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ASRB ARS / NET / SMS / STO Atmospheric Science & Agrometeorology: practice questions

ASRB ARS / NET / SMS / STO — Environmental Science and Agricultural Meteorology disciplines · India · Agricultural Scientists Recruitment Board (ICAR)

Original practice questions written to the published syllabus · not past papers
10worked here, free
42written to this exam in this subject
10syllabus units

What this subject covers

Radiation, heat, moisture and motion in the atmosphere, and how weather and climate are measured and used on farms. ICAR AICE subject 60 Agricultural Meteorology examines it in depth.

In the exams: ICAR AICE-JRF/SRF subject 60 (whole paper); ICAR AIEEA-PG 3.1 Agricultural Meteorology; GATE ES Section 6 (meteorology for dispersion)

Where the 42 questions fall

How our questions for ASRB ARS / NET / SMS / STO spread across the units of Atmospheric Science & Agrometeorology. Difficulty: 10 easy, 18 medium, 14 hard.

Evapotranspiration estimation and crop water needs9
Solar radiation, albedo and the surface energy balance5
Latent and sensible heat; temperature inversions5
Atmospheric humidity: vapour pressure, relative humidity, dew point, VPD5
Crop-weather relations: growing degree days, photoperiod, thermal stress5
Weather instruments and observatories; automatic weather stations4
Weather forecasting for agriculture and agromet advisories4
Climatic classification: Köppen and Thornthwaite3
Radiation laws: Planck, Stefan-Boltzmann, Wien, Kirchhoff, Beer-Lambert1
Clouds, precipitation processes and forms1

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Solar radiation, albedo and the surface energy balance · medium

Using the Ångström–Prescott equation with FAO-56 defaults (as = 0.25, bs = 0.50), extraterrestrial radiation Ra = 38.0 MJ m⁻² day⁻¹, actual sunshine n = 8 h and maximum possible N = 12 h, what is the estimated global radiation Rs?

  1. 28.5 MJ m⁻² day⁻¹
  2. 12.7 MJ m⁻² day⁻¹
  3. 22.2 MJ m⁻² day⁻¹
  4. 38.0 MJ m⁻² day⁻¹
Answer and working

C. 22.2 MJ m⁻² day⁻¹

Rs = (0.25 + 0.50 × 8/12) × 38.0 = 0.583 × 38.0 = 22.2. 28.5 assumes a cloudless day; 38.0 inverts n/N; 12.7 omits the 0.25 term.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

2 · Atmospheric humidity: vapour pressure, relative humidity, dew point, VPD · medium

For the same air (30 °C, dew point 16 °C), what is the vapour pressure deficit?

  1. 2.42 kPa
  2. 1.60 kPa
  3. 1.82 kPa
  4. 4.24 kPa
Answer and working

A. 2.42 kPa

VPD = es(30) − es(16) = 4.243 − 1.818 = 2.42 kPa. 4.24 is es alone; 1.60 computes es at the temperature difference; 1.82 is the actual vapour pressure.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

3 · Climatic classification: Köppen and Thornthwaite · medium

What is the central concept of Thornthwaite's 1948 climate classification?

  1. Mean wind speed and direction, used to define the exposure of each region
  2. The number of frost-free days, used to define the length of the growing season
  3. The distribution of natural vegetation, mapped from satellite images of land cover
  4. Potential evapotranspiration set against precipitation
Answer and working

D. Potential evapotranspiration set against precipitation

Thornthwaite based his scheme on PE estimated from temperature and day length, and on the water balance of precipitation against PE. Köppen's scheme was designed around vegetation; frost days and wind are not his basis.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

4 · Climatic classification: Köppen and Thornthwaite · medium

Why should a Thornthwaite moisture-index boundary always be quoted with the formula used?

  1. The index is defined only for India, so values from other countries cannot be compared
  2. The index uses wind speed, which is measured at different heights in different countries
  3. The 1948 index weights aridity by 0.6 and the 1955 Thornthwaite–Mather one does not
  4. The index was abandoned in 1948 and replaced entirely by the Köppen classification
Answer and working

C. The 1948 index weights aridity by 0.6 and the 1955 Thornthwaite–Mather one does not

Different formulations of Im exist (1948 with a 0.6 weighting of the aridity index; Thornthwaite–Mather 1955 without), so class boundaries and values are not interchangeable. The index is global, uses no wind, and remains in use.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

5 · Crop-weather relations: growing degree days, photoperiod, thermal stress · easy

Wheat (base temperature 5 °C) experiences Tmax = 30 °C and Tmin = 16 °C with 11.5 h of day length. What are the photothermal units for the day?

  1. 432 °C day h
  2. 207 °C day h
  3. 264.5 °C day h
  4. 18 °C day h
Answer and working

B. 207 °C day h

GDD = (30 + 16)/2 − 5 = 18; PTU = GDD × day length = 18 × 11.5 = 207. 18 is GDD alone; 264.5 forgets the base; 432 uses 24 hours instead of day length.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

6 · Crop-weather relations: growing degree days, photoperiod, thermal stress · hard

Why was photoperiod insensitivity a key trait of Green Revolution rice and wheat varieties?

  1. They flowered on time across latitudes and seasons, widening where they could grow
  2. It made them flower only in long days, limiting them to high-latitude temperate regions
  3. It made them resistant to all pests by shortening their exposure in the field
  4. It removed their need for water, allowing cultivation without irrigation
Answer and working

A. They flowered on time across latitudes and seasons, widening where they could grow

Insensitive varieties flower according to temperature and age, not day length, so they fit different latitudes and sowing dates. They still need water and are not inherently pest-resistant.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

7 · Evapotranspiration estimation and crop water needs · easy

Reference ETo is 6.0 mm day⁻¹ and the mid-season crop coefficient is 1.15. What is crop evapotranspiration?

  1. 7.15 mm day⁻¹
  2. 6.0 mm day⁻¹
  3. 6.9 mm day⁻¹
  4. 5.2 mm day⁻¹
Answer and working

C. 6.9 mm day⁻¹

ETc = Kc × ETo = 1.15 × 6.0 = 6.9. Dividing gives 5.2; adding gives 7.15; 6.0 ignores Kc.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

8 · Evapotranspiration estimation and crop water needs · hard

Wind speed measured at 10 m is 3.2 m s⁻¹. Using FAO-56, u₂ = uz × 4.87/ln(67.8z − 5.42), what is the wind speed at 2 m?

  1. 0.64 m s⁻¹
  2. 4.28 m s⁻¹
  3. 3.20 m s⁻¹
  4. 2.39 m s⁻¹
Answer and working

D. 2.39 m s⁻¹

ln(678 − 5.42) = 6.511; u₂ = 3.2 × 4.87/6.511 = 2.39. 3.20 ignores the height correction; 0.64 scales linearly with height; 4.28 inverts the factor.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

9 · Evapotranspiration estimation and crop water needs · easy

Which method is regarded as the reference standard for directly measuring crop evapotranspiration in the field?

  1. A Campbell–Stokes recorder
  2. A weighing lysimeter
  3. A tipping-bucket rain gauge
  4. A Class A pan
Answer and working

B. A weighing lysimeter

Weighing lysimeters measure the change in mass of an isolated soil–crop block. Pans estimate ETo indirectly; the other instruments measure sunshine and rain.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

10 · Weather forecasting for agriculture and agromet advisories · hard

A rain forecast verification table over 100 days shows 20 hits, 10 false alarms, 5 misses and 65 correct negatives. What is the probability of detection?

  1. 0.85
  2. 0.80
  3. 0.33
  4. 0.67
Answer and working

B. 0.80

POD = hits/(hits + misses) = 20/25 = 0.80. 0.33 is the false alarm ratio; 0.85 is the proportion correct; 0.67 is hits/(hits + false alarms), the success ratio.

Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there

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Atmospheric Science & AgrometeorologyThe full study manual, free ASRB ARS / NET / SMS / STOFormat, sections and how it has changed

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