Skip to main content

8th Science Power -Conservation of Plants and animals [Questions with solution]

Formative and Summative assessments 8th Science Power
Topic For Group discussion:
Q. Saving paper helps to conserve our natural resources.
Organize a seminar on the topic: Flora and fauna and its diversity.
Prepare a project report on the wild life which has reached on the verged of extinct.
General Questions:
1. Define these terms:
a. Flora:
The plants found in a particular area is called flora of that area
b. Wild life sanctuary
Wildlife sanctuaries is protected area reserved for wild animals, birds and plants
c. Biosphere reserve
The protected area in which multiple use of land is permitted to conserve biodiversity.
d. Fauna
The animals found in a particular area are called fauna of that area.
2. Differentiate between the following:
a. Endangered and Extinct species
b. Endemic and exotic species
3. Answer these Questions:
a. What is red Data book?
Answer: Red data book contain a record of all those species of plants and animal which are under the threat of extinction or are rare and vulnerable for extinction.
b. Define deforestation and Afforestation and Give consequences of deforestation?
Answer: Deforestation is the practice of cutting down of trees whereas the practice of planting more trees is called Afforestation.
Following are the consequences of deforestation:
a. Global warming      b. Climatic change         c. Desertification    d. Drought and flood
e. Soil erosion and loss of wild life
c. Mention the effect of deforestation on environments, wild animal and soil?
Answer: The effect of deforestation on environments: Deforestation increases the temperature and wind speed which causes global warming.
d. Why is tiger an endangered species?
Answer: Tiger is an endangered species because tiger will soon became extinct if the same causative factors continue.
e. What are the major threats to wild life?
Answer: Following are the major threats to wild life:
(i) Habitat loss (ii) Indiscriminate killing and poaching (iii) pollution of air water and soil pollution
f. Why biosphere reserve the best way of wild life conservation?
Answer: This is because biosphere reserves not only help in maintaining the biodiversity but also create awareness to maintain culture of that area.  
g. Human population is a threat to biodiversity? Justify?
Answer: Human population is a threat to biodiversity because more and more land of forest is cleared to fulfill the needed of growing population.
h. List major steps taken by government to conserve biodiversity in India?
Answer: Following are the major steps taken by government to conserve biodiversity in India:
(i) Creation of protected area like park sanctuaries and biosphere reserve
(ii) Making strict low to prevent poaching
(iii) Making captivity for breading
(iv) Starting project to save endangered animals
i. Write notes on (i) project tiger (ii) Gir lion project
Answer:
 (i) Project Tiger: Project tiger is a program launched by government that saves tiger from poaching by making 23 tiger reserve.
(ii) Gir lion project :  Gir lion project started by Gujarat government in 1972 to protect Asian lion.
j. What does IUCN Stands for?
Answer: IUCN Stands for The international Union for Conserve of Nature and Natural Resources.
k. What do you know about Chipko movement?
Answer: Chipko movement is started by Sundar Lal Bahuguna in Uttarakhand to save tree.  The women of village embraced the tree and protest cutting of tree.

Hots Questions:

1. Why rainfall reduced in Cherrapunji which once had the highest rain fall?
Answer: Rainfall in Cherrapunji greatly reduced by loss of forest cover to create space for growing population and for cement industries.
 2. How does over grazing lead to desertification?
Answer: over grazing reduce the grass cover of land and expose the soil to the atmospheric air. The Soil dries up and its humus layer is lost which slowly changed grass land into desert.
Q. Give any two examples of the endemic flora of Pachmarhi Biosphere reserve.
Answer.  Sal and wild mango.
Q. Why is species of some animal become endangered?
Answer. Some species of animal become endangered, because of the disturbances in their natural habitat.
Q. Why migratory birds migrated from one place 'to another?
Answer. Migratory birds migrate to overcome climatic changes. For example, Surkhab migrated from Malaysia.
8th  Conservation of Plants and Animals 

Conservation of Plants and Animals Download File

Check point[Formative Assessment] Download File

8th Conservation of Plants and Animals[Questions with Solution]

8th Conservation of Plants and animals : Science power Book for class 8th

8th Conservation of Plants and animals -Formative Assessment- Oral Questions - Fill in the blanks - Match the Columns - Think Zone – Puzzle /Quiz –and many more Download/Read
8th Science Power Solution - Conservation of Plants and animals Download/Read

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca