CSIR-UGC NET — Earth, Atmospheric, Ocean and Planetary Sciences · India · National Testing Agency for CSIR
Original practice questions written to the published syllabus · not past papersRadiation, 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)
CSIR-UGC NET — Earth, Atmospheric, Ocean and Planetary Sciences
Atmospheric thermodynamics, radiation, general circulation, monsoon and tropical cyclones, cloud physics and precipitation processes.
How our questions for CSIR-UGC NET spread across the units of Atmospheric Science & Agrometeorology. Difficulty: 3 easy, 17 medium, 3 hard.
| Radiation laws: Planck, Stefan-Boltzmann, Wien, Kirchhoff, Beer-Lambert | 5 | |
| Atmospheric stability and adiabatic processes | 5 | |
| Layers of the atmosphere and their characteristics | 4 | |
| Pressure systems, winds and general circulation | 4 | |
| Clouds, precipitation processes and forms | 2 | |
| Monsoons, air masses, fronts and cyclones | 2 | |
| Weather forecasting for agriculture and agromet advisories | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
What is the average environmental lapse rate in the troposphere?
C. About 6.5 °C per km
The standard atmosphere uses 6.5 °C per km. 9.8 °C per km is the dry adiabatic lapse rate of a rising unsaturated parcel, not the average environmental profile.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
Why is the tropopause higher over the equator than over the poles?
D. Strong surface heating and deep convection in the tropics lift the troposphere to about 16–18 km
Vigorous tropical convection mixes heat high, giving a tropopause near 16–18 km versus about 8 km at the poles. Rotation, ozone placement and gravity are not the explanation; gravity is in fact slightly weaker at the equator.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
A leaf surface at 300 K has an emissivity of 0.95. Using σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, how much longwave radiation does it emit?
A. 436 W m⁻²
E = εσT⁴ = 0.95 × 5.67 × 10⁻⁸ × 300⁴ = 436 W m⁻². 459 ignores emissivity; using T³ gives 1.45; using 27 °C instead of kelvin gives a negligible value.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
Kirchhoff's law states that, for a body in thermal equilibrium,
C. its emissivity at a wavelength equals its absorptivity at that wavelength
Kirchhoff's law links emissivity and absorptivity at the same wavelength. The other statements are the Stefan–Boltzmann law, Wien's law and the Beer–Lambert law.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
A solar beam strikes level ground at 60° from the vertical. Compared with a beam at normal incidence, the energy received per unit area is
C. half, by Lambert's cosine law
Irradiance ∝ cos(zenith angle); cos 60° = 0.5. The inverse-square law concerns distance from the source, and Stefan–Boltzmann concerns emission, not incidence angle.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
Surface air temperature is 30 °C and the dew point is 18 °C. Using the approximation of 125 m per °C of dew-point depression, at what height will a rising parcel reach its lifting condensation level?
D. 1500 m
LCL ≈ 125 × (30 − 18) = 1500 m. 3750 m uses the air temperature alone; 2250 m uses the dew point alone; 12 m forgets the 125 m factor.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
Using f = 2Ω sin φ with Ω = 7.292 × 10⁻⁵ s⁻¹, what is the Coriolis parameter at 30° latitude?
A. 7.292 × 10⁻⁵ s⁻¹
f = 2 × 7.292 × 10⁻⁵ × sin 30° = 2 × 7.292 × 10⁻⁵ × 0.5 = 7.29 × 10⁻⁵ s⁻¹. 1.46 × 10⁻⁴ is the polar value; 3.65 × 10⁻⁵ omits the factor 2; 0 is the equatorial value.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
Why do surface winds blow across isobars towards low pressure rather than parallel to them?
C. Friction slows the wind and weakens the Coriolis force, so the pressure gradient force wins partly
Above the friction layer the wind is near-geostrophic, parallel to isobars. Near the ground friction reduces speed and thus Coriolis deflection, so air crosses isobars towards low pressure. Coriolis is not strongest at the surface, and surface charts do show isobars.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
A sequence of cirrus, cirrostratus, altostratus and then nimbostratus with prolonged light rain typically signals the approach of
A. a warm front
Warm air rising gently over cold air ahead of a warm front gives a broad, lowering cloud sequence with prolonged rain. Cold fronts give a narrow band of cumulonimbus and showers; highs give clear skies.
Further reading: See the manual sources: FAO-56, WMO Guide to Agricultural Meteorological Practices, and the textbooks named there
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