Environmental Management Class 9 Science Chapter 9 Questions And Answers Maharashtra Board

Class 9 Science Chapter 9

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 9 Environmental Management Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 9 Environmental Management Question Answer Maharashtra Board

Class 9 Science Chapter 9 Environmental Management Question Answer Maharashtra Board

1. Match the items in column ‘A’ with the proper ones in coloum ‘B’ and explain their impact on the environment.

Column ‘A’ Column ‘B’
1. Harmful waste – a. Glass, rubber, carry bags, etc.
2. Domestic waste – b. Chemicals, pigments, ash, etc.
3. Biomedical waste – c. Radioactive material
4. Industrial waste – d. Left over food, vegetables, peelings of fruits.
5. Urban waste – e. Bandages, cotton, needles, etc.
Answer:
(1 – c),
(2 – d),
(3 – e),
(4 – b),
(5 – a)

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

2. Complete the statements using the given options and justify those statements. (Geographic favourability, climate, weather, observatory)

a. Of the abiotic factors that affect biodiversity by far the most important is ……………… .
Answer:
Climate.
Climate influences our basic needs like food, clothing and shelter as well as our occupations. Various factors of climate like temperature, atmospheric pressure, sunlight, rainfall, humidity etc. will decide which kind of plants and animals can survive there.

b. A description of the climatic conditions of short duration in a particular area is ……………… .
Answer:
Weather.
Atmospheric conditions at a specific time at a particular place are referred to as weather. Weather is related to a specific location and specific time whereas climate is related to a longer duration and larger area.

c. Irrespective of the progress of human beings, we have to think about ……………… .
Answer:
Geographical favourability.
Geographical favourability includes location, availability of water, availability of natural resources and climate of a region. This has a great impact on the progress of human beings. Countries that have more favourable factors will progress more.

d. Establishments where various climatic factors are recorded are called ……………… .
Answer:
Observatories
Most countries in the world have established meteorology departments for recording climatic factors. These departments have observatories which are equipped with modern instruments and technology.

3. Answer the following questions.

a. How is first aid provided to victims of disasters who are injured?
Answer:
(i) Bleeding: If the victim is injured and bleeding through the wound, the wound should be covered with an antiseptic pad and pressure applied on it for 5 minutes with either thumb or palm.

Fracture and impact on vertebrae: If any bone is fractured, it is essential that the fractured part be immobilized. It can be done with the help of any available wooden rods / batons / rulers. If there is an impact on the back or vertebral column; the patient should be kept immobile on a firm stretcher.

Bums: If victims have bum injuries, it is beneficial to hold the injured part under clean and cold flowing water for at least 10 minutes. How is Solid Waste classified? OR What are the sources of Solid Waste?

Domestic waste: Waste food, paper, plastic paper, plastic bags, vegetable waste, fruit skins, glass and sheet metal articles, etc. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Industrial waste: Chemicals, pigments, sludge, ash, metals, etc.

Hazardous waste: Chemicals generated in various industries, radioactive materials, explosives, infectious materials, etc. Farm/Garden waste: Leaves, flowers, branches of trees, crop residues like straw, animal urine and dung, pesticides, remains of various chemicals and fertilizers, etc.

Electronic waste: Non-functional TV sets, cell phones, music systems, computers and their parts, etc.

Biomedical waste: Bandages, dressings, gloves, needles, saline bottles, medicines, medicine bottles, test tubes, body parts, blood, etc. from clinics, hospitals, blood banks and laboratories. Urban waste: Waste generated through household industries and large commercial and industrial establishments, carry bags, glass, metal pieces and rods, threads, rubber, paper, cans from shops, vegetable and meat markets, construction waste, etc.

(viii) Radioactive waste: Radioactive materials like Strontium-10, Cerium-141, Barium-140 and heavy water, etc. generated from atomic energy plants, uranium mines, atomic research centres, nuclear weapons testing sites, etc.
(ix) Mining waste: Remains of heavy metals like lead, arsenic, cadmium, etc. from mines.

b. State the scientific and eco-friendly methods of waste management.
Answer:
Following are the scientific and eco-friendly methods of waste management:

  • Waste separation: In this method, dry and wet wastes are separated, stored and later collected for proper use.
  • Composting: Decomposition of degradable kitchen waste in small pits in the backyard, garden or terrace is called composting. Good quality manure can be produced by composting leftover food, peels of fruits, vegetables etc.
  • Vermicomposting: Converting solid waste into manure or compost with the help of earthworms is called vermicomposting.
  • Vermicompost manure is highly nutritious and can be used for agricultural purposes and garden plants.
  • Secured landfill: Solid waste is disposed off in secured landfills.
  • The site for secured landfill is selected minimum 2 km away from water bodies and human habitation.
  • Care is taken to see that the site does not fall in sensitive zone.
  • The landfills are layered with clay and plastic and then the garbage is spread and left to decompose naturally.
  • Pyrolysis: In this method, the waste is heated to a high temperature to obtain gas and electricity.
  • Semi-combustible waste is burnt in pyrolysis. It is suitable for municipal solid waste management.
  • Incineration: Biomedical waste is burnt in incinerators to kill the pathogens. Disinfection and sterilization is also done while treating the biomedical waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

c. Explain with suitable examples, the relationship between weather forecasting and disaster management.
Answer:

  1. Depending upon the factors such as storms, clouds, rainfall, etc., weather forecasts are made. It is useful in aviation, shipping, fishing, industries as well as during natural calamities like dust storms, sand storms, heavy rainfall, tsunami etc where proper predictions are made with well equipped satellites and highclass technology.
  2. Observatories at several locations are doing excellent work in the analysis of the information received from these satellites.
  3. This data can be used in prevention of losses and danger, improving tolerance, providing relief from disaster, minimising the intensity and extent of harm as well as preparation to face the disaster.

d. Why is e-waste harmful? Express your opinion about this.
Answer:

  • e-waste is electronic waste which includes non-functional TV Sets, cell phones, music systems, computers and their parts, etc.
  • E-waste contains heavy metals like lead, beryllium, mercury and cadmium.
  • These metals accumulate in the soil for long periods and thus affect the biodiversity of the soil.
  • Many a times e-waste like battery etc. contain acids which make the soil acidic.
  • e-waste can cause ground water pollution, which also affects living organisms when they drink this polluted water.
  • Thus, e-waste affects human health and soil microorganisms directly or indirectly.

e. How will you register individual your participation in solid waste management?
Answer:
We can register our individual participation in solid waste management in the following ways:

  • Following the 3R mantra: Reduce (reducing the waste), reuse (reuse of waste) and recycle (recycling of waste).
  • Throwing plastic wrappers of chocolates, ice-creams, biscuits, etc. into dust bins. Avoid littering.
  • Avoiding the use of plastic bags and instead using cloth bags or bags prepared from old sarees, bed-sheets, curtains, etc.
  • Using both sides of a paper for writing. Reusing greeting cards and gift papers.
  • Avoiding use of tissue paper and preferring to use one’s own handkerchief.
  • Using rechargeable batteries instead of lead batteries.
  • Implementing various programmes of solid waste management and educating, encouraging the family and society in this regard. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • Avoiding ‘use and throw’ type of articles like pens, canned cold drinks and tetra-packs etc.

4. Write notes.
Meteorology, Climatic factors, Monsoon model, Industrial waste, Plastic waste, Principles of first aid.

5. Give examples of the importance of climate in the living world with explanations, in your own words.
Answer:

  1. Daily weather as also long term climatic conditions influence human lifestyle directly or indirectly.
  2. Land, water bodies, plants and animals collectively form the natural environment on earth. This environment is responsible for the development of organisms.
  3. The climate of a particular region helps to determine the diet, clothing, housing, occupations and lifestyle of the people of that region. For example, the characteristic lifestyle of Kashmiri and Rajasthani people.
  4. Salinity of marine water, formation of oceanic currents, water cycle, etc. are all related to various weather and climatic factors.
  5. Various climatic factors bring about the weathering of rocks in the earth’s crust.
  6. Climate plays a very important role in the formation and enrichment of soil.
  7. Microbes in the soil play an important role in formation of organic materials. This process depends upon various climatic factors.

6. Explain with suitable examples, the care to be taken when using the methods of transporting patients.
Answer:

  • Cradle Method: This method is used for children and under-weight victims.
  • Carrying piggy back: This method is used for carrying patients who are unconscious.
  • Human crutch method: If one of the legs is injured, the victim should be supported with minimum load on the other leg. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • Pulling or lifting method: This method is used for carrying an unconscious patient, through a short distance.
  • Carrying on four-hand chair: This method is used when support is needed for the part below the waist.
  • Carrying on two-hand chair: This method is useful for those patients who cannot use their hands but can hold their body upright.
  • Stretcher: In an emergency, if a conventional stretcher is not available, then a temporary stretcher can be made using bamboo, blanket, etc.

7. Explain the differences.

a. Weather and climate
Answer:

Weather Climate
(i) Atmospheric conditions at a specific time at a particular place are referred to as weather. (i) The climate of a particular region is the average of daily readings of various weather-related parameters recorded for several years.
(ii) Weather can change continuously. (ii) Climate remains constant in a region for a long duration.
(iii) Weather is related to a specific location and specific time. (iii) Climate is related to a longer duration and larger area.
(iv) Changes in the weather may occur for short periods of time. (iv) Changes in the climate take place slowly over a much long duration.

b. Degradable and non-degradable waste
Answer:

Degradable Waste Non-degradable Waste
(i) This type of waste is easily degraded by microbes. (i) This type of waste is not easily degraded by microbes because it takes a very long period of time and the use of various techniques.
(ii) It includes kitchen waste (spoiled food, fruits, vegetables), ash, soil, dung, parts of the plants etc. (ii) It includes plastic, metal and other similar materials.
(iii) It is also called wet solid waste or wet garbage. (iii) It is also called dry solid waste or dry garbage.
(iv) If it is carefully decomposed, we can obtain compost and fuel of good quality from it. (iv) It can be recycled.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Class 9 Science Chapter 9 Environmental Management Intext Questions and Answers

Can you recall?

Environmental Management Class 9 Question 1.
Which natural calamities have you experienced? How did they affect the conditions in your surroundings?
Answer:
In July 2005, heavy rains caused flooding in many parts of Maharashtra including large areas of Mumbai.

They effects were as follows.

  • Many people lost life.
  • There was water clogging in most parts causing severe damage to private property and public property.
  • Hospitals were submerged causing severe inconvenience to patients.
  • The city incurred huge financial loss.

Environmental Management Class 9 Exercise  Question 2.
How will you make a plan to be safe from calamities or to minimize the damage?
Answer:

  1. As a responsible citizen we can contribute by educating the people about all the precautions to be taken when such a calamity occurs.
  2. We can also contribute by minimizing the occurrence of natural calamity like flood by taking measure to reduce land pollution, proper disposing waste material and recycling them.
  3. We can plan to be safe by keeping the following things handy-torch, first-aid kit, ample food supply and water storage, emergency numbers list and extra pair of clothes.

Thus by taking appropriate measures, we can be safe from calamities

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Class 9 Science Chapter 9 Environmental Management  Question 3.
How does the atmosphere affect our daily life?
Answer:
The atmosphere affect our daily life in following ways

  • If protects us from harmfull radiations of sun.
  • Atmosphere helps to sustain life on earth by providing oxygen for human beings and animal to breathe and carbon dioxide to plants.
  • It helps in keeping us warm due to green house effect.

Environmental Management Class 9 Answers Question 4.
Forecasts about which weather related factors are given during the news bulletins on Doordarshan and Akashvani?
Answer:
Forecasts about different weather related factors such as cloud cover, rain, snowfall, wind speed and temperature are given during the news bulletins on Doordarshan and Akashvani.

9 Environmental Management Exercises  Question 5.
What is meant by pollution?
Answer:
Contamination of natural environment that can harmfully affect the ecosystem is called as pollution.

9th Class Science Chapter 9 Environmental Management Exercise Question 6.
What is meant by solid waste?
Answer:
The waste materials generated through the various daily human activities are called solid waste.

Maharashtra Board Class 9 Science Solutions Question 7.
What are the different things included in solid waste?
Answer:
Domestic waste, industrial waste, hazardous waste, electronic waste, biomedical waste, urban waste, radioactive waste and mining waste are the different things included in solid waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 8.
Why is it necessary to recycle non-degradable waste?
Answer:
(i) non-degradable waste cannot be easily degraded because it takes a very long period of time and the use of various techniques.
(ii) Therefore, non-degradable waste should be recycled so that it does not accumulate and cause hazards to the environment.

Question 9.
Which materials are included in solid dry waste?
Answer:
Solid dry waste includes paper, plastic, metals, glass, cardboard, thermocol etc.

Answer the following

Question 1.
In which different ways do our surroundings get polluted?
Answer:

  • Air pollution is caused due to emissions from industries, vehicles, burning of fossil fuels, construction, mining and agriculture.
  • Water pollution is caused by domestic sewage and industrial waste water released into the water bodies.
  • Soil pollution is caused due to industrial wastes, domestic waste, chemical fertilizers, biomedical waste and pesticides.
  • Noise pollution is caused by machines, vehicular traffic, loudspeakers and household appliances.

Question 2.
Which factors are affected favourably or unfavourably by climate? What must we do to minimize the effect?
Answer:

  • Climate plays a very important role in our day to day life.
  • It influences our basic needs like food, clothing and shelter as well as our occupations.
  • Climate is especially important for an agrarian country like India. Also climatic factors like direction and speed of the winds, temperature, atmospheric pressure etc. are also considered during construction work.
  • The science of meteorology helps in predicting climatic conditions by satellites in different ways such as prediction of rainfall, air pollution, dust storms, hot and cold waves tsunamic etc.
  • so that all the citizens are well-equipped beforehand and can take preventive measures.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 3.
Into which two categories can the waste materials in the lists above be classified?
Answer:
The waste materials can be classified as biodegradable waste and non-biodegradable waste.

Answer the following:

Question 1.
Make a list of various waste materials and articles in your area and prepare a chart as follows:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 6

Question 1.
Nowadays, an electronic device – the cell phone – is very popular. From a mobile shop near your house, find out how they dispose off old and broken down cell phones.
Answer:
Old and broken down cell phones are sold to scrap dealers, who sell to a bigger dealer where reusable parts are taken out and useless parts are sent for recycling.

Question 2.
What are the different types of casualties that are seen to occur in different types of disasters?
Answer:

  • Death.
  • Injuries.
  • Loss of limbs or body parts.
  • Burns.
  • Diseases.
  • Fractures.
  • Bleeding.
  • People becoming unconscious.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 3.
Which waste management processes are used in your village/town / city?
Answer:
i. The Municipal Corporation or Municipality collects the various kinds of wastes like dry waste, solid waste, biomedical waste in different coloured containers and transports them to areas where they are treated and disposed off.

ii. Industrial waste is mostly recycled and biomedical waste is treated by the scientific methods mentioned below.

a. Solid waste is disposed off in secured land fills. The site for secured landfill is selected 2 km away from water bodies and away from human habitation. Care is taken to see that the site does not fall in sensitive zone. The landfills are layered with clay and plastic and then the garbage is spread and left.
b. Pyrolysis is done for semi combustible material. Semi combustible materials are heated to high temperature by gas arid electricity.
c. The municipality can also set up biogas plants where the solid waste is converted to biogas by anaerobic fermentation. The biogas can be used to generate power and also a good manure which can be used for agriculture.
(d) Biomedical waste is burnt in incinerators to kill the pathogens. Disinfection and sterilization is also done while treating biomedical waste by the Municipal Corporation.

These are some of the waste management processes used in village/town/city.

Answer the following questions:

Question 1.
Observe the garbage collected in the dustbin of your classroom and make a list of the various materials in it.
Discuss with your teacher, how these materials can be properly disposed off. Can we do the same with the garbage generated in our house? Think about it.
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 1
Answer:
(i) Garbage collected in classroom dustbin:

  • Waste paper and paper bits.
  • Pencil shavings
  • Wrappers of chocolates, biscuits etc.
  • Left over food from tiffins and fruit peels.
  • Empty ball pen refills.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

(ii) Out of these left over food from tiffins and fruit peels and pencil shaving can be used to make compost manure in school garden. The remaining waste can be sold to scrap dealers and sent for recycling.
(iii) Yes, we can do the same with the garbage generated in the house.

Question 2.
What is the main difference between what we see in the two pictures alongside (A and B).
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 2
Answer:

  • We see that in picture A there is lot of garbage spread around making the place very dirty, whereas in picture B, there is no garbage and the place is absolutely neat and clean.

Question 3.
What should we do to permanently maintain the condition seen in picture B?
Answer:
To permanently maintain the condition seen in picture B, we should follow the 3R mantra (Reduce, Reuse, Recycle). Also, we must avoid littering, throwing plastic bags, wrappers of chocolates, ice-creams, biscuits etc.

Class 9 Science Chapter 9 Environmental Management Additional Important Questions and Answers

Choose and write the correct option

Question 1.
is a long term predominant condition of the atmosphere.
(a) Climate
(b) Weather
(c) Pressure
(d) Biosphere
Answer:
(a) Climate

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.
Various climatic factors like are considered during construction of runways, seaports, huge bridges and skyscrapers, etc.
(a) Direction and speed of wind
(b) Temperature
(c) Atmospheric pressure
(d) All of these
Answer:
(d) All of these

Question 3.
The was established by the United Nations Organization on 23rd March 1950.
(a) World Health Organization
(b) National Oceanographic and Atmospheric Administration
(c) World Meteorological Organization
(d) Indian Institute of Tropical Meteorology
Answer:
(c) World Meteorological Organization

Question 4.
is/are related to various weather and climatic factors.
(a) Salinity of marine water
(b) Formation of ocean currents
(c) Water cycle
(d) All of these
Answer:
(d) All of these

Question 5.
is the founder of IMD.
(a) Dr. Vasantrao Govarikar
(b) H.F. Blanford
(c) Sir Gilbert Walker
(d) Dr. Radhakrishnan Nair
Answer:
(b) H. F. Blanford

Question 6.
The monsoon model based upon 16 worldwide parameters was developed by the initiative of
(a) Dr. Vasantrao Govarikar
(b) H.F. Blanford
(c) Sir Gilbert Walker
(d) Virghese Kurien
Answer:
(a) Dr. Vasantrao Govarikar

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 7.
In , forecasts are made taking into account the estimates of current weather related events and ongoing physical activity.
(a) Statistical model
(b) Holistic model
(c) Mathematical model
(d) All of these
Answer:
(c) Mathematical model

Question 8.
In , predictions are based upon those parameters used in other models which have the greatest effect on the monsoon.
(a) Holistic model
(b) Mathematical model
(c) Statistical model
(d) Scientific model
Answer:
(a) Holistic model

Question 9.
Radioactive materials, explosives and infectious materials are classified as waste.
(a) Industrial waste
(b) Biomedical waste
(c) Urban waste
(d) Hazardous waste
Answer:
(d) Hazardous waste

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question10.
are radioactive waste.
(a) Remains of heavy metals like arsenic, cadmium etc.
(b) Strontium-10, Cerium-141, Barium -140
(c) Waste from blood banks and laboratories
(d) All of these
Answer:
(b) Strontium-10, Cerium-141, Barium-140

Question 11.
Waste food, paper, plastic, vegetable and fruit waste etc. are classified as waste.
(a) Industrial waste
(b) Farm waste
(c) Domestic waste
(d) Urban waste
Answer:
(c) Domestic waste

Question 12.
is the 3 R mantra.
(a) Refuse, Research, Recycle
(b) Reduce, Reuse, Recycle
(c) Reduce, Reuse, Refuse
(d) Rethink, Recycle, Reuse
Answer:
(b) Reduce, Reuse, Recycle

Question 13.
is the largest producer of electricity from solid waste.
(a) India
(b) America
(c) Japan
(d) China
Answer:
(b) America

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 14.
Chemicals, pigments, sludge, ash, metals, etc. are classified as waste.
(a) Domestic waste
(b) Industrial waste
(c) Urban waste
(d) Hazardous waste
Answer:
(b) Industrial waste

Question 15.
Period of natural degradation for banana peels is
(a) 1 month
(b) 1-2 weeks
(c) 3-4 weeks
(d) 2 months
Answer:
(c) 3-4 weeks

Question 16.
Period of natural degradation for cloth bags is
(a) 2-3 weeks
(b) 1 month
(c) 5 months
(d) 1 year
Answer:
(b) 1 month

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 17.
Period of natural degradation for wood is
(a) 1 month
(b) 5 months
(c) 10-15 years
(c) 40-50 years
Answer:
(c) 10-15 years

Question 18.
Period of natural degradation for certain plastic bags is
(a) 50-100 years
(b) infinite duration
(c) 10 lakh years
(d) 1000 years

(c) 10 lakh years

Question 19.
Period of natural degradation for thermocol or Styrofoam cup is
(a) 10 lakh years
(b) infinite duration
(c) 200-250 years
(d) 1 year
Answer:
(b) infinite duration

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 20.
If any bone is fractured, it is essential that the fractured part be
(a) mobilized
(b) immobilized
(c) pulled
(d) massaged
Answer:
(b) immobilized

Question 21.
For transporting children and under-weight victims, method is used.
(a) carrying piggy back
(b) human crutch method
(c) cradle method
(d) stretcher
Answer:
(c) cradle method

Question 22.
method is useful to carry patients who are unconscious.
(a) Cradle Method
(b) Carrying piggy back
(c) Human crutch
(d) Carrying on four-hand chair
Answer:
(b) Carrying piggy back

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 23.
method is used for carrying an unconscious patient through a short distance.
(a) Carrying piggy back
(b) Cradle method
(c) Carrying on two-hand chair
(d) Pulling or lifting method
Answer:
(d) Pulling or lifting method

Question 24.
method is useful to carry patients who cannot use their hands but can hold their body upright.
(a) Carrying on four-hand chair
(b) Carrying on two-hand chair
(c) Carrying piggy back
(d) Cradle method
Answer:
(b) Carrying on two-hand chair

v 25.
method is used to carry patients when support is needed for the part below the waist.
(a) Carrying on four-hand chair
(b) Carrying on two-hand chair
(c) Stretcher
(d) Pulling or lifting method
Answer:
(a) Carrying on four-hand chair

Question 26.
For injuries like sprains, twisting and contusion, should be applied on the injured part.
(a) turmeric powder
(b) antiseptic pad
(c) ice-pack
(d) pressure
Answer:
(c) ice-pack

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Find the odd man out:

Question 1.
Waste food, paper, plastic, bandages.
Answer:
Bandages: It is a biomedical waste whereas the others are domestic waste.

Question 2.
Pesticides, fertilizers, crop residue, sludge.
Answer:
Sludge: It is an industrial waste whereas the others are farm waste.

Question 3.
Strontium -10, Cerium – 141, Barium – 140, Cadmium.
Answer:
Cadmium: It is a mining waste whereas the others are radioactive waste.

Question 4.
Banana peels, cloth bag, food waste, plastic bag.
Answer:
Plastic bag: It is a non-biodegradable waste whereas the others are degradable wastes.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 5.
Fruits, ash, metals, vegetables.
Answer:
Metal: It is a non-biodegradable waste whereas the others are degradable wastes.

Complete the analogy:

(1) Specific duration and specific time: Weather : : Longer duration and longer time :
(2) Mumbai: 5000 tons solid waste :: Pune:
(3) Kitchen waste, parts of plants : Wet solid waste :: Plastic, metals :
(4) Largest producer of electricity from solid waste : America : : Production of useful materials from banana peelings :
(5) Cloth bags : 1 month : : Rags :
(6) Tin cans : 50-100 years : : Aluminium cans :
(7) Wood : 10-15 years :: Styrofoam :
Answer:
(1) Climate
(2) 1700 tons solid waste
(3) Dry solid waste
(4) Japan
(5) 5 months
(6) 200-250 years
(7) Infinite duration

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Mining waste (a) Leaves, flowers, crop residue.
(2) Electronic waste (b) Remains of heavy metals like lead, arsenic, cadmium.
(3) Farm waste (c) Strontium-10,
Cerium-141,
Barium-140
(4) Radioactive waste (d) Cell phones, TV sets, Computers

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.

Column ‘A’ Column ‘B’
(1) Banana peels (a) 10 lakh years.
(2) Plastic bags (b) 200-250 years.
(3) Leather shoes (c) 3-4 weeks
(4) Aluminium cans (d) 40-50 years

Answer:
(1 – c),
(2 – a),
(3 – d),
(4 -b)

Question 3.

Column ‘A’ Column ‘B’
(1) Thermocol
(2) Tin cans
(3) Woollen socks
(4) Wood
(a) 1 year.
(b) Infinite duration.
(c) 10-15 years
(d) 50-100 years

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

State whether the following statements are true or false. Correct the false statements:

Question 1.
Climate is a long term predominant condition of the atmosphere.
Answer:
True.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.
Climate changes continuously.
Answer:
False. Climate does not change continuously. It remains constant in a region for a long duration.

Question 3.
If present climatic conditions are analysed with reference to the past climatic conditions, we can predict climatic changes of the future.
Answer:
True.

Question 4.
Forecasting is difficult for places where climatic changes are slow and of a limited nature.
Answer:
False. Forecasting is easy for places where climatic changes are slow and of a limited nature.

Question 5.
Climate plays a very important role in the formation and enrichment of soil.
Answer:
True.

Question 6.
Prediction maps are prepared once in 24 hours.
Answer:
False. Prediction maps are prepared twice in every 24 hours.

Question 7.
The first prediction of monsoon in India was made by Dr. Vasantrao Govariakar.
Answer:
False. The first prediction of monsoon in India was made my H.F. Blanford.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 8.
H.F. Blanford used the rainfall in Kerala as the parameter for prediction of monsoon in India.
Answer:
False. H.F. Blanford used the snowfall in Himalayas as a parameter for prediction of monsoon in India.

Question 9.
In Holistic model, predictions are made taking into account estimates of current weather-related events and ongoing physical interactions between them.
Answer:
False. In Holistic model, predictions are based upon those parameters used in other models which have the greatest effect on monsoon.

Question 10.
Any meteorological model depends upon the inter-relationships between parameters used in that model and the results expected from it.
Answer:
True.

Question 11.
Remains of heavy metals like lead, arsenic, cadmium, etc. from mines are industrial waste.
Answer:
False. Remains of heavy metals like lead, arsenic, cadmium etc. from mines are mining waste.

Question 12.
Use of ‘use and throw’ type of articles like pens, canned cold drinks, tetra packs should be encouraged in waste management.
Answer:
False. Use of ‘use and throw’ type of articles like pens, canned cold drinks, tetra packs should be strictly avoided in waste management.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 13.
We should use tissue paper instead of one’s own handkerchief for effective waste management.
Answer:
False. We should use one’s own handkerchief instead of tissue paper for effective waste management.

Question 14.
Certain plastic bags take 1 month to degrade.
Answer:
False. Certain plastic bags take 10 lakh years to degrade.

Question 15.
Banana peels can degrade in 3-4 weeks.
Answer:
True.

Question 16.
If the victim has burn injuries, it is beneficial to cover the burnt part with blanket.
Answer:
False. If the victim has burn injuries, it is beneficial to hold the injured part under clean and cold flowing water for at least 10 minutes.

Question 17.
CPR helps to bring the circulation to normal.
Answer:
True.

Question 18.
If breathing has stopped, the head should be held in backward sloping position.
Answer:
False. If breathing has stopped, the victim should be given artificial ventilation by mouth to mouth resuscitation.

Question 19.
Japan is the largest producer of electricity from solid waste.
Answer:
False. America is the largest producer of electricity from solid waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 20.
China has developed the projects of production of threads, paper and other useful materials from banana peels.
Answer:
False. Japan has developed the projects of production of threads, paper and other useful materials from banana peels.

Define the following:

Question 1.
Weather
Answer:
Atmospheric conditions at a specific time at a particular place are referred to as weather.

Question 2.
Climate
Answer:
The climate of a particular region is the average of daily readings of various weather-related parameters recorded for several years.

Question 3.
Meteorology
Answer:
The science that studies the inter-relationships between the various components of air, natural cycles, geological movements of earth and climate is called meteorology.

Question 4.
Solid waste
Answer:
Waste materials generated through daily human activities are called solid waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 5.
Urban waste
Answer:
Waste generated through household industries and large commercial and industrial establishments is called Urban waste.

Answer the following:

Question 1.
What is the necessity of solid waste management?
Answer:
Necessity of solid waste management:

  • For preventing environmental pollution and to keep the surroundings clean.
  • For energy as well as fertilizer production and through that to generate work and employment opportunities.
  • To reduce the strain on natural resources through treatment of solid waste.
  • To improve the health and quality of life and to maintain environmental balance.

Question 2.
What are the harmful effects of solid waste?
Answer:
Harmful effects of solid waste:

  • Effect on biodiversity.
  • Releases bad odour.
  • Produces toxic gases.
  • Leads to degradation of natural beauty.
  • Leads to pollution of air, water and soil.
  • Spreads diseases.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 3.
What first-aid should be given for injuries like sprains, twisting and contusion?
Answer:
For injuries like sprains, twisting and contusion, the ‘RICE’ remedy should be applied:

  • Rest: Allow the victim to sit in a relaxed position.
  • Ice: Apply an ice-pack to the injured part.
  • Compression: After the ice-pack treatment, the injured part should be massaged gently.
  • Elevate: The injured part should be kept in a raised/elevated position.

Question 4.
Why do meteorological models need to be changed continually?
Answer:

  1. Any meteorological model depends upon the inter-relationship between parameters used in that model and the results expected from it.
  2. However, as these inter-relationships with reference to the ocean and atmosphere are never constant, meteorological models need to be changed continually.

Question 5.
What is urban waste? What does it include?
Answer:

  1. Waste generated through household industries and large commercial and industrial establishments is called urban waste.
  2. It includes carry bags, glass, metal pieces and rods, threads, rubber, paper, cans from shops, waste from vegetable and meat markets, construction waste etc.

Question 6.
What does biomedical waste include?
Answer:
biomedical waste includes bandages, dressings, gloves, needles, saline bottle, medicines, medicine bottles, test tubes, body parts, blood etc. from clinics, hospitals, blood banks and laboratories.

Write short notes on:

Question 1.
Meteorology
Answer:

  • The science that studies the inter-relationship between the various components of air, natural cycles, geological movements of the earth and climate is called meteorology.
  • Meteorology includes the study of storms, clouds, rainfall, thunder, lightning etc.
  • Depending upon the study of such factors, weather forecasts are made.
  • They are useful to common people, farmers, fisheries, aviation services, water transport and various other organizations. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.
Climatic factors
Answer:

  • The climate of a particular region is the average of daily readings of various weather-related parameters recorded for several years.
  • Climatic factors include direction and speed of wind, temperature, atmospheric pressure, clouds, rainfall, humidity, visibility etc.
  • These factors influence our basic needs like food, clothing, shelter as well as our occupations.
  • Various climatic factors bring about the weathering of rocks in the earth’s crust.
  • Microbes in the soil play an important role in formation of organic materials. This process also depends upon various climatic factors.

Question 3.
Monsoon model
Answer:

  1. The tradition of forecasting the monsoon season in India is older than 100 years.
  2. After the famine of 1877, H.F. Blanford, the founder of IMD had made such a prediction for the first time taking the snowfall in the Himalayas as a parameter for this prediction.
  3. In the decade of the 1930’s, the then director of IMD, Sir Gilbert Walker had underlined the relationship between various worldwide climatic factors and the Indian monsoon and based on available observations and previous recordings related to this relationship, he put forth a hypothesis regarding the nature of the monsoon.
  4. With the initiative of Dr. Vasantrao Govarikar in the decade of the 1990’s, a monsoon model based upon 16 worldwide climatic parameters was developed. This model was in use from 1990 to 2002.
  5. Presently, new models are being developed at IITM. Work is in progress at two levels, namely designing new models and developing new technology.
  6. The main focus is on the development of the radar system and satellite technology.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 4.
Plastic waste
Answer:

  • Plastic waste is the accumulation of plastic products in the environment that adversely affects environment, humans and animals.
  • Plastic waste is excessively generated as plastic is inexpensive and durable.
  • Plastic is slow to degrade. It takes around 10 lakh years for certain plastic bags to degrade. .
  • Plastic waste affects land and water.
  • It also affects the health of animals, cattle unknowingly ingest these plastic bags leading to stomach cancer in them.
  • Plastic also releases toxic chemicals which are carcinogenic to humAnswer:
  • To avoid plastic waste, cloth bags should be used instead of plastic bags. Plastic articles should be recycled.

Question 5.
Industrial waste
Answer:

  • Industrial waste is the waste produced by industrial processes or activities.
  • There is a huge variety of industries producing different types of materials and articles. All of these use raw materials and give out a lot of waste.
  • There are hundreds of mines which extract copper, silver, gold, iron, coal etc. Huge quantities of waste are produced while processing them.
  • Cement industries give out solid, liquid and gaseous wastes.
  • While refining crude oil, a lot of poisonous gaseous and liquid wastes are produced.
  • Construction units produce huge quantities of waste stones, pebbles, broken bricks, wood waste etc. Mostly they are dumped in landfills.
  • It also includes chemicals, pigments, sludge, ash, metal, etc. given out from mining, textile, construction, chemical industries.

Question 6.
Principles of first aid
Answer:
Life and Resuscitation – ‘ABC’ is the Basic Principle of first aid which is provided to the victims of disaster.

  1. Airway: If the victim has difficulty in breathing, the head should be held in a backward sloping position or the chin should be raised so that the respiratory passage remains open.
  2. Breathing: If breathing has stopped, the victim should be given artificial ventilation by mouth to mouth resuscitation.
  3. Circulation: If the victim is unconscious, then after giving mouth to mouth respiration twice,

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 7.
Statistical Model
Answer:

  • In this model, current climatic observations in a region are compared with earlier parameters such as oceanic temperature, atmospheric” pressure and the nature of the monsoon rainfall for several years.
  • This data is comparatively analysed by statistical methods and predictions are made about the monsoon in the present conditions.

Question 9.
Holistic Model
Answer:

  • In this model, predictions are based upon those parameters used in other models which have the greatest effect on the monsoon.
  • Nowadays, predictions declared by IMD are the collective outcome of various model. This is called a holistic model.

Complete the following concept chart:

Question 1.
Weather-related climatic factors
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 3

Question 2.
Harmful effects of solid waste
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 4

Question 3.
Scientific and eco-friendly waste Management
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 5

Answer in detail:

Question 1.
What are the principles of solid waste management?
Answer:
Principles of solid waste management:

  • Reuse: After use, materials should be reused for some other proper purposes.
  • Refuse: Refusal to use articles made from non- degradable articles like plastic and thermocol.
  • Recycle: Production of useful articles by recycling solid wastes. For example, paper and glass can be recycled.
  • Rethink: Rethinking our habits, activities and their consequences in connection with the use of various articles of daily use. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • Reduce: Restricting the use of resources to avoid their wastage.
  • Old materials should be reused. One thing should be shared by many, use and throw type of objects should be avoided.
  • Research: Conducting research related to reuse of materials that are temporarily out of use.
  • Regulation and Public awareness: Following the laws and rules related to waste management and motivating others to do the same.

Question 2.
What is disaster management? What actions does it include?
Answer:
Disaster management is action implemented through proper planning, organized activity and co-ordination.

It includes the following:

  • Prevention of loss and danger.
  • Improving tolerance.
  • Providing relief from disaster, minimising the intensity and extent of harm.
  • Preparation to face the disaster.
  • Immediate action in the disaster situation.
  • Assessment of damages and intensity of the disaster.
  • Arranging for rescue work and help.
  • Rehabilitation and rebuilding.

Question 3.
Write a short note on : Indian Meteorological Department
Answer:

  • The Indian Meteorological Department was founded by the British in 1875 at Shimla.
  • Its head office is at Pune and its Regional offices are at Mumbai, Kolkata, Chennai, Nagpur and Delhi.
  • Maps are prepared every day which indicate the daily predictions about the weather.
  • Such maps are prepared and published twice in every 24 hours.
  • In this institute, research goes on continuously on various aspects like instruments for climatic readings, predictions made about climate using radar, predictions about climate related to seismology, predictions regarding rainfall by satellites, air pollution etc. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • The Indian Meteorological Department provides information regarding weather and climatic conditions to other departments like aviation, shipping, agriculture, irrigation, marine oil exploration and production etc.
  • Predictions regarding calamities like dust storms, sand storms, heavy rainfall, hot and cold waves, tsunami, etc. are communicated to various departments, all types of mass communication media and all citizens.
  • India has launched several satellites equipped with highclass technology.
  • Observatories at several locations are doing excellent work in the analysis of the information received from these satellites.

9th Std Science Questions And Answers:

Life Processes in Living Organisms Class 9 Science Chapter 15 Questions And Answers Maharashtra Board

Class 9 Science Chapter 15

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 15 Life Processes in Living Organisms Question Answer Maharashtra Board

Class 9 Science Chapter 15 Life Processes in Living Organisms Question Answer Maharashtra Board

1. Match the pairs and explain.

Column ‘A’ Column ‘B’
(1) Growth of pollen tube towards ovules
(2) Growth of shoot system
(3) Growth of root system
(4) Growth towards the water
(a) Gravitropic movement
(b) Chemotropic movement
(c) Phototropic movement
(d) Growth-irrelevant movement
(e) Hydrotropic movement

Answer:
(1 – b),
(2 – c),
(3 – a),
(4 – e).

Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms

2. Complete the paragraph.

The milk was on the stove. Rasika was engrossed in watching television. She smelled something burning. She ran towards the kitchen. The milk was boiling over. She held the vessel with her bare hands but, screaming, she let it go at once. This activity was controlled by ……………….. cells. Special ends of ……………….. in these cells collected the information, from where it was transferred to the ……………….. and then towards the terminal end of the ……………….. The chemicals produced at the terminal end passed through the minute space i.e. ………………. . In this way, ……………….. were conducted in the body and the process of ……………….. was completed by conducting the impulses from ……………….. to ………………

(Nerve, muscle cell, impulse, dendrite, synapse, axon, reflex action, cell body)
Answer:
nerve, dendrite, cell body, axon, synapse, impulses, reflex action, nerve, muscle cell.

3. Write notes on Root pressure, Transpiration, Nerve cell, Human brain, Reflex action

4. Name the hormones of the following endocrine glands and the function of each.
Pituitary, Thyroid, Adrenal, Thymus, Testis, Ovary.

Life Processes In Living Organisms Class 9 Question 1.
Pituitary:
Answer:

Hormone Function
(i) Growth hormone Stimulates growth of bones.
(ii) Adrenocorticotropic hormone Stimulates adrenal gland.
(iii) Thyroid-stimulating hormone Stimulates thyroid gland.
(iv) Prolactin Stimulates milk production.
(v) Follicle-stimulating hormone Controls growth of gonads.
(vi) Luteinizing hormone Controls menstrual cycle and ovulation.
(vii) Oxytocin Contracts uterus during parturition.
(viii) Antidiuretic hormone Regulates water-level in the body.

Class 9 Science Chapter 15 Life Processes In Living Organisms Question 2.
Thyroid
Answer:

Hormone Function
(i) Thyroxine Controls growth of body and metabolic activities.
(ii) Calcitonin Controls calcium metabolism and calcium level in blood.

Life Processes In Living Organisms Class 9 Exercise Question 3.
Adrenal
Answer:

Hormone Function
(i) Adrenalin and nor-adrenalin Controls behaviour during crisis and emotional situation.
(ii) Corticosteroid Maintains balance of Na+ and K+ and stimulates metabolism.

9th Class Science Chapter 15 Life Processes In Living Organisms Exercise Question 4.
Thymus
Answer:

Hormone Function
Thymosin Control the cells which give rise to immunity.

Question 5.
Testis
Answer:

Hormone Function
Testosterone Stimulates growth of secondary sexual characteristics like beard, mustache, hoarse voice etc. in men.

Std 9 Science Chapter 15 Life Processes In Living Organisms Question 6.
Ovary
Answer:

Hormone Function
(i) Oestrogen Stimulates growth of the endometrium. Stimulates growth of secondary sexual characteristics in women.
(ii) Progesterone Prepares the endometrium for conception and maintains pregnancy.

5. Draw and label the diagrams.
Human endocrine glands, Human brain, Nephron, Nerve cell, Human excretory system.

15. Life Processes In Living Organisms Question 1.
Human endocrine glands
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 1

15 Life Processes In Living Organisms Question 2.
Human brain
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 2

Life Processes In Living Organisms 9th Class Notes Question 3.
Nephron
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 3

Life Processes In Living Organisms Class 9 Questions And Answers Question 4.
Nerve cell
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 4

Life Processes In Living Organisms Class 9 Solutions Question 5.
Human excretory system
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 5

6. Answer the following.

a. Explain chemical co-ordination in humans and give the names and functions of some hormones.
Answer:
Chemical co-ordination in humans :

  • In humans chemical co-ordination is brought about with the help of certain chemical substances called hormones.
  • Hormones are secreted by endocrine glands. These glands are also called ductless glands.
  • These glands do not have any arrangement of their own to either store or carry their secretions. So, the hormones are directly released into the blood circulation.
  • Though the endocrine glands are present at specific locations in our body, their secretions reach all parts of the body via the blood.
  • Endocrine glands along with the nervous systems are responsible for control and co-ordination in our body.
    These two systems help each other to control and integrate the various activities of the body.
  • A marked difference between these two systems is that nerve impulses are fast but short-lived whereas the actions of hormones are very slow but long-lasting.
  • It is very important that hormones are secreted only in the required quantity and there is a special mechanism which controls the quantity and timing of hormone secretion.
  • For example, whenever there is an increase in blood glucose level, certain cells in the pancreas get stimulated and as a response, they release a greater quantity of insulin, thus bringing down the sugar level to normal.
Hormone Function
(1) Growth hormone Stimulates growth of bones.
(2) Luteinizing hormone Controls menstrual cycle and ovulation.
(3) Thyroxine Controls growth of body and metabolic activities.
(4) Insulin Stimulates liver to convert excess blood glucose to glycogen.
(5) Testosterone Stimulates growth of secondary sexual characters like beard, mustache, hoarse voice, etc. in men.

Names and functions of some hormones:

b. Explain the difference between the excretory system of humans and plants.
Answer:

Excretory system of humans Excretory system of plants
(i) In humans, the excretory system carries out the function of the removal of waste from the body. (i) In plants, there is no special system or organ for excretion.
(ii) Waste substances are generally eliminated out of the body (ii) Most of the waste substances are stored in the leaves, flowers, fruits and bark of the stem.
(iii) The excretory products are urea, uric acid, ammonia, etc. (iii) The excretory products are gum, resin, latex of rubber, etc.

c. Explain co-ordination in plants with the help of suitable examples.
Answer:

  • Plants do not have systems like the nervous system or muscular system.
  • In plants, movements are mainly in the form of responses given to the stimuli.
  • Plants show two types of movements – growth relevant movements and growth irrelevant movements.
  • Movement or growth of any part of the plant in response to an external stimulus is called tropism or tropic movement.
  • Examples of growth-relevant movements are :
  • The shoot system of any plant responds to the light stimulus i.e., it grows towards the source of light. This is called phototropic movement.
  • The root system of plants responds to stimuli like gravitation and water. These responses are called gravitropic movement and hydrotropic movement respectively.
  • Movement shown by plants in response to specific chemicals is called chemotropic movement.
  • For example, the growth of the pollen tube towards the ovule.
  • Some specific movements of the plants do not lead to the plant growth.
  • Such movements are called growth-irrelevant movements.
  • As a response to changes in the surroundings, plant hormones bring about various movements in plants.
  • Examples of growth irrelevant movements are closing of leaflets of Touch-me-not (Mimosa) plant on touch.
  • In the plant called Venus fly trap, when an insect visits the flower like trap, the trap closes up and the trapped insect is digested by the plant.
  • The lotus flower opens during day-time while that of the tuberose opens at night.
  • In plants like Touch-me-not (Mimosa), movement also occurs at places other than where it has been touched.
  • Hence, we can infer that the information about the touch must have been released within the plant from one place to another.
  • Plants use electrochemical impulses for transfer of information from one place to another.
  • Plant cells change their shape by increasing or decreasing the water content and thereby, bring about the movements of plants.

7. Explain in your own words with suitable examples.

a. What is meant by co-ordination?
Answer:

  • Several different organ systems function in multicellular organisms.
  • Their life goes on smoothly if there is co-ordination between different organ systems or organs and the stimuli in the surrounding.
  • Depending upon this, we can say that systematic regulation of different processes can be called control and bringing about the different processes in the proper sequence is called co-ordination.
  • If any activity in the body is to be completed successfully, proper co-ordination between different systems, and organs participating at different steps of that activity is necessary.
  • If due to lack of co-ordination, there is confusion at any step, the activity may not get completed. There should be no randomness at any step.
  • There should be proper co-ordination between internal activities of the body resulting from various factors like body temperature, water- level, enzyme-level, etc. or stimuli arising in the surrounding environment.
  • Proper co-ordination between various systems of an organism helps to maintain a state of equilibrium called homeostasis which is necessary for the optimal efficiency of the body.

b. How does excretion occur in human beings?
Answer:

  • Removal of wastes from the body is very important and it is the excretory system that carries out this function.
  • The human excretory system consists of a pair of kidneys, a pair of ureters, the urinary bladder and the urethra. Urine is formed by the kidneys by separating the waste and unwanted excess substances from the blood.
  • The functional unit of the kidney that performs the basic function of filtration is called a nephron. The urea produced in the liver comes into the blood.
  • When the urea-containing blood comes into the glomerulus, it is filtered through capillaries and urea and other similar substances are separated from it.
  • Water molecules and small molecules of some other substances can cross the semipermeable membrane of Bowman’s capsule.
  • The solution accumulated in the cavity of Bowman’s capsule passes into the tubular part of the nephron. Here, molecules of water and some other useful substances are reabsorbed into the blood.
  • Urine is formed from the remaining solution which is full of waste materials. The urine is carried by the ureters and stored in the urinary bladder. Afterwards, urine is given out through the urethra.

c. How is excretion in plants useful to human beings?
Answer:

  • Plants give out oxygen during photosynthesis by diffusion which is useful to human beings for respiration.
  • Some waste materials of plants for example, gum, resin, latex of rubber, etc. are useful to humans.

d. Describe the transportation system in plants.
Answer:

  • Plants have two types of conducting tissues i.e., xylem which conducts water and phloem which conducts food.
  • During transpiration, water is released into the atmosphere.
  • As a result, water level is the epidermal layer of the leaf decreases.
  • Water is brought up to the leaves through the xylem so as to compensate for the lost water.
  • Transpiration helps in absorption of water and minerals and distribution to all parts of the plant whereas root pressure preforms the important role of pushing the water up at night time.
  • The food produced by leaves is transported to each cell of the plant through phloem.
  • When the food material like sucrose is transported towards a part of the plant through phloem, using ATP, the water concentration in that part decreases.
  • As a result, water enters the cell by diffusion. The pressure on the cell wall increases due to increase in the cellular contents.
  • Due to the increased pressure, food is pushed into the neighboring cells where the pressure is low.
  • This process helps the phloem to transport the materials as per the need of the plant.

Class 9 Science Chapter 15 Life Processes in Living Organisms Intext Questions and Answers

Class 9 Science Chapter 15 Life Processes In Living Organisms Exercise  Question 1.
How do the digestive system and respiratory system work?
Answer:
(a) Digestive system:

  • The digestive system is responsible for digestion of food. i.e. breaking down complex insoluble food into simpler soluble form.
  • The digestive system is made up of alimentary canal and digestive glands.
  • The alimentary canal is a long and muscular tube of varying diameter which extends from the mouth to the anus.
  • The digestive glands associated with alimentary canal are salivary glands, gastric glands, liver and pancreas.
  • The process of digestion includes various stages wherein the food passes through the different organs of the alimentary canal and there it is digested part by part with the help of various enzymes secreted from the digestive glands.
  • After the food is digested, the important nutrients are absorbed into the blood and the undigested food and residue of digested food is removed out of the body through the anus.

(b) Respiratory system:

  • Respiration is the process of release of energy from the assimilated food.
  • The respiratory system consists of nose, pharynx, larynx, trachea, bronchi and lungs (alveoli).
  • Oxygen enters our body through breathing.
  • This oxygen is taken to the cells and with the help of oxygen, food is oxidized into the cells to release energy in the form of ATR
  • During this process, carbon dioxide and water vapour are given out as by-products which are given out during breathing.

Thus, the digestive and respiratory systems work in co-ordination with each other to produce energy from food.

Life Processes In Living Organisms Exercise Question 2.
Previously you have performed the activity of observing a branch covered in a plastic bag. What did you observe in that activity?
Answer:

  • When a branch is covered in a plastic bag, we observe that small droplets of water accumulate on the inner side of the plastic bag.
  • This shows that leaves release water in the form of vapour through the stomata during transpiration. These vapours get condensed to form water droplets and accumulate inside the bag.

Life Process In Living Organism Exercise Question 3.
Which are the sensory organs of an organism? What is their function?
Answer:

  • Sensory organs of an organism are eyes, ears, tongue, skin and nose.
  • The eyes help the organism to perceive the sensation of light to see the objects around it.
  • The ears help the organism to perceive the sensation of sound and also perceive the sensation of pressure.
  • The tongue helps to perceive different tastes and also in chewing the food.
  • The skin helps to perceive the sensation of touch, pain, pressure etc.
  • Some insects also possess antennae for sensing the changes in the environment.

15 Life Processes In Living Organisms Class 9 Question 4.
Where are the gustatory and olfactory nerves to be found?
Answer:

  • Gustatory nerves are concerned with sense of taste whereas olfactory nerves are concerned with the sense of smell.
  • Gustatory nerves are found in the tongue from where they carry impulses to the gustatory cortex in the brain.
  • Whereas the olfactory nerves are found in the nasal passage from where they carry impulses to the olfactory lobes in the brain.

Answer the following questions:

15.Life Processes In Living Organisms Question 1.
Why do we eat fruits and vegetables? Do the plants also need minerals like we do?
Answer:

  • We eat fruits and vegetables to get nutrients like carbohydrates, vitamins, minerals and fibers. These nutrients are important for the proper functioning of the body .
  • Plants too need minerals for their growth and development.

Life Processes In Living Organisms Class 9 Notes Question 2.
From where do plants get inorganic substances other than carbon dioxide and oxygen?
Answer:
Plants get inorganic substances like nitrogen, phosphorus, magnesium, manganese, sodium etc. from the soil. Soil is the nearest and richest source of these elements.

9th Std Science Chapter 15 Life Processes In Living Organisms Question 3.
At least a small quantity of garbage or waste is produced every day in each house. What will happen if you keep this garbage for many days in your house?
Answer:

  • If we keep garbage for many days in our house, it will start decaying due to the growth of decomposers like fungi and bacteria.
  • This will result in foul odour and will also lead to spread of diseases.

Life Processes In Living Organisms Exercise Class 9 Question 4.
Sometimes, while eating we bite our own finger or tongue by mistake.
Answer:

  • Sometimes, while eating we over-fill our mouth causing the bite to misalign, thus we bite our own finger or tongue by mistake.
  • If we are dehydrated, the muscles of the mouth and tongue do not function properly causing us to bite our own tongue.
  • If we are distracted with other activities while eating, it can cause us to bite our tongue or finger.
  • Thus, due to lack of co-ordination sometimes while eating, we bite our own finger or tongue by mistake.

9th Class Science 15 Chapter Life Processes In Living Organisms Question 5.
Sometimes, we choke while eating in a hurry.
Answer:

  • There is an elastic flap at the starting of the wind-pipe called epiglottis which opens whenever we breathe in air and closes to prevent the entry of food into the respiratory system.
  • While eating in hurry, due to lack of co-ordination, the epiglottis opens to take in air but food too enters in causing us to choke.

Life Processes In Living Organisms Class 9 Extra Questions Question 6.
An injury to the medulla oblongata can lead to death. Why?
Answer:

  • The medulla oblongata controls involuntary activities like the beating of the heart, blood circulation, breathing etc.
  • An injury to the medulla oblongata will stop these activities, leading to death.

Life Processes In Living Organisms Class 9 Maharashtra Board  Question 7.
You may have seen how a drunken person struggles to maintain his body balance. An excess of alcohol in the body causes one to lose control over it. Why does this happen? Find out the reason with help of the internet.
Answer:

  • The cerebellum co-ordinates voluntary movements and maintains balance of the body.
  • An excess of alcohol causes dysfunctioning of the cerebellum.
  • It results in primary loss of co-ordinated movements and loss of balance.

9th Life Processes In Living Organisms 5 Question 8.
Why does this happen?

(a) Leaves of plants fall off in a particular season.
Answer:

  • Plants lose water through their leaves by transpiration. So some plants shed their leaves in dry season to reduce transpiration and conserve water.
  • In plants, the waste substances are stored in the vacuoles of leaf-cells which are then shed along with the leaves in a specific season.

(b) Fruits, flowers fall off after a certain period of time.
Answer:

  • In plants, waste substances are stored in the vacuoles of fruits and flowers. So, they fall off after some time to excrete these wastes.
  • Also, when fruits and flowers mature, the supply of nutrients to mature fruits and flowers is blocked. This causes abscission in that part of the plant leading to their detachment from the plant body and falling off.

(c) Substances like resin, gum, etc. are given out of the plant body.
Answer:

  • Resin, gum etc. are the waste products of plants. They are given out after some time as a process of excretion in plants.
  • Many plants, particularly, woody plants produce resin and gum in response to injury. These resin and gum act as bandage protecting the plants from invading insects and pathogens.

15 Life Processes In Living Organisms Exercise Question 9.
Which waste products are produced in our body through metabolic activities?
Answer:
The waste products formed in our body through metabolic activities are urea, uric acid, ammonia, carbon dioxide etc.

Life Processes In Living Organisms 9th Class Exercise Question 10.
As compared to the monsoons and winter a very small quantity of urine is produced in the summer season. Why is it so?
Answer:
In summer, we sweat more due to excess heat. So, the body tries to conserve water by reducing urine formation. This is to prevent dehydration.

Question 11.
In adults, the process of urination is under their control but not in infants. Why is it so?
Answer:

  • In adults, the muscles of the bladder are under the voluntary control. So the process of urination is under control.
  • In infants, the bladder muscles are not mature until about two years of age. So, the process of urination is not under control.

Answer the following questions:

Question 1.
Observe the following figures carefully.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 9
Answer:
Observations:

  • Figure ‘A’ shows chemotropic movement. Pollen tube grows towards the ovule in response to specific chemicals released by the ovule.
  • Figure ‘B’ shows phototropic movement. Shoot grows towards the Sun in response to stimulus of light.
  • Figure ‘C’ shows thigmotropic movement. Tendrils of climbers are sensitive to touth. When they come in contact with an object, they entwine around the object and cling to it.
  • All the above movements are related to growth. Hence, they are called growth relevant movements.

Question 2.
Observe the pictures carefully and think about them.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 10
Answer:

  • Figure ‘A’ shows touch-me-not (Mimosa) plant. In this plant, the leaflets close in response to stimulus of touch.
  • Figure ‘B’ shows Venus fly trap. In this plant, there is a trap that appears and smells like a flower and deceives the insect. When an insect visits that flower-like trap, the trap closes up and the trapped insect is digested by the plant,
  • Figure ‘C’ shows lotus plant. The lotus flower opens during day-time.?
  • Figure ‘D’ shows Balsam plant. In this plant, the ripened fruit dehisces (bursts open) at the right time to disperse the seeds.

Question 3.
As you watch the match being played on your school ground, you will see the control and co-ordination among the movements of the players. Make a list of all such different actions.
Answer:
In an inter-house basketball match, control and co-ordination is seen in dodging the ball, jumping, running, passing it to the team mates and dribbling the ball.

Question 4.
Observe the figure carefully and as per the numbers in that figure, answer the following questions.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 11

(a) What is happening at 1 and 2?
Answer:
At 1 stimulus of heat is received by thermo-receptors in the skin. At 2, the hand is withdrawn instantly.

(b) Which nerve carried the impulse to the point marked 3? In which direction is it conducting the impulse?
Answer:
Sensory nerve carried the impulse to the point marked 3. The impulse is conducted from the skin towards the spinal cord.

(c) Which is the nerve shown by 4?
Answer:
The nerve shown by 4 is association neuron. It is present in the gray matter of the spinal cord. It receives the sensory impulse, interprets it and generates impulse.

(d) Which is the organ marked as 5?
Answer:
Spinal cord.

(e) At 6, which nerve is conducting the response impulse?
Answer:
Motor nerve.

(f) At 7, where has the impulse reached? What is its effect?
Answer:
At 7, the impulse has reached the muscles of the hand. As a result, the muscles contract and the hand is withdrawn instantly in response to the stimulus of heat.

Question 5.
Sketch and label:
Answer:

  • Stimulus
  • Hand is withdrawn
  • Sensory neuron
  • Association neuron
  • Spinal cord
  • Motor neuron
  • Muscles

(6) Observe the figure and list down functions for ‘A’ and ‘B’
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 8
Answer:

Functions of Left Brain Functions of right brain
(i) Analytical thinking Holistic thinking
(ii) Logical thinking Intuition
(iii) Language Creativity
(iv) Science and Mathematics Art and music

Answer the following questions:

Question 1.
Take a small plant like balsam or tuberose with its roots intact. Wash and clean its roots. As shown in the fig, keep it in the water containing a stain like safranin or eosin. Observe the stem and the veins of the leaves after 2-3 hours.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 12
Answer:

  • The stem and veins of the leaves appear pinkish in colour.
  • Due to the root pressure generated, eosin solution is continuously pushed upward.
  • As a result water reaches xylem and the water column is pushed forward.

Question 2.
Take a transverse section of the stem of a plant and observe the stained xylem under a compound microscope.
Answer:
The xylem elements appear pinkish as the eosin solution is conducted upwards through the xylem. From this we infer that the xylem is responsible for the conduction of water and minerals in the plant.

Question 3.
Observe your mother while she cuts the elephant’s foot (Amorphophallus) or arum leaves. Your hands may also begin to itch if you try to cut these leaves. Why does this happen? Try to find out. Ask your mother what she does to prevent the itching.
Answer:

  • In some plants like elephant foot (Amorphophallus) or arum, waste materials are in the form of crystals of calcium oxalate. They are called raphides.
  • As they are needle-shaped, they prickle and cause irritation of the skin leading to itching.
  • This can be prevented by applying tartaric acid (tamarind) or citric acid (lemon) on the hands while cutting these vegetables as the acids dissolve the crystals of calcium oxalate.

Class 9 Science Chapter 15 Life Processes in Living Organisms Additional Important Questions and Answers

Choose and write the correct options.

Question 1.
The transport of food produced in leaves to each cell in the plant body is called …………………………… .
(a) Transpiration
(b) Root pressure
(c) Translocation
(d) Ascent of sap
Answer:
(c) translocation

Question 2.
Kidneys filter blood about ……………………………. times every day.
(a) 100
(b) 200
(c) 300
(d) 400
Answer:
(d) 400

Question 3.
About ……………………………. of blood is sent through the dialysis machine at one time.
(a) 500 ml
(b) 5 l
(c) 200 ml
(d) 400 ml
Answer:
(a) 500 ml

Question 4.
The movement of plant towards the source of light is called ……………………………. movement.
(a) chemotropic
(b) hydrotropic
(c) phototropic
(d) gravitropic
Answer:
(c) phototropic

Question 5.
The movement of root system towards stimulus of gravity is called ……………………………. movement.
(a) phototropic
(b) gravitropic
(c) hydrotropic
(d) chemotropic
Answer:
(b) gravitropic

Question 6.
The movement of root system towards stimulus of water is called ……………………………. movement.
(a) Phototropic
(b) Gravitropic
(c) Hydrotropic
(d) Chemiotropic
Answer:
(c) hydrotropic

Question 7.
Hormone ……………………………. helps in stem elongation.
(a) Auxin
(b) Gibberellin
(c) Cytokinin
(d) Abscisic acid
Answer:
(b) Gibberellin

Question 8.
Hormone ……………………………. helps in cell division.
(a) Auxin
(b) Gibberellin
(c) Cytokinin
(d) Abscisic acid
Answer:
(c) Cytokinin

Question 9.
The hormone ……………………………. is effective in prevention and retardation of growth, leaf wilting, etc.
(a) Auxin
(b) Cytokinin
(c) Gibberellin
(d) Abscisic acid
Answer:
(d) Abscisic acid

Question 10.
……………………………. neurons conduct impulses from the sensory organs to the brain and the spinal cord,
(a) Sensory
(b) Motor
(c) Association
(d) All of these
Answer:
(a) Sensory

Question 11.
The brain of an adult human weighs about …………………………… .
(a) 500 – 600 grams
(b) 2 – 3 kilograms
(c) 1300 -1400 grams
(d) 1500 -1600 grams
Answer:
(c) 1300 – 1400 grams

Question 12.
……………………………. is the largest part of the brain.
(a) Cerebrum
(b) Cerebellum
(c) Medulla oblongata
(d) Spinal cord
Answer:
(a) Cerebrum

Question 13.
There are two triangular swollen structure called ……………………………. on the upper side of the medulla oblongata.
(a) Ventricles
(b) Meninges
(c) Pyramids
(d) Pons
Answer:
(c) Pyramids

Question 14.
Control of voluntary movements is done by …………………………… .
(a) Cerebrum
(b) Cerebellum
(c) Medulla oblongala
(d) Spinal cord
Answer:
(a) Cerebrum

Question 15.
Co-ordination of voluntary movements is done by …………………………… .
(a) cerebrum
(b) cerebellum
(c) medulla oblongata
(d) spinal cord
Answer:
(b) cerebellum

Question 16.
Control of involuntary activities like the beating of the heart, blood circulation, breathing etc. are controlled by …………………………… .
(a) Spinal cord
(b) Cerebrum
(c) Cerebellum
(d) Medulla oblongata
Answer:
(d) Medulla oblongata

Question 17.
The function of cerebrum is to …………………………… .
(a) control of voluntary movements
(b) co-ordination of voluntary movements
(c) control of involuntary activities
(d) maintaining the body balance
Answer:
(a) control of voluntary movements

Question 18.
The function of cerebellum is …………………………… .
(a) to control of voluntary movements
(b) to maintain the body balance
(c) to control of involuntary activities
(d) to co-ordinate of voluntary movements
Answer:
(b) to maintain the body balance

Question 19.
Hormone ……………………………. stimulates the liver to convert excess blood glucose into glycogen.
(a) Glucagon
(b) insulin
(c) Somatostatin
(d) Pancreatic polypeptide
Answer:
(a) Glucagon

Question 20.
Hormone ……………………………. controls behaviour during crisis and emotional situation.
(a) Thyroxine
(b) Parathormone
(c) Adrenalin and Nor adrenalin
(d) Somatostatin
Answer:
(c) Adrenalin and Nor adrenalin

Question 21.
……………………………. controls menstrual cycle and ovulation.
(a) Prolactin
(b) Oxytocin
(c) Luteinizing hormone
(d) Progesterone
Answer:
(c) Luteinizing hormone

Question 22.
……………………………. stimulates growth of secondary sexual characteristics in men.
(a) Thymosin
(b) Testosterone
(c) Oestrogen
(d) Thyroxine
Answer:
(b) Testosterone

Question 23.
……………………………. occurs through stomata.
(a) Translocation
(b) Perspiration
(c) Transpiration
(d) Conduction
Answer:
(c) Transpiration

Question 24.
……………………………. performs the important role of pushing the water up during the night time.
(a) Translocation
(b) Root pressure
(c) Transpiration pull
(d) None of these
Answer:
(b) Root pressure

Question 25.
In plants, gaseous substances are given out by …………………………….
(a) Root pressure
(b) Osmosis
(c) Diffusion
(d) Translocation
Answer:
(c) diffusion

Question 26.
Each kidney has approximately ……………………………. nephrons.
(a) 1 lakh
(b) 10,000
(c) 10 lakh
(d) 50 lakh
Answer:
(c) 10 lakh

Question 27.
Insulin is secreted by the ……………………………. of the pancreas.
(a) Alpha cells
(b) Beta cells
(c) Delta cells
(d) F cells
Answer:
(b) Beta cells

Select the odd man out:

Question 1.
Skin, Brain, Kidneys, Lungs.
Answer:
Brain – It is a controlling organ whereas the others are excretory orgAnswer:

Question 2.
Gum, Resin, Urea, Latex
Answer:
Urea – It is an excretory product of animals whereas the others are excretory products of plants.

Question 3.
Gibberellin, Auxin, Thyroxine, Cytokinin.
Answer:
Thyroxine – It is a hormone in animals whereas the others are hormones in plants.

Question 4.
Dendrite, Cell body, Axon, Synapse.
Answer:
Synapse – It is a minute space between two adjacent neurons whereas the others are parts of the neuron.

Question 5.
Thyroxine, Calcitonin, Insulin, Auxin.
Answer:
Auxin – It is a plant hormone whereas the others are hormones in human beings.

Complete the Analogy:

(1) Water : Xylem :: Food : …………………………….
(2) Movement of root system : Gravitropic :: Movement of shoot system : …………………………….
(3) Brain: Cranial Cavity :: Spinal Cord : …………………………….
(4) Cavities of the brain : Ventricles :: Cavities of the spinal cord : …………………………….
(5) Involuntary activities : Medulla oblongata : : Balance of the body : …………………………….
(6) Stem elongation: Gibberellin :: Cell division : …………………………….
(7) Control of voluntary movements : Cerebrum :: Co-ordination of voluntary movements : …………………………….
(8) Converts glucose to glycogen : Insulin :: Converts glycogen to glucose : …………………………….
(9) Secondary sexual characters in males : Testosterone :: Secondary sexual characters in females : …………………………….
Answer:
(1) Phloem
(2) Phototropic
(3) Vertebral column
(4) Central canal
(5) Cerebellum
(6) Cytokinin
(7) Cerebellum
(8) Glucagon
(9) Oestrogen

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Gibberellins
(2) Cytokinins
(3) Abscisic acid
(4) Auxin
(a) Enlargement of cells
(b) Helps in stem elogation
(c) Cell division
(d) Wilting of leaf

Answer:
(1 – b),
(2 – c),
(3 – d),
(4 – a).

Question 1.

Column ‘A’ Column B’
(1) Cerebrum
(2) Cerebelleum
(3) Medulla oblongata
(4) Spinal cord
(a) Control of involuntary activities
(b) Decision – making, memory and intellectual activities.
(c) Centre for co­ordination of reflex actions
(d) Maintaining body’s balance

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c).

State whether the following statements are true or false. Correct the false statements.

Question 1.
Plants need less energy as compared to animals.
Answer:
True

Question 2.
Xylem conducts food whereas phloem conducts water.
Answer:
False. Xylem conducts water whereas phloem conducts food.

Question 3.
The loss of water from the plants is known as translocation.
Answer:
False. The loss of water from the plants is known as transpiration.

Question 4.
Calcium oxalate crystals present in some plants are called resins.
Answer:
False. Calcium oxalate crystals present in some plants are called raphides.

Question 5.
Root pressure helps to push the water up in plants during night time.
Answer:
True

Question 6.
Translocation is carried out by phloem in downward direction.
Answer:
False. Translocation is carried out by phloem in both upward and downward direction.

Question 7.
In unicellular organisms, waste materials are directly eliminated across the cell surface.
Answer:
True

Question 8.
Urea is produced in the liver.
Answer:
True

Question 9.
Systematic regulation of different processes is called co-ordination.
Answer:
False. Systematic regulation of different processes is called control.

Question 10.
A hormone auxin produced in the apical part of the shoot helps in enlargement of cells.
Answer:
True

Question 11.
Gibberellin helps in prevention and retardation of growth, leaf wilting etc.
Answer:
False. Abscisic acid helps in prevention and retardation of growth, leaf wilting etc.

Question 12.
The movement shown by venus fly trap while trapping an insect is a growth relevant movement.
Answer:
False. The movement shown by venus fly trap while trapping an insect is a growth irrelevant movement.

Question 13.
The lotus flower opens during night time while that of tuberose (polyanthus) opens during day time.
Answer:
False. The lotus flower opens during day time while that of tuberose (polyanthus) opens during night time.

Question 14.
Sensory neurons conduct impulses from the brain and the spinal cord to effector organs like muscles or glands.
Answer:
False. Sensory neurons conduct impulses from the sensory organs to the brain and the spinal cord.

Question 15.
There are 31 pairs of cranial nerves.
Answer:
False. There are 12 pairs of cranial nerves.

Question 16.
Activities like heart beat, blood circulation, breathing, sneezing, etc. are controlled by cerebellum.
Answer:
False. Activities like heart beat, blood circulation, breathing, sneezing, etc. are controlled by cerebellum.

Question 17.
Spinal nerves are associated with head, thorax and abdomen.
Answer:
False. Spinal nerves are associated with arms, legs, skin and some other parts of the body.

Question 18.
Nerve impulses are slow and long lasting whereas action of hormones is fast and short lived.
Answer:
False. Nerve impulses are fast and short lived whereas action of hormones is slow and long lasting.

Question 19.
Reflex actions are controlled by cerebellum.
Answer:
False. Reflex actions are controlled by spinal cord.

Question 20.
Thyroid stimulating hormone is secreted by the pituitary gland.
Answer:
True

Question 21.
The hormone thymosin is secreted by the thyroid gland.
Answer:
False. The hormone thymosin is secreted by the thymus gland.

Question 22.
Parathormone controls metabolism of calcium and phosphorns.
Answer:
True

Question 23.
The right kidney is slightly lower than the left kidney.
Answer:
True

Name the following:

Question 1.
The structural and functional unit of the nervous system.
Answer:
Newtron.

Question 2.
The network of capillaries in the Bowman’s capsule.
Answer:
Glomerulus.

Question 3.
The minute space between two adjacent nervous.
Answer:
Synapse.

Question 4.
The special cells which bring about control and co-ordination.
Answer:
Neurons.

Question 5.
The cavities present in various parts of the brain.
Answer:
Ventricles.

Question 6.
The long tubular cavity of the spinal cord.
Answer:
Central Canal.

Question 7.
The thread like fibrous structure at the end of the spinal cord.
Answer:
Filum terminale.

Give scientific reasons:

Question 1.
It is necessary to remove harmful and waste substances from the body.
Answer:

  • Many harmful and waste substances like urea, uric acid, ammonia, etc. are produced in living organisms.
  • If these substances accumulate in the body or are retained in the body for long, it can lead to serious harm or even death.
  • Hence, it is necessary to remove such harmful and waste substances from the body.

Question 2.
In plants like Touch-me-not (Mimosa), movement also occurs at the places other than where it has been touched.
Answer:

  • In plants like Touch-me-not (Mimosa), the information about the touch is relayed within the plant from one place to another.
  • Plants use electro-chemical impulses for transfer of information from one place to another.
  • Plant cells change their shape by increasing or decreasing their water content and thereby, bring about the movements of plants.
  • Hence, in plants like Touch-me-not (Mimosa) movement also occurs at the places other than where it has been touched.

Question 3.
Endocrine glands are also called ductless glands.
Answer:

  • Hormones are secreted by endocrine glands.
  • These glands do not have any arrangement of their own to store or carry their secretions.
  • Hence, as soon as hormones are produced, they are directly released into the blood circulation.
  • Thus, though these endocrine glands are present at specific locations in our body, their secretions reach all parts of the body via blood.
  • Therefore, endocrine glands are also called ductless glands.

Define the following:

Question 1.
Transportation
Answer:
The process by which a substance either synthesized or absorbed in one part of the body reaches another is called as transportation.

Question 2.
Control
Answer:
Systematic regulation of different processes is called as control.

Question 3.
Coordination
Answer:
Bringing about the different processes in the proper sequence is called as co-ordination.

Question 4.
Tropism or Tropic movement
Answer:
Movement or growth of any part of the plant in response to an external stimulus is called as tropism or tropic movement.

Question 5.
Excretion
Answer:
Removal of waste or harmful substances from the body is called excretion.

Question 6.
Dialysis
Answer:
The process of separating the nitrogenous waste from the blood with the help of a man?made machine is called dialysis.

Question 7.
Reflex action
Answer:
An immediate and involuntary response given to a stimulus from the environment is called a reflex action.

Question 8.
Homeostasis
Answer:
Proper co-ordination between various systems of an organism helps to maintain a state of equilibrium called homeostasis.

Question 9.
Tropism
Answer:
Movement or growth of any part of the plant in response to an external stimulus is called tropism.

Distinguish between:

Question 1.
Nervous control and Chemical control
Answer:

Nervous Control Chemical Control
(i) Nervous control is brought about by special types of cells called neurons. (i) Chemical control is brought about by chemical substances called hormones.
(ii) It involves the nervous system i.e., brain, spinal cord and nerves. (ii) It involves the endocrine glands.
(iii) Nerve impulses are fast and short-lived. (iii) Actions of hormones are very slow and long-lasting.

Question 2.
Growth relevant movements and growth irrelevant movements
Answer:

Growth relevant movements Growth irrelevant movements
(i) Growth relevant movements result in growth of the plants. (i) Growth irrelevant movements do not result in growth of the plants.
(ii) They are also called as tropic movements. (ii) They are also called as nastic movements.
(iii) They are in response to stimulus of light, water, gravity, chemicals, etc. (iii) They are generally in response to stimulus of touch.
(iv) Examples : Bending of stem towards light, movement of roots towards gravity and water, etc. (iv) Example : Closing of leaflets of Mimosa plant, closing of Venus fly trap, etc.

Question 3.
Cerebrum and Cerebellum
Answer:

Cerebrum Cerebellum
(i) It is the largest part of the brain. (i) It is the smaller part of the brain.
(ii) It occupies two thirds of the brain. (ii) It is situated below the cerebrum at the back of the cranial cavity.
(iii) Its surface has deep, irregular ridges and grooves which are called convolutions. (iii) Its surface shows shallow grooves instead of convolutions.
(iv) It controls voluntary movements, concentration, planning, decision-making, memory, intelligence and intellectual activities. (iv) It co-ordinates voluntary movements and maintains the body’s balance.

Write short notes:

Question 1.
Root Pressure
Answer:

  • Root cells are in contact with water and minerals in the soil.
  • Water and minerals enter the cells of the root surface due to difference in concentration. As a result, these cells become turgid.
  • This is called root pressure.
  • Under the effect of this pressure, water and minerals reach the xylem of the roots and to reduce this difference in concentration, they are continuously pushed forward.
  • As a result of this continuous movement, a water column is formed, which is continuously pushed ahead.
  • This pressure, is sufficient to lift the water up in shrubs, small plants and small trees.

Question 2.
Transpiration
Answer:

  • Plants give out water in the form of water vapour through the stomata on their leaves. This is known as transpiration.
  • Two cells called guard cells are present around the stomata. They control the opening and closing of stomata.
  • Water is released into the atmosphere by leaves through the process of evaporation.
  • As a result, water in the epidermal layer of the leaf decreases.
  • Water is brought up to the leaves through the xylem to compensate for the lost water.
  • Transpiration helps in absorption of water and minerals and distribution to all parts of the plant.

Question 3.
Nerve cell
Answer:

  • Nerve cells also called as neurons are special types of cells which conduct impulses from one place to another in the body.
  • Neurons are the structural and functional units of the nervous system.
  • Nerve cells, the largest cell in the human body, may measure upto a few metres in length.
  • Nerve cells have the ability to generate and conduct electrochemical impulse.
  • The cells that support the nerve cells and help in their functioning are called neuroglia. Nerve cells and neuroglial cells together form nerves.

Question 4.
Reflex action
Answer:

  • An immediate and involuntary response given to a stimulus from the environment is called a reflex action.
  • Sometimes we react to an incident without any thinking on our part or control over the reaction.
  • This is a response given to a certain stimulus from the surroundings.
  • In such situations, proper control and co-ordination is achieved even without the intervention of the brain.
  • For example, when our hand touches a hot object, the receptor in the skin detects the stimulus.
  • The sensory neuron sends the impulse to the relay neuron in the spinal cord.
  • The impulses are then sent to motor neurons which cause the muscles of the hand to contract and move away from the hot object.
  • The spinal cord is the center of co-ordination for reflex actions.

Question 5.
Spinal cord
Answer:

  • The spinal cord is the part of the central nervous system and it is held within the vertebral column.
  • It is slightly thick but gradually tapers towards the end.
  • There is a thread like fibrous structure at its end. It is called the filum terminale.
  • The long tubular cavity of the spinal cord is called the central canal. It is filled with cerebrospinal fluid.
  • The spinal cord conducts impulses from the skin towards the brain and from the brain to the muscles and glands.
  • It functions as a centre of co-ordination of reflex actions.

Question 6.
Dialysis
Answer:

  • The efficiency of kidneys can be adversely affected by injury, infection or decreased blood supply.
  • If this happens, excess of toxic substances accumulates in the body and it can lead to death.
  • If kidneys fail, nitrogenous wastes are separated from the blood with the help of a man-made machine.
  • The process of separating the nitrogenous waste from the blood with the help of this machine is called dialysis.
  • About 500 ml of blood is sent at one time through this machine. Purified blood is reinfused into the body of the patient.

Draw neat and labelled diagrams:

Question 1.
Vertical Section of Kidney
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 6

Question 2.
Reflex action
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 7

Name the hormones of the following endocrine glands and the function of each:

Question 1.
Parathyroid
Answer:

Hormone Function
Parathormone Controls metabolism of calcium and phosphorus

Question 2.
Pancreas
Answer:

Hormone Function
(i) Glucagon Stimulates liver to convert glycogen into glucose.
(ii) Insulin Stimulates liver to convert excess blood glucose into glycogen.
(iii) Somatostatin Controls levels of insulin and glucagon.
(iv) Pancreatic Polypeptide Controls movement of intestine and thereby glucose absorption. Controls secretion of pancreatic juice.

Complete the following table of plant hormones with their functions:
Answer:

Plant hormone Functions
(i) Auxin Produced in the apical part of the shoot
Helps in enlargement of cells
(ii) Gibberellins Help in stem elongation
(iii) Cytokinins Help in cell division
(iv) Abscisic acid Effective in prevention and retardation of growth, leaf wilting, etc.

State the functions of different parts of the brain.
Answer:

Parts of the Brain Functions
Cerebrum Control of voluntary movements, concentration, planning, decision­making, memory, intelligence, and intellectual activities.
Cerebellum (i) Co-ordination of voluntary movements.
(ii)  Maintaining the body’s balance.
Medulla oblongata Control of involuntary activities like the beating of the heart, blood circulation, breathing, sneezing, coughing, salivation, etc.
Spinal cord (i) Conduction of impulses from the skin towards the brain.
(ii) Conduction of impulses from brain to muscles and glands.
(iii) Functions as centre of co­ordination of reflex actions.

Answer the following:

Question 1.
How do plants bring about movements in them?
Answer:

  • Plants use electro-chemical impulses for transfer of information from one place to another.
  • Plant cells change their shape by increasing or decreasing their water content and thereby bring about the movement of plants.

Answer the following:

Question 1.
What are growth-irrelevant movements? Give examples.
Answer:

  • Some specific movements of the plants do not lead to the plant growth. Such movements are called growth-irrelevant movements.
  • In the plant called venus fly trap, there is a trap that appears and smells like flowers and deceives insects. When an insect visits the flower-like trap, the trap closes up and the trapped insect is digested by the plant.
  • Fibrils present on the leaves of the insectivorous plant Drosera, bend inward as soon as an insect lands on the leaves and surround the insects from all sides.
  • In Balsam, the ripened fruit dehisces (bursts-open) at the right time to disperse the seeds.

Question 2.
Which are the different types of nerve cells or neurons?
OR
How are nerve cells classified according to their functions?
Answer:
According to their functions, nerve cells are classified into three types:

  • Sensory neurons: They conduct impulses from the sensory organs to the brain and the spinal cord.
  • Motor neurons : They conduct impulses from the brain or spinal cord to effector organs like muscles or glands.
  • Association neurons : Association neurons perform the function of integration in the nervous system.

Question 3.
What are the different parts of the human nervous system?
Answer:
The human nervous system can be divided into the following three parts.

  1. Central nervous system (CNS) : It consists of the brain and the spinal cord. It controls and regulates all activities of the body.
  2. Peripheral nervous system (PNS) : The peripheral nervous system consists of the nerves originating from the central nervous system which connect the central nervous system with all parts of the body. They are of two types cranial nerves and spinal nerves.
  3. Autonomic nervous system (ANS) : The autonomous nervous system consists of the nerves of involuntary organs like the heart, lungs, stomach etc. It is not under the control of our will.

Question 4.
Describe the structure of the central nervous system.
Answer:

  • The central nervous system consists of the brain and spinal card.
  • The organization of the brain is extremely delicate and highly evolved.
  • The brain is the main controlling part of the nervous system and is safely located in the cranial cavity. The spinal cord is protected by the vertebral column.
  • In the space between the delicate central nervous system and its bony covering are the protective layers called the meninges.
  • Cavities present in various parts of the brain are called ventricles whereas the long tubular cavity of the spinal cord is called the central canal.
  • The ventricles, central canal and spaces between the meninges are filled with cerebro?spinal fluid.
  • This fluid supplies nutrients to the central nervous system and protects it from shock.

Question 5.
Which are the two types of peripheral nerves?
Answer:
Peripheral nerves are of two types:
(a) Cranial nerves:

  • Nerves originating from the brain are called cranial nerves.
  • They are associated with various parts in the head.
  • There are 12 pairs of cranial nerves.

(b) Spinal nerves:

  • Nerves originating from the spinal cord are called spinal nerves.
  • They are associated with arms, legs, skin and some other parts of the body.
  • There are 31 pairs of spinal nerves.

Question 6.
How do plants get rid of their excretory products?
Answer:

  • Excretion is a simpler process in plants. There is no special organ or system for excretion in plants.
  • Gases are given out by diffusion.
  • Most of the waste substances of plants are stored in vacuoles of leaf-cells and in flowers, fruits and the bark of the stem. After some time these parts fall off.
  • Some other waste materials are stored in old and worn out xylem in the form of resin and gum.
  • Some waste materials are also given out through roots in the surrounding soil.

Answer in detail:

Question 1.
Explain how food and other substances are transported in plants?
OR
Explain translocation in plants.
Answer:

  • The food produced in leaves is transported to each cell in the plant body.
  • Excess food, except amino acids, is stored in roots, fruits and seeds. This process is called translocation of materials.
  • It is carried out in both the upward and downward directions by the phloem.
  • Translocation is not a simple process, it requires energy. This energy is obtained from ATP.
  • Whenever food material like sucrose is transported towards a part of a plant via the phloem with the help of ATP, the water concentration decreases in the part. As a result, water enters the cell by the process of diffusion.
  • The pressure on the cell wall increases due to increase in cellular contents.
  • Due to the increase pressure, food is pushed into the neighbouring cells where the pressure is low.
  • This process helps the phloem to transport the materials as per the need of the plant. During flowering season, the sugar stored in roots or stem is transported towards the floral buds to make them open and blossom.

Question 2.
What is tropic movement? Describe the different types of tropic movements.
OR
What are growth relevant movements? Describe the different types of growth relevant movements.
Answer:
Movement or growth of any part of the plant in response to an external stimulus is called tropic movement or tropism.

The different types of tropic movements are:

  • Phototropic movement : The shoot system of any plant responds to the light stimulus, i.e., it grows towards the source of light. The movement shown by plants towards the source of light is called phototropic movement.
  • Gravitropic movement : The root system of plants responds to stimulus of gravity. This is called gravitropic movement.
  • Hydrotropic movement : The root system of plants responds to the stimulus of water. This is called hydrotropic movement.
  • Chemotropic movement : Movement shown by plants in response to specific chemicals is called chemotropic movement. For example, the growth of the pollen tube towards the ovule.

All the above-mentioned movements of plants are related with growth, hence are called growth relevant movement.

Question 3.
Write short note on : Human Brain
Answer:

  • The brain is the main controlling part of the nervous system and is safely located in the cranial cavity.
  • The brain is covered by protective layers called the Meninges.
  • Cavities present in various parts of the brain are called ventricles.
  • The ventricles are filled with cerebrospinal fluid. This fluid supplies nutrients to the brain and also protects it from shock.
  • The brain of an adult human weighs about 1300 -1400 grams and consists of approximately 100 million neurons.
  • The left side of our brain controls the right side of our body and right side of our brain controls left side of the body.
  • In addition, the left side of the brain controls our speech and conversation, writing, logical thinking etc., whereas the right side controls artistic abilities.

Human brain is made up of following parts:

(a) Cerebrum:

  • This is the largest part of the brain and consists of two cerebral hemispheres. These hemispheres are joined with each other with the help of tough fibres and nerve tracts.
  • The cerebrum occupies two-thirds of the brain. Hence, it is also called as large brain.
  • Its surface has deep, irregular ridges and grooves which are called convolutions, convolution increases the surface area of the cerebrum and therefore a large number of nerve cells can be accommodated.

(b) Cerebellum:

  • This is the smaller part of the brain situated below the cerebrum at the back of the cranial cavity.
  • Its surface shows shallow grooves instead of deep convolutions.

(c) Medulla Oblongata:

  • This is the hind-most part of the brain.
  • There are two triangular swollen structures called pyramids on the upper side of medulla oblongata.
  • The medulla oblongata continues downwards the spinal cord.

(6) How are nerve impulses conducted in the body?
Answer:

  • Special types of cells which conduct impulses from one place to another in the body are called neurons. Nerve cells have the ability to generate and conduct electro-chemical impulses.
  • The cells that support the nerve cells and help in their functioning are called neuroglia.
  • All the information about our surroundings is collected by the ends or dendrites of the neuron.
  • The chemical process begins at those ends and electric impulses are generated which are conducted from the dendrites to the cell body, from the cell body to the axon and from the axon to its terminal.
  • These impulses are then to be transferred from this nerve cell to the next.
  • Now the impulse that reaches the terminal of an axon, stimulates the nerve cell to secrete certain chemicals.
  • These chemicals pass through a minute space, called the synapse, between two adjacent neurons and generate the impulses in the dendrites of the next neuron.
  • In this way, impulses are conducted in the body and these impulses are finally conveyed by nerve cells to muscles cells or glands.

9th Std Science Questions And Answers:

Current Electricity Class 9 Science Chapter 3 Questions And Answers Maharashtra Board

Class 9 Science Chapter 3

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 3 Current Electricity Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 3 Current Electricity Question Answer Maharashtra Board

Class 9 Science Chapter 3 Current Electricity Question Answer Maharashtra Board

1. The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 26
A. By which method are the appliances connected?
Answer:
Appliances are connected in parallel.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

B. What must be the potential difference across individual appliances?
Answer:
The potential difference across all appliances is same in parallel connection.

C. Will the current passing through each appliance be the same? Justify your answer.
Answer:
No, as every appliance has a different load (resistance), the current flowing through each appliance will be different.

D. Why are the domestic appliances connected in this way?
Answer:
The domestic appliances are connected in parallel as the potential difference remains same.

E. If the T.V. stops working, will the other appliances also stop working? Explain your answer.
Answer:
No, the other devices will not stop working as the current flowing through them is along different paths.

2. The following figure shows the symbols for components used in the accompanying electrical circuit.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 1
Which law can you prove with the help of the above circuit?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 2
(b) This circuit can be used to prove Ohm’s law.
(c) V = 1R is the expression of Ohm’s law

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

3. Umesh has two bulbs having resistances of 15 W and 30 W. He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.

A. Which method should he use to connect the bulbs?
B. What are the characteristics of this way of connecting the bulbs depending on the answer of A above?
C. What will be the effective resistance in the above circuit?

4. The following table shows current in Amperes and potential difference in Volts.

a. Find the average resistance.
b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
c. Which law will the graph prove? Explain the law.

5. Match the pairs

‘A’ Group – ‘B’ Group
1. Free electrons – a. V/ R
2. Current – b. Increases the resistance in the circuit
3. Resistivity – c. Weakly attached
4. Resistances in series – d. VA/LI

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

6. The resistance of a conductor of length x is r. If its area of crosssection is a, what is its resistivity? What is its unit?

7. Resistances R1, R2, R3 and R4 are connected as shown in the figure. S1 and S2 are two keys. Discuss the current flowing in the circuit in the following cases.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 3

a. Both S1 and S2 are closed.
b. Both S1 and S2 are open.
c. S1 is closed but S2 is open.
Answer:
(a) When both S1 and S2 are dosed, the effective resistance of the circuit decreases and hence, current will increase.
(b) When both S1 and S2 are open, the effective resistance of the rircuit increases and hence, current will decrease.
(c) When S2 is closed and S2 is open, the effective resistance of the tircuit decreases and hence current will increase. [Current will be more than case (b) but less than in case (a)]

8. Three resistances x1, x2 and x3 are connected in a circuit in different ways. x is the effective resistance. The properties observed for these different ways of connecting x1, x2 and x3 are given below. Write the way in which they are connected in each case. (I-current, V-potential difference, x-effective resistance)

a. Current I flows through x1, x2 and x3
b. x is larger than x1, x2 and x3
c. x is smaller than x1, x2 and x3
d. The potential difference across x1, x2and x3 is the same
e. x = x1 + x2 + x3
\(\text { f. } x=\frac{1}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}\)

9. Solve the following problems.

A. The resistance of a 1m long nichrome wire is 6Ω. If we reduce the length of the wire to 70 cm. what will its resistance be? (Answer : 4.2Ω)
Answer:
The resistance of 70cm wire will be 4.2 Ω

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

B. When two resistors are connected in series, their effective resistance is 80Ω. When they are connected in parallel, their effective resistance is 20Ω. What are the values of the two resistances? (Answer : 40Ω, 40Ω)
Answer:
The values of the two resistances R1 and R2 are 40Ω and 40Ω.

C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current? (Answer : 1.4 A)
Answer:
Given: Electric charge (Q) = 420 C
Time (t) = 5 min = 5 x 60
= 300 sec.
To find: Electric current (1) = ?
Formula:
\(I=\frac{Q}{t}\)
Solution:
\(I=\frac{Q}{t}\)
The current in the circuit is 1.4 A.

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Can you recall?

Question 1.
You must have seen a waterfall. Which way does the water flow?
Answer:
Water flows from a certain height of a mountain towards the ground.

Question 2.
Material: Copper and aluminium wires, glass rod, rubber.
Make connection as shown in figure 3.8. First connect a copper wire between points A and B and measure the current in the circuit. Then in place of the copper wire, connect the aluminium wire, glass rod, rubber, etc one at a time and measure the current each time. Compare the values of the current in different cases.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 4
Also take different metal strips (Iron, Copper, Zinc, and Aluminium) and connect it in slot AB. Now observe the difference in the resistance using Ohm meter.
Answer:
When copper and aluminium wires are connected to the circuit, current flows through it, as both are good conductors of electricity. When glass rod or rubber was connected to the circuit, current does not flow through it, as both are bad conductors of electricity.

Copper displays lowest resistance while the resistance increases with aluminium, zinc and iron respectively.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 3.
Set up the experiment as shown in figure. Then remove the clamp from the rubber tube.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 27

(a) What happens when the clamp is removed?
Answer:
When the clamp is removed, water flows from higher level to lower level.

(b) Does the water stop flowing? Why?
Answer:
Yes, the water stops flowing. This happens when the level of water becomes equal in both the bottles, i.e., there is no difference in the water levels.

(c) What will you do to keep the water flowing for a longer duration?
Answer:
The difference in the water level has to be maintained till that time. The difference must never be zero.

Question 4.
Point out the mistakes in the figure below:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 28
Answer:
A: Wire is broken at the negative terminal. Bulb will not glow as the circuit is incomplete.
B: Wire is disconnected at the negative terminal. Bulb will not glow as the circuit is incomplete.
C: The circuit is complete. Therefore, bulb will glow.
D: Rubber is a bad conductor of electricity. Hence, it will not allow current to flow and the bulb will not glow.

Question 5.
Why are the bulbs in Figures B, C and D not lighting up?
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 29
Answer:

  • In B, the blue wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
  • In C, the red wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
  • In D, both wires are connected to the same terminal. Hence, there is no potential difference and current does not flow. Therefore, bulb will not light up.

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
1mA = …………… A.
(a) 103
(b) 10-3
(c) 106
(d) 10-6
Answer:
(a) 103

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
To increase the effective resistance in a circuit the resistors are connected in ………….. .
(a) Series
(b) Parallel
(c) Both ways
(d) None of these
Answer:
(a) series

Question 3.
1 kilowatt hr = …………… joules.
(a) 4.6 x 106
(b) 3.6 x 106
(c) 30.6 x 106
(d) 3.6 x 1O5
Answer:
(b) 3.6 x 106

Question 4.
The voltage difference in India between the live and neutral wires is about ………….. .
(a) 110 V
(b) 220 V
(c) 440 V
(d) 60 V
Answer:
(b) 220 V

Question 5.
Resistivity is the specific property of a ………….. .
(a) Area of cross-section
(b) Temperature
(c) Length
(d) Material
Answer:
(d) material

Question 6.
If a P.D. of 12 V is applied across a 3Ω resistor then the current passing through it is ………….. .
(a) 36 A
(b) 4 A
(c) 0.25 A
(d) 15 A
Answer:
(b) 4 A.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 7.
In order to measure the electric current flowing through a circuit, we connect …………… with the circuit.
(a) a voltmeter in parallel
(b) a voltmeter in series
(c) an ammeter in parallel
(d) an ammeter in series
Answer:
(d) an ammeter in series

Question 8.
P and Q are two wires of same length and different cross-sectional areas and made of same material. Name the property which is same for both the wires.
(a) Resistivity
(b) Resistance
(c) Current
(d) Both (a) and (b)
Answer:
(a) Resistivity

Question 9.
The following is true for identical bulbs connected in parallel.
(a) All bulbs glow with unequal brightness.
(b) If one bulb is non-functional, all will stop working.
(c) All bulbs glow with equal brightness.
(d) Bulbs function for longer time.
Answer:
(c) All bulbs glow with equal brightness

Question 10.
The …………… wire is either yellow or green in colour.
(a) Live
(b) Neutral
(c) Earth
(d) Fuse
Answer:
(c) earth

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 11.
A current flows through a circuit due to the difference in …………… between two points in the conductor.
(a) Gravity
(b) Potential
(c) Resistance
(d) Fuse
Answer:
(b) potential

Question 12.
…………… is the amount of charge flowing through a particular cross sectional area in unit time.
(a) Electric current
(b) Ampere
(c) Volt
(d) Force
Answer:
(a) Electric current

Question 13.
The flow of …………… constitutes the electric current in a wire.
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) electrons

Question 14.
The conventional direction of flow of current is from …………… terminal to …………… terminal.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(c) positive, negative

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 15.
Current stops flowing when potential difference between two ends of a wire becomes ………….. .
(a) Zero
(b) Positive
(c) Negative
(d) Higher
Answer:
(a) zero

Question 16.
Resistances are connected in …………… so as to pass the same current through them.
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(a) series

Question 17.
To decrease the effective resistance in a circuit, the resistances are connected in ………….. .
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(b) parallel

Question 18.
1μV = …………… V
(a) 102
(b) 10-6
(c) 106
(d) 103
Answer:
(b) 10-6

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 19.
Good conductors contain a large number of ………….. .
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) free electrons

Question 20.
Electrons flow from …………… terminal to …………… terminal in a conductor when a potential difference is applied.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(a) negative, positive

Find the odd one out:

Question 1.
Voltmeter, Ammeter, Galvanometer, Thermometer
Answer:
Thermometer

Question 2.
Rubber, Silver, Copper, Gold
Answer:
Rubber

Question 3.
Wood, Glass, Steel, Rubber
Answer:
Steel

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
Graphite, Diamond, Fullerenes, Coal
Answer:
Fullerenes

Distinguish between:

Question 1.
Voltmeter and Ammeter
Answer:

Voltmeter Ammeter
(i) It is an instrument used to measure the potential difference between two terminals of a cell. (i) It is an instrument to measure the electric current flowing through a circuit.
(ii) It is connected in parallel with the cell. (ii) It is connected in series with the cell.
(iii) It has a very high resistance. (iii) It has a very low resistance.
(iv) Voltmeter has range of volts. (iv) Ammeter has range of amps.

Question 2.
Ohmic conductors and Non-Ohmic conductors
Answer:

Conductors Insulators
(i) Substances which have very low electrical resistances are called conductors. (i) Substances which have extremely high electrical resistances are called Insulators.
(ii) They contain a large number of free electrons. (ii) They contain practically no free electrons.
(iii) Conductors are mostly metals. (iii) Insulators are mostly non metals.
(iv) Conductor example iron, copper. (iv) Insulator example rubber, plastic.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 3.
Conductors and Insulators
Answer:

Resistance Resistivity
(i) The hindrance to the flow of electrons is called resistance. (i) Resistivity is the specific property of the material of a conductor.
(ii) The S.I. unit of resistance is ohm (Q). (ii) The S.I. unit of resistivity is ohm-metre (Q – m).
(iii) It depends on temperature, area of cross-section, length of conductor and material of the conductor. (iii) It depends on material of the conductor.
(iv) Resistance can be changed as it depends of external factor as well. (iv) Resistivity cannot be changed as it depends of internal factors.

Question 4.
Resistance in Series and Resistance in Parallel
Answer:

Resistance in Series Resistance in Parallel
(i) Effective resistance of the resistors is equal to the sum of their individual resistances. (i) Inverse of the effective resistance is equal to the sum of the inverse of individual resistances.
(ii) The same current flows through each resistor. (ii) The total current flowing through the circuit is the sum of the currents flowing through individual resistors.
(iii) The effective resistance is larger than each of the individual resistances. (iii) The effective resistance of resistors connected in parallel is less than the least resistance of individual resistors.
(iv) This arrangement is used to increase the resistance in a circuit. (iv) This arrangement is used to decrease the resistance in a circuit.

Question 5.
Answer:

Electric current Potential difference
(i) The flow of electric charge per unit time is called electric current. (i) The difference in potential between the positive and negative terminal of a cell is the potential difference of that cell.
(ii) The S.I. unit of electric current is ampere. (ii) The S.I. unit of potential difference volt.
(iii) Ammeter is used to measure electric current. (iii) Voltmeter is used to measure electric current.
(iv) Current is represented by: \(\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}\) (iv) Potential difference is represented by: \(\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}\)

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Make pair:

Question 1.
Copper : Conductor :: Rubber : ……………….
Answer:
Insulator

Question 2.
Aluminium : ………………. :: Indium oxide : Super Insulator
Answer:
Super conductor

Question 3.
Parallel Connection : \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\) :: Series Connection : ……………….
Answer:
Rs = R1 + R2

(4) Electric Current : ………………. :: Electric charge : Coulomb
Answer:
Ampere

(5) Electric resistance : Ohm :: Potential difference : ……………….
Answer:
Volt

State whether the following statements are true or false. Correct the false statements:

(1) The SI unit of charge is volt.
(2) Voltmeter is always connected in series with the device.
(3) The conventional direction of flow of current is from positive terminal to negative terminal.
(4) Silver and copper are good conductors.
(5) Resistivity of pure metals is more than alloys.
(6) Resistance in series arrangement is used to decrease resistance of circuit.
(7) A conducting wire offers less resistance to flow of electrons.
(8) Charges are measured in ampere.
(9) The unit of potential difference is ampere.
(10) Resistance of a conductor is inversely proportional to the length of the conductor.
(11) Ammeter is connected in parallel to the cell to measure current.
(12) Fuse is made of wire having high melting point.
Answer:
(1) False. The SI unit of charge is coulomb.
(2) False. Voltmeter is ahvays connected in parallel with the device.
(3) True
(4) True
(5) False. Resistivity of pure metals is less than alloys.
(6) False. Resistance in series arrangement is used to increase resistance of circuit.
(7) True
(8) False. Charges are measured in coulomb.
(9) False. The unit of potential difference is volt.
(10) False. Resistance of a conductor is directly proportional to the length of the conductor.
(11) False. Ammeter is connected in series to the cell to measure current.
(12) False. Fuse is made of wire having low melting point.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Answer the following in one sentence:

Question 1.
Which is the unit used to measure large voltages?
Answer:
Kilovolts and Megavolts are the units used to measure large voltages.

Question 2.
What is the SI unit of potential difference?
Answer:
The SI unit of potential difference is volt (V).

Question 3.
What is lightning?
Answer:
Lightning is the electric discharge travelling from clouds at high potential to earth’s surface which is at zero potential.

Question 4.
What is the unit of resistivity.
Answer:
The unit of resistivity is ohm metre (Qm).

Question 5.
Which substances are called conductors of electricity?
Answer:
Those substances which have very low electrical resistance are called conductors of electricity.

Question 6.
What is Earth wire?
Answer:
Earth wire is generally yellow or green colour, it is connected to a metal plate buried deep underground near the house and is for safety purpose.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Write formula:
(1) E1ectriccurrent \(=\frac{Q}{t}\)
(2) Electric charge = It
(3) Potential difference = IR
(4) Electric resistance \(=\frac{V}{I}\)
(5) Current \(=\frac{V}{R}\)
(6) Resistivity \(=\frac{RA}{L}\)

Give scientific reasons:

Question 1.
Free electrons are required for conduction of electricity.
Answer:

  • Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to the nucleus.
  • These are called free electrons. These electrons can easily move from one part of a conductor to its other parts. The negative charge of the electrons also gets transferred as a result of this motion.
  • The free electrons in a conductor are the carriers of negative charge. Hence, free electrons are required for conduction of electricity.

Question 2.
Wood and glass are good insulators.
Answer:

  • Those substances which have infinitely high electrical resistance are called insulators.
  • Wood and glass have high resistance and negligible free electrons for conduction of electricity.
  • Hence, wood and glass are good insulators.

Question 3.
Connecting wires in a circuit are made of copper and aluminium.
Answer:

  • Copper and aluminum are good conductors of electricity.
  • They have low electrical resistance and large number of free electrons.
  • As they are malleable and ductile, they can be drawn into thin wires. Hence, connecting wires in a circuit are made of copper or aluminum.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
A thick wire has a low resistance.
Answer:

  • The resistance (R) of a wire is inversely proportional to the cross-sectional area (A) of a wire. i.e., R i
  • Thus, greater is the cross-sectional area of a conductor (wire), lower is its resistance. Hence, a thick wire has a low resistance.

Question 5.
A series combination of resistances is used to increase the resistance of a circuit.
Answer:

  • When resistances are connected in series, the effective resistance of the resistors is equal to the sum of their individual resistances. Rs = R1 + R2 ………….. Rn
  • The effective resistance is larger than each of the individual resistances. Hence, This arrangement is used to increase the resistance in a circuit.

Question 6.
A parallel combination of resistances decreases the effective resistance of the circuit.
Answer:

  • In a parallel combination, the inverse of the effective resistance is equal to the sum of the inverses of individual resistances. \(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \ldots \ldots \cdot \frac{1}{\mathrm{R}_{\mathrm{n}}}\)
  • The effective resistance of resistors connected in parallel is less than the individual resistors.
  • Due to this, any addition of an individual resistance in parallel combination will decrease the overall resistance of the circuit. Hence, a parallel combination of resistance decreases the effective resistance of the circuit.

Question 7.
Lightning occurs from sky to earth.
Answer:

  • Lightning is the electric discharge travelling from clouds at high potential to the earth’s surface, which is at zero potential.
  • The earth is always at lower potential as compared to the clouds.
  • Hence, lightning occurs from sky to earth.

Question 8.
In streetlights, bulbs are connected in parallel.
Answer:

  1. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows through the other paths, and the rest of the bulbs light up.
  2. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually. Hence, streetlights are connected in parallel.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Numerical:
Numericals based on the formula: (1) Q= It (2) W= VQ

Question 1.
A current of 0.4 A flows through a conductor for 5 minutes. How much charge would have passed through the conductor?
Answer:
Given: Current (I) = 0.4 A
Time (t) = 5 min = 5 x 60 = 300 s
To find: Charge (Q) =?
Formula: Q = 1 x
Solution: Q = 0.4 x 300
Q= 120 C.
Charge passing through the conductor is 120

Question 2.
Find the amount of work done if 3 C of charge is moved through a potential difference of 9 V.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 5
The work done is 27 joule.

Question 3.
The resistance of the filament of a bulb is 1000Ω. It is drawing a current from a source of 230 V. How much current is flowing through it?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 6
The current flowing through the filament of bulb is 0.23 A.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 7
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 8
The resistivity of the wire is 4.71 x 10-5 Qm.

Question 5.
A current of 0.24 A flows through a conductor when a potential difference of 24 V is applied between its two ends. What is its resistance?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 9
The resistance of a conductor is 100Ω.

Question 6.
If three resistors 15Ω, 3Ω and 4Ω each are connected in series, what is the effective resistance in the circuit?
Answer:
Given:
R1 =15Ω
R2 = 3Ω
R3 = 4Ω
Effective resistance in series (Rs) = ?
Rs = R1 + R2 + R3
Rs = 15 + 3 + 4
Rs = 22Ω
The effective resistance in the circuit is 22Ω.

Question 7.
Three resistances 15Ω, 20Ω and 10Ω are connected in parallel. Find the effective resistance of the circuit.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 10
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 11
The effective resistance of the circuit is 4.615 Ω. It is less than the least of the three i.e., 10Ω.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Write a note on the following:

Question 1.
Electric current
Answer:
An electric current is the flow of electrons through a conductor. Quantitatively, current
(I) is defined as the charge passing through a conductor in unit time.
\(T=\frac{Q}{t}\)

Question 2.
1 ampere
Answer:
One ampere current is said to flow in a conductor if one coulomb charge flows through it every second.
\(1 \mathrm{~A}=\frac{1 \mathrm{C}}{1 \mathrm{~s}}\)

Question 3.
1 volt
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
\(1 \mathrm{~V}=\frac{1 \mathrm{~J}}{1 \mathrm{C}}\)

Question 4.
Potential Difference
Answer:
The amount of work done to carry a unit positive charge from point A to point B is called the electric potential difference between the two points.
\(V=\frac{W}{Q}\)

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 5.
Conductor
Answer:
Those substances which have very low resistance are called conductors. Current can flow easily through such materials.

Question 6.
Insulators
Answer:
Those substances which have extremely high resistance and through which current cannot flow are called insulators.

Question 7.
1 ohm
Answer:
If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.
\(\frac{1 \text { Volt }}{1 \text { Ampere }}=1 \mathrm{Ohm}\)

Question 8.
Potential
Answer:
The level of electric charge present is known as potential.

Question 9.
Ohm’s Law
Answer:
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.
V = IR

Question 10.
Superconductors
Answer:
The resistance of some conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Such conductors are called superconductors.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 11.
Non-ohmic conductors
Answer:
Conductors which do not obey Ohm’s law are called non-ohmic conductors.

Complete the flow charts:

(1) Protection from Electricity
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 12

(2) Resistance
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 13

(3) Resistivity
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 14

Write properties/characteristics/advantages of the following:

Superconductors
Answer:
The resistance of these conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Aluminium is an example of Super Conductor. Superconductors can be used in space missions to increase/ boost the signal strength. They are also used i in the data fibres to increase the speed of data transfer.

Give explanations of the given statements:

Question 1.
Safety precautions are to be taken while using electricity.
Answer:

  • Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
  • Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
  • One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • As rubber is an insulator, it prevents the current from flowing’ through our body, thereby protecting it.
  • If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
  • If not, then you should use a wooden pole to push the person away from the electric wire.

Question 2.
In a domestic circuit colour code is followed while setting up electrical wiring.
Answer:

  • The electricity in our homes is brought through the main conducting cable either from the electric pole or from underground cables.
  • Usually, there are three wires in the cable.
    (a) Live wire which brings in the current. It has a red or brown insulation.
    (b) Neutral wire through which the current returns. It is blue or black.
    (c) Earth wire is of yellow or green colour. This is connected to a metal plate buried deep underground near the house and is for safety purposes.
  • In India, the voltage difference between the live and neutral wires is about 220 V.
  • Live and neutral wires are connected to the electric meter through a fuse.
  • They are connected through a main switch, to all the conducting wires inside the home so as to provide electricity to every room.
  • In each separate circuit, various electrical appliances are connected between the live and neutral wires.
  • The different appliances are connected in parallel and the potential difference across every appliance is the same.

Question 3.
Fuse used in electrical circuit can save electrical objects from damage.
Answer:

  • Fuse wire is used to protect domestic appliances.
  • It is made of a mixture of substances and has a specific melting point.
  • It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • The circuit gets broken and the flow of current stops, thus protecting the appliance.
  • This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1 A, 2 A, 3 A, 4 A, 5 A and 10 A are used.

Question 4.
Bulbs arranged in parallel glow brighter than bulbs arranged in series.
Answer:

  • The amount of light given out by bulbs in parallel combination will be more than that in series combination.
  • In parallel combination the resistance of the overall circuit decreases whereas in series it increases, so the current flowing through the bulbs in parallel circuit is more.
  • Due to this, intensity of light given out by bulbs in parallel combination is more than the bulbs in series combination.

Complete the following table:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 15

Solve the numerical:

Question 1.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 16
The resistivity of the wire is 4.71 x 10-5 Ωm.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
Determine the current that will flow when a potential difference of 33 V is applied between two ends of an appliance having a resistance of 110 Ω. If the same current is to flow through an appliance having a resistance of 500 Ω, how much potential difference should be applied across its two ends?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 17
The current is 0.3 A and potential difference to be applied is 150 V.

Question 3.
Determine the resistance of a copper wire having a length of 1 km and diameter of 0.5 mm.
Answer:
Given: Resistivity of copper (p)
= 1.7 x 10-8 Ω m
Converting all measures into metres.
Length of wire (L) = 1 km
= 1000 m = 103 m
Diameter of wire (d) = 0.5 mm
= 0.5 x 10-3m
To find: Resistance of wire (R) = ?
Formula:
\(R=\rho \frac{L}{A}\)
Solution:
If d is the diameter of the wire then, its area of cross-section
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 18
The resistance of a copper wire is 85Ω and area of cross section is 0.2 x 10-6 m2

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 5.
Two resistors having resistance of 16 and 14 are connected in series. If a potential difference of 18 V Is applied across them, calculate the current flowing through the circuit and the potential difference across each individual resistor.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 19
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 20
The current in the circuit is 0.6 A and potential across 16 Ω retor is 9.6 volt and 14 Ω resistor is 8.4 voIt.

Question 6.
If the resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel to battery of 12 V, find the effective resistance in the circuit. Calculate the total current and current in each resistor.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 21
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 22
(a) The total current is 4 A and current in each resistor is 2.4 A, 1.2 A and 0.4 A respectively.
(b) The effective resistance in the circuit is 3Ω.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Complete the diagram and answer the questions:

Question 1.
The following figure shows the symbols for components used in the accompanying electrical circuit.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 23
(a) Place them at proper places and complete the circuit.
(b) Which law can you prove with the help of the above circuit?
(c) State expression for Ohm’s law
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 24
(b) This circuit can be used to prove Ohm’s law.
(c) V = IR is the expression of Ohm’s law

Question 3.
Explain with the help of a diagram, what are free electrons and how they move through the conductor?
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 25
Answer:

  • Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to nucleus.
  • These are called free electrons.
  • These electrons can easily move from one part of a conductor to its other parts.

Complete the paragraph:

Question 1.
If resistors are connected in series,
Answer:
The same current flows through each resistor. The effective resistance of the resistors is equal to the sum of their individual resistances. The potential difference between the two extremes of the arrangement is equal to the sum of the potential differences across individual resistors. The effective resistance is larger than each of the individual resistances. This arrangement is used to increase the resistance in a circuit. This type of connection is used in electrical heating equipment like geysers, iron, and hair dryers.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
If a number of resistors are connected in parallel,
Answer:
The inverse of the effective resistance is equal to the sum of the inverses of individual resistances. The current flowing through an individual resistor is proportional to the r inverse of its resistance and the total current flowing through the circuit is the sum of the currents flowing through individual resistors. The potential difference across all r resistors is the same. The effective resistance of resistors connected in parallel is less than r the least resistance of individual resistors.

This arrangement is used to reduce the resistance in a circuit. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows, through the other paths, and the rest of the bulbs light up. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually.

Read the paragraph and answer the questions.

Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket. Before cleaning an electrical appliance it should be switched off and its plug removed from the socket. One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. As rubber is an insulator, it prevents | the current from flowing through our body, thereby protecting it. If a person gets an electric shock, you should not touch that person. You should switch off the main switch and if the switch is too far or you do not know where it is located, then you should remove the plug from the socket if possible. If not, then you should use a wooden pole to push the person away from the electric wire.

(i) Why should the electrical sockets be fitted at a certain height?
Answer:
Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside.

(ii) Why plug wires should not be pulled out while removing any electrical device?
Answer:
Plug wires should not be pulled out while removing any electrical device as it may cause the wire to break causing short circuit which can lead to fire or death.

(iii) Why should a person wear footwear with rubber soles while handling electrical appliances. .
Answer:
As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it. Hence a person should wear footwear with rubber soles while handling electrical appliances.

(iv) Saee is touching an electrical button socket with wet hands what will you advise her and why?
Answer:
We will advise her to dry her hands before touching any electrical sockets or devices as water on the hands can cause an electrical short circuit producing shock to the person touching it.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

(v) Sneha is getting an electrical shock what will you do the save her life?
Answer:
We should switch off the main switch and if the switch is too far or we do not know where it is located, then we should remove the plug from the socket if possible. If not, then we should use a wooden pole to push the person away from the electric wire.

(vi) Give a title to the above passage.
Answer:
Precautions to be taken while using electricity

Answer the questions in details:

Question 1.
Find the expression (i.e., derive the expression) for the resistors connected in series.
Answer:
Expression for the resistance connected in series:
(i) Let R1, R2 and R3 be three resistances connected in series between C and D.
(ii) Let Rs be the effective resistance in circuit and V1, V2 and V3 be the potential difference across R1, R2 and R3 respectively.
(iii) Let the potential difference across CD be V.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 30
(iv) In series combination.
v = v1 + v2 + v3 ……………………(i)
By using Ohm’s law
V = IRs
∴ V1= IR1, V2 = IR2 and V3 = 1R3
Substituting these values in equation (j) we get
IRs = IR1 + IR2 + IR3
∴ Rs = R1 + R2 + R3
For ‘n’ number of resistors conneded in series we get
Rs = R1 + R2 + R3 + …………………. + Rn

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
Find the expression (i.e., derive the expression) for the resistors connected in parallel.
Answer:
Expression for the resistance connected in parallel.
(i) Let R1, R2 and R3 be the three resistances connected in parallel combination between points C and D and let R be their effective resistance.
(ii) Let I1, I2 and I3 be the currents flowing through resistances R1, R2 and R3 respectively. Let I be the current flowing through the circuit and V be the potential difference of the cell.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 31
(iii) For parallel combination of resistances,
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 32
(iv) Substituting the values of (I, I1, I2 and I3) in equation (i) we get
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 33

Question 3.
Find the expression for resistivity of a material.
Answer:
(i) At a given temperature, the resistance (R) of a conductor depends on its length (L), area of cross-section (A) and the material it is made of. If the resistance of a conductor is R, then
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 34
(ii) p is the constant of proportionality and is called the resistivity of the material.
(iii) The unit of resistivity in SI units is Ohm metre (Ω m).
(iv) Resistivity is a specific property of a material and different materials have different resistivity. ’

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Make the concept diagram and explain:

Question 1.
Make the concept diagram of an electrical circuit and explain the working of a fuse.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 35
Answer:

  • Fuse wire is used to protect domestic appliances.
  • It is made of a mixture of substances and has a specific melting point.
  • It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. The circuit gets broken and the flow of current stops, thus protecting the appliance.
  • This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1A, 2A, 3A, 4A, 5A, and lO Aareused.

(2) Show motion of electrons in an circuit and explain precautions while using an electrical device.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 37

  • Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
  • Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
  • One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
  • If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
  • 1f not, then you should use a wooden pole to push the person away from the electric wire.

Q.4.4.Complete the incomplete figure and give an explanation:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 38
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 39
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.

I α V
I = kV (k = constant of proportionality)
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 40

This is known as Ohm’s law.
We can obtain the SI unit of resistance from the above formula, Potential difference and current are measured in Volts and Amperes respectively. The unit o resistance is called Ohm. It is indicated by the symbol Ω.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 41
The resistance of one Ohm : If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.

9th Std Science Questions And Answers:

Introduction to Biotechnology Class 9 Science Chapter 17 Questions And Answers Maharashtra Board

Class 9 Science Chapter 17

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 17 Introduction to Biotechnology Question Answer Maharashtra Board

Class 9 Science Chapter 17 Introduction to Biotechnology Question Answer Maharashtra Board

1. Each of the following statements is wrong. Rewrite them correctly by changing either one or two words.

a. Simple squamous epithelium is present in the respiratory tract.
b. Glandular epithelium is present in kidneys.
c. Chlorenchyma helps the plant to float in water.
d. Striated muscles are also called involuntary muscles.
e. Chloroplast is present in permanent tissue.
Answer:
a. False. Ciliated columnar epithelium is present in respiratory tract.
b. False. Cuboidal epithelium is present in kidneys.
c. False. Aerenchyma helps the plant to float in water.
d. False. Striated muscles are also called voluntary muscles.
e. False. Chlorenchyma is present in permanent tissue.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

2. Identify the odd word and explain why it is odd.

a. Xylem, phloem, permanent tissue, meristematic tissue.
Answer:
Meristematic tissue. It has the ability to divide whereas the others have lost the ability to divide.

b. Epithelium, Muscle fibre, nerve fibre, the epidermis.
Answer:
Epidermis. It is a plant tissue whereas the others are animal tissues.

c. Cartilage, bone, tendon, cardiac muscle.
Answer:
Cardiac muscle. It is a muscular tissue whereas the others are connective tissues.

3. Write the names of the following tissues.

a. Tissue lining inner surface of mouth.
Answer:
Squamous epithelium.

b. Tissue joining muscles and bones.
Answer:
Tendon

c. Tissue responsible for increasing height of plants.
Answer:
Apical meristem.

d. Tissue responsible for increasing girth of stem.
Answer:
Lateral meristem.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

4. Write the differences.
Simple tissue and complex tissues in plants.
Answer:

Simple Tissue in plants Complex Tissues in plants
(i) They are made up of only one type of cells. (i) They are made up of more than one type of cells.
(ii) They are found in all parts of the plant. (ii) They are found in the vascular regions of the plant.
(iii) They perform different functions like storage (iii) They mainly perform the function of
of food, support, giving strength etc. conduction of water and food.
(iv) Examples – Parenchyma, Collenchyma, Sclerenchyma. (iv) Example – Xylem and phloem.

5. Write short notes.

a. Meristematic tissue.
Answer:

  • Meristematic tissue is present in specific parts of a plant where growth takes place.
  • Cells of meristematic tissue contain thick cytoplasm, a conspicuous nucleus and a thin cell wall and are compactly packed together.
  • Vacuoles are usually absent in these cells and they are highly active.
  • The main function of meristematic tissue is to bring about plant growth.
  • According to the location, meristematic tissues are of three types: Apical meristem, intercalary meristem and lateral meristem.

b. Xylem
Answer:

  • Xylem is a complex permanent tissue in plants.
  • It consists of thick-walled dead cells.
  • The type of cells in xylem are trachieds, vessels, xylem fibres (dead cells) and xylem parenchyma (living cells).
  • Its structure is like interconnected tubes conduct water and minerals only in upward direction.

c. Striated muscles.
Answer:

  • The cells of striated muscles are long, cylindrical, multinucleate and have no branches.
  • These are alternate dark and light bands on these muscles.
  • As they are attached to bones, they are also called skeletal muscles.
  • They move as per our will, hence they are called voluntary muscles.
  • Striated muscles bring about movements of arms and legs, running, speaking etc.

d. Agro-complementary business.
Answer:
(i) The business that are complementary to agriculture and generate supplementary income for the farmers are called agro complementary business.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

(ii) These include:

  • Animal Husbandry: It is practiced for milk production and for using the cattle as help in farming operations, e.g. cows and buffaloes are raised for milk whereas bulls and male buffaloes for pulling heavy loads.
  • Poultry farming: It is the rearing of egg and meat yielding chickens. Chickens raised for eggs are called layers while those raised for meat are called broilers.
  • Sericulture: It is the rearing of silkworms (moths) for silk production. The silk fibres obtained are processed, reeled and then woven into fabric.

e. Genetic engineering
Answer:

  • Genetic engineering is the deliberate modification of the characteristics of an organism by manipulating its genetic material.
  • An organism that is generated through genetic engineering is called a genetically modified organism (GMO).
  • Genetic engineering is applied in many fields like research, agriculture, industrial biotechnology and medicine.
  • In agriculture, genetic engineering is used in the production of cash crops, improvement in varieties of cash crops, increase in ability of plants to withstand environmental stresses.
  • In medicine, genetic engineering is used for vaccine production, early diagnosis of congenital disease, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

f. Sericulture
Answer:

  • Sericulture is the rearing of silkworms (moths) for production of silk.
  • Bombyx mori is the most commonly used variety of silkworm for this purpose.
  • The life cycle of silkworm consist of four stages namely egg, larva, pupa and adult.
  • Thousands of eggs deposited by female moths are incubated artificially to shorten the incubation period.
  • Larvae hatching out of eggs are released on mulberry plants.
  • Larvae are nourished by feeding on mulberry leaves.
  • After feeding for 3-4 days, larvae move to branches of mulberry plant.
  • The silk thread is formed from the secretion of their salivary glands.
  • Larvae spin this thread around themselves to form a cocoon. The cocoon may be spherical in nature.
  • Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
  • Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
  • These fibres are unwound, processed and reeled. Various kinds of fabric is woven from silk threads.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

6. Explain the meaning of biotechnology and its impact on agricultural management with suitable examples.
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called biotechnology. Impact of biotechnology on agricultural management:

  • Genetically modified crops (GM crops) are being produced by introducing changes in DNA of natural crops.
  • Normally such varieties are not found in nature.
  • Thus, new varieties are produced artificially Different useful characters are introduced in such varieties.
  • Some naturally occurring varieties cannot withstand environmental stress like frequently changing temperature.
  • Wet and dry famines, changing climates etc. However, GM crops can grow in any of such adverse conditions.
  • As GM crops are resistant to insect pests, pathogens, chemical weedicides, etc. the use of harmful chemicals like pesticides can be avoided.
  • Due to use of seeds of GM crops, there is improvement in nutritive value and decrease in loss of crops.

Draw neat and labelled diagrams of the following:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 2
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 3

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

7. Which two main techniques are used in biotechnology? Why?
Answer:

  • The two main techniques used in biotechnology are – Genetic engineering and tissue culture.
  • These techniques are used to bring about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings.
  • Its uses are in the production of cash crops, improvement in varieties of cash crops, early diagnosis of congenital diseases, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

8. Discuss ‘Agritourism’ in the class and write a project on an agrotourism centre nearby. Present it in the class in groups.
Answer:

  • In agritourism, plantlets of flowering, medicinal, ornamental, vegetable plants and fruit trees are produced on a large scale by tissue culture technique.
  • By growing some of the plants fully, an agritourism centre can be developed.
  • If sufficient land is available, the emerging field of agritoursim would be a good business.
  • An agritourism centre consists of following:
    (a) Mango, chikoo (sapota), guava, coconut, custard apple and some other regional fruit trees.
    (b) Shade giving local or exotic attractive plants.
    (c) Ornamental and flowering plants.
    (d) Butterfly garden.
    (e) Medicinal plant garden.
    (f) Organic vegetables and fruits.
  • People visit places with such attraction in large numbers.
  • Selling plantlets/seedlings, fruits, vegetables at such places can be quite profitable.

9. Define the term tissue and explain the concept of tissue culture.
Answer:
A group of cells having the same origin, same structure and same function is called tissue. Concept of tissue culture:

  • Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium is called tissue culture.
  • Nowadays, a complete organism can be developed from a single cell or from tissue with the help of the tissue culture technique. Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology
  • A liquid, solid or gel-like medium prepared from agar, which supplies nutrients and energy necessary for tissue culture is used in this technique.
  • Tissue culture can be used to grow plants on a large scale, which bear flowers, fruits of excellent quality, in shorter durations and are disease free.
  • The various processes involved in tissue culture are:

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 7

10. ‘Rearing of sheep is a livestock’. Justify this statement.
Answer:

  • The term livestock refers to animals reared for profit or for use.
  • Sheep provides us with wool, skin, meat and milk.
  • Therefore, rearing of sheep is a livestock.

Class 9 Science Chapter 17 Introduction to Biotechnology Intext Questions and Answers

Question 1.
Which components bring about important processes in the living organisms?
Answer:
Cells, tissues, organs, organ systems bring out important processes in living organisms.

Question 2.
Which is the smallest structural and functional unit of the body of living organisms?
Answer:
Cell is the smallest structural and functional unit of the body of living organisms.

Question 3.
Which type of muscle is the diaphragm of the respiratory system?
Answer:
Diaphragm is a skeletal muscle. It is an involuntary muscle that regulates breathing, although some voluntary control can be achieved.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 4.
What is the main difference between the growth of animals and plants?
Answer:

  • Growth in animals is uniform whereas the growth in plants occurs in specific parts of the body.
  • Growth in animals occurs for a certain period of life whereas the growth in plants occur throughout their life.

Question 5.
Why does the growth of a plant occur only at specific parts of the plant body?
Answer:
Growth of a plant occurs only at specific parts of the body due to the presence of meristematic tissues which contain dividing cells.

Question 6.
Suppose you want to grow a garden like the one shown in the picture, around your home or school. What would you do to achieve that? By which methods will you cultivate the seedlings?
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 4
Answer:

  • To grow a garden around the school or home, one must cultivate seedlings of different plants and those seedlings must be watered regularly and nurtured properly.
  • Seedlings can be cultivated by the following methods:
    (a) Seed sowing: Seeds are sown, watered and allowed to germinate.
    (b) Transplantation: Seeds are sown somewhere else and then the seedlings are transplanted in the garden.

Question 7.
You must have seen flowers of same variety but of 2 or 3 different colours borne by same plant. How is this possible?
Answer:

  1. The different coloured flowers borne by the same plant are due to the pigments like anthocyanins, carotenoids etc. according to the genetic makeup of the plant.
  2. Also, by using the latest techniques of biotechnology and tissue culture, it is possible to manipulate the genes for flower colour and get the desired flower colour.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 8.
What keeps the various organs and organ systems separate from each other? Why?
Answer:
The epithelial tissue keeps the various organs and organ systems separate from each other by forming a layer on the organs.

Question 9.
Why are epithelial tissues said to be simple tissues?
Answer:
Epithelial tissues are said to be simple tissues as they are made up of only one type of cells.

Question 10.
Why do slim persons feel more cold in winter than those who are obese?
Answer:

  • Below the skin, there is a tissue called adipose tissue.
  • The cells of these tissue are filled with fat droplets.
  • Due to storage of fat, it acts as an insulator and helps to retain heat in the body.
  • Obese people have more fat deposited in the adipose tissue as compared to slim persons. Therefore, slim persons feel more cold in winter than those who are obese.

Question 11.
Why can bones not be folded?
Answer:

  • Bone cells called osteocytes are embedded in solid ground substance made up of calcium phosphate.
  • This makes the bone hard, rigid and non- flexible. Therefore, bones cannot be folded.

Question 12.
Which other industries can be started as an extension of the plant nursery business?
Answer:
Businesses like Agritourism, Ecotourism, forest resorts and organic fruit gardens, yoga and meditation centres can be started as an extension of the plant nursery business.

Question 13.
To which places do people choose to go on vacation in order to relax when they are tired of crowds and stressful life?
Answer:
People choose to go to hill stations, beaches, forest resorts and places where there is lot of greenery to relax when they are tired of crowds and stressful life.

Question 14.
What is the inter-relationship between the two questions (7 and 8) above?
Answer:
The inter-relationship between the above two questions is that businesses like agritourism, ecotourism, forest resorts etc. can fetch a huge profit as people look for these kind of places to relax, to be away from the hustle and bustle of city life and feel close to nature.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 15.
Why are the cocoons transferred to boiling water before the pupa develops into an adult?
Answer:

  • Once the pupa develops into an adult, it will secrete a fluid to dissolve the silk and emerge out.
  • Hence the cocoons are transferred to boiling water before the pupa develops in to an adult to kill the silkworms before transformation is complete.

Question 16.
Why we cannot see our organs like the heart, blood vessels and intestines?
Answer:
We cannot see our organs like the heart, blood vessels and intestines as they are located inside the body.

Question 17.
What is meant by white revolution? Who was its pioneer? What benefits did it bring?
Answer:

  • White revolution was the programme launched by the National Dairy Development Board to increase the milk production in India.
  • Dr. Verghese Kurien was the pioneer of white revolution.
  • It transformed India from a milk-deficient nation to the largest producer of milk. It also helped the dairy farmers in directing their own development and empowering them.

It also helped to reduce the malpractices carried out by milk traders and merchants.

Answer the following questions:

Question 1.
Collect information about GM varieties of crops in your area and make a note of them. Also find out if there are adverse effects of GM crops on human beings and environment.
Answer:

  • Some of the GM crops are:
    (a) Maize: MON 810, MON 863
    (b) Rice: Golden rice
    (c) Brinjal: BT brinjal
    (d) Cotton: BT cotton
  • Effects of GM Crops on human beings: GM crops may cause toxicity and allergic reactions in humAnswer:
  • Effects of GM Crops on environment:
    (a) GM crops may be toxic to pollinators and non-target species like butterflies.
    (b) Many GM crops may be a threat to soil ecosystem as they secrete their toxins into the soil.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
As shown in the figure, place an onion on each gas jar in such a way that its base (roots) will remain dipped in water. Measure and record the length of the roots of both onions on the first, second and third day. On the fourth day, cut off 1 cm of the roots of the onion in flask B Measure the length of the roots of both onions for the next five days and record your observations in the following table.
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 6

Length (cm) Day 1 Day 2 Day 3 Day 4 Day 5
Flask A
Flask B

Answer:
(Students are expected to record their observation in the given table).

(a) Which onion has longer roots? Why?
Answer:
The onion in jar Ahas longer roots as it continues to grow due to the presence of meristematic tissue at root tip which had dividing cells.

(b) Why did the roots of the onion in jar B stop growing?
Answer:
The roots of onion in jar B stopped growing as the meristematic tissue present in the root tips were cut off.

Question 3.
Take a fresh and fleshy leaf of Rhoeo, lily or any other plant. Pull and press it, tearing it obliquely in such a way that its transparent epidermis will be visible at the cut margins. Take the transparent epidermis with the forceps and keep it in dilute safranin solution for 1 minute. Spread it on a slide, cover it with a cover-slip and observe it under a compound microscope.
Answer:

  • The cells of the epidermis are flat and polygonal and there are no intercellular spaces between them.
  • A single continuous layer is formed.
  • The epidermis is covered by a waxy cuticle.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 4.
Establish your own plant nursery near your school or home. Prepare the seedlings of flowering plants, fruit plants, and ornamental plants being grown in your area. Can you start a business in the future with the help of this activity? Think it over.
Answer:
(Students are expected to do this activity on their own.)

Yes, we can start a business of plant nursey in future. If sufficient land is available, the emerging field of agritourism would be a good business.

Question 5.
Observe the skin of the back of your hand with the help of a magnifying lens. Do you see the closely attached squarish and pentagonal shapes?
Answer:
Yes, we can see the squarish and pentagonal shapes. This is the stratified squamous epithelium present on the outer layer of skin.

Question 6.
Observe a permanent slide of blood smear under a compound microscope. What did you see?
Answer:

  • A permanent slide of blood smear under a compound microscope shows different types of cells like RBCs, WBCs and platelets.
  • The different types of WBCs that can be seen are eosin mphocytes.

Question 7.
Visit a modern cowshed nearby and record the following points – The number of cattle, their variety, total milk production, cleanliness in cattle-shed, arrangements for health care of cattle.
Answer:
Students are expected to do this activity on their own.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 8.
Collect more information about animal husbandry.
Answer:

  • The branch of agriculture which deals with the feeding, shelter, health and breeding of domestic animals is called animal husbandry.
  • It is a scientific and systematic management of livestock.
  • Animal husbandry practices include:
    (a) Proper food and clean drinking water.
    (b) Proper shelter
    (c) Proper methods of breeding
    (d) Prevention and cure of disease.
  • Animal husbandary practices serve as an alternative income for the farmers and help to satisfy the need of food for man.

Question 9.
Find out from the internet the average daily milk yield from local and exotic varieties of cow.
Answer:

  • Milk yield from local cow : 12-15 litres/day.
  • Milk yield from exotic cow: 15-30 litres/day.

Question 10.
Are the structure and functions of the bodies of plants and animals the same?
No. The structure and functions of the bodies of plants and animals are not the same. Some differences between plants and animals:
Answer:

Plants Animals
Plants continue to grow throughout their life. Animals grow only for a specific period of life.
Growth in plants in not uniform. Growth in animals is uniform.
Plants have dividing and non-dividing tissues. Animals do not have different dividing and non­dividing tissues.
Plants are sedentary. Animals generally move from place to place in search of food, shelter and partners.
Energy needs of plants are less. Energy needs of animals are greater.
Plants can prepare their own food by photosynthesis. Animals cannot prepare their own food. They depend on plants and other animals for their food.

Class 9 Science Chapter 17 Introduction to Biotechnology Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
Protective coverings in the animal body are called tissues.
(a) meristematic
(b) muscular
(c) epithelial
(d) bone
Answer:
(c) epithelial

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Epithelial tissue is present in
(a) skin
(b) inner surface of blood vessels
(c) walls of the alveoli
(d) all of these
Answer:
(d) all of these

Question 3.
helps in selective transport of substances.
(a) Squamous epithelium
(b) Stratified epithelium
(c) Cuboidal epithelium
(d) Columnar epithelium
Answer:
(a) Squamous epithelium

Question 4.
helps in secretion of digestive juice, absorption of nutrients.
(a) Glandular epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Ciliated epithelium
Answer:
(b) Columnar epithelium

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
supports internal organs
(a) Areolar tissue
(b) Adipose tissue
(c) Lymph
(d) Tendon
Answer:
(a) Areolar tissue

Question 6.
tissue helps in insulation, supply of
Answer:
(b) Adipose tissue

Question 7.
connect muscles to bones.
(a) Ligaments
(b) Tendons
(c) Cartilages
(d) Nerves
Answer:
(b) Tendons

Question 8.
join two bones to each other.
(a) Cartilages
(b) Tendons
(c) Ligaments
(d) Muscles
Answer:
(c) Ligaments

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 9.
meristem increases the length of root and stem.
(a) Intercalary
(b) Apical
(c) Lateral
(d) Permanent
Answer:
(b) Apical

Question 10.
showed that cells and tissues can be grown ex vivo.
(a) Frederick Campion Steward
(b) Gregor Johann Mendel
(c) Frederick Miescher
(d) Robert Brown
Answer:
(a) Frederick Campion Steward

Question 11.
is an example of genetically modified maize.
(a) Vaishali
(b) Vistive Gold
(c) MON 810
(d) Amflora
Answer:
(c) MON 810

Question 12.
is an exotic variety of cow.
(a) Holstein
(b) Gir
(c) Sahiwal
(d) Dangi
Answer:
(a) Holstein

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 13.
is a layer chicken.
(a) Brahma
(b) Leghorn
(c) Cochin
(d) Aseel
Answer:
(b) Leghorn

Question 14.
is a broiler chicken.
(a) Leghorn
(b) Lehman
(c) Aseel
(d) Minorca
Answer:
(c) Aseel

Question 15.
is a broiler chicken.
(a) Minroca
(b) Ancona
(c) Leghorn
(d) Cochin
Answer:
(d) Cochin

Question 16.
Cartilage is found
(a) all around the cells in body
(b) in nose, ear, larynx, trachea
(c) between the muscles and skin
(d) around the blood vessels
Answer:
(b) in nose, ear, larynx, trachea

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 17.
Which of the following is an exotic variety of COW?
(a) Plymouth Rock
(b) New Hampshire
(c) Black Rock
(d) Brown Swiss
Answer:
(d) Brown Swiss

Question 18.
epithelium is present in the inner surface of respiratory tract.
(a) Stratified
(b) Columnar
(c) Ciliated
(d) Cuboidal
Answer:
(c) Ciliated

Question 19.
Bones consist of osteocytes embedded in solid ground substance made up of
(a) calcium bicarbonate
(b) calcium carbonate
(c) calcium phosphate
(d) calcium sulphate
Answer:
(c) calcium phosphate

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 20.
helps in growth of branches, formation of leaves and flowers.
(a) Lateral meristem
(b) Apical meristem
(c) Intercalary meristem
(d) Xylem
Answer:
(c) Intercalary meristem

Question 21.
tissue is present in the hard coat of seeds, outer covering of coconut.
(a) Parenchyma
(b) Xylem
(c) Collenchyma
(d) Sclerenchyma
Answer:
(d) Sclerenchyma

Question 22.
The cells of tissue are dead.
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma
(d) All of these
Answer:
(c) Sclerenchyma

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 23.
Cuboidal epithelium is found in
(a) inner surface of mouth, blood vessels
(b) inner layer of skin
(c) tubules of kidney, salivary gland
(d) inner surface of respiratory tract
Answer:
(c) tubules of kidney, salivary gland

Find the odd man out:

Question 1.
Cell body, muscle fibre, axon, dendrites.
Answer:
Muscle fibre. It is a muscle cell whereas the others are parts of nerve cell.

Question 2.
Tracheids, vessels, companion cells, xylem fibres.
Answer:
Companion cells. It is an element of phloem whereas the others are elements of xylem.

Question 3.
Sieve tubes, tracheids, phloem parenchyma, phloem fibres.
Answer:
Tracheids. It is an element of xylem whereas the others are elements of phloem.

Question 4.
Holstein, Brown swiss, Bombyx mori, Jersey.
Answer:
Bombyx mori. It is a variety of silkworm whereas the others are varieties of cows.

Complete the analogy:

(1) Inner surface of mouth : Squamous epithelium :: Inner surface of intestine :
(2) Glandular epithelium : Secretion of sweat, oil :: Cuboidal epithelium:
(3) Respiratory tract: Ciliated columnar epithelium :: Kidney tubules:
(4) Outer layer of skin : Stratified epithelium :: Inner layer of skin :
(5) Muscular tissue : Movement :: Nervous tissue :
(6) Tendons : Join muscles to bones :: Ligaments :
Answer:
(1) Columnar epithelium
(2) Secretion of saliva
(3) Cuboidal epithelium
(4) Glandular epithelium
(5) Conduction of excitation
(6) Join two bones to each other

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Squamous epithelium (a) Secretion of digestive juice
(2) Cuboidal epithelium (b) Selective transport of substances
(3) Columnar epithelium (c) Protection of organs
(4) Stratified epithelium (d) Secretion of saliva

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

Question 2.

Column ‘A’ Column ‘B’
(1) Brahma (a) Local variety of cow
(2) Jersey (b) Layer chicken
(3) Devin (c) Exotic variety of cow
(4) Lehman (d) Broiler chicken

Answer:
(1 – d),
(2 – C),
(3 – a),
(4 – h)

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 3.

Column ‘A’ Column ‘B’
(1) Potato (a) Vistive Gold
(2) Maize (b) Amflora
(3) Soybean (c) Vaishali
(4) Tomato (d) MON 863

Answer:
(1-b),
(2-d),
(3-a),
(4-c)

Answer the following:

Question 1.
Explain the different types of muscular tissues.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 5

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Explain the types of complex permanent tissues.
Answer:

Name of tissue Xylem Phloem
Characteristics Consists of thick-walled dead cells Consists of cytoplasm containing living cells.
Types of cells Tracheids, vessels and xylem fibres – dead cells. Xylem parenchyma – living cells. Sieve tubes, companion cells, phloem parenchyma – living cells. Phloem fibres – dead cells.
Function A structure like interconnected tubes, conduct water and minerals only in upward direction. Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in upward and downward direction

State whether the following statements are true or false. Correct the false statement:

Question 1.
Lateral meristem increase girth(diameter) of the root and stem.
Answer:
True

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Due to use of seeds of GM crops, there is decrease in nutritive value and increase in loss of crops.
Answer:
False. Due to use of seeds of GM crops there is improvement in nutritive value and decrease in loss of crops.

Question 3.
In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
Answer:
True

Question 4.
Rare and endangered plants can be grown by tissue culture and can be protected from extinction.
Answer:
True

Question 5.
The silk thread is formed from the secretion of the sweat glands of silkworm.
Answer:
False. The silk thread is formed from the secretion of the salivary glands of silkworm.

Question 6.
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
Answer:
True

Question 7.
Phloem consists of thick-walled dead cells.
Answer:
False. Phloem consists of living cells containing cytoplasm.

Question 8.
Stratified squamous epithelium is present in inner surface of mouth, oesophagus, blood vessels and alveoli.
Answer:
False. Stratified squamous epithelium is present in the outer layer of skin.

Question 9.
Broiler chickens are raised for eggs.
Answer:
False. Broiler chickens are raised for meat.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 10.
Non-striated muscles bring about contraction and relaxation of heart.
Answer:
False. Cardiac muscles bring about contraction and relaxation of heart.

Name the following:

Question 1.
Tissue responsible for growth of branches, formation of leaves and flowers.
Answer:
Intercalary meristem.

Question 2.
Tissue that helps in insulation, supply of energy, storage of fats.
Answer:
Adipose tissue.

Question 3.
Tissue present in outer layer of skin.
Answer:
Glandular epithelium.

Question 4.
Tissue present in nose, ear, larynx, trachea.
Answer:
Cartilage.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
Dead cells in xylem.
Answer:
Tracheids, vessels and xylem fibres.

Question 6.
Living cells in xylem.
Answer:
Xylem parenchyma.

Question 7.
Living cells in phloem.
Answer:
Sieve tubes, companion cells, phloem parenchyma.

Question 8.
Dead cells in phloem.
Answer:
Phloem fibres.

Question 9.
GM crops.
Answer:
BT cotton, Amflora, Golden Rice.

Question 10.
Local Indian varieties of cow.
Answer:
Sahiwal, Sindhi, Lai kandhari, Devni, Khillari, Dangi.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 11.
Exotic varieties of cow.
Answer:
Jersey, Brown swiss, Holstein.

Question 12.
Layers.
Answer:
Leghorn, Minorca, Ancona, Lehman.

Question 13.
Broilers.
Answer:
Brahma, Long, Cochin, Aseel.

Question 14.
Chickens reared for both eggs and meat.
Answer:
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock.

Give scientific reason.

Question 1.
Blood is a complex tissue.
Answer:

  • A complex tissue is made up of more than one type of cells.
  • In blood, cells of different types, colour and shapes are mixed together.
  • Blood contains erythrocytes (RBCs), leucocytes (WBCs) and platelets in a liquid plasma.
  • Therefore, blood is a complex tissue.

Define the following:

Question 1.
Tissue
Answer:
A group of cells having the same origin, same structure and same function is called tissue.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Differentiation
Answer:
When cells acquire a specific structure, shape and location and perform a specific function, it is called differentiation.

Question 3.
Biotechnology
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called ‘Biotechnology’.

Question 4.
Tissue culture
Answer:
Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is called tissue culture.

Write short notes

Question 1.
Animal husbandry
Answer:

  • In India, animal husbandry is practised for milk production and for using the cattle as help in farming operations.
  • Example – Cows and buffaloes are raised for milk and bulls and male buffaloes for pulling the heavy loads.
  • Local Indian varieties of cows like Sahiwal, Sindhi, Gir, Lai kandhari, Devni, Khillari, Dangi, etc. and exotic varieties like Jersey, Brown swiss, Holstein, etc. are kept for their milk.
  • Proper care of cattle is necessary for a clean and high yield of milk which includes:
    (a) A balanced diet, i.e. all constituents of food should be given to cattle. It must include fibre- rich coarse food, fodder, and sufficient water.
    (b) The cattle-shed should be clean and dry with proper ventilation and a roof.
    (c) Cattle should be regularly vaccinated.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Poultry farming.
Answer:

  • Rearing of egg and meat yielding chickens is called poultry farming.
  • Chickens raised for laying eggs are called layers . e.g. Leghorn, Minorca, Ancona, Lehman.
  • Chickens raised for meat are called broilers, e.g. Brahma, Long, Cochin, Aseel.
  • Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
  • The objectives behind development of new hybrid varieties from a cross between Indian varieties like Aseel and exotic varieties like Leghorn are to produce good quality chickens in large numbers.
  • To develop the ability to withstand high temperature, to use by-products of agriculture as poultry feed, etc.

Distinguish between:

Question 1.
Xylem and Phloem
Answer:

Xylem Phloem
(i) Consists of mostly thick-walled dead cells. (i) Consists of mostly living cells containing cytoplasm.
(ii) The types of cells include dead cells- tracheids, vessels and xylem fibres and living cells-Xylem parenchyma. (ii) The types of cells include living cells – Sieve tubes, companion cells, phloem parenchyma and dead cells – phloem fibres.
(iii) Structure like interconnected tubes, conduct water and minerals only in upward direction. (iii) Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in both upward and downward direction.

Question 2.
Striated muscles and Non-striated muscles.
Answer:

Striated muscles Non-striated muscles
(i) Muscle cells are long, cylindrical, multinucleate and have no branches. (i) Muscle cells are short, spindle-shaped, uninucleate and have no branches.
(ii) There are alternate dark and light bands on these muscles and they are attached to bones. (ii) Dark and light bands are absent. Not attached to bones.
(iii) They move as per our will, hence they are called voluntary muscles. (iii) They are not under the control of our will, hence they are called involuntary muscles.
(iv) These muscles bring about movements of arms and legs, running, speaking etc. (iv) These muscles bring about movement of eyelids, passage of food through alimentary canal, contraction and relaxation of blood vessels etc.

Question 3.
Cartilage and Bone
Answer:

Cartilage Bone
(i) They are present in nose, ear, larynx, trachea. (i) They form the skeleton of the body.
(ii) They contain cells supported by fibrous, flexible, jelly-like ground substance. (ii) They contain osteocytes embedded in solid ground substance made up of calcium phosphate.
(iii) Lubricates the surface of bones, gives support and shape to organs. (iii) Supports and protects different organs, helps in movement.

Answer the following questions:

Question 1.
Close your eyes and try to identify different objects by feeling them with your hand. Why is it possible for you to identify things like a note-book, text-book, bench, compass-box, etc. only by touching them?
Answer:

  • We can identify objects just by touching them because of the memory that we retain in our brain.
  • The nervous tissue enables us to respond to the stimuli of touch.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Answer in detail:

Question 1.
Write the location, structure and function of following epithelial tissues:
(a) Squamous epithelium
(b) Stratified epithelium
(c) Glandular epithelium
(d) Columnar epithelium
(e) Ciliated epithelium
(f) Cuboidal epithelium
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 8

Question 2.
Write the structure and function of the following connective tissues:
(a) Blood
(b) Lymph
(c) Areolar tissue
(d) Adipose tissue
(e) Cartilage
(f) Bones
(g) Tendons and ligaments
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 9

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Describe the structure of nervous tissue with the help of a neat labelled diagram.
Answer:

  • Nervous tissue enables us to respond to stimuli like touch, sound, odour, colour, etc.
  • Cells of the nervous tissue are specifically made to become excited and conduct the excitation from one part of the body to another.
  • The main part of the nerve cell is the cell body which contains the cytoplasm and the nucleus.
  • Numerous, small, branched fibres called dendrites arise from the cell body.
  • One of the fibres, is extremely long and is called
    the axon.
  • The length of the nerve cell may be up to one metre.
  • Many nerve cells are bound together with the help of connective tissue to form a nerve.
  • Nervous tissue is present in the brain, spinal cord and the network of nerves spread all throughout the body.
  • In most animals, action in response to a stimulus occurs due to the integrated functioning of the nervous tissue and muscular tissue.
    Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 1

Question 3.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Question 4.
Explain types, location and function of Meristematic tissue in tabular form:
Answer:

Types Apical Intercalary Lateral
Location At tip of the root and stem At the base of the petiole of leaves and branches At sides of root and stem
Function Increase the length of the root and stem Growth of branches, the formation of leaves and flowers Increases diameter of the root and stem

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Name of Tissue Parenchyma Collenchyma Sclerenchyma
Struc­ture of cells Living cells with thin cell wall and intercellular spaces. Elongated living cells with thickened cell wall at corners due to cellulose and pectin. Dead and fibrous cells with tapering ends, cell wall contains lignin.
Loca­tion All parts like roots, stem, leaves, flowers and seeds. At the base of leaf petiole, branches and stem Stem, veins of leaves, hard coats of seeds, outer covering of coconut.
Func­tions Support, storage of food and filling vacant spaces. Support and flexibility to various parts. Give strength and rigidity to parts of the plants.
Sub­ types Chlorenchyma: Leaves, perform photosynthesis. Aerenchyma: Helps aquatic plants, leaves and stem to float.

Question 6.
Write down the applications of biotechnology (tissue culture) in floriculture, nurseries and forestry.
Answer:
Applications of biotechnology (tissue culture) in floriculture, nurseries and foresty:

  • Tissue culture can be used to grow those plants on a large scale which bear flowers, fruits of excellent quality.
  • Fully grown plants can be produced in shorter durations.
  • Plants can be grown on a large scale even if means of pollination or germinating seeds are not available.
  • For example, orchids or pitcher plants do not germinate but these plants can easily be produced by means of tissue culture.
  • In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
  • Bioreactors are useful for producing plantlets on a very large scale.
  • A large number of seedlings/plantlets can be produced in a short time using minimum resources and materials.
  • Usually, plants produced by tissue culture and genetic modification techniques are disease-free.
  • Plantlets produced by tissue culture technique of the meristem are virus-free.
  • Embryos produced using conventional hybridization technique between two or more varieties may not grow fully for some reasons. Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology
  • However, embryos produced by tissue culture technique always complete their growth.
  • Rare and endangered plants can be grown using tissue culture technique and can thus be protected from extinction.
  • Similarly, various parts and seeds of such plants can be preserved by tissue culture and those varieties can be protected.

9th Std Science Questions And Answers:

Classification of Plants Class 9 Science Chapter 6 Questions And Answers Maharashtra Board

Class 9 Science Chapter 6

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 6 Classification of Plants Question Answer Maharashtra Board

Class 9 Science Chapter 6 Classification of Plants Question Answer Maharashtra Board

1. Match the proper terms from columns A and C with the description in column B.
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 10

2. Complete the sentences by filling in the blanks and explain those statements.
(angiosperms, gymnosperms, spore, Bryophyta, thallophyta, zygote)
a. ……………….. plants have soft and fiber-like body.

b. ……………….. is called the ‘amphibian’ of the plant kingdom.
Answer:
Bryophyta plant is called the ‘amphibian’ of the plant kingdom

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

c. In pteridophytes, asexual reproduction occurs by ……………….. formation and sexual reproduction occurs by ………………..formation.
Answer:
Spore, zygote: Pteridophyta plants show alteration of generation. One generation reproduces by spore-formation and the next generation reproduces sexually by zygote formation.

d. Male and female flowers of ……………….. are borne on different sporophylls of the same plant.
Answer:
Gymnosperms bear their male and female flowers on different sporophylls of the same plant

3. Answer the following questions in your own words.

a. Write the characteristics of subkingdom Phanerogams.
Answer:

  • Plants which have special structures for reproduction and produce seeds are called Phanerogams.
  • In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food.
  • During the germination of the seed, the stored food is used for the initial growth of the embryo.
  • Depending upon whether seeds are enclosed in a fruit or not phanerogams are classified into gymnosperms and angiosperms.

b. Distinguish between monocots and dicots.
Answer:

Dicots Monocots
Seed Two cotyledons Single cotyledon
Root Well developed, primary root (Taproot) Fibrous roots
Stem Strong, hard. e.g. Banyan tree Hollow, e.g. Bamboo
False, e.g. Banana
Disc-like, e.g. Onion.
Leaf Reticulate venation Parallel venation
Flower Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous) Flowers with 3 parts or in multiples of three (trimerous).

c. Write a paragraph in your own words about the ornamental plants called ferns.
Answer:

  • Ferns belong to the group of plants called Pteridophyta.
  • They have well-developed roots, stem and leaves but do not bear flowers and fruits.
  • They have separate tissues for the conduction of food and water.
  • They reproduce with the help of spores formed along the back or posterior surface of their leaves.
  • They reproduce asexually by spore formation and sexually by zygote formation.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

d. Sketch, label and describe the Spirogyra.

e. Write the characteristics of the plants belonging to division Bryophyta.
Answer:

  • Bryophyta group of plants are called the amphibians of the plant kingdom because they grow in moist soil but need water for reproduction.
  • These plants are thalloid, multicellular and autotrophic.
  • They reproduce by spore-formation.
  • Their plant body structure is flat, ribbon-like, long, without true roots, stem and leaves.
  • Instead, they have stem-like or leaf-like parts and root-like rhizoids.
  • They do not have specific tissues for the conduction of food and water.
  • Examples: Moss (Funaria), Anthoceros, Riccia etc.?

4. Sketch and label the figures of the following plants and explain them into brief.
Marchantia, Funaria, Fern, Spirogyra.

Question 1.
Spirogyra.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 5

  • Spirogyra belongs to the division thallophyta. They are called as algae.
  • It grows mainly in water.
  • It does not have specific parts like root-stem- leaves-flowers but are autotrophic due to the presence of chlorophyll.
  • The plant body of Spirogyra is soft and fibre-like.
  • It has spirally arranged chloroplasts in its cell.

Question 2.
Funaria and Marchantia (Bryophyta)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 6

  • These plants are called ‘amphibians’ of the plant kingdom because they grow mostly in soil and need water for reproduction.
  • They do not have specific tissues for the conduction of food and water.
  • The plant body is fiat, ribbon-like long, without true roots, stem and leaves
  • Instead, they have stem-like or leaf-like parts and root like rhizoids.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Fern (Pteridophyta):
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 7

  • They have well-developed roots, stem and leaves for the conduction of food and water.
  • They do not bear flowers and fruits.
  • They reproduce with the help of spores present along the back or posterior surface of the leaves.

5. Collect a monocot and dicot plant available in your area. Observe the plants carefully and describe them in scientific language.

6. Which criteria are used for the classification of plants? Explain with reasons.
Answer:
Criteria for classification of plants:

  • If plants do not bear flowers, fruits and seeds, they are non-seed bearing plants. If they bear flowers, fruit and seeds, they are seed-bearing plants.
  • Presence or absence of conducting tissues- Plants such as pteridophytes, gymnosperms and angiosperms which possess conducting tissues are included in vascular plants whereas thallophytes and bryophytes which do not possess conducting tissues are included under non-vascular plants.
  • Depending upon whether the seeds are enclosed in fruit or not, plants are classified as gymnosperms (naked-seeds) and angiosperms (seeds covered by fruit) Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants
  • Depending upon the number of cotyledons in seeds, plants are classified into dicotyledons and monocotyledons

Class 9 Science Chapter 6 Classification of Plants Intext Questions and Answers

Can you recall?

Question 1.
How have living organisms been classified?
Answer:
(i) Organisms have been classified based on the following:

  • Cell structure
  • Body Organisation
  • Mode of nutrition
  • Reproduction

(ii) Organisms are also classified at kingdom level and groups and subgroups.

Activity-based questions

Question 1.
You may have seen a lush green soft carpet on old walls, bricks and rocks in the rainy season. Scrape it gently with a small ruler, observe it under a magnifying lens and discuss.
Answer:

  • It shows considerable tissue complexity and is differentiated into two main parts: a root and a shoot.
  • They have a variety of specialized tissues within these two regions of the body.
  • Same kind of cells are seen throughout the whole body except reproductive cells.

Question 2.
You may have seen ferns among the ornamental plants in a garden. Take a leaf of a fully grown fern and observe it carefully.
Answer:

  • New leaves typically expand by the unrolling in a tight spiral manner.
  • The anatomy of fern leaves can either be simple or highly divided.
  • They show the presence of spores formed along the back or posterior surface of their leaves.

Question 3.
Observe all garden plants like Cycas, Christmas tree, Hibiscus, Lily, etc. and compare them. Note the similarities and differences among them. Which differences did you notice in gymnosperms and angiosperms?
Answer:
Cycas and Christmas tree are gymnosperms, whereas Hibiscus and lily are angiosperms.

  1. Similarities: These plants have special structures for reproduction and produce seeds. During the germination of the seed, the stored food is used for the initial growth of the embryo.
  2. Differences: In gymnosperms, reproductive organs have cones whereas in angiosperms reproductive organs have flowers.
  3. In gymnosperms, seeds are without natural coverings whereas in angiosperms seeds are enclosed in natural coverings called fruits.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 4.
Soak the seeds of corns, beans, groundnut, tamarind, mango, wheat, etc. in water for 8 to 10 hrs. After they are soaked, check each seed to see whether it divides into two equal halves or not and categorize them accordingly.
Answer:
Monocots: com, wheat (it cannot be divided into equal halves)
Dicots: beans, groundnut, tamarind and mango (it can be divided into two equal halves)

Class 9 Science Chapter 6 Classification of Plants Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The five-kingdom classification was proposed b7
(a) Robert Whittaker
(b) Robert Hooke
(c) Eichler
(d) Louis Pasteur
Answer:
(a) Robert Whittaker

Question 2.
In 1883, classified plants into two sub-kingdoms.
(a) Robert Whittaker
(b) Alexander Fleming
(c) Eichler
(d) Robert Hooke
Answer:
(c) Eichler

Question 3.
Ulothrix, ulva, sargassum belong to
(a) Bryophyta
(b) Thallophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(b) Thallophyta

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 4.
is a bryophyte.
(a) Ulva
(b) Nephrolepis
(c) Funaria
(d) Equisetum
Answer:
(c) Funaria

Question 5.
In the seeds are naked.
(a) Pteridophyta
(b) Angiosperms
(c) Gymnosperms
(d) Bryophyta
Answer:
(c) Gymnosperms

Question 6.
In the flowers are reproductive organs.
(a) Angiosperms
(b) Gymnosperms
(c) Pteridophyta
(d) Bryophyta
Answer:
(a) Angiosperms

Question 7.
In the flowers are tetramerous or pentamerous.
(a) Monocotyledons
(b) Dicotyledons
(c) Gymnosperms
(d) Pteridophyta
Answer:
(b) Dicotyledons

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 8.
In monocotyledonous plants, the stem is
(a) hollow
(b) false
(c) disc-like
(d) all of these.
Answer:
(d) all of these

Question 9.
Lycopodium belongs to
(a) Thallophyta
(b) Bryophyta
(c) Gymnosperms
(d) Pteridophyta
Answer:
(d) Pteridophyta

Question 10.
Leaves of show reticulate venation.
(a) Bamboo
(b) Banana
(c) Onion
(d) Banyan
Answer:
(d) Banyan

Question 11.
Various types of fungi like yeasts and moulds are included in the group
(a) Thallophyta
(b) Halophyte
(c) Xenophyta
(d) Angiosperms
Answer:
(a) Thallophyta

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 12.
Bryophytes have a root-like structure called
(a) Nodes
(b) Rhizoids
(c) Nodules
(d) Aerenchyma
Answer:
(b) Rhizoids

Question 13.
reproduce with the help of spores formed along the back or posterior surface of their leaves.
(a) Halophyta
(b) Pteridophyta
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Pteridophyta

Question 14.
In ……………………….., the reproductive organs cannot be seen.
(a) Pteridophyta
(b) Cryptogams
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Cryptogams

Question 15.
are mostly evergreen, perennial and woody.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 16.
Gymnosperms bear male and female flowers on different of the same plant.
(a) Branches
(b) Roots
(c) Sporophylls
(d) Flowers
Answer:
(c) Sporophylls

Question 17.
In the seeds are not enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 18.
In the seeds are enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(d) Angiosperms

Question 19.
The plants whose seeds cannot be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(d) Monocotyledons

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 20
The plants whose seeds can be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(c) Dicotyledons

Find the odd one out:

Question 1.
Ulothrix, Ulva, Nephrolepis, Sargassum
Answer:
Nephrolepis: It belongs to division Pteridophyta whereas the others belong to division thallophyta.

Question 2.
Funaria, Marchantia, Anthoceros, Spirogyra
Answer:
Spirogyra:

Question 3.
Marsilea, Pteris, Lycopodium, Riccia
Answer:
Riccia:

Question 4.
Cycas, Mango, Apple, Banyan
Answer:
Cycas:

Question 5.
Onion, Papaya, Wheat, Green peas
Answer:
Green peas:

Complete the analogy:

(1) Spirogyra : Thallophyta : : Riccia :
(2) Moss : Bryophyta : : Selaginella :
(3) Nephrolepis : Pteridophyta :: Ulothrix :
(4) Pteridophyta : Roots :: Bryophyta :
(5) Gymnosperms : naked seeds : : Angiosperms :
(6) Dicotyledon : Reticulate venation : : Monocotyledon:
(7) Bamboo stem: Hollow:: Onion Stem:
(8) Monocotylendon : Tap root:: Dicotyledon :
Answer:
(1) Bryophyta
(2) Pteridophyta
(3) Thallophyta
(4) Rhizoids
(5) Covered seeds
(6) Parallel venation
(7) Disc like
(8) Fibrous roots

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Difference between:

Question 1.
Thallophyta and Bryophyta
Answer:

Thallophyta Bryophyta
These plants grow mainly in water They grow in moist soil but need water for reproduction

Question 2.
Gymnosperms and Angiosperms
Answer:

Gymnosperms Angiosperms
No natural covering on seeds Seeds are formed in fruits

Question 3.
Algae and Moss
Answer:

Algae Moss
These plants mainly grow in water. These plants need water for reproduction.

State whether the following statements are true or false. Correct the false statements:

(1) Thallophyta are called as the amphibians of the plant kingdom.
(2) Fungi like yeasts and moulds are included in division bryophyta.
(3) Moss (Funaria) belongs to division bryophyta.
(4) Bryophyta have specific tissues for conduction of food and water.
(5) Plants belonging to Thallophyta group are only unicellular.
(6) Pteridophytes have well developed roots, stems and leaves.
(7) Pteridophytes reproduce with the help of spores formed along the back or posterior surface of their leaves.
(8) Nephrolepis belongs to division Pteridophyta.
(9) Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into monocots and dicots.
(10) Gymnosperms are mostly evergreen, perennial and woody.
(11) Gymnosperms bear male and female flowers on different sporophylls of different plants.
(12) In Angiosperms, the seeds are covered by fruits.
(13) Dicotyledonous plants show reticulate venation.
(14) Moncotyledonous plants have trimerous flowers.
(15) In dicotyledonous plants, the stem is strong and hard.
Answer:
(1) False. Thallophyta plants grow mainly in water.
(2) False. Fungi like yeasts and moulds are included in division thallophyta.
(3) True
(4) False. Bryophyta do not have specialised tissuesfor conduction of food and water.
(5) False. Plants belonging to thallophyta group may be unicellular or multicellular.
(6) True
(7) True Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants
(8) True
(9) False. Depending whether seeds are enclosed in. a fruit or not, angiosperms are classified into monocots and dicots.
(10) True
(11) False. Gymnosperms bear male and female flowers on different sporophylls of the same plant.
(12) True
(13) False. Dicotyledonous plants show parallel venation.
(14) True
(15) True.

Give name

Question 1.
What are ornamental plants are called?
Answer:
Ferns

Question 2.
Plants with two cotyledons are called.
Answer:
Dicots

Question 3.
Plants with single cotyledon are called.
Answer:
Monocots

Question 4.
Type of venation showed by hibiscus plant leaves
Answer:
Reticulate venation

Question 5.
Type of venation showed by lily plant leaves
Answer:
Parallel venation

One line answers

Question 1.
Which plants are mostly evergreen, perennial and woody?
Answer:
Gymnosperms are mostly evergreen, perennial and woody.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Which type of venation showed by dicot plants?
Answer:
Leaves of dicot plants show reticulated venation.

Question 3.
Which type of venation showed by monocot plants?
Answer:
Leaves of monocot plants show parallel venation

Question 4.
How are angiosperms classified into monocot and dicot?
Answer:
Depending whether seeds and enclosed in fruit or not, angiosperms are classified into monocot and dicot

Question 5.
In which division are fungi like moulds and yeast classified?
Answer:
Fungi like moulds and yeast classified in division thallophyta.

Question 6.
Plants belonging to which group may be unicellular or multicellular?
Answer:
Plants belonging to thallophyta group may be unicellular or multicellular

Give scientific reason

Question 1.
Thallophyta plants have thin and fibre like body
Answer:
Thallophyta: These plants grow mainly in water i.e. fresh water as well as in saline water, therefore they usually have a soft and fibre-like (filamentous) body.

Question 2.
Bryophyta plants are called the amphibian plants.
Answer:
Bryophyta: They grow in moist soil but need water for reproduction. Therefore, they are called ‘amphibians of plant kingdom’.

Question 3.
Gymnosperms bear their male and female flowers on different sporophylls of the same plant
Answer:
Gymnosperms: As these plants do not take the assistance of pollinators i.e. vectors, the male and female flowers are present on the different sporophyll of the same plant for successful fertilisation.

Write note on

Question 1.
August W. Eichler
Answer:
In 1883, Eichler, a botanist, classified the Kingdom Plantae into two subkingdoms. As a result, two subkingdoms, cryptogams and phanerogams were considered for plant classification.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Thallophyta
Answer:
These plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 3.
Bryophyta
Answer:
This group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 4.
Pteridophyta
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 5.
Phanerogams
Answer:
Plants which have special structures for reproduction and produce seeds are called phanerogams. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 6.
Gymnosperms
Answer:
Gymnosperms are mostly evergreen, perennial and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 7.
Angiosperms
Answer:
The flowers these plants bear are their reproductive organs. Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Complete the flow chart.

Question 1.
Living Organisms
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 1
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 2

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Kingdom: Plantae
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 4

Distinguish between:

Question 1.
Bryophyta and Pteridophyta:
Answer:

Bryophyta Pteridophyta
Bryophytes grow in soil but need water for reproduction. Pteridophytes grow in soil.
Plant body is without specific parts like true roots, stem and leaves. Plant body is differentiated into true roots, stem and leaves.
Conducting tissues for food and water absent. Conducting tissues for food and water present.
Examples: Moss (Funaria), Marchantia, Anthoceros, etc. Examples: Nephrolepis, Marsilea, Pteris, Adiantum, Lycopodium etc.

Question 2.
Angiosperms and Gymnosperms.
Answer:

Angiosperms Gymnosperms
(i) In Angiosperms, the stems have branches. (i) In Gymnosperms, the stems are without branches.
(ii) Reproductive organs are flowers. (ii) Reprodcutive organs are cones.
(iii) Seeds are enclosed in natural coverings, i.e., fruits. (iii) Seeds are not enclosed in natural coverings.
(iv) Examples: Mango, Bamboo, etc. (iv) Examples: Cycas, Picea etc.

Question 3.
Cryptogams and Phanerogams.
Answer:

Cryptogams Phanerogams
(iii) Their reproductive organs are hidden. (iii) Their reproductive organs are exposed.
(iii) They reproduce by forming spores. (iii) They reproduce by forming seeds.
(iii) They are less evolved plants. (iii) They are highly evolved plants.
(iv) They are divided into Thallophyta, (iv) They are divided into Gymnosperms and
Bryophyta, Pteridophyta. Angiosperms.

Distinguish between:

Question 1.
Thallophyta

Answer:
Spirogvra, Ulothrix, Ulva, Sargassum

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Bryophyta
Answer:
Moss (Funaria), Marchantia, Anthoceros, Riccia

Question 3.
Pteridophyta
Answer:
Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium

Question 4.
Gymnosperms
Answer:
Cycas, Picca (Christmas tree), Thuja (Morpankhi), Pinus (Deodar)

Question 5.
Angiosperms
Answer:
Tamarind, Mango, Apple, Lemon

Question 6.
Monocot plants
Answer:
Bamboo, bananas, corn, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

Question 7.
Dicot plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

Observe the figure and answer the questions

1. Dicot Plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 8

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Well developed, primary root (Tap root)

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Reticulate Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Strong and hard

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Two cotyledons

Question 6.
Give example of the following types of plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

2. Monocot Plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 9

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Fibrous roots

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 3 parts or in multiples of three (trimerous).

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Parallel Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Hollow, False or Disc-like

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Single cotyledons

Question 6.
Give example of the following types of plants
Answer:
Bamboo, bananas, com, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

3. Spirogyra
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 11

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division I Thallophyta.

Question 2.
Where does this plant grow?
Answer:
These plants grow mainly in water.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Are these types of plants unicellular or multicellular?
Answer:
They may be unicellular or multicellular and microscopic or large.

Question 4.
Are these types of plant autotropic?
Answer:
They are autotrophic due to the presence of chlorophyll but types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 5.
Do these plants have a root-stem-leaves-flowers system?
Answer:
They do not have specific parts like root-stem- leaves-flowers.

Question 6.
How is the body of these types of plants?
Answer:
These plants usually have a soft and fibre-like body.

4. Funaria
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 12

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division II Bryophyta.

Question 2.
Where does this plant grow?
Answer:
They grow in moist soil but need water for reproduction.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
What are these group of plants called in the plant kingdom?
Answer:
This group of plants is called the ‘amphibians’ of the plant kingdom.

Question 4.
Are these types of plant autotropic?
Answer:
They reproduce by spore formation.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves.

Question 6.
What do these plants have instead of roots?
Answer:
They have root like rhizoids.

5. Fern
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 13

Question 1.
Which division of plants does this plant come? under?
Answer:
This plant come under Division III Pteridophy ta.

Question 2.
Where does this plant grow?
Answer:
They grow in soil.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
How do these plants reproduce?
Answer:
These plants reproduce asexually by spore- formation and sexually by zygote formation.

Question 4.
Do these plants produce flowers and fruits?
Answer:
They do not bear flowers and fruits.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water.

Question 6.
Where are the spores formed in the plants body?
Answer:
The spores formed along the back or posterior surface of their leaves.

6. Cycas
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 14

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division III Phanerogams Division I Gymnosperms.

Question 2.
Explain structure of these types of plants?
Answer:
Gymnosperms are mostly evergreen perennial and woody.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
How is stem and leaves of these types of plants?
Answer:
Their stems are without branches and the leaves form a crown.

Question 4.
Where are the male and female flowers located?
Answer:
These plants bear male and female flowers on different sporophylls of the same plant.?

Question 6.
Give some examples of these types of plants?
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

7. Monocot and Dicot plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 15

Question 1.
Which division of plants does this plant come under?
Answer:
ThisplantcomeunderDivisionlllPhanerogams Division II Angiosperms

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
How are the seeds of these types of planis?
Answer:
The seeds are formed within fruits thus these seeds are covered

Question 3.
How can we classify the plants according to their seeds in this division?
Answer:
The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Question 4.
How the venations are present on the leaves of these types of plants?
Answer:
These plants bear parallel or reticulated venations on the leaves.

Question 5.
How is the root system of these types of plants?
Answer:
The root systems of these types of plant are tap roots or fibrouš roots.

Complete the paragraph

Question 1.
Thallophyta plants grow mainly in …………….. . This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of …………….., is called algae. Algae show great diversity. They may be unicellular or …………….., and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a …………….. and fibre-like body. Various types of …………….. like yeasts and moulds which do not have …………….. are also included in this group.
Answer:
Thallophyta plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
…………….. group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need …………….. for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by …………….. formation. The structure of the plant body of bryophytes is flat, ribbon?like long, without true …………….., stem and leaves. Instead, they have stem-like or leaf?like parts and root-like ……………. . They do not have specific …………….. for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.
Answer:
Bryophyta group of plants is called the ‘amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 3.
Plants from Pteridophyta group have well developed roots, stem and leaves and separate …………….. for conduction of food and water. But,
they do not bear …………….. and ……………… They reproduce with the help of …………….. formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce …………….. by spore-formation and sexually by …………….. formation. They have a well-developed conducting system.
Answer:
Plants from Pteridophyta group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 4.
Phanerogams plants which have special structures for …………….. and produce …………….. In these plants, after the process of reproduction, seeds are formed which contain the …………….. and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in …………….. a or not, phanerogams are classified into …………….. and ……………. .
Answer:
Phanerogams plants which have special structures for reproduction and produce seeds. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 5.
Gymnosperms are mostly …………….., perennial and woody. Their stems are without …………….. The leaves form a …………….. These plants bear male and female flowers on different …………….. of the same plant …………….. of these plants do not have natural coverings, i.e. these plants do not form …………….. and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.
Answer:
Gymnosperms are mostly evergreen, perennial • and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 6.
The flowers of Angiosperms plants bear are their …………….. organs Flowers develop into …………….. and seeds are formed within …………….. . Thus, these seeds are ……………..; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called …………….. plants and those whose seeds cannot be divided into equal parts are called …………….. plants.
Answer:
The flowers of Angiosperms plants bear are their reproductive orgAnswer: Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Answer the questions in detail:

Question 1.
Write the characteristics of Thallophyta.
Answer:

  • Thallophyta plants grow mainly in water.
  • The group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll are called algae.
  • Algae show great diversity They may be unicellular or multicellular and microscopic or large.
  • Some of these are found in freshwater while some are found in saline water.
  • Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.
  • Examples: Spirogyra, Ulothrix, Ulva, etc.

Question 2.
Write the characteristics of Gymnosperms.
Answer:

  • Gymnosperms are mostly evergreen, perennial and woody.
  • Their stems are without branches.
  • The leaves form a crown.
  • These plants bear male and female flowers on different sporophylls of the same plant.
  • Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms (gmnos: naked, sperms: seeds)
  • Examples: Cycas, Picea (christmas tree), Thuja, Pinus (deodar), etc.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Make concept diagram

Question 1.
Plant classification
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 16

Question 2.
Taxonomy of carnivorous 1ant
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 17

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Taxonomy of mango plant
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 18

9th Std Science Questions And Answers:

Useful and Harmful Microbes Class 9 Science Chapter 8 Questions And Answers Maharashtra Board

Class 9 Science Chapter 8

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 8 Useful and Harmful Microbes Question Answer Maharashtra Board

Class 9 Science Chapter 8 Useful and Harmful Microbes Question Answer Maharashtra Board

1. Complete the statements using the proper option from those given below. Explain the statements. (mycotoxins, budding, rhizobium)

a. Yeast reproduces asexually by the …………………….. method.
Answer:
The yeast cells develop small round bodies on the parent cell. These are called buds. New daughter cells develop from these buds.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

b. Toxins of fungal origin are called …………………….. .
Answer:
Mycotoxins are poisonous chemicals released into the food by fungi. This makes the food poisonous.

c. Leguminous plants can produce more proteins due to …………………….. .
Answer:

  • (i) Nitrogenous compounds are required to produce proteins.
  • (ii) Rhizobia produce nitrogenous compounds by fixing atmospheric nitrogen and make it available for their host plants like leguminous plants.

2. Write the names of microbes found in the following food materials.
yogurt, bread, root nodules of leguminous plants, idli, dosa, spoiled potato curry.
Answer:

Food materials Microbes
Yogurt Lactobacilli
Bread Yeast
Root nodules of leguminous plants Rhizobium
Idli Yeast, bacteria
Dosa Yeast, bacteria
Spoiled potato curry Clostridium

3. Identify the odd word out and say why it is the odd one?

a. Pneumonia, diphtheria, chicken pox, cholera.
Answer:
Chickenpox. It is caused by a virus, whereas others are caused by bacteria.

b. Lactobacilli, rhizobia, yeast, clostridia.
Answer:
Yeast. It is a fungus, whereas the rest are bacteria.

c. Root rot, rust (tambura), rubella, mozaic.
Answer:
Rubella. It is a disease of humans, whereas the rest are diseases of plants.

4. Give scientific reasons.

a. Foam accumulates on a the surface of ‘dal’ kept for a long time in summer.
Answer:

  • Dal is rich in proteins.
  • During summer, bacteria attack the dal and cause fermentation resulting in the production of carbon dioxide gas.
  • Therefore, foam accumulates on the surface of the ‘dal’ kept for long time in summer.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

b. Why are naphthalene balls kept with clothes to be put away.
Answer:

  • Naphthalene balls are balls of chemical pesticide and deodorant.
  • They help to kill or repel insects such as moths, cockroaches, mice etc.
  • Therefore, naphthalene balls are kept with clothes to be put away to prevent clothes from getting damaged.

5. Write down the modes of infection and the preventive measures against fungal diseases.
Answer:

  • Mode of infection: Contact with infected person or his/her belongings like clothes.
  • Preventive measure: Personal hygiene and avoid contact with infected person.

6. Match the pairs.

‘A’ group ‘B’ group
1. Rhizobium a. Food poisoning
2. Clostridium b. Nitrogen fixation
3. Penicillium c. Bakery products
4. Yeast d. Production of antibiotics
Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

7. Answer the following questions.

a. Which vaccines are given to infants? Why?
Answer:

  • Hepatitis A and B, DTP (Diphtheria, Tetanus, Pertussis.) Polio, MMR (Measles, Mumps, Rubella), Chicken pox, Influenza, Tetanus, BCG, Rotavirus, etc.
  • Vaccines consist of dead or weakend microbes. When these are swallowed or injected, the body produces antibodies to fight them.
  • These antibodies remain in the body and protect it from any future attack of the disease causing microbes.
  • Therefore, vaccines are given to infants for preventing diseases.

b. How is a vaccine produced?
Answer:

  • Vaccines are made using the disease causing bacteria or virus but in a form that will not harm the human beings.
  • Vaccine is made from dead or weakened microbes or their toxins.
  • Vaccine stimulates the immune system to produce antibodies which give life-long protection against the disease.
  • There are specific vaccines for specific diseases.

c. How do antibiotics cure disease?
Answer:
Antibiotics cure diseases by destroying or preventing the growth of harmful micro-organisms.

d. Are the antibiotics given to humans and animals the same? Why?
Answer:

  • Generally, antibiotics work against any harmful bacteria, whether it is attacking humans or animals.
  • But some of them are better suited to humans while some are better for animals. This is due to the adverse effects they show in different species. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Also, the dosages of antibiotics for humans and animals differ.

e. Why is it necessary to safely store the pathogens of a disease against which vaccines are to be produced?
Answer:

  • Pathogens are microbes which can cause diseases in us.
  • For the preparation of a vaccine, a particular pathogen is cultured and grown in a laboratory.
  • If these pathogens are not safely stored, they many get modified due to environmental factors, resulting in decrease in the efficiency of the vaccine.
  • Also, the live pathogens may escape and cause diseases in us.

8. Answer the following questions in brief.

a. What are ‘broad-spectrum antibiotics’?
b. What is fermentation?
Answer:

  • Yeast uses sugar for food.
  • Yeast grows and multiplies rapidly due to the carbon compounds in the sugar solution.
  • In the process of obtaining nutrition, yeast cells convert the carbohydrates in the food into alcohol and carbon dioxide.
  • Also, the bacteria Lactobacilli convert lactose, the sugar in milk into lactic acid.
  • This process is called fermentation.

c. Define ‘Antibiotic’.
Answer:

  • Carbon compounds obtained from some bacteria and fungi for destroying or preventing the growth of harmful micro -organisms are called antibiotics.
  • Antibiotics, a discovery of the 20th century, have brought a revolution in the field of medicine.
  • Antibiotics mainly act against bacteria. Some antibiotics can destroy protozoa.
  • Some antibiotics are useful against a wide variety of bacteria they are called broad-spectrum antibiotics. Examples – Ampicillin, amoxicillin, tetracycline, etc.
  • When the pathogen cannot be identified even though the symptoms of the disease are visible, broad-spectrum antibiotics are used. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Whenever a pathogenic micro-organism is definitely known, then narrow-spectrum antibiotics are used. Examples: Penicillin, gentamycin, erythromycin, etc.

Class 9 Science Chapter 8 Useful and Harmful Microbes Intext Questions and Answers

Can you recall?

Question 1.
What is meant by microbes? What are their characteristics?
Answer:
Microbes are tiny microscopic organisms which cannot be seen with the unaided eye.

Characteristics of Microbes.

  • They are the smallest organisms on earth.
  • They are composed of prokaryotic or eukaryotic cells.
  • They can be seen only with the help of a microscope.
  • They are found in any kind of environment ranging from coolest polar regions to hottest of deserts. Also found in soil, water and air.
  • Some of them are useful, whereas some of them are harmful micro-organisms.

Question 2.
How do you observe microbes?
Answer:
Microbes are observed using a microscope.

Answer the following.

Question 1.
Why are wineries located near Nashik in Maharashtra?
Answer:

  • Nashik in Maharashtra is the leading grape producer in the country as it has the soil suitable for the production of grapes. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Glucose and fructose, the sugars present in grape juice are fermented with the help of yeast to produce wines. Therefore, wineries are located near Nashik in Maharashtra.

Question 2.
Find out the uses of fungi to plants and animals?
Answer:

  • Fungi decompose the bodies of dead animals and convert them into simple carbon compounds. These substances easily mix with air, water and soil from where they are again absorbed by plants and enter the food chain.
  • Some fungi living in symbiotic association with plants help to absorb water and inorganic compounds like nitrate and phosphate.
  • Fungi are also used to derive antibiotics like penicillin which are useful to animals.
  • Ants grow fungi in their anthill and obtain food from it.
  • Some species of wasps and insects lay their eggs in the fungal bodies growing on trees, thus ensuring a food supply for their larvae.

Question 3.
What is the structure of lichen, a condiment? Where else is it used?
Answer:

  • Lichen is a symbiotic association between a fungus and an algae (Cyanobacterium).
  • Lichens are sensitive to environmental disturbances and are used in assessing air pollution in an area.
  • Lichens are also used in making dyes, perfumes and in traditional medicines.
  • A few lichen species are eaten by insects or animals such as reindeer.

Open-ended questions

Answer the following questions:

Class 9 Science Chapter 8 Useful And Harmful Microbes Exercise Question 1.
How is yoghurt made from milk? What exactly happens in this process?
Answer:

  • Milk contains sugar called lactose which is broken down with help of Lactobacilli.
  • Lactobacilli converts lactose into Lactic Acid. This process is called fermentation. As a result, the pH of milk decreases causing coagulation of milk proteins.
  • Thus, milk proteins are separated from other constituents of milk and milk changes to yoghurt.

Useful And Harmful Microbes Class 9 Exercise Question 2.
Sometimes, you may notice a black powder or white discs floating on the pickle or murabba, when a jar is opened after a long time. What exactly is this? Why are such food items not good to eat?
Answer:

  • A black powder or white disc floating on the pickle or murabba are fungi.
  • Different fungal species depend on host (pickle and murabba) for their growth and reproduction.
  • During this process, fungi secretes mycotoxins which are poisonous chemicals; which ultimately spoil the food. Consuming such food can cause food poisoning. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Therefore, such food items are not good to eat.

8 Useful And Harmful Microbes Class 9 Question 3.
How many different industries depend upon the Lactobacilli bacteria?
Answer:
Industries like milk products, cider, cocoa, pickles, pharmaceuticals depend on Lactobacilli bacteria.

Useful And Harmful Microbes Class 9 Question 4.
Which types of cottage industries and factories can be started in areas with abundant milk production?
Answer:
Cottage industries like the manufacture of milk products like ghee, cheese, paneer, curd, shrikhand, etc. and chocolate making can be started in areas with abundant milk production.

9th Class Science Chapter 8 Useful And Harmful Microbes Exercise Question 5.
Which changes do you notice in leather articles and gunny (jute) bags during the rainy season?
Answer:
In rainy season we can notice whitish-green cotton-like growth or black powder or white discs on leather articles and gunny (jute) bags during the rainy season as these articles are infected by fungus.

Class 9th Science Chapter 8 Useful And Harmful Microbes Exercise Question 6.
For how long afterwards can you use those articles?
Answer:
Those articles cannot be used for long as they wear out and do not last long.

Question 7.
Why do these articles not get spoilt during the summer or winter?
Answer:

  • Spores of fungi can germinate when there is sufficient moisture.
  • During summer or winter the weather is hot and dry and so fungus cannot grow in such weather.
  • Also microbes cannot survive extreme hot or cold temperatures of summer or winter. Therefore, these articles do not get spoilt during summer or winter.

Question 8.
Why do doctors advise you to take yoghurt or buttermilk if you have indigestion or abdominal discomfort?
Answer:

  • The Lactobacilli present in yoghurt or buttermilk help to restore the natural microbial flora in the intestine, thus helping in digestion and absorption of nutrients.
  • Also buttermilk helps to cool down the stomach and works as a laxative to ease the congestion during abdominal discomfort.

Question 9.
Sometimes, yoghurt becomes bitter and froths up. Why does this happen?
Answer:

  • Sometimes yoghurt becomes bitter due to excess fermentation by bacteria.
  • Excess amount of lactic acid is produced making the curd bitter.

Question 10.
Which different milk products are obtained at home by fermentation of the cream from the milk?
Answer:
Yoghurt, buttermilk, ghee, cheese, shrikhand, sour cream, etc.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 11.
Recently, it has been made compulsory in India and some other countries to mix 10% ethanol with fuels like petrol and diesel. What is the reason for this?
Answer:

  • Ethanol is a smokeless and high quality fuel. So it helps to reduce pollution when mixed with petrol or diesel.
  • As petrol or diesel is a fossil fuel less consumption of it will lead to resourceful use of it and making the country self-efficient by moving towards sustainable fuel like ethanol.

Question 12.
Chapattis made from wheat only swell up but bread becomes spongy, soft and easy to digest. Why is it so?
Answer:

  • The chapatti dough has water, which on heating converts into steam and tries to escape.
  • While doing so, it lifts up the upper layer of the chapatti. Therefore, the chapatti swells up.
  • Bread is made by adding yeast to the flour.
  • In the process of obtaining nutrition, the yeast cells convert the carbohydrates into alcohol and carbon dioxide.
  • When this dough is baked, the carbon dioxide escapes out making the bread spongy, soft and easy to digest.

Question 13.
Salt is applied on the inner surface of pickle jars and the pickle is covered with oil. Why is this done?
Answer:

  • Salt acts as a preservative. It prevents the growth of bacteria by forcing the microbes to lose water by osmosis. Hence, salt is applied on the inner surface of pickle jars.
  • Pickle is covered with oil as oil acts as preservative. It seals off the air from the item that is being pickled and provides an environment in which microbes cannot grow.

Question 14.
Which preservatives are mixed with ready to eat foods to prevent them from spoiling?
Answer:
Common salt, sugar, sodium benzoate, citric acid, sodium meta-bi-sulfite etc. are some 1 of the preservatives mixed with ready-to-eat foods to prevent them from spoiling.

Question 15.
Which plant and animal diseases are caused by micro-organisms and what are the 1 measures to be taken against them? Answer:
Plant diseases:

  • Citrus canker is a bacterial disease that affects 1 trees of citrus fruits.
  • Rust of wheat is a fungal disease that affects wheat crops.
  • Yellow vein mosaic is a viral disease which affects vegetables like bhindi (okra).

Preventive Measures:

  • Seeds which are healthy and disease-free should be selected for sowing.
  • Infected plants should be removed.
  • Plants should be sprayed with fungicides and germicides to prevent diseases.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Animal diseases:

  • Anthrax is a disease that affects cattle. It is caused by a bacterium.
  • Foot and mouth is a dangerous disease in cattle caused by a virus.
  • Rabies is a viral disease that affects animals.

Preventive Measures:

  • The place where animals are kept should be washed with germicides.
  • Animals should be dewormed regularly.
  • The animals should be treated with necessary antibiotics for infectious diseases.
  • They should be regularly vaccinated.
  • Take the animals to a veterinary hospital for proper treatment and vaccination.

Answer the following questions:

Question 1.
Bring ‘active dry yeast’ from the market. Mix a spoonful of yeast, two spoonfuls sugar with a sufficient quantity of lukewarm water in a bottle. Fix a colourless, transparent balloon on the mouth of that bottle.

What changes do you observe after 10 minutes? Mix limewater with the gas accumulated in the balloon. Collect that limewater in a beaker and observe it. What do you notice?
Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes 3
Answer:

  • After 10 minutes, the balloon is filled with a gas and gets inflated.
  • Lime water turns milky thus proving that the gas accumulated is carbon dioxide.

Class 9 Science Chapter 8 Useful and Harmful Microbes Additional Important Questions and Answers

Select the correct option:

Question 1.
The rod-shaped bacteria found in milk or buttermilk are called ……………………. .
(a) Rhizobium
(b) Clostridium
(c) Lactobacilli
(d) Saccharomyces
Answer:
(c) Lactobacilli

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Yoghurt has a specific sour taste due to ……………………. .
(a) lactic acid
(b) citric acid
(c) acetic acid
(d) alcohol
Answer:
(a) lactic acid

Question 3.
Bacteria found in the root nodules of leguminous plants are ……………………. .
(a) clostridium
(b) streptococcus
(c) Lactobacilli
(d) Rhizobium
Answer:
(d) Rhizobium

Question 4.
A mutually beneficial relationship is called ……………………. .
(a) symbiosis
(b) parasitism
(c) autotropism
(d) none of these
Answer:
(a) symbiosis

Question 5.
Carbon compounds obtained from bacteria and fungi for destroying or preventing the growth of harmful micro-organisms are called ……………………. .
(a) probiotics
(b) antibiotics
(c) antibodies
(d) antigens
Answer:
(b) antibiotics

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 6.
Fungi release ……………………. into the food, making the food poisonous.
(a) cyanotoxins
(b) dinotoxins
(c) mycotoxins
(d) cytotoxins
Answer:
(c) mycotoxins

Question 7.
……………………. produce bottle-shaped endospores in adverse conditions.
(a) Lactobacilli
(b) Clostridium
(c) Yeast
(d) Rhizobium
Answer:
(b) Clostridium

Question 8.
……………………. conducted important research on the toxin responsible for gas gangrene and the antitoxin responsible for treating it.
(a) Ida Bengston
(b) Van Ermengem
(c) Louis Pasteur
(d) Alexander Fleming
Answer:
(a) Ida Bengston

Question 9.
……………………. is a smokeless and high quality fuel.
(a) Methanol
(b) Ethanol
(c) Petrol
(d) Diesel
Answer:
(b) Ethanol

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 10.
Antibiotics mainly act against ……………………. .
(a) bacteria
(b) viruses
(c) algae
(d) fungi
Answer:
(a) Bacteria

Question 11.
……………………. is a broad-spectrum antibiotic.
(a) Gentamycin
(b) Penicillin
(c) Amoxicillin
(d) Erythromycin
Answer:
(c) Amoxicillin

Question 12.
……………………. is a narrow-spectrum antibiotic.
(a) Ampicillin
(b) Amoxicillin
(c) Tetracyclin
(d) Penicillin
Answer:
(d) Penicillin

Question 13.
Antibiotic penicillin was discovered by ……………………. .
(a) Louis Pasteur
(b) Alexander Fleming
(c) Ida Bengston
(d) Van Ermengem
Answer:
(b) Alexander Fleming

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 14.
……………………. proved that the anaerobic bacterium Clostridium botulinum is responsible for food poisoning.
(a) Louis Pasteur
(b) Ida Bengston
(c) Alexander Fleming
(d) Van Ermengem
Answer:
(d) Van Ermengem

Question 15.
AIDS is caused by ……………………. .
(a) virus
(b) bacteria
(c) protozoa
(d) fungi
Answer:
(a) Virus

Question 16.
Dengue is caused by ……………………. .
(a) droplets spread in air
(b) contact with infected person
(c) mosquito bite
(d) contaminated water and food
Answer:
(c)mosquitobite

Question 17.
Pneumonia is caused by ……………………. .
(a) virus
(b) bacteria
(c) fungi
(d) protozoa
Answer:
(b) bacteria

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 18.
……………………. can be prevented by vaccination.
(a) Malaria
(b) AIDS
(c) Leprosy
(d) Chicken pox
Answer:
(d) Chicken pox

Question 19.
Malaria is caused by ……………………. .
(a) protozoa
(b) bacteria
(c) fungi
(d) virus
Answer:
(a) protozoa

Question 20.
Bird flu (H7N9) and swine flu (H1N1) are caused by ……………………. .
(a) bacteria
(b) protozoa
(c) fungi
(d) virus
Answer:
(d) virus

Question 21.
The Lactobacilli convert lactose, the sugar in the milk, into ……………………. .
(a) lactic acid
(b) acetic acid
(c) alcohol
(d) citric acid
Answer:
(a) lactic acid

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 22.
The ……………………. destroys harmful microbes present in the milk.
(a) high pH
(b) neutral pH
(c) low pH
(d) none of these
Answer:
(c) low pH

Question 23.
Lactobacilli kill the harmful bacteria like ……………………. present in the alimentary canal.
(a) Rhizobium
(b) Saccharomyces
(c) Clostridium
(d) Alcanivorax
Answer:
(c) Clostridium

Question 24.
During fermentation, yeast cells convert carbohydrates into ……………………. .
(a) glucose and fructose
(b) alcohol and carbon dioxide
(c) proteins and fats
(d) fatty acids and amino acids
Answer:
(b) alcohol and carbon dioxide

Question 25.
Molasses is fermented with the help of yeast called ……………………. .
(a) Yarrowia lipolytica
(b) Alcanivorax
(c) Rhizobia
(d) Saccharomyces
Answer:
(d) Saccharomyces

Question 26.
A yeast ……………………. is used to absorb the toxins released during the production of palm oil.
(a) Yarrowia lipolytica
(b) Alcanivorax
(c) Saccharomyces cerevisiae
(d) Penicillium
Answer:
(a) Yarrowia lipolytica

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 27.
The bacteria which spoil cooked food are ……………………. .
(a) Saccharomyces
(b) Lactobacilli
(c) Clostridium
(d) Rhizobium
Answer:
(c) Clostridium

Question 28.
……………………. can grow and reproduce only in living cells.
(a) Bacteria
(b) Viruses
(c) Fungi
(d) Protozoa
Answer:
(b) Viruses

Find the odd man out:

Question 1.
AIDS, Hepatitis, Leprosy, Dengue.
Answer:
Leprosy. It is caused by bacteria, whereas the rest are caused by viruses.

Question 2.
Cholera, Leprosy, Pneumonia, Influenza.
Answer:
Influenza. It is caused by a virus, whereas the rest are caused by bacteria.

Question 3.
Ampicillin, Amoxycillin, Penicillin, Tetracycline.
Answer:
Penicillin. It is a narrow-spectrum antibiotic, whereas others are broad-spectrum antibiotics.

Question 4.
Tetracycline, Penicillin, Gentamycin, Erythromycin.
Answer:
Tetracycline. It is a broad-spectrum antibiotic, whereas others are narrow-spectrum antibiotics.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Complete the analogy:

Question 1.
(1) Dengue : Virus :: Malaria : …………………………. .
(2) Hepatitis : Virus :: Pneumonia : …………………………. .
(3) Cholera : Bacteria :: Swine flu : …………………………. .
(4) Swine flu : HJNJ : : Bird Flu : …………………………. .
(5) Measles : Virus :: Ringworm : …………………………. .
(6) Yoghurt: Lactobacilli : : Bread : …………………………. .
(7) Oil spills: Alcanivorax :: Absorption of arsenic : …………………………. .
(8) Rhizobium : Nitrogen fixation : : Clostridium : …………………………. .
Answer:
(1) Protozoa
(2) Bacteria
(3) Virus
(4) HyN9
(5) Fungi
(6) Yeast
(7) Saccharomyces cerevisiae
(8) Food poisoning.

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Leprosy (a) Virus
(2) Ringworm (b) Fungi
(3) Influenza (c) Protozoa
(4) Malaria (d) Bacteria

Answer:
(1 – d),
(2 – b),
(3 – a),
(4 – c)

State whether the following statements are true or false. Correct the false statements:

(1) Lactobacilli are aerobic bacteria.
(2) Lactobacilli converts lactose sugar into alcohol.
(3) Yoghurt has a specific sour taste due to acetic acid.
(4) The bacteria Clostridium are present in the root . nodules of leguminous plants.
(5) Yeast cell is a prokaryotic cell.
(6) The use of Rhizobium has helped to reduce the use of chemical fertilizers.
(7) Ethanol is a smokeless and high quality fuel.
(8) A yeast, Saccharomyces cerevisiae is used for absorbing toxins released during palm oil production.
(9) Gentamycin is a narrow-spectrum antibiotic.
(10) Antibiotics mainly act against bacteria. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
(11) Oil spills in oceans are cleared with the help of Clostridium bacteria.
(12) Tetracycline is a narrow-spectrum antibiotic.
(13) Amoxicillin is a broad-spectrum antibiotic.
(14) Penicillin is a group of antibiotics obtained from a fungus Saccharomyces.
(15) Antibiotic Penicillin was discovered by Alexander Fleming.
(16) The bacteria Lactobacilli cause food-poisoning.
(17) Clostridium bacteria grow in aerobic conditions.
(18) AIDS is caused by a virus.
(19) Influenza is caused by a bacteria.
(20) Antibiotics useful to one person can be suggested to others also.
(21) Dengue is caused by a bacteria.
(22) Dandruff and ringworm are caused by fungi.
(23) Pneumonia is spread through droplets spread in air by infected person.
(24) Chicken pox spread due to contaminated food and water.
(25) Ida Bengston was honoured with the Typhoid Medal’ in 1947.
Answer:
(1) False. Lactobacilli are anaerobic bacteria.
(2) False. The Lactobacilli converts lactose sugar into lactic acid.
(3) False. Yogurt has a specific sour taste due to lactic acid.
(4) False. The bacteria Rhizobium are present in the root nodules of leguminous plants.
(5) False. Yeast cell is a eukaryotic cell.
(6) True. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
(7) True.
(8) False. A yeast, Yarrowia lipolytic is used to absorb the toxins released during the production of palm oil.
(9) True.
(10) True.
(11) False. Oil spills in oceans are cleared with the help of Alcanivorax bacteria.
(12) False. Tetracycline is a broad-spectrum antibiotic.
(13) True.
(14) False. Penicillin is a group of antibiotics obtained from a fungus Penicillium.
(15) True.
(16) False. The bacteria Clostridium cause food? poisoning.
(17) False. Clostridium bacteria grow in anaerobic conditions.
(18) True.
(19) False. Influenza is caused by a virus.
(20) False. Antibiotics useful to one person cannot be suggested to others as different diseases require different antibiotics.
(21) False. Dengue is caused by a virus.
(22) True.
(23) True.
(24) False. Chicken pox spread due to contact with infected person.
(25) False. Ida Bengston was honoured with the ‘Typhus Medal’ in 1947.

Complete the statements using the proper option from those given below. Explain the statements: (mycotoxins, budding, Rhizobium, molasses, endospores, broad-spectrum, Lactobacilli)

Question 1.
Lactobacilli bacteria are used for making yoghurt.
Answer:
The lactobacilli convert lactose, the sugar in the milk, into lactic acid. As a result, the pH of milk decreases causing a coagulation of milk proteins. Milk changes into yogurt.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
The use of Rhizobium has helped to reduce the use of chemical fertilizers.
Answer:
Rhizobium bacteria are found in the root nodules of leguminous plants. They help to convert atmospheric nitrogen into nitrogen compounds and provide it to the plants. This helps to reduce the use of chemical fertilizers and their adverse effects.

Question 3.
Amoxicillin is a broad-spectrum antibiotic.
Answer:
This antibiotic is useful against a wide variety of bacteria. It is used against pathogens which cannot be identified during symptoms of a disease.

Question 4.
Ethanol is produced by the fermentation of molasses.
Answer:
Molasses is produced from sugarcane juice. It is rich in carbohydrates. When it is fermented with the help of the yeast called Saccharomyces, ethanol (C2H5OH) is produced.

Question 5.
Clostridium bacteria produce bottle-shaped endospores.
Answer:
These endospores help them to survive in adverse conditions.

Give scientific reasons:

Question 1.
Lactobacilli are used for making yoghurt from milk.
Answer:

  • Lactobacilli convert lactose, the sugar in the milk, into lactic acid. This process is called fermentation.
  • As a result, the pH of milk decreases causing coagulation of milk proteins.
  • Thus, milk proteins are separated from other constituents of milk and milk changes into yoghurt.
  • Yoghurt has a specific sour taste due to lactic acid. The low pH destroys harmful microbes present in the milk. Therefore, Lactobacilli are used for making yoghurt from milk.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Antibiotics should be taken only when prescribed by a doctor.
Answer:

  • Antibiotics are a group of medicines used to kill disease-causing bacteria and certain protozoa.
  • The doctor selects and prescribes the antibiotic best suited for our disease.
  • If taken in extra dose, they can kill the useful bacteria present in our body.
  • If the course of antibiotics is not completed, the bacteria develop resistance to that antibiotic making it ineffective.
  • Therefore, antibiotics should be taken only when prescribed by a doctor.

Question 3.
Nowadays, seeds are coated with Rhizobial solution or powder before sowing.
Answer:

  • When seeds coated with Rhizobial solution or powder are sown, Rhizobia enter the plantlets.
  • This is called Rhizobial inoculation.
  • Rhizobia can produce nitrogenous compounds from atmospheric nitrogen.
  • This experiment has helped in the supply of nitrogen to cereal and other crops, besides leguminous crops.
  • Therefore, nowadays seeds are coated with Rhizobial solution or powder before sowing.

Question 4.
Antibiotics are not effective against common cold or influenza.
Answer:

  • Antibiotics are a group of medicines used to control inflections caused by bacteria.
  • Common cold or influenza is caused by a virus.
  • Antibiotics are not effective against viruses.
  • Therefore, antibiotics are not effective against common cold or influenza.

Question 5.
Cotton fabrics, gunny bags, leather items and wooden items do not last long.
Answer:

  • Microscopic spores of fungi are present in the air.
  • If there is sufficient moisture, spores germinate on cotton fabric, gunny bags, leather, wooden items etc.
  • The fungal hyphae (fibres of the fungus) penetrate deep into the material to obtain nutrition and to reproduce.
  • This causes the materials to wear and become weak.
  • As a result, cotton fabric, gunny bags, leather and wooden items do not last long.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 6.
Food on which fungi has grown cannot be eaten.
Answer:

  • Various species of fungi grow on food items like pickles, murabba, jam, sauce, chutney etc.
  • They use the nutrients in these food items for growth and reproduction.
  • During this activity, fungi release mycotoxins, certain poisonous chemicals, into the food and thus food becomes poisonous.
  • Hence, the food on which fungi have grown cannot be eaten.

Write short notes:

Question 1.
Rhizobial inoculation.
Answer:

  • Nowadays, seeds are coated with rhizobial solution or powder before sowing.
  • After sowing, Rhizobia enter the plantlets.
  • This is called Rhizobial inoculation.
  • This experiment has helped in the supply of nitrogen to cereal and other crops, besides leguminous crops.

Question 2.
Bio-remediation.
Answer:

  • Bio-remediation is a technique that involves the use of organisms to break down environmental pollutants.
  • Generally, fungi like yeast and bacteria are used for bio-remediation.
  • A yeast, Yarrowia lipolytica is used to absorb the toxins released during the production of palm oil and the heavy metals and minerals released in some other industrial processes.
  • Saccharomyces cerevisiae is used for absorption of a pollutant, arsenic.
  • Oil spills in oceans are cleaned with the help of Alcanivorax bacteria.

Question 4.
Clostridium.
Answer:

  • Clostridium are the bacteria that spoil food.
  • Out of about 100 different species of this bacterium, some are free living in the soil whereas some live in the alimentary canals of humans and other animals.
  • These bacteria are rod-shaped and produce bottle-shaped endospores in adverse conditions.
  • One special characteristic of these bacteria is that they cannot withstand the normal oxygen level of the air because they grow in anaerobic conditions.

Write down the mode of infection and preventive measures for the following:

Question 1.
Write down the causative pathogen, mode of infection and preventive measures of AIDS.
Answer:

  • Causative Pathogen: Virus.
  • Mode of infection: Through blood and semen of infected person and milk of mother suffering from AIDS.
  • Preventive measure: Safe sexual contact, avoid resuse of needles and injections.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Write down the modes of infection and preventive measures against Bird Flu (H7N9) and Swine Flu (HjN.,).
Answer:

  • Mode of infection: Contact with infected birds and animals.
  • Preventive measure: Personal hygiene, properly cooked meat.

Question 3.
Write down the modes of infection and preventive measures against Malaria and dengue.
Answer:

  • Mode of infection: Mosquito bite, unclean surroundings.
  • Preventive measure: Cleanliness of surroundings, preventing stagnation of water, controlling mosquitoes.

Question 4.
Write down the modes of infection and preventive measures against Pneumonia.
Answer:

  • Mode of infection: Droplets spread in air by infected person.
  • Preventive measure: Vaccination, avoiding contact with infected person.

Question 5.
Write down the modes of infection and preventive measures for leprosy.
Answer:

  • Mode of infection: Long term contact with infected person.
  • Preventive measure: Avoiding contact with infected persons and their belongings.

Question 5.
What are the modes of infection and preventive measures for Hepatitis?
Answer:

  • Mode of infection: Contaminated water and food.
  • Preventive measure: Use clean and filtered water, proper storage of food.

Question 6.
What are the modes of infection and preventive measures for Influenza.
Answer:

  • Mode of infection: Contact with infected person.
  • Preventive measure: Personal hygiene and avoiding contact with infected person.

Answer the following questions:

Question 1.
How is bread made?
Answer:

  • Bread is made using flour, yeast, salt and water. The yeast uses sugar as food.
  • In the process of obtaining nutrition, yeast cells convert the carbohydrates into alcohol and carbon dioxide. This process is called fermentation. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • The bubbles of carbon dioxide given off cause the dough to rise.
  • This dough can be used to make bread.
  • When this dough is baked, more bubbles of carbon dioxide reformed due to heat. As the gas escapes, the bread rises and becomes soft and fluffy.

Question 2.
What is the advantage of Rhizobium to farmers?
Answer:

  • The use of Rhizobium has helped to reduce the use of chemical fertilizers and thereby their adverse effects.
  • It has also helped to reduce expenses on fertilizers and thus benefited the farmers.

Question 3.
How can we observe Lactobacilli in buttermilk?
Answer:

  • Smear a drop of fresh buttermilk on a glass slide.
  • Stain it with methylene blue and put a coverslip over it.
  • Observe the smear under the 10X objective of a compound microscope and then with the more powerful 60X objective.
  • The blue rod-shaped organisms moving about are Lactobacilli.

Question 4.
What is symbiosis? Give example.
Answer:

  • Symbiosis is a mutually beneficial relationship.
  • Example: Rhizobium living in root nodules of leguminous plants supply nitrates, nitrites and amino acids to that plant and in exchange get energy in the form of carbohydrates.

Complete the following table:

Question 1.
The spread and prevention of disease
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes 1
Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes 2

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Different species of Clostridium bacteria and the diseases caused by them.
Answer:

Species of Clostridium Diseases
Clostridium tetani Tetanus
Clostridium perfringens Food poisoning
Clostridium botulinum Botulism (Paralysis of muscles)
Clostridium difficile Colitis (Infection of the intestine)

Answer in detail:

Question 1.
How is alcohol produced?
Answer:

  • Alcohol is often produced along with sugar in sugar factories.
  • Molasses is produced from sugarcane juice. It is rich in carbohydrates.
  • Molasses is fermented with the help of the yeast Saccharomyces.
  • In this process, ethanol (C2H5OH) alcohol is produced as a primary product and ester and other alcohols are produced as secondary products.
  • Besides molasses, maize, barley and other grains are also used for industrial production of alcohol.
  • Glucose and fructose, the sugars present in grape juice are also fermented with the help of yeast to produce alcohol which is used to make wines.

Question 2.
Give the uses of Lactobacilli.
Answer:
Uses of Lactobacilli:

  • Various milk products like yoghurt, buttermilk, ghee, cheese, shrikhand, etc. can be obtained by fermentation of milk.
  • Lactobacilli fermentation is useful for large scale production of cider, cocoa, pickles of vegetables etc.
  • Lactobacilli and some other useful microbes taken together are used to treat abdominal discomfort.
  • Leavened fodder offered to domestic cattle like cows and buffaloes is fodder fermented with the help of lactobacilli.
  • The Lactobacilli fermentation process is used to make wine and some types of bread.

Question 3.
What is Penicillin? What is it used for?
Answer:

  • Penicillin is a group of antibiotics obtained from a fungus, penicillium.
  • It is used for controlling the infections caused by bacteria like Staphylococci, Clostridia, Streptococci, etc.
  • Medicines containing Penicillin are useful to treat certain bacterial infections of the ear, nose, throat and skin as well as diseases like Pneumonia and scarlet fever.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 4.
How was the antibiotic penicillin discovered?
Answer:

  1. Alexander Fleming, a professor of microbiology at St. Mary’s Hospital had cultured varieties of bacteria and fungi in petri dishes in his laboratory.
  2. On 3rd September 1928, while observing Staphylococci cultures, he made an interesting observation in one petri dish.
  3. In that petri dish, fungal colonies had grown but the area around those colonies was clean and clear, i.e. the bacteria had actually been destroyed.
  4. After further studies, he confirmed that the fungus growing there was Penicillium and its secretion had destroyed the bacterial colonies.
  5. Thus, the first antibiotic – penicillin had been discovered accidentlly and this formed the basis to find cures for incurable diseases.

Question 5.
What are the precautions to be followed while taking antibiotics?
Answer:

  • Antibiotics should be taken only when prescribed by a doctor.
  • Don’t purchase any antibiotic from medical stores without a prescription from a doctor.
  • Don’t consume antibiotics on your own to treat common diseases like a throat infection, common cold or influenza.
  • Even if you feel well before completing of the prescribed course of the antibiotics, you must continue and complete it.
  • Don’t suggest to others the antibiotics which were useful to you.

Question 6.
How can we observe Rhizobium bacteria in the roots of leguminous plant?
Answer:

  • Take a plantlet of fenugreek, groundnut or any other bean and sterilize it with a 3 to 5% solution of hydrogen peroxide. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Afterwards, keep it in a 70% solution of ethyl alcohol for 4 to 5 minutes.
  • Clean the roots with sterile water and take thin sections of the root nodules.
  • Select a good section and place it an a solution of saffranin for 2 to 3 minutes.
  • Place the stained section on a glass slide, cover it with a coverslip and observe it under the compound microscope. The pinkish rod-shaped organisms are the Rhizobium bacteria.

9th Std Science Questions And Answers:

Measurement of Matter Class 9 Science Chapter 4 Questions And Answers Maharashtra Board

Class 9 Science Chapter 4

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 4 Measurement of Matter Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 4 Measurement of Matter Question Answer Maharashtra Board

Class 9 Science Chapter 4 Measurement of Matter Question Answer Maharashtra Board

1. Give examples.

a. Positive radicals
Answer:
Na+– Sodium ion, K+ – Potassium ion

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

b. Basic radicals
Answer:
Na+ – Sodium ion, K+ – Potassium ion, Ag+ – Silver ion

c. Composite radicals
Answer:
\(\mathrm{SO}_{4}^{2-}, \mathrm{NH}_{4}^{+}\)

d. Metals with variable valency
Answer:
(a) Iron (Ferrum)
(i) Fe2+ – Ferrous [Iron – II]
(ii) Fe3+ – Ferric [Iron – III]

(b) Copper (Cuprum)
(i) Cu+ – Cuprous [Copper -1]
(ii) Cu2+ – Cupric [Copper – II]

(c) Mercury (Hydragyrum)
(i) Hg+ – Mercurous [Mercury -1]
(ii) Hg2+ – Mercuric [Mercury – II]

e. Bivalent acidic radicals
Answer:
O2- – Oxide, S2- – Sulphide, \(\mathrm{CO}_{3}^{2-}\) – Carbonate

f. Trivalent basic radicals
Answer:
Al3+ – Aluminium, Cr3+ – Chromium, Fe3+ – Ferric.

2. Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.
Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 24

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

3. Write the steps in deducing the chemical formulae of the following compounds.
Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide
Answer:
In order to write the chemical formulae of compounds, it is necessary to know the symbols and valency of various radicals.

1. Sodium Sulphate:
Step – 1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
\(\mathrm{Na} \quad \mathrm{SO}_{4}\)
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 40
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 41
Step – 4 : To write down the chemical formula of the compound.
Na2 SO4
(Sodium sulphate)

2. Potassium Nitrate:
Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
K NO3
Step – 2 : To write the valency below the respective radical.
\(\begin{array}{cc}
\mathrm{K} & \mathrm{NO}_{3} \\
1 & 1
\end{array}\)
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 42
Step – 4 : To write down the chemical formula of the compound.
KNO3
(Potassium nitrate)

3. Ferric phosphate:
Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
Fe PO4
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 43
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 44
Step – 4 : To write down the chemical formula of the compound.
FePO4
(Ferric phosphate)

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

4. Calcium oxide:
Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radicals on the right)
Ca O
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 60
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 45
Step – 4 : To write down the chemical formula of the compound.
CaO
(Calcium oxide)

5. Aluminium hydroxide:
Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radical on the right)
Al OH
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 61
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 46
Step – 4 : To write down the chemical formula of the compound.
Al(OH)3
(Aluminium hydroxide)

6. Calcium carbonate:
Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radicals on the right)
Ca CO3
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 47
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 48
Step – 4 : To write down the chemical formula of the compound.
CaCO3
(Calcium Carbonate)

7. Sodium dichromate:
Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radical on the right)
Na Cr2O7
Step – 2 : To write the valency below the respective radical.
\(\begin{array}{cc}
\mathrm{Na} & \mathrm{Cr}_{2} \mathrm{O}_{7} \\
1 & 2
\end{array}\)
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 49
Step – 4 : To write down the chemical formula of the compound.
Na2Cr2O7
(Sodium dichromate)

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

4. Write answers to the following questions and explain your answers.

a. Explain how the element sodium is monovalent.

Answer:

  1. The number of protons or electrons (atomic number) in Sodium (Na) atom is 11. Therefore the electronic configuration of sodium atom is (2, 8,1).
  2. In chemical reaction, sodium atom has the capacity to give away le_ from its outermost orbit to form Na+ ion with stable electronic configuration (2, 8).
  3. As sodium atom gives away le- and a cation of sodium is formed, hence the valency of sodium is 1 and therefore, the element sodium is monovalent.

b. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate.
Answer:
M is a bivalent metal. Following are the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate:

(i) Compound of metal ‘M’ with radical sulphate
Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
M SO4
Step – 2: To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 21
Step – 3: To cross multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 21
Step – 4: To write down the chemical formula of the compound.
M SO4

(ii) Compound of metal ‘M’ with radical phosphate.
Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
M PO4
Step – 2: To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 23
Step – 3: To cross multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 51
Step – 4: To write down the chemical formula of the compound.
M3 (PO4)2

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

c. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.
Answer:

  • The mass of an atom is concentrated in its nucleus and it is due to the protons (p) and neutrons (n) in it.
  • Since an atom is very very tiny, it was not possible to measure atomic mass accurately. Therefore, the concept of relative mass of an atom was formed.
  • To express relative mass of an atom, reference of atom is considered. The two reference atoms were as follows:

(a) Hydrogen (H) atom: The hydrogen atom is the lightest. The relative mass of a hydrogen atom is 1 which has only 1 proton in its nucleus. On this scale, the relative atomic mass of many elements comes out to be fractional. Therefore, carbon was selected as a reference atom.

(b) Carbon (C) atom: The carbon atom is selected as reference atom. In this scale, the relative mass of a carbon atom is accepted as 12.

  • The relative atomic mass of 1 hydrogen (H) atom compared to the carbon (C) atom becomes

d. What is meant by Unified Atomic Mass.
Answer:

  • During earlier time, relative mass of an atom was considered for measuring the mass of an atom directly. But since the founding of unified mass, relative mass is not accepted henceforth.
  • Unified atomic mass is the unit of atomic mass called as Dalton.
  • Its symbol is ‘u’. lu = 1.66053904 x 10-27 kg.

e. Explain with examples what is meant by a ‘mole’ of a substance.
Answer:

  • A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.
  • For example: Atomic mass of oxygen atom (O) is 16u. Thus, the molecular mass of oxygen molecule (O2) is 16 x 2 = 32u. Therefore, 32 g of oxygen is 1 mole of oxygen.

5. Write the names of the following compounds and deduce their molecular masses.
Na2SO4, K2CO3, CO2, MgCl2, NaOH, AlPO4, NaHCO3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 5

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

6. Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:
‘sample m’ mass : 7g
Mass of constituent oxygen : 2g
Mass of constituent calcium : 5g
‘sample n’ mass : 1.4g
Mass of constituent oxygen : 0.4g
Mass of constituent calcium : 1.0g

Which law of chemical combination does this prove? Explain.
Answer:
(i) The expected proportion by weight of the constituent elements of quick lime that is calcium oxide would be from its known molecular formula CaO. The atomic mass of Ca and O are 40 and 16 respectively. This means, the proportion by weight of the constituent elements Ca and O in the compound CaO is 40 :16 which is 5 : 2.

(ii) Now, for the given sample’m’ of CaO = 5 g
mass of given sample = 7 g
mass of constituent Ca in sample’m’ = 5 g
mass of constituent O in sample’m’ = 2 g

(iii) This means that 7 g of calcium oxide contairis 5 g of calcium (Ca) and 2 g of oxygen (O); apd the proportion by weight of calcium and oxygen in it is 5 : 2.

(iv) Now, for the given sample ‘n’ of CaO mass of given sample CaO = 1.4 g
Mass of constituent Ca in sample ‘n’ = 1.0 g
Mass of constituent O in sample ‘n’ = 0.4 g
This means that 1.4g of calcium oxide contains 1.0 g of calcium (Ca) and 0.4 g of oxygen (O); and the proportion by weight of calcium and oxygen in it is 5 : 2.

(v) Above samples’m’ and ‘n’ of calcium oxide (CaO) shows that the proportion by weight of the constituent elements in different samples of a compound is always constant that is the proportion by weight of calcium (Ca) and oxygen (O) in different samples of calcium oxide (CaO) is constant.

(vi) The experimental value of proportion by weight of the constituent elements matched with the expected proportion calculated by molecular mass. This proves and verifies the law of constant proportion.

The law states that ‘The proportion by weight of the constituent elements in the various samples of a compound is fixed’.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

7. Deduce the number of molecules of the following compounds in the given quantities.
32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine.

Class 9 Science Chapter 4 Measurement Of Matter Notes Question 1.
32g oxygen
Answer:
Given : Mass of oxygen (O2) m = 32g
To find : Number of molecules in 32g of oxygen.
Solution : Atomic mass of oxygen (O) = 16
∴ Molecular mass of oxygen (O2) M = 16 x 2 = 32
According to the formula, Number of moles in the given O2 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 12
1 mol of O2 contains 6.022 x 1023 molecules that is 32 g of 02 contains 6.022 * 1023 molecules of O2.
32g of oxygen contains 6.022 x 1023 molecules of oxygen.

Class 9 Science Chapter 4 Measurement Of Matter Answers Question 2.
90g water
Answer:
Given : Mass of water (H2O) m = 90g.
To find : Number of molecules in 90g of water.
Solution : Molecular mass of (H2O) M = (Atomic mass of H) x 2 + (Atomic mass of O) x 1
∴ Molecular mass of (H2O) M = 1 x 2 +16
∴ Molecular mass of (H2O) M = 18
According to the formula,
Number of moles in the given H2O (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 13
1 mol of H2O contains 6.022 x 1023 molecules.
5 mol of H2O contains 5 x 6.022 x 1023 molecules. = 30.11 x 1023 molecules, that is 90g of H2O contains 30.11 x 1023 molecules of H20.
90g of water contains 30.11 x 1023 molecules of water.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

4 Measurement Of Matter Exercise Question 3.
8.8g carbon dioxide
Answer:
Given : Mass of Carbon dioxide (CO2)m = 8.8g.
To find : Number of molecules in 8.8g of carbon dioxide.
Solution : Molecular mass of (CO2)M = (Atomic mass of C) x 1 + (Atomic mass of O) x 2
∴ Molecular mass of (CO2)M = 12 x 1 + 16 x 2 = 12 + 32
Molecular mass of (CO2)M = 44
According to the formula, Number of moles in the given CO2 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 14
∴ 1 mol of CO2 contains 6.022 x 1023 molecules.
∴ 0.2 mol of CO2 contains 0.2 x 6.022 x 1023 molecules.
= 1.2044 x 1023 molecules,
that is 8.8g of CO2 contains 1.2044 x 1023 molecules of CO2.
8.8g of CO2 contains 1.2044 x 1023 molecules of CO2.

Class 9 Science Solutions Maharashtra Board Question 4.
7.1g chlorine
Answer:
Given : Mass of Chlorine (Cl2)m = 7.1g.
To find : Number of molecules in 7.1g of chlorine.
Solution : Atomic mass of (Cl) = 35.5
∴ Molecular mass of chlorine (Cl2)M = 35.5 x 2 = 71
According to the formula, Number of moles in the given Cl2 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 15
∴ 1 mol of Cl2 contains 6.022 x 1023 molecules.
∴ 0.1 mol of Cl2 contains 0.1 x 6.022 x 1023 molecules.
= 0.6022 x 1023 molecules,
that is 7.1g of Cl2 contains 0.6022 x 1023 molecules of Cl2.
7.1g of Cl2 contains 0.6022 x 1023 molecules of chlorine.

8. If 0.2 mol of the following substances are required how many grams of those substances should be taken? Sodium chloride, magnesium oxide, calcium carbonate
Answer:
Given : Number of moles of sodium chloride (NaCl) n = 0.2 mol
To find : Mass in grams of 0.2 mol of NaCl
Solution:
Molecular mass of (NaCl)M = (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1
= 23 x 1 + 35.5 x 1
= 23 + 35.5
Molecular mass of (NaCl)M = 58.5
According to the formula,
Number of moles in the given NaCl (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 16
Mass of NaCl in grams (m) = 0.2 x 58.5
Mass of NaCl in grams (m) = 11.7 g
Mass of 0.2 mole of NaCl is 11.7g

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Maharashtra State Board Class 9 Science Solutions Question 1.
What is the type of chemical bond in NaCl and MgCl2?
Answer:

  • The type of chemical bond in NaCl and MgCl2 is ionic bond.

9th Class Science Chapter 4 Measurement Of Matter Question 2.
Determine the valencies of H, Cl, O and Na from the molecular formulae H2, HC1, H2O and NaCl.
Answer:
(i) In the molecular formula HCl
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 3
∴ The valency of H is 1 and Cl is 1.

(ii) In the molecular formula H2O
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 2
∴ The valency of H is 1 and O is 2.

(iii) In the molecular formula NaCl
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 50
∴ The valency of Na is 1 and Cl is 1.
∴ From all the above, the valencies of the given elements are as follows : H = 1, Cl = 1, O = 2 and Na = l.

Measurement Of Matter Class 9 Exercise Answers Question 3.
How is an element indicated in Chemistry?
Answer:
In chemistry an element is indicated by its symbol.

Question 4.
Write down the symbols of the elements you know.
Answer:
Symbols of some elements are

  • Hydrogen – H
  • Helium – He
  • Boron – B
  • Carbon – C
  • Aluminium – A1

Question 5.
Write down the symbols for the following elements.
Antimony, Iron, Gold, Silver, Mercury, Lead, Sodium
Answer:
The symbols of given elements are as follows:

  • Antimony – Sb
  • Iron – Fe
  • Gold – Au
  • Silver – Ag
  • Mercury – Hg
  • Lead – Pb
  • Sodium – Na

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Following are atomic masses of a few elements in Daltons and the molecular formulae of some compounds. Deduce the molecular masses of those compounds:

Atomic masses – H(l), 0(16), N(14), C(12), K(39), S(32) Ca(40), Na(23), Cl(35.5), Mg(24), Al(27)

Question 1.
Molecular formula – NaCl
Answer:
Molecular mass of NaCl (M)
= (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1
= (23 x 1) + (35.5 x 1)
= 23 + 35.5
= 58.5
∴ Molecular mass of NaCl (M) = 58.5

Question 2.
Molecular formula – MgCl2
Answer:
Molecular mass of MgCl2 (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of Cl) x 2
= (24 x 1) + (35.5 x 2)
= 24 + 71
= 95
∴ Molecular mass of MgCl2 (M) = 95?

Question 3.
Molecular formula – KNO3
Answer:
Molecular mass of KNO3 (M)
= (Atomic mass of K) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (39 x 1) + (14 x 1) + (16 x 3)
= 39 + 14 + 48
= 101
Molecular mass of KNO3 (M) = 101

Question 4.
Molecular formula – H2O2
Answer:
Molecular mass of H2O2 (M)
= (Atomic mass of H) x 2 + (Atomic mass of O) x 2
= (1 x 2) + (16 x 2)
= 2 + 32
= 34
∴ Molecular mass of H2O2 (M) = 34.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 5.
Molecular formula – A1C13
Answer:
Molecular mass of A1C13 (M)
= (Atomic mass of Al) x 1 + (Atomic mass of Cl) x 3
= (27 x 1) + (35.5 x 3)
= 27 + 106.5
= 133.5
∴ Molecular mass of A1C13 (M) = 133.5

Question 6.
Molecular formula – Ca(OH)2
Answer:
Molecular mass of Ca(OH)2 (M)
= (Atomic mass of Ca) x 1 + (Atomic mass of O + Atomic Mass of H) x 2
= (40 x 1) + (16 + 1) x 2
= 40 + (17 x 2)
= 40 + 34
= 74
∴ Molecular mass of Ca(OH)2 (M)
= 74

Question 7.
Molecular formula – MgO
Answer:
Molecular mass of MgO (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of 0)xl
= (24 x 1) + (16 x 1)
= 24 + 16
= 40
Molecular mass of MgO (M) = 40

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 8.
Molecular formula – H2S04
Answer:
Molecular mass of H2S04 (M)
= (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
= (1 x 2) + (32xl) + (16×4)
= 2 + 32 + 64
= 98
Molecular mass of H2S04 (M) = 98

Question 9.
Molecular formula – HN03
Answer:
Molecular mass of HN03 (M)
= (Atomic mass of H) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (lxl)+ (14xl)+ (16×3)
= 1 + 14 + 48
= 63
Molecular mass of HNOs (M) = 63

Question 10.
Molecular formula – NaOH
Answer:
Molecular mass of NaOH (M)
= (Atomic mass of Na) x 1 + (Atomic mass of O) x 1 + (Atomic mass of H) x 1
= (23 x 1) + (16 x 1) + (l x l)
= 23 + 16 + 1
= 40
Molecular mass of NaOH (M) = 40

Question 11.
How many molecules of water are there in 36 g water?
Answer:
Given : Mass of water (H2O) m = 36g
To find : Number of molecules in 36g of water
Solution :
Molecular mass of (H2O) M = (Atomic mass of H) x 2 + (Atomic mass of O) x 1 Molecular mass of (H2O) M
= (1 x 2) + 16 x 1
Molecular mass of (H2O) M = 18
According to the formula,
Number of moles in the given H2O (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 19
1 mol of H2O contains 6.022 x 1023 molecules.
∴ 2 mol of H2O contains 2 x 6.022 x 1023 molecules.
= 12.044 x 1023 molecules, that is 36g of H2O contains 12.044 x 1023 molecules of H2O.
36 g of water contains 12.044 x 1023 molecules of water.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 12.
How many molecules of H2S04 are there in a 49 g sample?
Answer:
Given : Mass of Sulphuric acid (H2SO4) m = 49g
To find : Number of molecules in 49g of H2SO4
Solution:
Molecular mass of (H2SO4) M = (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
Molecular mass of (H2SO4)M = (1 x 2) + (32 x 1) + (16 x 4)
= 2 + 32 + 64
= 98.
According to the formula,
Number of moles in the given H2SO4 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 20
∴ 1 mol of H2SO4 contains 6.022 x 1023 molecules.
∴ 0.5 mol of H2SO4 contains 0.5 x 6.022 x 1023 molecules.
= 3.011 x 1023 molecules,
that is 49g of H2SO4 contains 3.011 x 1023 molecules of H2SO4.
49 g of Sulphuric acid contains 3.011 x 1023 molecules of H2SO4.

Question 13.
Fill the following tables.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 25

Question 14.
Complete the following chart.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 27

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 15.
The relative atomic masses of some elements in the chart below are given. You have to find the relative atomic masses of the others.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 28

Question 16.
Classify the following radicals into simple radicals and composite radicals: (Use your brain power;
\(\begin{array}{l}
\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cl}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{Fe}^{2+}, \mathrm{ClO}_{3}^{-}, \mathrm{NH}_{4}^{+}, \mathrm{Br}^{-} \\
\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Cu}^{+}
\end{array}\)
Answer:

Simple radicals Composite radicals
Ag+ \(\mathrm{SO}_{4}^{2-}\)
Mg2+ \(\mathrm{ClO}_{3}^{-}\)
Cl \(\mathrm{NH}_{4}^{+}\)
Fe2+ \(\mathrm{NO}_{3}^{-}\)
Br
Na+
Cu+

Question 17.
Which are the basic radicals and which are the acidic radicals among the following?
\(\begin{array}{l}
\mathrm{Ag}^{+}, \mathrm{Cu}^{2+}, \mathrm{Cl}^{+}, \mathrm{I}, \mathrm{SO}_{4}^{2-}, \mathrm{Fe}^{3+}, \mathrm{Ca}^{2+}, \mathrm{NO}_{3} ; \mathrm{S}^{2}, \mathrm{NH}_{4}^{+} \\
\mathrm{K}^{+}, \mathrm{MnO}_{4}, \mathrm{Na}^{+}
\end{array}\)
Answer:

Basic Radical Acidic Radical
(i)Ag+ (i) Cl
(ii) Cu2+ (ii) I
(iii) Fe3+ \(\text { (iii) } \mathrm{SO}_{4}^{2-}\)
(iv) Ca2+ \(\text { (iv) } \mathrm{NO}_{3}^{-}\)
Wnh; \(\text { (v) } \mathrm{S}^{2-}\)
(vi) K+ \(\text { (vi) } \mathrm{MnO}_{4}^{-}\)
(vii) Na+

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Give examples:

Question 1.
Make a list of elements in the monoatomic and in the diatomic molecular state. (Make a list and discuss;
Answer:

  • Elements in the monoatomic molecular state are: Helium (He), Neon (Ne), Argon (Ar), Sodium (Na), Copper (Cu),
  • Elements in the diatomic molecular state are:
    Oxygen (O2), Nitrogen (N2), Hydrogen (H2), Chlorine (Cl2), Fluorine (F2).

Problem-based questions

Answer the following questions:

Question 1.
Is it possible to weigh one molecule using a weighing balance?
Answer:
No, it is not possible to weigh one molecule using a weighing balance.

Question 2.
Will the number of molecules be the same in equal weights of different substances?
Answer:
No, the number of molecules will not be the same in equal weights of different substances.

Question 3.
If we want equal number of molecules of different substances, will it work to take equal weights of those substances.
Answer:
No, if we want equal number of molecules of different substances, it will not work to take equal weights of those substances.

Answer the following:

Question 1.
What is the Dalton’s atomic theory?
Answer:
Dalton’s Atomic theory-

  • All matter is made of atoms. Atoms are indivisible and indestructible.
  • All atoms of a given element are identical in mass and properties.
  • Compounds are formed by a combination of two or more different kinds of atoms.
  • A chemical reaction is a rearrangement of atoms.

Question 2.
How are compounds formed?
Answer:
Compounds are formed by a chemical combination of two or more different kinds of atoms.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 3.
What are the molecular formulae of salt, slaked lime, water, lime, limestone?
Answer:
The molecular formulae for
Salt – Sodium chloride – NaCl
Slaked lime – Calcium hydroxide Ca(OH)2
Water – H2O
Lime – Calcium oxide – CaO
Lime stone – Calcium carbonate – CaCO3

Question 4.
From which experiments was it discovered that atoms have an internal structure? When?
Answer:

  • In 1911, Earnest Rutherford conducted a well known experiment called as ‘Gold foil experiment’.
  • From this experiment it was discovered that atoms have internal structure.

Question 5.
What are the two parts of an atom? What are they made up of?
Answer:
The two parts of atoms are nucleus and extra nuclear part. Nucleus is made up of positively charged protons and electrically neutral neutrons and the extra nuclear part is made up of negatively charged electrons revolving around the nucleus in different orbits.

Open-ended questions

Q.3. 2. Answer the following questions:

Question 1.
How will the compounds, MgCl2 and CaO be formed from their elements?
Answer:
(1) Magnesium Chloride (MgCl2)
Magnesium atom (Mg): Electronic configuration
\((2,8,2) \stackrel{-2 e^{-}}{\longrightarrow}\) Magnesium ion Mg2+ (2,8).
Chlorine atom (Cl). Electronic configuration \((2,8,7) \stackrel{+1 e^{-}}{\longrightarrow}\) Chloride ion Ch (2,8,8).
∴ Mg2+ + 2CT → MgCl2 (Magnesium Chloride)

  • A Magnesium atom gives away 2e and a cation of Magnesium (Mg2+) is formed, hence, the valency of magnesium is two.
  • Two chlorine atoms takes le each and forms two anions of chlorine (2Cl) (chloride), and thus, the valency of chlorine is one.
  • After the give and take of electrons is over, the electronic configuration of all the resulting ions has a complete octet.
  • Due to the attraction between the unit but opposite charges on all the ions, one chemical bond known as ionic bond is formed between Mg2+ and 2C1 each and the compound MgCl2 is formed.

(2) Calcium Oxide (CaO)
Calcium atom (Ca): Electronic configuration
\((2,8,8,2) \stackrel{-2 e^{-}}{\longrightarrow}\) Calcium ion Ca2+ (2,8,8).
Oxygen atom (O). Electronic configuration (2,6)
\(\stackrel{+2 e^{-}}{\longrightarrow}\) Oxygen ion O2- (2,8).
∴ Ca2+ + O2- → CaO

  • A calcium atom gives away 2er and a cation of calcium (Ca2+) is formed, hence, the valency of calcium is two.
  • An oxygen atom takes 2e and forms anions of oxygen (O2-) (oxide), and thus, the valency of oxygen is two.
  • After the give and take of electrons is over, the electronic configuration of both the resulting ions has a complete octet.
  • Due to the attraction between the unit but opposite charges on the two ions, one chemical bond known as ionic bond is formed between Ca2+ and O2- and the compound CaO is formed.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 2.

  • Take 56 g calcium oxide in a large conical flask and put 18 g water in it.
  • Observe what happens.
  • Measure the mass of the substance formed.
  • What similarity do you find? Write your inference.

Answer:
(i) When 18 g of water is added to 56 g of calcium oxide, calcium oxide combines with water to form calcium hydroxide Ca(OH)2

(ii) The mass of calcium hydroxide formed is 74 g.?

(iii) In this activity the total mass of reactants, Calcium oxide + Water = 56 g +18 g = 74 g.
It is equal to the mass of the product formed. Ca(OH)2 = 74g.

This activity verifies the Law of Conservation of Matter, i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

Question 3.

  • Take a solution of calcium chloride in a conical flask and a solution of sodium sulphate in a test tube.
  • Tie a thread to the test tube and insert it in the conical flask.
  • Seal the conical flask with an airtight rubber cork.
  • Weigh the conical flask using a balance.
  • Now tilt the conical flask so that the solution in the test tube gets poured in the conical flask.
  • Now weigh the conical flask again.

Answer:

  • In this activity, a white precipitate of CaS04 in NaOl is seen in the conical flask after the reaction.
  • There is no change in the weight of the flask before and after the reaction.
  • This activity verifies the Law of Conservation of Matter i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

Question 4.
Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds : Calcium carbonate, Sodium bicarbonate, Silver chloride, Calcium hydroxide, Magnesium oxide, Ammonium phosphate, Cuprous bromide, Copper sulphate, Potassium nitrate, Sodium dichromate.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 39
Answer:
Calcium carbonate – CaCO3 Sodium bicarbonate – NaHCO3 Silver chloride – AgCl, Calcium hydroxide – Ca(OH)2, Magnesium oxide – MgO, Ammonium phosphate – (NH4)3PO4, Cuprous bromide – CuBr, Copper sulphate – CuSO4, Potassium nitrate – KNO3, Sodium dichromate – Na2Cr2O7.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

(A) Select the correct option:

Question 1.
The proportion by weight of hydrogen and oxygen in water is ……………………….. .
(a) 8 : 1
(b) 2 : 1
(c) 1 : 2
(d) 1 : 8
Answer:
(d) 1: 8

Question 2.
The proportion by weight of carbon and oxygen in carbon dioxide is ……………………….. .
(a) 8 : 3
(b) 3 : 8
(c) 3 : 2
(d) 2 : 3
Answer:
(b) 3 : 8

Question 3.
A nucleus of an atom is made up of positively charged ………………………… and electrically neutral ……………………….. .
(a) protons; neutrons
(b) electrons; neutrons
(c) neutrons; protons
(d) neutrons; electrons
Answer:
(a) protons; neutrons

Question 4.
The size of an atom is determined by its ……………………….. .
Answer:
radius

Question 5.
Atomic radius is expressed in ……………………….. .
(a) milimetres
(b) centimetres
(c) nanometres
(d) picometres
Answer:
(c) nanomet res

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 6.
The atomic size depends on the number of ………………………… in the atom.
(a) protons
(b) nucleus
(c) neutrons
(d) electron orbits
Answer:
(d) electron orbits

Question 7.
The mass of an atom is concentrated in its ……………………….. .
(a) protons
(b) nucleus
(c) neutrons
(d) electrons
Answer:
(b) nucleus

Question 8.
The total number of protons and neutrons in the atomic nucleus is called the ……………………….. .
(a) atomic number
(b) electronic configuration
(c) atomic mass number
(d) valency
Answer:
(c) atomic mass number

Question 9.
A ………………………… is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.
(a) mole
(b) dalton
(c) dozen
(d) gross
Answer:
(a) Mole

Question 10.
Avogadro’s number is denoted by the symbol ……………………….. .
(a) NG
(b) Nv
(c) NA
(d) ND
Answer:
(c) NA

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 11.
A mole of any substance stands for ………………………… molecules.
(a) 60.22 x 1023
(b) 6.022 x 1022
(c) 6.022 x 1023
(d) 60.22 x 1022
Answer:
(a) 60.22 x 1023

Question 12.
The capacity of an element to combine is called its ……………………….. .
(a) valency
(b) electronic configuration
(c) atomic number
(d) volence electrons
Answer:
(a) valency

Question 13.
Electronic configuration of sodium atom is ……………………….. .
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8,1)
Answer:
(d) (2,8,1)

Question 14.
Electronic configuration of chlorine atom is ……………………….. .
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8, 1)
Answer:
(b) (2, 8, 7)

Question 15.
Positively charged ions are called as ……………………….. .
(a) cations
(b) anions
(c) nucleous
(d) protons
Answer:
(a) cations

Question 16.
Negatively charged ions are called as ……………………….. .
(a) cations
(b) anions
(c) nucleus
(d) electrons
Answer:
(b) anions

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 17.
Iron (Fe) exhibits the variable valencies ……………………….. .
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 2 and 4
Answer:
(b) 2 and 3

Question 18.
Cationic radicals are called as ………………………… radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer:
(a) basic

Question 19.
Anionic radicals are called as ………………………… radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer:
(b) acidic

Question 20.
The unit Dalton is used to express …………………………
(a) atomic mass
(b) atomic radius
(c) atomic number
(d) mass number
Answer:
(a) atomic mass

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 21.
The valency of element with electronic configuration ………………………… is 2.
(a) (2,5)
(b) (2, 4)
(c) (2, 6)
(d) (2, 7)
Answer:
(c) (2, 6)

Question 22.
The symbol of Avogadro’s number is ……………………….. .
(a) ND
(b) N0
(c) NB
(d) NA
Answer:
(d) NA

Question 23.
………………………. is bicarbonate radical.
\((a) \mathrm{HCO}_{3}^{2-} (b) \mathrm{CO}_{3}^{-}
(c) \mathrm{HCO}_{3}^{-}
(d) \mathrm{CO}_{3}^{2-}\)
Answer:
\(\text { (c) } \mathrm{HCO}_{3}^{-}\)

Question 24.
Molecular formula of sodium sulphate is ……………………….. .
(a) Na(SO4)2
(b) Na2SO4
(c)Na2(SO4)2
(d)NaSO4
Answer:
(b) Na2SO4

Question 25.
………………………… is a composite radical.
(a) Fe3+
(b) Ca2+
(c) NH4+
(d) S2-
Answer:
(c) NH

Question 26.
A mole of any substance stands for ………………………… molecules.
(a) 6.022 x 1023
(b) 6.022 x 1022
(c) 60.22 x 1023
(d) 60.22 x 1022
Answer:
(a) 6.022 x 1023

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 27.
The mass of an atom is concentrated in its ………………………… .
(a) nucleus
(b) electrons
(c) extranuclear part
(d) protons
Answer:
(a) nucleus

Question 28.
………………………… g of water make 1 mole of water.
(a) 32
(b) 33
(c) 16
(d) 18
Answer:
(d) 18

Complete the analogy:

(1) Electron : extra nuclear part:: Neutron ………………………… .
(2) Sodium: (2, 8, 1):: Chlorine:: ………………………… .
(3) K : basic radical :: Br : ………………………… .
(4) Cut: simple radical:: NH4+ : ………………………… .
(5) Sodium sulphate: Na2SO4:: Potassium Sulphate: ………………………… .
(6) Mercurous: Hg+:: Mercuric : ………………………… .
(7) Positively charged ion : cation:: Negatively charged ion : ………………………… .
(8) 12: 1 dozen :: 144 : ………………………… .
(9) Hydrogen : \(\odot\) :: Copper : ………………………… .
(10) Law of constant proportions : J. L. Proust::
Law of conservation of matter : ………………………… .
Answer:
(1) nucleus
(2) (2, 8, 7)
(3) acidic radical
(4) composite radical
(5) K2SO4
(6) Hg2+
(7) anion
(8) 1 gross
(9) ©
(10) Antoine Lavoisier.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Match the columns:

Column A’ Column ‘B’
Example Atomic radius (in metres)
(1) Water molecule
(2) Haemoglobin molecule
(3) Hydrogen atom
(a) 10-10
(b) 10-9
(c) 10-8

Answer:
(1-b),
(2- c),
(3 – a)

Column ‘A’ Column ‘B’
Element Atomic mass
(1) Neon (a) 35.5
(2) Silicon (b) 32
(3) Chlorine (c) 28
(4) Sulphur (d) 20

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b)

Column ‘A’ Column ‘B’
Molecule Molecular mass in grams
(1) h2 (a) 32 g
(2) H2O (b) 34 g
(3) O2 (c) 2 g
(4) H2O2 (d) 18 g

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Column ‘A’ Column B’
Radicals Names
(1) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (a) Carbonate
(2) \(\mathrm{ClO}_{3}^{-}\) (b) Chromate
(3) \(\mathrm{CO}_{3}^{2-}\) (c) Dichromate
(4) \(\mathrm{CrO}_{4}^{2-}\) (d) Chlorate

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Answer the following in one sentence:

Question 1.
What are valence electrons?
Answer:
The electrons present in the outermost orbit of an atom are called valence electrons.

Question 2.
Give the formula to determine the number of moles of a substance.
Answer:
The formula to determine the number of moles of a substance is as given below.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 4

Question 3.
What are basic radicals? Give examples.
Answer:
The radicals which are formed by removal of electrons from the atoms of metals are called as basic radicals, e.g., Na+, Cu2+

Question 4.
What are acidic radicals? Give examples.
Answer:
The radicals which are formed by adding electrons to the atoms of non-metals are called as acidic radicals, e.g., CT, S2-

State whether the following statement is ‘True’ or ‘False’. Correct the false statement.

(1) Molecular state of oxygen is monoatomic.
(2) The capacity of an element to combine is called its valency.
(3) Anionic radicals are basic radicals.
(4) The magnitude of charge on any radical is its atomic number.
(5) In a chemical reaction, mass of original matter and mass of matter newly formed as a result of chemical change are equal.
(6) The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 5.
(7) Relative mass of hydrogen is 1.
(8) The number of molecules in a given quantity of a substance is determined by its atomic mass.
(9) Avogadro’s number is 6.022 x 1023
(10) Valency of sodium is 2.
Answer:
(1) False. Molecular state of oxygen is diatomic:
(2) True
(3) False. Anionic radicals are acidic radicals.
(4) False. Magnitude of charge on any radical is its valency.
(5) True
(6) False. The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 8.
(7) True
(8) False. The number of molecules in a given quantity of a substance is determined by its molecular mass.
(9) True
(10) False. Valency of sodium is 1.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Name the following:

Question 1.
Scientist who gave Law of Conservation of Matter.
Answer:
Antoine Lavoisier

Question 2.
Scientist who gave Law of Constant Proportion.
Answer:
J. L. Proust

Question 3.
What are protons and neutrons present in nucleus together called as?
Answer:
Nucleons

Question 4.
Unit used to express atomic radius.
Answer:
Nanometre

Question 5.
The number (p + n) in the atomic nucleus is called as?
Answer:
Atomic mass number

Question 6.
Name the unit of atomic mass.
Answer:
Dalton (u)

Question 7.
Write molecular formula of two ionic compounds containing chlorine.
Answer:
NaCl, MgCl2

Question 8.
Give two monoatomic radicals.
Answer:
Na+, Cl

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 9.
Give two examples of simple radicals.
Answer:
Ag+, O2-

Give scientific reasons:

Question 1.
An atom is electrically neutral though it contains charged particles.
Answer:

  • An atom is made up of a nucleus and an extranuclear part. Protons and neutrons are present in the nucleus.
  • The nucleus is positively charged. The extranuclear part is made up of negatively charged electrons.
  • Protons are positively charged, electrons are negatively charged and neutrons are without any charge.
  • The magnitude of their charges is the same when they are equal in number.
  • Hence, the negative charge on all the extra, nuclear electrons together balances the positive charge on the
  • nucleus.
  • Therefore, an atom is electrically neutral though it contains charged particles.

Question 2.
Neon is chemically inert element.
Answer:

  • Atomic number of neon is 10, so its electronic configuration is (2, 8). There are 8 electrons in its 2nd shell, fulfilling its capacity.
  • Thus, neon has a complete octet.
  • It has a stable orbit therefore, it does not indulge in chemical reactions. Hence, neon is a chemically inert element.

Question 3.
The valency of sodium (Na) is one.
Answer:

  • The electronic configuration of sodium (Na) is (2, 8,1). It has 1 electron in its 3rd orbit.
  • It tends to give up this electron so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.
  • The loss of one electron leads to the formation of sodium ion (Na+) which is positively charged as it has lost one electron.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 4.
The valency of chlorine (Cl) is one.
Answer:

  • The electronic configuration of chlorine (Cl) is (2, 8, 7). It has 7 electrons in its 3rd orbit.
  • It tends to take one electron from another atom so that it has 8 electrons in the outermost orbit with electronic configuration (2,8,8) with stable state.
  • The gaining of one electron leads to formation of chloride ion (Cl) which is negatively charged as it has gained one electron.

Question 5.
The valency of Magnesium (Mg) is two.
Answer:

  • The electronic configuration of Magnesium (Mg) is (2,8,2), it has 2 electrons in its 3rd orbit.
  • It tends to give these ‘2’ electrons so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.
  • The loss of two electrons leads to the formation of Magnesium ion (Mg2+) which is double positively charged as it has lost two electrons.

Question 6.
Valency is always a whole number.
Answer:

  • The number of electrons that an atom of an element gives away, takes up or shares forming a bond is called the valency of that element.
  • These electrons are always in whole numbers and not in fractions.
  • Therefore, valency is always a whole number.

Question 7.
Atomic size of potassium is bigger than atomic size of sodium.
Answer:

  • The atomic size of an element depends on the number of electron orbits in the atom of that element.
  • The greater the number of orbits, the larger the size.
  • Atomic number of potassium (K) is 19. Hence, its electronic configuration is (2, 8, 8,1). While atomic number of sodium (Na) is 11. Hence its electronic configuration is (2, 8,1)
  • Number of orbits in potassium atom is 4, while that in sodium atom is 3.
  • Hence, atomic size of potassium is bigger than atomic size of sodium.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 8.
The atomic size of sodium is bigger than atomic size of Magnesium.
Answer:

  • The atomic size of an element depends on the number of electron orbits in the atom of that element.
  • If 2 atoms have the same outermost orbit, then the atom having the larger number of electrons in the outermost orbit is smaller than the one having fewer electrons in the same outermost orbit.
  • Atomic number of sodium (Na) is 11. Hence, its electronic configuration is (2, 8, 1) while atomic number of magnesum (Mg) is 12 and hence its electronic configuration is (2, 8, 2).
  • As compared to sodium atom Magnesum atom has larger number of electrons n its electronic configuration.
  • Therefore, atomic size of sodium is bigger than atomic size of Magnesium.

Write the names of the following compounds and deduce their molecular masses:

Atomic masses : H(1), 0(16), N(14), C(12), K(39), S(32), Ca(40), Na(23), C1(35.5), Mg(24), A1(27), P(31)

Question 1.
Molecular mass of K2CO3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 6

Question 2.
Molecular mass of CO2
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 7

Question 3.
Molecular mass of MgCl2
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 8

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 4.
Molecular mass of NaOH
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 8

Question 5.
Molecular mass of AIPO4
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 10

Question 6.
Molecular mass of NaHCO3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 11

Numerical.

Question 1.
Magnesium Oxide:
Answer:
Given : Number of moles of Magnesium oxide (MgO)n = 0.2 mol
To find : Mass in grams of 0.2 mol of MgO
Solution:
Molecular mass of (MgO)M
= (Atomic mass of Mg) x 1 + (Atomic mass of O) x 1
= 24 x 1 + 16 x 1
= 24 + 16
Molecular mass of (MgO)M = 40
According to the formula Number of moles in the given MgO (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 17
Mass of MgO in grams (m) = 0.2 x 40
Mass of MgO in grams (m) = 8 g.
Mass of 0.2 mole of MgO is 8 g

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 2.
Calcium Carbonate:
Answer:
Given : Number of moles of Calcium carbonate (CaCO3) n = 0.2 mol
To find : Mass in grams of 0.2 mol of CaCO3
Solution:
Molecular mass of (CaCO3) M
= (Atomic mass of Ca) x 1 + (Atomic mass of C) x 1 + (Atomic mass of O) x 3
= (40 x l) + (12 x 1) +(16 x 3)
= 40+ 12+ 48
Molecular mass of (CaCO3) M = 100
According to the formula Number of moles in the given CaCO3 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 18
∴ Mass of CaCO3 in grams (m) = 0.2 x 100
∴ Mass of CaCO3 in grams (m) = 20 g
Mass of 0.2 mole of CaCO3 is 20 g

State laws/Define the following:

Question 1.
Law of Conservation of Matter.
Answer:
In a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to chemical reaction.

Question 2.
Law of Constant Proportion.
Answer:
The proportion by weight of the constituent elements in the various samples of a compound is fixed.

Question 3.
Molecular Mass:
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance.

Question 4.
Mole
Answer:
A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 5.
Valency
Answer:
The capacity of an element to combine is called its valency.

Question 6.
Electronic definition of Valency
Answer:
The number of electrons that an atom of an element gives away or takes up while forming an ionic bond is called valency of that element.

Question 7.
Radicals
Answer:
The positively or negatively charged ions that take part independently in chemical reactions are called radicals.

Question 8.
Atomic size determination
Answer:
The size of an atom is determined by its radius. The atomic radius of an isolated atom is the distance between the nucleus of an atom and its outermost orbit.

Question 9.
Atomic mass number
Answer:
The number of protons and neutrons in the atomic nucleus is called the atomic mass number.

Question 10.
Unified mass
Answer:
Unified mass is the standard unit of atomic mass that quantifies mass on an atomic or molecular scale. Its symbol is ‘u’.
1 u = 1.66053904 x 10-27 kg.

Question 11.
Molecular mass of a substance
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance. Like atomic mass, molecular mass is also expressed in the unit Dalton (u).

Answer the following questions:

Question 1.
What is variable valency?
Answer:

  • Under different conditions, the atoms of some elements give away or take up a different number of electrons.
  • In such cases, those elements exhibit more than one valency.
  • This property of elements is called variable valency.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Complete the following table:

Question 1.
Write down the cations and anions obtained from the compounds in the following chart.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 26

Answer the following questions:

Question 1.
Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds:

(a) Calcium carbonate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 29
∴ Chemical formula of Calcium carbonate is CaCO3

(b) Sodium bicarbonate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 30
∴ Chemical formula of Sodium bicarbonate is NaHCO3

(c) Silver chloride
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 31
∴ Chemical formula of Silver chloride is AgCl

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

(d) Calcium hydroxide Answer: Symbol Ca OH
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 52
∴ Chemical formula of Calcium hydroxide is Ca(OH)2

(e) Magnesium oxide
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 53
∴ Chemical formula of Magnesium oxide is MgO

(f) Ammonium phosphate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 54
∴ Chemical formula of Ammonium phosphate is (NH4)3PO4

(g) Cuprous bromide
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 55
∴ Chemical formula of Cuprous bromide is CuBr.

(h) Copper sulphate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 57
∴ Chemical formula of Copper sulphate is CuSO4.

(i) Potassium nitrate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 58
∴ Chemical formula of Potassium nitrate is KNO3.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

(j) Sodium dichromate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 59
∴ Chemical formula of Sodium dichromate is Na2Cr2O7.

9th Std Science Questions And Answers:

Reflection of Light Class 9 Science Chapter 11 Questions And Answers Maharashtra Board

Class 9 Science Chapter 11

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 11 Reflection of Light Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 11 Reflection of Light Question Answer Maharashtra Board

Class 9 Science Chapter 11 Reflection of Light Question Answer Maharashtra Board

1. Answer the following questions.

a. Explain the difference between a plane mirror, a concave mirror and a convex mirror with respect to the type and size of the images produced.
Answer:

Plane mirror Concave mirror Convex mirror
Type of image Virtual and Erect Virtual (erect) as well as Real (inverted) Virtual and Erect
Size of image Same size Diminished, Same size and magnified Diminished

b. Describe the positions of the source of light with respect to a concave mirror in
1. Torch light
2. Projector lamp
3. Floodlight
Answer:
(a) Torch light: The source of light is placed at the focus.
(b) Projector lamp : The source of light is placed at the centre of curvature.
(c) Flood light : The source of light is placed just beyond the centre of curvature.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

c. Why are concave mirrors used in solar devices?
Answer:

  • Solar devices like solar cooker or solar water heater use solar energy to cook food or heat water.
  • When sun rays fall on the concave mirror, they converge and come together in the focal plane.
  • Due to convergence, the intensity of sun rays increases and the food or water is heated faster. Hence, concave mirrors are used in solar- devices.

d. Why are the mirrors fitted on the outside of cars convex?
Answer:

  • A convex mirror is used as rear view mirror because they form erect, virtual, and diminished images.
  • This, allows the driver to view a large area in a small mirror.

e. Why does obtaining the image of the sun on a paper with the help of a concave mirror burn the paper?
Answer:

  • When sunrays fall on the concave mirror, they converge and come together in the focal plane.
  • Due to convergence, the intensity of sunrays increases.
  • Hence, image of the sun on a paper with the help of concave mirror bums the paper.

f. If a spherical mirror breaks, what type of mirrors are the individual pieces?
Answer:

  • When a spherical mirror breaks into smaller pieces, the radius of curvature and focal length does not change.
  • Hence, it will continue to behave like a spherical mirror only.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

2. What sign conventions are used for reflection from a spherical mirror?
Answer:
According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.

  1. The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
  2. All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
  3. The distance measured vertically upwards from the principal axis are taken to be positive.
  4. The distance measured vertically downwards from the principal axis are taken to be negative.
  5. The focal length of a concave mirror is negative while that of a convex mirror is positive.

3. Draw ray diagrams for the cases of images obtained in concave mirrors as described in the table on page 122.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 1
Answer:
(a) A ray diagram for object at infinity for a concave mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 3

Image position Nature of image
At focus Real, inverted and point image

(b) A ray diagram for object beyond centre of curvature for a concave mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 4
An object beyond centre of curvature for a concave mirror

Image position Nature of image
Between the centre of curvature and focus. Real, inverted and diminished.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

(c) A ray diagram for object at the centre of curvature for a concave mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 5
Object at centre of Curva fu re be a concave mirror.

Image position Nature of image
At the centre of curvature. Real, inverted and same size

(d) A ray diagram for object between F and C for a concave mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 6
Object between F & C for a concave mirror

Image position Nature of image
Beyond the centre of curvature. Real, inverted and magnified.

(e) A ray diagram for obj ect at focus for a concave mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 7
Object at focus for a concave mirror.

Image position Nature of image
At infinity. Real, inverted and highly magnified.

(f) A ray diagram for object between pole and focus for a concave mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 8

Image position Nature of image
Behind the mirror. Virtual, erect and magnified.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

4. Which type of mirrors are used in the following?
Periscope, floodlights, shaving mirror, kaleidoscope, street lights, headlamps of a car.

Answer:

Objects Type of Mirror
Periscope Plane mirror
Floodlights Concave mirror
Shaving mirror Concave mirror
Kaleidoscope Plane mirror
Street lights Convex mirror
Head lamps of car Concave mirror

5. Solve the following examples

a. An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?
Solution:
Given: Object size (h1) = 7 cm
Object distance (u) = -25 cm
Focal length (f) = -15cm
To find: Image distance (u) = ?
Image size (h2) = ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 9
The screen should be kept 373 cm in front of the mirror. The image is real.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 10
The height of the image is 10.5 cm, it is an inverted and enlarged image.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

b. A convex mirror has a focal length of 18 cm. The image of an object kept in front of the mirror is half the height of the object. What is the distance of the object from the mirror?
Solution:
Given: Image size (h2) = 1/2 h1
Focal length (f) = 18 cm
To find: Object distance (u) ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 11
The object is placed in front of the convex mirror at a distance of 18 cm.

c. A 10 cm long stick is kept in front of a concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Solution:
Given: Object size (h1) = 10 cm
Object distance (u) = -20 cm
Focal length (f) = -10 cm
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 12
The height of the image is 10 cm and it is a real and inverted image.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

6. Three mirrors are created from a single sphere. Which of the following:
pole, centre of curvature, radius of curvature, principal axis – will be common to them and which will not be common?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 30

  • Centre of curvature and Radius of curvature will be common for all three pieces.
  • Pole and Principal axis will not be common.

Class 9 Science Chapter 11 Reflection of Light Intext Questions and Answers

Class 9 Science Chapter 11 Reflection Of Light Question 1.
What is light
Answer:
Light is a form of electromagnetic radiation that produces the sensation of vision.

9th Class Science Chapter 11 Reflection Of Light Exercise Question 2.
What is a mirror?
Answer:
A mirror is a reflecting surface which reflects light and creates clear images.

9th Class Science Chapter 11 Reflection Of Light Exercise Answer Question 3.
Principal Focus of Concave and Convex Mirror.
Answer:

Principal Focus of the Concave Mirror Principal Focus of the Convex Mirror
(i) Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror.
(ii) It is formed in front of the mirror.
(iii) Focus of concave mirror is real.
(i) Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror.
(ii) It is formed behind the mirror.
(iii) Focus of convex mirror is virtual.

9th Class Science Chapter 11 Reflection Of Light Notes Question 4.
If we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror. Why does it happen?
Answer:

  • When we hold a page of a book in front of the mirror, the image of the words appear laterally inverted.
  • The image of every point of the word is formed behind the mirror at the same distance from the mirror
  • Because of this the left and right side of the image is interchanged.
  • Hence, if we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

11 Reflection Of Light Exercise Question 5.
Which letters of the English alphabet form images that look the same as the original letters?
Answer:
A, H, I, M, O, T, U, V, W, X, Y

9th Science Chapter 11 Reflection Of Light Exercise Question 6.
When a person stands in front of a plane mirror, how is the image formed? What is the nature of the image?
Answer:

  • The image of a person is formed from every point of the source, thereby forming an extended image of the whole source.
  • The image formed would be virtual, upright and left-right reversed.

Answer the following questions:

Reflection Of Light Class 9 Questions And Answers Question 1.
Place two plane mirrors at an angle of 90a to each other. Place a small object between them. Images will be formed in both mirrors. How many images do you see? Now change the angle between the mirrors as given in the following table and count the number of images each time. How is this number related to the measure of the angle?
Answer:
The Relation between the angle between the mirrors and the number of images formed is given by
\(n=\frac{360^{\circ}}{\mathrm{A}}-1\)
n = number of images,
A = angle between the mirrors

Angle Number of images
120° 2
90s 3
60® 5
45® 7
30® 11

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 13

Class 9 Science Chapter 11 Reflection of Light Additional Important Questions and Answers

Can you recall?

9th Class Science Chapter 11 Reflection Of Light Question 1.
What is meant by reflection of light and what are the types of reflection?
Answer:
The bouncing back of light when it hits an opaque surface is called reflection of light. The two types of reflection are regular and irregular reflection.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Reflection Of Light Class 9 Exercise Answers Question 2.
What are the laws of reflection.
Answer:

  • The incident ray, reflected ray and normal all lie in the same plane at the point of incidence.
  • The angle of incidence is equal to the angle of reflection.
  • The incident ray and the reflected ray lie on opposite sides of the normal.

Choose and the correct option:

Class 9th Science Chapter 11 Reflection Of Light Question Answer Question 1.
If the reflected rays do not actually meet, such an image is called as image.
(a) real
(b) virtual
(c) magnified
(d) inverted
Answer:
(b) virtual

Class 9 Science Chapter 11 Reflection Of Light Exercise Solutions Question 2.
In a plane mirror, the perpendicular distance of the image from the mirror is equal to
(a) the perpendicular distance of the source from the object.
(b) the perpendicular distance of the source from the mirror.
(c) the parallel distance of the source from the object.
(d) the parallel distance of the source from the mirror.
Answer:
(b) the perpendicular distance of the source from the mirror

Reflection Of Light Class 9 Notes Pdf Maharashtra Board Question 3.
The image formed in a convex mirror is always
(a) virtual, smaller and behind the mirror
(b) virtual, smaller and in front of the mirror
(c) real, smaller and behind the mirror
(d) real, smaller and in front of the mirror
Answer:
(a) virtual, smaller and behind the mirror

Reflection Of Light Class 9 Solutions Question 4.
images can be displayed on a screen.
(a) Virtual
(b) Real
(c) Virtual and erect
(d) Virtual and inverted
Answer:
(b) Real

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

9th Class Science Chapter 11 Reflection Of Light Exercise Pdf Question 5.
A concave mirror is also called as a mirror.
(a) converging
(b) diverging
(c) plane
(d) outward curved
Answer:
(a) converging

9th Science Chapter 11 Reflection Of Light Question 6.
The centre of the mirror surface is called its
(a) pole
(b) centre of curvature
(c) principal axis
(d) focus
Answer:
(a) pole

Class 9 Science Chapter 11 Question Answer Reflection Of Light Question 7.
According to the new sign convention, the of the mirror is taken as origin.
(a) focus
(b) pole
(c) optical centre
(d) centre of curvature
Answer:
(b) pole

Class 9 Science Chapter 11 Reflection Of Light Exercise Question 8.
A convex mirror is also called as a mirror.
(a) converging
(b) plane
(c) diverging
(d) inward curved
Answer:
(c) diverging

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Reflection Of Light Class 9 Maharashtra Board Question 9.
In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be
(a) same height as that of the person
(b) double the height of the person
(c) half the height of the person
(d) quarter the height of the person
Answer:
(c) half the height of the person

Reflection Of Light Exercise 9th Class Question 10.
If the inner surface of the spherical mirror is reflecting, then it is a mirror, and if the outer surface is reflecting then it is mirror.
(a) convex, concave
(b) convex, plane
(c) concave, plane
(d) concave, convex
Answer:
(d) concave, convex

9th Std Science Chapter 11 Reflection Of Light Question 11.
The image formed by a concave mirror is
(a) always virtual and erect
(b) always virtual and inverted
(c) virtual if the object is placed between the pole and the focus
(d) virtual if the object is beyond the focus
Answer:
(c) virtual if the object is placed between the pole and the focus

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 12.
No matter how far you stand from a spherical mirror, your image appears erect. The mirror may be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer:
(d) either plane or convex

Question 13.
In case of a concave mirror, an erect image is
(a) real and enlarged
(b) real and diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer:
(d) virtual and enlarged

Question 14.
A rear view mirror of a car is
(a) plane mirror
(b) concave mirror
(c) convex mirror
(d) cylindrical mirror
Answer:
(c) convex mirror

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 15.
An image of an object placed at infinite distance from a concave mirror is formed at
(a) the focus of the mirror
(b) behind the mirror
(c) centre of curvature
(d) infinity
Answer:
(a) the focus of the mirror

Question 16.
A ray of light parallel to principal axis after reflection from concave mirror passes through
(a) centre of curvature
(b) focus
(c) pole
(d) optical centre
Answer:
(b) focus

Question 17.
The image made by a plane mirror is a image.
(a) real
(b) virtual
(c) inverted
(d) diminished
Answer:
(b) virtual

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 18.
The size of the image of an object placed at the focus of a concave mirror is
(a) erect
(b) very large
(c) same size
(d) diminished
Answer:
(b) very large

Question 19.
For virtual images, the height is while for real images, it is
(a) positive, positive
(b) negative, positive
(c) negative, negative
(d) positive, negative
Answer:
(d) positive, negative

Find the odd man out:

Question 1.
Torches, flood lights, head lamps of vehicles, rear view mirror.
Answer:
Rear view mirror – In rear view mirrors, convex . mirror is used. Concave mirrors are used in the rest.

Question 2.
Side mirrors of cars, parking mirrors, flood lights, mirror fitted in shops.
Answer:
Flood lights – In flood lights concave mirror is used. Convex mirrors are used in the rest.

Question 3.
Virtual and enlarged, virtual and diminished, real and inverted, real and magnified
Answer:
Virtual and diminished type of image is not formed by a concave mirror. All the other types of images are formed by a concave mirror.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 4.
Image is laterally inverted, image is of same size, image is at same distance, image is diminished.
Answer:
Image is diminished is not a characteristic of image formed in a plane mirror. Rest of them are characteristics of plane mirror.

Answer the following in one sentence:

Question 1.
What kind of mirror will a doctor use to concentrate on teeth, eyes, ears etc.?
Answer:
The doctor will use a concave mirror to concentrate on teeth, eyes, ears etc.

Question 2.
What do the nature, position and size of the image depend on?
Answer:
The nature, position and size of the image depend upon the distance of the object from the reflecting surface.

Question 3.
Give the expression for mirror formula.
Answer:
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Question 4.
State any four uses of concave mirror.
Answer:
Concave mirrors are used in torches, headlights, shaving mirrors, dentists’ mirrors, solar devices etc.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 5.
What are the two types of spherical mirror?
Answer:
Convex mirror and concave mirror are the two types of spherical mirror.

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Plane mirror (a) Rear view mirror
(2) Concave mirror (b) At laughing gallery
(3) Convex mirror (c) At a hair dresser
(4) Irregular

curved mirror

(d) At a dentist

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Question 2.

Column ‘A’ Column ‘B’
(1) Plane mirror (a) Virtual and diminished image
(2) Concave mirror (b) Virtual and same size image
(3) Convex mirror (c) Real and inverted image

Answer:
(1 – b),
(2 – c),
(3 – a)

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

State whether the following statements are true or false. Correct the false statements:

(1) If the mirrors are kept at right angle to each other, then the number of images formed will be 4.
(2) A convex mirror is used in flood lights.
(3) A concave mirror always forms a magnified image.
(4) Images formed by convex mirrors are always virtual.
(5) The distance between the focus and the pole is called the radius of curvature.
(6) Reflection from a spherical mirror obeys laws of reflection.
(7) The reflecting surface of a concave mirror is curved.
(8) Distances measured in the direction of the incident light are taken as positive.
(9) If the image is erect, the height of the image is negative.
(10) A real image can be displayed on a screen.
(11) A concave mirror always forms a real and inverted image.
(12) Doctors use diverging beam of light to study teeth, ears and eyes.
Answer:
(1) False. if the mirrors are kept at right angle to each other then the number of images formed will be 3.
(2) False. a concave mirror is used in flood lights.
(3) False. a concave mirror can sometimes form a diminished image as well.
(4) True
(5) False. the distance between the focus and the pole is called the focal length.
(6) True
(7) True
(8) True
(9) False. if the image is erect, the height of the image is positive.
(10) True
(11) False. a concave mirror can also form a virtual and erect image.
(12) False. doctors use a converging beam of light to study teeth, ears and eyes.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Give scientific reasons:

Question 1.
A concave mirror is called a converging mirror.
Answer:

  • When rays of light parallel to the principal axis are incident on concave mirror, they converge.
  • After convergence, they meet at one point on the principal axis, hence concave mirror is called converging mirror.

Question 2.
Concave mirrors are used in torches and in car headlights.
Answer:

  1. Concave mirrors are used in torches and car headlights because when a source of light is placed at the focus of a concave mirror, a parallel beam of light rays is obtained.
  2. This helps us to see things upto a considerable distance in the darkness.

Question 3.
A dentist uses a concave mirror while examining teeth.
Answer:

  • A concave mirror produces an erect, virtual and magnified image of an object placed between its pole and focus.
  • A dentist uses this principle to get a clear and distinct image of teeth, hence, a dentist uses a concave mirror.

Solve the following numerlcals.

Tips for solving numerical:

  • Object distance (u) is always -ve
  • If Image distance (u) is +ve then image is behind the mirror and virtual. if u is -ve then image is in front of the mirror and real.
  • Object height (h1) is always +ve since it is erect.
  • Image height (h2) can be +ve for virtual and -ve for real.

Type – A

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 14

Question 1.
A bird is sitting in front of two plane mirrors inclined at an angle of 600 to each other. How many images does the bird see in the mirror?
Solution:
Given : Angle between mirror A = 600
To find: Number of images formed n = ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 15
The brrd sees 5 images in the mirror.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 2.
A coin is kept in front of two plane mirrors inclined to each other. If 3 images of the coin are seen then what is the angle A between the mirrors?
Solution:
Given: no. of images formed n =3
To find: Angle between mirror A =?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 16
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 17
The mirrors are inclined atan angle of 900 to each other.

Question 3.
An image is formed 5 cm behind a convex mirror of focal length 10 cm. At what distance is the object placed from the mirror?
Solution:
Given: Image distance (u) = 5 cm
Focal length (f) = 10 an
To find: Object distance (u) = ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 18
The object is placed at a distance of 10 cm in front of the mirror.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 4.
An object placed 20 cm in front of a convex mirror is found to have an image 15cm behind the mirror. Find the focal length of the mirror.
Solution:
Given: Object distance (u) = -20 cm
Image distance (u) = 15 cm
To find: focal length (f) = ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 19
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 20
The focal length of the convex mirror is 60 cm.

Numerical For Practice

Question 5.
An object is placed at a distance of 36 cm from a concave mirror of focal length 12 cm. Find the image distance.
Answer:
-18 cm

Question 6.
An arrow is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the image distance.
Answer:
11.1 cm

Type – B

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 21

Question 1.
An object 4cm in height is placed at a distance of 36 cm from a concave mirror. The image is formed 18 cm in the front of the mirror. Find the height of the image.
Solution:
Given: Object height (h1) = 4 cm
Image distance (u) = -18 cm
Object distance (u) = -36 cm
To find: Height of image (h2) = ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 22
The height of the image is 2 cm and it is inverted.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 2.
An object 2 cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3 cm high. Find the image distance.
Solution:
Given: Object height (h1) = 2 cm
Object distance (u) = -16 cm
Image height (h2) = -3 cm
To find: Image distance (u) = ?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 23
The image is formed at a distance of 24 cm in front of the mirror.

Numericals For Practice

Question 3.
An object 10cm in height is placed at a distance of 36 cm from a concave mirror. 1f the image is formed at a distance of 18 cm in front of the mirror, find the height of the image.
Answer:
-5cm

Question 4.
A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror. Find the image distance.
Answer:
-80cm

Type – C

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 24

Question 1.
Rajashree wants to get an inverted image of height 5 cm of an object kept at a distance of 30 cm from a concave mirror. The focal length of the mirror is 10 cm. At what distance from the mirror should she place the screen? What will be the type of the image, and what is the height of the object?
Solution:
Given:
Focal length = f = -10 cm,
Object distance = u = -30 cm
Height of the image = h2 = 7 cm
To find: Height of the object = h1 = ?
Image distance = u =?
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 25
Rajashree has to place the screen 15 cm to the left of the mirror.
Magnification formula
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 26
The height of the object is 10 cm. Thus, the image will be real and diminished.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 2.
A 10 cm long stick is kept horizontally in front of the concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?

Solution:
The stick is kept parallel to the Principal axis. Distance between A and P is 20 cm. Say u1 = 20 cm.
Hence, the other end of the stick is at distance, u2 = (u1 + 10) = 30 cm from pole of the mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 27
Using mirror formula for concave mirror,
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 28
Here, negative signs indicate that images are formed on the left of the mirror.
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 29

The length of the image formed ?s given by, u = u2 – u1 = 15 – (-20) = 5cm.
The length of the image is 5 cm.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Numerical For Practice

Question 3.
An object 2 cm in height is placed at a distance of 16 cm from a concave mirror. If the focal length of the mirror is 9.6 cm., find the image distance, nature and size of the image.
Answer:
u = -24 an, h2 = -3 cm; real, inverted and enlarged.

Question 4.
An arrow of 2.5cm height is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.
Answer:
v = 11.1cm, h1 = 1.1cm; virtual and in dimirrished form.

Define the following:

Question 1.
Centre of curvature of mirror (C)
Ans.
The centre of the sphere of which the mirror is a parte is called the centre of curvature of the mirror.

Question 2.
The radius of curvature (R)
Answer:
The radius of the sphere of which the mirror is a part, is called the radius of curvature of the mirror.

Question 3.
Pole (P)
Answer:
The centre of the mirror surface is called its pole.

Question 4.
The principal axis of a mirror
Answer:
The straight line passing through the pole and centre of curvature of the mirror is called its principal axis.

Question 5.
The focus of a concave mirror (F)
Answer:
Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror.

Question 6.
Focus of a convex mirror (F)
Answer:
Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 7.
Focal length of a mirror (f)
Answer:
The distance (f) between the pole and the principal focus of the mirror is called the focal length. This distance is half of the radius of curvature of the mirror. \(f=\frac{R}{2}\)

Answer the following in short:

Question 1.
What are the rules for drawing ray diagrams for the formation of image by spherical mirror?
Answer:
The rules are as follows :

  • If an incident ray is parallel to the principal axis, then the reflected ray passes through the principal focus.
  • If an incident ray passes through the principal focus of the mirror, the reflected ray is parallel to the principal axis.
  • If an incident ray passes through the centre of curvature of the mirror, the reflected ray traces the same path back.

Distinguish between:

Question 1.
Convex mirror and Concave mirror
Answer:

Convex mirror Concave mirror
(i) In a convex mirror, the reflecting surface is on the outer side.
(ii) It is called a diverging mirror.
(iii) The focus of a convex mirror is virtual.
(iv) It can form only a virtual image.
(v) It can form only a diminished image.
(i) In a concave mirror, the reflecting surface is on the inner side.
(ii) It is called a converging mirror.
(iii) The focus of a concave mirror is real.
(iv) It can form a real as well as a virtual image.
(v) It can form an enlarged, diminished as well as the same size image.

Question 2.
Real image and Virtual image
Answer:

Real image Virtual image
(i) A real image is formed only when the reflected rays actually meet at a point.
(ii) Real images can be obtained on a screen.
(iii) All real images are inverted.
(i) A virtual image is formed only when the reflected rays appear to meet at a point.
(ii) Virtual images cannot be obtained on a screen.
(iii) All virtual images are erect.

Answer the following questions:

Question 1.
If we keep the mirrors parallel to each other, how many images will we see ?
Answer:
When two mirrors are kept parallel to each other infinite images are formed, this is because light gets reflected infinite times.

Answer in detail:

Question 1.
What sign conventions are used for reflection from a spherical mirror?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 2
According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.

  • The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror. Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light
  • All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
  • Distance measured vertically upwards from the principal axis are taken to be positive.
  • Distance measured vertically downwards from the principal axis are taken to be negative.
  • The focal length of a concave mirror is negative while that of a convex mirror is positive.

Question 2.
Draw ray diagrams for the image obtained in convex mirrors.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 31

Image position Nature of image
Behind the mirror. (A) Virtual,
(B) Erect
(C) Diminished

Question 3.
In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be half the height of the person. Explain.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 32

Proof:

  1. In the figure, the point at the top of the head, the eyes and a point at the feet of a person are indicated by H, E and F respectively.
  2. R and S are midpoints of HE and EF respectively.
  3. The mirror PQ is at a height of NQ from the ground and is perpendicular to it. PQ is the minimum height of the mirror in order to obtain the full image of the person.

For this, RP and QS must be perpendicular to the mirror.

Minimum height of the mirror
PQ = RS
= RE + ES
\(=\frac{\mathrm{HE}}{2}+\frac{\mathrm{EF}}{2}=\frac{\mathrm{HF}}{2}\)
= Half of the person’s height.

Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light

Question 4.
Determine the sign of magnification in each of the 6 cases in the table and verify that they are same using formulae
\(\mathbf{M}=\frac{h_{2}}{h_{1}}\) and \(\mathbf{M}=\frac{-v}{u}\)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 33

Question 5.
Explain the images formed by concave mirrors with respect to position of the image and object and also the Nature and size of image
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 11 Reflection of Light 34

9th Std Science Questions And Answers:

Study of Sound Class 9 Science Chapter 12 Questions And Answers Maharashtra Board

Class 9 Science Chapter 12

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 12 Study of Sound Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 12 Study of Sound Question Answer Maharashtra Board

Class 9 Science Chapter 12 Study of Sound Question Answer Maharashtra Board

1. Fill in the blanks and explain.

a. Sound does not travel through ……………………….……….. .
b The velocity of sound in steel is ……………………….………… than the velocity of sand in water.
c. The incidence of ……………………….………… in daily life shows that the velocity of sound is less than the velocity of light.
d. To discover a sunken ship or objects deep inside the sea, ……………………….………… technology is used.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

2. Explain giving scientific reasons.

a. The roof of a movie theatre and a conference hall is curved.
Answer:

  • Sound waves get reflected from the walls and roof of a room multiple times. This causes a single sound to be heard not once but continuously. This is called reverberation.
  • Due to reverberation, some auditoriums or some particular seats in an auditorium have inferior sound reception. This can be compensated with curtains.
  • Ceilings of these halls are made curved so that sound after reflecting from the ceiling, reaches all parts of the hall and the quality of sound improves.

b. The intensity of reverberation is higher in a closed and empty house.
Answer:

  • Reverberation occurs due to multiple reflections of sound.
  • The furniture in the house acts as a sound-absorbing material.
  • So if the house is closed and empty, a reflection of sound will be maximum and hence, intensity of reverberation is higher.

c. We cannot hear the echo produced in a classroom.
Answer:

  • For distinct echoes, the minimum distance of the reflecting surface from the source of sound must be 17.2 m.
  • Benches in the classroom are sound absorbing materials which prevent echo of sound.
  • Because of these two reasons echo is not heard in a classroom.

3. Answer the following questions in your own words.

a. What is an echo? What factors are important to get a distinct echo?
Answer:

  • An echo is the repetition of the original sound because of reflection by some surface.
  • At 22°C, the velocity of sound in air is 344 m/s.
  • Our brain retains a sound for 0.1 seconds Thus, for us to be able to hear a distinct echo, the sound should take more than 0.1 seconds after starting from the source to get reflected and. come back to us.
  • We know that,
    Distance = speed x time
    = 344 m/s x 0.1 s
    = 34.4 m
  • Thus, to be able to hear a distinct echo, the reflecting surface should be at a minimum distance of half of the above, i.e. 17.2 m.
  • As the velocity of sound depends on the temperature of air, this distance depends on the temperature.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

b. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.
Answer:

  • Goighumat with a height of 51 metres and diameter of 37 metres with 3 metres thick walls is spread over approximately 1700 square metres.
  • This meets the conditions for echo i.e. : 17.2 metres minimum.
  • The dome of the golghumat is curved and hence, sound reflects multiple times before reaching the observer.
  • This is the reason for multiple echoes being produced.

c. What should be the dimensions and the shape of classrooms so that no echo can be produced there?
Answer:

  1. Dimensions: The distance between opposite walls in a classroom must be less than 17.2 m so that the reflected sound returns to the observer within 0.1 s.
  2. Shape: The classrooms should have curved ceilings and walls so that the reflected sound is directed towards the observer instantly within 0.1 s

4. Where and why are sound-absorbing materials used?
Answer:
The sound absorbing materials are used in :

  • School, cinema hall, concert hall, houses or places where quality of sound is important.
  • In the absence of sound absorbing material the sound will undergo multiple reflection causing reverberation of sound.

5. Solve the following examples.

a. The speed of sound in air at O °C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?
Answer:
Given:
Initial speed of sound at 0°C 332 m/s.
Final speed of sound -344 m/s.
Rate of increase per degree rise in temp. = 0.6m/s
To find:
Temperature when speed is 344m/s
Formulae:
Increase in temperature
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 1
Temperature when the speed of sound is 344 m/sis 20°C

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her? (The velocity of sound in air is 340 m/s?)
Answer:
Given : Speed of sound (v) = 340 m/s
Time taken (f) = 4 sec
To find : Distance (s) = ?
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 2
The lightning has struck at a distance of 1360 m from the observer.

c. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.
1. What is the velocity of sound in air?
2. What is the distance between the two walls? (Ans: 330 m/s; 1650 m)
Answer:
Given:
Distance of the closer wall (S1) = 660 m
Time of echo from closer wall = 4 sec
∴ Time taken (t1) = 4/2 sec = 2 sec
Time of echo from distant wall = 6 sec
∴ Time taken (t2) = 6/2 sec = 3 sec
To find :
Velocity of sound in air (y) =?
Distance between two walls (S1 + S2) = ?
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 3
The velocity of sound in air is 330 mIs and the distance between two walls is 1650 m.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How may times as fast as the other? (Ans: In A; Twice)
Answer:
In A; Thrice

e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw? (Ans: Temperature of B is 4 times the temperature of A.)
Given:
Mass of Helium in bottle A = (mA) = 10gm
Mass of Helium in bottle B = (mB) = 40gm
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 4
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 5
The temperature of B is 4 times the temperature of A

Class 9 Science Chapter 12 Study of Sound Intext Questions and Answers

Study Of Sound Class 9 Notes Maharashtra Board Question 1.
How does the velocity of sound depend on its frequency?
Answer:
The velocity of sound is directly proportional to its frequency
ν = υ λ
when ν = velocity
υ = frequency
λ = wavelength

9th Class Science Chapter 12 Study Of Sound Exercise Question 2.
The molecular weight of oxygen gas (O2) is 32 while that of hydrogen gas (H2) is 2. Prove that under the same physical conditions, the velocity of sound in hydrogen is four times that in oxygen.
Answer:
Given:
Molecular wt of Oxygen (Mo) =32
Molecular wt of hydrogen (MH) = 2
To Find:
VH = 4 vo
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 6
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 7
Hence, proved that velocity of sound in hydrogen is four times that in oxygen.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Answer the following:

Study Of Sound Class 9 Maharashtra Board  Question 1.
How will you reduce reverberation in public halls or buildings?
Answer:
(i) Reverberation in public halls or buildings will be reduced by using sound absorbing materials like curtains on wall, carpets on the floor.
(ii) By keeping the windows open, as sound will not get reflected.

12 Study Of Sound 9th Class Exercise  Question 2.
How is ultrasound used in medical science?
Answer:

  • Sonography: Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
  • Echocardiography: Echocardiography is a test that uses ultrasonic sound waves to produce live images of your heart.

9th Science Chapter 12 Study Of Sound Exercise Question 3.
To hear the echo distinctly, will the distance from the source of sound to the reflecting surface be same at all temperatures? Explain your answer.
Answer:

  • No,the distance from the source of sound to the reflecting surface will not be the same at all temperatures.
  • Velocity of sound is directly proportional to the square root of temperature.
  • One of the conditions of echo is that the time interval between the original and reflected sound should be more than 0.1 sec.
  • So if the temperature increases, the velocity of sound increases and the reflected sound reaches in less than 0.1 sec.
  • So for echo to be heard the distance between the observer and the reflecting surface has to increase.

9th Science Chapter 12 Study Of Sound Question 4.
When is the reflection of sound harmful?
Answer:

  • Reflected sound of high intensity called as noise is disturbing and harmful to the ears.
  • When sound reverberates i.e it undergoes multiple reflections, poor quality of sound is produced.

9th Class Science Chapter 12 Study Of Sound Question Answer Question 5.
What kind of waves are created when a stone is dropped in water ?
Answer:

  • When a stone is dropped in water, the particles of water oscillate up and down.
  • These oscillations are perpendicular to the direction of propagation of the wave, such waves are called transverse waves.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Answer the following question:

12 Study Of Sound 9th Class Question 1.
Observe the graph/ diagram and discuss your observation.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 19

  1. Fig. A shows changes in density. The region where particles are crowded is called compression and where they are far apart are rarefaction.
  2. Fig. B show change in pressure. The lines represent layers of air. The regions when lines are crowded are high pressure regions while when they are far apart are of low pressure.
  3. Fig. C shows changes in density or pressure. The crest represents high pressure region while trough represents low pressure region.

Answer the following question:

Study Of Sound Class 9 Question Answer Question 1.
How are the frequencies of notes sa, re, ga, ma, pa, dha, ni related to each other?
Answer:
The frequencies of notes sa, re, ga, ma, pa, dha, ni are related in the ratio.
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 20
i.e if first Sa is 240Hz then the next Sa will be 240 x 2 = 480Hz

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Class 9 Science Chapter 12 Study Of Sound Exercise Question 2.
What is the main difference between the frequencies of the voice of a man and that of a woman?
Answer:

  • Voice of a woman is high pitch i.e shorter wavelength and higher frequency
  • Voice of man is low pitch i.e larger wavelength and smaller frequency.

Question 3.
Try this;
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 21

(a) In the above activity, what will happen if you lift one of the tubes to some height?
Answer:
If one of the tubes is lifted, angle of incidence will not be equal to angle of reflection, hence, the sound will not be clearly audible.

(b) Measure the angle of incidence 01 and the angle of reflection 02. Try to see if they are related in any way.
Answer:
Angle of incidence is same as the angle of reflection.

Class 9 Science Chapter 12 Study of Sound Additional Important Questions and Answers

Can you recall?

12.Study Of Sound Question 1.
How is the direction of the oscillation of the particles of the medium related to the direction of propagation if the sound wave?
Answer:

  • Sound travels as a longitudinal wave.
  • In a longitudinal wave, the particle of the medium oscillate parallel to the direction of propagation of the wave.

Choose and write the correct option:

Class 9 Science Chapter 12 Study Of Sound Question 1.
The unit of frequency is ……………………………… .
(a) Hertz
(b) m/s2
(c) Decibels
(d) m/s
Answer:
(a) Hertz

Study Of Sound Class 9 Exercise Question 2.
The normal hearing range for humans is ……………………………… .
(a) 0 Hz to 20 Hz
(b) greater than 20,000 Hz
(c) 20 Hz to 20,000 Hz
(d) none of these
Answer:
(c) 20 Hz to 20,000 Hz

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Class 9th Science Chapter 12 Study Of Sound Question Answer  Question 3.
Sound will not travel through ……………………………… .
(a) Vacuum
(b) Liquid
(c) Solid
(d) Gases
Answer:
(a) vacuum

Class 9 Science Chapter 12 Study Of Sound Question Answer Question 4.
SI unit of ………………………………. is Hertz (Hz).
(a) Wavelength
(b) Frequency
(c) Speed of wave
(d) Velocity
Answer:
(b) frequency

Reflection Of Sound Class 9 Question 5.
The velocity of sound is inversely proportional to the ……………………………… .
(a) Pressure
(b) Square root of temperature
(c) Square root of density
(d) Humidity
Answer:
(c) square root of density

Question 6.
Sound waves with frequency greater than 20 kHz are called ……………………………… .
(a) Infrasound
(b) Ultrasound
(c) Sonic
(d) Damped sound
Answer:
(b) ultrasound

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 7.
The loudness of a sound depends upon ……………………………… .
(a) Amplitude
(b) Speed
(c) Density
(d) Wavelength
Answer:
(a) Amplitude

Question 8.
……………………………… are used in sonography.
(a) High frequency ultrasound
(b) Stationary waves
(c) High frequency infrasound
(d) High frequency micro waves
Answer:
(a) High frequency ultrasound

Question 9.
The ……………………………… receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve.
(a) Cochlea
(b) Tympanic cavity
(c) Stapes
(d) Pinna
Answer:
(a) Cochlea

Find the odd one out:

Question 1.
Bats, rats, cats, dolphins
Answer:
Cats: cannot produce ultrasonic sound.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 2.
Clothes, paper, curtains, mirror
Answer:
Mirror: is a good reflector of sound, while others are poor reflectors.

Question 3.
Submarines, icebergs, internal organ, sunken ships.
Answer:
Internal organ: sonography is used , while for others sonar system is used.

Question 4.
Temperature, density, molecular weight, pressure
Answer:
Pressure: for a fixed temperature, the speed of sound does not depend on the pressure of the gas, all other factors affect speed of sound.

Answer in one sentence:

Question 1.
How can one produce sound?
Answer:
Vibration set up in an object produces sound (or) sound is produced when an object is disturbed and starts vibrating.

Question 2.
What is velocity of sound wave ?
Answer:
The distance covered by a point on the wave in unit time is the velocity of the sound wave.

Question 3.
What is the minimum distance of the reflecting surface to hear an echo ?
Answer:
To be able to hear a distinct echo, the reflecting surface should be at a minimum distance of 17.2 m.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Match the columns:

Question 1.

Column ‘A’ Column B’ Column C
(1) Transverse wave (a) Particles oscillate parallel to direction of propagation (i) Wave produced in a slinky
(2) Longitudinal wave (b) Particles oscillate perpendicular to direction of propagation (ii) Frequency less than 20 Hz
(3) Ultrasound (c) Echo formation is heard under particular conditions (iii) Wave produced in string
(4) Infrasound (d) High frequency waves (iv) Frequency between 20 Hz to 20000 Hz
(5) Audible frequency (e) Low frequency waves (v) Frequency greater than 20000 Hz

Answer:
(1-b- iii),
(2a- i),
(3 – d – v),
(4 – e – ii),
(5 -c- iv)

Question 2.

Column A’ Column ‘B’ Column C
(1) Amplitude (a) T (i) Pitch of sound
(2) Frequency (b) A (ii) Loudness of sound
(3) Wavelength (c) υ (iii) Reciprocal of frequency
(4) Time period (d) λ (iv) v/υ

Answer:
(1 -b – ii),
(2 -c – i),
(3-d – iv),
(4 – a – iii)

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Name the following:

Question 1.
A form of energy which produces sensation of hearing in our ears.
Answer:
Sound energy

Question 2.
Repetitions of sound due to reflection .
Answer:
Echo

Question 3.
The audible range of sound for human being.
Answer:
20 Hz to 20,000 Hz

Question 4.
A method to obtain images of internal organs of the human body.
Answer:
Sonography

Question 5.
The matter or substance through which sound gets transmitted.
Answer:
Solid, liquid, gases

Question 6.
Three major parts of the ear.
Answer:
External ear, the middle ear and the inner ear.

Question 7.
Any two examples in which infrasound is produced.
Answer:
Pendulum, earthquake.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 8.
Name the living beings that can produce ultrasound.
Answer:
Bats, dolphins, mice.

Give scientific reasons:

Question 1.
Bats can navigate in dark.
Answer:

  • The ultrasonic sound produced by bats, gets reflected on hitting an obstacle.
  • This reflected sound is received by their ears and they can locate the obstacle and estimate its distance even in the dark.
  • Hence, bats can navigate in dark.

Question 2.
A SONAR system is installed in a ship.
Answer:

  • A SONAR system determines the depth of the sea.
  • It locates underwater hills, valleys, icebergs, submarines and sunken ships. It also locates the positions of other ships or submarines.
  • Hence a SONAR system is installed in a ship.

Question 3.
Sound travels faster in iron than in air.
Answer:

  • Sound requires a material medium for its propagation and travels in the form of a longitudinal wave.
  • The denser the medium, faster is the propagation of sound.
  • Hence, sound travels faster in iron than in air.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Solve the following:

Type – A

Formula:
\(\text { (i) Velocity }=\frac{\text { distance }}{\text { time }}\)

Question 1.
Ultrasonic waves are transmitted downwards into the sea with the help of a SONAR. The reflected sound is received after 4 s. What is the depth of the sea at that place? (Velocity of sound in seawater = 1550 m/s)
Answer:
Given:
Time to hear echo = 4 sec
Time taken by sound waves to reach the bottom 4 of sea (t) = 4/2 sec = 2 sec
Velocity of sound in sea water (v) = 1550 m/s
To find:
Depth of sea(s) = ?
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 8
The depth of the sea at that place is 3100 m.

Question 2.
A person standing near a hill fires a gun and hears the echo after 1 second. If speed of sound in air is 340 m/s. Find the distance between the hill and the person.
Answer:
Given:
Time to hear echo = 1 sec 1
Time taken (t) = 1/2 sec
Velocity of sound (v) = 340 m/s
To find:
Distance (s) = ?
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 9
Distance between the person and hill is 170 m.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Numerical For Practice

Question 3.
If you hear the thunder 20 seconds after you see the flash of lightning, how far from you has the lightning occurred? (Speed of sound in air = 340 m/s)
Answer:
6800m

Question 4.
Aboy observes smoke from a cannon 3 seconds before he hears the bang. If the cannon is 1020 m from the observer, find the velocity of sound.
Answer:
340 rn/s

Question 5.
A soldier standing between the two buildings fires a gun. He heard the echo of the sounds from the first building after 2 seconds and echo from the second building after 3 seconds. Find the distance between two buildings. (Speed of sound in air = 340 m/s)
Answer:
850m

Type – B

\(Formula:
(i) Velocity = Frequency \times Wavelength
(ii) Velocity =\frac{\text { distance }}{\text { time }}\)

Question 1.
Sound waves of wavelength 1 cm have a velocity of 340 mIs in air. What is their frequency? Can this sound be heard by the human ear?
Answer:
Given:
wave length (λ) = 1cm = 1/100
Velocity of sound (v) = 340 m/s
To fInd :
frequency (u) = ?
Formulae:
ν = υ λ
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 10
The frequency of the sound waves is 34000 Hz. The frequency is higher than 20000 Hz and therefore, this sound cannot be heard by the human ear.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 2.
How long will it take for a sound wave of 25 cm wavelength and 1.5 kHz frequency, to travel a distance of 1.5 km?
Answer:
Given:
frequency (u) = 1.5 kHz = 1500 Hz
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 11
\(\begin{array}{l}
=\frac{1500}{375} \\
=4 \mathrm{sec}
\end{array}\)
The sound wave takes 4 sec to travel the distance of 1.5 km.

Question 3.
Calculate distance travelled by a sound wave having frequency 1000 Hz and wavelength 0.25 m, if it travels for 5 seconds in a certain medium.
Answer:
Given:
frequency (u) = 1000 Hz
wavelength (λ) = 0.25 m
time (t) = 5 seconds
To find :
Distance (d) =?
Formulae:
ν = υ λ
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 12
The distance travelled by the sound wave is 1250 m.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 4.
The audible range of sound is 20 Hz to 20000 Hz. At 22°C in air speed of sound is 344 mIs. Express the range of sound in terms of wavelength by calculating the respective values.
Answer:
Given:
frequency (u1) 20 Hz
frequency ( u2) = 20,000 Hz
velocity (v) = 344 rn/s
To find :
Wavelengths λ1 and λ2 = ?
Formulae:
ν = υ λ
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 13
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 14
Audible range of wavelength of sound is from 17.2 x 10-3 m to 17.2 m.

Numerical For Practice

Question 5.
A sound wave has frequency 320 Hz and wavelength 0.25 m. How much distance will it travel in 10 second?
Answer:
The distance travelled is 800 m.

Type – C

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 15

Question 1.
Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gin respectively. In which bottle will sound travel faster? How many timés as fast as the other?
Answer:
Given:
Mass of hydrogen in bottle A (mA) = 12gm
Mass of hydrogen in bottle B(mB) = 48gm
To find:
In which bottle sound travels faster.
Formulae:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 16
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 17
Since both bottles are identical hence, the volume is the same, i.e. v
Dividing (j) and (ii),
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 18

(i) Vivacity of sound will be more in bottle A.
(ii) Velocity of sound in bottle A (VA) is twice of that in bottle B (vB)

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Numerical For Practice

Question 2.
Argon gas is filled in two identical bottles X and Y. The mass of the gas in the two bottles is 5 gm and 25gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw?
Answer:
(Temperature of Y is 5 times the temperature of X.)

Type – D

Numerical For Practice

Question 1.
Velocity of sound in air at 0°C is 332nVs. It increases by 0.6ni/s for each °Celsius rise in temperature. At what temperature of ait the velocity will be 359m1s?
Answer:
45°C

Question 2.
Velocity of sound In air at 0°C is 332m/s It increases by 0.6mIs for each degree Celsius rise In temperature. What will be the velocity of sound at 60°C?
Answer:
368 rn/s

Define the following:

Question 1.
Wave length (λ)
Answer:
The distance between two consecutive compressions (or crests) or two consecutive rarefactions (or troughs) is called the wavelength.

Question 2.
Amplitude (A)
Answer:
The maximum value of pressure or density is called amplitude.

Question 3.
Frequency (υ)
Answer:
The frequency of a sound wave is defined as the number of complete oscillations of density (or pressure of the medium) per second.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 4.
Time Period (T)
Answer:
The time taken for one complete oscillation of pressure or density at a point in the medium is called the time period.

Question 5.
Echo
Answer:
An echo is the repetition of the original sound because of reflection by some surface.

Question 6.
Transverse waves
Answer:
Oscillations of the particles of the medium vibrate at right angles to the direction of propagation of the wave are called transverse waves.

Question 7.
longitudinal waves
Answer:
The particles of the medium oscillate about their central or mean position in a direction parallel to the propagation of wave is called as longitudinal waves.

Question 8.
Velocity of wave
Answer:
The distance covered by a point on the wave (for example the point of highest density or lowest density) in unit time is the velocity of the sound wave.’

Distinguish between:

Question 1.
Infrasound and Ultrasound
Answer:

Infrasound Ultrasound
(i) Longitudinal waves whose are below 20 Hz are called Infrasound waves. frequencies Infrasonic or (i) Longitudinal waves whose frequencies lie- above 20,000 Hz are called Ultrasonic or ultrasound waves.
(ii) Whales, elephants produce sound in the infrasound range. (ii) Bats produce (30 kHz to 50 kHz) frequency and dolphins produce ultrasound (100 kHz).

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 2.
Transverse waves and Longitudinal waves
Answer:

Transverse waves Longitudinal waves
(i) Particles of the medium vibrate at right angles to the direction of propagation of the wave.
(ii) They produce crests and troughs.
(iii) For transverse waves, a wavelength is made up of one crest and one trough.
(i) Particles of the medium vibrate parallel to the direction of propagation of the wave.
(ii) They produce compression and rarefaction.
(iii) For longitudinal waves, a wavelength is made up of one compression and one rarefaction.

Question 3.
Consider two cases
(A) whistle of train (B) roar of a lion

(I) In which case the sound is high pitch?
Answer:
Whistle of a train is high pitch as compared to roar of a lion, as the frequency is higher.

(II) What is the real cause of sound production? Explain with examples.
Answer:

  • Vibrations in the object are responsible to produce a sound.
  • Vibration is a rapid to and fro motion of an object.
  • Sometimes the vibrations may be strong enough to be seen by eyes, e.g. string vibrations in string instruments, vibration on mobile phone, blowing air in the cap of your pen by holding it near the lips.

(III) Three sounds 5 Hz, 500 Hz and 50,000 Hz are produced by different sources.
(a) Which sound will be heard by humans?
(b) Which sounds may be produced by bats?
(c) Which sounds may be produced by elephants?
Answer:
(a) 500 Hz – Humans can hear sounds in the range of 20 Hz-20,000 Hz
(b) 50,000 Hz – Bats produce ultrasonic sounds above 20,000 Hz
(c) 5 Hz – Elephants can produce infrasonic sounds below 20 Hz

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 6.
Suppose you and your friend are on the moon. Will you be able to hear any sound
Answer:
Sound waves need a material medium for their propagation. Since there is no atmosphere on the moon, we cannot hear any sound on the moon.

Answer in detail:

Question 1.
What are the factors on which velocity of sound in gaseous medium depend?
Answer:
The velocity of sound in a gaseous medium depends on the physical conditions i.e. the temperature, density of the gas and its molecular weight.

  1. Temperature (T): The velocity of sound is directly proportional to the square root of the temperature of the medium. This means that increasing the temperature four times doubles the velocity.
    \(\text { v } \alpha \sqrt{\mathrm{T}}\)
  2. Density(p): The velocity of sound is inversely proportional to the square root of density. Thus, increasing the density four times, reduces the velocity to half its value.
    \(\mathrm{v} \alpha \frac{1}{\sqrt{\rho}}\)
  3. Molecular weight (M): The velocity sound is inversely proportional to the square root of molecular weight of the gas. Thus, increasing the molecular weight four times, reduces the velocity to haff its value.
    \(\mathrm{v} \alpha \frac{1}{\sqrt{\mathrm{M}}}\)

Question 2.
What are the uses of ultrasonic sound?
Answer:
Uses of ultrasonic sound are as follows:

  • For communication between ships at sea.
  • To join plastic surfaces together.
  • To sterilize liquids like milk by killing the bacteria in it so that the milk keeps for a longer duration.
  • Echocardiography which studies heartbeats, is based on ultrasonic waves (Sonography technology).
  • To obtain images of internal organs in a human body.
  •  In industry to clean intricate parts of machines where hands cannot reach.
  • To locate the cracks and faults in metal blocks.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 3.
Explain with the help of a neat labelled diagram the working of human ear.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 22

  • The ear is an important organ of the human body.
  • When sound waves fall on the eardrum, it vibrates and these vibrations are converted into electrical signals which travel to the brain through nerves.
  • The ear can be divided into three parts:
    (a) Outer ear
    (b) Middle ear
    (c) Inner ear.

(a) Outer ear or Pinna
The outer ear collects the sound waves and passes them through a tube to a cavity in the middle ear. Its peculiar funnel like shape helps to collect and pass sounds into the middle ear.

(b) Middle ear
There’ is a thin membrane in the cavity of the middle ear called the eardrum. When a compression in a sound wave reaches the eardrum, the pressure outside it increases and it gets pushed inwards. The opposite happens when a rarefaction reaches there. The pressure outside decreases and the membrane gets pulled outwards. Thus, sound waves cause vibrations of the membrane.

(c) Inner ear
The auditory nerve connects the inner ear to the brain. The inner ear has a structure resembling the shell of a snail. It is called the cochlea. The cochlea receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve. The brain analyses these signals.

Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound

Question 4.
Write a short note on SONAR
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound 23
(i) SONAR is the short form for Sound Navigation and Ranging. It is used to determine the direction, distance and speed of an underwater object with the help of ultrasonic sound waves. SONAR has a transmitter and a receiver, which are fitted on ships or boats.

(ii) The transmitter produces and transmits ultrasonic sound waves. These waves travel through water, strike underwater objects and get reflected by them. The reflected waves are received by the receiver on the ship.

(iii) The receiver converts the ultrasonic sound into electrical signals and these signals are properly interpreted. The time difference between transmission and reception is noted. This time and the velocity of sound in water give the distance from the ship, of the object which reflects the waves.

(iv) SONAR is used to determine the depth of the sea. SONAR is also used to search underwater hills, valleys, submarines, icebergs, sunken ships etc.

Question 5.
Write a short note on Sonography. How is it misused?
Answer:

  • Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
  • This is useful in finding out the cause of swelling, infection, pain, condition of the heart, the state of the heart after a heart attack as well as the growth of foetus inside the womb of a pregnant woman.
  • This technique makes use of a probe and a gel.
  • The gel is used to make proper contact between the skin and the probe so that the full capacity of the ultrasound can be utilized. Maharashtra Board Class 9 Science Solutions Chapter 12 Study of Sound
  • High-frequency ultrasound is transmitted inside the body with the help of the probe.
  • The sound reflected from the internal organ is again collected by the probe and fed to a computer which generates the images of the internal organ.
  • As this method is painless, it is increasingly used in medical practice for correct diagnosis.
  • This technique is used by many people to find out gender of an unborn baby and this often leads to the incidence of female foeticide.

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Question 1.
Pair up with your partner and answer the following questions.
(a) Do you like to travel?
(b) Where have you traveled?
(c) Have you heard about the wonders of the world?
(d) Have you heard about UNESCO?
(e) What does UNESCO stand for?
(f) What is meant by Heritage?
(g) Do you know of any Heritage building in your own city?
Answer:
(a) yes, no, sometimes, etc.
(b) Discuss the places, the mode of travel, the sights, etc.
(c) Yes, I have.
(d) Yes, I have, but not much.
(e) UNESCO stands for United Nations Educational, Scientific and Cultural Organization.
(f) Heritage is something that is valued and preserved because of its historical/cultural/natural importance
(g) Students can find out the heritage buildings in their own towns/cities.

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Question 2.
Your teacher will explain the ‘Heritage Cycle’.
→ By understanding (cultural heritage) people value it.
→ By valuing it people want to care for it.
→ By caring for it, it will help people enjoy it.
→ From enjoying it, comes a thirst to understand.
→ By understanding it …………………
Maharashtra Board Class 10 English Solutions Unit 4.3 World Heritage 1

Question 3.
Do you know the difference between :
(a) World Heritage
(b) Cultural Heritage
Refer to a dictionary/encyclopedia/reference book/the internet to get the detailed meaning.
Answer:
(a) World heritage means a global heritage that belongs to everyone.
(b) Cultural heritage is the legacy of physical artifacts and intangible attributes of society that are inherited from past generations, maintained in the present and preserved for the benefit of futurft generations. Cultural heritage includes tangible culture (such as buildings, monuments, landscapes, books, works of art, and artifacts), intangible culture (such as folklore, traditions, language, and knowledge), and natural heritage (including culturally significant landscapes, etc.)

Question 4.
With the help of your partner complete the information in the table.

Tourist spot Favourite Why?
Park
Mountain
Beach
Sea
Forest
Countryside/Rural site

Question 5.
Heritage Sites – Rank these with your partner. Put the best at the top.

  • Great Barrier Reef
  • Mount Fuji
  • Grand Canyon
  • The Pyramids
  • Panda Sanctuaries
  • Machu Picchu
  • Vatican City
  • Great Wall of China

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World Heritage Class 10 English Workshop Questions and Answers Maharashtra Board

Question 1.
Answer the following questions.
(a) What do heritage sites in the world include?
Answer:
A World Heritage Site should have significant cultural or natural importance to humanity.

(b) What kind of sites are protected and maintained by the International World Heritage Programme?
Answer:
Sites which have significant cultural or natural importance to humanity are protected and maintained by the International World Heritage Programme. They may include forests, lakes, monuments, buildings and cities. They can also be a combination of cultural and natural areas.

(c) How many world Heritage sites were there around the world?
Answer:
World Heritage Sites include forests, lakes, monuments, buildings and cities. They cari also be a combination of cultural and natural areas.

(d) What is the role of World Heritage Committees?
Answer:
World Heritage Sites are those that are significant culturally and naturally. People are interested in seeing such sites and learning the history behind them; hence they are a major tourist attraction, and they draw tourists to a country, boosting the country’s revenue.

(e) What is the texture of World Heritage Committee?
Answer:
The tenure of the World Heritage Committee is six years.

(f) What are our duties towards preservation/conservation of any historical site?
Answer:
We should see that the historical sites are not in any danger due to pollution, tourism, uncontrolled urbanization, etc. Whenever we visit the site we must maintain the cleanliness and purity of the place. We must obey whatever orders and guidelines are put in place by the authorities. We should motivate others to do so too.

(g) Why should we preserve the World Heritage Sites?
Answer:
World Heritage Sites promote tourism. They are a part of the culture of the world, a part of the past. It is essential to preserve the past in order to learn from it. Natural beauty too should be preserved so that everyone today and in future can enjoy it. Hence, we should preserve World Heritage Sites.

(h) What is the role of World Heritage Sites in developing tourism in any country?
Answer:
World Heritage Sites are those that are significant culturally and naturally. People are interested in seeing such sites and learning the history behind them; hence they are a major tourist attraction, and they draw tourists to a country, boosting the country’s revenue.

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Question 2.
Complete the following.

Year Establishment Role
1954 ………………………………………………… …………………………………………………
1959 ………………………………………………… …………………………………………………
1965 ………………………………………………… …………………………………………………
1968 ………………………………………………… …………………………………………………
1972 ………………………………………………… …………………………………………………

Answer:

Year Establishment Role
1954 Egypt starts plans to build Aswan High Dam Started the momentum for the protection of cultural and natural heritage sites around the world.
1959 International campaign by UNESCO to protect temples and artifacts likely to be destroyed by Aswan High Dam; a draft convention initiated for the same Protection of natural and cultural sites around the world
1965 A White House Conference in the US called for a World Heritage Trust’. To protect not only the historic and cultural sites but also the significant natural and scenic sites around the world.
1968 The International Union for Conservation of Nature Development of goals similar to those of the White House Conference
1972 Convention concerning the protection of World Cultural and Natural Heritage adopted by UNESCO’s General Conference. To protect not only Are histone and cultural sites but also the significant natural and scenic sites around the world.

Question 3.
Complete the following by giving reasons why World Heritage Sites are in danger.
Maharashtra Board Class 10 English Solutions Unit 4.3 World Heritage 2
Answer:
Maharashtra Board Class 10 English Solutions Unit 4.3 World Heritage 5

Question 4.
Choose the correct alternative and complete the given sentences.
(a) Mount Huangshan is situated in …………………………. .
(i) Japan
(ii) China
(iii) Philippines
Answer:
(ii) China

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(b) The famous dam situated in Egypt on River Nile is …………………………. .
(i) Buzwaa High Dam
(ii) Rizwa High Dam
(iii) Aswan High Dam
Answer:
(iii) Aswan High Dam

(c) The conference based on Human Environment was held at …………………………. .
(i) Athens, Greece
(ii) Mascow, Russia
(iii) Stockholm, Sweden
Answer:
(iii) Stockholm, Sweden

(d) The highest number of world Heritage Sites are located in …………………………. .
(i) France
(ii) Italy
(iii) Germany
Answer:
(ii) Italy

(e) …………………………. can delete/exclude a site from World Heritage list.
(i) World Heritage Committee
(ii) World Peace Committee
(iii) World Health Committee.
Answer:
(i) World Heritage Committee

(f) The tenure of World Heritage Committee is …………………………. years.
(i) Four
(ii) Five
(iii) Six.
Answer:
(iii) Six.

Question 5.
Match the pairs to define different roles of the World Heritage Sites.

A B
(i) Identify cultural and natural sites a green, local based, stable and decent jobs.
(ii) Identify sites of b of outstanding universal value across countries.
(iii) Identify sites that represent c preserve outstanding sites and natural resources.
(iv) UNESCO seeks to d tourism.
(v) World Heritage Sites should have e an asset for economic development and investment.
(vi) World Heritage Sites should f protect these sites.
(vii) World Heritage Sites serve as g best examples of world’s cultural and/or natural heritage.
(viii) World Heritage Sites should ensure h special importance for everyone.
(ix) It should at large develop i relevant development plan policies.

Answer:

‘A’ ‘B’
(1) Identify cultural and natural sites (a) green, local based, stable and decent.
(2) Identify sites of (b) of outstanding universal values across countries.
(3) Identify sites that represent (c) preserve outstanding sites and natural resources.
(4) UNESCO seeks to (d) tourism.
(5) World Heritage Sites should have (e) an asset for economic development and investment.
(6) World Heritage should (f) protect these sites.
(7) World Heritage Sites serve as
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(g) best examples of world’s cultural and or natural heritage.
(8) World Heritage Site should ensure (h) special importance for everyone.
(9) It should at large develop (i) relevant development plan policies.

Question 6.
Look at the words and their meanings. Choose the correct alternative.
(a) determine :
(i) think over
(ii) decide
(iii) ask for
(iv) look over
Answer:
(ii) decide

(b) monument :
(i) statue
(ii) pillar
(iii) memorial
(iv) fort
Answer:
(iii) memorial

(c) significant:
(i) clever
(ii) effective
(iii) systematic
(iv) important
Answer:
(iv) important

(d) disaster:
(i) problem
(ii) incident
(iii) calamity
(iv) accident
Answer:
(iii) calamity

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Question 7.
Find from the text minimum 8 words related to cultural heritage and make a word register. Arrange them in alphabetical order.
Cultural heritage: Abu Simbel Temples, artifacts, buildings, cilles, Histqric Center of Vienna, Machu Picchu, monuments, Mount Huangshan, Sydney Opera House.

Question 8.
Complete the following information from the text.
Maharashtra Board Class 10 English Solutions Unit 4.3 World Heritage 3
Answer:
(a) Preservation of any historical site (Duties):
(i) International World Heritage Programme
(ii) UNESCO World Heritage Committee.

(b) Preserving World Heritage Sites (Reasons) :
(i) They have significant cultural and natira Importance to humanity.
(ii) They promote tourism.

(C) Promoting tourism (Role) :
(i) They attract tourists.
(ii) They help one to understand the history of a place.

Question 9.
Write a report on any tourist place/historical place/World Heritage Site you recently visited using the points given below.
(i) Title
(ii) Place/Location
(iii) Background/History
(iv) Features/Specialities
(v) Security policies
(vi) Sign boards and discipline
(vii) Overall scenario.
Answer:
Ajanta – A Masterpiece In Rock
– Rani Iyer
Mumbai, 12 February: The Ajanta Caves, situated in the Deccan in Maharashtra and about 110 km from Aurangabad, are a UNESCO World Heritage Site. A visit to the place reveals a large slice of history.

These rock – cut cave temples and monasteries of Buddhists date back to around 2nd century BCE. Since they were in the jungles, cut into a mountainside, the external world did not know about them for a long time. They were discovered only in the 19th century, The caves include paintings and rock – cut sculptures described as among the finest surviving examples of ancient Indian art. The paintings are expressive and present emotion through gesture, pose and form, According to UNESCO, these are masterpieces of Buddhist religious art that influenced the Indian art I that followed.

Two new visitor centers provide extensive information about the heritage site using audiovisual media. Local staff is employed for security purposes. Though photography is allowed at certain places, with fees for the use of a camera, use of tripods and flash is prohibited. Signboards and brochures/leaflets provide information about the care to be taken to preserve this UNESCO protected heritage site. “The caves are now being looked after by a private company under the Indian government’s ‘Adopt a Heritage Site’ program,” said a senior official.

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Question 10.
(A) Use ‘not only but also’ in the following sentences.
(a) UNESCO and the International Council on monuments initiated a draft convention to create an international organisation responsible for protecting cultural heritage.
(b) The state parties are responsible for identifying and nominating new sites.
(c) Opera House in Australia and the Historic Center of Vienna in Austria are cultural sites of World Heritage.
(d) World Heritage Sites can also be a combination of both cultural and natural areas.
Answer:
(a) Not only UNESCO but also the International Council on Monuments and Sites initiated a draft convention to create an international organization responsible for protecting cultural heritage.
(b) The State Parties are responsible for not only identifying but also nominating new sites.
(c) Not only the Opera House in Australia but also the Historic Center of Vienna in Austria are cultural sites of World Heritage.
(d) World Heritage Sites can be a combination of not only cultural but also natural areas.

(B) Rewrite the following sentences replacing ‘as soon as’ by ‘No sooner than’
(Note : The phrase, ‘No sooner’ must always be followed by an auxiliary (helping verb).)
For example : As soon as he came, they all gave an applause.
No sooner did he come, than they all gave an applause.

(a) As soon as the Bill is passed, it will become an Act.
(b) As soon as the thief escaped, the family informed the police.
(c) As soon as you have finished, you can submit your answer- papers.
(d) As soon as they can manage, they should change their house.
(e) As soon as the bell rings, the School Assembly will start.
Answer:
(a) No sooner is the Bill passed, than it will become an Act.
(b) No sooner did the thief escape, than the family informed the police.
(c) No sooner do you finish, than you can submit your answer papers.
(d) No sooner can they manage, than they should change their house.
(e) No sooner does the bell ring, than the School Assembly will start.

Question 11.
Identify whether the following sentences are Simple (One Subject + One Predicate or Complex (One Main Clause + One or more Dependent Clauses) or Compound (Combination of 2 or more Independent/ Co-ordinate Clauses).
(1) They vary in type but they include forests, monuments etc.
(2) The mountain is significant because of its characteristics.
(3) To protect the temples and artefacts, UNESCO launched an international campaign.
(4) If the site meets with this criteria, it can be inscribed on the World Heritage List.
(5) There are 890 World Heritage Sites that are located in 148 countries.
(6) The project cost about US $ 80 million and $ 40 million came from 50 different countries.
Answer:
(1) Compound
(2) Simple
(3) Simple
(4) Complex
(5) Complex
(6) Compound

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Question 12.
Following are the ways to preserve ‘World Heritage Sites.’ Transfer these points into a small paragraph. Suggest a title.

  • Practical conservation of posterity
  • Human/animal trespassing to be prohibited
  • Monitored/controlled/restricted access
  • Threat of local administrative negligence to be removed

Answer:
Ways to Preserve World Heritage Sites
World Heritage Sites need to be preserved for posterity. The future generations must know about their history and culture. For this purpose, steps have to be taken to preserve these sites. First and foremost, trespassing by either humans or animals should be strictly prohibited. Access to these sites for tourists, vendors, officials and the general public should be restricted or monitored carefully. Another important step is to keep a check on the local administration of the site, so that negligence in the performance of the duties at this level is ruled out.

Question 13.
Read the points given in column ‘A’ specifying Dos towards prevention of any historical site. Write Don’ts in column ‘B’ specifying things that should not be done.

Dos Don’ts
(i) Obey rules and regulations.
(ii) Maintain discipline and order.
(iii) Maintain cleanliness.
(iv) Use dustbins and garbage bags.
(v) Observe silence.
(vi) Maintain environmental safety.
(vii) Protect our country’s heritage.

Answer:

Dos Don’ts
1. Obey rules and regulations. 1. Don’t dirty the surroundings.
2. Maintain discipline and order. 2. Don’t write on the walls/trees.
3. Maintain cleanliness. 3. Don’t pluck flowers/ destroy plants.
4. Use dustbins and garbage bags. 4. Don’t smoke/ drink alcohol.
5. Observe silence. 5. Don’t play loud music or make a loud noise.
6. Maintain environmental safety. 6. Don’t defecate in the open.
7. Protect our country’s heritage. 7. Don’t trespass.

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Question 14.
Complete the following flow chart by choosing the option given below to show how any site of any country can become a World Heritage Site.
Maharashtra Board Class 10 English Solutions Unit 4.3 World Heritage 4
Options :
1. Inclusion of the name of site from tentative list to the nomination file.
2. Name of the site is inscribed on the World Heritage List after meeting the criteria.
3. Inclusion of the name of site for the nomination in a tentative list after an inventory in country or State.
4. Decision of the World Heritage Committee after review of the nominated file.
5. A review of the included file by the advisory bodies.
Answer:
Step 1 : Inclusion of the name of the site for nomination in a tentative list after inventory in country or state.
Step 2 : Inclusion of the name of site from tentative list to the nomination file.
Step 3 : A review of the included file by the advisory bodies.
Step 4 : Decision of the World Heritage Committee after review of the nominated file.
Step 5 : Name of the site inscribed on the World Heritage List after meeting the criteria.

Question 15.
Projects :
(a) Make a list of sites from our State which are included in the World Heritage Sites. Try to visit one of them. Write the importance of this World Heritage Site. Also write your impression of it in your notebook.

(b) Write a ‘tourism leaflet’ on any one of the following :
(i) Your home town
(ii) A historical place
(iii) A place of natural beauty
(iv) A place of pilgrimage

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→ Make use of the following points :
(i) How to reach there?
(ii) Accommodation facilities
(iii) What to see and visit?/Attractions of the place
(iv) Places of interest nearby
(v) Best time to visit
(vi) Shopping Attractions
Add your own points.
Answer:
MATHERAN: ‘UNSPOILT NATURE’
(1) How to get there:
(a) By train: From Mumbai or Pune to Neral Junction.. From Neral to Matheran by narrow – gauge train that depart at fixed times.
(b) By road: Shared taxis or minibuses from Neral to MTDC Holiday Camp. Cars and other vehicles not allowed beyond a certain point.
(2) Accommodation facilities: Plenty of hotels offering full or half board. MTDC Tourist Camp provides dorm – beds at affordable rates. The ‘camp’ is near the Dasturi car park. You can eat at one of the many thali joints along MG Road.
(3) Best Time of the year to visit: All the year round except the rainy season.
(4) What to see: Viewpoints such as Porcupine, Louisa and Echo have the finest views. On a clear day, one can see Mumbai lying afar from ‘Hart Point’; perpendicular cliffs plunge into steep ravines; monkeys and squirrels.
(5) Shopping Attractions: Locally made chappals, home – made chocolates and chikkis, caps and other items for campers; walking sticks.
(6) Special features: Greenery, nature at its best, trekking, horse riding; good for a day’s group picnic.

(c) Vocabulary Extension – Choose several words from the text. Use a dictionary or internet to build up more associations/collocations of each word.
Answer:
(1) Associations:

(2) Collocations:

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(d) World Heritage Sites – Make a poster about World Heritage Sites, specifying some details and specialties about each of them.

(e) Letter – Write a letter to an expert on the environment. Ask him/her five questions about your concern/doubts about World Heritage sites. Give him/her three ideas on how to protect them.
Answer:
Amit Awte
102, Riddhi Vihar
R.N. Road
Aurangabad – 431 001,
12th November, 2020

To
Mr. Avinash Ranade
‘Nisarg’
Shastri Marg
Aurangabad – 431 002.
Sub: Protection of World Heritage Sites

Sir
It was with great interest that I read in our local newspaper about your interest in the protection of World Heritage Sites. I would like to ask you a few questions which have haunted me for some time.

They are:
(1) Does India have enough experts to restore the damage done to our sites by nature/visitors etc.?
(2) Is there enough security to protect the sites from vandalism?
(3) Can there be a restriction on the number of visitors to the sites?
(4) Are the funds allocated to the preservation of the sites by the World Heritage Committee sufficient?
(5) Is there enough information about the sites on tourist websites/ Internet?

I would like to give some suggestions. Can we not impose a limit on the number of tourists visiting these places? We can also have sessions on how to maintain cleanliness and prevent vandalism. Stiff fines must be imposed on all those who break the rules.

I hope to receive a reply from you, as I am also very concerned about the preservation of our heritage sites.

Yours faithfully,
Amit Awte

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(f) Article – Write an article for a magazine about the World Heritage sites at risk. Include imaginary interviews with people who are trying to save them. Read your article in front of your class.

Question 16.
The project of dismantling and moving the temples in the valley to higher ground cost $80 million.
(i) $40 million
(ii) $50 million
(iii) $80 million.
Answer:
(iii) $80 million.

Question 17.
Complete the following: (The answers are given directly and underlined.)
(1) A World Heritage Site should have significant …………………
(2) World Heritage Sites include ……………………..
(3) They can also be a ……………………..
(4) The convention concerning the protection of  Cultural Natural Herttage was adopted by ……………………
Answer:
(1) cultural or natural importance to humanity.
(2) combination of cultural and natural areas.
(3) forests, lakes, monuments, buildings and cities.
(4) UNESCO’s General Conference on November 16, 1972.

Question 18.
(a) UNESCO launched an international campaign in 1959.
Answer:
In 1954, Egypt started plans to build the Aswan High Dam. The initial plans for the dam’s construction would have flooded the valley containing the Abu Simbel Temples and scores of ancient Egyptian artifacts. To protect the temples and artifacts, UNESCO launched an international campaign in 1959, that called for the dismantling and movement of the temples to higher ground.

(b) A White House Conference in the United States called for a ‘World Heritage Trust’.
Answer:
A White House Conference in the United States called for a World Heritage Trust’ to protect the world’s historic and cultural sites as well as the significant natural and scenic sites.

(c) convention:
(i) typical
(ii) agreement
(iii) old – fashioned
(iv) persuade
Answer:
(d) agreement

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Question 19.
Pick out 4 adjectives from the passage ending with the suffix ‘ – al’.
Answer:
1. educational,
2. cultural,
3. natural,
4. historical.

Question 20.
The World Heritage Committee meets once a year.
(i) once a year
(ii) twice a year
(iii) three times a year
Answer:
(i) once a year

Question 21.
A nominated site has to be first included in a Tentative List.
(i) World Heritage List
(ii) Nomination File
(iii) Tentative List
Answer:
(iii) Tentative List

Question 22.
Explain what the World Heritage Committee is responsible for.
Answer:
The World Heritage Committee is the main group responsible for establishing which sites will be listed as UNESCO World Heritage Sites.

Question 23.
Who makes recommendations to the World Heritage Committee?
Answer:
The recommendations to the World Heritage Committee are made by two Advisory Bodies, the International Council on Monuments and Sites and the World Conservation Union.

Question 24.
(1) Choose the correct noun forms from those given in the brackets:
(1) inscribed (inscription/inscribtion)
(2) responsible (responsive/responsibility)
(3) nominated (nominative/nomination)
(4) included (inclusion/inclution)
Answer:
(1) inscription
(2) responsibility
(3) nomination
(4) inclusion.

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Question 25.
Complete the following by choosing the correct nouns from the box:
Bodies Parties List File
(1) Tentative ……………….
(2) Nomination ……………….
(3) Advisory ……………….
(4) State ……………….
Answer:
(1) Tentative List
(2) Nomination File
(3) Advisory Bodies
(4) State Parties.

Question 26.
World Heritage Committee can delete/exclude a site from the World Heritage List
(a) World Heritage Committee
(b) World Peace Committee
(c) World Health Committee
Answer:
(a) World Heritage Committee

Question 27.
There are 890 World Heritage Sites around the World.
(a) 689
(b) 890
(c) 36
Answer:
(b) 890

Question 28.
36 – sites have been included from India.
(a) 176
(b) 44
(c) 36
Answer:
(c) 36

Question 29.
Match the places with the countries:

‘A’ ‘B’
(1) Sydney Opera House (a) Vienna
(2) Historic Center (b) Peru
(3) Grand Canyon National Park (c) Australia
(4) Machu Pichhu (d) United States

Answer:

‘A’ ‘B’
(1) Sydney Opera House (c) Australia
(2) Historic Center (a) Vienna
(3) Grand Canyon National Park (d) United States
(4) Machu Pichhu (b) Peru

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Question 30.
Complete the following:
(1) Peru’s Machu Picchu is an example of a ………………….
(2) The World Heritage Committee can choose to delete a site from the list if ……………….
Answer:
(1) mixed site, both cultural as well as natural.
(2) the site loses the characteristics which allowed for it to originally be included on the World Heritage List.

Question 31.
Explain the role of the World Heritage Committee.
Answer:
The World Heritage Committee allocates resources from the World Heritage Fund to a World Heritage Site which is in danger due to any reason or in need of protection or restoration. If a site loses the characteristics which allowed for it to originally be included on the World Heritage List, the World Heritage Committee can choose to delete the site from the list.

Question 32.
Choose the correct meanings:
(a) allocate:
(i) allow
(ii) distribute
(iii) catch
(iv) understand
Answer:
(b) distribute

Question 33.
Fill in the blanks with the correct words from the passage: (The answers are given directly and underlined.)
(a) Many small towns of the previous century have become crowded cities now due to rapid ………………..
(b) We should be well – prepared for …………….. problems like global warming.
Answer:
(a) urbanization.
(b) environmental

Question 34.
(1) Pick out a gerund from the lesson and use – it in your own sentence.
(2) Find out five hidden words from the given word: international
(3) Use the following phrase in your own sentence: a corhbination of
(4) Spot the error/errors and rewrite the correct sentence: The State Parties is then responsible for identifying and nominating new sites.
(5) Identify the type of sentence: Get out of my way.
(6) Punctuate: if however a site loses the characteristics which allowed for it to be originally included on the world heritage list the world heritage committee can choose to delete the site from the list.
(7) From the following verbs, pick out the verb which forms its present and past participle by doubling the last letter. run, lose, trip, quit
(8) Arrange the following words in alphabetical order: Sydney, Vienna, Austria, Peru, Australia, Egypt, China
Answer:
(1) protecting: We should think of various ways of protecting our environment.
(2) internatIonal : natIonal, nation, train, trail, trial.
(3) The rainbow is a combination of seven colours.
(4) The State Parties are then responsible for identifying and nominating new sites.
(5) Imperative sentence.
(6) If however, a site loses the characteristics which allowed for it to be originally included on the World Heritage List, the World Heritage Committee can choose to delete the site from the list.
(7) trip : tripped, trIpping.
(8) Australia, Austria, China, Egypt, Peru, Sydney, Vienna

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Question 35.
(1) Use the following word and its homophone in two separate sentences : rain
(2) UNESCO launched an International campaign In 1959. (Rewrite beginning ‘An lnternattonal…..)
(3) DIfferent plans are put Into place. (Rewrite using the future perfect tense of the verb.)
Answer:
(1) (i) There was light rain yesterday evening near my house.
(ii) In the past, a cõuntry progressed dutlng the reign of good kings.
(2) An international campaign was launched by UNESCO in 1959.
(3) Different plans will have been put into place.

Question 36.
(1) Italy has the highest number of World Heritage Sites. (Rewrite using the comparative form.)
(2) The World Heritage Committee meets once a year to review these recommendations. (Rewrite as a compound sentence.)
Answer:
(1) Italy has a higher number of World Heritage Sites than any other country.
(2) The World Heritage Committee meets once a year and (it) reviews these recommendations.

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