Physical Geography

261 questions

Question 61Question

Match each pedogenic process involved in soil profile formation on the left with its corresponding chemical characteristic and profile manifestation on the right.

Click a left item, then click its matching right item

Items

Podsolization
Lateritization (Ferrallitisation)
Calcification
Gleization

Matches

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Answer

Podsolization matches with intense leaching producing an ash-grey silica-rich E horizon. Lateritization matches with silica removal yielding reddish iron and aluminum sesquioxide accumulation. Calcification matches with precipitation of calcium carbonate in subsoil under moisture deficit. Gleization matches with anaerobic reduction of ferric iron into a bluish-grey reduced horizon.
Each pedogenic process corresponds directly to its specific climatic, vegetation, and biochemical mechanism: podsolization forms ash-grey leached upper horizons under cool coniferous litter; lateritization concentrates red sesquioxides via tropical desilication; calcification precipitates calcium carbonates in semi-arid subsoils; and gleization reduces iron under anaerobic waterlogged conditions to give a bluish-grey horizon.

Step-by-Step Solution

1
Analyze Podsolization
Identified as typical of boreal coniferous forests producing acidic chelation and ash-grey eluvial horizons.
Acid mor humus mobilizes sesquioxides downwards, leaving quartz-rich E horizons.
2
Analyze Lateritization
Identified as humid tropical weathering characterized by desilication.
High temperatures and abundant moisture dissolve silica while leaving insoluble iron and aluminum oxides.
3
Analyze Calcification
Identified as arid/semi-arid pedogenesis where evapotranspiration exceeds rainfall.
Restricted leaching causes calcium carbonate accumulation in horizon B.
4
Analyze Gleization
Identified as hydromorphic pedogenesis in saturated soils.
Lack of oxygen drives microbial reduction of iron from ferric to bluish-grey ferrous state.

Key Concept

Pedogenic processes controlling soil profile development across global climatic zones.
Question 62Question

Match each Köppen climate classification code on the left with its correct characteristic rainfall and temperature regime on the right.

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Items

Af
BWh
Cs
ET

Matches

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Answer

Af pairs with Tropical wet climate with high temperatures and rainfall year-round; BWh pairs with Hot desert climate with extreme aridity; Cs pairs with Mediterranean climate featuring dry summers and rainy winters; ET pairs with Tundra climate with cool summers and cold winters.
Each Köppen climate code corresponds to specific temperature and precipitation thresholds: Af represents constantly wet equatorial regions, BWh denotes hyper-arid hot deserts, Cs represents dry-summer Mediterranean zones, and ET represents cold polar tundras.

Step-by-Step Solution

1
Identify the primary temperature/moisture group represented by each capital letter.
A = Tropical humid, B = Dry/Arid, C = Mild temperate, E = Polar.
The first capital letter in the Köppen system defines the broad global thermal and moisture zone.
2
Interpret the lower-case and second capital modifier letters.
f = no dry season, W = desert (Wüste), h = hot, s = dry summer, T = tundra.
Secondary letters refine the climate type by seasonal rainfall pattern and thermal intensity.
3
Match each combined code to its corresponding description.
Af to Equatorial rainforest, BWh to Hot desert, Cs to Mediterranean, ET to Tundra.
Connecting the code meanings to geographic rainfall and temperature regimes establishes the correct pairs.

Key Concept

Köppen Climate Classification Scheme
Question 63Question

At a weather observation station in Jos, Nigeria, a meteorologist recorded the following daily temperature readings using a Six's maximum and minimum thermometer over a four-day period:

- Day 1: Maximum = 31.0C31.0^\circ\text{C}, Minimum = 19.0C19.0^\circ\text{C}
- Day 2: Maximum = 33.5C33.5^\circ\text{C}, Minimum = 17.5C17.5^\circ\text{C}
- Day 3: Maximum = 28.0C28.0^\circ\text{C}, Minimum = 16.0C16.0^\circ\text{C}
- Day 4: Maximum = 29.5C29.5^\circ\text{C}, Minimum = 18.5C18.5^\circ\text{C}

Calculate the mean diurnal (daily) temperature range for this four-day period in degrees Celsius (C^\circ\text{C}).

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Answer: 12.75

Answer

The mean diurnal temperature range over the four-day period is 12.75C12.75^\circ\text{C}.
The mean diurnal range measures the average difference between daily peak heat and night cooling over a given period. Subtracting minimum from maximum for each day yields 12.0C12.0^\circ\text{C}, 16.0C16.0^\circ\text{C}, 12.0C12.0^\circ\text{C}, and 11.0C11.0^\circ\text{C}. The average of these four values is 12.75C12.75^\circ\text{C}.

Step-by-Step Solution

1
Find the diurnal range for each individual day
Day 1: 12.0C12.0^\circ\text{C}, Day 2: 16.0C16.0^\circ\text{C}, Day 3: 12.0C12.0^\circ\text{C}, Day 4: 11.0C11.0^\circ\text{C}
The diurnal temperature range is defined as the difference between the maximum and minimum temperatures recorded in a single 24-hour period.
2
Calculate the sum of all daily ranges
51.0C51.0^\circ\text{C}
To compute an average over multiple days, the individual daily ranges must first be aggregated.
3
Compute the arithmetic mean across the 4 days
12.75C12.75^\circ\text{C}
Dividing the aggregate sum by the total number of observation days (4) yields the mean diurnal range.

Key Concept

Diurnal Temperature Range and Mean Calculation
Question 64Question

In humid tropical regions with heavy seasonal rainfall, intense chemical alteration of granite bedrocks produces deep, clay-rich regolith layers in situ. Following a prolonged rainstorm, a large volume of this water-saturated regolith suddenly loses cohesion and moves rapidly downhill under the direct influence of gravity. Which of the following processes accounts for this rapid downhill movement of the saturated regolith?

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Answer: Mudflow

Answer

Mudflow
The scenario describes mass wasting where gravity pulls heavy, water-saturated weathered material down a slope. A mudflow specifically refers to the rapid movement of fine-grained, highly saturated regolith down steep slopes after intense rainfall.

Step-by-Step Solution

1
Distinguish between weathering and mass wasting mechanisms described in the scenario.
Chemical weathering (hydrolysis) formed the clay-rich regolith in situ, whereas gravity caused the subsequent downhill movement.
Weathering involves stationary breakdown, while mass wasting involves downslope displacement driven by gravity.
2
Identify the specific mass wasting type matching the movement characteristics.
A rapid movement of heavily saturated, fluid-like clay regolith following heavy rainfall is categorized as a mudflow.
Mudflows occur when fine-grained weathered material becomes saturated with water and rapidly flows down slopes.

Key Concept

Mass wasting processes vs. in-situ weathering
Estimated Time:1m 15s
Question 65Question

Arrange the sequential stages involved in the process of frost shattering (freeze-thaw weathering) in chronological order, starting from the initial entry of moisture to the final disintegration of the rock face.

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Answer

The correct sequence begins with liquid water percolating into rock joints during warmer temperatures, followed by water freezing and expanding by about 9%9\% as temperatures fall below 0C0^\circ\text{C}. This frost wedging progressively widens rock fractures over repeated freeze-thaw cycles, eventually causing angular rock fragments to dislodge and form scree slopes at the base of the rock face.
The process of frost shattering follows a clear mechanical progression: liquid water must first occupy pre-existing fractures in the rock face; sub-zero temperatures then cause the trapped water to freeze and expand by roughly 9%9\%, creating intense lateral pressure; recurrent freeze-thaw cycles continuously strain and widen these micro-fractures; and ultimately, angular fragments detach from the parent cliff and accumulate downslope as scree deposits.

Step-by-Step Solution

1
Identify the initial moisture entry requirement.
Water must first collect inside pre-existing rock joints and fissures.
Physical freeze-thaw weathering cannot take place without trapped liquid water inside open rock spaces.
2
Determine the physical change triggered by freezing conditions.
Water turns to ice below 0C0^\circ\text{C} and expands by 9%9\%.
The anomalous expansion of freezing water exerts immense outward pressure on fracture walls.
3
Trace the structural deterioration over time.
Repeated thermal cycling widens and extends internal fractures.
Continuous pressure fluctuations weaken the cohesive strength of the rock along lines of weakness.
4
Establish the end product of the weathering process.
Angular rock fragments break free and accumulate as talus or scree at the mountain base.
Complete mechanical failure occurs when fractures sever the fragment from the main outcrop.

Key Concept

Mechanism and Stages of Frost Shattering (Freeze-Thaw Weathering)
Estimated Time:1m 15s
Question 66Question

Which climatic factor is primarily responsible for lower atmospheric temperatures recorded in highland regions compared to nearby lowlands situated at the same latitude?

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Answer: Altitude

Answer

Altitude
Altitude is the main climatic control involved. As altitude increases, atmospheric density and pressure decrease, reducing the atmosphere's capacity to absorb heat re-radiated from the Earth's surface. Consequently, temperatures decline at an average normal lapse rate of about 6.5C6.5^\circ\text{C} per 1,000 m1,000\text{ m}.

Step-by-Step Solution

1
Identify the variable factor between the two locations described.
The two places share the same latitude, meaning solar angle is identical, but one location is higher in elevation than the other.
This isolates elevation (altitude) as the primary physical control responsible for the temperature difference.
2
Apply the concept of the Normal Lapse Rate.
Within the troposphere, temperature decreases with height at an average rate of approximately 6.5C6.5^\circ\text{C} per 1,000 meters1,000\text{ meters} (3.5F3.5^\circ\text{F} per 1,000 feet1,000\text{ feet}).
Higher altitude air is less dense and absorbs less terrestrial radiation, resulting in lower temperatures.

Key Concept

Altitude as a Climatic Control and Environmental Lapse Rate
Estimated Time:45s
Question 67Question

Inside a Stevenson screen at a meteorological station, a dry-bulb thermometer reads 28C28^\circ\text{C} while a wet-bulb thermometer reads 22C22^\circ\text{C}, yielding a wet-bulb depression of 6C6^\circ\text{C}. Which atmospheric variable is determined using this temperature difference, and which weather instrument would be mistakenly selected if an observer attempted to measure this variable by recording atmospheric pressure instead?

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Answer: Relative humidity is determined from the temperature difference; a barometer would be mistakenly selected because it measures atmospheric pressure.

Answer

Relative humidity is derived from the depression of the wet-bulb thermometer (28C22C=6C28^\circ\text{C} - 22^\circ\text{C} = 6^\circ\text{C}) using psychrometric tables. Mistakenly using a barometer targets atmospheric pressure instead of atmospheric moisture.
Relative humidity is the ratio of actual water vapour present in the air to the maximum amount the air can hold at that temperature. It is calculated using the depression of the wet bulb (the difference between dry-bulb and wet-bulb thermometer readings). A barometer is designed specifically to measure atmospheric pressure, so choosing a barometer to assess atmospheric humidity represents an instrument-parameter mismatch.

Step-by-Step Solution

1
Identify the weather element evaluated by a wet-and-dry bulb psychrometer.
The dry-bulb thermometer records ambient air temperature (28C28^\circ\text{C}), while evaporation from the moist muslin sheath lowers the wet-bulb temperature to 22C22^\circ\text{C}.
The rate of evaporation and resulting cooling (wet-bulb depression of 6C6^\circ\text{C}) depends directly on the relative humidity of the air.
2
Identify the mistaken instrument association.
A mercury or aneroid barometer measures atmospheric pressure (force exerted per unit area by the atmosphere).
Selecting a barometer when intending to measure relative humidity confuses pressure measurement with atmospheric moisture measurement.

Key Concept

Weather Elements and Instruments: Hygrometer/Psychrometer for Relative Humidity vs. Barometer for Atmospheric Pressure
Estimated Time:1m 30s
Question 68Question

Town X is located on longitude 15E15^\circ\text{E}. If the local time at the Greenwich Meridian (00^\circ) is 12:00 noon, what is the local time at Town X?

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Answer: 1:00 pm

Answer

The local time at Town X is 1:00 pm.
Earth rotates 360360^\circ in 24 hours, which means 1515^\circ equals 1 hour. Because Town X is located at 15E15^\circ\text{E}, it is ahead of Greenwich (00^\circ) by 1 hour. Adding 1 hour to 12:00 noon gives 1:00 pm.

Step-by-Step Solution

1
Calculate the longitudinal difference between Greenwich Meridian (00^\circ) and Town X (15E15^\circ\text{E}).
150=1515^\circ - 0^\circ = 15^\circ.
The rate of Earth's rotation is 1515^\circ per hour (360/24 hours=15/hour360^\circ / 24\text{ hours} = 15^\circ/\text{hour}).
2
Convert the longitudinal difference into time difference.
15÷15/hour=1 hour15^\circ \div 15^\circ/\text{hour} = 1\text{ hour}.
Every 1515^\circ of longitude corresponds to a time difference of 11 hour.
3
Determine whether to add or subtract time based on direction relative to GMT.
12:00 noon + 1 hour = 1:00 pm.
Places located East of the Greenwich Meridian are ahead in time ('East gain, West lose').

Key Concept

Local time calculation using longitude differences and Earth rotation rate
Question 69Question

Ecosystems and human land-use patterns are strongly influenced by global atmospheric circulation and surface thermal regimes. Match each world climate type on the left with its defining meteorological mechanism and moisture characteristic on the right.

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Items

Mediterranean Climate (Cs)
Tropical Monsoon Climate (Am)
Mid-Latitude Steppe Climate (BSk)
Tundra Climate (ET)

Matches

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Answer

Mediterranean Climate (Cs) pairs with summer drought under subtropical high subsidence; Tropical Monsoon Climate (Am) pairs with seasonal wind reversal and heavy summer rain; Mid-Latitude Steppe Climate (BSk) pairs with semi-arid continental interior conditions; Tundra Climate (ET) pairs with warmest month mean temperatures between 0C0^\circ\text{C} and 10C10^\circ\text{C}.
Each climate category is uniquely defined by its primary driver: Mediterranean (Cs) by subtropical high subsidence in summer; Tropical Monsoon (Am) by wind reversal; Mid-latitude Steppe (BSk) by continental rain shadows and distance from seas; and Tundra (ET) by polar air masses limiting summer temperatures to between 0C0^\circ\text{C} and 10C10^\circ\text{C}.

Step-by-Step Solution

1
Analyze the pressure system and precipitation dynamics for Mediterranean Climate (Cs).
Identified that Cs climates experience summer desiccation from sinking air in subtropical high-pressure belts and winter rain from rain-bearing westerlies.
Atmospheric pressure belt migration is the primary control for dry-summer subtropical climates.
2
Evaluate the atmospheric mechanism behind Tropical Monsoon Climate (Am).
Matched Am with seasonal land-sea wind shifts that supply heavy summer precipitation.
Monsoons are fundamentally driven by pressure gradients caused by thermal contrast between landmasses and oceans.
3
Examine the geographic location and moisture envelope of Mid-Latitude Steppe Climate (BSk).
Linked BSk with semi-arid continental interiors suffering from rain shadows and continentality.
Isolation from moisture-laden maritime air masses creates semi-arid grassland conditions in mid-latitudes.
4
Determine the thermal threshold for Tundra Climate (ET).
Paired ET with the polar thermal boundary of 0C0^\circ\text{C} to 10C10^\circ\text{C} mean temperature for the warmest month.
Köppen classification specifies 10C10^\circ\text{C} as the upper mean temperature limit for tundra vegetation growth.

Key Concept

Köppen Climate Classification and Climatic Controls
Estimated Time:2m 0s
Question 70Question

In Southeastern Nigeria, intense rainfall on fragile soils often leads to severe slope degradation and gully erosion. Which of the following statements correctly distinguishes between the initial weathering of rocks on these slopes and the mass wasting process that follows?

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Answer: Weathering is the in-situ disintegration of rock materials without transport, whereas mass wasting is the downslope movement of rock and soil debris under the influence of gravity.

Answer

Weathering is the in-situ disintegration of rock materials without transport, whereas mass wasting is the downslope movement of rock and soil debris under the influence of gravity.
Weathering involves the weakening and disintegration of rocks in place (in-situ) without moving them. Once rocks and soil are loosened, mass wasting takes place as gravity pulls the unattached earth materials down slopes, contributing to severe landscape degradation such as gullies in region like Southeastern Nigeria.

Step-by-Step Solution

1
Define weathering in physical geography
Identify that weathering refers to the static, in-situ (in place) breakdown or alteration of rocks and minerals by physical, chemical, or biological agents without significant transport.
Establishing the static nature of weathering isolates it from transport processes.
2
Define mass wasting in the context of slope degradation
Identify that mass wasting involves the bulk downward movement of soil, regolith, and rock debris under the direct force of gravity.
Gravity is the primary driving mechanism of mass wasting.
3
Compare the definitions to evaluate slope degradation mechanisms in Southeastern Nigeria
Distinguish between static rock breakdown (weathering) and gravity-driven movement down slopes (mass wasting).
This clear distinction directly addresses the core geography concept tested.

Key Concept

Distinction between Weathering and Mass Wasting
Estimated Time:45s
Question 71Question

Match each weathering or mass wasting process in Column A with its correct driving mechanism or characteristic environment in Column B.

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Items

Carbonation
Frost Shattering
Solifluction
Rockfall

Matches

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Answer

Carbonation pairs with chemical dissolution of carbonate rocks; Frost Shattering pairs with mechanical disintegration from freeze-thaw cycles; Solifluction pairs with slow downslope flow of water-saturated soil over permafrost; Rockfall pairs with rapid free-fall of detached bedrock fragments down steep cliffs.
Each weathering and mass wasting process matches its unique mechanism: Carbonation is a chemical reaction involving carbonic acid and limestone; Frost Shattering is physical breakdown caused by expanding ice in joints; Solifluction is saturated regolith moving slowly over permafrost; and Rockfall is direct gravitational free-fall down steep cliffs.

Step-by-Step Solution

1
Analyze chemical weathering processes
Carbonation involves rain absorbing carbon dioxide to form weak carbonic acid, which chemically dissolves limestone into soluble calcium bicarbonate.
Chemical weathering involves alterations to the chemical structure of minerals in rock.
2
Analyze mechanical weathering processes
Frost shattering operates purely mechanically as diurnal freeze-thaw cycles exert immense pressure inside joints and fissures.
Mechanical weathering breaks rocks into smaller fragments without changing mineral composition.
3
Differentiate between mass wasting types
Solifluction describes saturated soil creep over frozen permafrost, whereas rockfall describes extremely rapid free-fall movement under gravity.
Mass wasting processes are classified by movement velocity, material moisture, and environmental setting.

Key Concept

Distinction between Mechanical Weathering, Chemical Weathering, and Mass Wasting Mechanisms
Question 72Question

If the local time at the Greenwich Meridian (00^\circ) is 12:00 noon, what is the longitude of a city where the local time is 4:00 PM on the same day? Express your answer as a number in degrees East.

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Answer: 60

Answer

The longitude of the city is 60E60^\circ\text{E}.
The time difference between 12:00 noon and 4:00 PM is 4 hours. Since the Earth rotates 1515^\circ per hour, 4 hours corresponds to 4×15=604 \times 15^\circ = 60^\circ. Because the local time at the city is ahead of Greenwich Mean Time (GMT), the city lies to the East of the Greenwich Meridian at longitude 60E60^\circ\text{E}.

Step-by-Step Solution

1
Determine the time difference between the two locations.
Time difference = 4 hours
Subtract 12:00 noon from 4:00 PM (16:00).
2
Convert the time difference into degrees of longitude.
Angular distance = 6060^\circ
Earth rotates 360360^\circ in 24 hours, which equals 1515^\circ per hour (4 hours×15=604\text{ hours} \times 15^\circ = 60^\circ).
3
Determine the direction (East or West) relative to the Prime Meridian.
60E60^\circ\text{E}
Locations with time ahead of Greenwich are to the East.

Key Concept

Calculation of longitude from local time difference relative to Greenwich Meridian
Question 73Question

A meteorological station situated at latitude 14N14^\circ\text{N} records the annual climatic summary shown below:

ParameterJanFebMarAprMayJunJulAugSepOctNovDec
Mean Temp (C^\circ\text{C})222428313331282728292623
Precipitation (mm)00210451102102301152000

Which Köppen climate group symbol and dominant atmospheric control correctly account for this station's temperature and rainfall regime?

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Answer: Aw climate, controlled by the seasonal migration of the Inter-Tropical Convergence Zone (ITCZ)

Answer

Aw climate, controlled by the seasonal migration of the Inter-Tropical Convergence Zone (ITCZ)
The station displays a tropical thermal profile where every monthly mean temperature exceeds 18°C. The precipitation pattern shows a severe winter drought (November to March) and a distinct summer maximum (June to September). This pattern matches the Tropical Savanna (Aw) classification, which is governed by the seasonal poleward migration of the Inter-Tropical Convergence Zone (ITCZ) during the high-sun season.

Step-by-Step Solution

1
Analyze temperature values to determine the main Köppen climate group.
The minimum monthly mean temperature is 22°C (December/January), which is well above the 18°C threshold required for Köppen's Tropical 'A' climate group.
Tropical climates (Group A) require every month to have a mean temperature of 18°C or higher.
2
Analyze precipitation distribution to determine seasonal sub-classification.
Total annual precipitation is 742 mm, with five distinct dry months (0–2 mm) in winter and heavy rainfall concentrated between June and September.
A distinct winter dry season with moderate total precipitation characterizes the Tropical Wet-and-Dry or Savanna climate (Aw).
3
Identify the primary atmospheric controlling mechanism for this rainfall distribution.
The summer rain belt corresponds to the northward movement of the ITCZ, bringing moist maritime tropical air, while winter dryness corresponds to the subtropical high pressure belt.
Latitudes around 10°–15°N experience alternating tropical rain belts (ITCZ) in summer and dry trade winds (subtropical high) in winter.

Key Concept

Köppen Tropical Savanna Climate (Aw) and ITCZ Migration Mechanism
Estimated Time:2m 0s
Question 74Question

In a karst landscape, percolating carbonated groundwater dissolves and expands the vertical joints of exposed limestone bedrock, creating a characteristic pavement structure of flat-topped blocks separated by deep vertical fissures. Which geomorphological term specifically identifies these deep solution fissures?

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Answer: Grikes

Answer

Grikes are the deep vertical solution fissures formed in a limestone pavement.
Grikes is the correct term for the deep vertical channels or fissures formed when carbonated rainwater percolates into limestone joints and dissolves the surrounding rock. This creates the classic karst surface feature known as a limestone pavement.

Step-by-Step Solution

1
Analyze the landform process described in the stem
Rainwater absorbing carbon dioxide forms weak carbonic acid (H2CO3H_2CO_3), which dissolves calcium carbonate (CaCO3CaCO_3) in well-jointed limestone through carbonation.
Chemical solution along joints creates a limestone pavement consisting of two distinct alternating features: blocks and fissures.
2
Distinguish between surface solution features of a limestone pavement
The widened vertical joints/fissures are called grikes, while the raised, flat limestone slabs left standing between them are called clints.
Precise geomorphological classification requires identifying whether the feature represents the eroded trench or the remnant block.

Key Concept

Karst Surface Solution Features (Limestone Pavement)
Estimated Time:1m 0s
Question 75Question

Pedogenesis involves progressive biochemical and physical alterations of parent material over extended periods. Which sequence correctly places the following stages of soil development in chronological order, from initial bedrock weathering to the establishment of a mature zonal soil profile?

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Answer

The correct chronological sequence of soil formation is: bedrock breakdown into regolith, pioneer organism colonization and humification, topsoil eluviation, subsoil illuviation, and final zonal profile stabilization.
The correct sequence follows the natural progression of soil development: primary weathering creates parent regolith, biological colonization adds organic material, downward percolating water removes fine minerals via eluviation, these leached materials accumulate in the subsoil via illuviation, and sustained pedogenic action produces a mature zonal profile with distinct horizons.

Step-by-Step Solution

1
Identify the initial physical process required before soil development can begin
Weathering of solid parent bedrock into loose mineral regolith occurs first.
Soil cannot form without a substrate of loose mineral matter derived from parent rock disaggregation.
2
Determine the role of biological inputs in soil maturation
Pioneer organisms colonize the regolith, producing organic humus.
Humification transforms bare mineral regolith into a true soil substrate capable of supporting distinct horizons.
3
Trace the movement of water and materials through the developing profile
Eluviation leaches fine clays and soluble materials downward from the topsoil, followed by illuviation where these materials accumulate in the subsoil.
Eluviation (removal) must precede illuviation (deposition) in the chronological sequence of horizon differentiation.
4
Identify the final equilibrium state of pedogenesis
A mature zonal soil profile with stable O, A, E, B, and C horizons is established.
Horizonation reaches equilibrium over long periods under stable climatic and vegetation conditions.

Key Concept

Chronological stages of pedogenesis and soil profile differentiation
Question 76Question

Arrange the following sequential stages in the retreat of a waterfall and the subsequent formation of a river gorge, from the initial bedrock arrangement to the final landscape feature.

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Answer

The correct order of stages in waterfall retreat and gorge formation begins with a river flowing over hard rock overlying soft rock, followed by undercutting of the softer rock to form an overhang, then collapse of the unsupported overhang into the plunge pool, and finally continuous headward retreat leaving behind a steep-sided gorge.
The evolution of a river gorge begins when running water encounters alternating rock strata, where a resistant rock layer overlies softer rock. Hydraulic action and abrasion erode the softer stratum beneath, creating an overhang above a plunge pool. When the overhang loses support, it collapses under gravity. As this sequence of undercutting and collapse repeats over time, the waterfall retreats upstream, leaving behind a narrow, steep-sided valley termed a gorge.

Step-by-Step Solution

1
Identify the initial geological precondition.
A resistant caprock overlying less resistant rock layer provides the differential erosion setup.
Differential erosion cannot occur without contrast in rock hardness.
2
Trace the process of mechanical river erosion at the base.
Undercutting creates a plunge pool and an unsupported caprock overhang.
Hydraulic action and abrasion preferentially hollow out the weaker underlying stratum.
3
Determine the structural breakdown stage.
The overhang collapses due to gravity.
Once the undercutting removes critical support beneath the hard caprock, collapse is inevitable.
4
Identify the long-term geomorphic outcome.
Headward retreat creates a gorge.
Repeated cycles of undercutting and collapse cause the waterfall to migrate upstream.

Key Concept

Waterfall Retreat and Gorge Formation
Question 77Question

In a limestone region, a surface stream suddenly disappears underground through an opening formed by carbonation and solution, leaving the river valley downstream dry. Which karst landform is formed at the exact point where the stream sinks underground?

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Answer: Swallow hole

Answer

A swallow hole (also known as a ponor or sinkhole shaft) is the feature formed where a surface stream disappears subterraneanly into carbonated limestone bedrock.
In limestone regions, surface water containing carbonic acid dissolves vertical joints in the bedrock. Over time, these joints widen into funnel-shaped openings or vertical shafts known as swallow holes (or ponors). When a surface stream reaches a swallow hole, it plunges underground into a subterranean drainage system, leaving the valley lower downstream completely dry.

Step-by-Step Solution

1
Analyze the geomorphic process described in the stem
Groundwater carbonation dissolves joints in limestone bedrock, allowing surface runoff to carve vertical passages underground.
Rainwater containing dissolved carbon dioxide (H2CO3\text{H}_2\text{CO}_3) acts on calcium carbonate (CaCO3\text{CaCO}_3) to dissolve limestone bedrock along vertical joints.
2
Identify the specific landform created at the point of river disappearance
The entrance or shaft through which a river flows into subterranean cavern systems is known as a swallow hole or ponor.
As the swallow hole enlarges, the entire river is diverted underground, leaving the dry valley downstream without water flow.

Key Concept

Karst Drainage Features and Stream Disappearance
Question 78Question

To control the rapid advancement of desertification and soil degradation across the Sudano-Sahelian belt of Northern Nigeria, the establishment of shelterbelts of drought-resistant trees is widely promoted. Which of the following best explains how shelterbelts primarily mitigate land degradation in this vulnerable ecological zone?

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Answer: By reducing surface wind velocity to minimize mechanical deflation and retain soil moisture

Answer

Shelterbelts primarily mitigate land degradation by reducing surface wind velocity to minimize mechanical deflation and retain soil moisture.
Shelterbelts consist of dense linear plantings of trees positioned perpendicular to prevailing wind directions. In semi-arid regions like Northern Nigeria, they significantly decrease surface wind speed, preventing wind deflation (the stripping away of fine topsoil particles) and reducing evapotranspiration, thereby preserving soil fertility and moisture.

Step-by-Step Solution

1
Identify the primary environmental hazard and degradation mechanism in Northern Nigeria's Sudano-Sahelian zone.
The region is predominantly affected by wind erosion (deflation) and desertification driven by strong dry winds acting on dry, sparse soil.
Sparse vegetative cover and high wind speeds make topsoil vulnerable to detachment and aerial transport.
2
Analyze the physical function of tree shelterbelts in environmental management.
Planted rows of trees perpendicular to prevailing winds create a friction barrier that reduces wind kinetic energy close to the ground.
Lowering wind velocity directly reduces soil particle lifting and evaporation rates.
3
Distinguish the primary mechanism from incorrect geological or administrative processes.
Mechanical wind reduction directly targets topsoil deflation, whereas weathering involves rock decay, fossil fuels form over geological time, and EIAs are regulatory planning steps.
Option analysis confirms that physical wind velocity reduction is the accurate functional mechanism of shelterbelts.

Key Concept

Desertification Control and Wind Erosion Mitigation via Shelterbelts
Estimated Time:2m 0s
Question 79Question

City P is located at longitude 38E38^\circ\text{E} where the local time is 4:20 p.m. At the exact same moment, the local time at City Q is 9:40 a.m. on the same day. What is the longitude of City Q in degrees West of the Greenwich Meridian?

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Answer: 62

Answer

City Q is located at longitude 62W62^\circ\text{W}.
The calculation demonstrates that a time difference of 6 hours and 40 minutes equals 100100^\circ of longitude (400 minutes÷4 minutes/degree400 \text{ minutes} \div 4 \text{ minutes/degree}). Because City Q is earlier in time (9:40 a.m.) than City P (4:20 p.m.), City Q is located 100100^\circ to the west of 38E38^\circ\text{E}. Subtracting 3838^\circ to reach 00^\circ leaves 6262^\circ in the Western Hemisphere, giving a final position of 62W62^\circ\text{W}.

Step-by-Step Solution

1
Find the local time difference between City P and City Q
Local time difference is 6 hours and 40 minutes (400 minutes)
Converting both times to 24-hour format (16:20 and 09:40) allows direct subtraction: 16:2009:40=6 h 40 min16:20 - 09:40 = 6\text{ h } 40\text{ min}.
2
Convert time difference into angular longitude difference
Longitude difference is 100100^\circ
Earth rotates 11^\circ every 4 minutes. Dividing 400 minutes by 4 yields 100100^\circ of total longitude separation.
3
Determine the relative direction of City Q from City P
City Q is located 100100^\circ west of City P
Places with earlier local times lie to the west because the Earth rotates from west to east.
4
Calculate the final longitude position west of Greenwich (00^\circ)
Longitude of City Q is 62W62^\circ\text{W}
Moving 100100^\circ west from 38E38^\circ\text{E} covers 3838^\circ to reach 00^\circ, and the remaining 6262^\circ extends into the Western Hemisphere (10038=62100^\circ - 38^\circ = 62^\circ).

Key Concept

Calculation of longitude position using local time differences across the Prime Meridian
Question 80Question

Match each landform created by running water or underground water with its corresponding course stage or morphological characteristic.

Click a left item, then click its matching right item

Items

Interlocking spurs
Oxbow lake
Stalactite
Polje

Matches

Show answer & explanation

Answer

Interlocking spurs correspond to the upper course feature formed around hard rock projections; Oxbow lake corresponds to the lower course crescent-shaped water body from a breached meander; Stalactite corresponds to the calcite deposit hanging from a cavern ceiling; Polje corresponds to the extensive flat-floored karst solution depression.
Each feature is correctly matched to its primary geomorphic environment: Interlocking spurs occur in upper river courses due to vertical erosion. Oxbow lakes develop in mature lower courses via meander cutoffs. Stalactites hang from underground limestone cave ceilings, and poljes represent large flat-floored karst surface basins formed by extensive chemical solution.

Step-by-Step Solution

1
Identify fluvial surface landforms and their corresponding river course stages.
Interlocking spurs belong to the upper course characterized by steep vertical headward erosion, whereas oxbow lakes belong to the depositional lower course floodplain environment.
Energy levels and dominant erosion types (vertical vs lateral) define river course features.
2
Identify underground water (karst) features and their formation sites.
Stalactites hang downward from cave roofs due to dripping mineralized groundwater, while poljes are large-scale surface solution depressions in limestone terrains.
Precipitation of calcium carbonate forms cavern speleothems, whereas surface solution and cavern collapse generate broad depressions.

Key Concept

Classification of Fluvial and Karst Landforms
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