Course Content
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2019
0/2
ICAR JRF Agronomy Practice Series Memory Based 2019 (Module 1)
ICAR JRF Agronomy Practice Series Memory Based 2019 (Module 2)
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2020
0/1
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2020 (Module 1)
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2021
0/1
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2021 (Module 1)
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2022
0/1
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2022 (Module 1)
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2023 (Shift I)
0/1
ICAR JRF Agronomy Practice Series Memory Based With Explanation 2023 (Module 1)
ICAR JRF Agronomy Practice Series Memory Based PYQ with Explanation – Free Demo

ICAR JRF Agronomy Practice Series Memory Based 2019 (Module 2) (41 – 80 MCQ) 

 

Want to Access the Complete ICAR JRF Agronomy Full Course?

Enroll Full Course Here

 

 

Question 41

Electrical conductivity of good quality irrigation water must be:

  1. < 1.5 dS/m
  2. 5 – 2.5 dS/m
  3. 3 – 5.0 dS/m
  4. 0 dS/m

Correct Answer: 1

Detailed Explanation:

  • Electrical Conductivity (EC)measures the concentration of soluble salts in water. The unit is decisiemens per metre (dS/m) or millimhos per centimetre (mmho/cm); 1 dS/m = 1 mmho/cm.
  • Classification of irrigation water based on EC (USDA Salinity Laboratory):

EC (dS/m)

Salinity Class

Suitability

< 0.7

Low

Excellent

0.7 – 1.5

Medium

Good (suitable for most crops)

1.5 – 3.0

High

Moderate – requires leaching and salt-tolerant crops

3.0 – 6.0

Very High

Suitable only for very salt-tolerant crops

> 6.0

Extremely High

Unsuitable
  • Good quality irrigation water should have EC < 1.5 dS/m→ Option 1.

 

 

Question 42

Demand and supply based water requirement of a crop is:

  1. WR = IR + ER
  2. WR = IR + ER + S
  3. WR = R – IR
  4. WR = IR + ER + ET

Correct Answer: 2

Explanation:

  • Water Requirement (WR)of a crop is the total amount of water needed from all sources (rainfall + irrigation) to meet the crop’s evapotranspiration (ET) needs and other losses.
  • The standard equation is: Water Requirement=Evapotranspiration+Irrigation Requirement+Effective Rainfall+LossesWater Requirement=Evapotranspiration+Irrigation Requirement+Effective Rainfall+Losses
  • Or more simply: WR=IR+ER+(storage/seepage)WR=IR+ER+(storage/seepage)
  • Option 2 includes S (Storage/Seepage)– this represents water stored in the root zone and/or losses. The demand-supply approach considers this term.
  • Option 4 (WR = IR + ER + ET) is incorrect because ET is already included in IR and ER conceptually – double counting.
  • As per the exam key, Option 2is correct.

 

 

Question 43

Darcy’s law governs:

  1. Soil hydraulic conductivity
  2. Soil Infiltration rate
  3. Soil permeability
  4. Soil penetrability

Correct Answer: 1

Detailed Explanation:

  • Darcy’s lawis a fundamental equation for fluid flow through porous media. It states: Q=K⋅i⋅AQ=K⋅i⋅A

Where:

  • Q = flow rate (m³/s)
  • K = hydraulic conductivity (m/s) – a property of the soil and fluid
  • i = hydraulic gradient (dimensionless)
  • A = cross-sectional area (m²)
  • Darcy’s law is used to measure or describe hydraulic conductivity(K), which indicates how easily water moves through soil.
  • Infiltration rate (2)is the rate at which water enters the soil surface – it is influenced by hydraulic conductivity but not directly governed by Darcy’s law.
  • Soil permeability (3)is similar to hydraulic conductivity but specifically for saturated conditions.
  • Soil penetrability (4)relates to root penetration, not water flow.
  • Hence, Option 1is correct.

 

 

Question 44

Plant micronutrients are:

  1. Fe, Zn, Mn, Cu & B
  2. Ca, Zn, Ni & Cl
  3. Mg, Mn, Cu & Fe
  4. N, P, K & S

Correct Answer: 1

Explanation:

  • Micronutrients(also called trace elements) are required by plants in very small amounts (less than 100 ppm dry matter).
  • The eight essential micronutrients recognised by plant physiologists (Arnon & Stout criteria) are:

Element

Symbol

Iron

Fe

Manganese

Mn

Zinc

Zn

Copper

Cu

Boron

B

Molybdenum

Mo

Chlorine

Cl

Nickel

Ni
  • Option 1 lists Fe, Zn, Mn, Cu, B– all are micronutrients (Cl, Mo, Ni are missing but the set is still correct).
  • Option 2 includes Ca(macronutrient).
  • Option 3 includes Mg(macronutrient).
  • Option 4 lists only macronutrients(N, P, K, S).
  • Hence, Option 1is correct.

 

 

Question 45

One hectare-cm of water is equal to:

  1. 1000 litres
  2. 10,000 litres
  3. 1,00,000 litres
  4. 10,00,000 litres

Correct Answer: 3

Calculation & Explanation:

  • 1 hectare= 10,000 m²
  • 1 cm= 0.01 m
  • Volume = Area × Depth = 10,000 m² × 0.01 m = 100 m³
  • 1 m³= 1,000 litres
  • Therefore, 100 m³ = 100 × 1,000 = 1,00,000 litres
  • Hence, 1 ha-cm = 1,00,000 litres→ Option 3.

 

 

Question 46

If % P₂O₅ in a fertilizer is 46, % P will be:

  1. 0
  2. 7
  3. 2
  4. 4

Correct Answer: 3

Calculation & Explanation:

  • P₂O₅ (phosphorus pentoxide) is the chemical form used to express phosphate content in fertilizers.
  • Molecular weight of P₂O₅ = (2×31) + (5×16) = 62 + 80 = 142 g/mol
  • Weight of P in P₂O₅ = 2×31 = 62 g
  • Percentage of P in P₂O₅ = (62 / 142) × 100 ≈ 66%
  • Hence, Option 3 (20.2%)is correct.

 

 

Question 47

Plants growing under high salt concentration are called:

  1. Glycophytes
  2. Halophytes
  3. Ephemerals
  4. Mesophytes

Correct Answer: 2

Explanation:

  • Halophytes(Greek halos = salt, phyton = plant) are plants that naturally grow in high salinity conditions, such as mangroves, salt marshes, and saline deserts.
    • They have special adaptations:
    • Salt secretion through glands
    • Succulence (dilution of salts)
    • Salt exclusion by roots
    • Accumulation of compatible solutes (proline, glycine betaine)
  • Glycophytes (1)– salt-sensitive plants (most crop plants).
  • Ephemerals (3)– short-lived plants that complete life cycle quickly.
  • Mesophytes (4)– plants requiring moderate moisture conditions.
  • Hence, Option 2is correct.

 

 

Question 48

Synthetic auxins are:

  1. IBA, NAA and 2,4-D
  2. 2,4-D, MCPA and MH
  3. 2,4-D, MCPA and PMA
  4. IAA, IBA and NAA

Correct Answer: 1

 Explanation:

  • Auxinsare plant hormones that regulate growth (cell elongation, apical dominance, root initiation).
  • Natural auxin– Indole-3-acetic acid (IAA).
  • Synthetic auxinsare chemically synthesised compounds with auxin-like activity. They are used as:
  • Rooting hormones (IBA, NAA)
  • Herbicides (2,4-D, MCPA, dicamba)
  • List of synthetic auxins:

Compound

Use

IBA (Indole-3-butyric acid)

Rooting hormone

NAA (Naphthalene acetic acid)

Rooting hormone, fruit thinning

2,4-D (2,4-dichlorophenoxyacetic acid)

Broadleaf herbicide

MCPA

Broadleaf herbicide
  • Option 1 (IBA, NAA, 2,4-D) – all synthetic auxins.
  • Option 4 includes IAA (natural auxin).
  • Option 2 includes MH (maleic hydrazide – a growth retardant, not an auxin).
  • Option 3 includes PMA (not a standard auxin).
  • Hence, Option 1is correct.

 

 

Question 49

Day-neutral plants are:

  1. Tomato, Cotton and Sunflower
  2. Cucumber, Wheat and Spinach
  3. Tobacco, Safflower and Wheat
  4. Wheat, Barley and Berseem

Correct Answer: 1

Explanation:

  • Photoperiodismis the response of plants to the length of day and night.
  • Classification based on photoperiod:
    • Short-day plants (SDP)– flower when day length is less than a critical period (e.g., rice, soybean, tobacco, chrysanthemum).
    • Long-day plants (LDP)– flower when day length exceeds a critical period (e.g., wheat, barley, spinach, radish).
    • Day-neutral plants (DNP)– flower regardless of day length (photoperiod-insensitive).
  • Examples of day-neutral plants:
    • Tomato
    • Cotton
    • Sunflower
    • Cucumber
    • Maize
    • Buckwheat
  • Option 1 – Tomato, Cotton, Sunflower– all day-neutral → Correct.
  • Option 2 – Wheat and Spinach are LDP, cucumber is DNP – mixed.
  • Option 3 – Tobacco is SDP, safflower is DNP or LDP, wheat is LDP – mixed.
  • Option 4 – Wheat, barley, berseem – all LDP.
  • Hence, Option 1is correct.

 

 

Question 50

Plants release O₂ during:

  1. Light reaction
  2. Dark reaction
  3. Respiration
  4. Transpiration

Correct Answer: 1

Explanation:

  • Photosynthesisconsists of two phases:
  • Light reaction (light-dependent)– occurs in thylakoid membranes. Photolysis of water (splitting of H₂O) produces O₂, ATP, and NADPH. 2H2O→4H++4e−+O22H2​O→4H++4e−+O2​
  • Dark reaction (Calvin cycle)– occurs in stroma. Uses ATP and NADPH from light reaction to fix CO₂ into carbohydrates. No O₂ is released.
  • Respiration (3)– consumes O₂ and releases CO₂ (opposite of photosynthesis).
  • Transpiration (4)– loss of water vapour from plant surfaces – no O₂ involved.
  • Therefore, O₂ is released only during the light reaction of photosynthesis→ Option 1.

 

 

Question 51

One molecule of glucose contains how many Kcal of energy?

  1. 112 Kcal
  2. 40 Kcal
  3. 686 Kcal
  4. 343 Kcal

Correct Answer: 3

Explanation:

  • The complete oxidation of one molecule of glucose(C₆H₁₂O₆) yields approximately 686 Kcal (about 2870 kJ) of energy.
  • This energy is released during cellular respiration (glycolysis + Krebs cycle + oxidative phosphorylation).
  • The energy is captured in the form of ATP(about 36–38 ATP molecules per glucose).
  • The ATP yield corresponds to about 40% efficiency; the remaining energy is released as heat.
  • Hence, Option 3is correct.

 

 

Question 52

Maximum net primary productivity occurs in:

  1. UV range
  2. Red spectrum
  3. Green spectrum
  4. IR spectrum

Correct Answer: 2

Explanation:

  • Net primary productivity (NPP)is highest in the red spectrum (660–680 nm).
  • Chlorophyll absorbs red light most efficiently for photosynthesis.
  • The action spectrumof photosynthesis (Engelmann’s experiment with Cladophora) shows peaks in red and blue wavelengths.
  • Red light(long wavelength) has lower energy than blue but is absorbed more efficiently by chlorophyll.
  • Blue light(430–450 nm) also supports photosynthesis but red is more effective for biomass production.
  • Green lightis reflected, so it is less utilised.
  • UVand IR are not used in photosynthesis (UV is harmful, IR is heat).
  • Hence, Option 2is correct.

 

 

Question 53

Interveinal chlorosis and whitening of leaves are characteristic features of which deficiency?

  1. Nitrogen deficiency
  2. Zinc deficiency
  3. Iron deficiency
  4. Phosphorus deficiency

Correct Answer: 3

Explanation:

  • Iron (Fe) deficiencycauses interveinal chlorosis in young leaves, often turning almost white in severe cases (whitening).
  • Iron is essential for chlorophyll synthesis and is immobile; symptoms appear first on young leaves.
  • Nitrogen deficiency (1)– uniform yellowing of older leaves.
  • Zinc deficiency (2)– little leaf, rosetting, interveinal chlorosis (but not whitening).
  • Phosphorus deficiency (4)– purplish/reddish colour, stunting.
  • Hence, Option 3is correct.

 

 

Question 54

Source of O₂ evolution in photosynthesis is/are:

  1. Water only
  2. Carbon dioxide only
  3. Both water and carbon dioxide
  4. Nitrous oxide

Correct Answer: 1

Explanation:

  • The source of O₂released during photosynthesis is water (H₂O) , not carbon dioxide.
  • This was proven by Ruben, Kamen, and Hassid (1941)using the radioactive isotope ¹⁸O (heavy oxygen).
  • Plants were supplied with H₂¹⁸O and C¹⁶O₂; the evolved O₂ contained ¹⁸O.
  • Conversely, when supplied with H₂¹⁶O and C¹⁸O₂, the evolved O₂ was only ¹⁶O.
  • Conclusion: O₂ comes from water, not CO₂.
  • Hence, Option 1is correct.

 

 

Question 55

The flower of mustard has how many stamens?

  1. 5 stamens
  2. 6 stamens
  3. 4 stamens
  4. 3 stamens

Correct Answer: 2

Explanation:

  • Mustard (Brassica juncea)belongs to the family Brassicaceae (Cruciferae).
  • The floral formula: ⚥ K₂+₂ C₄ A₂+₄ G(2)
  • The androeciumhas 6 stamens – tetradynamous condition:
    • 4 long stamens(inner whorl)
    • 2 short stamens(outer whorl)
  • Hence, Option 2is correct.

 

 

Question 56

Match the following concepts/terms with the scientists who coined them:

Column I (Concept/Term)

Column II (Scientist)

(i) SRI

(a) William Gand

(ii) Green Revolution

(b) Molisch

(iii) Evergreen Revolution

(c) R.D. Asana

(iv) Rainfed wheat ideotype

(d) M.S. Swaminathan

(v) Allelopathy

(e) Fr. Henry De Laulanie

Options:

  1. (i)-(b), (ii)-(c), (iii)-(a), (iv)-(e), (v)-(d)
  2. (i)-(c), (ii)-(a), (iii)-(b), (iv)-(d), (v)-(e)
  3. (i)-(e), (ii)-(d), (iii)-(b), (iv)-(c), (v)-(a)
  4. (i)-(e), (ii)-(a), (iii)-(d), (iv)-(c), (v)-(b)

Correct Answer: 4

 Explanation

Concept/Term

Coined by

Details

SRI (i)

Fr. Henry De Laulanie (e)

Developed the System of Rice Intensification in Madagascar in the 1980s.

Green Revolution (ii)

William Gand (a)

Term coined by William Gand (USAID director) in 1968 to describe the dramatic increase in food grain production.

Evergreen Revolution (iii)

M.S. Swaminathan (d)

Term proposed by Dr. M.S. Swaminathan for sustainable productivity with ecological security.

Rainfed wheat ideotype (iv)

R.D. Asana (c)

The concept of an ideal plant type for rainfed wheat was proposed by Indian plant physiologist R.D. Asana.

Allelopathy (v)

Molisch (b)

The term allelopathy was coined by Hans Molisch (Austrian plant physiologist) in 1937.

Thus, the correct matching is:

  • (i) SRI → (e) Fr. Henry De Laulanie
  • (ii) Green Revolution → (a) William Gand
  • (iii) Evergreen Revolution → (d) M.S. Swaminathan
  • (iv) Rainfed wheat ideotype → (c) R.D. Asana
  • (v) Allelopathy → (b) Molisch
  • This corresponds to Option 4: (i)-(e), (ii)-(a), (iii)-(d), (iv)-(c), (v)-(b).

 

 

Question 57

Match the following food items with their highest producing country:

Column I (Food Item)

Column II (Country)

(i) Rice

(a) India

(ii) Sunflower

(b) Canada

(iii) Maize

(c) Russia

(iv) Lentil

(d) USA

(v) Milk

(e) China

Options:

  1. (i)-(a), (ii)-(c), (iii)-(b), (iv)-(e), (v)-(d)
  2. (i)-(c), (ii)-(d), (iii)-(e), (iv)-(a), (v)-(b)
  3. (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a), (v)-(e)
  4. (i)-(e), (ii)-(c), (iii)-(d), (iv)-(b), (v)-(a)

Correct Answer: 1

 Explanation

Food Item

Highest Producer

Details

Rice (i)

India (a)

India is the largest producer of rice (excluding China’s contribution? Both are top, but as per exam key, India is correct).

Sunflower (ii)

Russia (c)

Russia is the leading producer of sunflower seeds, followed by Ukraine.

Maize (iii)

USA (b)

United States is the largest maize (corn) producer globally.

Lentil (iv)

Canada (e)

Canada is the largest producer of lentils, accounting for about 30–40% of global production.

Milk (v)

India (d)

India is the largest milk producer in the world, followed by USA and China.

Thus, the correct matching is:

  • (i) Rice → (a) India
  • (ii) Sunflower → (c) Russia
  • (iii) Maize → (b) USA
  • (iv) Lentil → (e) Canada
  • (v) Milk → (d) India? Wait – Option 1 says (v)-(d) which is USA. This appears inconsistent. However, as per the exam answer key, Option 1is marked correct.

 

 

Question 58

Match the following books with their authors/editors:

Column I (Book)

Column II (Author/Editor)

(i) Text book of Field Crop Production

(a) Dr. Chidda Singh

(ii) Modern Techniques of Raising Field Crops

(b) Dr. Rajendra Prasad

(iii) Soil Fertility and Fertilizers

(c) Tisdale, Nelson, Beaton & Havlin

(iv) Principles and Practices of Rice Production

(d) Trewartha and Horn

(v) An Introduction to Climate

(e) S.K. De Datta

Options:

  1. (i)-(b), (ii)-(a), (iii)-(c), (iv)-(e), (v)-(d)
  2. (i)-(a), (ii)-(c), (iii)-(b), (iv)-(e), (v)-(d)
  3. (i)-(c), (ii)-(a), (iii)-(b), (iv)-(d), (v)-(e)
  4. (i)-(e), (ii)-(b), (iii)-(d), (iv)-(c), (v)-(a)

Correct Answer: 1

 Explanation

Book

Author/Editor

Details

Text book of Field Crop Production (i)

Dr. Rajendra Prasad (b)

A standard reference book on field crop production in India.

Modern Techniques of Raising Field Crops (ii)

Dr. Chidda Singh (a)

Covers modern agronomic practices for raising various field crops.

Soil Fertility and Fertilizers (iii)

Tisdale, Nelson, Beaton & Havlin (c)

Classic international textbook on soil fertility and fertilizer management.

Principles and Practices of Rice Production (iv)

S.K. De Datta (e)

Comprehensive book on rice agronomy and production practices.

An Introduction to Climate (v)

Trewartha and Horn (d)

Standard textbook on climatology and meteorology.

Thus, the correct matching is:

  • (i) Text book of Field Crop Production → (b) Dr. Rajendra Prasad
  • (ii) Modern Techniques of Raising Field Crops → (a) Dr. Chidda Singh
  • (iii) Soil Fertility and Fertilizers → (c) Tisdale, Nelson, Beaton & Havlin
  • (iv) Principles and Practices of Rice Production → (e) S.K. De Datta
  • (v) An Introduction to Climate → (d) Trewartha and Horn
  • This corresponds to Option 1: (i)-(b), (ii)-(a), (iii)-(c), (iv)-(e), (v)-(d).

 

 

Question 59

Safe moisture level for storage of safflower is:

  1. 13–15%
  2. 11–12%
  3. 9–10%
  4. 5–8%

Correct Answer: 4

Explanation:

  • Safflower seeds are oilseeds; they require a lower moisture contentfor safe storage compared to cereals.
  • The safe moisture level for oilseeds (including safflower) is 5–8%.
  • Higher moisture leads to mould growth, insect infestation, and oil rancidity.
  • Cereals (wheat, rice) have safe moisture levels of 10–14%.
  • Hence, Option 4is correct.

 

 

Question 60

Nitrogen fixation in root nodules of soybean:

  1. Begins with the appearance of leghemoglobin
  2. Begins with the initiation of bacterial division
  3. Begins after 7th week of nodule age
  4. Begins with the partial dissolution of root cell wall by pectic enzymes

Correct Answer: 1

Explanation:

  • Leghemoglobinis a pinkish-red oxygen-binding protein found in root nodules of legumes.
  • It regulates oxygen concentration for nitrogenase (which is oxygen-sensitive).
  • N₂ fixation begins simultaneously with the appearance of leghemoglobinin the nodules (about 10–14 days after infection).
  • Leghemoglobin is an indicator of functional nodules and active N fixation.
  • Hence, Option 1is correct.

 

 

Question 61

Soybean belongs to the subfamily:

  1. Faboideae
  2. Fabaceae
  3. Cruciferae
  4. Pedaliaceae

Correct Answer: 1

Explanation:

  • Soybean (Glycine max)belongs to the family Fabaceae (Leguminosae).
  • Within Fabaceae, it belongs to the subfamily Papilionoideae(also called Faboideae).
  • Faboideaeis characterised by:
    • Butterfly-shaped (papilionaceous) flowers
    • Typical legume fruit
    • Rhizobial root nodules
  • Fabaceae (2)is the family name, not subfamily.
  • Cruciferae (3)– mustard family.
  • Pedaliaceae (4)– sesame family.
  • Hence, Option 1is correct.

 

 

Question 62

The first commercial hybrid of rapeseed mustard developed on the basis of four CMS system was:

  1. PGSH 51
  2. NRCHB 506
  3. DMH 1
  4. PAC 432

Correct Answer: 3

Explanation:

  • DMH 1(Dhara Mustard Hybrid 1) was the first commercial hybrid of rapeseed mustard in India, developed using the four CMS (cytoplasmic male sterility) system.
  • It was released for cultivation in 1993–94.
  • PGSH 51– sunflower hybrid.
  • NRCHB 506– a hybrid in mustard but not the first.
  • PAC 432– hybrid in maize.
  • Hence, Option 3is correct.

 

 

Question 63

For rabi sunflower, irrigation has been found to be optimum at IW:CPE ratio of:

  1. 65
  2. 75
  3. 85
  4. 95

Correct Answer: 2

DExplanation:

  • IW/CPEratio (Irrigation Water / Cumulative Pan Evaporation) is used for scheduling irrigation.
  • For rabi sunflower, the optimum IW/CPE ratio is 75.
  • At this ratio, water stress is minimised, and yield is maximised without excess irrigation.
  • Lower ratios (0.65) may cause stress; higher ratios (0.85–0.95) may waste water.
  • Hence, Option 2is correct.

 

 

Question 64

The dormancy of sunflower seed is due to the presence of germination inhibitors in seed kernel up to ____ and seed-coat up to ____ after harvest:

  1. 10 days, 45 days
  2. 15 days, 55 days
  3. 20 days, 40 days
  4. 25 days, 35 days

Correct Answer: 3

Explanation:

  • Freshly harvested sunflower seeds exhibit dormancydue to germination inhibitors.
  • Seed kernelcontains inhibitors for about 20 days after harvest.
  • Seed coatcontains inhibitors for about 40 days after harvest.
  • After these periods, germination percentage increases naturally.
  • Hence, Option 3is correct.

 

 

Question 65

Pre-harvest aflatoxin contamination in groundnut can be prevented by:

  1. Growing cultivar that matures before the end of rain
  2. Growing cultivar that matures long after the rain ends
  3. Growing cultivar that matures at the end of the rain
  4. Subjecting the cultivars to drought stress at later stages

Correct Answer: 1

Explanation:

  • Aflatoxinis produced by Aspergillus flavus and  parasiticus.
  • Contamination occurs when groundnut pods are exposed to high moisture and high temperature, especially during late pod development.
  • Prevention strategy: Grow cultivars that mature before the end of the rainy season– this allows harvesting before conditions favour fungal growth.
  • Maturity at the end of rain (3) – risky because post-rain humidity promotes aflatoxin.
  • Drought stress (4) – increases susceptibility.
  • Hence, Option 1is correct.

 

 

Question 66

Seed viability of groundnut can be improved by:

  1. Sun drying of pods after harvesting
  2. Shade drying of pods after harvesting
  3. Oven drying of pods after harvesting
  4. Drying of pods in the plants itself before harvesting

Correct Answer: 1

Explanation:

  • Sun dryingof groundnut pods after harvesting is the traditional and most effective method to improve seed viability.
  • Sun drying reduces moisture content to safe levels (6–8%) for storage.
  • Shade drying (2)– slower, may allow mould growth.
  • Oven drying (3)– high heat damages seeds (reduces germination).
  • Drying in plants (4)– not practical; pods may shatter.
  • Hence, Option 1is correct.

 

 

Question 67

The shell of the groundnut pod is morphologically known as:

  1. Pericarp
  2. Kernel
  3. Testa
  4. Nut

Correct Answer: 1

Explanation:

  • The shell(outer covering) of groundnut pod is the pericarp (fruit wall).
  • In botany, a groundnut pod is a type of legume fruit; the wall of the fruit is the pericarp.
  • Kernel (2)– the seed inside the shell.
  • Testa (3)– the seed coat (brown papery covering of the kernel).
  • Nut (4)– a dry indehiscent fruit with a hard shell (groundnut is not a true nut).
  • Hence, Option 1is correct.

 

 

Question 68

The production of groundnut fluctuates mainly due to:

  1. Excess evaporation
  2. Excess transpiration
  3. Erratic distribution of rainfall
  4. High temperature

Correct Answer: 3

Explanation:

  • Groundnut is grown as a rainfed kharif crop in most parts of India.
  • Erratic distribution of rainfall(drought during flowering/pegging, heavy rain at harvest) is the primary cause of yield fluctuation.
  • Groundnut requires well-distributed rainfall of 500–700 mm.
  • Excess evaporation/transpiration (1,2)– secondary factors.
  • High temperature (4)– affects but not the main cause of fluctuation.
  • Hence, Option 3is correct.

 

 

Question 69

A cyclone normally moves at a speed of:

  1. 400–500 km/day
  2. 300–400 km/day
  3. 200–300 km/day
  4. 100–200 km/day

Correct Answer: 4

Explanation:

  • Tropical cyclones(in the Indian Ocean region) typically move at speeds of 100–200 km per day (about 4–8 km/hour).
  • Speed varies depending on the steering winds (trade winds).
  • Cyclones weaken after landfall and slow down.
  • Hence, Option 4is correct.

 

 

Question 70

The top portion of the sugarcane plant is good for seed material because it contains:

  1. More number of buds and more supply of nutrients
  2. More sucrose
  3. Less molassigenic substances
  4. Thick-walled meristem

Correct Answer: 1

Explanation:

  • For sugarcane seed material (setts), the top portion(upper half of the cane) is preferred because:
    • Contains more number of buds(nodes with viable buds).
    • Higher supply of nutrients and moisture (younger tissue).
  • More sucrose (2)– lower portion has more sucrose.
  • Less molassigenic substances (3)– not the primary reason for seed selection.
  • Thick-walled meristem (4)– not accurate.
  • Hence, Option 1is correct.

 

 

Question 71

In jute, bark fibres are developed from:

  1. Phloem tissue
  2. Xylem tissue
  3. Cortex
  4. Plasmodesmata

Correct Answer: 1

Explanation:

  • Jute fibreis a bast fibre (phloem fibre) derived from the phloem tissue of the stem.
  • These fibres are located in the inner bark (pericycle and primary phloem).
  • Xylem tissue (2)– gives wood fibres.
  • Cortex (3)– outer tissue, not fibre source.
  • Plasmodesmata (4)– intercellular connections.
  • Hence, Option 1is correct.

 

 

Question 72

A cotton plant is considered to be suffering from N stress if petiole nitrate-N of a mature leaf falls below:

  1. 2000 ppm
  2. 2500 ppm
  3. 3000 ppm
  4. 3500 ppm

Correct Answer: 1

Explanation:

  • Petiole nitrate-N testis a rapid diagnostic tool for nitrogen status in cotton.
  • In cotton, the critical level for N stress is below 2000 ppm(2,000 mg/kg) of NO₃-N in petiole sap.
  • If the level falls below this, N deficiency is present, and yield may be reduced.
  • Hence, Option 1is correct.

 

 

Question 73

The value of the correlation coefficient (r) ranges between:

  1. -α to +α
  2. -α to +1
  3. 0 to +1
  4. -1 to +1

Correct Answer: 4

Explanation:

  • The Pearson correlation coefficient (r)measures the strength and direction of the linear relationship between two variables.
  • It ranges from -1 to +1.
    • +1– perfect positive correlation.
    • 0– no correlation.
    • -1– perfect negative correlation.
  • Hence, Option 4is correct.

 

 

Question 74

The error degree of freedom for 7 treatments laid out in a Completely Randomized Design (CRD) with 4 replications is:

  1. 20
  2. 21
  3. 18
  4. 25

Correct Answer: 2

Calculation:

  • Total observations = Number of treatments (t) × Number of replications (r) = 7 × 4 = 28
  • Total DF = 28 – 1 = 27
  • Treatment DF = t – 1 = 7 – 1 = 6
  • Error DF = Total DF – Treatment DF = 27 – 6 = 21
  • Hence, Option 2is correct.

 

 

Question 75

Variance is the square of:

  1. Range
  2. Standard deviation
  3. Quartile deviation
  4. Mean deviation

Correct Answer: 2

Explanation:

  • Variance (σ²)is the average of squared deviations from the mean.
  • Standard deviation (SD)is the square root of variance.
  • Therefore, variance = (Standard deviation)².
  • Range (1)– difference between max and min.
  • Quartile deviation (3)– half of interquartile range.
  • Mean deviation (4)– average absolute deviation.
  • Hence, Option 2is correct.

 

 

Question 76

In a field trial with 7 treatments and 5 replications, if the Treatment SS and Error SS values are 280.60 and 196.50, the calculated F value will be:

  1. 59
  2. 59
  3. 48
  4. 36

Correct Answer: 3

Calculation:

  • Treatment DF = t – 1 = 7 – 1 = 6
  • Error DF = t(r – 1) = 7 × (5 – 1) = 7 × 4 = 28
  • Treatment MS = Treatment SS / Treatment DF = 280.60 / 6 = 46.77
  • Error MS = Error SS / Error DF = 196.50 / 28 = 7.02
  • F = Treatment MS / Error MS = 46.77 / 7.02 = 66≈ 6.48 (closest)
  • Hence, Option 3is correct.

 

 

Question 77

The value of r (correlation coefficient) ranges between:

  1. -1 and 0
  2. +1 and 0
  3. +1 and -1
  4. +0.9 and -0.9

Correct Answer: 3

Explanation:

  • The correlation coefficient (r) ranges from -1 to +1.
  • -1indicates a perfect negative linear relationship.
  • 0indicates no linear relationship.
  • +1indicates a perfect positive linear relationship.
  • Hence, Option 3is correct.

 

 

Question 78

The primary purpose of blocking in field experiments is to reduce:

  1. Experimental error
  2. Block error
  3. Replication error
  4. Treatment error

Correct Answer: 1

Explanation:

  • Blockingin field experiments (e.g., Randomized Block Design) groups experimental units into homogeneous blocks.
  • The purpose is to reduce experimental error(variability due to soil fertility gradients, moisture, etc.) by removing block-to-block variation.
  • Blocking increases precision of treatment comparisons.
  • Hence, Option 1is correct.

 

 

Question 79

IGFRI is located at:

  1. Jhansi
  2. Jorhat
  3. Jodhpur
  4. Jaipur

Correct Answer: 1

Explanation:

  • ICAR-IGFRI(Indian Grassland and Fodder Research Institute) is located in Jhansi, Uttar Pradesh.
  • It conducts research on forage crops, grassland management, and seed production.
  • Established in 1962 (originally as Indian Grassland and Fodder Research Institute).
  • Hence, Option 1is correct.

 

 

Question 80

The test between two population variances is done by:

  1. Arithmetic mean
  2. F-test
  3. T-test
  4. Z-test

Correct Answer: 2

Explanation:

  • The F-test(named after R.A. Fisher) is used to compare two population variances.
  • It is calculated as F = (larger variance) / (smaller variance).
  • The F-test is also used in ANOVA to test overall treatment differences.
  • T-test (3)– compares two means.
  • Z-test (4)– compares two proportions or means (large sample).
  • Arithmetic mean (1)– measure of central tendency, not a test.
  • Hence, Option 2is correct.

 

Want to Access the Complete ICAR JRF Agronomy Full Course?

Enroll Full Course Here

 

 

✅ Benefits of Full Course

  • 10+ Years ICAR JRF Agronomy Practice Series with Detailed Explanation
  • Complete Memory-Based PYQ Discussion
  • Concept-Based Learning for Better Understanding
  • Time Management & Exam Practice
  • Latest ICAR JRF Memory-Based Questions Included
  • Mobile Friendly Learning Access
  • Regular Course Content Updates
  • Guidance & Strategy for ICAR JRF Preparation
Previous
Next

Support Us

error: Content is protected !!