be the change in velocity of a body required to increase its kinetic? energy to four times of its initial value?
Answer:
The velocity of the body should doubled at constant mass.

Question 9.
Under what conditions the work done by a force is zero inspite of displacement being taking place?
Answer:
When displacement is in a direction perpendicular to the applied force.

Question 10.
What is the power of a machine which does 2000 joules of work in 10 seconds?
Solution:
Power of a machine = \(\frac{\text { Work done }}{\text { time }}\)
= \(\frac {2000J}{10s}\)
= 200 W

Question 11.
What is the SI unit of kinetic energy?
Answer:
Joule.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Question 12.
Water flows down the mountains to the plains. What happens to the potential energy of water?
Answer:
Potential energy of water will decrease. It will change to kinetic energy of water.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Punjab State Board PSEB 9th Class Science Important Questions Chapter 10 Gravitation Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Long Answer Type Questions:

Question 1.
Define Newton’s universal law of gravitation and establish mathematical formula for force of attraction between two objects.
Answer:
Newton’s universal law of gravitation-Every particle in this universe attracts each and every particle, the force of attraction is:
1. directly proportional to the product of both the masses.
2. Inversely proportional of the square of the distance between the two. This force always acts along the line joining the two masses.
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 1
Derivation of Mathematical Formula: From the fig. Jet there are two balls A and B having masses m1 and m2 distance between them is ‘r’.
According to Newton’s third law of motion ball A exerts a force FAB on the ball B and ball B exerts a force FBA on the ball A. These forces are equal and opposite
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 2
where G is universal gravitational constant. This is named so since its numerical value remains constant in whole of the universe and the formula is known as Newton’s universal law’ of gravitation.
Value of G
G = 6.67 × 10-11 Nm2/kg2

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 2.
Write Kepler’s law in context with the motion of planets.
Answer:
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 3
By the 16th centuary, a lot of data on the motion of planets had been collected by many astronomers. Johannes Kepler derived three law s based on these data. These are known as Kepler’s laws. These are:
1. Law of orbits (First laws): The orbit of a planet is an ellipse and the sun is at one of the foci, as shown in fig. In this fig. Sun is shown at O.

2. Law of area (Second laws): The line joining the planet and sun sweep equal areas in equal interv als of time. Thus, time taken for the motion from A to B is same as for the motion from C to D and area OAB and OCD are equal.

3. Law of time period (Third law): The cube of mean distance of a planet from the sun is directly proportional to square of its orbital time period
T2 ∝ r3
But Kepler could not give a theory which explained the motion of planets.
Newton showed that due to motion of planets sun exerts force of gravtation on them.
i.e. T2/r3 = constant.

Question 3.
How did Robert Boyle proved experimentally that all bodies fall in vacuum with same acceleration?
Or
How did Robert Boyle show experimentally that a coin and a piece of paper when dropped simultaneously from same height in vacuum fall with same acceleration?
Answer:
Boyle’s Experiment: As shown in fig. Robert Boyle took a long glass tube. A heavy coin and a piece of paper were placed inside the tube. The ends of the tube were closed. Air from the tube was removed with the help of a vacuum pump.
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 4
When the tube was quickly inverted it was seen that coin and a piece of paper hit the bottom of the tube at the same time. Now the experiment was repeated with air inside the glass tube. This time it was observed that a piece of paper falls slowly whereas the coin immediately hit the bottom of glass tube. This experiment proves that in vacuum all bodies irrespective of their masses (both light and heavy) fall towards earth with same acceleration

Question 4.
(a) Prove that acceleration due to gravity is independent of mass.
(b) Find the value of ‘g’.
Answer:
(a) Consider a body of mass ‘m’ lying on the surface of earth. Suppose M and R are respectively mass and radius of earth.
Let F be the force of gravity acting on the body
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 5
This acceleration is called acceleration due to gravity
∴ g = \(\frac{G M}{R^{2}}\)
This equation is free from mass ‘m’. This shows that due to force of gravity the acceleration produced in an object is independent of its mass.

(b) To find the value of ‘g’
G = 6.67 × 10 “n Nm2/kg2
R = 6400 km
= 6400 × 1000 m
= 64 × 105 m
= 6.4 × 106 m
Mass of the earth(M) = 6 × 1024 kg
But g = \(\frac{G M}{R^{2}}\)
= \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{\left(6.4 \times 10^{6}\right)^{2}}\)
or g = 9.8 ms-2

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 5.
Describe those factors which are responsible for variation in the value of acceleration due to gravity ‘g’?
Answer:
Variation in the value of acceleration due to following factors:
Variation in g with altitude-Value of ‘g’ is maximum on the surface of earth. As we move in the upward direction (higher altitudes) value of ‘g’ goes on decreasing. We can calculate value of ‘g at altitude using mathematical formula. If
ge = acceleration due to gravity on the surface of earth
gh = acceleration due to gravity at height h
R = radius of earth
we know gh = ge\(\left[\frac{\mathrm{R}^{2}}{(\mathrm{R}+h)^{2}}\right]\)
If we know value of h, we can calculate gh.
Special case-If ‘h’ is half the value of radius of earth i.e. h = \(\frac {R}{2}\) then
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 6
Another interesting fact that value of ‘g’ is zero at the centre of earth.

2. Variation in g due to shape of earth-
Earth is not exactly spherical. It is somewhat egg-shaped. Its radius at poles is less than that at equator as shown in fig.
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 7
As shown in fig. radius of earth at equator is 6378 km and radius of earth at poles is 6357 km. Thus, value of g at poles is gp = 9.831 m s-2 (maximum) and minimum value at equator is ge = 9.782 m s-2. We use average value of g calculated from ge, and gp.

3. Effect of depth: If we go deep under the earth e.g. in caves or in mines, value of ‘g’ i’ decreased. Value of g goes on decreasing, if depth goes on increasing. At the centre of earth, value of g is zero.
\(\frac{\mathrm{W}_{d}}{\mathrm{~W}_{e}}=\frac{g_{d}}{g_{e}}=\left[1-\frac{d}{\mathrm{R}}\right]\)

Question 6.
Establish the relation between ‘g’ and ‘G’.
Or
Deduce an expression for it in terms of mass of the earth ‘M’ and universal gravitational constant ‘G’.
Or
Show that the acceleration due to gravity of an object is independent of its mass.
Answer:
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 8
Relation between ‘g’ and ‘G’. Suppose the earth is a sphere of mass M and radius R, Consider a body of mass m lying at a distance ‘r’ from the centre of the earth. According to Newton’s law of gravitation, the force of attraction between the earth and the body,
F = \(\frac{\mathrm{GM} m}{r^{2}}\) ……………… (i)
Tills force of gravity produces an acceleration ‘g’, is the body of mass m.
Hence, from Newton’s second law,
F = Mass × Acceleration
F = m × g ……………..(ii)
From equations (i) and (ii) we get
mg = \(\frac{\mathrm{GM} m}{r^{2}}\)
or g = \(\frac{\mathrm{GM}}{r^{2}}\) ………..(iii)
This equation gives relation between acceleration due to gravity ‘g’ at points far away from the earth and gravitational constant ‘G’.
Since (3) does not involve’m’ it is therefore eviden t that ‘g’ does not depend upon the mass of the body.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 7.
What is Archimedes principle? How can you verify it experimentally? Also write applications of Archimedes principle.
Answer:
Archimedes Principle: According to this principle when a solid body is immersed either completely or partially in a liquid, it experiences an upward thrust which is equal to the weight of the liquid displaced by the immersed part of the body.”

Experimental Verification: Take a stone piece and tie it to the hook of a spring balance. Hold the spring balance in your hand or support it from the stand so as to suspend the stone piece as shown in fig.
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 9
Note, the reading on the spring balance to know the weight of the stone. Now pour some water a beaker with and slowly immerse the stone in the water while remaining tied with the hook of spring balance as shown in fig. (b). You will see that spring balance shows a decreased reading i.e. weight of stone is decreased after it is immersed in the water.

Subtract the two readings to know the decrease in the weight of stone. As soon as the stone is immersed in the water it displaces water equal to its own volume. Collect this water in other beaker and measure its weight. You will find that this weight of water is equal to the decrease of weight of the stone. Thus Archimedes principle is proved.

Applications of Archimedes’ Principle:

  1. Archimedes principle is used in designing ships and submarines.
  2. Lactometers are constructed on Archimedes’ Principle which are used to measure puritv of a sample of milk.
  3. Hydrometers used to measure density of liquid are also based on Archimedes principle.

Short Answer Type Questions:

Question 1.
Which one is greater-the gravitational force of the earth on 1 kg iron or the force of gravitation applied by 1 kg on earth?
Answer:
According to Newton’s law of Gravitation there exists of mutual force of attraction between two objects. Since the mass of iron is less than the mass of the earth, therefore earth attracts 1 kg mass towards it with a greater force which is noticeable.

Question 2.
Why is G called universal gravitational constant?
Answer:
G is called universal gravitational constant because its numerical value is same in the whole universe. This value is G = 6.67 x 10-11 Nm2/kg2.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 3.
Is the value of ‘g’ at a given place same for different bodies or it is variable?
Answer:
From the relation between the acceleration due to gravity ‘g’ and universal gravitational constant G, we know g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
From this equation it is evident that the value of ‘g’ does not depend upon the mass of the body. Thus we reach a conclusion that at a given place the value of g is same for different bodies.

Question 4.
Why does a body becomes weightless at the centre of earth?
Answer:
We know value of g goes on decreasing as we continue moving deep into the earth and value of acceleration due to gravity (g) = 0 at the centre of earth. Thus, a body whose mass is m,
Weight of body = m × 0 = 0
Therefore, body becomes weightless.

Question 5.
A tennis ball jumps higher at hills than at planes. Explain.
Answer:
Value of g decreases with altitude, thus gravitational force acting on ball at hills is less as a result ball jumps higher at hills than at planes.

Question 6.
The weight of an object on the surface of earth is 9.8 N. What does this statement mean?
Answer:
We know that value of ‘g’ on the surface of earth is 9.8 m/s2 and the relation for weight of an object is:
w = m × g
9.8 = m × 9.8
or m = \(\frac {9.8}{9.8}\)
∴ m = 1 kg
The given statement therefore means the mass of the object on earth is 1 kg.

Question 7.
What type of motion a freely falling body execute under gravity?
Answer:
A body falling freely under gravity executes uniform accelerated motion. If bodies with different masses and different shapes are allowed to fall freely in vacuum, they all will have same acceleration due to gravity.

Question 8.
Give points of difference between Acceleration due to gravity (g) and Universal gravitational constant'(G).
Answer:
Difference between g and G:

Acceleration due to gravity (g) Gravitational constant (G)
1. It represents acceleration acquired by the body due to gravity. It represents force of attraction between two masses of 1 kg each lying 1 m apart.
2. Its value is different at different places on earth surface. Its value is constant at all places. Thus, it is called universal constant.
3. Its value at the surface of earth is 9.8 m/s2. Its value is 9.67 × 10-11 Nm2kg-2.
4. It is a vector quantity. It is a scalar quantity.

Question 9.
You buy W weight of sugar at a place situated on equitorial line and then take it to Antarctica. Will that sugar weigh same there? If not whether it would be more or less.
Answer:
The value of ‘g’ at Antarctica is not same as on equator. The value of ‘g’ increases on Antarctica therefore, sugar bought at any place on equitorial line when taken to Antarctica would have more weight but its mass will remain the same because mass is a constant quantity.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 10.
Why cannot we move a finger without disturbing all the stars?
Answer:
When we move our finger, then distance between finger and all other things in the universe change and as a result force of gravitation also changes. Thus, all the things get disturbed, although this disturbance is negligible.

Question 11.
Distinguish between Gravitation and Gravity.
Answer:

  • Gravitation: Gravitation is the force of attraction between any two bodies in the universe. The attraction between the sun and the earth, the attraction between a table and a chair are examples of gravitation.
  • Gravity: Gravity is a special case of gravitation when one of the two bodies is the earth. Gravity is the attraction between the earth and any object lying on or near its surface. A ball thrown upward falls back on the surface of the earth due to earth’s force of gravity.

Question 12.
Explain why a small piece of stone is not attracted towards another big place of stone on the earth’s surface?
Answer:
Because of very small value of G, the force of attraction between any two such ordinary sized bodies is so small that it cannot produce motion in them.

Question 13.
The earth attracts an apple. Does the apple also attract with earth? If it does, why does the earth not move towards the apple?
Answer:
The apple also attracts the earth with an equal and opposite force. The mass of the earth is very large compared to that of apple. So, the acceleration produced in earth is very small as compared to that in the apple. Hence, the motion of the earth towards the apple is not appreciable and therefore, is not noticeable.

Question 14.
If the force of gravity somehow vanishes today, why would we be sent being in space?
Answer:
In the absence of force of gravity, the centripetal force required to keep us rotating along the earth would not be available. As a result would fly off along the tangent to with into the space.

Question 15.
What is meant by density and relative density?
Answer:
Density of a substance is defined as the mass of a substance contained in a unit volume.
Density = \(\frac{\text { Mass }}{\text { Volume }}\)
Its SI unit is kg m-3.

Relative density or specific gravity is the ratio of density of substance to the density of same volume of water
R.D or S.G = \(\frac{\text { Density of substance }}{\text { Density of water }}\)

Density of water is 1,000 kg m-3 in SI.
According to Archimedes’ principle, when a body is immersed in water, the loss of weight in water is equal to the weight of an equal volume of water, i.e., weight of water displaced. Hence R.D. can be written as
R.D = \(\frac{\text { Weight of substance }}{\text { Loss of weight of body in water }}\)

Question 16.
What do you mean by buoyancy and centre of buoyancy? In which direction does the buoyant force on an object immersed in a liquid act?
Answer:
Buoyancy: When a body is immersed partially or wholly in a fluid (liquid or gas), it displaces fluid. The displaced fluid exerts an upward force on the body.

The upward force acting on a body imnu rsed in a fluid is called upthrust or force of buoyancy and the phenomenon is called buoyancy. The buoyancy acts through the centre of gravity of the displaced fluid which is called centre of buoyancy.

Question 17.
State Archimedes’ principle.
Answer:
Archimedes’ Principle. When a solid body is immersed completely or partially in a fluid, it experiences an upward thrust which is equal to the weight of the fluid displaced by the immersed part of the body.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 18.
Name two forces which act on a body immersed in a liquid. Give the directions in which they act.
Answer:
Two forces which act on a body immersed in a liquid are:

  • Weight of the body acting downward.
  • An upthrust due to displaced liquid.

Question 19.
How is submarine able to move on water surface as well as go under water?
Answer:
To move submarine under water, water is pumped in its special tanks. This causes a net increase in weight and thus submarine goes down in water.

To bring the submarine upon the surface of sea, the water from the tanks is thrown out by pumps. When water from the tanks of submarine is taken out, the net weight of the submarine decreases without any change in its volume and it comes up on the surface of water.

Question 20.
Give reason when Big buildings and dams have wide foundations for safety.
Answer:
Dams and big buildings have wide foundations for safety, because they may not collapse under high pressure of the building.

Question 21.
A steel needle sinks in water but a steel ship floats. Explain how?
Answer:
A steel needle sinks in water because it displaces less weight of water which provides less buoyant force than the actual weight of the needle. On the other hand, steel ship floats because it displaces a large weight of water which provides a greater buoyant force to keep it a float.

Question 22.
Give reasons for the following :
(a) A sharp blade is more effective in cutting an object than a blunt blade.
(b) A cork piece floats but an iron piece sinks in water.
Answer:
(a) We know Pressure = \(\frac{\text { Thrust }}{\text { Area }}\). For the given thrust, pressure ∝ \(\frac{\text { 1 }}{\text { Area }}\). Thus the effect of the same magnitude of force is more when the area of surface in contact is less. Hence a sharp blade is more effective in cutting an object than a blunt blade.
(b) The density of the cork piece is less than the density of water, and the. density of the iron piece is more than the density of water. Hence, the cork piece floats but the iron piece sinks in water.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 23.
Explain the following:
(a) Swimmers are provided with an inflated rubber jacket.
(b) It is easier of swim in sea water than in river water.
Answer:
(a) Swimmers are provided with an inflated rubber jacket to increase the volume such that density is decreased, which is helpful for swimming.
(b) It is easier to swim in sea water than in river water because density of sea water is more than water of the needle sinks in a river while a ship float on it.

Question 24.
Why is the pressure on ground more when a man is walking than when he is standing?
Answer:
When the man stands he exerts a force on the ground equal to his own weight. But when he walks, he pushes the ground backward and exerts an additional force on the ground. That is why pressure on the ground is more when a man is walking than when he is standing.

Question 25.
Why a bucket of water is lighter when in water than when it is taken out of water?
Answer:
Inside water, the bucket experiences upthrust exerted by displaced water, so its apparent weight becomes less than the actual weight. When bucket is taken out of water, upthrust on the bucket disappears and it feels heavier.

Question 26.
If a fresh egg is put into a beaker filled with water, it sinks. On dissolving a lot of salt in the water, the egg begins to rise and floats. Why?
Answer:
The average density of a fresh egg is more than that of pure water but less than that water in which salt is dissolved. So a fresh egg sinks in pure water while it floats in salty water.

Important Formulae:

  1. F = G\(\frac{\mathrm{M} m}{\mathrm{R}^{2}}\)
  2. F = mg
  3. g = G\(\frac{\mathrm{M}}{\mathrm{R}^{2}}\)
  4. υ = u +gt
  5. υ2 = u2 + 2gh
  6. h = ut + \(\frac {1}{2}\)gt2
  7. gm = \(\frac {1}{6}\)ge
    (gm = Acceleration due to gravity on moon , ge = Acceleration due to gravity on earth)

Necessary Data-
Earth
Mass = 6 × 1024 Kg
Radius = 6.4 × 106 m (6400 Km)
Distance from Sun = 1.5 × 1011 m
Moon
Mass = 7.3 × 1022 Kg

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Numerical Problems (Solved):

Question 1.
Two spheres of 1 kg mass each are separated by 3 m. Calculate the gravitational force between then. Given G = 6.67 × 10-11 Nm2/kg2.
Solution:
Here m1= m2 = 1 kg
Distance between two spheres, (r) = 3 m
Gravitational constant (G) = 6.67 × 10-11 Nm2/kg2
We know
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 10

Question 2.
The radius of moon is 1.7 × 106 m and its mass is 7.35 × 1022 kg. What is the acceleration due to gravity on the surface of moon? Given G = 6.67 × 10-11 Nm2/kg2
Solution:
Radius of moon, (R) = 1.7 × 106 m
Mass of moon, (M) = 7.35 × 1022 kg
Gravitational constant, (G) = 6.67 × 10-11 Nm2/kg2 (Given)
We know, Acceleration due to gravity on moon
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 11

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 3.
Find the change in weight percentage of a body when it is taken from equator to poles. Polar radius is 6357km and equitorial radius is 6378km.
Solution:
Polar radius, (r) = 6357 km
Equitorial radius (R) = 6378 km
∴ h = R – r
⇒ h = (6378 – 6357)km
h = 21km
R = 6400 km(Approx)
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 12

Question 4.
At what height above the earth surface, the acceleration due to gravity will be half that on the surface of earth? Suppose R is the radius of earth.
Solution:
Let h be the height above earth surface where
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 13
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 14

Question 5.
A ball is dropped from top of 40 m high tower. What will be its velocity after covering a distance of 20 m ? What will be its velocity on striking the earth?
Answer:
1. Here, height of the tower, (h) = 40 m
Initial velocity, (u) = 0
Acceleration due to gravity, (g) = 10m/s2
Distance covered (S) = 20m
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 15
i.e. Velocity of ball when it strikes the earth. = 20 m/s

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 6.
If weight of an object is 49 N then what will be its mass?
Solution:
Weight of the object W = 49 N
g = 9.8 m s-2
We know, weight = Mass × acceleration due to gravity
W = m × g
49 = m × 9.8
or m = \(\frac {49}{9.8}\)
∴ m = 5 kg

Question 7.
Ah object is projected vertically upward with a velocity of 50 m/s. After what time the object will attain the maximum height.
Solution:
Initial velocity of the object (u) = 50 m/s [upward direction]
Acceleration due to gravity, (g) = – 10 ms-2 [object comes to rest]
Velocity of the object at maximum height,
(υ) = 0
Time taken (t) =?
We know, υ = u + gt
0 = 50 + (-10) × t
0 = 50 – 10t
or 10t = 50
∴ t = \(\frac {50}{10}\) = 5s
So the object will attain its maximum height after 5 seconds.

Question 8.
A stone is dropped from the edge of a rooftop. If it crosses 2 m high window in 0.1 second then what is the distance between the upper end of window and the roof?
Solution:
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 16
Distance covered in crossing the window
(S) = 2 m
g = 9.8 ms-2
t = 0.1 s
Let υ be the velocity of the stone when it reaches the upper edge A of the window.
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 17

Question 9.
A helicopter drops food packet for people caught in stationary boat. This is moving at a height of 20 m with a horizontal velocity of 2 m/s. When food packet is dropped the nearest end of the boat is just below the helicopter. If boat is 5 m long shall people caught in the boat receive the dropped food packets?
Solution:
h = 20 m, u = 2 m/s g = 10ms-2
Vertical Range (R) = \(\frac{2 \sqrt{2 h}}{g}\)
= \(\frac{2 \sqrt{2 \times 20}}{10}\)
= 2\( \sqrt{{4}} \)
= 2 × 2 = 4 m
But lenght of boat, = 5m
So people caught in the boat will receive food packets.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 10.
An object of volume V is immersed in a liquid of density ρ. Calculate the magnitude of buoyant force acting on the object due to the liquid.
Solution:
By Archimedes’ principle, the magnitude of buoyant force,
U.F. = Weight of liquid displaced = Volume × Density × g or
or U.F. = Vρg

Question 11.
The pressure exerted by the weight of a cubical block of side 4 cm on the surface is 10 pascal. Calculate the weight of the block.
Solution:
Here pressure exerted, P = 10 pascal = 10 Nm-2,
Area, A = 4 cm × 4 cm
= 16 × 10-4 m2
As P = \(\frac {F}{A}\)
= \(\frac {W}{A}\)
W = P × A
= 10 Nm-2 × 16 × 10-4m2
= 1.6 × 10-2 N

Question 12.
The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm-3? What will be the piass of the water displaced by this packet?
Solution:
Mass of packet = 500 g
Volume of packet = 350 cm3.
Density of packet = \(\frac{\text { Mass }}{\text { Volume }}\)
= \(\frac{500 \mathrm{~g}}{350 \mathrm{~cm}^{3}}\)
= 1.43 g cm-3
Since its density (1.43 g cm-3) is more than that of water (1 g cm-3) thus, the sealed packet will sink in water
Volume of water displaced = Volume of packet
= 350 cm3
Mass of water displaced = Volume × Density
= 350 cm3 × 1gcm-3
= 350 g

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 13.
A block of wood is kept on the table top. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.
Solution:
Here, M = 5kg and g = 9.9 ms-2
Weight = force exerted on table top
= mg
= 5 × 9.8
= 49 N
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 18

Question 14.
A solid body floating in water has \(\frac {1}{6}\)th of its volume above the surface.
What fraction of its volume will project upwards if it floats in a liquid of density 1,020 km m-3?
Solution:
Let V be the volume of the body
Volume inside water = V – \(\frac {V}{6}\)
= (1 – \(\frac {1}{6}\))V
= \(\frac {5}{6}\)V
Upward thrust in water = \(\frac {5}{6}\) × 1,000 × g N ……………..(i)
where density of water is 1,000 kg m-3
Let υ be the volume of the body outside the liquid of density, 1,020 kg m-3.
∴ Volume inside the liquid = (V – υ)
Upward thrust in liquid = (V – υ) × 1,020 × g N ……….(ii)
Upward thrust in two cases must be the same and must be equal to the weight of the body.
PSEB 9th Class Science Important Questions Chapter 10 Gravitation 19

Very Short Answer Type Questions:

Question 1.
What is g on the moon as compared to that on earth?
Answer:
Nearly \(\frac {1}{6}\)th of its value on earth.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 2.
What is unit of G?
Answer:
It is Nm2 kg-2.

Question 3.
What is the mass of the earth?
Answer:
It is about 6 × 1024 kg.

Question 4.
What is essential property of matter-mass or weight?
Answer:
Mass is the basic and essential property. It is constant everywhere. Weight of a body varies from place to place.

Question 5.
What is SI unit of weight of a body?
Answer:
It is newton (N).

Question 6.
The earth’s gravitational force causes an acceleration of 5 m s-2 on a 1 kg mass somewhere in the space. How much will be the acceleration of a 3 kg mass at that place?
Answer:
Same i.e., 5ms-2 since g at a place is independent of mass of the body.

Question 7.
Why one can jump higher on the surface of moon than on the earth?
Answer:
The g at moon surface is nearly 1/6th of that at the surface of earth. Hence one can jump six times higher on the moon with a given initial velocity.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 8.
Give the value of universal gravitational constant in S.I. units.
Answer:
(G) = 6.67 × 10-11 Nm2/kg2
The value of gravitational constant.

Question 9.
The value of ‘G’ on the surface of earth is 6.67 × 10-11 Nm2/kg2. What is its value on the surface of moon?
Answer:
Since G is universal constant so its value on moon surface will be same as on the earth surface i.e. G = 6.67 × 10-11 Nm2/kg2 .

Question 10.
State two factors on which the gravitational force between two objects depends.
Answer:
The gravitational force between two objects depends on: (i) their masses (ii) distance between them.

Question 11.
Write the formula to find the magnitude of gravitational force between the earth and an object on the surface of the earth.
Answer:
F = \(\mathrm{G} \frac{\mathrm{Mm}}{r^{2}}\)

Question 12.
Can the mass of a body be zero.
Answer:
No, mass of body can never be zero.

Question 13.
Mass of an object on the earth is 600g. What will be its mass on moon?
Answer:
Mass of the object on moon will be same i.e. 600 g.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 14.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
Mass is a constant quantity. So, it can not be more or less than 42 kg.

Question 15.
How does the value of ‘g’ vary from equator to poles?
Answer:
The value of ‘g’ increases as we move from equator to poles.

Question 16.
What will be the weight of an object on the earth whose mass is 10 kg?
Answer:
Weight of the object on earth
(W) = m x g
= 10 kg × 10 m s-2
= 100 N

Question 17.
Write the S.I. unit of G.
Answer:
Nm2/kg2.

Question 18.
When does an object float when placed on the surface of water?
Answer:
If the density of object is less than water, it will float.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 19.
While swimming why do we feel light?
Answer:
The swimmer experiences an upward force by water, this causes buoyancy and makes the swimmer feel light.

Question 20.
Why does a truck or a motor-bus has much wider tyres?
Answer:
So that pressure acting on the road due to weight of truck or motor-bus may be small.

Question 21.
An army tank weighing more than a hundred tonne move conveniently on an earthen road. How?
Answer:
The army tank rests upon a continuous broad chain. So, the total surface area is large and pressure on road due to weight of tank is not very high.

Question 22.
What is the unit of relative density? Why?
Answer:
Relative density has no unit because it is a ratio of two terms having same units. So relative density is expressed in numbers only.

PSEB 9th Class Science Important Questions Chapter 10 Gravitation

Question 23.
The weight of an object on the moon is ………………. of its weight on the earth.
Answer:
The weight of an object on the moon is 1/6th of its weight on the earth.

PSEB 12th Class Sociology Important Questions in Punjabi English Medium

PSEB 12th Class Sociology Important Questions

Punjab State Board Syllabus PSEB 12th Class Sociology Important Questions Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class Sociology Important Questions in Punjabi English Medium

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Punjab State Board PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Long Answer Type Questions:

Question 1.
What are the different types of Forces? Explain each with the help of example
Answer:
Types of forces. There are two types of forces:
1. Balanced forces
2. Unbalanced Forces.
1. Balanced Forces: When several forces are acting simultaneously on a body and their resultant is zero, the forces are said to be balanced forces.

In the case of balanced forces if some body is at rest then it will remain at rest and if it is moving with uniform speed then it will continue to move with the same speed, as if no force is doing any work. In this way with the effect of balanced forces. There does not take place any change in position of the body.

Balanced forces change the shape of the objects, e.g. if a rubber ball is pressed between palms by applying equal and opposite forces then the shape of the ball changes. This ball no longer remains round and instead becomes flat.
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 1
Example: In a tug of war, when both teams pull the rope with equal force, then resultant forces becomes zero. Therefore, both teams remain in their places. In this situation the forces applied by the two teams are equal and opposite so get balanced.
Condition for forces to be balanced.
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 2
Two equal forces acting in the opposite direction become balanced if they act along the same line and their magnitudes are equal.
Effect of Balanced Forces: Forces acting on any object if do not change its state of rest or its motion then these do change the shape of the object.

2. Unbalanced Forces: If the resultant of the several forces acting on a body is not zero, the forces are said to be unbalanced forces. Unbalanced forces produce change in the direction of uniform motion of the body or its state of rest.
Example:
In a tug-of-war, when one of the two teams pulls the rope with a larger force, it is able to pull the weaker team towards it. Here two forces are not balanced. Therefore, it results in the motion of the weaker team towards the larger force along the rope.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 2.
State and explain the Newton’s First Law of Motion.
Answer:
Newton’s First Law of Motion. This law states that “In this universe all bodies will remain in their state of rest or of uniform motion in a straight line until some external force is applied to bring about change in their state.
According to this law, motion can be divided into two parts:
1. First part says that a body at rest continues in its state of rest unless some external force is applied to bring change in its state of rest. We find a book lying on the table keeps lying in the same state unless someone applies force on it to pick it up.

2. Second part says that a body jn uniform motion continues moving in straight line path with a uniform speed unless someone applies external force to stop it. But in our daily life it appears but different.

As for example when we stop pedalling a bicycle the moving bicycle stops. After studying minutely it is found that between tyres of bicycle and ground there acts a force of friction which is an external force which opposes the motion. The resistance of air also opposes the motion of bicycle. So, moving bicycle stops moving due to these two forces.

Question 3.
What is inertia? What are its different types? Give examples for each one of them.
Answer:
Inertia: It is the property of matter by virtue ofivhich an object is unable to change by itself its state of rest or of uniform motion in a straight line.
Because of this property Newton’s First Law of Motion is also called the law of Inertia.

Types of Inertia Inertia is of three types:
1. Inertia of Rest: It means a body at rest tends to remain in its position of rest. i.e. a body at rest opposes the forces which try to move it. It can be understood clearly by the following example.
A man standing in a stationary bus or train falls backward when the bus or train suddenly starts moving forward. When the bus moves, the lower part (limbs) of his body begins to move along with the bus while his upper-part tends remain at rest due to inertia.

2. Inertia of Motion: This means a body in motion continues to move with uniform
motion in a straight line. i.e. it is the tendency of a body to remain in its state of uniform motion in a straight line.

Example: 1. A person sitting in moving bus falls forward when the moving bus suddenly stops. It is because as the bus stops the lower part of his body comes to rest along with the bus while upper part of his body continues to remain in motion due to inertia of motion.

2. An athelete runs for some distance before taking a jump so that his inertia of motion may help his muscular force to a longer jump.

3. Inertia of Direction. It is the property of a body which helps to maintain its direction i.e. it is inability of a body to change by itself its direction of motion.
Example: The mudguard is fitted in the wheel of a bicycle to protect from mud particles and water drops sticking to its wheel leaving it tangentially.
Imagine a stone tied to the end of a thread moving in a horizontal circle, while doing so, if the thread breaks then due to inertia of direction the stone flies off tangentially in a straight line.
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 3

Question 4.
State and explain Newton’s Second Law of motion with the help of this law how can we measure force.
Answer:
Newton’s Second Law of Motion. Newton’s second Law of motion helps to calculate the force required to bring a body in motion. It consists of two parts
1. The rate of change of momentum of a body is directly proportional to the applied unbalanced force and
2. the change in momentum due to the external applied force takes place in the direction of force.
i.e. According to this law “the external force applied in the body is directly proportional to the product of mass of the body and the acceleration produced in the direction of force.

Explanation: When a force acts on a body, it produces change in its momentum. If the force is doubled then the change in momentum also gets doubled. The more the force applied; the more is the change in momentum produced. Momentum is the product of mass of the body and its velocity. Generally no change in mass occurs. Therefore, the rate of change of momentum is actually the rate of change of velocity. Thus the applied force is proportional to the acceleration.

When an external force acts on a body at rest, it begins to move in the direction of force. When force acts in the direction of motion of the body then its momentum gets increased. But when force acts in a direction opposite to the direction of motion, its momentum gets reduced.
Force (F) ∝ mass (m) × acceleration (a)
or F = k × m × a ….(i)
(where k is a constant of proportionality)
If we choose the unit of force in such a way that as unit of force produces unit acceleration in a body of unit mass, then
Substituting F = 1, m = 1 and a = 1 in (1)
1 = k × 1 × 1
or k = 1
∴ from equation (i), F = 1 × m × a
or F = m × a
Force = Mass × Acceleration

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 5.
Derive the mathematical relation for magnitude of force from Newton’s Second Law of Motion.
Answer:
Measurement of Magnitude of Force from Newton’s Law of Motion. Suppose force F acts on a body of mass’m’ for Y seconds which changes its velocity from u to υ, then
Initial momentum of body, p1 = mu
Final momentum of body, p1 = mυ
Now because final velocity (υ) is more than the initial velocity (u), therefore final momentum (p2) will be more than the initial momentum (p1) change in momentum,
p = p2 – p1
= mυ – mu
= m (υ – u)
According to second law of motion,
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 4

Question 6.
State and explain Newton’s third Law of motion.
Answer:
Newton’s third Law of Motion. This law states that “Every action has equal and opposite reaction” According to this law, there does not exist one force in isolation. Force always exists in pair i.e. forces of action and reaction always act on different bodies.

Explanation: (1) Consider two similar spring balances attached to each other through their hooks. One end of spring balance A is fastened to the fixed support. Pull the free end of spring balance B to right side. Note the readings of both the spring balances. Both will read the same as shown in fig. It is because read.
A and B pull each other in opposite direction with same force.
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 5
(2) Keep two balls A and B on a table at some distance from each other. When you push the ball. A towards B then the ball A acts on ball B. This force is represented by
FA→B According to Newton’s third law of motion, the ball B also reacts and applies
force on ball A. This force is represented as FB→A. if both the balls are similar then the magnitudes of action and reaction will be equal.
∴ FA→B = – FB→A
Action and reaction forces always act in the direction opposite to each other.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 7.
What is meant by the Law of Conservation of Momentum? Deduce this law mathematically with the help of Newton’s second and third law of motion.
Answer:
Law of Conservation of Momentum. For a system of bodies, the total vector sum of momenta of all the bodies due to mutual action and reaction remain unchanged so long as no external force acts on the system.”
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 6
Mathematical Derivation: Consider two rubber balls of masses m1, m2 and initial velocities u1, u2 respectively. Let these collide and their velocities after collisions be υ1 and υ2 respectively. If A applies a force F on B also for time t; B also applies a force F on A for same time t.
From Newton’s Second Law of Motion:
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 6
i.e. total momentum of balls A and B before collision
= Total momentum of balls A and B after collision
This proves the law of conservation of momentum that is the total momentum remains conserved.

Short Answer Type Questions:

Question 1.
What is force? Give its units.
Answer:
Force. Force is an agent which

  • produces or tends to produce motion in the body
  • stops the moving body or tends to stop
  • increases the speed of the body or tries to increase the speed therefore,

Force may be defined is as physical cause (a push or a pull) which changes or tends to change the state of rest or uniform motion or direction of motion of a body. The force exerted by the engine makes the train to move from its actual position of rest while the force exerted by the brakes slows down or stops the moving train. The force exerted on the steering wheel of a car changes its direction of motion.
Force is a vector quantity.
Units of force: The unit of force depends upon the unit of mass and acceleration. S.I. unit of force is Newton and C.G.S. unit is Dyne
1N = 105 Dynes

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 2.
Why does the horse rider fall forward when a running horse suddenly stops?
Answer:
When horse is running, both the horse and the rider are in motion. When the horse suddenly stops, the lower part of the rider and horse come in the state of rest while the upper part of the rider remains in motion so that he falls forward.

Question 3.
When a horse suddenly gallops, the rider falls backward. Why?
Answer:
The horse and the rider form one system. Initially both are at rest. When the horse suddenly gallops then the horse and the lower part of the rider come into motion in the forward direction while the upper part of the rider’s body tends to remain at rest. Therefore, the rider falls backward.

Question 4.
Why does a passenger fall forwaid when he alights from the moving bus?
Answer:
While alighting from the moving bus the passenger falls forward because when the feet of the passenger touch the ground, his lower part suddenly comes to rest while the upper part still remains in motion. In this way the passenger falls forward.

Question 5.
Define momentum of a body. Also give its units.
Answer:
Momentum. It is defined as the quantity of motion possessed by the body. It is measured by the product of the mass and velocity of the body.
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 8

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 6.
A fast-moving bullet when hits the windowpane makes a round hole while a stone strikes and shatters it, why?
Answer:
If stone piece strikes window pane glass, it gets shattered while a fast-moving bullet when strikes the windowpane a round hole is formed. The reason is that small particle of glass around and near the hole do not move due to inertia along with the bullet and threfore, do not scatter.

Question 7.
Explain how a dirty blanket becomes dust-free if it is jerked once or twice?
Answer:
If a dirty blanket is beaten with a stick or is given a jerk then dust particles get separated from it because when we beat or give a jerk to the blanket it comes in motion but due to inertia of rest, dust particles remained at rest and get separated and the blanket becomes dust-free.

Question 8.
Why a fan continues to rotate for sometime even after it is switched off?
Answer:
When a fan is rotating, then because of inertia of motion it continues it rotation for some time even if we switch if off. Due to friction on resistance of air it comes to rest after few seconds.

Question 9.
Why does a gun recoil when a bullet is fired from the gun? Explain.
Answer:
When bullet is not fired, then gun and bullet both are at rest, thus total momentum of both together is zero. When a bullet is fired from the gun the bullet moves with very high speed in the forward direction i.e. its momentum is very high. Now according to law of conservation of momentum, total momentum should still be zero as was before the firing of the bullet. Thus to balance the momentum of bullet in the forward direction gun recoil.

Question 10.
Why a cricket player lowers his hand while taking of catch of cricket ball?
Answer:
A lot of force is needed to stop a fast moving ball. If we lower our hands while catching the ball, acceleration of the ball is decreased and we have to apply less force of stop the ball.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 11.
Write differences between balanced and unbalanced forces.
Answer:
Differences between balanced forces and unbalanced forces:

Balanced forces Unbalanced forces
1. When balanced forces act simultaneously on a body then their net resultant is zero. When unbalanced forces act simultaneously on a body then their net resultant is not zero.
2. Balanced forces are acting on a body which is at rest, these can not bring it in motion. Unbalanced forces act on a body at rest, these can bring it in motion.
3. These forces cannot bring a change in speed or direction of the motion. This force can bring change in the speed of direction of motion.
4. This force can make a change in the shape of the body. This force can not make a change in the shape of the body.

Question 12.
Why a boatman exists a force on water with his oars in the opposite direction to move forward?
Answer:
That force, which causes motion in any direction, is the reaction of the applied force. To move the boat in forward direction boatman exerts a force on water with his oars in the opposite direction. As a reaction to this force boat moves in forward direction because action and reaction are equal and opposite to each other.

Important Formulae:

  1. Force (F) = m × a
  2. Acceleration (a) = \(\frac {F}{m}\)
  3. Acceleration (a) = \(\frac {υ – u}{t}\)
  4. Momentum(p) = m × υ
  5. Pressure (P) = \(\frac {F(force)}{A(Area)}\)

Numerical Problems (Solved):

Question 1.
What acceleration will be produced in a body of mass 3 kg, when a force of 12 N is applied?
Solution:
Here, Force (F) = 12 N
Mass (m) = 3 kg
Acceleration (a) =?
We know F = m × a
12 = 3 × a
or a = \(\frac {12}{3}\)
∴ a = 4 m s-2

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 2.
How much force will be required to produce an acceleration of 4 m s-2 in a ball of mass 6 kg?
Solution:
Mass of the ball (m) = 6 kg
Acceleration produced in the ball (a) = 4ms-2
Force (F) = ?
We know, F = m × a
= 6 × 4
Force (F) = 24 N

Question 3.
A man throws a ball of mass 0.5 kg vertically upwards with a velocity of 10 m s-1. What will be its initial momentum? What would be its momentum at the highest point of its reach?
Solution:
Mass of the ball (m) = 0.5 kg
Initial velocity of ball (u) = 10 m s-1
Final velocity of the ball (v) = 0 (At highest point the ball comes to rest)
Initial momentum of the ball = m × u
= 0.5 × 10
= 5 kg – m/s
Final momentunvof the ball = m × υ
= 0.5 × 0 = 0

Question 4.
A steam engine of mass 3 × 104 kg pulls two wagons each of mass 2 × 104 kg with an acceleration of 0.2 m s-2. Neglecting frictional force, calculate the:
1. force exerted by the engine.
2. force experienced by each wagon.
Solution:
Mass of steam engine (m1) = 3 × 104 kg
Mass of two wagons (m2) = 2 × (2 × 104 kg)
Total mass of the engine and wagons (m) = m1 + m2
= 3 × 104 + 4 × 104
= (3 + 4) × 104 kg
= 7 × 104 kg
Acceleration (a) = 0.2 ms-2
1. We know, F = m × a
= 7 × 104 × 0.2
= 1.4 × 104 N
2. Force experienced by each wagon = 1.4 × 104 N

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 5.
A bullet of mass 20 g moving with a speed of 500 ms-1 strikes a wooden block of mass 1 kg and gets embedded in it. Find the speed with which block moves along with the bullet.
Solution:
Suppose the final velocity of the bag with bullet embedded in it is υ.
For Bullet, m1 = 20 g = 0.02 kg, u1 = 500 m s-1, υ1 = υ
For Block, m2 = 1 kg, u2 = 0, υ2 = υ
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 9

Question 6.
A car travelling at the speed of 108 km/h takes 4 s to stop on applying brakes. Calculate the force acting on the car after applying brakes. Total mass of the car (including passengers) is 1000 kg.
Solution:
Inital velocity of the car (u) = 108 km/h
= 108 × \(\frac {5}{18}\)
= 30 ms-1
Final velocity of the car (υ) = 0 ms-1
Total mass of the car (m) = 1000kg
Time taken to stop the car (t) = 4s
Force ‘F’ due to application of brakes = m\(\frac {υ – u}{t}\)
= 1000kg × \(\frac {0 – 30}{4s}\)ms-1
= – 750 kg – ms-2
= – 750 N
Negative sign shows that the force applied by the brakes is acting in a direction opposite to the direction of motion of the car.

Question 7.
Which one requires more force, a body of mass 2 kg accelerated at the rate 5 m s-2 or a body of mass 4 kg accelerated at 2 m s-2
Solution:
Given:
m1 = 2kg, a1 = 5 ms-2
m2 = 4kg, a2 = 2 ms-2
Force required for first body, F1 = m1 × a1
= 2kg × 5ms-2
Force required for second body, F2 = m2 × a2
= 4kg × 2ms-2 = 8N
F1 > F2
From this, it is clear that the first body would require more force.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 8.
A bullet of mass 0.03 kg is fired from a gun of mass 3 kg which leaves the muzzle of the gun write a velocity of 100 ms-1. If bullet takes 0.003 s to come out of the gun then calculate the force acting on the gun.
Solution:
Mass of the gun (m1) = 3 kg
Mass of the bullet (m2) = 0.03 kg
Before firing, both the gun and the bullet are at rest
∴ Initial velocity of the gun (u1) = 0
Initial velocity of the bullet (u2) = 0
After firing. Final velocity of the gun (υ1) =?
Final velocity of the bullet (υ2) = 100 m s-1
According to the law of conservation of momentum
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 11
Negative sign shows that the gun moves backward and experiences a force of 1000 N

Question 9.
From a rifle of mass 5000 g a bullet of 20 g is fired with a velocity of 500 ms-1 with respect to the ground. Find the velocity of recoil of the rifle.
Solution:
According to law of conservation of momentum,
MV + mυ = 0
V = – \(\frac {mυ}{M}\)
Now m = 0.02 kg, υ = 500 ms-1 and M = 5 kg
∴ \(\frac {20×500}{5000}\)
or V = – 2ms-1
Negative sign shows recoiling of the rifle.

Question 10.
A girl of 40 kg mass jumps with a horizontal velocity of 5 ms-1 on a stationary trolley. The wheels of the skate are frictionless. What will be the velocity of the girl in the position of start of the trolley. Suppose no unbalanced force is acting in the horizontal direction.
Solution:
Suppose in the initial motion of the trolley, the velocity of the trolley and the girl is υ
Total initial momentum of the girl and trolley = 40 kg × 5 m s-1 + 3 kg × 0 ms-1
= 200 kg ms-1 + 0
= 200 kg – ms-1
Total momentum of the girl and trolley after she jumps on the trolley.
= 40 kg × υ ms-1 + 3 kg × υ m s-1
= (40 + 3) × υ kg – ms-1
= 43 υ kg – ms-1
According to the law of conservation of momentum.
43υ = 200
υ = \(\frac {200}{43}\)
= 4.65 ms-1
The girl boarding on the trolley will move with a velocity of 4.65 m s-1 in the direction of jump.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 11.
The following is the distance-time table of an object in motion.

Time in seconds Distance in metres
0 0
1 1
2 8
3 27
4 64
5 125
6 216
7 343

(a) What conclusion can you draw about acceleration? Is it constant, increasing, decreasing or zero?
(b) What do vou infer about forces acting on obiect?
Solution:
PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion 10
(a) Table shows that the motion is accelerated and acceleration is increasing uniformly with time i.e. acceleration is increasing by 6 ms-2 every second.
(b) Since acceleration is increasing uniformly, the force is also increasing uniformly with time.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 12.
Two persons manage to push a motor car of mass 1,200 kg at uniform velocity on the road. The same motor can be pushed by three persons to produce an acceleration of 0.2 ms-2. With what force does each person push the motor car? Assume that all persons push motor car with same muscular effort.
Solution:
Here, mass of the car (m) = 1200 kg
Acceleration produced in the car (a) = 0.2 ms-2
Acceleration produced by first two persons in the car = 0
It is clear that when third person pushes the car, an unbalanced force acts on the car which produces an acceleration of 0.2 ms-2
∴ Force applied by three persons together (F) = m × a
= 1200 × 0.2 = 240 N
Now because three persons together push the car using their muscular force to produce an acceleration of 0.2 ms-2
∴ Push (Force) applied by each person = \(\frac{F}{3}\)
= \(\frac{240 \mathrm{~N}}{3}\)
= 80 N

Question 13.
A hammer of mass 500 g moving at 50 ms-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Solution:
Mass of the hammer (m) = 50 gm
= \(\frac{500}{1000}\)kg
= \(\frac{1}{2}\)kg
Initial velocity (u) = 50ms-1
Final velocity (v) = 0ms-1
Time (t) = 0.01s
Force (F) = ?
We know, υ = u + at
0 = 50 + a × 0.01
– 50 = a × \(\frac{1}{100}\)
a = – 50 × 100
∴ a = – 5000 ms-2
Here, negative sign indicates that there is retardation.
Now force applied by the nail on the hammer (F) = m × a
= \(\frac{1}{2}\) × (- 5000)
= – 2500 N
∴ Force = 2500 N

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 14.
A motorcar of mass 1,200 kg is moving along a straight line with uniform velocity of 90 km h-1. Its velocity is slowed down to 18 km h-1 in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
Here, mass of the car (m) = 1200 kg
Time (t) = 4 s
Initial velocity of the car (u) = 90 km/h
= \(\frac{90 \times 5}{18}\)ms-1
= 25ms-1
Final velocity of the car (υ) = 18 km/h
= 18 × \(\frac{5}{18}\)ms-1 = 5ms-1
Acceleration of the car (a) = ?
Change in momentum of the car = ?
Magnitude of the force acting on the car (F) = ?
We know, υ = u + at
5 = 25 + a × 4
– 20 = 4a
or a = \(\frac{20}{4}\)
Initial momentum = mu = 1,200 × 25 = 30,000kg ms-1
Final momentum = 1,200 × 5 = 6,000kg ms-1
Change in momentum = Final momentum – Initial momentum
= mυ – mu
= m × (υ – u)
= 1200 × (5 – 25)
= 1200 × (- 20)
= – 24000 kg ms-1
= 24000kg – ms-1 (decrease)
F = \(\frac{m×(υ – u)}{t}\)
= \(\frac{1,200×(5 – 25)}{4}\)
F = – 6,000 N
Magnitude of the force (F) = 6000N

Very Short Answer Type Questions:

Question 1.
To bring a body into motion, what is required to be done?
Answer:
It is required to be pulled, pushed or kicked.

Question 2.
Why does an object fall down?
Answer:
Due to unbalanced gravitational force.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 3.
Which type of force is required to change the direction of motion of the body – a balanced or unbalanced force?
Answer:
An unbalanced force is required.

Question 4.
Why does a body stop after rolling down for some time?
Answer:
Due to frictional force.

Question 5.
How can force of friction be reduced?
Answer:
By polishing/smearing the surface with a lubricant.

Question 6.
Which scientist postulated the three laws of motion?
Answer:
Newton.

Question 7.
By which other name the first law of motion is known?
Answer:
Law of Inertia.

Question 8.
Of heavy and light objects, which have more inertia?
Answer:
Heavier objects have more inertia.

Question 9.
What is the S.I unit of momentum?
Answer:
The S.I. unit of momentum is kg – ms-1.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 10.
Why is talcom powder sprinkled on carrom board while playing?
Answer:
In order to reduce friction.

Question 11.
Why does an athelete run before taking a high jump?
Answer:
To increase inertia in order to take high leap.

Question 12.
What is law of conservation of momentum?
Answer:
Law of conservation of momentum. The sum total of momentum of two objects before and after collision remains same unless some external force is applied.

Question 13.
A bus and a ball are moving with the same speed. To stop which one would require more force?
Answer:
Due to more inertia of bus, more force is required to be applied for stopping it.

Question 14.
A vehicle stops on applying brakes. During this activity, what happens to its momentum?
Answer:
In this activity the major part of momentum of the vehicle is transferred to the ground while the remaining part is transferred to the air molecules.

PSEB 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Question 15.
1 kg wt is equal to how many newtons?
Answer:
1 kg wt = 9.8 Newtons.

Question 16.
1 newton is equal to how many kg wt?
Answer:
1 newton = 0.102 kg wt.

Question 17.
On which physical quantity inertia of an object depends?
Answer:
On mass of the object.

Question 18.
On which Newton’s law of motion, the working principle of rocket is based?
Answer:
On Newton’s third law of motion.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Punjab State Board PSEB 9th Class Science Important Questions Chapter 8 Motion Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Long Answer Type Questions:

Question 1.
Derive mathematically the equations of motion.
Answer:
Equations of motion under uniform acceleration: The following are the equations of motion of an object moving in a straight line with uniform acceleration.
1. υ = u + at
2. S = ut + \(\frac{1}{2}\)at2
3. υ2 – u2 = 2aS
1. First Equation of Motion, υ = u + at
Let u be the initial velocity and ‘a’ be the uniform acceleration of a body in motion so that after every 1 sec its velocity increases by ‘a’.
Increase in velocity of the body after 1 sec = a
Secondary Increase in velocity of the body after 2 sec = a + a = 2 a
∴ Increase in velocity after t sec = a × t = at
Therefore, velocity of an object after t secs = Initial velocity + increase in velocity of the object after ‘t’ secs
or υ = u + at

2. Second equation of motion S = ut + \(\frac{1}{2}\)at2
Let ‘u’ be the initial velocity of the body in motion having uniform acceleration ‘a’ and after ‘t’ second the final velocity be ‘υ’.
∴ Increase in velocity of the body after 1sec = a
Velocity of the body after 1 sec from start of its motion = (u + a)
Velocity of the body 1 second before the end of its motion = (υ – a)
PSEB 9th Class Science Important Questions Chapter 8 Motion 1
PSEB 9th Class Science Important Questions Chapter 8 Motion 2

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 2.
For uniform accelerated motion, draw by graphical method establish the following equations of motion:
1. υ = u + at
2. S = ut + \(\frac{1}{2}\)at2
3. υ2 = u2 + 2aS
Answer:
Consider an object moving with initial velocity ‘u’ along a straight line and uniform acceleration ‘a’. Suppose its final velocity becomes υ after time t and during this time it travels a distance S. If time is represented along X-axis and velocity along Y-axis, then velocity-time graph obtained will be BA, an ipchned straight line. With its help we can derive the various equators of motion.
Here OA = ED = u,
OC = EB = υ and OE = t = AD
(a)Equation for velocity-time relation. We know that
Acceleration = Slope of velocity-time graph AB or
PSEB 9th Class Science Important Questions Chapter 8 Motion 3
This proves the first equation of motion.

(b) Equation for position-time relation
We know that area below velocity-time graph and time axis gives the distance covered by the body.
PSEB 9th Class Science Important Questions Chapter 8 Motion 3
(S = ut + \(\frac{1}{2}\)at2)
Distance travelled by the object in time t is
S = Area of the trapezium OABE
= Area of rectangle OADE + Area of triangle ADB
= OA × OE + \(\frac{1}{2}\)DB × AD
a = \(\frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{DB}}{t}\)
or DB = at
= u × t + \(\frac{1}{2}\)at × t
or S = ut + \(\frac{1}{2}\)at2
This proves the second equation of motion.

(c) Equation for position-velocity relation (υ2 – u2 = 2aS).
The distance travelled by object in time t is:
S = Area of trapezium OABE
= \(\frac{1}{2}\)(EB + OA) × OE
= \(\frac{1}{2}\)(EB + ED) × OE …….. (1)
PSEB 9th Class Science Important Questions Chapter 8 Motion 5

Question 3.
Draw velocity-time graph for a body moving with uniform velocity. Hence show that the area under the velocity-time graph gives the distance travelled by the body in a given time interval.
Answer:
Distance covered as area under the velocity-time graph. In Fig. straight line PQ is the velocity-time graph of a body moving with a uniform velocity, ‘υ’ represented by OP.
PSEB 9th Class Science Important Questions Chapter 8 Motion 6
Area of rectangle ABCD
= AD × AB
= OP × AB
= υ × (t2 – t1)
= Velocity × time
= Distance travelled in time interval (t2 – t1)
Hence, the area under the velocity-time graph gives the distance travelled by the body in the given time interval.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 4.
What is meant by Angular velocity? How is it related to linear velocity? Derive the relation.
Answer:
Angular velocity: The angle traced out by a body moving along a circular path in unit time is called Angular velocity. It is usually represented by a greek letter ω (omega). If θ is the angle covered by a body in time ‘t’, then
Angular velocity (ω) = \(\frac{Angle traced out by the moving body(θ)}{t}\)
∴ ω = \(\frac{θ}{t}\)
or θ = ωt

Relation between Linear velocity and Angular velocity:
Suppose a body in motion in a circular path of radius ‘r’ having linear velocity υ. Suppose ‘S’ is the linear distance covered by the body in time ‘t’ and relative to this, ‘θ’ is the angular displacement produced, then
PSEB 9th Class Science Important Questions Chapter 8 Motion 7

Short Answer Type Questions:

Question 1.
Define rest and motion. Give one example for each.
Answer:

  • Rest: A body is said to be at rest if it does not change its position with respect to its surroundings. A book lying on the table in a room is at rest with respect to the other objects in the room.
  • Motion: A body is said to be in motion if it changes its position with time with respect to its surroundings. A car running on the road is in motion with respect to the electricpole, trees along the roadside.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 2.
Show that rest and motion are relative terms.
Or
Can an object be af rest as well as in motion at the same time?
Answer:
An object may be at rest relative to one surrounding object and at the same time it may be in motion with respect to some other object. For example, a passenger sitting in moving train is at rest relative to other co-passengers. But since he is sharing the motion of the train, so he is in motion relative to the outside trees, lamp posts etc. Thus, rest and motion are relative terms.

Question 3.
Give some points of differences between distance and displacement.
Or
Give two differences between distance and displacement.
Answer:
Differences between distance and displacement:

Distance Displacement
1. Distance is the length of the actual path covered by a body, irrespective of its direction of motion. Displacement is the shortest distance between the initial and final positions of a body in a given direction.
2. Distance between two given points may be same or different for different paths chosen. Displacement between two given points is always same.
3. It is a scalar quantity. It is a vector quantity.
4. Distance covered is always positive or zero. Displacement covered may be positive, negative or zero.

Question 4.
What is meant by uniform motion?Give an example.
Answer:
Uniform motion. If an object covers equal distances in equal intervals of time, hoxvever small the time interval may be, then the motion of the object is said to be uniform motion. For example, suppose a bus moves 10 km in the first 15 minutes, 10 km in second 15 minutes, 10 km in third 15 minutes and so on. Then one can say that the bus is in uniform motion.

Question 5.
Define the term velocity. What is its SI unit?Is it a scalar or vector quantity?
Answer:
Velocity: Velocity of a body is defined as the displacement produced per unit time. It is also defined as the speed of a body in a given direction.
Velocity = Displacement/Time
If ‘S’ is the distance travelled by a body in a given direction and t is the time taken to travel that distance, then the velocity ‘υ’ is given by
υ = \(\frac{S}{t}\)
The SI unit of velocity is m s-1.
Velocity is a vector quantity because it requires both magnitude and direction for its complete description.

Symbols of Physical Quantities and Important Formulae:

(A) Symbols of Physical Quantities

  • Time = t
  • Speed = v
  • Distance = S
  • Initial Velocity = u
  • Final Velocity = v
  • Acceleration = a
  • Average Velocity = Vav

(B) Important Formulae:

If initial velocity u at time t velocity υ and uniform acceleration a Then the under given relations are called Equations motion:
υ = u + at (Velocity at time t)
S = ut + \(\frac{1}{2}\)at2(Displacement in time ‘t’)
υ2 = u2 + 2aS (Velocity square relation)
Displacement in n th The Second of time (Sn) = u + (2n – 1)

PSEB 9th Class Science Important Questions Chapter 8 Motion

Numerical Problems (Solved):

Question 1.
A police car running on a highway with a speed of 30 km/h fires on the vehicle of thiefs running in the same direction at a speed of 192 km/h. If the velocity of the bullet is 150 m/s then with what velocity the bullet will hit the thiefs?
Solution:
Velocity of the bullet fired = 150 m/s
= 150 × \(\frac{18}{5}\)km/h = 540 km/h
∴ Total speed = Speed of police vehicle + Speed of bullet
= 30 km/h + 540 km/h = 570 km/h
Speed of the bullet relative to the speed of thief’s vehicle in the same direction
= (570 – 192) km/h
= 378 km/h
= \(\frac{378 \times 1000}{60 \times 60}\) m/s
= 105 m/s

Question 2.
A train 50 m long travels on a plain and level track and reached a post in 5 secs. Find
1. speed of the train
2. the time train will take to cross 450 m long bridge.
Solution:
1. Since the train takes 5 secs to reach the post, it covers a distance equal to its own length, then
speed of train = \(\frac{Total distanve travelled}{Total time taken}\)
= \(\frac{50m}{5s}\)
= 10 m/s

2. Distance travelled by the train to cross the bridge
= Length of the bridge + Length of train
= 450 m + 50 m
= 500 m

∴ Time taken to cross the bridge = \(\frac{Total distance covered}{speed of the train}\)
= \(\frac{500m}{10m/s}\)
= 50 s

Question 3.
A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h upto distances less than 500 m. If a cheetah spots his prey at a distance of 100 m. What is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h.
Solution:
Here, υ = 90 km/h
= \(\frac{90 \times 1000 \mathrm{~m}}{3600 \mathrm{~s}}\)
= 25 m/s
s = 100m
Minimum time, t = \(\frac{s}{υ}\)
= \(\frac{100}{25}\) = 4s.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 4.
A car travels a certain distance with a speed of 50 km/h and returns with a speed of 40 km/h. Calculate the average speed for the whole journey.
Solution:
Let the one-way distance = x km
PSEB 9th Class Science Important Questions Chapter 8 Motion 8

Question 5.
On a 100 km track, a train travels the first 30 km at a uniform speed of 30 km h-1. How fast must the train travel the next 70 km so as to average 40 km h-1 for the entire trip?
solution:
PSEB 9th Class Science Important Questions Chapter 8 Motion 9
PSEB 9th Class Science Important Questions Chapter 8 Motion 10

Question 6.
A railway train 50 m long passes over a bridge 250 m long with uniform velocity of 10 m s-1. How long will it take to completely, pass over the bridge?
Solution:
PSEB 9th Class Science Important Questions Chapter 8 Motion 11
To completely pass over the bridge, the train will have to cover the length of the bridge, as well as its own length i.e., AB + BC
∴ Total distance to be covered, S = 50 + 250 = 300 m
Velocity of train, u = 10 m s-1
Here velocity is uniform, hence
a = 0 m s-2
Using S = ut + \(\frac{1}{2}\)at2
300 = 10t + 0
or t = 30 s
Hence train will pass completely over the bridge in 30 s.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 7.
A car starts from rest and moves with uniform acceleration for 40 s. It then moves with uniform velocity attained by it and is finally brought to rest in 30 s under uniform retardation. The car covers a total distance of 1,380 m in 2.5 min. Calculate constant speed, acceleration and retardation.
Solution:
Let x be the constant speed. Now the total distance covered is equal to total area between velocity-time graph and time axis. Therefore
Distance covered = area of ∆ABC + area of rectangle BCDE + area of ADEF
PSEB 9th Class Science Important Questions Chapter 8 Motion 12

Question 8.
The graph shown in Fig. indicates the position of body at different positions. Calculate the speed of the body as it moves from (i) A to B; (ii) B to C and (iii) C to D.
Solution:
PSEB 9th Class Science Important Questions Chapter 8 Motion 13
Speed from A to B = Slope of AB
= \(\frac{\mathrm{BF}}{\mathrm{AF}}=\frac{3-0}{5-2}\) = 1ms-1
(ii) Speed from B to C is zero since slope of BC is zero. From B to C, the body is at rest.
(iii) Speed between C and D = Slope of CD
= \(\frac{D G}{C G}=\frac{7-3}{9-7}\)
= \(\frac{4}{2}\) = 2ms-1

Very Short Answer Type Questions:

Question 1.
What is motion?
Answer:
Motion: When an object changes its position with time then it is said to be in motion.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 2.
What is displacement of object?
Answer:
Displacement. The change in position of an object in a given direction is known as displacement. It is measured by the shortest distance moved by an object from its initial position to final position.

Question 3.
Which device shows the speed of vehicles?
Answer:
Odometer.

Question 4.
What is uniform motion?
Answer:
Uniform motion. When a body covers equal distances in equal intervals of time then its motion is called uniform motion.

Question 5.
Give two examples of non-uniform motion.
Answer:
1. A car moving on the road.
2. A man doing exercise in park.

Question 6.
Define speed.
Answer:
Speed. The distance travelled by a body in a unit time is called its speed.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 7.
What is the SI unit of speed?
Answer:
S.I. unit of speed is ms-1.

Question 8.
How is average speed obtained?
Answer:
Average speed. It is the total distance travelled by an object divided by the time taken to travel that distance.
Average Speed = \(\frac{Total distance travelled}{Total time taken}\)

Question 9.
What is velocity?
Answer:
Velocity: The speed in a particular direction is called velocity.
Or
It is the distance travelled by an object in a particular direction.

Question 10.
What is acceleration?
Answer:
Acceleration. It is defined as the rate of change of velocity with time.
Acceleration = \(\frac{Change of velocity}{Total time taken}\)

Question 11.
What is the SI unit of acceleration?
Answer:
ms-2

Question 12.
A cricket player tosses the ball upward and again catches it. What is the total displacement?
Answer:
Total displacement is zero.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 13.
Is displacement a scalar or vector quantity?
Answer:
Displacement is a vector quantity because it needs both magnitude and direction for its complete representation.

Question 14.
What would be acceleration of a body if its velocity-time graph is line parallel to the time axis?
Answer:
Zero, as the body possesses uniform velocity.

Question 15.
A body is moving along the boundary of a square plot of land with constant speed. Does its velocity remain unchanged?
Answer:
No. because its velocity changes due to change in direction.

Question 16.
What will be the position-time graph of a city bus standing at rest at a depot?
Answer:
A straight line parallel to the time axis.

Question 17.
What is the nature of the distance-time graph for an object moving uniformly along a straight long road?
Answer:
A straight line inclined to the time axis.

PSEB 9th Class Science Important Questions Chapter 8 Motion

Question 18.
Does the speedometer of a car measure its average speed?
Answer:
No. It measures only instantaneous speed.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Punjab State Board PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Long Answer Type Questions:

Question 1.
Give a broad outline classification of kingdom plantae.
Answer:
Classification of kingdom Plantae:

  1. Kingdom Plantae is divided into five subdivisions i.e. Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms.
  2. Thallophyta includes Algae.
  3. Algae are found in water and moist places.
  4. Fungi are non-green, heterotrophic (parasitic and saprophytic) thallophytes.
  5. In Bryophytes, plant body is thallus-like (thalloid) or leaf-like (foliose). Body differentiated to form stem and leaf-like structure. Riccia, Marchantia and Funaria are examples.
  6. In pteridophytes plant body is differentiated into root, stem and leaves. There is specialised tissue for conduction of water and other materials. This group includes Marselia, Ferns, and horse tails.
  7. In Gymnosperms, seeds are not covered with fruit wall or pericarp. Cycas, Pine, and deodar are examples.
  8. In Angiosperms, seeds are covered with fruit wall or pericarp to form fruits.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 2.
What are phanerogams? Write characters of gymnosperms and angiosperms.
Answer:
Phanerogamae. These are seed-bearing plants. Vascular tissue is present. They are found on land. This division is composed of two main sub-divisions i.e. Gymnosperms and Angiosperms.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 1
Characters of gymnosperms:

  1. They are flowering plants and seeds develop inside fruit.
  2. Cotyledons are the plant embryos present inside the seed.
  3. Plant body is divided into root system and shoot system. Examples: Mango, Rose, Wheat, Grasses.

Question 3.
Give an outline classification of animal kingdom with two characters of each phylum with example.
Answer:
Classification of kingdom Animalia:

Name of phylum Characters Examples
1. Porifera Pore bearing body, spongocoel present. Leucosolenia, Sycon, Spongilla
2. Coelenterata Diploblastic, radial symmetry, nematocysts present. Hydra, Obelia, Physalia, Aurelia
3. Platyhelminthes Triploblastic, flat body, acoelomate Fasciola, Taenia, Turbellaria
4. Nematehelminthes Pseudocoelomate, cylindrical body Ascaris, Dracunculus
5. Annelida coelomate, segmented body, setae or suckers or parapodia present. Pheretima, Nereis, Leech
6. Arthropoda Jointed appendages, chitinous exoskeleton, haemocoel present Crab, prawn, centipede, spider, cockroach
7. Mollusca soft-bodied, shell present Pila, Unio, Chiton, Dentalium
8. Echinodermata spiny skin, water vascular system Sea Lily, Seastar, Sea urchin, Sea cucumber
9. Hemichordata Notochord in proboscis, Tomaria larva Balanoglossus
10. Chordata Notochord present, dorsal hollow nerve tube, pharynx perforated by gill slits Fishes, Frog, Toad, Snakes, Lizard, Birds, Mammals

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 4.
Write characters of phylum protozoa and phylum porifera.
Answer:
Characters of Phylum Protozoa:
1. Very minute, one-celled microscopic organisms. Cell itself is an organism therefore, they are also called acellular organisms.
2. Structure is very simple. The body consists merely of a mass of protoplasm. There is no tissue or organ formation.
3. Generally, there is no skeleton. However, locomotory and feeding organelles such as pseudopodia, flagella or cilia may be present.
Example. Amoeba, Entamoeba, Paramecium, Trypanosoma.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 2
Characters of Phylum Porifera:
1. They are primitive, sessile animals with specialized cells but no tissues or organs.
2. Pores ali over the body, no digestive tract.
3. Collar cells filter out food particles from water current flowing through canal system.
4. Mostly marine.
5. They are called sponges.
Examples: Euplectella, Sycon, Spongilla.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 3

Question 5.
Give a brief account of Whittaker’s five kingdoms of life.
Answer:
Whittaker’s Five Kingdoms are: Monera, Protista, Fungi, Plantae and Animalia.

  1. Kingdom Monera: Monerans are prokaryotes. Examples: Bacteria, Blue-green algae.
  2. Kingdom Protista: Protists are unicellular eukaryotes having widely diverse lifestyles. Protists form tine precursors from which higher forms-Plantae, Fungi and Animalia-arose on photosynthetic, assimilative and ingestive lines. Examples: Mastigophores, Ciliates, Sarcodines, Sporozoans, Slime moulds, Diatoms.
  3. Kingdom Fungi: Fungi are eukaryotic, heterotrophic organisms with assimilative nutrition. Examples: Rhizopus, Agarieus,
  4. Kingdom Plantae: Plantae includes eukaryotic, multicellular, photosynthetic organisms. It includes green multicellular plants.
  5. Kingdom Animalia: Animals are eukaryotic, multicellular consumer characterised by ingestive nutrition. It includes non-chordates and chordates.

Question 6.
Differentiate between plants and animals.
Answer:
Differences between plants and animals:

Characters Plants Animals
1. Mode of nutrition Autotrophs, prepare food from inorganic substances. Heterotrophs, feed on com­plex organic compounds.
2. Photosynthesis Photosynthesize. Do not photosynthesize.
3. Chlorophyll Possess chlorophyll. Have no chlorophyll.
4. Locomotion Are fixed but move parts of the body. Move the whole body.
5. Branching Branching body. Compact body.
6. Response Are less sensitive and respond slowly. Are highly sensitive and re­spond quickly.
7. Cell wall Cellulose cell walls usually present. Have no cellulose cell walls.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 7.
Enlist the peculiar features of phylum Arthropoda.
Answer:

  1. These have organ-system organization and bilateral symmetry.
  2. These are triploblastic and haemocoelomates (have hemocoel-a body cavity with blood).
  3. Their body is externally covered by a sclerotized chitinous exoskeleton formed of plates, called sclerites, and are formed of chitin.
  4. Body is divided into 2 (cephalothorax and abdomen) or 3 parts (head, thorax and abdomen).
  5. Mouth is surrounded by mouthparts which are of different types in different arthropods.
  6. These have jointed appendages which are of different types to perform different functions.
  7. These have complete and coiled gut.
  8. Respiration occurs by gills or book lungs or tracheae (air tubes).
  9. Blood is colourless and is called hemolymph.
  10. Excretion occurs by green glands or coxal glands or malpighian tubules.
  11. Examples: Insects Periplaneta (cockroach), Butterfly, Dragonfly, Housefly, Crustaceans, Crab, Centipede, Palaemon (Prawn), Millipede and Scorpion (Arachnid).

Question 8.
Write distinct features of phylum Echinodermata.
Answer:

  1. Spiny skin of these animals contain dermal calcareous plates. Spines may be movable (c.g. sea urchin).
  2. Pentaradial symmetry in adult but larva is bilaterally symmetrical.
  3. Head absent. But divisible into central disc and arms.
  4. Water vascular system (Ambulacra! system) is present. It consists of an array of radiating and tube-like appendages called tube feet. Locomotory organs are tubular tube feet or podia and also aid in capturing prey.
  5. Body has oral surface bearing mouth and aboral surface bears anus.
  6. Tube feel, and dermal branchiae aid in gaseous exchange.
  7. All are marine, mostly gregarious and free living, creep slowly on the sea bottom.
    Examples: Star fish, Sea urchin. Sea cake, Sea lily and Sea cucumber.

Question 9.
Give diagnostic characters of phylum chordata.
Answer:
Diagnostic characters of Phylum Chordata:

  1. Presence of notochord, at any stage of life history. Notochord is stiff rod-like structure which is retained throughout life in lower chordates. It is replaced by vertebral column in higher chordates.
  2. Presence of dorsal, hollow nervous system.
  3. Pharynx is perforated by gill slits at any stage of life history.
  4. Tail is present at any stage of life history.
  5. Respiratory pigment haemoglobin is present in the R.B.C.

Short Answer Type Questions:

Question 1.
Briefly describe diversity of life.
Answer:
Diversity of life (also called biological diversity or biodiversity) is the variety of living systems. It may refer to extinct organisms, but also to their diversity in the past. It usually covers multiple levels of biological diversity. About 2 million species have been described upto date. About 17000 – 19000 new species are added every year to the list. Exact number of species present on earth is unknown, current estimates favouring figures in the range of 8-12 million.

Thus biological diversity includes all forms of life including micro-organisms, plants and animals. Biological diversity is constantly changing.

Question 2.
Differentiate taxonomy and systematics.
Answer:
Difference between Taxonomy and Systematics:

Taxonomy Systematics
1. It is the science of identification, nomenclature and classification.

2. It deals with rules and principles of classification.

1. Systematics is the science of identification, nomenclature, description and classification.

2. It brings out unique properties at every level of classification

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 3.
Describe the importance of classification.
Answer:
Importance of classification:

  1. Recognition: Classification of living beings recognises the basic taxonomic units or species.
  2. Description: Classification of organisms is responsible for description of species.
  3. Relationship: It tells a possible way for grouping these units on the basis of their resemblances and relationships.
  4. Phylogeny: It gives us most of the information permitting a reconstruction of phylogeny of life. It helps in understanding the evolution of organisms.
  5. Applied biology: It has great role in applied biology (agriculture, public health, environmental biology) also.
  6. The introduction of harmful plants and animals can be checked. Exact identification of harmful pests, a disease vector, a pathogen and their control is made possible with knowledge of applied biology.
  7. Human health: Exact identification of insects helps in controlling the epidemic diseases like malaria, filaria, dengue fever, kala-azar etc.
  8. Horticulture: Several ornamentals have been introduced due to proper identification and nomenclature.

Question 4.
Why are local names not sufficient to recognize the organisms? What are the advantages of keeping scientific names?
Answer:
Local names are not understood elsewhere. The common name of a species often varies with language and region of world. In Biology, we deal with a very large number of species. It would not be possible to refer to them unless each one of them had a separate name for itself.
Advantages of scientific names:

  1. The scientific names are same all over the world.
  2. They are uniformly binomial.
  3. They are definite and accepted universally.
  4. They are descriptive.
  5. They indicate general relationship.

Question 5.
Write different categories of classification.
Answer:

  1. Species: It is defined as a dynamic group of organisms, which:
    • resemble each other in all essential respects, i.c. structure and function.
    • Interbreed among themselves to produce fertile young ones of their own kind.
  2. Genus: It forms the taxonomic higher category than species. It is a group of closely related species.
  3. Family: A number of genera having several common characters constitute a family.
  4. Order: A number of families having many common characters are placed in an order.
  5. Class: The class is the basic category. Similar orders are grouped together in the common class. A class is generally a subdivision of a phylum.
  6. Phylum: A number of classes having common features constitute a phylum.
  7. Division: In the case of plants, many classes constitute a division which corresponds to the phylum of animal kingdom.
  8. Kingdom: All the animals are included in Kingdom Animalia, while all plants are included in the Kingdom Plantae. It is the highest taxonomic category.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 6.
Classify tiger and mango.
Answer:

Category Tiger Mango
Kingdom Animalia Plantae
Phylum/Division Chordata Tracheophyta
Sub-Phylum Vertebrata
Class Mammalia Mangoliopsida
Order Carnivora Spindales
Genus Panther a Mangifera
Species Tigris Indica

Question 7.
List differences between monocot plants and dicot plants.
Answer:
Apart   below:

Monocot plants Dicot plants
1. Roots are generally adventitious. 1. They usually possess tap root.
2. Venation in leaves is parallel. 2. Venation is reticulate.
3. Flowers of monocots are trimerous i.e. floral leaves in each whorl are either three or a multiple of three. 3. Flowers in dicots are either tetramerous or pentamerous.
4. The calyx and corolla are not differentiated and the outer two whorls are exactly alike to form perianth. 4. The flower has distinct calyx and corolla.
5. Vascular bundles in stem are scattered and closed. 5. The vascular bundles in dicot stem arranged in-ring and are open.

Question 8.
What are the basis of Whittaker’s system of classification?
Answer:
Whittaker based his classification on the following three criteria:

  1. The prokaryotic versus eukaryotic structure of cell.
  2. The unicellular versus multicellular organization.
  3. The three different modes of nutrition i.e. autotrophic, absorptive and holozoic (ingestive).

Question 9.
What are the advantages of five-kingdom system of classification?
Answer:
Advantages of five-kingdom system:

  1. The subdivisions of two kingdoms have been redistributed among additional kingdoms. Such an arrangement reflects the phylogeny evolutionary history of different life styles in a better way.
  2. This system allows us to visualize the increase of complexity with evolutionary time.
  3. Five kingdom arrangement allows us to visualize the divergence of their modes of nutrition in the multicellular organisms.
  4. An added advantage of Whittaker’s system lies in the coherence and definable characters of a kingdom as a unit of classification.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 10.
Name the five kingdoms as proposed by R.H. Whittaker. Give at least one example in each case.
Answer:
Robert H. Whittaker of Cornell University organised the living organisms into five kingdoms as follows:

  1. Kingdom Monera (Bacteria, Blue-green algae)
  2. Kingdom Protista (Amoeba, Euglena)
  3. Kingdom Fungi (Mushroom)
  4. Kingdom Plantae (Green Plants)
  5. Kingdom Animalia (Animals)

Question 11.
Give general characters of kingdom Monera.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 4
Characters of Kingdom Monera:

  1. Body is formed of single-cell but the cells lack nuclear membrane and membrane-bound organelles such as mitochondria, chloroplasts etc.
  2. They may be motile or non-motile. Flagella, if present, have a solid core and formed of flagellin protein.
  3. Monerans have various forms of nutrition: some are autotrophs capable of photosynthesis, others are capable of deriving energy from inorganic chemical reaction (chemosynthetic).
  4. Most species absorb organic nutrients from their environment.
  5. Reproduction is asexual.
  6. Some organisms may have cell wall.
  7. Monerans are the important decomposers in the biosphere. Examples are Bacteria, Blue-Green algae (Cyanobacteria) and Mycoplasma.

Question 12.
What kinds of organisms are grouped under Protista?
Answer:
Characters of Kingdom Protista:

  1. This kingdom includes diverse kinds of mostly unicellular and primarily aquatic eukaryotes.
  2. They are eukaryotic organisms having typical eukaryotic cell organelles such as nucleus, mitochondria, ER, Golgi bodies, plastids etc.
  3. Mode of nutrition may be absorptive, ingestive or photo-autotrophic.
  4. Mostly bear eukaryotic flagella or cilia composed of 9 + 2 internal microtubular structure.
  5. Reproduction asexual as well as sexual. Examples: Protozoan (Amoeba), Euglena, Diatoms.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 13.
Write the general characters of kingdom Fungi.
Answer:
Tire kingdom of multicellular decomposers is Fungi. This kingdom includes diverse kinds of eukaryotic, predominantly multicellular, heterotrophic organisms.
Characters of Kingdom Fungi:
1. Body organization mycelial or secondarily unicellular, composed of hyphae, coenocytic or septate.
2. They are non-green as chlorophyll is absent.
3. They are heterotrophic in nutrition and obtain food from dead and decaying organic matter by absorption.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 5
4. Cell wall is chitinous cellulose
5. Asexual reproduction by spore formation.
6. Sexual reproduction also occurs.
Examples: Bread moulds, mushrooms, puff balls and bracket fungi, Penicillium, yeast and certain parasitic fungi.

Question 14.
Make a list of characteristics of kingdom Plantae.
Answer:
Kingdom Plantae. This kingdom includes multicellular producers. The characteristics are:

  1. Complex multicellular plants adapted for photosynthesis.
  2. The plant cells are rigid because of cellulose cell wall.
  3. Mostly the cells are rigid and cannot contract and relax like animal cells do. Plants are immobile and do not exhibit the phenomenon of locomotion.
  4. Plant synthesize all organic constituents from water, C02 and inorganic forms of essential elements using light energy trapped by chlorophyll and accessory pigments.
  5. They have unlimited growth.

Question 15.
List the general characters of kingdom Animalia.
Answer:
Characters of Kingdom Animalia:

  1. Members of the group are multicellular eukaryotes with tissue differentiation.
  2. Nutrition hetero trophic, ingestive mode of intake of food.
  3. Muscle cells well developed which provide mobility.
  4. Nervous system well developed.
  5. Ecologically animals are consumers.
  6. Sexual reproduction occurs. Examples: Sponges, insects, molluscs, fishes, birds, reptiles and mammals.

Question 16.
What is binomial nomenclature? Illustrate with an example.
Answer:
In binomial nomenclature, name of every organism is composed of two components-first one is genus (generic) and second species name (specific).
Example: Scientific name of human is Homo sapiens where Homo is generic name and sapiens is specific name.

Question 17.
Give some examples of binomial nomenclature.
Answer:

  • Scientific name of potato is Solanum tuberosum Linn. Where Solanum is genus, tuberosum is species and Linn is the scientist.
  • Man is scientifically called as Homo sapiens Mill. Here Homo is a generic name and sapiens is specific name.
  • Mill is the scientist. Sometimes specific names can be given after a country or locality e.g. Rosa indica, Rumcx nepalensis.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 18.
Write some advantages of binomial nomenclature.
Answer:
(a) The biological names are same all over the world.
(b) They are uniformly binomial.
(c) They are definite and accepted universally.
(d) They are descriptive.
(e) They indicate general relationship.
(f) All newly discovered plants and animals can be named and described easily.

Question 19.
Write distinguishing features of division Thallophyta.
Answer:
Characters of Division Thallophyta:
(a) Plants belonging to this group are the simplest and primitive.
(b) The plants are made up of single cell or group of cells.
(c) The plant body is thallus i.e. it is not differentiated into stem, root and leaves. id) The reproductive organs are unicellular and thus unjacketed.
(e) The zygote formed after fertilization, gives rise to either plant body directly or produces spores.
(f) Vascular tissue absent.

Question 20.
Write three characters of Algae. Give examples.
Answer:
Characters of Algae
(a) Green in colour due to chlorophyll.
(b) They are photoautotrophs.
(c) Reserve food material is starch.
Example: Chlamydomonas, Volvox, Ulothrix, Spirogyra.

Question 21.
Sketch Aspergillus, Penicillium and Agaricus.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 6

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 22.
What are lichens?
Answer:
Lichens: They are organisms formed of intimate combination of an alga and fungus found in all types of habitat. It is an example of mutualism. Mycobiont is the fungal component and phycobiont an algal partner.
Example: Graphis, Parmelia and Usnea.

Question 23.
Write important characters of division Bryophyta.
Answer:
Characters of Division Bryophyta
(a) The plant body is not differentiated into root, stem and leaves (thalloid).
(b) Absorbing and anchoring organs are rhizoids.
(c) Vascular tissue and mechanical tissue absent.
(d) Vegetative reproduction is very common.
(e) Sexual reproduction is by gamete formation.
(f) Sex organs are multicellular and jacketed.
(g) Formation of sporogonium takes place.
Examples: Riccia, Porella, Punaria (Moss), Marchantia (Liverwort) and Anthoceros.

Question 24.
List four important features of Pteridophytes.
Answer:
Important features of Pteridophytes:

  1. The dominant plant body in ferns is sporophyte. It is differentiated into root, stem and leaves.
  2. The leaves are large (megaphyllous), variously-shaped and look like the branches. They are called as fronds.
  3. The stem may be an underground rhizome or a trunk as in tree ferns while the roots are adventitious.
  4. Ferns bear special spore-bearing leaves called the sporophylls. The spores are produced in sporangia.
  5. Spores in ferns, germinate each forming an independent, small gametophyte, the prothallus. The latter bear the sex organs, antheridia (male) and archegonia (female).

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 25.
Make a few diagrams of common pteridophytes.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 7

Question 26.
What are the two main divisions of angiosperms on the basis of number of cotyledons.
Answer:
Classification according to cotyledons
The number of cotyledons in seed of a flowering plant is either one or two. The angiosperms are divided into two groups-1. Monocotyledons and 2. Dicotyledons.
1. Monocotyledons are the plants in which the seed contains a single cotyledon; e.g. Wheat, Rice, Maize, Sugarcane, Grasses. There are about 50,000 species of them.
2. Dicotyledons are those flowering plants in which the seed contains two cotyledons.
e.g. Solarium sp, Mango, Beans, Castor, Gram, Sunflower.

Question 27.
Write two differences between gymnosperms and angiosperms.
Answer:
Differences between gymnosperms and angiosperms:

Gymnosperms Angiosperms
1. The seeds are naked.

2. These are cone-bearing plants.

1. The seeds are present within fruits.

2. They bear flowers.

Question 28.
What characters of seed plants make them specially adapted to life on land?
Answer:
Characters of seed plants which make them specially adapted to life on land are:

  1. Presence of vascular tissue.
  2. The development of seed habit removed the liquid medium as an essential feature for fertilization.
  3. Presence of cuticle in the leaves.

Question 29.
Write three features of kingdom Animalia.
Answer:
Features of Kingdom Animalia:

  1. Eukaryotic cells lack cell wall.
  2. They do not perform photosynthesis instead take readymade food, thus heterotrophs.
  3. They have power of locomotion.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 30.
Write unique features of Coelenterates (cnidarians).
Answer:
Unique Features of Coelenterates (cnidarians)

  1. Tissue level of organisation of the body.
  2. Special stinging cells, the cnidoblasts, for defence and offence.
  3. Incomplete digestive tract bounded by gastrodermis of body wall.
  4. Simple gonads without gonoducts.
  5. Show polymorphism.

Examples: Hydra, Jellyfish and Sea Anemone.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 8

Question 31.
Write characters of flatworms. Give example.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 9
Characters of Flatworms:

  1. Dorsoventrally flattened triploblastic, bilaterally symmetrical animals.
  2. They are acoelomate.
  3. Mostly parasite, a few are free living.
  4. Incomplete branched or unbranched alimentary canal.
  5. Bisexual or hermaphrodite. Life history is complicated. Examples: Planaria, Fasciola and Taenia.

Question 32.
Write features of roundworms.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 10
Features of Roundworms:

  1. Unsegmented, bilaterally symmetrical cylindrical worms also called as roundworms.
  2. They are triploblastic and pseudo- coelomatic.
  3. Mostly free living.
  4. Complete alimentary canal present.
  5. Sexes are separate and show sexual dimorphism. Fertilized egg has a thick wall and survive adverse conditions. Examples: Ascaris (Roundworm), Ancylostoma (Hookworm), Dracunculus (Guinea worm), Rhabditis.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 33.
List unique features of phylum Annelida. Give examples.
Answer:
Unique Features of Phylum Annelida:
1. Metameric segmentation i.e. body divided externally by grooves into metameres or segments and internally bv septa into compartments.
2. Nephridia for excretion and osmoregulation.
3. Closed circulatory system with respiratory pigment dissolved in the plasma. Examples : Nereis, Pheretima (Earthworm), Ilirudinaria (Leech), Aphrodite (Sea-mouse).
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 11

Question 34.
List three important distinguishing characters of phylum Arthropoda.
Answer:
Distinguishing Characters of Phylum Arthropoda
1. Body covered with chitinous exoskeleton.
2. Body bears jointed appendages.
3. One or two pairs of jointed antennae present.
4. There is an open circulatory system, and so the blood does not flow in well defined blood vessels.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 12

Question 35.
Write unique features of phylum Mollusca. Give three examples.
Answer:
Unique Features of Phylum Mollusca:
1. Three body regions head, visceral mass and foot.
2. A glandular fold, the mantle, over the body.
3. Mantle cavity with anal, excretory and genital apertures in it.
4. Calcareous shell around the body in most cases.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 13
5. A rasping organ, the radula in the buccal cavity.
Examples: Pila, Sepia, Octopus.

Question 36.
What are Echinoderms?
Answer:
1. Phylum Echinodennata includes starfish, sea cucumbers, sea urchins, sea lilies etc.
2. The animals show pentaradial symmetry.
3. There is an exoskeleton of calcareous plates and spines.
4. Locomotion is by tube feet.
5. There occurs a peculiar water vascular system. Echinoderms are confined to sea.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 14

Question 37.
Write five examples of echinoderms.
Answer:
Examples of echinoderms

  1. Asterias (Star fish)
  2. Sea urchin
  3. Antedon (Sea lily)
  4. Sea Cucumber
  5. Sea cake

Question 38.
List three characters of phylum Hemichordata. Give one example.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 15
Characters of Phylum Hemichordata

  1. Body divided into proboscis, collar and trunk.
  2. Respiration through gills.
  3. Bilateral symmetry.
  4. Example: Balanoglossus (Tongue worm).

Question 39.
Give the general characteristics of the vertebrates.
Answer:
General Characters of Vertebrates:

  1. The body is divided into three regions – a head with an internal cranium, trunk and postanal tail.
  2. The notochord is replaced by vertebral column during life history.
  3. Body contains cartilaginous or bony endoskeleton.
  4. The post-anal part called tail is usually present.
  5. There is a complex brain with special sense organs.
  6. There is well developed ventral heart with two, three or four chambers.
  7. Two pairs of lateral appendages, fins or limbs present.
  8. The excretory organs are kidneys.
  9. Respiration occurs through gills or lungs.
  10. The sexes are separate.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 40.
Differentiate exoskeleton and endoskeleton.
Answer:
Differences between exoskeleton and endoskeleton:

Exoskeleton Endoskeleton
1. It is present outside the body.

2. Examples: chitinuous cuticle of insects, calcareous shell of mollusc, hard plates of echinoderm, scales, feathers, hair, nails, horns of vertebrates.

1. It is present inside the body.

2. Bones and cartilages constitute the endoskeleton.

Question 41.
Write characters of class Pisces.
Answer:
Characters of Class Pisces:

  1. All fish varieties belong to this class which are exclusivusingely water living animals.
  2. Their skin is covered with scales/plates.
  3. They obtain oxygen dissolved in water by gills.
  4. The body is streamlined and a muscular tail is used for movement.
  5. They are cold-blooded and their hearts have only two chambers.
  6. Some fish varieties have their skeletons made entirely of cartilage, such as Sharks and some with a skeleton made of both bone and cartilage, such as Tuna or Rohu.

Question 42.
List distinguishing characters of cartilaginous fishes.
Answer:
Chondrichthyes (Cartilaginous fishes):
1. Skeleton cartilaginous
2. Mouth and nares ventral
3. Gill slits uncovered
4. Tail fin asymmetrical
5. No swim bladder
6. Intestine with scroll valve
7. Males with claspers
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 16
Examples: Scoliodon, Narcine, Torpedo, Trygon, Angler fish, Lionfish

Question 43.
List distinguishing features of bony fishes.
Answer:
Osteichthyes (Bony fishes):
1. Skeleton bony
2. Mouth anterior
3. Gill slits covered by opercula
4. Tail symmetrical
5. Swim bladder present
6. Intestine without scroll valve
7. Copulatory organs claspurs absent.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 17
Examples: Labeo, Rita, Exocoetus, Hippocampus, Solea, Muraena, Lophius, Anabas, Protopterus.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 44.
Give a few features of class Amphibia.
Answer:
Features of Class Amphibia:
Amphibia (amphibians): First land vertebrates, evolved from lobe-finned bony fishes, skin naked and moist for respiration, have four limbs, digits without claws, sac-like lungs, 3-chambered heart, eggs laid in water, tailed, gill-breathing larva undergoes metamorphosis, embryonic membranes not formed.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 18
Examples: Ichthyophis, Ram (Frog), Bufo (Toad), Salamander.

Question 45.
Write distinguishing features of class Reptilia.
Answer:
Distinguishing characters of Class Reptilia
1. Dry homy scale-covered skin present.
2. Body divisible into head, neck, trunk and tail,
3. Heart divisible into incomplete four-chambered heart.
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 19

Question 46.
Write distinct features of class Aves.
Answer:
Features of Class Aves:

  1. Body covered with feathers.
  2. Forelimbs modified into wings.
  3. jaws absent, beak present.
  4. Lungs with air sacs.

Question 47.
Write distinguishing features of class Mammalia.
Answer:
Distinguishing features of Class Mammalia:

  1. Body is covered with hair.
  2. Presence of rnilk produced by mammary glands.
  3. Presence of pinnae or external ear.
  4. Presence of diaphragm.
  5. They give birth to young ones.

Question 48.
Name at least five invertebrates belonging to different groups which are of economic importance.
Answer:

  1. Pearl oyster helpful in pearl industry.
  2. Silkmoth spins silk.
  3. Earthworm helps the farmer as an aid in ploughing.
  4. Palaemon is consumed as food.
  5. Bath sponge is used as bath sponge and also for making cushions.

Question 49.
Write differences between cartilaginous fishes and bony fishes.
Answer:
Differences between cartilaginous and bony fishes:

Cartilaginous Fish Bony Fish
1. Alwavs marine. 1. May be marine or fresh water.
2. Skin covered by small placoid scales. 2. Scales are large either cycloid or ganoid.
3. Mouth is subterminal or ventral. 3. Mouth terminal.
4. Gill slits not covered by operculum. 4. Gills covered by operculum.
5. Swim bladder absent. 5. Presence of swim bladder.
6. Endoskeleton entirely cartilaginous. 6. Endoskeleton bony.
7. Tail fin asymmetrical. 7. Tail fin symmetrical.
8. Usually viviparous. 8. Oviparous.
Examples-Sharks, Rays, Sea horse. Examples-Labeo, Catfish, Flat fish.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 50.
Draw a few examples of common birds.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 20

Question 51.
Make a flow chart of classification of Kingdom Animalia.
Answer:
PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms 21

Question 52.
Write differences between chardates and non-chardates.
Answer:
Differences between chordates and non-chordates:

Chordates Non-Chordates
1. Notochord is present in some stage of the life cycle.

2. Tail is present at some stage.

3. A living endoskeleton is present.

4. Pharyngeal gill slits are present at some stages of life.

5. Digestive tract is complete.

6. Heart is ventral.

1. The notochord is absent in non- chordates.

2. Tail is absent in most cases.

3. Endoskeleton if present, is non-living.

4. Pharyngeal gill slits are absent.

5. Digestive tract may be complete, incomplete or absent.

6. Heart is absent, if present, then on dorsal or lateral side.

Very Short Answer Type Questions:

Question 1.
What is Classification?
Answer:
The method of placing organisms into groups and such groups depending upon similarities and differences.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 2.
Who is the father of taxonomy?
Answer:
Carl Linnaeus.

Question 3.
Define species.
Answer:
Species. It is defined as a dynamic group of organisms, which:
1.  resemble each other in all essential respects, i.e. structure and function.
2. interbreed freely under natural conditions to produce fertile young ones of their own kind.

Question 4.
Name the two groups according to old system of classification.
Answer:
1. Plant kingdom
2. Animal kingdom.

Question 5.
Who devised binomial nomenclature?
Answer:
Carolus Linnaeus.

Question 6.
Write two sub-groups of plant kingdom.
Answer:
1. Cryptogamae
2. Phanerogamae.

Question 7.
Why are algae green in colour?
Answer:
Green colour of algae is due to the presence of chlorophyll pigment in their cells.

Question 8.
How do fungi obtain their food?
Answer:
Fungi obtain their food from dead decaying matter by absorption.

Question 9.
Name two kinds of angiospermic plants.
Answer:
1. Monocotyledonous plants
2. Dicotyledonous plants.

Question 10.
Name the respiratory organ of fishes.
Answer:
Gills.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 11.
How many chambers are present in the heart of fishes?
Answer:
Two chambers.

Question 12.
Give examples of bony fishes.
Answer:
Labeo (Rahu), Anabas, Exocoetus, Hippocampus (Sea horse).

Question 13.
Write examples of phylum Echinodermata.
Answer:
Starfish, Sea cucumber, Sea Lily, Sea urchin, Antedon (Feather star).

Question 14.
Write examples of phylum Urochordata.
Answer:
Herdmania, Doliolum, Pyrosoma.

Question 15.
Give one example of cartilaginous fish.
Answer:
Scoliodon.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 16.
Name the excretory structures of flatworms.
Answer:
Flame cells

Question 17.
Name the body cavity of Nematodes.
Answer:
Pseudocoel.

Question 18.
Give the habitat of Ascaris.
Answer:
An endoparasite of intestine of man, especially children.

Question 19.
Name any two mammals.
Answer:
1. Elephant
2. Monkey

Question 20.
What are the different forms with respect to size in which life occurs on earth?
Answer:
Microscopic bacteria a few micrometres in size, the ~30 metre long blue whale and -100 metre tall redwood trees of California.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 21.
Name the plant having life span of more than thousand years.
Answer:
Pine trees.

Question 22.
Which division of biology helps us in exploring the diversity of life forms?
Answer:
Taxonomy.

Question 23.
Who made the first attempt to classify animals?
Answer:
Aristotle.

Question 24.
Name any five marine animals.
Answer:
Whale, Octopus, Star fish, Shark, Sea horse.

Question 25.
Name the organisms which carry out photosynthesis.
Answer:
Green plants.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 26.
What do you mean by characteristics?
Answer:
Characteristic is a particular form or function of a living organism.

Question 27.
What is a prokaryotic cell?
Answer:
Prokaryotic cell. A cell without clearly demarcated nucleus and membrane-bound organelles.

Question 28.
Give examples of prokaryotes.
Answer:
Prokaryotes (monerans) includes bacteria, blue-green algae.

Question 29.
What is evolution?
Answer:
Descent with modification.

Question 30.
Who wrote the book ‘Origin of species’?
Answer:
Charles Darwin (1859).

Question 31.
What are primitive animals?
Answer:
The group of animals which have ancient body divisions and have not changed very much. They are also called lower organisms.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 32.
Define advanced organisms.
Answer:
The group of organisms that have acquired their particular body design relatively recently, also called ‘higher’ organisms.

Question 33.
Name five kingdoms as proposed by R.H. Whittaker.
Answer:
Monera, Protista, Fungi, Plantae, Animalia.

Question 34.
What are basis of five kingdom classification as proposed by Whittaker?
Answer:
1. Nature of cells
2. Number of cells
3. Mode of nutrition

Question 35.
Name the two divisions as proposed by Karl Woose.
Answer:
1. Archaebacteria
2. Eubacteria

Question 36.
Name the various taxonomic categories.
Answer:
Species, genus family, order, class, phylum and kingdom.

Question 37.
Name the tnree aspects of systematics.
Answer:
Identification, nomenclature and classification.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 38.
What is highest taxonomic category?
Answer:
Kingdom.

Question 39.
What is the nature of cells of monerans?
Answer:
Prokaryotic cell.

Question 40.
Name the kind of mode of nutrition in kingdom Monera,
Answer:
Autotrophic and Saprotrophic.

Question 41.
Give examples of kingdom Monera.
Answer:
Bacteria, Cyanobacteria and Mycoplasma.

Question 42.
Write locomotory structures of kingdom Protista.
Answer:
1. Cilia
2. Flagella

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 43.
Write examples of Protista.
Answer:
Single-celled algae, Protozoa (Fuglena, Amoeba) and Diatoms.

Question 44.
Give examples of Fungi.
Answer:
Yeast, Mushroom, Mucor, Rhizopus.

Question 45.
What is the mode of nutrition of Fungi?
Answer:
Fungi obtain their food from dead decaying organic matter.

Question 46.
What is symbiosis?
Answer:
An inter-relationship between two different species e.g. Lichens.

Question 47.
Write few characters of kingdom Animalia.
Answer:
1. Eukaryotic
2. Multicellular
3. Heterotrophic mode of nutrition

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 48.
What is thallus-like plant body?
Answer:
When there is no differentiation of the plant body into root, stem and leaves.

Question 49.
Give few examples of Bryophytes.
Answer:
Riccia, Mardmntia, Funaria (Moss).

Question 50.
Name the division which is called ‘amphibians of plant kingdom’.
Answer:
Bryophyta.

Question 51.
What is hypha?
Answer:
In fungi, thallus is made up of colourless filamentous structure called hypha.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 52.
What is the mode of nutrition in Fungi?
Answer:
Absorptive-saprophytic, Heterotrophic.

Question 53.
Name the group in which seeds are naked.
Answer:
Gymnosperms.

Question 54.
Name the group in which reproductive organs are flowers.
Answer:
Angiosperms.

Question 55.
Name the class of angiosperms in which reticulate venation is present.
Answer:
Dicots.

Question 56.
Name the class of organisms in which leaves show parallel venation.
Answer:
Monocots.

Question 57.
Name the reproductive organ of Angiosperms.
Answer:
Flower.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 58.
What are cryptogams?
Answer:
Cryptogams are flowerless, seedless, lower plants.

Question 59.
What are phanerogams?
Answer:
Phanerogams are seed bearing plants.

Question 60.
What are Algae?
Answer:
These are green, autotrophic thallophytes.

Question 61.
Write examples of dicot plants.
Answer:
Pea, Gram, Rose.

Question 62.
How do oviparous and viviparous animals differ from each other?
Answer:
Oviparous animals lay eggs e.g. birds, while viviparous animals give birth to young ones e.g. most of mammals.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 63.
What are hermaphrodite animals?
Answer:
Animals which have both male and female reproductive organs e.g. earthworm, leech, etc.

Question 64.
Name the skeletal elements of sponges.
Answer:
Spicules (needles) or spongin fibres or both.

Question 65.
What are cnidoblasts? Give their function.
Answer:
These are stinging cells present on the tentacles of coelenterates like Hydra. These inject the hypnotoxin and paralyze the prey.

Question 66.
Name the cavity present in the body of Coelenterates.
Answer:
Coelenteron.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 67.
Mention two characters of sponges.
Answer:
Presence of two types of pores (dermal ostia and osculum) and collar cells.

Question 68.
Which characters appeared for the first time in the flatworms?
Answer:
Triploblastic, bilateral symmetry and organ-system organisation.

Question 69.
Write two examples of phylum Annelida.
Answer:
Earthworm, Leech.

Question 70.
Write two examples of phylum Coelenterata.
Answer:
Hydra, Obelia.

Question 71.
Name five animals belonging to kingdom Arthropoda.
Answer:
Prawn, Butter fly, Housefly, Spider, Crab, Body Louse.

Question 72.
Write two examples of phylum Mollusca.
Answer:
Pila, Unio, Cuttlefish.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 73.
Give two examples of phylum Echinodermata.
Answer:
Star Fish, Sea lily.

Question 74.
What is primary unit of classification?
Answer:
Species.

Question 75.
What were the two most outstanding contributions of Linnaeus to the modem science of taxonomy?
Answer:
The outstanding contributions of Linnaeus were his method of grouping species in a hierarchy and his binomial method of nomenclature.

Question 76.
What is a genus?
Answer:
Group of related species.

PSEB 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Question 77.
Why do species included in a genus resemble in many features?
Answer:
Because they have originated from a common ancestor.