Practical Geography Alıştırma Soruları — JAMB UTME — Sayfa 6

Soru 101Soru →

A topographical map extract shows a series of streams radiating outward from a central dome, with subsequent tributaries capturing flow along concentric ring valleys formed on eroded sedimentary layers. Which drainage pattern is illustrated by this stream arrangement?

Annular pattern

Radial pattern

Centripetal pattern

Trellis pattern

Cevabı ve açıklamayı göster

Cevap: Annular pattern

Cevap

Annular drainage pattern

Annular drainage develops on maturely dissected domes where alternating concentric bands of hard and soft rock are exposed. Main streams flow outward from the dome, while tributary streams carve circular, ring-like channels along the softer rock belts.

Adım Adım Çözüm

1

Analyze the structural landform and elevation description in the stem.

Identified a dissected dome structure with elevated center and concentric belts of contrasting rock hardness.

Geological structure dictates the structural control governing drainage network development.

2

Evaluate stream geometry and flow direction.

Streams flow outwards from the high central region, while tributary streams follow curved concentric bands of weaker rock.

Differential erosion creates circular valleys where tributaries flow along rings around the central dome.

3

Match the observed stream geometry to standard genetic drainage patterns.

The ring-like or circular pattern around a central upland is classified as annular drainage.

Annular pattern is defined by concentric ring-like stream paths formed on breached domes or maturely dissected structural domes.

Anahtar Kavram

Annular Drainage Pattern and Structural Controls

Tahmini Süre:1m 15s

Soru 102Soru →

Drainage networks develop in response to specific rock structures and surface slopes. How do the following drainage pattern types match with their primary geological controls?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

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Cevap

Dendritic pattern pairs with uniform lithology and flat-lying strata; Radial pattern pairs with volcanic domes or central peaks; Trellis pattern pairs with alternating bands of hard and soft folded rocks; Centripetal pattern pairs with inward-sloping interior basins.

Each drainage pattern reflects specific surface geology: dendritic networks form on uniform rocks with equal erosion resistance, radial networks diverge outward from central domes, trellis networks align with alternating bands of folded rocks, and centripetal networks converge inward toward central basins.

Adım Adım Çözüm

1

Identify the characteristic geological setting for a dendritic pattern

Dendritic drainage exhibits random branching on homogeneous rock types.

Equal resistance of underlying rock allows streams to flow in any direction without structural restriction.

2

Identify the structural control of a radial pattern

Radial drainage radiates outward from elevated peaks or domes.

High central relief forces water to flow downhill in all outward directions.

3

Identify the structural control of a trellis pattern

Trellis drainage follows alternating weak and resistant folded strata.

Main streams carve long parallel valleys in soft strata while short tributaries join at right angles across hard ridges.

4

Identify the structural control of a centripetal pattern

Centripetal drainage converges inward into a central depression.

Topography slopes downward toward a common interior low point.

Anahtar Kavram

Geological Controls on Drainage Patterns

Soru 103Soru →

On a topographical map drawn to a scale of $1 : 25,000$ , Point P is situated at a trigonometric station with an elevation of $520\text{ m}$ , while Point Q lies at a stream confluence at an elevation of $370\text{ m}$ . The measured straight-line distance between Point P and Point Q on the map is $15\text{ cm}$ . What is the slope gradient between Point P and Point Q, expressed as the denominator $N$ in the ratio $1 \text{ in } N$ ?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The denominator

N

of the slope gradient ratio (

1 \text{ in } N

) is 25.

The Vertical Interval (VI) between Point P and Point Q is

520\text{ m} - 370\text{ m} = 150\text{ m}

. Using the map scale of

1 : 25,000

, the Horizontal Equivalent (HE) is

15\text{ cm} \times 25,000 = 375,000\text{ cm} = 3,750\text{ m}

. Dividing the VI by the HE gives

\frac{150}{3,750} = \frac{1}{25}

. Therefore, the gradient expressed as a ratio is

1 \text{ in } 25

, and the denominator

N

is 25.

Adım Adım Çözüm

1

Determine the Vertical Interval (VI)

\text{VI} = 520\text{ m} - 370\text{ m} = 150\text{ m}

Vertical Interval is the difference in elevation between the higher and lower points.

2

Calculate the Horizontal Equivalent (HE) in ground units (meters)

\text{HE} = 15\text{ cm} \times 25,000 = 375,000\text{ cm} = 3,750\text{ m}

Horizontal Equivalent is obtained by converting the measured map distance to actual ground distance using the representative fraction scale.

3

Compute the Gradient Ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{150\text{ m}}{3,750\text{ m}} = \frac{1}{25}

Gradient is the ratio of Vertical Interval to Horizontal Equivalent, both expressed in identical units.

Anahtar Kavram

Slope and Gradient Calculation using Representative Fraction Map Scale

Tahmini Süre:1m 30s

Soru 104Soru →

Match each specified drainage pattern with its underlying geological control or characteristic landform surface.

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Cevabı ve açıklamayı göster

Cevap

Trellis drainage pattern matches with folded strata containing alternating hard and soft rocks; Rectangular pattern matches with faulted and jointed bedrock; Radial pattern matches with outward flow from a central dome or volcano; Dendritic pattern matches with tree-like branching on uniform rock resistance.

Each pattern corresponds directly to its bedrock control: Trellis requires alternating soft/hard folded belts, Rectangular follows structural joints and faults, Radial descends from a central highland dome or cone, and Dendritic branches randomly over uniform lithology.

Adım Adım Çözüm

1

Analyze Trellis Drainage Pattern

Trellis drainage features long main streams parallel to strike valleys with short tributaries entering at right angles, characteristic of folded, tilted strata.

Differential erosion along parallel belts of soft and hard rock forces tributaries into strike valleys.

2

Analyze Rectangular Drainage Pattern

Rectangular drainage is characterized by right-angled bends in main streams and tributaries along lineaments.

Bedrock fractures, joint systems, and faults direct the line of weakest resistance for stream incision.

3

Analyze Radial Drainage Pattern

Radial drainage streams flow in all cardinal directions downward from a central peak.

Topographic highs like domes and volcanic summits direct water outward along radial slopes.

4

Analyze Dendritic Drainage Pattern

Dendritic drainage exhibits a random branching tree-like structure without structural alignment.

Uniform rock resistance (lithology) allows equal erosion in all directions.

Anahtar Kavram

Geological Controls on Drainage Network Patterns

Soru 105Soru →

On a topographical map, the vertical difference between two points, X and Y, is $100\text{ m}$ . If the horizontal distance separating them on the ground is $2\text{ km}$ , what is the gradient of the slope between Point X and Point Y?

1 in 20

1 in 2

1 in 200

20 to 1

Cevabı ve açıklamayı göster

Cevap: 1 in 20

Cevap

The gradient between Point X and Point Y is 1 in 20.

The vertical interval is

100\text{ m}

and the horizontal equivalent is

2\text{ km}

. Converting

2\text{ km}

into metres gives

2,000\text{ m}

. Applying the slope gradient formula

\frac{\text{VI}}{\text{HE}} = \frac{100\text{ m}}{2,000\text{ m}} = \frac{1}{20}

gives a gradient of

1\text{ in } 20

.

Adım Adım Çözüm

1

Identify the Vertical Interval (VI) and Horizontal Equivalent (HE)

\text{VI} = 100\text{ m}

and

\text{HE} = 2\text{ km}

Gradient calculation requires both vertical elevation change and horizontal ground distance.

2

Convert Horizontal Equivalent into metres so both measurements share identical units

\text{HE} = 2\text{ km} \times 1,000 = 2,000\text{ m}

Gradient is a unitless ratio, requiring numerator and denominator to be in the same unit.

3

Calculate gradient using the formula

\text{Gradient} = \frac{\text{VI}}{\text{HE}}

\text{Gradient} = \frac{100\text{ m}}{2,000\text{ m}} = \frac{1}{20}

Simplifying the fraction yields a ratio of

1\text{ in } 20

.

Anahtar Kavram

Slope Gradient Calculation on Topographical Maps

Soru 106Soru →

Match each topographic map slope calculation scenario on the left with its corresponding calculated gradient representation on the right.

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Cevap

The correct pairings are: (1) A vertical rise of 50 m across 4 cm on a 1:25,000 map matches 1 in 20 (5.0%); (2) An elevation change from 120 m to 320 m over 5 cm on a 1:50,000 map matches 1 in 12.5 (8.0%); (3) A rise of 150 m over 6 cm on a 1:25,000 map matches 1 in 10 (10.0%); (4) Crossing 4 contour intervals of 20 m across 4 cm at 1:100,000 scale matches 1 in 50 (2.0%).

Each scenario is accurately paired by converting the map distance to actual ground distance in meters using the given scale, determining the vertical interval, and computing the gradient as a ratio (1 in N) and percentage.

Adım Adım Çözüm

1

Convert the map distance for each scenario into the ground horizontal distance (Horizontal Equivalent, HE) in meters using the designated map scale.

Scenario 1:

4\text{ cm} \times 25,000 / 100 = 1,000\text{ m}

. Scenario 2:

5\text{ cm} \times 50,000 / 100 = 2,500\text{ m}

. Scenario 3:

6\text{ cm} \times 25,000 / 100 = 1,500\text{ m}

. Scenario 4:

4\text{ cm} \times 100,000 / 100 = 4,000\text{ m}

.

Map scale Representative Fractions convert map measurements to real ground horizontal distances, which must be expressed in meters to match the unit of Vertical Interval.

2

Calculate the total height difference (Vertical Interval, VI) in meters for each terrain scenario.

Scenario 1:

\text{VI} = 50\text{ m}

. Scenario 2:

\text{VI} = 320\text{ m} - 120\text{ m} = 200\text{ m}

. Scenario 3:

\text{VI} = 150\text{ m}

. Scenario 4:

\text{VI} = 4 \text{ intervals} \times 20\text{ m} = 80\text{ m}

.

Vertical Interval represents the net vertical relief difference between the start and end points of the transect.

3

Compute the gradient ratio (VI / HE) and express it both as a simple ratio (1 in N) and as a percentage slope.

Scenario 1:

\frac{50}{1,000} = \frac{1}{20} \rightarrow 1 \text{ in } 20

(

5.0\%

). Scenario 2:

\frac{200}{2,500} = \frac{1}{12.5} \rightarrow 1 \text{ in } 12.5

(

8.0\%

). Scenario 3:

\frac{150}{1,500} = \frac{1}{10} \rightarrow 1 \text{ in } 10

(

10.0\%

). Scenario 4:

\frac{80}{4,000} = \frac{1}{50} \rightarrow 1 \text{ in } 50

(

2.0\%

).

Gradient formula is

\text{Gradient} = \frac{\text{Vertical Interval (VI)}}{\text{Horizontal Equivalent (HE)}}

.

Anahtar Kavram

Slope and Gradient Calculation from Topographical Maps

Tahmini Süre:3m 0s

Soru 107Soru →

A cross-section is constructed from a topographic map drawn at a horizontal scale of $1 : 50\,000$ . If the vertical scale of the cross-section is set such that $1\text{ cm}$ represents $100\text{ m}$ of elevation, what is the vertical exaggeration of the cross-section?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The vertical exaggeration of the cross-section is 5.

To find the vertical exaggeration, both scales must be in representative fraction format. The horizontal scale is

1 : 50\,000

. The vertical scale (

1\text{ cm}

to

100\text{ m}

) equals

1 : 10\,000

. Dividing the horizontal denominator (

50\,000

) by the vertical denominator (

10\,000

) gives a vertical exaggeration of 5.

Adım Adım Çözüm

1

Convert the vertical scale to representative fraction form

Vertical Scale = 1 : 10,000

Both scales must be expressed as dimensionless ratios (RF) in matching units to determine vertical enlargement.

2

Divide the horizontal scale denominator by the vertical scale denominator

VE = 50,000 / 10,000 = 5

Vertical exaggeration quantifies how many times larger the vertical scale is compared to the horizontal scale.

Anahtar Kavram

Vertical Exaggeration in Topographic Profiles

Soru 108Soru →

On a topographical map with a scale of $1 : 50,000$ , two hilltops are located at elevations of $750\text{ m}$ and $450\text{ m}$ respectively. If the straight-line distance separating the two hilltops on the map is $6\text{ cm}$ , what is the average gradient between them?

1\text{ in }10

1\text{ in }50

1\text{ in }100

1\text{ in }4

Cevabı ve açıklamayı göster

Cevap: $1\text{ in }10$

Cevap

The average gradient between the two hilltops is

1\text{ in }10

.

The vertical interval (VI) is the difference in elevation:

750\text{ m} - 450\text{ m} = 300\text{ m}

. The horizontal equivalent (HE) is the ground distance:

6\text{ cm} \times 50,000 = 300,000\text{ cm} = 3,000\text{ m}

. The gradient is

\frac{\text{VI}}{\text{HE}} = \frac{300}{3,000} = \frac{1}{10}

, expressed as

1\text{ in }10

.

Adım Adım Çözüm

1

Calculate the Vertical Interval (VI)

\text{VI} = 750\text{ m} - 450\text{ m} = 300\text{ m}

Gradient requires the difference in elevation between the two specified points.

2

Calculate the Horizontal Equivalent (HE) on the ground in meters

\text{HE} = 6\text{ cm} \times 50,000 = 300,000\text{ cm} = 3,000\text{ m}

Map distance must be converted to actual ground distance using the representative fraction scale.

3

Calculate the gradient using the formula

\text{Gradient} = \frac{\text{VI}}{\text{HE}}

\text{Gradient} = \frac{300\text{ m}}{3,000\text{ m}} = \frac{1}{10}

or

1\text{ in }10

Dividing the vertical interval by the horizontal equivalent in identical units yields the ratio slope.

Anahtar Kavram

Topographic gradient calculation from map scale and elevation differences

Tahmini Süre:1m 30s

Soru 109Soru →

On Map X, drawn at a scale of $1 : 25,000$ , a planned agricultural settlement occupies a square grid block measuring $4\text{ cm}$ by $4\text{ cm}$ . Map X is reduced to produce Map Y such that the same agricultural settlement occupies a reduced area of $1\text{ cm}^2$ . What is the Representative Fraction (R.F.) scale of Map Y?

1 : 6,250

1 : 50,000

1 : 100,000

1 : 400,000

Cevabı ve açıklamayı göster

Cevap: $1 : 100,000$

Cevap

The Representative Fraction scale of Map Y is

1 : 100,000

.

The original grid block has an area of

16\text{ cm}^2

on Map X. Since the reduced area on Map Y is

1\text{ cm}^2

, the area reduction factor is

16

. The linear reduction factor is the square root of the area factor, which is

\sqrt{16} = 4

. Reducing linear dimensions by a factor of

4

makes the map scale smaller by a factor of

4

, increasing the scale denominator from

25,000

to

100,000

. Thus, Map Y has a scale of

1 : 100,000

.

Adım Adım Çözüm

1

Calculate the original area of the settlement on Map X.

\text{Area on Map X} = 4\text{ cm} \times 4\text{ cm} = 16\text{ cm}^2

.

Map area for a rectangular or square feature is length multiplied by width.

2

Determine the area scale change ratio.

\text{Area Ratio} = \frac{\text{Area on Map Y}}{\text{Area on Map X}} = \frac{1\text{ cm}^2}{16\text{ cm}^2} = \frac{1}{16}.

Area scale ratio is the ratio of final map area to initial map area.

3

Determine the linear scale reduction factor.

\text{Linear Factor} = \sqrt{\frac{1}{16}} = \frac{1}{4}.

Linear scale factor is the square root of the area scale factor (

n = \sqrt{n^2}

).

4

Calculate the new Representative Fraction scale denominator for Map Y.

\text{New Scale Denominator} = 25,000 \times 4 = 100,000 \Rightarrow \text{Scale of Map Y} = 1 : 100,000.

When a map is reduced, its scale denominator increases proportionally by the reciprocal of the linear scale factor.

Anahtar Kavram

Relationship between linear scale factor and area scale factor in map reduction

Tahmini Süre:2m 0s

Soru 110Soru →

Match each river basin morphometric parameter with its corresponding quantitative definition and hydrological significance.

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Cevabı ve açıklamayı göster

Cevap

Bifurcation Ratio (

R_b

) pairs with the ratio of stream numbers between successive orders (

N_u / N_{u+1}

). Drainage Density (

D_d

) pairs with the total stream length divided by basin area. Stream Frequency (

F_s

) pairs with the total number of stream segments per unit basin area. Form Factor (

R_f

) pairs with the ratio of basin area to the square of basin length.

Each morphometric parameter correctly maps to its distinct mathematical definition and hydrological role: Bifurcation Ratio relates channel counts across orders (

N_u / N_{u+1}

); Drainage Density measures stream length per area (

\sum L / A

); Stream Frequency measures individual stream counts per area (

N / A

); and Form Factor evaluates areal shape (

A / L^2

).

Adım Adım Çözüm

1

Identify the formula and definition for Bifurcation Ratio (

R_b

)

Bifurcation Ratio is defined as

R_b = \frac{N_u}{N_{u+1}}

, representing the ratio between stream segment counts of order

u

and order

u+1

.

This establishes the relationship between successive stream orders in Strahler's morphometric ordering.

2

Identify the formula and definition for Drainage Density (

D_d

)

Drainage Density is defined as

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

, where

\sum L

is the total channel length and

A

is the total basin area.

It quantifies channel spacing and regional runoff dynamics per unit area.

3

Identify the formula and definition for Stream Frequency (

F_s

)

Stream Frequency is defined as

F_s = \frac{N}{A}

, counting the number of stream channels (

N

) per unit area (

A

).

It expresses channel density in terms of individual segment counts rather than channel length.

4

Identify the formula and definition for Form Factor (

R_f

)

Form Factor is calculated as

R_f = \frac{A}{L^2}

, comparing basin area to the square of maximum basin length.

It measures circularity versus elongation of a river basin to predict flood hydrograph shapes.

Anahtar Kavram

Morphometric Analysis of River Basins

Soru 111Soru →

On a topographical map drawn to a scale of $1 : 40,000$ , a communications cable line is planned between Trigonometrical Station $P$ at an elevation of $680\text{ m}$ and Spot Height $Q$ at an elevation of $200\text{ m}$ . If the straight-line distance measured between $P$ and $Q$ on the map is $12\text{ cm}$ , what is the average gradient of the slope between these two points expressed as a percentage?

10\%

0.1\%

14.17\%

4\%

Cevabı ve açıklamayı göster

Cevap: $10\%$

Cevap

The average gradient of the slope expressed as a percentage is

10\%

.

The slope gradient is calculated by dividing the Vertical Interval (

480\text{ m}

) by the Horizontal Equivalent (

4,800\text{ m}

). Multiplying the ratio

\frac{480}{4,800} = 0.10

by

100

yields

10\%

.

Adım Adım Çözüm

1

Calculate the Vertical Interval (VI)

\text{VI} = 680\text{ m} - 200\text{ m} = 480\text{ m}

Vertical Interval is the difference in elevation between the two points.

2

Calculate the Horizontal Equivalent (HE) in meters

\text{Ground distance} = 12\text{ cm} \times 40,000 = 480,000\text{ cm} = 4,800\text{ m}

Multiply the map distance by the scale factor and convert from centimeters to meters (

100\text{ cm} = 1\text{ m}

).

3

Compute the slope gradient as a percentage

\text{Gradient (\%)} = \left( \frac{\text{VI}}{\text{HE}} \right) \times 100 = \left( \frac{480\text{ m}}{4,800\text{ m}} \right) \times 100 = 10\%

Gradient percentage is defined as the ratio of Vertical Interval to Horizontal Equivalent multiplied by 100.

Anahtar Kavram

Slope and Gradient Calculation on Topographic Maps

Soru 112Soru →

A topographical map, Map A, drawn to a scale of $1 : 100,000$ , displays a rectangular reservoir measuring $5\text{ cm}$ by $8\text{ cm}$ . When Map A is reduced to create Map B, the reservoir covers an area of $10\text{ cm}^2$ on Map B. What is the representative fraction (R.F.) scale of Map B?

1 : 50,000

1 : 200,000

1 : 400,000

1 : 25,000

Cevabı ve açıklamayı göster

Cevap: $1 : 200,000$

Cevap

The representative fraction scale of Map B is

1 : 200,000

.

The area of the reservoir on Map A is

5\text{ cm} \times 8\text{ cm} = 40\text{ cm}^2

. On Map B, the area is reduced to

10\text{ cm}^2

, giving an area reduction ratio of

\frac{10}{40} = \frac{1}{4}

. Since area scale change is the square of the linear scale change (

n^2

), the linear reduction factor is

\sqrt{\frac{1}{4}} = \frac{1}{2}

. Reducing a map by a linear factor of 2 doubles its scale denominator from

100,000

to

200,000

, resulting in a new scale of

1 : 200,000

.

Adım Adım Çözüm

1

Calculate the area of the reservoir on Map A.

\text{Area on Map A} = 5\text{ cm} \times 8\text{ cm} = 40\text{ cm}^2.

The surface area of a rectangular section on a map is obtained by multiplying length by width.

2

Determine the area reduction factor between Map A and Map B.

\text{Area Ratio} = \frac{\text{Area on Map B}}{\text{Area on Map A}} = \frac{10\text{ cm}^2}{40\text{ cm}^2} = \frac{1}{4}.

Dividing the area on the new map by the area on the original map gives the factor of areal change.

3

Calculate the linear reduction factor.

\text{Linear Ratio} = \sqrt{\text{Area Ratio}} = \sqrt{\frac{1}{4}} = \frac{1}{2}.

The change in surface area follows the square of the linear scale change (

n^2

), so taking the square root gives the linear factor (

n

).

4

Calculate the new Representative Fraction (R.F.) for Map B.

\text{New Scale Denominator} = 100,000 \times 2 = 200,000

, yielding a scale of

1 : 200,000$.

Map reduction results in a smaller scale, which increases the scale denominator proportionally to the linear reduction factor.

Anahtar Kavram

Map Reduction and Area Scale Relationship

Soru 113Soru →

During chain surveying, a field surveyor needs to mark the precise endpoint of each completed chain length along a baseline on the ground. Which of the following elementary surveying tools is used specifically for this purpose?

Arrows (chaining pins)

Prismatic compass

Abney level

Offset rod

Cevabı ve açıklamayı göster

Cevap: Arrows (chaining pins)

Cevap

Arrows (chaining pins) are inserted into the ground at the end of each chain length to keep accurate count of distances measured.

Arrows, also known as chaining pins, are standard chain surveying accessories made of heavy steel wire pointed at one end. They are inserted directly into the ground at the forward end of the chain to mark each completed chain length, allowing the surveyor to keep an accurate record of total distance measured.

Adım Adım Çözüm

1

Identify the fieldwork task described in the stem.

The task requires marking the exact endpoint of each measured chain length on the ground during linear survey work.

Accurate distance counting in chain surveying requires temporary physical markers at every chain interval.

2

Evaluate the function of elementary surveying instruments.

Arrows (steel chaining pins) are carried by the leader surveyor and stuck into the ground at the forward handle of the chain when stretched, marking the point for tallying.

This prevents errors in counting total chain lengths and ensures continuity along the baseline.

Anahtar Kavram

Identification and primary function of elementary chain surveying accessories

Soru 114Soru →

A drainage basin outlined on a topographic map with a scale of $1:50,000$ has a total stream network length of $45\text{ cm}$ and a basin area of $36\text{ cm}^2$ measured directly from the map sheet. What is the actual drainage density of the basin in $\text{km}/\text{km}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 2.5

Cevap

The actual drainage density of the river basin is

2.5\text{ km}/\text{km}^2

.

To calculate the true drainage density, map linear measurements and area measurements must first be converted to ground units using the scale

1:50,000

(

1\text{ cm} = 0.5\text{ km}

,

1\text{ cm}^2 = 0.25\text{ km}^2

). Total ground stream length is

45 \times 0.5 = 22.5\text{ km}

and total ground basin area is

36 \times 0.25 = 9\text{ km}^2

. Dividing stream length by basin area yields

2.5\text{ km}/\text{km}^2

.

Adım Adım Çözüm

1

Convert the measured total stream network length from map units (cm) to real-world kilometers.

Ground stream length

L = 45\text{ cm} \times 0.5\text{ km/cm} = 22.5\text{ km}

.

At a scale of

1:50,000

,

1\text{ cm}

on the map represents

50,000\text{ cm} = 0.5\text{ km}

on the ground.

2

Convert the measured basin area from square centimeters to square kilometers.

Ground basin area

A = 36\text{ cm}^2 \times (0.5\text{ km})^2 = 36 \times 0.25\text{ km}^2 = 9\text{ km}^2

.

The areal scale factor is the square of the linear scale factor (

1\text{ cm}^2 = 0.25\text{ km}^2

).

3

Divide the total ground stream length by the total ground basin area.

Drainage density

D_d = \frac{22.5\text{ km}}{9\text{ km}^2} = 2.5\text{ km}/\text{km}^2

.

Drainage density measures stream channel length per unit area within a river basin.

Anahtar Kavram

Drainage Density and Scale Conversion in Basin Analysis

Tahmini Süre:2m 0s

Soru 115Soru →

On a topographical map with a scale of $1 : 20,000$ , a proposed electric transmission line extends from Point $K$ situated on a ridge contour of $650\text{ m}$ to Point $L$ located near a stream contour of $450\text{ m}$ . If the measured distance between Point $K$ and Point $L$ on the map is $8.0\text{ cm}$ , what is the gradient of the slope between these two points?

1 \text{ in } 8

1 \text{ in } 4

1 \text{ in } 12.5

1 \text{ in } 125

Cevabı ve açıklamayı göster

Cevap: $1 \text{ in } 8$

Cevap

The gradient between Point

K

and Point

L

is

1 \text{ in } 8

.

The slope gradient is calculated using the formula

\text{Gradient} = \frac{\text{Vertical Interval}}{\text{Horizontal Equivalent}}

. The Vertical Interval is

650\text{ m} - 450\text{ m} = 200\text{ m}

. The Horizontal Equivalent is calculated by multiplying the map measurement (

8.0\text{ cm}

) by the representative fraction scale factor (

20,000

), yielding

160,000\text{ cm}

, which converts to

1,600\text{ m}

. Placing these in the gradient ratio yields

\frac{200}{1,600} = \frac{1}{8}

, expressed as

1 \text{ in } 8

.

Adım Adım Çözüm

1

Calculate the Vertical Interval (VI)

\text{VI} = 650\text{ m} - 450\text{ m} = 200\text{ m}

The Vertical Interval is the difference in elevation between the highest and lowest contour points.

2

Calculate the Horizontal Equivalent (HE) in ground units

\text{HE} = 8.0\text{ cm} \times 20,000 = 160,000\text{ cm} = 1,600\text{ m}

Map distance must be multiplied by the scale factor and converted into meters to match the Vertical Interval unit.

3

Compute the slope gradient ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{200\text{ m}}{1,600\text{ m}} = \frac{1}{8}

Gradient is expressed as the ratio of Vertical Interval to Horizontal Equivalent.

Anahtar Kavram

Slope and Gradient Calculation from Topographic Maps

Tahmini Süre:2m 0s

Soru 116Soru →

A morphometric analysis of a river basin provides the quantitative data shown below:

Morphometric Parameter	Value
Total Basin Surface Area ( $A$ )	$50\text{ km}^2$
Cumulative Stream Length ( $L$ )	$125\text{ km}$

Based on these hydrological parameters, what is the drainage density of the basin, and what does its magnitude imply about the basin's surface runoff response?

2.5\text{ km/km}^2

, indicating a high drainage density associated with rapid surface runoff and high flood potential.

0.4\text{ km/km}^2

, indicating a low drainage density associated with slow surface runoff and dominant groundwater recharge.

2.5\text{ km/km}^2

, indicating a low drainage density because stream channels are widely spaced over highly permeable terrain.

6,250\text{ km/km}^2

, indicating an ultra-dense drainage network resulting from extreme vertical relief mismatch.

Cevabı ve açıklamayı göster

Cevap: $2.5\text{ km/km}^2$ , indicating a high drainage density associated with rapid surface runoff and high flood potential.

Cevap

The drainage density of the basin is

2.5\text{ km/km}^2

, which indicates a high drainage density characterized by rapid surface runoff and high peak flood potential.

The correct answer properly calculates drainage density (

D_d

) as the total stream length divided by basin area (

125\text{ km} / 50\text{ km}^2 = 2.5\text{ km/km}^2

). Hydrologically, a high drainage density (

2.5\text{ km/km}^2

) signifies a closely knit stream network that collects and transmits storm runoff rapidly, resulting in swift hydrological response and high flood potential.

Adım Adım Çözüm

1

Identify the formula for drainage density (

D_d

).

D_d = \frac{\text{Total Cumulative Stream Length } (L)}{\text{Total Basin Surface Area } (A)}

Drainage density measures the total length of stream channels per unit area of a drainage basin.

2

Substitute the given values into the formula.

D_d = \frac{125\text{ km}}{50\text{ km}^2} = 2.5\text{ km/km}^2

Dividing

125\text{ km}

by

50\text{ km}^2

yields the linear stream length per square kilometer of basin area.

3

Interpret the hydrological significance of the calculated drainage density value.

A drainage density of

2.5\text{ km/km}^2

represents a relatively high density, implying impermeable surface rocks/soils, steep slopes, minimal infiltration, rapid runoff delivery into main channels, and high flood responsiveness.

Higher drainage density values mean precipitation reaches stream channels quickly, increasing surface runoff rates.

Anahtar Kavram

Drainage Density (

D_d

) and Basin Hydrological Response

Soru 117Soru →

Pair each geographical representation method listed below with the specific type of spatial data or phenomenon it is best suited to display.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Dot Map matches with discrete point distribution using uniform symbols; Choropleth Map matches with administrative data shown via progressive shading; Flow Line Map matches with direction and volume of movement; Isoline Map matches with continuous environmental variables.

Each mapping technique aligns directly with specific spatial data properties: Dot Maps depict discrete distributions using uniform symbols; Choropleth Maps summarize regional averages per administrative unit using shading; Flow Line Maps measure linear movement volume; and Isoline Maps map continuous environmental fields by connecting points of equal value.

Adım Adım Çözüm

1

Analyze the characteristic of a Dot Map.

Dot maps utilize uniform point symbols where each dot represents a fixed quantitative value to show spatial distribution.

This method is suitable for discrete distributions such as population or crops.

2

Analyze the characteristic of a Choropleth Map.

Choropleth maps apply varying shades to administrative divisions according to average data values.

Shading intensities correspond to ranges of density or ratios within defined political boundaries.

3

Analyze the characteristic of a Flow Line Map.

Flow line maps use arrows or bands of varying thickness along travel routes.

The line width directly indicates the volume of movement such as trade or passenger traffic.

4

Analyze the characteristic of an Isoline Map.

Isoline maps connect geographic locations having identical numeric values.

This technique represents continuous spatial fields such as elevation, temperature, or rainfall.

Anahtar Kavram

Cartographic selection and statistical mapping representations in practical geography.

Soru 118Soru →

On a topographical map of a river basin drawn to a Representative Fraction (R.F.) scale of $1 : 40,000$ , the distance along a proposed drainage channel between two agricultural settlements measures $14.5\text{ cm}$ . What is the actual ground distance of the drainage channel in kilometers?

Cevabı ve açıklamayı göster

Cevap: 5.8

Cevap

The actual ground distance of the drainage channel is

5.8\text{ km}

.

The correct answer of

5.8\text{ km}

is obtained by multiplying the map distance of

14.5\text{ cm}

by the R.F. denominator

40,000

(

14.5 \times 40,000 = 580,000\text{ cm}

) and then converting centimeters to kilometers by dividing by

100,000

(

580,000 \div 100,000 = 5.8\text{ km}

).

Adım Adım Çözüm

1

Calculate the total ground distance in centimeters

580,000\text{ cm}

According to the R.F. scale of

1 : 40,000

,

1\text{ cm}

on the map represents

40,000\text{ cm}

on the ground.

2

Convert the ground distance from centimeters to kilometers

5.8\text{ km}

Since

1\text{ km} = 100,000\text{ cm}

, divide the total distance in centimeters by

100,000

.

Anahtar Kavram

Ground distance calculation using Representative Fraction (R.F.) scale

Tahmini Süre:1m 15s

Soru 119Soru →

On a topographical map extract, a major river stream crosses a valley where the V-shaped contour lines point towards the north-east. In which cardinal direction is the river flowing?

North-east, because the river flows in the exact direction that the V-shaped contour apexes point.

South-west, because contour V-shapes point upstream towards higher elevation, meaning the river flows downstream in the opposite direction.

North-west, because river flow always aligns parallel to the strike of contour lines regardless of gradient.

South-east, because water in a valley basin flows towards areas of increasing contour values.

Cevabı ve açıklamayı göster

Cevap: South-west, because contour V-shapes point upstream towards higher elevation, meaning the river flows downstream in the opposite direction.

Cevap

South-west, because contour V-shapes point upstream towards higher elevation, meaning the river flows downstream in the opposite direction.

In practical map reading, whenever contour lines cross a river valley, they form a 'V' pattern pointing toward higher ground (upstream). If the apex of the 'V' points north-east, the elevation increases toward the north-east. Because rivers flow downhill from higher to lower ground, the flow direction must be toward the south-west.

Adım Adım Çözüm

1

Identify the law of V-shaped contours in valley landforms.

V-shaped contour lines crossing a stream or valley always point upstream (towards the source or higher elevation).

Topography rises as you move up a river valley, causing contour lines of higher elevation to bend upstream.

2

Determine the direction of flow based on the orientation of the contour V-apex.

Since the V-shapes point north-east (upstream), the river must flow downstream towards the south-west.

Water naturally flows from higher elevation to lower elevation, which is directly opposite to the upstream-pointing apex of the contour V-shapes.

Anahtar Kavram

Contour V-Rule for River Valley Flow Direction

Soru 120Soru →

In a morphometric analysis of a river basin, a hydrologist counts $32$ first-order streams and $8$ second-order streams using Strahler's stream ordering system. What is the bifurcation ratio between the first-order and second-order streams?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The bifurcation ratio between the first-order and second-order streams is 4.

The bifurcation ratio (

R_b

) is obtained by dividing the number of streams of a given order (

N_1 = 32

) by the number of streams of the next higher order (

N_2 = 8

). Therefore,

R_b = \frac{32}{8} = 4

.

Adım Adım Çözüm

1

Identify the number of streams of order

u

and order

u+1

N_1 = 32

and

N_2 = 8

The bifurcation ratio measures the ratio of the number of stream segments of a given order to the number of segments of the higher order.

2

Calculate the bifurcation ratio using

R_b = \frac{N_1}{N_2}

R_b = \frac{32}{8} = 4

Dividing the count of first-order streams by the count of second-order streams gives the ratio of stream branching.

Anahtar Kavram

Bifurcation Ratio in River Basin Morphometry

Practical Geography

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