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Tamil Nadu PSC Remote Sensing, GIS & GNSS for Land & Water: practice questions

Tamil Nadu PSC — Agricultural Engineering (Code 164, degree standard) · India · Tamil Nadu Public Service Commission

Original practice questions written to the published syllabus · not past papers
10worked here, free
92written to this exam in this subject
10syllabus units

What this subject covers

Mapping 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

What Tamil Nadu PSC asks from it

Tamil Nadu PSC Agricultural Engineering (Code 164), Unit III

Degree-standard questions on watershed development using GIS and remote sensing: FCC interpretation, land use mapping, drainage network and morphometric analysis, and site selection for conservation structures.

Where the 92 questions fall

How our questions for Tamil Nadu PSC spread across the units of Remote Sensing, GIS & GNSS for Land & Water. Difficulty: 29 easy, 48 medium, 15 hard.

Image classification12
Sensors, platforms and resolutions10
Image interpretation10
GIS data models10
Spatial analysis9
QGIS/ArcGIS practice9
The electromagnetic spectrum8
Vegetation indices8
Digital elevation models and watershed delineation8
Applications to watershed, irrigation and erosion mapping8

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 · The electromagnetic spectrum · medium

By the Stefan–Boltzmann law, if the absolute temperature of a blackbody doubles, its total emitted energy per unit area becomes

  1. 8 times the original value
  2. 4 times the original value
  3. 16 times the original value
  4. 2 times the original value
Answer and working

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

2 · The electromagnetic spectrum · medium

On a near-infrared image, a lake appears much darker than the surrounding cropland. The main reason is that

  1. lakes are cooler than cropland and cooler surfaces reflect less near-infrared radiation
  2. water absorbs nearly all near-infrared energy while crop canopies scatter it strongly
  3. water scatters near-infrared energy sideways so that little reaches the sensor overhead
  4. suspended sediment in the lake absorbs near-infrared energy that clear water would reflect
Answer and working

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

3 · Sensors, platforms and resolutions · medium

An image band has 30 m pixels. A classified paddy area contains 1200 pixels. What is the paddy area?

  1. 1080 ha
  2. 10.8 ha
  3. 108 ha
  4. 36 ha
Answer and working

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

4 · Sensors, platforms and resolutions · medium

An engineer must map paddy transplanting across a monsoon district where cloud cover persists for weeks. Which sensor type is most suitable?

  1. A thermal infrared sensor, because clouds are transparent in the 8–14 µm window
  2. A hyperspectral sensor, because narrow bands can separate cloud from standing water
  3. A high-resolution optical sensor, because finer pixels see between the cloud gaps
  4. Synthetic aperture radar, because active microwave imaging works through cloud
Answer and working

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

5 · Vegetation indices · medium

For a partly bare field with NIR reflectance 0.30 and red reflectance 0.10, what is the soil-adjusted vegetation index with L = 0.5?

  1. 0.50
  2. 0.75
  3. 0.33
  4. 0.22
Answer and working

C. 0.33

SAVI = (1 + L)(NIR − Red)/(NIR + Red + L) = 1.5 × 0.20/(0.30 + 0.10 + 0.5) = 0.30/0.90 = 0.33. 0.50 is the NDVI (L = 0); 0.22 omits the (1 + L) factor; 0.75 omits L from the denominator (1.5 × 0.2/0.4).

Further reading: Huete (SAVI); Jensen, Remote Sensing of the Environment

6 · GIS data models · medium

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?

  1. 1.4 times
  2. 2 times
  3. 8 times
  4. 4 times
Answer and working

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)

7 · Applications to watershed, irrigation and erosion mapping · hard

A GIS-based groundwater prospect map rates a village as a 'very good' zone. What does this mean for siting a borewell?

  1. It shows the measured depth to the water table, so the well depth can be set from the map
  2. It guarantees a yield above the state norm, so drilling can proceed without further survey
  3. It indicates higher likelihood; a geophysical survey or test drilling must confirm the aquifer
  4. It means the aquifer is recharged by canals, so yields will not decline in dry years
Answer and working

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

8 · Image interpretation · medium

What is the approximate forward overlap between successive photographs in a standard strip of vertical aerial photography for stereo mapping?

  1. About 20 percent
  2. About 90 percent
  3. About 60 percent
  4. About 40 percent
Answer and working

C. About 60 percent

About 60% forward overlap ensures every point appears on at least two photographs, giving stereo coverage. About 20–30% is the usual sidelap between adjacent strips. 90% is wasteful for manned photography; 40% is the advance, not the overlap.

Further reading: Lillesand, Kiefer and Chipman, Remote Sensing and Image Interpretation

9 · GIS data models · medium

What is a triangulated irregular network (TIN)?

  1. A network of stream lines coded by Strahler order for hydrological analysis
  2. A raster of equal square cells, each storing a single height value for its area
  3. A surface model of non-overlapping triangles joining irregularly spaced height points
  4. A set of contour lines stored as closed polygons with a height attribute each
Answer and working

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

10 · Spatial analysis · medium

Reclassifying a slope raster into classes 1 (0–3%), 2 (3–8%) and 3 (above 8%) is an example of

  1. a local operation that changes each cell's value independently of its neighbours
  2. a global operation that needs every cell to compute the value of each cell
  3. a focal operation that uses a moving 3 × 3 window around each cell
  4. a zonal operation that summarises values inside separate polygon zones
Answer and working

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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Read the subject

Remote Sensing, GIS & GNSS for Land & WaterThe full study manual, free Tamil Nadu PSCFormat, sections and how it has changed

More Tamil Nadu PSC practice

Surveying & LevellingSoil & Water Conservation EngineeringIrrigation Engineering (Soil-Water-Plant, Conveyance, Surface Methods)Fluid Mechanics & Open Channel HydraulicsWater Resources Management, Floods & Canal SystemsBuilding Materials, Theory of Structures & ConstructionWatershed Planning & ManagementWatershed HydrologySprinkler & Micro-Irrigation Systems

Remote Sensing, GIS & GNSS for Land & Water for other exams

ICAR AICE-JRF/SRF · IndiaNCEES PE Agricultural and Biological Engineering · USA

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