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Geographical Techniques Questions and Answers

61.

Match List-I and List-II and select the correct answer using the codes given below :

List-IList-II
(a) Lorenz curve(1) A synthetic graph which is a combination of graphs and diagrams.
(b) Hydrographs(2) The stage-discharge curve for any channel
(c) Ergograph(3) Simple graphical method for comparing distribution on two-dimensional surface.
(d) Rating curve(4) Flow rate of streams as a function of time.

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The correct matching is a-3, b-4, c-1, d-2.

  1. The Lorenz curve compares a distribution with perfect equality on a two-dimensional graph.
  2. A hydrograph shows the flow rate of a stream as a function of time.
  3. An ergograph is a synthetic graph combining graphs and diagrams, for example of climate and the farming calendar.
  4. A rating curve is the stage-discharge curve of a channel.

Option B gives hydrographs the rating curve's meaning, and options C and D misplace the Lorenz curve.

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Answer: B

A 100 percent slope means the horizontal distance is the same as the vertical difference.

Percentage slope is the vertical rise (rise) divided by the horizontal distance (run), multiplied by 100.

\[ \text{slope} = \frac{\text{rise}}{\text{run}} \times 100 \]

\[ \frac{100}{100} \times 100 = 100\% \]

So a 100 percent slope rises 100 m in 100 m of horizontal distance, an angle of 45°. A horizontal distance double the rise gives 50 percent, half the rise gives 200 percent, and ten times the rise gives 10 percent, so the other options are wrong.

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63.

The sensing and recording ability of the sensor in different wavelength of electromagnetic radiation is known as :

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The ability is spectral resolution.

Spectral resolution describes how many bands of the electromagnetic spectrum a sensor records and how narrow they are. A multispectral sensor records a few broad bands; a hyperspectral sensor records hundreds of narrow ones.

The other three are different properties. Spatial resolution is the size of the smallest ground area a pixel represents. Radiometric resolution is the sensitivity to differences in brightness, expressed as the number of bits per pixel. Temporal resolution is how often the sensor revisits the same place.

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64.

Which of the following are used for calculating drainage density ?

(1) Stream length
(2) Stream order
(3) Basin perimeter
(4) Basin area

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

Drainage density uses (1) and (4): stream length and basin area.

Horton defined drainage density as the total length of all the streams in a basin divided by the area of the basin:

\[ D_d = \frac{\sum L}{A} \]

It is expressed in kilometres per square kilometre, and it is high on impermeable rock and weak, sparsely vegetated ground.

Stream order is used for bifurcation ratio and stream frequency, not drainage density, and basin perimeter is used for shape measures such as the circularity ratio. So every option including (2) or (3) is wrong.

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65.

Which of the following profiles represents only an outline of the hill ranges ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The key gives the composite profile.

Composite, superimposed and projected profiles are all drawn from a set of parallel serial profiles across a map. A composite profile combines them into a single line by taking, along the section, only the highest points of all of them. The result is one summit-level outline of the relief, as seen in a horizontal plane from a great distance, which is why it shows only the outline of the hill ranges.

A cross profile and a longitudinal profile are single sections across and along a valley. A projected profile keeps the visible part of each serial profile in turn, so it shows several lines one behind another rather than a single outline.

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66.

Method of data collection changes primarily with :

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The key gives scope of enquiry.

The reading that fits is that the method of collecting data follows how wide the enquiry is. A wide-area study points to census or sample surveys and secondary sources, while a narrow, site-specific study can rely on focused fieldwork. Research methods texts treat the nature and scope of the enquiry as the main factor in choosing a collection method.

What could not be confirmed: no source found states the sentence in these exact terms, so the key is kept as a well-supported principle and not a formal rule. The subject of investigation (A) also influences the method, so A is a rival reading that the key does not take. A statistical technique (C) is chosen after the data are collected, and the study area (D) is part of the scope rather than a separate determinant.

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67.

Which type of regions can be best analysed by "Principal Component Analysis" ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

Principal component analysis is best suited to formal regions.

A formal (uniform or homogeneous) region is one whose areas share similar characteristics, often across many variables at once. Principal component analysis reduces a large set of correlated variables, such as income, literacy, employment and housing, to a few independent components, and areas can then be grouped by their component scores into regions that are homogeneous on many measures together.

A functional (nodal) region is defined by flows and interaction with a centre and is analysed with flow and gravity methods. A planning region is defined for administration, and 'non-developed region' is not a class defined by multivariate similarity.

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68.

Which of the following options is true for a positively skewed distribution ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

For a positively skewed distribution, Mean \(\gt\) Median \(\gt\) Mode.

A positively skewed distribution has a long tail to the right. The mode sits at the peak, on the left. The median, which divides the frequencies into halves, lies to the right of the mode. The mean is pulled farthest to the right by the few very large values in the tail, so the order is mean greater than median greater than mode. Income distribution is the usual example.

Mean = Median = Mode holds only in a symmetrical distribution. Mean \(\lt\) Median \(\lt\) Mode is the negatively skewed case, and the order in option A is not possible for a unimodal skewed distribution, since the median always lies between the mean and the mode.

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69.

Match List-I and List-II and select the correct answer by using the codes given below :

List I (Satellite)List II (Major Application)
(a) ASTER(1) Used for cartographic application, urban, rural and coastal land use and other land information system
(b) MODIS(2) High resolution commercial satellite used for land use studies, natural resource, disaster management etc.
(c) Cartosat(3) High resolution multi-spectral stereo images for digital terrain model, change detection of land surfaces
(d) IKONOS(4) Used for large scale global cloud cover, radiation budget and oceanic studies

Codes : (a) (b) (c) (d)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The correct matching is (3) (4) (1) (2).

  • ASTER, on NASA's Terra satellite, records multispectral images with along-track stereo, widely used for digital elevation models and land surface change.
  • MODIS, on Terra and Aqua, images the whole Earth every one to two days for global cloud cover, radiation budget and ocean studies.
  • Cartosat, India's stereo mapping satellite series, is designed for cartographic applications and land information systems.
  • IKONOS, launched in 1999, was the first high-resolution commercial imaging satellite.

Option A gives ASTER the cartographic role of Cartosat, and options B and C misplace MODIS.

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70.

Match List-I and List-II and select the correct answer by using the codes given below :

List IList II
(a) Isohypse(1) Equal cloudiness
(b) Isohel(2) Equal height and pressure lines
(c) Isocrymes(3) Equal sunshine
(d) Isonephs(4) Equal frost

Codes : (a) (b) (c) (d)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The correct matching is (2) (3) (4) (1).

  • An isohypse joins points of equal height; on upper-air charts it joins points where a pressure level lies at the same height, hence 'equal height and pressure lines'.
  • An isohel joins places with equal sunshine (Greek helios, sun).
  • An isocryme joins places with equal frost, or the same lowest mean temperature (Greek kryos, cold).
  • An isoneph joins places with equal cloudiness (Greek nephos, cloud).

Option A pairs isohypse with cloudiness, and options B and C reverse other pairs.

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Tags: Geographical Techniques Questions and Answers