IX CBSE - Formative Assessment-II IX Biology - Improvement in Food Resources

IX CBSE - Formative Assessment-II [SET-1]

Q.1. What are green revolution and white revolution?

Q.2. What nutrients do the following provide:

(i) Pea (ii) Linseed (iii) Vegetables (iv) pigeon pea

(v) Wheat (vi) Castor (vii) Spices (viii) fruits.

Q.3. Differentiate between Kharif and Rabi crops.

Q.4. What is interspecific hybridization?

Q.5. How does change in maturity duration of crop help the framers?

Q.6. Categorize the following as macronutrients and micronutrients:

(i) Calcium (ii) Molybdenum (iii) Phosphorus

(iv) Iron (v) Boron (vi) Magnesium.

Q.7. What is organic farming?

Q.8. What is vermicomposting?

Q.9. Differentiate between mixed and inter cropping.

Q10. What are river lift systems?

Q.11. Why are weeds considered as unwanted plants in cultivated lands?

Q.12. In what three ways do the pests attack plants?

IX CBSE - Formative Assessment-II [SET-2]

Q.1. What qualities are expected in a cross-breed variety of fowl?

Q.2. Differentiate between capture fishing and culture fishing.

Q.3. What is composite fish culture system?

Q.4. Which Italian bee variety is used in India to yield honey?

Q.5. What is the use of Satellites and echo sounders in marine fishery?

Q.6. _________, ________ and _______ are some biotic factors responsible for storage losses.

Q.7. _________ is the scientific management of animal livestock.

Q.8. Milk-producing animals are called ________________________

Q.9. What qualities of Brown Swiss and Sahiwal are considered to use them for crossbreeding?

=============================================

Value Based Questions

Q.10. Seema and her mother went to a party and were appalled by the amount of food wasted by the
people. Sema‟s mother told her to take only that much food in her plate that she could eat. After the party
got over, Seema‟s mother asked the host to call the local NGO that would distribute the unused food to
poor people.

(i) State two methods to increase the production of food crops.

(ii) Why is storage of grains considered a very important aspect of agriculture?

(iii) What values are shown by Seema‟s mother?

IX Biology : Improvement in Food Resources

CBSE solved ,unsolved test papers,Notes, Assignments and Guess Papers : First Term:  Improvement in
Food Resources. Plant and animal breeding and selection for quality improvement and management; use of fertilizers, manures; protection from pests and diseases; organic farming.

IX- Improvement in Food Res Solved Questions
Download
IX- Imp. Food Resources:Revision assignment
Download
IX- Food Resources Quick revision notes by KV
Download

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 d...

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 ...

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

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 ...

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution: Mass of the reactants = 10g ; Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g. In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....

CBSE ADDA

Search This Blog