Practical Geography

175 questions

Question 141Question

A wildlife sanctuary covers an area of 24 cm224\text{ cm}^2 on Map P, which is drawn to a scale of 1:60,0001 : 60,000. If Map P is enlarged to produce Map Q with a scale of 1:20,0001 : 20,000, what is the area of the sanctuary on Map Q in cm2\text{cm}^2?

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Answer: 216

Answer

The area of the sanctuary on Map Q is 216 cm2216\text{ cm}^2.
When a map is enlarged from a scale of 1:60,0001 : 60,000 to 1:20,0001 : 20,000, the linear dimension increases by a factor of 60,00020,000=3\frac{60,000}{20,000} = 3. Because area is a two-dimensional measurement, the area scale factor is the square of the linear scale factor (32=93^2 = 9). Thus, the new area on Map Q is 24 cm2×9=216 cm224\text{ cm}^2 \times 9 = 216\text{ cm}^2.

Step-by-Step Solution

1
Find the linear scale enlargement factor (kk)
k=Old Scale DenominatorNew Scale Denominator=60,00020,000=3k = \frac{\text{Old Scale Denominator}}{\text{New Scale Denominator}} = \frac{60,000}{20,000} = 3
Enlarging a map scale from 1:60,0001 : 60,000 to 1:20,0001 : 20,000 increases linear dimensions by a factor of 3.
2
Determine the area scale multiplier
\text{Area Scale Factor} = k^2 = 3^2 = 9
Area changes according to the square of the linear scale change ratio.
3
Calculate the new map area
24\text{ cm}^2 \times 9 = 216\text{ cm}^2
Multiplying the original area on Map P by the area scale factor yields the enlarged area on Map Q.

Key Concept

Map Enlargement Area Calculation
Question 142Question

Arrange the following sequential steps involved in conducting a chain survey fieldwork operation in the correct chronological order from start to finish.

Drag items to arrange them in the correct order

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Answer

The correct operational order for carrying out a chain survey is: first, carrying out a preliminary reconnaissance survey; second, driving station pegs and positioning ranging poles; third, measuring distances along survey lines and taking perpendicular offsets; fourth, entering measurements and sketches into the field book; and fifth, plotting the recorded measurements to scale in the office.
Chain surveying follows a strict procedural sequence: reconnaissance survey occurs first to inspect terrain and select stations; ranging poles and pegs are then set up to establish straight baselines; chaining along lines and offset measurements follow; measurements are recorded directly in the field book during the process; and finally, office plotting synthesizes the field notes into a scaled map.

Step-by-Step Solution

1
Identify the initial site assessment step required before surveying.
A preliminary reconnaissance survey is performed to inspect the site and select main station points.
Site inspection is essential to choose obstruction-free lines and suitable station points.
2
Determine the ground setup procedure for main lines.
Station pegs are driven in and ranging poles are sighted to mark straight survey lines.
Ranging defines line alignment before distance measuring tools are deployed.
3
Identify the core field measurement operation.
Chaining along baselines and taking perpendicular offsets to nearby features is conducted.
Physical length and feature position gathering form the main fieldwork task.
4
Identify the data recording requirement.
All distances and offset measurements are entered into the field book.
Field booking must happen on site during measurement to maintain accurate records.
5
Identify the final office map production phase.
The recorded data is plotted to scale on paper in the office to make the final plan.
Final map plotting occurs after all field data has been collected and verified.

Key Concept

Standard procedural stages in chain survey fieldwork
Estimated Time:1m 30s
Question 143Question

A hydrologist conducts a morphometric analysis on a river basin using a topographic map drawn at a scale of 1:25,0001:25,000. Stream ordering reveals 2828 first-order streams, 77 second-order streams, and 11 third-order stream. If the total basin area measured on the map is 32 cm232\text{ cm}^2, what is the stream frequency (FsF_s) of the basin in streams per km2\text{streams per km}^2?

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Answer: 18

Answer

The stream frequency of the river basin is 18 streams per km218\text{ streams per km}^2.
To determine stream frequency (FsF_s), first calculate the total stream count (N=28+7+1=36N = 28 + 7 + 1 = 36). Next, convert the map area of 32 cm232\text{ cm}^2 to actual ground area using the scale 1:25,0001:25,000. Since 1 cm1\text{ cm} on the map represents 0.25 km0.25\text{ km}, 1 cm21\text{ cm}^2 represents 0.0625 km20.0625\text{ km}^2. Multiplying 32 cm232\text{ cm}^2 by 0.0625 km2/cm20.0625\text{ km}^2/\text{cm}^2 gives a basin area of 2.0 km22.0\text{ km}^2. Finally, divide the total number of streams by the basin area (36/2.0=18 streams per km236 / 2.0 = 18\text{ streams per km}^2).

Step-by-Step Solution

1
Sum the number of streams of each order to find the total stream count (NN).
N=28+7+1=36 streamsN = 28 + 7 + 1 = 36\text{ streams}.
Stream frequency considers the entire stream network comprising all stream orders in the basin.
2
Convert map area to ground area using the map scale (1:25,0001:25,000).
1 cm=0.25 km    1 cm2=0.0625 km21\text{ cm} = 0.25\text{ km} \implies 1\text{ cm}^2 = 0.0625\text{ km}^2. Actual Area A=32×0.0625=2.0 km2A = 32 \times 0.0625 = 2.0\text{ km}^2.
Stream frequency must be expressed per unit of actual ground area in km2\text{km}^2 rather than map area.
3
Divide total number of streams by actual basin area.
Fs=362.0=18 streams/km2F_s = \frac{36}{2.0} = 18\text{ streams/km}^2.
Stream frequency (FsF_s) is defined as the ratio of total number of streams to the drainage basin area.

Key Concept

Stream Frequency (FsF_s) Calculation in River Basin Morphometry
Estimated Time:2m 0s
Question 144Question

During elementary fieldwork, a geography student needs to mark intermediate stations along a baseline to ensure a perfectly straight line over long distances. Which surveying instrument is designed for maintaining this straight alignment?

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Answer: Ranging pole

Answer

The ranging pole is the elementary surveying instrument used to mark stations and maintain straight alignment during fieldwork.
The correct answer is the ranging pole because its primary purpose in chain and compass surveying is to mark survey stations and facilitate the process of 'ranging'—aligning intermediate points along a straight survey line.

Step-by-Step Solution

1
Identify the fieldwork task described in the stem.
The task requires marking intermediate points to align survey lines straight over a distance.
Maintaining a straight baseline is essential in chain surveying before linear distance measurements are recorded.
2
Match the required function with the appropriate instrument.
The ranging pole (or ranging rod) is designed specifically for sighting and alignment (ranging) along survey lines.
Ranging poles are painted with alternating colors (red/white or black/white) to make them clearly visible from a distance.

Key Concept

Function of elementary surveying instruments (ranging poles in chain surveying)
Estimated Time:45s
Question 145Question

During a chain survey fieldwork operation, why is the double-line method standard practice for recording entries in a field book?

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Answer: It provides a central column representing the chain line to separate main baseline chainage figures from side offset measurements and feature sketches.

Answer

The double-line method is standard practice because it provides a central column representing the chain line to clearly separate main baseline chainage figures from side offset measurements and feature sketches.
In elementary chain surveying, field notes must be unambiguous. The double-line system represents the main survey chain line as a central column between two ruled lines. Mainline cumulative distances (chainages) are entered strictly within this column, while offset distances to boundary walls, trees, or buildings are recorded on the respective left or right outer margins. This prevents confusion between chainage points and offset measurements during map plotting.

Step-by-Step Solution

1
Identify the primary purpose of a field book in chain surveying.
Recognize that field booking is a systematic method for recording linear distances along the main chain line and lateral offsets to nearby features.
Clear separation of data prevents misinterpretation when plotting the final map in the drawing office.
2
Analyze the structural design of the double-line field book system.
Two parallel lines (usually 1.5 cm to 2 cm apart) are ruled longitudinally down the middle of each page. The space inside the double line represents the baseline itself.
All main chainage readings are written inside this central column, while offset distances, feature symbols, and side notes are booked on the outer left or right margins corresponding to their physical position relative to the line.
3
Evaluate the distractors against survey principles.
Field book entry methods organize recorded data; they do not perform physical measurement functions, angular alignment, or physical error corrections.
Instruments like cross-staffs measure right angles, compasses measure bearings, and mathematical formulas correct physical tape errors.

Key Concept

Field Booking Conventions in Chain Surveying (Double-Line System)
Question 146Question

During a fieldwork measurement along a baseline, a geography student uses a steel tape with a nominal length of 30.00 m30.00\text{ m}. Due to high ambient temperature, the actual length of the tape has expanded to 30.15 m30.15\text{ m}. If the uncorrected recorded distance for the baseline is 450.00 m450.00\text{ m}, what is the true ground distance and the nature of the measurement error?

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Answer: The true ground distance is 452.25 m452.25\text{ m}, representing a cumulative negative error.

Answer

The true ground distance is 452.25 m452.25\text{ m}, representing a cumulative negative error.
The correct answer states that the true ground distance is 452.25 m452.25\text{ m} and represents a cumulative negative error. When a 30.00 m30.00\text{ m} tape expands to 30.15 m30.15\text{ m}, every time the tape is laid out, 30.15 m30.15\text{ m} of ground is covered, but only 30.00 m30.00\text{ m} is added to the tally. Multiplying 450.00 m450.00\text{ m} by the ratio 30.15/30.0030.15 / 30.00 gives a true ground distance of 452.25 m452.25\text{ m}. Because the recorded distance is shorter than the actual ground distance, the error is negative, and because it acts consistently in one direction throughout the survey, it is cumulative.

Step-by-Step Solution

1
Determine the total number of tape applications made during the baseline measurement.
Number of applications N=450.00 m30.00 m=15N = \frac{450.00\text{ m}}{30.00\text{ m}} = 15.
The recorded distance is measured in standard 30.00 m30.00\text{ m} units.
2
Calculate the true ground distance using the actual expanded length of the tape.
True Distance =15×30.15 m=452.25 m= 15 \times 30.15\text{ m} = 452.25\text{ m} (or using Ltrue=Lrecorded×ll=450.00×30.1530.00=452.25 mL_{\text{true}} = L_{\text{recorded}} \times \frac{l'}{l} = 450.00 \times \frac{30.15}{30.00} = 452.25\text{ m}).
Each application of the expanded tape spans 30.15 m30.15\text{ m} of ground, while only 30.00 m30.00\text{ m} is recorded.
3
Classify the nature and sign of the survey error.
The recorded measurement is 2.25 m2.25\text{ m} less than the true distance (450.00 m<452.25 m450.00\text{ m} < 452.25\text{ m}). The error in the recorded distance is negative and accumulates proportionally with line length, making it a cumulative negative error.
When a measuring tape is too long, the recorded distance is always less than the true distance, causing a cumulative negative error.

Key Concept

Tape Length Error Corrections & Cumulative vs. Compensatory Errors
Estimated Time:2m 0s
Question 147Question

A cartographer constructs a quantitative dot map to represent the distribution of livestock across a region. On the map, a total of 6464 dots represent a total population of 160,000160,000 sheep. If Province X has a sheep population of 35,00035,000, how many dots should be plotted to represent Province X on this map?

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Answer: 14

Answer

14 dots are required to represent Province X.
To find the dot count for Province X, first determine the scale value of a single dot: 160,000 sheep64 dots=2,500 sheep per dot\frac{160,000\text{ sheep}}{64\text{ dots}} = 2,500\text{ sheep per dot}. Next, divide the population of Province X by the dot value: 35,000 sheep2,500 sheep/dot=14 dots\frac{35,000\text{ sheep}}{2,500\text{ sheep/dot}} = 14\text{ dots}.

Step-by-Step Solution

1
Determine the quantity represented by one dot.
Each dot represents 2,5002,500 sheep.
Dividing the overall sheep population (160,000160,000) by the total number of dots (6464) establishes the dot value.
2
Calculate the number of dots for Province X.
14 dots.
Dividing the provincial population (35,00035,000) by the single-dot value (2,5002,500) yields the exact number of dots required for visual representation.

Key Concept

Dot Map Scale and Quantifier Calculation
Question 148Question

An aerial reconnaissance map of a catchment area displays primary tributaries flowing parallel to each other in elongated valleys, with short secondary streams joining them almost at right angles along structural strike lines of tilted, alternating hard and soft rock strata. Which drainage pattern is depicted, and what main geological control governs its formation?

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Answer: Trellis pattern, controlled by dipping or folded sedimentary rocks with alternating bands of hard and soft strata

Answer

Trellis pattern, controlled by dipping or folded sedimentary rocks with alternating bands of hard and soft strata
The correct answer identifies a trellis pattern governed by dipping or folded sedimentary rocks. In folded terrain (such as ridge-and-valley topography formed by anticlines and synclines), main streams erode along weak strata in parallel valleys, while smaller tributaries flow down ridge slopes to join the main streams at nearly 90-degree angles.

Step-by-Step Solution

1
Analyze the stream geometric arrangement described in the scenario
Primary streams flow parallel to each other in main valleys, and secondary streams join them at approximately 90-degree right angles.
Identifying the characteristic right-angle tributary junction pattern narrows down the pattern type.
2
Examine the underlying geological structure described
The bedrock consists of tilted/dipping strata with alternating bands of hard (resistant) and soft (non-resistant) rocks.
The differential erosion of soft bands forms main strike valleys, while resistant ridge slopes direct short streams straight down at right angles.
3
Match stream geometry and geological structure to the correct drainage pattern classification
The combination of parallel main streams, right-angle tributary junctions, and folded/tilted rock belts defines a trellis drainage pattern.
Trellis patterns form specifically under these structural controls, distinguishing them from dendritic, radial, or centripetal patterns.

Key Concept

Trellis Drainage Pattern and Structural Geological Control
Estimated Time:1m 30s
Question 149Question

On a topographical map drawn at a scale of 1:100,0001 : 100,000, a communications mast is positioned at point X with an elevation of 180 m180\text{ m}, while a valley lookout at point Y has an elevation of 420 m420\text{ m}. If the terrain between these two points maintains a constant gradient of 1 in 251 \text{ in } 25, what is the straight-line distance between point X and point Y on the map?

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Answer: 6.0 cm6.0\text{ cm}

Answer

The distance between point X and point Y on the map is 6.0 cm6.0\text{ cm}.
The vertical elevation change between point X (180 m180\text{ m}) and point Y (420 m420\text{ m}) is 240 m240\text{ m}. Applying the slope gradient ratio of 1 in 251 \text{ in } 25 yields a horizontal ground distance (HEHE) of 240 m×25=6,000 m240\text{ m} \times 25 = 6,000\text{ m} (6 km6\text{ km}). Given the map scale of 1:100,0001 : 100,000, 1 cm1\text{ cm} on the map represents 100,000 cm100,000\text{ cm} (1 km1\text{ km}) on the ground. Dividing 6 km6\text{ km} by 1 km/cm1\text{ km/cm} gives a map distance of 6.0 cm6.0\text{ cm}.

Step-by-Step Solution

1
Calculate the Vertical Interval (VI) between points X and Y.
VI=420 m180 m=240 mVI = 420\text{ m} - 180\text{ m} = 240\text{ m}
The vertical interval represents the elevation difference between the two given points.
2
Determine the Horizontal Equivalent (HE) on the ground using the gradient ratio.
HE=VI×25=240 m×25=6,000 m=6 kmHE = VI \times 25 = 240\text{ m} \times 25 = 6,000\text{ m} = 6\text{ km}
Gradient is defined as Gradient=Vertical IntervalHorizontal Equivalent\text{Gradient} = \frac{\text{Vertical Interval}}{\text{Horizontal Equivalent}}, so HE=Vertical Interval×Gradient denominatorHE = \text{Vertical Interval} \times \text{Gradient denominator}.
3
Convert the ground distance to map distance using the map scale 1:100,0001 : 100,000.
\text{Map distance} = \frac{6,000\text{ m}}{1,000\text{ m/cm}} = 6.0\text{ cm}
A scale of 1:100,0001 : 100,000 means 1 cm1\text{ cm} on the map represents 100,000 cm100,000\text{ cm} (1,000 m1,000\text{ m} or 1 km1\text{ km}) on the ground.

Key Concept

Topographic Gradient and Map Scale Distance Calculation
Question 150Question

Relate each statistical mapping technique on the left to its governing cartographic principle or inherent analytical limitation on the right.

Click a left item, then click its matching right item

Items

Choropleth mapping
Isopleth mapping
Dot density mapping
Flow line mapping

Matches

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Answer

Choropleth mapping corresponds to shading pre-defined administrative units with potential area-size visual bias; Isopleth mapping connects points of equal value across continuous surfaces via interpolation; Dot density mapping utilizes uniform point symbols to portray spatial density without exact coordinates; Flow line mapping varies line thickness along routes to illustrate movement volume.
Each statistical mapping method follows specific cartographic logic: Choropleth maps shade administrative zones; Isopleth maps connect continuous equal-value interpolated lines; Dot density maps use uniform dots to illustrate distribution without pinpointing exact locations; and Flow line maps scale line thickness to show movement volume along transit corridors.

Step-by-Step Solution

1
Analyze the structural characteristics and limitations of Choropleth maps.
Identified that choropleth maps rely on administrative boundaries and quantitative shading, which can visually overemphasize larger geographical areas.
Choropleth mapping groups data into discrete areal classes defined by political or administrative boundaries.
2
Examine the continuous surface requirements for Isopleth maps.
Matched isopleths to lines connecting equal values calculated through interpolation of point measurements across space.
Isopleths represent continuous spatial phenomena (such as rainfall or elevation) rather than bounded areal counts.
3
Evaluate the spatial representation rules of Dot density maps.
Determined that dots represent aggregated quantities within an area and do not denote exact physical addresses or locations.
Dot maps distribute dots randomly within enumeration zones to convey density while maintaining uniform dot size.
4
Assess the visual metric used in Flow line maps.
Linked flow maps to variable line widths representing traffic, cargo, or population movement along specific pathways.
Flow lines use vector width proportional to magnitude to signify spatial interaction intensity.

Key Concept

Classification, principles, and limitations of thematic statistical maps
Question 151Question

During a field survey across undulating terrain, a surveyor must determine the vertical angle of inclination of a hillside to calculate the horizontal distance between two stations. Which of the following instruments is designed specifically for measuring this slope angle?

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Answer: Clinometer

Answer

Clinometer
The clinometer (or Abney level) is an elementary surveying instrument fitted with a graduated arc or pendulum specifically calibrated for sighting and measuring vertical angles of slope or elevation.

Step-by-Step Solution

1
Identify the surveying requirement stated in the stem.
The surveyor needs an instrument to measure vertical slope angles on sloping terrain.
Converting slope distance to horizontal equivalent requires knowing the angle of elevation or depression.
2
Evaluate the primary function of each listed surveying instrument.
The clinometer directly measures inclination angles; the prismatic compass measures horizontal bearings; the optical square sets 9090^\circ offsets; the plumb bob ensures accurate vertical alignment.
Matching instrument functions prevents parameter mismatches during field operations.

Key Concept

Function and Application of Elementary Surveying Instruments
Question 152Question

On a topographical map representing a mountainous terrain, a lower index contour line is marked at 500 m500\text{ m} above sea level and a higher index contour line is marked at 900 m900\text{ m} above sea level. If there are 44 equal contour intervals between these two index contours, what is the elevation in metres of the third contour line above the 500 m500\text{ m} index contour?

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Answer: 800

Answer

The elevation of the third contour line above the 500 m500\text{ m} index contour is 800 m800\text{ m}.
The vertical difference between the 500 m500\text{ m} and 900 m900\text{ m} index contours is 400 m400\text{ m}. Dividing this difference by the 44 contour intervals gives a uniform vertical interval of 100 m100\text{ m} per contour line. Therefore, three contour lines above the 500 m500\text{ m} line correspond to an elevation of 500 m+3(100 m)=800 m500\text{ m} + 3(100\text{ m}) = 800\text{ m}.

Step-by-Step Solution

1
Find the total elevation difference between the two known index contours
900 m500 m=400 m900\text{ m} - 500\text{ m} = 400\text{ m}
This establishes the cumulative elevation change across the four intervals.
2
Calculate the Vertical Interval (V.I.) between adjacent contour lines
400 m4=100 m\frac{400\text{ m}}{4} = 100\text{ m}
The contour interval is uniform across the map, so dividing the vertical change by the number of spaces yields the elevation change per line.
3
Calculate the elevation at the third contour line above the base index contour
500 m+(3×100 m)=800 m500\text{ m} + (3 \times 100\text{ m}) = 800\text{ m}
Moving up three contour lines from 500 m500\text{ m} increases the elevation by three times the vertical interval.

Key Concept

Vertical Interval and Contour Line Elevation Determination
Question 153Question

In satellite remote sensing, sensor performance and data utility are defined by distinct resolution dimensions. Match each resolution type on the left with its correct definition on the right.

Click a left item, then click its matching right item

Items

Spatial Resolution
Spectral Resolution
Radiometric Resolution
Temporal Resolution

Matches

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Answer

Spatial Resolution matches the minimum ground area dimensions per pixel; Spectral Resolution matches the number and width of electromagnetic wavelength bands; Radiometric Resolution matches sensor sensitivity to variations in energy intensity; Temporal Resolution matches satellite revisit frequency over the same location.
Each resolution type targets a specific dimension of remote sensing data: spatial for ground distance/pixel size, spectral for waveband detail, radiometric for signal energy sensitivity, and temporal for revisit frequency.

Step-by-Step Solution

1
Identify spatial measurement characteristics in remote sensing.
Spatial resolution determines ground cell size (pixel resolution).
Pixel dimensions govern spatial detail.
2
Identify spectral characteristics.
Spectral resolution relates to electromagnetic band sampling.
It defines how spectral bands are isolated.
3
Identify radiometric characteristics.
Radiometric resolution relates to energy levels and bit depth.
It measures sensitivity to subtle signal differences.
4
Identify temporal characteristics.
Temporal resolution defines time between successive satellite imaging passes.
Temporal relates directly to time frequency.

Key Concept

Four Resolution Dimensions in Remote Sensing
Estimated Time:45s
Question 154Question

On a regional planning map drawn to a Representative Fraction (R.F.) scale of 1:80,0001 : 80,000, a proposed bypass road between two industrial hubs measures 16.5 cm16.5\text{ cm}. What is the actual ground distance of the bypass road in kilometers?

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Answer: 13.2 km13.2\text{ km}

Answer

The actual ground distance of the bypass road is 13.2 km13.2\text{ km}.
Multiplying the map distance of 16.5 cm16.5\text{ cm} by the scale denominator 80,00080,000 yields 1,320,000 cm1,320,000\text{ cm}. Converting centimeters to kilometers by dividing by 100,000100,000 correctly yields 13.2 km13.2\text{ km}.

Step-by-Step Solution

1
Calculate the ground distance in centimeters using the Representative Fraction scale.
Ground Distance (cm)=16.5 cm×80,000=1,320,000 cm\text{Ground Distance (cm)} = 16.5\text{ cm} \times 80,000 = 1,320,000\text{ cm}
The R.F. scale of 1:80,0001 : 80,000 indicates that 1 cm1\text{ cm} on the map represents 80,000 cm80,000\text{ cm} on the ground.
2
Convert the ground distance from centimeters to kilometers.
Ground Distance (km)=1,320,000 cm100,000 cm/km=13.2 km\text{Ground Distance (km)} = \frac{1,320,000\text{ cm}}{100,000\text{ cm/km}} = 13.2\text{ km}
There are 100,000 cm100,000\text{ cm} in 1 km1\text{ km} (100 cm/m×1,000 m/km100\text{ cm/m} \times 1,000\text{ m/km}).

Key Concept

Ground distance calculation from Representative Fraction (R.F.) map scale
Question 155Question

Match each elementary surveying instrument listed on the left with its primary fieldwork function on the right.

Click a left item, then click its matching right item

Items

Gunter's Chain
Abney Level
Surveying Arrows
Plumb Bob

Matches

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Answer

Gunter's Chain matches measuring linear distances for land area units; Abney Level matches measuring vertical slope angles; Surveying Arrows match marking completed full chain lengths; Plumb Bob matches centering instruments precisely over ground station pegs.
Each instrument is correctly paired based on standard practical geography fieldwork procedures: Gunter's chain is calibrated for land area (acreage) calculations; the Abney level measures vertical slope angles; surveying arrows tally completed chain intervals along the baseline; and the plumb bob provides precise vertical centering of equipment over ground station pegs.

Step-by-Step Solution

1
Identify the primary structural design and historical purpose of Gunter's Chain
Gunter's chain measures 66 ft66\text{ ft} (4 rods4\text{ rods}), providing a direct link between linear chaining and land area units in acres.
It was specifically constructed so that 10 square chains=1 acre10\text{ square chains} = 1\text{ acre}.
2
Determine the operational function of an Abney Level
The instrument is used for vertical angle measurement and slope determination during field reconnaissance.
It incorporates a handheld sighting tube with a index arm and spirit level attached to a vertical arc.
3
Distinguish between auxiliary chain survey tools (Arrows vs. Plumb Bob)
Surveying arrows mark completed chain intervals, while a plumb bob ensures vertical alignment over a ground mark.
The follower collects arrows inserted by the leader to tally chain lengths, whereas a plumb bob uses gravity for accurate station centering.

Key Concept

Elementary Surveying Instruments and Fieldwork Functions
Estimated Time:2m 0s
Question 156Question

An environmental planning team in Nigeria is integrating multispectral satellite imagery with ground survey data in a Geographic Information System (GIS) to monitor vegetation health across the Guinea Savanna belt. When analyzing the reflectance values of healthy green crops, the sensor records high reflectance in the near-infrared (NIR) spectrum and high absorption in the red visible spectrum. Which fundamental remote sensing principle explains this specific spectral reflectance behavior?

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Answer: Chlorophyll pigments absorb visible red light for photosynthesis while the internal mesophyll cell structure strongly scatters near-infrared energy.

Answer

Chlorophyll pigments absorb visible red light for photosynthesis while the internal mesophyll cell structure strongly scatters near-infrared energy.
The correct response accurately identifies that active leaf chlorophyll absorbs red visible light (around 0.66 µm) for energy conversion, while the refractive boundaries of healthy mesophyll cells reflect and scatter near-infrared radiation (0.7–1.1 µm). This sharp contrast creates the characteristic 'red edge' in vegetation spectral signatures.

Step-by-Step Solution

1
Identify the optical response of vegetation across different electromagnetic spectrum bands.
Healthy vegetation exhibits a distinct spectral signature characterized by low reflectance in the visible red region and high reflectance in the near-infrared region.
Plant physiological properties interact distinctly with specific wavelengths of solar radiation.
2
Analyze the biochemical cause of visible red light absorption.
Chlorophyll a and b inside leaf chloroplasts absorb blue and red light primary wavelengths for photosynthesis.
Absorbed photon energy drives photosynthetic chemical reactions.
3
Analyze the structural cause of near-infrared scattering.
The spongy mesophyll cellular structure inside leaves refracts and scatters up to 50% of incoming NIR radiation to prevent thermal overheating.
Unabsorbed NIR energy passes through cell wall interfaces and reflects outward toward satellite sensors.

Key Concept

Spectral Signatures of Land Cover Types in Remote Sensing
Estimated Time:1m 30s
Question 157Question

On a topographical map extract drawn at a scale of 1:50,0001:50,000, the total length of all stream channels within a river basin is measured as 18 cm18\text{ cm}, while the total area of the basin on the map is 12 cm212\text{ cm}^2. What is the drainage density of this river basin in km/km2\text{km/km}^2?

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Answer: 3

Answer

The drainage density of the river basin is 3 km/km23\text{ km/km}^2.
Drainage density is determined by dividing total real stream length by total real area of the drainage basin. Given a map scale of 1:50,0001:50,000, 18 cm18\text{ cm} of stream length translates to 9 km9\text{ km}, and 12 cm212\text{ cm}^2 of area translates to 3 km23\text{ km}^2. The resulting quotient gives a drainage density of 3 km/km23\text{ km/km}^2.

Step-by-Step Solution

1
Convert linear map measurement to actual distance in kilometers
18 cm on map=18×0.5 km=9 km18\text{ cm} \text{ on map} = 18 \times 0.5\text{ km} = 9\text{ km}
At a scale of 1:50,0001:50,000, 1 cm1\text{ cm} represents 50,000 cm50,000\text{ cm} or 0.5 km0.5\text{ km} on the ground.
2
Convert map area measurement to actual area in square kilometers
12 cm2 on map=12×(0.5 km)2=12×0.25 km2=3 km212\text{ cm}^2 \text{ on map} = 12 \times (0.5\text{ km})^2 = 12 \times 0.25\text{ km}^2 = 3\text{ km}^2
Area scale ratio is the square of the linear scale ratio ((0.5 km/cm)2=0.25 km2/cm2(0.5\text{ km/cm})^2 = 0.25\text{ km}^2/\text{cm}^2).
3
Calculate drainage density using the formula Dd=LAD_d = \frac{\sum L}{A}
Dd=9 km3 km2=3 km/km2D_d = \frac{9\text{ km}}{3\text{ km}^2} = 3\text{ km/km}^2
Drainage density is defined as the total length of streams per unit drainage area.

Key Concept

Drainage Density and Scale Conversion
Estimated Time:1m 30s
Question 158Question

Match each statistical mapping technique listed on the left with the specific spatial phenomenon or cartographic data representation it is best suited to display on the right.

Click a left item, then click its matching right item

Items

Choropleth mapping
Dot distribution mapping
Isopleth mapping
Flow line mapping

Matches

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Answer

Choropleth mapping pairs with density ratios displayed within predefined administrative unit boundaries. Dot distribution mapping pairs with absolute population numbers or livestock distributions using fixed quantitative point symbols. Isopleth mapping pairs with continuous spatial phenomena such as mean annual precipitation or temperature gradients. Flow line mapping pairs with volume, direction, and movement routes of international trade cargo or passenger traffic.
Choropleth mapping is designed for administrative ratio data; dot distribution mapping visually renders absolute discrete counts; isopleth mapping connects lines of equal continuous values; and flow line mapping depicts direction and volume of movement along transport routes.

Step-by-Step Solution

1
Identify the data characteristic of Choropleth mapping.
It relies on spatial units (states/districts) to display derived ratios/densities via shading.
Administrative boundary categorization is the defining feature of choropleth technique.
2
Identify the data characteristic of Dot distribution mapping.
It uses dots representing absolute quantities (e.g., 1 dot = 1,000 cattle) to depict spatial density.
Point-based quantitative representations accurately show spatial clustering of discrete counts.
3
Identify the data characteristic of Isopleth mapping.
It connects points of equal value over continuous surfaces like atmospheric pressure, elevation, or temperature.
Isopleths require smooth, continuous spatial data distributions rather than abrupt administrative boundaries.
4
Identify the data characteristic of Flow line mapping.
It represents linear movements with line width scaled to movement volume.
Flow lines are specialized for network movement and transportation dynamics.

Key Concept

Selection of Statistical Maps and Graphical Representation Methods
Question 159Question

A field surveyor needs to determine the horizontal distance between two stations located on steeply sloping ground using a steel measuring tape and a plumb bob via the direct stepping method. Arrange the following field procedure steps in the correct chronological order from start to finish.

Drag items to arrange them in the correct order

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Answer

The correct procedural order is: (1) Insert ranging poles at both stations to establish the line, (2) Hold the tape zero mark at the upper station and extend the tape horizontally, (3) Suspend a plumb bob from the tape end to transfer the point vertically down, (4) Pin an arrow at the plumb bob tip on the ground, and (5) Advance down the slope repeating the steps progressively.
The stepping method requires first establishing a straight line between the stations using ranging poles. Next, short horizontal segments are extended from the higher elevation toward the lower elevation. A plumb bob is suspended from the elevated end of the horizontal tape to transfer the point vertically to the ground surface, where an arrow is pinned. The team then advances to repeat this sequence progressively down the slope.

Step-by-Step Solution

1
Establish the baseline path between stations.
Ranging poles fixed at both higher and lower stations line up the direction of measurement.
Clear line alignment prevents off-line measurement errors.
2
Stretch the tape horizontally from the upper station.
The tape forms a horizontal segment over the sloping terrain.
The direct horizontal stepping method requires holding convenient short lengths of tape parallel to the horizontal plane.
3
Use a plumb bob to project the tape mark to the slope surface.
The vertical projection point on the sloping ground is identified.
A plumb bob ensures true vertical transfer under the influence of gravity.
4
Mark the transferred ground position.
An arrow is inserted at the plumb bob tip.
This establishes a reliable temporary station for the next measurement interval.
5
Advance continuously down the slope.
Subsequent horizontal sections are measured and summed.
Accumulating the individual horizontal step lengths gives the exact horizontal distance between the end stations.

Key Concept

Direct Stepping Method on Sloping Ground
Question 160Question

A proposed industrial estate is represented as a square measuring 6 cm6\text{ cm} by 6 cm6\text{ cm} on a topographical map drawn to a scale of 1:40,0001 : 40,000. If the map is reduced to a scale of 1:120,0001 : 120,000, what is the new area of the industrial estate on the reduced map?

Show answer & explanation

Answer: 4 cm24\text{ cm}^2

Answer

The new area of the industrial estate on the reduced map is 4 cm24\text{ cm}^2.
When a map is reduced from a scale of 1:40,0001:40,000 to 1:120,0001:120,000, the linear dimensions are reduced by a factor of 40,000120,000=13\frac{40,000}{120,000} = \frac{1}{3}. Since area is two-dimensional, the area changes by the square of the linear ratio: (13)2=19\left(\frac{1}{3}\right)^2 = \frac{1}{9}. Multiplying the original area of 36 cm236\text{ cm}^2 by 19\frac{1}{9} gives 4 cm24\text{ cm}^2.

Step-by-Step Solution

1
Calculate the original area of the feature on the map.
Original area = 6 cm×6 cm=36 cm26\text{ cm} \times 6\text{ cm} = 36\text{ cm}^2.
The feature is rectangular/square, so area on map is length multiplied by width.
2
Determine the linear scale change ratio.
Linear scale factor = Old Scale DenominatorNew Scale Denominator=40,000120,000=13\frac{\text{Old Scale Denominator}}{\text{New Scale Denominator}} = \frac{40,000}{120,000} = \frac{1}{3}.
Changing scale from 1:40,0001:40,000 to 1:120,0001:120,000 reduces all linear dimensions to 1/31/3 of their original size.
3
Calculate the area scale factor and final area.
Area scale factor = (13)2=19\left(\frac{1}{3}\right)^2 = \frac{1}{9}. New Area = 36 cm2×19=4 cm236\text{ cm}^2 \times \frac{1}{9} = 4\text{ cm}^2.
Areal change is proportional to the square of the linear scale ratio.

Key Concept

Relationship between linear scale ratio and area change in map reduction
Estimated Time:1m 30s
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