Practical Geography

175 soru

Soru 21Soru

On a topographical map drawn to a scale of 1:50,0001 : 50,000, the straight-line distance between Town X and Town Y is measured as 8 cm8\text{ cm}. What is the actual ground distance between the two towns in kilometers?

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Cevap: 4

Cevap

The actual ground distance between Town X and Town Y is 4 km4\text{ km}.
To determine the actual ground distance, multiply the measured map distance (8 cm8\text{ cm}) by the representative scale ratio denominator (50,00050,000), giving 400,000 cm400,000\text{ cm}. Dividing this by 100,000100,000 converts the distance into kilometers, yielding 4 km4\text{ km}.

Adım Adım Çözüm

1
Calculate the ground distance in centimeters using the map scale denominator.
Ground distance in cm = 8 cm×50,000=400,000 cm8\text{ cm} \times 50,000 = 400,000\text{ cm}.
A Representative Fraction of 1:50,0001 : 50,000 indicates that 1 unit1\text{ unit} on the map represents 50,000 units50,000\text{ units} of the same measurement on the actual ground.
2
Convert centimeters into kilometers.
Ground distance in km = 400,000100,000=4 km\frac{400,000}{100,000} = 4\text{ km}.
Since 1 m=100 cm1\text{ m} = 100\text{ cm} and 1 km=1,000 m1\text{ km} = 1,000\text{ m}, 1 km1\text{ km} contains 100,000 cm100,000\text{ cm}.

Anahtar Kavram

Ground distance calculation using Representative Fraction (RF) scale
Tahmini Süre:45s
Soru 22Soru

On a topographical map drawn to a scale of 1:50,0001 : 50,000, the straight-line distance between two settlements is measured as 14 cm14\text{ cm}. What is the actual ground distance between the two settlements in kilometers?

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Cevap: 7.0 km7.0\text{ km}

Cevap

The actual ground distance between the two settlements is 7.0 km7.0\text{ km}.
The scale 1:50,0001 : 50,000 means that 1 cm1\text{ cm} on the map represents 50,000 cm50,000\text{ cm} (or 0.5 km0.5\text{ km}) on the ground. Therefore, a distance of 14 cm14\text{ cm} on the map corresponds to 14×0.5 km=7.0 km14 \times 0.5\text{ km} = 7.0\text{ km} on the ground.

Adım Adım Çözüm

1
Identify the scale ratio and map distance
Map scale =1:50,000= 1 : 50,000; Map distance =14 cm= 14\text{ cm}.
Establishes the given measurements needed for calculation.
2
Convert the scale denominator from centimeters to kilometers
50,000 cm=50,000100,000 km=0.5 km50,000\text{ cm} = \frac{50,000}{100,000}\text{ km} = 0.5\text{ km}.
Since 1 km=100,000 cm1\text{ km} = 100,000\text{ cm}, 1 cm1\text{ cm} on the map represents 0.5 km0.5\text{ km} on the ground.
3
Calculate the ground distance
Ground distance =14 cm×0.5 km/cm=7.0 km= 14\text{ cm} \times 0.5\text{ km/cm} = 7.0\text{ km}.
Multiplying map distance by the ground distance per centimeter yields total real-world distance.

Anahtar Kavram

Map Scale and Distance Calculation
Soru 23Soru

A regional planning map with a statement scale of 2 cm to 5 km2\text{ cm to } 5\text{ km} is enlarged so that its linear dimensions are increased by a factor of 2.52.5. If the distance between two agricultural centers on the enlarged map measures 16 cm16\text{ cm}, what is the actual ground distance between them?

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Cevap: 16.0 km16.0\text{ km}

Cevap

The actual ground distance between the two agricultural centers is 16.0 km16.0\text{ km}.
The original statement scale of 2 cm to 5 km2\text{ cm to } 5\text{ km} simplifies to 1 cm to 2.5 km1\text{ cm to } 2.5\text{ km}. Enlarging the map's linear dimensions by a factor of 2.52.5 means that details become 2.52.5 times larger, so 1 cm1\text{ cm} on the new map represents 2.5 km2.5=1.0 km\frac{2.5\text{ km}}{2.5} = 1.0\text{ km}. Therefore, a measured distance of 16 cm16\text{ cm} on the enlarged map corresponds to 16 cm×1.0 km/cm=16.0 km16\text{ cm} \times 1.0\text{ km/cm} = 16.0\text{ km} on the ground.

Adım Adım Çözüm

1
Determine the unit scale of the original map.
Since 2 cm2\text{ cm} represents 5 km5\text{ km}, 1 cm1\text{ cm} represents 5 km2=2.5 km\frac{5\text{ km}}{2} = 2.5\text{ km}.
Establishing the distance represented by 1 cm1\text{ cm} on the original map provides the baseline linear scale.
2
Calculate the new linear scale after map enlargement.
The new scale denominator is 2.5 km2.5=1.0 km\frac{2.5\text{ km}}{2.5} = 1.0\text{ km} per centimeter (1 cm to 1 km1\text{ cm to } 1\text{ km}).
Enlarging a map by a linear factor of 2.52.5 means each centimeter on the new map represents a smaller ground distance by a factor of 2.52.5.
3
Compute the actual ground distance using the enlarged map measurement.
\text{Ground Distance} = 16\text{ cm} \times 1.0\text{ km/cm} = 16.0\text{ km}.
Multiplying the measured distance on the enlarged map by the new ground equivalence yields the total actual distance.

Anahtar Kavram

Map Scale Conversion during Linear Enlargement
Tahmini Süre:2m 0s
Soru 24Soru

On a topographical map with a Representative Fraction (RF) of 1:50,0001 : 50,000, a straight coastline segment between two coastal lighthouses measures 14.4 cm14.4\text{ cm}. What is the actual ground distance between the two lighthouses in kilometers?

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Cevap: 7.2

Cevap

The actual ground distance between the two lighthouses is 7.2 km.
To find actual ground distance from a Representative Fraction (1:50,0001 : 50,000), multiply the measured map distance (14.4 cm14.4\text{ cm}) by the scale denominator (50,00050,000) to get 720,000 cm720,000\text{ cm}. Converting to kilometers by dividing by 100,000100,000 yields 7.2 km7.2\text{ km}.

Adım Adım Çözüm

1
Calculate the ground distance in centimeters using the Representative Fraction denominator.
14.4 cm×50,000=720,000 cm14.4\text{ cm} \times 50,000 = 720,000\text{ cm}
The scale ratio 1:50,0001 : 50,000 means 1 cm1\text{ cm} on the map represents 50,000 cm50,000\text{ cm} on the ground.
2
Convert the ground distance from centimeters to meters.
720,000 cm÷100=7,200 m720,000\text{ cm} \div 100 = 7,200\text{ m}
There are 100 cm100\text{ cm} in 1 meter1\text{ meter}.
3
Convert the ground distance from meters to kilometers.
7,200 m÷1,000=7.2 km7,200\text{ m} \div 1,000 = 7.2\text{ km}
There are 1,000 meters1,000\text{ meters} in 1 kilometer1\text{ kilometer}.

Anahtar Kavram

Converting map distance to actual ground distance using Representative Fraction (RF) scale
Tahmini Süre:1m 0s
Soru 25Soru

A surveyor measures a straight stretch of road between two agricultural settlements on a topographical map drawn to a scale of 1:75,0001 : 75,000. If the distance between the two settlements on the map is 16 cm16\text{ cm}, what is the actual ground distance in kilometers?

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Cevap: 12

Cevap

The actual ground distance between the two settlements is 12 km12\text{ km}.
A Representative Fraction scale of 1:75,0001 : 75,000 means 1 cm1\text{ cm} on the map represents 75,000 cm75,000\text{ cm} on the ground. Converting 75,000 cm75,000\text{ cm} into kilometres gives 75,000100,000=0.75 km\frac{75,000}{100,000} = 0.75\text{ km}. Multiplying the measured map length of 16 cm16\text{ cm} by 0.75 km/cm0.75\text{ km/cm} yields an actual distance of 12 km12\text{ km}.

Adım Adım Çözüm

1
Convert scale denominator from centimetres to kilometres
1 cm on map=75,000 cm on ground=0.75 km1\text{ cm on map} = 75,000\text{ cm on ground} = 0.75\text{ km}
Dividing centimetres by 100,000100,000 (100 cm/m×1,000 m/km100\text{ cm/m} \times 1,000\text{ m/km}) converts centimetres to kilometres.
2
Calculate the actual ground distance
16 cm×0.75 km/cm=12 km16\text{ cm} \times 0.75\text{ km/cm} = 12\text{ km}
Multiplying the measured distance on the map by the ground distance represented per centimetre gives the total ground distance.

Anahtar Kavram

Map Scale Conversion and Ground Distance Calculation
Soru 26Soru

A map drawn at a scale of 1:50,0001 : 50,000 is enlarged to twice its original linear dimensions. Calculate the denominator of the new Representative Fraction (R.F.) scale.

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Cevap: 25000

Cevap

The denominator of the new Representative Fraction scale is 25,000.
When a map is enlarged linearly by a factor of 2, its scale increases by a factor of 2. Since Representative Fraction scale is expressed as 1 divided by the denominator, doubling the scale halves the denominator value from 50,000 to 25,000.

Adım Adım Çözüm

1
Identify the given scale denominator and enlargement factor
Original denominator = 50,000; Enlargement factor = 2
Enlarging a map makes features larger on paper, meaning 1 unit on the map represents a smaller ground distance, which corresponds to a larger scale (smaller denominator).
2
Calculate the new scale denominator
50,000 / 2 = 25,000
New Scale Denominator = Original Scale Denominator / Linear Enlargement Factor

Anahtar Kavram

Map enlargement decreases the denominator of the Representative Fraction proportionally by the linear factor of enlargement.
Soru 27Soru

A topographical map with a Representative Fraction (RF) of 1:100,0001 : 100,000 undergoes map reduction such that its new surface area is 14\frac{1}{4} of the original map area. If the straight-line distance between two towns measured on the reduced map is 15 cm15\text{ cm}, what is the actual ground distance between the two towns in kilometers?

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Cevap: 30.0 km30.0\text{ km}

Cevap

The actual ground distance between the two towns is 30.0 km30.0\text{ km}.
Reducing a map's area to 14\frac{1}{4} of its original size means its linear dimensions are reduced by a factor of 14=12\sqrt{\frac{1}{4}} = \frac{1}{2}. The new scale denominator becomes 100,000×2=200,000100,000 \times 2 = 200,000, resulting in a Representative Fraction of 1:200,0001 : 200,000. Multiplying the measured map distance of 15 cm15\text{ cm} by 200,000200,000 gives 3,000,000 cm3,000,000\text{ cm}, which equals 30.0 km30.0\text{ km} on the ground.

Adım Adım Çözüm

1
Determine the linear scale reduction factor from the areal reduction ratio
Linear reduction factor = 14=12\sqrt{\frac{1}{4}} = \frac{1}{2}
Linear scale changes as the square root of surface area changes.
2
Calculate the Representative Fraction (RF) of the reduced map
New scale denominator = 100,000×2=200,000100,000 \times 2 = 200,000, giving a new scale of 1:200,0001 : 200,000
Reducing linear dimensions by half doubles the scale denominator of the representative fraction.
3
Calculate the actual ground distance using the new map distance and new scale
Ground distance = 15 cm×200,000=3,000,000 cm=30.0 km15\text{ cm} \times 200,000 = 3,000,000\text{ cm} = 30.0\text{ km}
Converting map distance in centimeters to ground distance in kilometers by dividing 3,000,000 cm3,000,000\text{ cm} by 100,000 cm/km100,000\text{ cm/km}.

Anahtar Kavram

Linear versus Areal Map Scale Transformation and Ground Distance Calculation
Tahmini Süre:2m 0s
Soru 28Soru

On a topographical map drawn to a scale of 1:40,0001 : 40,000, the measured length of a railway line is 17.5 cm17.5\text{ cm}. If the map is enlarged so that a statement scale of 1 cm to 250 m1\text{ cm to } 250\text{ m} applies, what is the length of the railway line on the new enlarged map, in centimeters?

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Cevap: 28

Cevap

The length of the railway line on the enlarged map is 28 cm28\text{ cm}.
To find the distance on the enlarged map, first calculate the actual ground distance by multiplying the original map distance (17.5 cm17.5\text{ cm}) by the scale denominator (40,00040,000), yielding 700,000 cm700,000\text{ cm} or 7,000 m7,000\text{ m}. Next, divide this ground distance by the ground distance represented per centimeter on the new map (250 m250\text{ m}), which yields 28 cm28\text{ cm}.

Adım Adım Çözüm

1
Calculate the actual ground distance using the original map scale and measured distance.
Ground distance = 17.5 cm×40,000=700,000 cm=7,000 m17.5\text{ cm} \times 40,000 = 700,000\text{ cm} = 7,000\text{ m}.
Multiplying the map distance by the scale factor gives the true ground distance.
2
Calculate the corresponding distance on the enlarged map using the new statement scale.
New map distance = 7,000 m250 m/cm=28 cm\frac{7,000\text{ m}}{250\text{ m/cm}} = 28\text{ cm}.
Dividing the total ground distance by the ground distance represented per centimeter on the new map yields the map measurement.

Anahtar Kavram

Ground distance calculation and map scale conversion during map enlargement.
Soru 29Soru

A proposed power transmission line measures 14.4 cm14.4\text{ cm} on Map X, which is drawn to a Representative Fraction (RF) scale of 1:250,0001 : 250,000. The same power line is subsequently transferred to Map Y, where its total path is measured across two contiguous sections: a straight stretch of 8.0 cm8.0\text{ cm} and a curved section of 10.0 cm10.0\text{ cm}. What is the denominator of the Representative Fraction (RF) scale of Map Y?

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Cevap: 200000

Cevap

The denominator of the Representative Fraction (RF) scale of Map Y is 200,000 (representing a scale of 1:200,0001 : 200,000).
Because ground distance remains constant, the actual distance is 14.4 cm×250,000=3,600,000 cm14.4\text{ cm} \times 250,000 = 3,600,000\text{ cm}. On Map Y, the total route length is 8.0 cm+10.0 cm=18.0 cm8.0\text{ cm} + 10.0\text{ cm} = 18.0\text{ cm}. Dividing 3,600,000 cm3,600,000\text{ cm} by 18.0 cm18.0\text{ cm} gives a scale denominator of 200,000200,000, which corresponds to a scale of 1:200,0001 : 200,000.

Adım Adım Çözüm

1
Calculate the actual ground distance using the length on Map X and its scale
Ground distance = 14.4 cm×250,000=3,600,000 cm14.4\text{ cm} \times 250,000 = 3,600,000\text{ cm} (or 36 km36\text{ km})
Map scale defines the relationship between map distance and ground distance: Ground distance=Map distance×Scale denominator\text{Ground distance} = \text{Map distance} \times \text{Scale denominator}.
2
Determine the total measured length of the route on Map Y
Total distance on Map Y = 8.0 cm+10.0 cm=18.0 cm8.0\text{ cm} + 10.0\text{ cm} = 18.0\text{ cm}
The entire feature consists of two contiguous sections that must be added together.
3
Calculate the RF scale denominator of Map Y
RF denominator = 3,600,000 cm18.0 cm=200,000\frac{3,600,000\text{ cm}}{18.0\text{ cm}} = 200,000
The scale denominator is the ratio of ground distance to map distance when both are expressed in the same unit.

Anahtar Kavram

Ground distance conservation across maps of differing scales and multi-segment scale denominator calculation.
Tahmini Süre:2m 30s
Soru 30Soru

A proposed pipeline route between two water reservoirs measures 18 cm18\text{ cm} on a regional map drawn to a statement scale of 5 cm to 2 km5\text{ cm to } 2\text{ km}. If this map is reduced to a Representative Fraction (RF) scale of 1:100,0001 : 100,000, what will be the length of the pipeline route on the reduced map?

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Cevap: 7.2 cm7.2\text{ cm}

Cevap

7.2 cm7.2\text{ cm}
The original statement scale of 5 cm to 2 km5\text{ cm to } 2\text{ km} means each centimeter on the map corresponds to 0.4 km0.4\text{ km} on the ground. An 18 cm18\text{ cm} line therefore represents a total ground distance of 18×0.4 km=7.2 km18 \times 0.4\text{ km} = 7.2\text{ km}. On the new map with an RF scale of 1:100,0001 : 100,000, 1 cm1\text{ cm} represents 100,000 cm100,000\text{ cm} (or 1 km1\text{ km}). Converting the 7.2 km7.2\text{ km} ground distance onto the new map gives 7.2 km1 km/cm=7.2 cm\frac{7.2\text{ km}}{1\text{ km/cm}} = 7.2\text{ cm}.

Adım Adım Çözüm

1
Determine the ground distance represented by 1 cm1\text{ cm} on the original map scale.
Since 5 cm=2 km5\text{ cm} = 2\text{ km}, 1 cm=2 km5=0.4 km1\text{ cm} = \frac{2\text{ km}}{5} = 0.4\text{ km} (or 40,000 cm40,000\text{ cm}, giving an RF scale of 1:40,0001 : 40,000).
Converting statement scale to ground distance per centimeter establishes the actual linear distance on Earth.
2
Calculate the actual ground distance of the pipeline route.
\text{Ground Distance} = 18\text{ cm} \times 0.4\text{ km/cm} = 7.2\text{ km}(or (or 720,000\text{ cm}$).
Multiplying the map measurement by the ground equivalence per centimeter yields the true distance.
3
Calculate the new map length on the reduced map of scale 1:100,0001 : 100,000.
On an RF scale of 1:100,0001 : 100,000, 1 cm=100,000 cm=1 km1\text{ cm} = 100,000\text{ cm} = 1\text{ km}. Therefore, \text{New Map Length} = \frac{7.2\text{ km}}{1\text{ km/cm}} = 7.2\text{ cm}$.
Dividing the true ground distance by the new scale factor gives the updated map distance.

Anahtar Kavram

Map scale conversion between statement scale and Representative Fraction (RF), and linear distance scaling.
Soru 31Soru

On a topographical map with a scale of 1:50,0001:50,000, Point X is situated on a contour line marked 350 m350\text{ m} and Point Y is situated on a contour line marked 500 m500\text{ m}. If the measured distance between Point X and Point Y on the map is 6 cm6\text{ cm}, what is the gradient of the slope between Point X and Point Y?

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Cevap: 1 in 201\text{ in } 20

Cevap

The gradient of the slope between Point X and Point Y is 1 in 201\text{ in } 20 (or 1:201:20).
The slope gradient is calculated by dividing the Vertical Interval (difference in elevation: 500 m350 m=150 m500\text{ m} - 350\text{ m} = 150\text{ m}) by the Horizontal Equivalent (actual ground distance: 6 cm×50,000=300,000 cm=3,000 m6\text{ cm} \times 50,000 = 300,000\text{ cm} = 3,000\text{ m}). Converting both parameters into meters gives 1503000=120\frac{150}{3000} = \frac{1}{20}, which represents a slope ratio of 1 in 201\text{ in } 20.

Adım Adım Çözüm

1
Calculate the Vertical Interval (VI)
VI=500 m350 m=150 m\text{VI} = 500\text{ m} - 350\text{ m} = 150\text{ m}
The Vertical Interval is the difference in height between the two contour points.
2
Calculate the Horizontal Equivalent (HE) in ground measurement
HE=6 cm×50,000=300,000 cm=3,000 m\text{HE} = 6\text{ cm} \times 50,000 = 300,000\text{ cm} = 3,000\text{ m}
The Representative Fraction scale of 1:50,0001:50,000 means 1 cm1\text{ cm} on the map represents 50,000 cm50,000\text{ cm} (500 m500\text{ m}) on the ground.
3
Calculate the slope gradient formula Gradient=VIHE\text{Gradient} = \frac{\text{VI}}{\text{HE}}
Gradient=150 m3,000 m=120\text{Gradient} = \frac{150\text{ m}}{3,000\text{ m}} = \frac{1}{20}
Dividing the vertical height by the horizontal distance in identical units expresses the slope as a ratio.

Anahtar Kavram

Topographic Slope Gradient Calculation
Tahmini Süre:2m 0s
Soru 32Soru

On a topographical map, a series of V-shaped contour lines have their apexes (pointed ends) pointing towards higher elevation. Which of the following relief features is represented by this contour pattern?

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Cevap: A river valley

Cevap

A river valley
When contour lines cross a river or valley, they bend upstream into a V-shape. Because the river flows from higher to lower ground, the V-shape points toward higher elevation (upstream). Therefore, V-shaped contours pointing toward higher ground represent a river valley.

Adım Adım Çözüm

1
Examine the shape and orientation of the contour lines.
The contour lines form V-shapes pointing toward higher elevation.
Water flows downhill, so river valleys cut into higher land, causing contour lines to bend upstream (towards higher elevation) where they cross a stream.
2
Identify the corresponding topographic landform.
V-shaped contours pointing uphill represent a river valley or ravine.
The apex of the 'V' points up the valley toward the river source.

Anahtar Kavram

Interpretation of contour patterns for landform identification
Tahmini Süre:45s
Soru 33Soru

A topographical map with a scale of 1:100,0001 : 100,000 depicts a reservoir that covers an area of 16 cm216\text{ cm}^2 on the map. If the map is reduced to a new scale of 1:250,0001 : 250,000, what is the area of the reservoir on the new reduced map in square centimeters (cm2\text{cm}^2)?

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Cevap: 2.56

Cevap

The area of the reservoir on the new map is 2.56 cm22.56\text{ cm}^2.
To calculate the new map area after reduction, first find the linear scale factor k=100,000250,000=0.4k = \frac{100,000}{250,000} = 0.4. Because area changes in proportion to the square of linear dimensions, the area scale factor is (0.4)2=0.16(0.4)^2 = 0.16. Multiplying the original map area (16 cm216\text{ cm}^2) by 0.160.16 gives the correct new area of 2.56 cm22.56\text{ cm}^2.

Adım Adım Çözüm

1
Determine the linear scale factor (kk)
k=Original Scale DenominatorNew Scale Denominator=100,000250,000=0.4k = \frac{\text{Original Scale Denominator}}{\text{New Scale Denominator}} = \frac{100,000}{250,000} = 0.4
Map reduction changes linear dimensions proportionally to the ratio of original scale denominator to new scale denominator.
2
Calculate the area scale factor (k2k^2)
k2=(0.4)2=0.16k^2 = (0.4)^2 = 0.16
Map area varies with the square of the linear scale factor.
3
Compute the new map area
New Area=Original Area×k2=16 cm2×0.16=2.56 cm2\text{New Area} = \text{Original Area} \times k^2 = 16\text{ cm}^2 \times 0.16 = 2.56\text{ cm}^2
Applying the area scale factor to the original map area gives the reduced surface area on the new map.

Anahtar Kavram

Map Reduction and Area Scale Transformation
Soru 34Soru

On a topographical map drawn to a scale of 1:100,0001:100,000, a hill summit is marked with a spot height of 520 m520\text{ m}. The terrain features uniform contour intervals of 20 m20\text{ m}. Point XX is located on the fifth contour line directly below the summit. Point YY is situated at a lower elevation such that the vertical difference in height between Point XX and Point YY is 240 m240\text{ m}. If the straight-line distance measured between Point XX and Point YY on the map is 6 cm6\text{ cm}, calculate the average gradient between Point XX and Point YY expressed in the ratio 1:n1 : n. What is the value of nn?

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Cevap: 25

Cevap

The value of nn is 25, corresponding to an average gradient of 1:251 : 25 (or 1 in 25).
To find the denominator nn of the gradient ratio 1:n1:n, the vertical interval (VI = 240 m240\text{ m}) is compared to the horizontal equivalent (HE = 6,000 m6,000\text{ m}). Dividing 6,000 m6,000\text{ m} by 240 m240\text{ m} yields 2525, meaning the slope rises 1 meter vertically for every 25 meters horizontally.

Adım Adım Çözüm

1
Determine the elevation of Point X from the contour interval and spot height.
Elevation of Point X = 520 m(5×20 m)=420 m520\text{ m} - (5 \times 20\text{ m}) = 420\text{ m}.
Point X lies on the fifth contour line below the 520 m520\text{ m} summit with a 20 m20\text{ m} interval.
2
Identify the Vertical Interval (VI) between Point X and Point Y.
VI=240 m\text{VI} = 240\text{ m}.
The vertical elevation difference is directly specified in the problem statement.
3
Convert map distance to actual ground horizontal distance (Horizontal Equivalent, HE) using the map scale.
HE=6 cm×100,000=600,000 cm=6,000 m\text{HE} = 6\text{ cm} \times 100,000 = 600,000\text{ cm} = 6,000\text{ m}.
Unit conversions must align VI and HE in meters before computing the ratio.
4
Express the gradient as a ratio 1:n1 : n by dividing HE by VI.
n=HEVI=6,000 m240 m=25n = \frac{\text{HE}}{\text{VI}} = \frac{6,000\text{ m}}{240\text{ m}} = 25.
Gradient ratio is defined as Vertical IntervalHorizontal Equivalent=1n\frac{\text{Vertical Interval}}{\text{Horizontal Equivalent}} = \frac{1}{n}.

Anahtar Kavram

Gradient Calculation and Unit Alignment from Topographical Maps
Tahmini Süre:2m 30s
Soru 35Soru

A topographic cross-section is constructed from a map with a horizontal scale of 1:500001 : 50{}000. If the vertical scale of the cross-section is set at 1 cm1\text{ cm} to represent 100 m100\text{ m}, what is the vertical exaggeration of the cross-section?

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Cevap: 5

Cevap

The vertical exaggeration of the cross-section is 5.
To find vertical exaggeration, both scales must be in representative fraction (R.F.) form. A vertical scale of 1 cm1\text{ cm} to 100 m100\text{ m} equals 1:100001 : 10{}000. Dividing the horizontal scale denominator (5000050{}000) by the vertical scale denominator (1000010{}000) gives a vertical exaggeration of 55.

Adım Adım Çözüm

1
Convert the vertical scale to a representative fraction
Vertical Scale = 1 : 10,000
Units must be consistent (centimetres to centimetres) to obtain a non-dimensional ratio.
2
Divide the horizontal scale denominator by the vertical scale denominator
50,000 / 10,000 = 5
Vertical exaggeration measures how many times greater the vertical scale is than the horizontal scale.

Anahtar Kavram

Vertical Exaggeration of Cross-Sections
Soru 36Soru

On a topographical map drawn to a scale of 1:25,0001:25,000, a road ascends continuously from Point P at a contour elevation of 180 m180\text{ m} to Point Q at a contour elevation of 430 m430\text{ m}. If the average gradient along this section of the road is 1 in 201\text{ in } 20, what is the distance between Point P and Point Q on the map in centimeters?

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Cevap: 20.0 cm20.0\text{ cm}

Cevap

The distance between Point P and Point Q on the map is 20.0 cm20.0\text{ cm}.
The height difference (Vertical Interval) between the two points is 430 m180 m=250 m430\text{ m} - 180\text{ m} = 250\text{ m}. With a gradient ratio of 1 in 201\text{ in } 20, the horizontal ground distance (Horizontal Equivalent) is 250 m×20=5,000 m250\text{ m} \times 20 = 5,000\text{ m}, which equals 500,000 cm500,000\text{ cm}. Applying the map scale of 1:25,0001:25,000 gives a map measurement of 500,000 cm/25,000=20.0 cm500,000\text{ cm} / 25,000 = 20.0\text{ cm}.

Adım Adım Çözüm

1
Calculate the Vertical Interval (VI) between Point P and Point Q
VI=430 m180 m=250 m\text{VI} = 430\text{ m} - 180\text{ m} = 250\text{ m}
Vertical Interval represents the height difference between the two contour elevations.
2
Calculate the Horizontal Equivalent (HE) on the ground using the given gradient
HE=VI×20=250 m×20=5,000 m\text{HE} = \text{VI} \times 20 = 250\text{ m} \times 20 = 5,000\text{ m}
Gradient is defined as VIHE\frac{\text{VI}}{\text{HE}}. Therefore, HE=VI/Gradient=250 m×20\text{HE} = \text{VI} / \text{Gradient} = 250\text{ m} \times 20.
3
Convert the ground distance (HE) to centimeters
HE in cm=5,000 m×100 cm/m=500,000 cm\text{HE in cm} = 5,000\text{ m} \times 100\text{ cm/m} = 500,000\text{ cm}
Map distance calculations require identical linear units for numerator and denominator.
4
Determine the map distance using the Representative Fraction scale (1:25,0001:25,000)
Map Distance=500,000 cm25,000=20.0 cm\text{Map Distance} = \frac{500,000\text{ cm}}{25,000} = 20.0\text{ cm}
Dividing the actual ground distance by the scale factor gives the equivalent measurement on the map.

Anahtar Kavram

Topographic Gradient and Map Scale Conversion
Tahmini Süre:2m 0s
Soru 37Soru

A morphometric quantitative analysis of a river basin using Strahler's stream ordering method yields the following stream segment counts:
- N1N_1 (1st-order streams) = 4040
- N2N_2 (2nd-order streams) = 1010
- N3N_3 (3rd-order streams) = 44
- N4N_4 (4th-order streams) = 11

Based on these morphometric data, what is the mean bifurcation ratio (RbR_b) of this drainage basin?

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Cevap: 3.5

Cevap

The mean bifurcation ratio (RbR_b) of the drainage basin is 3.5.
The mean bifurcation ratio (RbR_b) is obtained by computing the ratio of stream segments between successive orders: N1/N2=4.0N_1/N_2 = 4.0, N2/N3=2.5N_2/N_3 = 2.5, and N3/N4=4.0N_3/N_4 = 4.0. Averaging these three values gives (4.0+2.5+4.0)/3=3.5(4.0 + 2.5 + 4.0) / 3 = 3.5.

Adım Adım Çözüm

1
Calculate the individual bifurcation ratios (RbR_b) between consecutive stream orders using the formula Rb=NuNu+1R_b = \frac{N_u}{N_{u+1}}.
Rb(12)=4010=4.0R_{b(1-2)} = \frac{40}{10} = 4.0, Rb(23)=104=2.5R_{b(2-3)} = \frac{10}{4} = 2.5, and Rb(34)=41=4.0R_{b(3-4)} = \frac{4}{1} = 4.0.
Bifurcation ratio measures the ratio of the number of stream segments of a given order to the number of segments of the next higher order.
2
Sum the calculated individual bifurcation ratios.
Sum =4.0+2.5+4.0=10.5= 4.0 + 2.5 + 4.0 = 10.5.
To find the average across all order transitions, the sum of all calculated ratios must first be determined.
3
Divide the total sum by the number of order transitions (k=3k = 3).
Mean Rb=10.53=3.5R_b = \frac{10.5}{3} = 3.5.
There are 3 transitions between the 4 stream orders, so dividing the sum by 3 gives the arithmetic mean bifurcation ratio.

Anahtar Kavram

Bifurcation Ratio and Stream Order Analysis in River Basins
Soru 38Soru

Match each topographical contour line pattern with the correct relief feature or slope characteristic it depicts on a map.

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Öğeler

Contours widely spaced at higher elevations and closely spaced at lower elevations near the base
Contours closely spaced at higher elevations and widely spaced at lower elevations near the base
V-shaped contours with their apexes (pointed ends) pointing toward lower elevation values
Asymmetrical contour arrangement with very tight spacing on one face and wide spacing on the opposite face

Eşleşmeler

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Cevap

The correct pairings are: Widely spaced at top / closely spaced at base matches Convex slope; Closely spaced at top / widely spaced at base matches Concave slope; V-shaped contours pointing downhill match Spur; Asymmetrical contour spacing on opposite sides matches Escarpment.
Each contour pattern corresponds to a fundamental geometric principle of map work: contour density reflects gradient steepness (wide = gentle, close = steep), and V-shaped orientation indicates direction of slope projection.

Adım Adım Çözüm

1
Analyze contour spacing relative to slope profiles.
Wide spacing indicates a gentle gradient, whereas close spacing indicates a steep gradient.
Topographical slope profiles (convex vs. concave) are determined by how gradient changes from high elevation to low elevation.
2
Differentiate between convex and concave slopes.
Convex slope = gentle top (wide spacing) to steep base (close spacing). Concave slope = steep top (close spacing) to gentle base (wide spacing).
Curvature of terrain determines where gradient transitions occur along elevation contours.
3
Analyze V-shaped contour orientation for relief features.
V-shapes pointing toward lower elevation represent spurs projecting downhill; V-shapes pointing toward higher elevation represent river valleys.
Contour lines bend around highland projections (spurs) differently than valley depressions.
4
Evaluate asymmetrical ridge patterns.
A steep slope on one side paired with a gentle slope on the other represents an escarpment (scarp and dip slopes).
Escarpments arise from tilted rock strata, resulting in contrasting contour densities across the ridge crest.

Anahtar Kavram

Interpretation of Relief and Slope Profiles from Topographical Contour Configurations
Soru 39Soru

On a topographical map, Point X is situated at an elevation of 200 m200\text{ m} and Point Y is situated at an elevation of 500 m500\text{ m}. If the horizontal ground distance between the two points is 6 km6\text{ km}, what is the gradient of the slope from Point X to Point Y expressed as a ratio?

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Cevap: 1:201 : 20

Cevap

The gradient of the slope between Point X and Point Y is 1:201 : 20.
The correct ratio 1:201 : 20 is obtained by subtracting the lower elevation (200 m200\text{ m}) from the higher elevation (500 m500\text{ m}) to obtain a vertical interval of 300 m300\text{ m}, converting the horizontal distance of 6 km6\text{ km} into 6,000 m6,000\text{ m}, and simplifying the fraction 3006000\frac{300}{6000} to 120\frac{1}{20}.

Adım Adım Çözüm

1
Calculate the Vertical Interval (VI)
VI=500 m200 m=300 m\text{VI} = 500\text{ m} - 200\text{ m} = 300\text{ m}
Vertical interval is the difference in elevation between the higher and lower points.
2
Convert the Horizontal Equivalent (HE) to the same units as VI
HE=6 km×1,000=6,000 m\text{HE} = 6\text{ km} \times 1,000 = 6,000\text{ m}
Both vertical difference and horizontal distance must be expressed in identical units (meters).
3
Compute the slope gradient as a ratio
Gradient=VIHE=300 m6,000 m=120=1:20\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{300\text{ m}}{6,000\text{ m}} = \frac{1}{20} = 1 : 20
Gradient is calculated by dividing Vertical Interval by Horizontal Equivalent and simplifying to a ratio of 1 in N.

Anahtar Kavram

Calculating topographic gradient using Vertical Interval (VI) and Horizontal Equivalent (HE)
Tahmini Süre:45s
Soru 40Soru

A topographic map with an original scale of 1:50,0001 : 50,000 is reduced to half its linear size. On the reduced map, the measured distance between Point X (elevation 750 m750\text{ m}) and Point Y along a slope is 5.0 cm5.0\text{ cm}. If Point Y lies 88 contour intervals below Point X on a map with a contour interval of 25 m25\text{ m}, what is the gradient between Point X and Point Y expressed as a ratio?

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Cevap: 1 in 251 \text{ in } 25

Cevap

The gradient between Point X and Point Y is 1 in 251 \text{ in } 25.
The gradient expressed as 1 in 251 \text{ in } 25 is correct because reducing a 1:50,0001 : 50,000 map to half its linear size yields a scale of 1:100,0001 : 100,000. At this scale, 5.0 cm5.0\text{ cm} corresponds to a Horizontal Equivalent (HE) of 5,000 m5,000\text{ m}. With 88 contour intervals at 25 m25\text{ m} each, the Vertical Interval (VI) is 200 m200\text{ m}. Dividing VI by HE gives 2005,000=125\frac{200}{5,000} = \frac{1}{25}.

Adım Adım Çözüm

1
Determine the new map scale after linear reduction
New Scale = 1:100,0001 : 100,000
Reducing a map to half its linear size doubles the scale denominator (50,000×2=100,00050,000 \times 2 = 100,000).
2
Calculate the Horizontal Equivalent (HE) ground distance
HE=5.0 cm×100,000=500,000 cm=5,000 m\text{HE} = 5.0\text{ cm} \times 100,000 = 500,000\text{ cm} = 5,000\text{ m}
Multiply map distance by the new scale factor and convert centimeters to meters.
3
Calculate the Vertical Interval (VI) between Point X and Point Y
VI=8×25 m=200 m\text{VI} = 8 \times 25\text{ m} = 200\text{ m}
Multiply the number of contour intervals by the contour interval value.
4
Compute the gradient ratio
Gradient=VIHE=200 m5,000 m=125\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{200\text{ m}}{5,000\text{ m}} = \frac{1}{25} or 1 in 251 \text{ in } 25
Divide Vertical Interval by Horizontal Equivalent in identical units to find the simple ratio.

Anahtar Kavram

Topographic Gradient and Map Scale Reduction
Tahmini Süre:3m 0s
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