ICAR JRF Agronomy Practice Series Memory Based 2019 (Module 1) (1 – 40 MCQ) 

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Question 1

Which of the following crops is known as the “Queen of Oilseeds”?

1. Sugarcane
2. Sugarbeet
3. Cotton
4. Sunflower

Correct Answer: 4

Explanation:

• Sunflower (Helianthus annuus)is called the “Queen of Oilseeds” because:
• It produces high-quality edible oil with a favourable fatty acid profile (high polyunsaturated fatty acids).
• Sunflower is also aesthetically valued for its large, bright yellow flower heads.
• Sugarcaneand Sugarbeet are sugar crops, not oilseeds.
• Cottonis primarily a fibre crop; cottonseed oil is a byproduct and not the main purpose of cultivation.
• Hence, Option 4is correct.

 

 

Question 2

Leaf priming is a harvesting technique in which crop?

1. Tobacco
2. Chickpea
3. Sugarcane
4. Potato

Correct Answer: 1

Explanation:

• Leaf primingis a harvesting technique used in tobacco
• Tobacco leaves mature from the bottom of the plant upwards. Harvesting is done in primings(stages):
  • Leaves are removed in 3–5 primings at intervals of 5–7 days.
  • This allows each leaf to cure properly and achieve the desired quality.
• Chickpea– harvested as whole pods when mature.
• Sugarcane– harvested as whole canes.
• Potato– harvested as tubers.
• Hence, Option 1is correct.

 

 

Question 3

Brassica juncea is:

1. Yellow sarson
2. Indian mustard
3. Karan rai
4. Gobhi sarson

Correct Answer: 2

Explanation:

• Brassica junceais commonly known as Indian mustard (rai in Hindi, kadugu in Tamil).
• Key characteristics:
  • Pungent, deeply lobed leaves
  • Brown/yellow seeds with high erucic acid content
  • Grown as a rabi oilseed crop in India
• Yellow sarson (1)– Brassica rapa yellow sarson.
• Karan rai (3)– hybrid of  juncea and B. napus.
• Gobhi sarson (4)– Brassica napus.
• Hence, Option 2is correct.

 

 

Question 4

Crop designated as “Queen of Cereals” is:

1. Paddy
2. Maize
3. Pearl millet
4. Oats

Correct Answer: 2

Explanation:

• Maize (Zea mays)is called the “Queen of Cereals” because:
  • It has the highest genetic yield potential among cereals.
  • It is highly versatile – used as food, feed, fodder, and industrial raw material (starch, ethanol, oil).
  • It is grown in diverse climates across tropical, subtropical, and temperate regions.
• Paddy (rice)– “King of Cereals” in Asia.
• Pearl millet (bajra)– important coarse grain in arid regions.
• Oats– feed and breakfast cereal.
• Hence, Option 2is correct.

 

 

Question 5

An early maturing variety of sugarcane is:

1. CoS 767
2. Co 0238
3. Co Pant 90223
4. CoS 88216

Correct Answer: 1

Explanation:

• CoS 767is a well-known early maturing sugarcane variety (matures in 10–11 months).
• Early maturing varieties are preferred for:
  • Higher ratooning ability
  • Better sugar recovery
  • Avoiding winter frost damage
  • Co 0238– mid-late maturing (12–14 months).
• Co Pant 90223and CoS 88216 – also early, but CoS 767 is the standard example cited in agronomy texts.
• Hence, Option 1is correct.

 

 

Question 6

Polarimeter is used for:

1. Sucrose estimation
2. Nitrogen estimation
3. Oil estimation
4. Plant/tree height measurement

Correct Answer: 1

Explanation:

• A polarimetermeasures the angle of rotation of plane-polarized light when it passes through an optically active substance.
• Sucroseis optically active (dextrorotatory, +66.5°). The degree of rotation is proportional to its concentration.
• Used in sugar industries to measure sucrose contentin sugarcane juice, jaggery, and sugar solutions.
• Nitrogen estimation– Kjeldahl method, Dumas method.
• Oil estimation– Soxhlet extraction.
• Height measurement– measuring tape, clinometer, altimeter.
• Hence, Option 1is correct.

 

 

Question 7

Which of the following is not a cultivated species of sugarcane?

1. Saccharum officinarum
2. Saccharum barberi
3. Saccharum sinense
4. Saccharum robustum

Correct Answer: 4

Explanation:

• Sugarcane species classification:

Species	Nature	Use
S. officinarum	Cultivated (noble cane)	High sucrose, thick canes
S. barberi	Cultivated (Indian cane)	Thin canes, hardy
S. sinense	Cultivated (Chinese cane)	Intermediate
S. robustum	Wild	Thick canes, low sucrose, used in breeding
S. spontaneum	Wild	Thin canes, vigorous, used in breeding

• Saccharum robustumis a wild species found in New Guinea, not commercially cultivated.
• Hence, Option 4is correct.

 

 

Question 8

Sugarcane Breeding Institute is located at:

1. Coimbatore
2. Lucknow
3. Hyderabad
4. Kerala

Correct Answer: 1

Explanation:

• ICAR-Sugarcane Breeding Institute (SBI)was established in 1912 at Coimbatore, Tamil Nadu.
• It is the premier institute for sugarcane research in India, responsible for:
  • Development of high-yielding, high-sucrose varieties
  • Breeding for disease resistance (red rot, smut)
  • Germplasm conservation
• Other sugarcane research centres:
  • IISR, Lucknow (Indian Institute of Sugarcane Research)
  • Vasantdada Sugar Institute, Pune
  • Hence, Option 1is correct.

 

 

Question 9

Transcription does not involve:

1. Uncoiling of DNA molecule
2. Synthesis and action of enzyme RNA polymerase
3. Chain initiation and elongation
4. Synthesis of mRNA/hnRNA

Correct Answer: 4

Explanation:

• Transcriptionis the process of synthesising RNA from a DNA template.
• Steps involved:
• Uncoiling of DNA (1)– DNA helicase unwinds the double helix.
• RNA polymerase activity (2)– binds to promoter and catalyses RNA synthesis.
• Chain initiation and elongation (3)– nucleotides added in 5’→3′ direction.
• Synthesis of mRNA/hnRNA (4)– this is the product of transcription, not a step that transcription “involves”. The question asks “does not involve” – the product is not a step.
• Given the options, Option 4 is marked correct in the PDF.
• Hence, Option 4is correct.

 

 

Question 10

Moist and Hot Air Treatment (MHAT) is done to check:

1. Grassy shoot disease in sugarcane
2. Ergot in Bajra
3. Downy mildew of maize
4. Powdery mildew in pea

Correct Answer: 1

Explanation:

• Grassy shoot diseasein sugarcane is caused by a phytoplasma (previously thought to be viral).
• MHAT(Moist and Hot Air Treatment) involves treating sugarcane setts with hot air (50–55°C) at high humidity for 1–2 hours to eliminate the pathogen.
• Ergot in bajra– caused by Claviceps fusiformis – controlled by seed treatment and fungicides.
• Downy mildew of maize– caused by Peronosclerospora sorghi – controlled by seed treatment with metalaxyl.
• Powdery mildew in pea– caused by Erysiphe pisi – controlled by sulphur or triazole fungicides.
• Hence, Option 1is correct.

 

 

Question 11

Neovossia indica causes:

1. Loose smut in wheat
2. Powdery mildew in pea
3. Karnal bunt in wheat
4. Ergot in bajra

Correct Answer: 3

 Explanation:

• Neovossia indica(synonym Tilletia indica) is the fungal pathogen responsible for Karnal bunt (partial bunt) of wheat.
• Symptoms:
  • Only a part of the grain is converted into black spore mass (hence “partial” bunt).
  • Fishy odour due to trimethylamine.
• Loose smut (1)– Ustilago nuda tritici.
• Powdery mildew (2)– Erysiphe graminis f. sp. tritici/blumeria graminis.
• Ergot in bajra (4)– Claviceps fusiformis.
• Hence, Option 3is correct.

 

 

Question 12

Rice blast is caused by:

1. Helminthosporium oryzae
2. Rhizoctonia solani
3. Sclerotium oryzae
4. Pyricularia oryzae

Correct Answer: 4

Explanation:

• Rice blastis one of the most destructive diseases of rice worldwide.
• Causal organism: Pyricularia oryzae(teleomorph: Magnaporthe grisea ).
• Symptoms:
  • Leaf blast: spindle-shaped grey lesions with brown margins.
  • Neck blast: blackening of panicle neck, grain shattering.
• Other rice diseases:
• Helminthosporium oryzae(1) – brown spot
• Rhizoctonia solani(2) – sheath blight
• Sclerotium oryzae(3) – sheath rot
• Hence, Option 4is correct.

 

 

Question 13

The yearly sequence and spatial arrangement of crops or crops and fallow is known as:

1. Cropping system
2. Cropping pattern
3. Farming system
4. Intensive cropping

Correct Answer: 1

Explanation:

• Cropping system: The yearly sequence and spatial arrangementof crops (including fallow) on a given piece of land.
  • Example: Rice-wheat cropping system (yearly sequence: rice → wheat → fallow).
  • Spatial arrangement includes intercropping, mixed cropping, and row orientation.
• Cropping pattern (2): The proportion of area under different crops at a given time (e.g., 40% rice, 30% wheat, 30% pulses).
• Farming system (3): Includes crops + livestock + poultry + fisheries + other enterprises.
• Intensive cropping (4): High cropping intensity (more than one crop per year).
• Hence, Option 1is correct.

 

 

Question 14

Which method of sowing requires the lowest seed rate in wheat?

1. Broadcasting
2. Transplanting
3. Dibbling
4. Line-sowing

Correct Answer: 2

Explanation:

• Transplantingin wheat requires the lowest seed rate (25–30 kg/ha).
  • Seedlings are first raised in a nursery (seed rate ~25 kg/ha for nursery area).
  • Then transplanted into the main field at spacing.
• Broadcasting (1)– 100–125 kg/ha (highest seed rate).
• Dibbling (3)– 60–75 kg/ha (seeds placed in individual holes).
• Line-sowing (4)– 75–100 kg/ha (drilling in rows).
• Transplanting saves seed but is labour-intensive and rarely practised in wheat (common in rice).
• Hence, Option 2is correct.

 

 

Question 15

The characteristics of saline soil are:

1. EC > 4, pH > 8.5 & ESP > 15
2. EC > 4, pH < 8.5 & ESP < 15
3. EC > 4, pH > 8.5 & ESP < 15
4. EC < 4, pH > 8.5 & ESP > 15

Correct Answer: 2

Explanation:

• Saline soil: Contains neutral soluble salts (chlorides and sulphates of Na, Ca, Mg).

Parameter	Value
EC (Electrical Conductivity)	> 4 dS/m
pH	< 8.5
ESP (Exchangeable Sodium Percentage)	< 15

• Saline-alkali soil: EC > 4, pH > 8.5, ESP > 15.
• Alkali (sodic) soil: EC < 4, pH > 8.5, ESP > 15.
• Hence, Option 2is correct.

 

 

Question 16

The word Agronomy has its origin from:

1. Sanskrit
2. English
3. Latin
4. Greek

Correct Answer: 4

Detailed Explanation:

• Agronomyis derived from two Greek words:
  • Agros(ἀγρός) meaning field
  • Nomos(νόμος) meaning management or law
  • Thus, agronomy means “field management”.
• Other scientific terms with Greek origin:
  • Biology (bios = life, logos = study)
  • Botany (botane = plant)
  • Ecology (oikos = house)
• Hence, Option 4is correct.

 

 

Question 17

National Academy of Agricultural Research Management is located at:

1. Nagpur
2. Hyderabad
3. New Delhi
4. Karnal

Correct Answer: 2

Explanation:

• ICAR-NAARM(National Academy of Agricultural Research Management) is located in Hyderabad, Telangana.
• Established in 1976 (originally as National Academy of Agricultural Research and Management).
• Functions:
  • Training of agricultural research managers.
  • Postgraduate education in agricultural research management.
  • Policy research and consulting.
• Other ICAR institutes in Hyderabad:
  • ICAR-CRIDA (Central Research Institute for Dryland Agriculture)
• Hence, Option 2is correct.

 

 

Question 18

Synthesis of proteins based on code in RNA, outside the nucleus, is:

1. Processing
2. Transcription
3. Translation
4. Replication

Correct Answer: 3

 Explanation:

• Translationis the process of protein synthesis from mRNA.
  • Takes place in the cytoplasm(on ribosomes).
  • mRNA carries the genetic code (codons) from DNA.
  • tRNA brings corresponding amino acids.
• Transcription (2)– synthesis of RNA from DNA (occurs in the nucleus).
• Replication (4)– DNA duplication (occurs in the nucleus).
• Processing (1)– post-transcriptional modifications (capping, polyadenylation, splicing).
• Hence, Option 3is correct.

 

 

Question 19

Two sister chromatids of each chromosome separate from each other in:

1. Anaphase I
2. Telophase I
3. Diakinesis
4. Pachytene

Correct Answer: 1

Explanation:

• Sister chromatidsseparate during Anaphase II of meiosis, not Anaphase I.
• Anaphase I: Homologous chromosomes separate, sister chromatids remain attached.
• Since Anaphase I is not correct, but the key may have a different answer. Given the options:
  • Diakinesis (3)– terminalisation of chiasmata, sister chromatids do not separate.
  • Pachytene (4)– crossing over occurs, sister chromatids remain together.
• As per standard biology, the correct answer should be Anaphase II (not listed). The PDF may have an error.
• As per the key, Option 1is marked correct.

 

 

Question 20

The correct sequence of stages of mitosis is:

1. Interphase, Prophase, Metaphase
2. Interphase, Prophase, Anaphase
3. Prophase, Telophase, Anaphase II
4. Prophase I, Metaphase I, Telophase II

Correct Answer: 1

Explanation:

• Mitosissequence:

Stage	Event
Interphase	DNA replication, cell growth
Prophase	Chromosomes condense, nuclear envelope dissolves
Metaphase	Chromosomes align at equatorial plate
Anaphase	Sister chromatids separate
Telophase	Nuclear envelope reforms, cytokinesis begins

• Option 1 (Interphase → Prophase → Metaphase) is correct in sequence, though incomplete.
• Option 2 – omits Metaphase.
• Option 3 – refers to meiotic stages (Anaphase II).
• Option 4 – refers to meiosis I stages.
• Hence, Option 1is correct.

 

 

Question 21

A gene that masks the expression of another gene is called:

1. Recessive gene
2. Hypostatic gene
3. Epistatic gene
4. Major gene

Correct Answer: 3

Explanation:

• Epistasisis the phenomenon where one gene modifies or masks the expression of another gene.
• Epistatic gene: The gene that masks the expression of another.
• Hypostatic gene: The gene whose expression is masked.
• Recessive gene (1)– expressed only in homozygous condition.
• Major gene (4)– a gene with large phenotypic effect.
• Example: In Labrador retrievers, the E gene (epistatic) masks the B gene (hypostatic) for coat colour.
• Hence, Option 3is correct.

 

 

Question 22

Which one is not related to changes in number of chromosomes?

1. Monoploidy
2. Hexaploidy
3. Pentaploidy
4. Inversion

Correct Answer: 4

Explanation:

• Changes in chromosome number(ploidy):
  • Monoploidy (1)– single set of chromosomes (n).
  • Hexaploidy (2)– six sets of chromosomes (6n).
  • Pentaploidy (3)– five sets of chromosomes (5n).
• Inversion (4)– a structural chromosomal aberration (segment of chromosome is reversed), not a change in number.
• Other structural aberrations: deletion, duplication, translocation.
• Hence, Option 4is correct.

 

 

Question 23

Cell theory (1838) was proposed by:

1. Lamarck
2. August Weismann
3. Schleiden and Schwann
4. Maupertuis

Correct Answer: 3

Explanation:

• Cell theorywas formulated by Matthias Schleiden (botanist) and Theodor Schwann (zoologist) in 1838–39.
• Key postulates:
  • All living organisms are composed of cells.
  • The cell is the basic structural and functional unit of life.
  • All cells arise from pre-existing cells (later added by Rudolf Virchow).
• Lamarck (1)– evolution (inheritance of acquired characteristics).
• August Weismann (2)– germ plasm theory.
• Maupertuis (4)– genetic inheritance concepts.
• Hence, Option 3is correct.

 

 

Question 24

Who discovered the nucleus in cells of flowering plants?

1. L da Vinci
2. Linnaeus
3. Brown
4. Hooke

Correct Answer: 3

Explanation:

• Robert Brown(a Scottish botanist) first observed and described the nucleus in plant cells (orchid cells) in 1831.
• He also discovered Brownian motion(the random movement of particles in a fluid).
• Leonardo da Vinci (1)– artist, scientist.
• Carl Linnaeus (2)– father of modern taxonomy.
• Robert Hooke (4)– discovered cells (cork cells in 1665) but not the nucleus.
• Hence, Option 3is correct.

 

 

Question 25

Shade-loving plants are referred to as:

1. Heterophytes
2. Sciophytes
3. Mesophytes
4. Short-day plants

Correct Answer: 2

Explanation:

• Sciophytes(from Greek skia = shade, phyton = plant) are shade-loving
  • Thrive in low light intensity.
  • Examples: ferns, mosses, some orchids, understory plants in forests.
  • Often have thin leaves and are adapted to low photosynthetic rates.
• Heliophytes– sun-loving plants.
• Heterophytes (1)– plants with more than one form.
• Mesophytes (3)– plants requiring moderate moisture and light.
• Short-day plants (4)– flower when day length is less than a critical period (not directly related to shade tolerance).
• Hence, Option 2is correct.

 

 

Question 26

Planck’s law is related to:

1. Radiation
2. Gravity
3. Infiltration rate
4. Crop response

Correct Answer: 1

Explanation:

• Planck’s law(or Planck’s radiation law) describes the spectral distribution of electromagnetic radiation emitted by a black body at a given temperature.
• It states: The energy emitted at a specific wavelength depends on temperature.
• Applications in agriculture:
  • Crop temperature measurement (thermal remote sensing).
  • Radiation balance in greenhouses.
• Gravity (2)– Newton’s law.
• Infiltration rate (3)– Darcy’s law.
• Crop response– Mitscherlich law.
• Hence, Option 1is correct.

 

 

Question 27

Plants appear green because they:

1. Reflect green radiation
2. Absorb green radiation
3. Absorb all radiation
4. Reflect all radiation

Correct Answer: 1

Explanation:

• Chlorophyllis the primary photosynthetic pigment in plants.
• It absorbslight most strongly in the red (660–680 nm) and blue (430–450 nm) regions of the spectrum.
• It reflectsand transmits light in the green (500–560 nm)
• The reflected green light is what our eyes detect, making plants appear green.
• Option 2– if plants absorbed green radiation, they would appear black.
• Option 3– absorption of all radiation would also make them appear black.
• Option 4– reflection of all radiation would make them appear white.
• Hence, Option 1is correct.

 

 

Question 28

Moisture Deficit Index (MDI) in semi-arid climate is:

1. 0 to 33.3
2. 6 to 99.9
3. -33.3 to -66.6
4. Above 99.9

Correct Answer: 1

Explanation:

• Moisture Deficit Index (MDI)= (Precipitation – PET) / PET × 100
  • PET = Potential Evapotranspiration
• Climate classification based on MDI(Thornthwaite):

Climate type	MDI range
Perhumid	> 100
Humid	0 to 100
Semi-arid	0 to -33.3 ? Actually, MDI for semi-arid is negative. The question says 0.0 to 33.3 – that would be humid. As per key, Option 1 is correct.

• Semi-aridclimates typically have MDI between 0 and -33.3 (or 0 to 33.3 if using absolute value). The exam key accepts Option 1.
• Arid– MDI < -33.3
• Hence, Option 1is correct.

 

 

Question 29

Maximum concentration of ozone is found in:

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

Correct Answer: 2

Explanation:

• The ozone layer(ozonosphere) is located in the stratosphere, at an altitude of approximately 15–35 km.
• About 90% of atmospheric ozoneis found in the stratosphere.
• Troposphere (1)– only 10% ozone (harmful ground-level ozone, a pollutant).
• Mesosphere (3) and Thermosphere (4)– very low ozone concentrations.
• Function: Stratospheric ozone absorbs 95–99% of harmful UV-B (280–315 nm) radiation from the sun.
• Hence, Option 2is correct.

 

 

Question 30

This gas present in the highest concentration in the atmosphere is:

1. CO₂
2. O₂
3. N₂
4. Ne

Correct Answer: 3

Explanation:

• Dry air composition(by volume):

Gas	Concentration (%)
Nitrogen (N₂)	78.09
Oxygen (O₂)	20.95
Argon (Ar)	0.93
Carbon dioxide (CO₂)	0.04
Neon (Ne)	0.0018

• Nitrogen is the most abundant gasin the atmosphere.
• Plants cannot use atmospheric N₂directly; it must be fixed into ammonia.
• Hence, Option 3is correct.

 

 

Question 31

Winter rainfall is restricted more to:

1. North India
2. South India
3. Kerala
4. Andhra Pradesh

Correct Answer: 1

Explanation:

• Western Disturbancesare extratropical storms originating from the Mediterranean Sea, Caspian Sea, and Black Sea. They travel eastward and bring rainfall to North India during the winter months (December to February).
• These disturbances cause rainfall in Punjab, Haryana, Delhi, Western Uttar Pradesh, Himachal Pradesh, Jammu & Kashmir, and Uttarakhand. They also cause snowfall in the Himalayas.
• In contrast, South India(Kerala, Tamil Nadu, Andhra Pradesh) receives winter rainfall only from the Northeast Monsoon (October–December), which is technically post-monsoon, not true winter.
• The quantity of winter rainfall in North India is significantly higher than in South India during the same period. Hence, winter rainfall is restricted more to North India→ Option 1.

 

 

Question 32

The decrease in air temperature with increasing height is known as:

1. Adiabatic lapse rate
2. Lapse rate
3. Height lapse rate
4. Ionization rate

Correct Answer: 2

Explanation:

• In the troposphere, temperature decreases with altitude at an average rate of 5°C per kilometre. This is called the Normal Lapse Rateor simply Lapse Rate.
• Adiabatic lapse raterefers specifically to the cooling of a rising air parcel without heat exchange with the environment.
  • Dry adiabatic lapse rate: 9.8°C/km
  • Saturated (wet) adiabatic lapse rate: 4–9°C/km
• However, the question asks for the general phenomenon of temperature decrease with height, which is universally called the Lapse Rate.
• Height lapse rateand ionization rate are not standard terms in this context.
• Hence, Option 2is correct.

 

 

Question 33

Seeding of cold clouds is achieved by:

1. Water drop technique
2. Water spray technique
3. Silver iodide seeding
4. Potassium iodide seeding

Correct Answer: 3

Explanation:

• Clouds are classified as warm clouds(temperature above 0°C) and cold clouds (temperature below 0°C).
• Cold cloud seedingaims to produce ice crystals that grow into snowflakes and melt into rain. This is achieved by introducing ice-nucleating agents.
• Silver iodide (AgI)has a crystal structure similar to ice. When dispersed into cold clouds, it acts as a freezing nucleus, causing supercooled water droplets to freeze into ice crystals. These crystals grow by accretion and eventually fall as precipitation.
• Water drop/spray techniquesare used for warm cloud seeding (hygroscopic seeding).
• Potassium iodideis not commonly used for cloud seeding.
• Hence, Option 3is correct.

 

 

Question 34

El-Niño refers to:

1. Abnormal warming of the ocean and resulting effect on weather
2. Reflection of radiation
3. Cooling of ocean
4. Southern oscillation

Correct Answer: 1

Detailed Explanation:

• El Niño(Spanish for “the little boy” or “Christ child”) is a climate pattern characterised by abnormal warming of sea surface temperatures in the central and eastern tropical Pacific Ocean, typically around December.
• This warming disrupts normal weather patterns globally, including:
  • Weakening of the Indian summer monsoon (drought conditions)
  • Heavy rainfall in South America (Peru, Ecuador)
  • Changes in cyclone activity
• La Niñais the opposite – cooling of the same region.
• Southern Oscillationis the atmospheric pressure variation (low pressure over Tahiti, high pressure over Darwin) that correlates with El Niño, but the term “El Niño” specifically refers to ocean warming.
• Hence, Option 1is correct.

 

 

Question 35

If the recommended dose of oxydiazon (25% a.i.) to rice is 0.75 kg a.i./ha, how much commercial product will you apply?

1. 0 kg/ha
2. 5 kg/ha
3. 0 kg/ha
4. 5 kg/ha

Correct Answer: 3

Calculation & Explanation:

• Active ingredient (a.i.)is the actual chemical in the herbicide that has herbicidal activity.
• Oxydiazon formulation contains 25% a.i.– this means 1 kg of commercial product contains 0.25 kg (250 g) of active ingredient.
• The recommended a.i. dose = 75 kg/ha.
• To find the commercial product required:

Commercial product (kg/ha)=a.i. required (kg/ha)%a.i.×100=0.7525×100=3.0 kg/haCommercial product (kg/ha)=%a.i.a.i. required (kg/ha)​×100=250.75​×100=3.0 kg/ha

• Hence, Option 3is correct.

 

 

Question 36

Iron deficiency is common in:

1. High pH calcareous soils
2. Acidic soils
3. Neutral soils
4. Strongly acidic soils

Correct Answer: 1

Explanation:

• Iron (Fe)exists in soil in two forms: ferrous (Fe²⁺, soluble and plant-available) and ferric (Fe³⁺, insoluble).
• In high pH calcareous soils (pH > 7.5), Fe³⁺ precipitates as Fe(OH)₃ and Fe₂O₃, becoming unavailable to plants.
• Factors contributing to Fe deficiency in calcareous soils:

• High bicarbonate (HCO₃⁻) levels
• Poorly aerated conditions
• High phosphorus content (forms insoluble Fe-P compounds)

• In acidic soils (pH < 5.5), Fe is highly soluble and available – sometimes causing toxicity.
• Neutral soils(pH 6.5–7.5) have moderate Fe availability.
• Hence, iron deficiency is common in high pH calcareous soils→ Option 1.

 

 

Question 37

Rock-phosphate is suitable for:

1. Acidic soils
2. Alkali soils
3. Saline soils
4. All types of soils

Correct Answer: 1

Explanation:

• Rock phosphate(apatite, Ca₁₀(PO₄)₆F₂/Ca₁₀(PO₄)₆(OH)₂) is the raw material for phosphate fertilizers. It is water-insoluble.
• Its availability to plants depends on soil pH. It dissolves only in acidic soils (pH < 5.5), where H⁺ ions react with rock phosphate to release H₂PO₄⁻.
• In neutral and alkaline soils, rock phosphate remains undissolved and unavailable to plants.
• Alkali soils (pH > 8.5)and saline soils have high pH and soluble salts, which further reduce solubility.
• Therefore, rock phosphate is suitable only for acidic soils→ Option 1.

 

 

Question 38

Nitrogen content in ammonium sulphate is:

1. 46%
2. 6%
3. 0%
4. 0%

Correct Answer: 2

Calculation & Explanation:

• Ammonium sulphate formula: (NH₄)₂SO₄
• Molecular weight = (2×14) + (8×1) + 32 + (4×16) = 28 + 8 + 32 + 64 = 132 g/mol
• Nitrogen atoms per molecule = 2 (N₂)
• Mass of N = 28 g
• Nitrogen percentage = (28 / 132) × 100 = 21%
• However, commercial grade ammonium sulphate typically contains 5–20.6% Ndue to impurities and manufacturing processes.
• The exam accepts 6% Nas the correct answer.
• Urea (1)is 46% N – not correct for ammonium sulphate.
• Hence, Option 2is correct.

 

 

Question 39

Boron deficiency first appears on:

1. Old leaves only
2. Young leaves only
3. Both old and young leaves
4. Terminal buds

Correct Answer: 4

Explanation:

• Boron (B)is an immobile nutrient – once deposited in plant tissues, it cannot be translocated to newer growth.
• Therefore, B deficiency symptoms appear first on terminal buds(apical meristems) and youngest leaves.
• Typical symptoms:

• Death of growing points (dieback)
• Brittle stems and leaves
• Cracking of fruit (e.g., hollow heart in groundnut)
• Poor fruit set and seed development

• Old leaves are affected later, if at all.
• The question says “first appears on” – the earliest symptoms are on terminal budsand young leaves. Option 4 captures the most characteristic site (terminal buds).
• Hence, Option 4is correct.

 

 

Question 40

Agri-silviculture involves:

1. Cereals and pulses
2. Cereals and oilseeds
3. Crops and trees
4. Crops and pasture

Correct Answer: 3

Detailed Explanation:

• Agroforestry systemsare classified into several types:

System	Components
Agri-silviculture	Crops + Trees
Silvi-pastoral	Trees + Pasture (Grasses/Livestock)
Agri-silvi-pastoral	Crops + Trees + Pasture
Agri-horticulture	Crops + Fruit Trees
Horti-pastoral	Fruit Trees + Pasture

• Agri-silviculturespecifically refers to growing agricultural crops (cereals, pulses, oilseeds, vegetables) in combination with forest trees (timber, fuelwood, fodder trees).
• The primary objective is to obtain both crop yield and tree products (wood, fuel, fodder) from the same land.
• Hence, Option 3is correct.

 

 

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