ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersDesigning greenhouses and controlling the climate inside them, with the plastics and soilless systems that go with them.
In the exams: ICAR AICE subject 40 Unit 9 and subject 43 Unit 5; NCEES PE greenhouses; skill modules and electives in the ICAR 6th Deans' Committee B.Tech
ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 40 (SWCE) and 43 (IDE)
Subject 40 Unit 9 and subject 43 Unit 5 name protected cultivation: greenhouse types and cladding properties, heat balance and ventilation, fan-pad cooling, fertigation equipment, hydroponics and plastic mulch and lining. Expect quick recall and one-step calculations at PhD-entrance depth.
How our questions for ICAR AICE-JRF/SRF spread across the units of Protected Cultivation & Greenhouse Engineering. Difficulty: 8 easy, 40 medium, 23 hard.
| Hydroponics and aeroponics | 9 | |
| Glazing and cladding materials | 8 | |
| Plastics in agriculture: mulch and lining | 8 | |
| Micro-irrigation and fertigation in greenhouses | 7 | |
| CO2 enrichment | 6 | |
| Greenhouse heat balance | 5 | |
| Natural ventilation | 5 | |
| Automation of climate control | 5 | |
| Fan-pad and fogging cooling | 4 | |
| Greenhouse heating | 4 | |
| Shading | 4 | |
| Greenhouse types and orientation | 3 | |
| Forced ventilation | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Which statement about structural design codes for commercial greenhouses (for example EN 13031-1 in Europe and IS 14462 in India) is correct?
D. Codes allow film houses a shorter design life than permanent buildings, which lowers the design wind speed
A shorter reference period for light film structures reduces the characteristic wind speed used in design. Crop loads from hanging vine crops are a named load case, wind pressure depends on the site's wind climate, and anchorage against uplift is a central check.
Further reading: EN 13031-1; IS 14462
What is the purpose of horizontal air-flow (HAF) fans hung in the canopy of a closed greenhouse?
A. To stir the inside air so that temperature and humidity are even and leaves dry faster
HAF fans circulate air inside the house to remove warm and humid pockets and reduce disease; they do not exchange air with outside, so they cannot replace exhaust fans. They are not used to pressurise the house or to remove CO2.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
A 30% shade net and a 50% shade net are used together, one above the other. What is the combined shade?
D. 65%
Transmissions multiply: (1 − 0.3) × (1 − 0.5) = 0.35, so shade = 1 − 0.35 = 65%. Adding the ratings gives 80%; 35% is the light that passes, not the shade; 15% multiplies the shade fractions.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
Low tunnels of thin transparent film over hoops are used in cold winters to raise crops such as cucurbits ahead of the field season. How are they managed as days warm?
C. Holes are cut or the edges raised for ventilation, and the tunnels are removed when outside temperatures allow
Unventilated low tunnels overheat quickly in sunshine, so growers open them progressively and then take them off. Sealing, painting or fogging them would cause heat or disease damage.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
In the nutrient film technique (NFT), the roots of the crop
C. lie in shallow sloping channels with a thin film of solution flowing past them
NFT uses a thin, flowing film so that the upper root mat is in air and the lower part in solution. The misted chamber is aeroponics, the floating raft is deep-water culture, and dripper-fed slabs are substrate culture.
Further reading: Resh, Hydroponic Food Production; Nelson, Greenhouse Operation and Management
Which soilless system gives the crop the most time to survive if the circulation pump fails on a hot day?
B. Deep-water floating-raft culture, since the large aerated tank buffers water and temperature
A deep tank holds a large volume of solution around the roots, so plants survive a pump failure for much longer. Roots in aeroponics and NFT dry within minutes to hours, which is why those systems need back-up power and alarms.
Further reading: Resh, Hydroponic Food Production; Nelson, Greenhouse Operation and Management
A 2,000 m² greenhouse has a crop water use of 5 mm/d and a drip application efficiency of 90%. There are 6,000 plants, each on one 1.6 L/h dripper. How long must the system run each day?
B. 69 min
Volume = 2,000 × 0.005 / 0.9 = 11.11 m³/d. Flow = 6,000 × 1.6 = 9,600 L/h = 9.6 m³/h. Time = 11.11 / 9.6 = 1.157 h = 69 min. 63 min ignores efficiency; 1.2 min is hours mislabelled as minutes; 694 min is a factor-of-ten slip.
Further reading: FAO Irrigation and Drainage Paper 56; Nelson, Greenhouse Operation and Management
How is fruit set in greenhouse tomatoes usually ensured, since there is no wind to shake the flowers?
D. By bumblebee colonies in the house or by vibrating the flower trusses
Tomato flowers self-pollinate when shaken; bumblebees buzz-pollinate them and electric vibrators do the same by hand. Water sprays, leaf removal and 'male plants' (tomato flowers are bisexual) do not achieve this.
Further reading: Nelson, Greenhouse Operation and Management
In a greenhouse integrated pest management programme, the parasitic wasp Encarsia formosa is released to control
B. greenhouse whitefly
Encarsia formosa parasitises whitefly scales and is one of the oldest biological controls in greenhouses. Spider mite is controlled by the predatory mite Phytoseiulus persimilis; thrips and nematodes need other agents.
Further reading: Nelson, Greenhouse Operation and Management
A square farm pond is 30 m × 30 m at the top, 2.5 m deep, with side slopes of 2:1 (horizontal:vertical). What area of liner covers the bottom and the four sloping faces, before anchor trench and overlaps?
C. 959 m²
Bottom side = 30 − 2 × 2 × 2.5 = 20 m. Slant length = 2.5 × √(1 + 2²) = 5.590 m. Each face = (30 + 20)/2 × 5.590 = 139.8 m². Total = 400 + 4 × 139.8 = 959 m². 900 m² is the plan area at the top; 650 m² uses the vertical depth instead of the slant length; 1,459 m² uses the top instead of the bottom area.
Further reading: Orzolek (ed.), Plastics in Agriculture
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