Skip to main content

Crop production and Management Science Mission Part 02 Questions answer Class 08 Book

Science Mission Part 02 Questions answer Class 08 Book

Chapter 01 Crop production and Management


Theoretical Questions

A. Short answer type questions.

1. Differentiate between the following:

i. Compost and green manure

Compost: Compost is the decomposed plant and animal waste which may include waste food, excreta, dried leaves and other organic matter. It is made in a compost pit using microbes.

Green Manure: The green leguminous plants with their root nodules is ploughed along with the soil. The green plants and the root nodules left in the soil decompose to form green manure.

ii. Compost and fertilizers

Compost is a manure prepared by a biological decomposition of organic waste. It takes longer period to absorbed by plants. It is lack of nitrogen, Phosphorous and potash

Fertilizer is manure prepared in factories from chemicals (mineral). It is easily absorbed by soil. It is rich in nitrogen, Phosphorous and potash

iii. Pesticides and weedicides

Pesticides are substances that are meant to control pests, may be fungus, bacteria or animal germs (harmful insects). Weedicides are the chemicals which are sprayed over field to get rid of weeds. Weedicides are poisonous for weeds but not for the crop.

iv. Nitrification and denitrification

Conversion of dead decaying matter into ammonia and then into nitrites and nitrates by nitrifying microbes is called nitrification.

Conversion of nitrites and nitrates from decaying organic matter into gaseous nitrogen by denitrifying microbes is called denitrification.

v. Crops and intercrop

A crop is a plant or animal product that can be grown and harvested extensively for profit or subsistence.

Growing of Trifolium or any other leguminous crop (as green manure) during intervening period in between Kharif and Rabi is intercropping.

2. List the tasks a farmer needs to perform for the cultivation of a crop.

Main tasks to be performed by Farmers before cultivation of crops are

(a) Preparation of soil. (b) Procuring and selection of seeds, (c) Improving soil fertility (d) Irrigation (e) Weeding (f) Protection of crops (g) Harvesting (h)Storage of farm produce.

3. Name one plant each from a kharif crop, rabi crop and the season (month) in which each one of them is (i) sown and (ii) harvested?

Ans: Kharif crop: paddy (rice); It is summer crop. Sown in summer season (May/June and harvested well before onset of winter(September).

Rabi crop: wheat. It is winter crop. Sown well before winter (November end) and harvested in March/April that before onset of summer.

4. What are the different foods which we get from animals?

Ans: Some of the food products obtained from animals are milk, meat including fish, eggs and honey.

5. List some animal sources of sea food.

Ans : Some animals as source of sea food are prawn, crab, oyster etc.

B. Long answer type questions.

1. How and why did humans start practicing agriculture?

Ans: Man was basically a food gatherer. At times when the weather got rough, he and his family had to eat  stale food or had to go without food. This made him to change his habit to domesticating animals and growing plants along his dwelling units. This was the beginning of farming by man.

2. How do we separate healthy seeds from a mixture of seeds?

Ans: Healthy seeds are separated from the mixture of seeds by following :

a. Manually by seeing the physical condition of the seed, damaged seeds are often broken or damaged in other ways.

B. by soaking them in water, damaged seeds are not able for germination process after soaking them.

3. What is intercropping? Name a few crops cultivated as intercrops.

Ans: Planting a leguminous crop in between two major crops in a field such as planting of Trifoliumin between Kharif and Rabi is called intercropping. This way, soil is enriched with nitrogenous nutrients for the next crop in addition to obtaining fodder for the farm animals rich in nutrients.

Intercropping is also a system of cropping where two or more crops are grown in proximity along ridges (as in mixed cropping).

Non-leguminous plant (such as one of Maize (corn), sorghum, millets, potato, cabbage, cauliflower, brinjal) and leguminous plant (such as one of pulses, beans including groundnut) are cultivated alternately as intercrops.



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