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 papersMapping land and water from satellites and in GIS: images, indices, elevation models and watershed delineation.
In the exams: ICAR AICE subject 40 Unit 5 (RS and GIS in SWC); TNPSC Unit III; NCEES PE GIS; skill modules in the ICAR 6th Deans' Committee B.Tech
ICAR AICE-JRF/SRF (PhD) — Soil and Water Conservation Engineering (subject 40), Unit 5
Principles of remote sensing and GIS applied to soil and water conservation: spectral signatures, sensors and resolutions, image classification and accuracy, vegetation indices, DEM-based watershed delineation and morphometry, RUSLE and runoff modelling in GIS, and watershed prioritisation. Expect recall of definitions mixed with short calculations.
How our questions for ICAR AICE-JRF/SRF spread across the units of Remote Sensing, GIS & GNSS for Land & Water. Difficulty: 32 easy, 57 medium, 30 hard.
| Image classification | 14 | |
| Sensors, platforms and resolutions | 12 | |
| GNSS | 12 | |
| Image interpretation | 11 | |
| GIS data models | 11 | |
| Spatial analysis | 11 | |
| The electromagnetic spectrum | 10 | |
| Vegetation indices | 10 | |
| Digital elevation models and watershed delineation | 10 | |
| Applications to watershed, irrigation and erosion mapping | 9 | |
| QGIS/ArcGIS practice | 9 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
By the Stefan–Boltzmann law, if the absolute temperature of a blackbody doubles, its total emitted energy per unit area becomes
C. 16 times the original value
M = σT⁴, so doubling T multiplies M by 2⁴ = 16. Four times would be a T² law; eight times a T³ law; twice a linear law. (Wien's law, by contrast, halves the peak wavelength.)
Further reading: Lillesand, Kiefer and Chipman, Remote Sensing and Image Interpretation
On a near-infrared image, a lake appears much darker than the surrounding cropland. The main reason is that
B. water absorbs nearly all near-infrared energy while crop canopies scatter it strongly
Clear water absorbs NIR within a few centimetres, so NIR reflectance is near zero, whereas vegetation reflects NIR strongly. Reflected NIR has nothing to do with surface temperature (that governs thermal emission). Sediment actually raises visible reflectance of water; it does not create the NIR darkness.
Further reading: Lillesand, Kiefer and Chipman, Remote Sensing and Image Interpretation
An image band has 30 m pixels. A classified paddy area contains 1200 pixels. What is the paddy area?
C. 108 ha
Each pixel covers 30 × 30 = 900 m² = 0.09 ha, so 1200 × 0.09 = 108 ha. 36 ha has no consistent derivation (using 30 m in place of 900 m² gives 1200 × 30 = 36,000 m², which is only 3.6 ha); 1080 and 10.8 ha are factor-of-ten slips in converting m² to ha (1 ha = 10,000 m²).
Further reading: Lillesand, Kiefer and Chipman, Remote Sensing and Image Interpretation
An engineer must map paddy transplanting across a monsoon district where cloud cover persists for weeks. Which sensor type is most suitable?
D. Synthetic aperture radar, because active microwave imaging works through cloud
Microwaves pass through cloud and SAR supplies its own illumination, so it images whatever the weather; flooded paddy fields also have a distinctive backscatter. Clouds are opaque to thermal infrared and to all optical wavelengths, whatever the spatial or spectral resolution.
Further reading: Lillesand, Kiefer and Chipman, Remote Sensing and Image Interpretation
Why is a single-date image often insufficient to separate two summer crops, and what is the usual remedy?
B. Their spectra can overlap on one date; multi-date data capture their different seasonal curves
Different crops at similar growth stages can look alike on one date, but their sowing, peak and harvest dates differ, so time series or multi-date stacks separate them. More bits, thermal data or dropping bands do not address the phenological overlap.
Further reading: Jensen, Introductory Digital Image Processing
On the Mercator projection, the scale factor at latitude φ is sec φ. How much is area exaggerated at 60° latitude compared with the equator?
D. 4 times
Linear scale = sec 60° = 2 in every direction, so area scale = sec² 60° = 4. 2 times is the linear scale only; 1.4 is sec 45°; 8 would require a cubic relationship.
Further reading: Snyder, Map Projections: A Working Manual (USGS PP 1395)
A surveyor notices that PDOP rises sharply when most visible satellites are clustered low in the eastern sky. Why?
B. Clustered satellites give poor geometry, so range errors map into large position errors
DOP expresses how satellite geometry amplifies range errors; when satellites are close together in the sky, their lines of sight are nearly parallel and the position is poorly determined. Clock accuracy and frequency sharing are not the cause, and low satellites are actually farther away.
Further reading: Hofmann-Wellenhof, Lichtenegger and Wasle, GNSS
A GIS-based groundwater prospect map rates a village as a 'very good' zone. What does this mean for siting a borewell?
C. It indicates higher likelihood; a geophysical survey or test drilling must confirm the aquifer
Prospect maps combine indirect evidence such as geology, geomorphology, lineaments and drainage density by weighted overlay; they rank likelihood only. Yield and depth need hydrogeological investigation, and the map says nothing about recharge source or sustainability.
Further reading: Hydrogeology and RS/GIS application literature
What is a triangulated irregular network (TIN)?
C. A surface model of non-overlapping triangles joining irregularly spaced height points
A TIN connects irregularly spaced points (often by Delaunay triangulation) into triangles, placing more points where terrain changes quickly. A grid of cells is a raster DEM; contours are lines, not triangles; a stream network is a different dataset.
Further reading: Chang, Introduction to Geographic Information Systems
Reclassifying a slope raster into classes 1 (0–3%), 2 (3–8%) and 3 (above 8%) is an example of
A. a local operation that changes each cell's value independently of its neighbours
Reclassification replaces each cell's value using a lookup rule based only on that cell, so it is local. Computing slope itself is focal; summarising by zones is zonal; distance surfaces are global.
Further reading: Chang, Introduction to Geographic Information Systems
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