Physical Geography

261 questions

Question 81Question

A navigator on Ship P observes local solar noon (12:00 noon) when a Greenwich Mean Time (GMT) chronometer reads 1:40 PM. If Ship Q is located at a position where the local time is exactly 4 hours ahead of Ship P, what is the longitude of Ship Q?

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Answer: 35E35^\circ\text{E}

Answer

35E35^\circ\text{E}
Ship P is at 25W25^\circ\text{W} because its local time is 1 hour 40 minutes (100 minutes) behind GMT (100 minutes / 4 min per degree = 25W25^\circ\text{W}). Since Ship Q is 4 hours ahead in time, it lies 6060^\circ to the east (4×154 \times 15^\circ). Measuring 6060^\circ east from 25W25^\circ\text{W} covers 2525^\circ to reach the Greenwich Meridian (00^\circ), leaving 3535^\circ into the Eastern Hemisphere, resulting in 35E35^\circ\text{E}.

Step-by-Step Solution

1
Calculate the longitude of Ship P using local time and GMT
Time difference = 1 hour 40 minutes = 100 minutes. Longitude of Ship P = 100 minutes/4 minutes per degree=25W100 \text{ minutes} / 4 \text{ minutes per degree} = 25^\circ\text{W} (since local time is behind GMT).
The Earth rotates 11^\circ every 4 minutes, and areas behind GMT are located in the Western Hemisphere.
2
Determine the time difference and direction from Ship P to Ship Q
Time difference = 4 hours ahead. Direction = East of Ship P.
Local time increases as one moves east.
3
Convert the 4-hour time difference into angular distance in degrees
Angular distance = 4 hours×15/hour=604 \text{ hours} \times 15^\circ/\text{hour} = 60^\circ.
Earth rotates at a rate of 1515^\circ per hour.
4
Calculate the longitude of Ship Q by moving 6060^\circ east from 25W25^\circ\text{W}
Distance to Greenwich Meridian (00^\circ) = 2525^\circ east. Remaining distance east into Eastern Hemisphere = 6025=35E60^\circ - 25^\circ = 35^\circ\text{E}.
Crossing the Prime Meridian changes the hemisphere designation from West to East.

Key Concept

Longitude calculation using Greenwich Mean Time (GMT) and local time adjustments across meridians
Estimated Time:2m 0s
Question 82Question

A solar observation station located at longitude 23W23^\circ\text{W} records local solar noon (12:00 PM12:00\text{ PM}) at a specific instant. At that exact moment, a research vessel at sea notes its local solar time as 7:16 PM7:16\text{ PM} (19:1619:16) on the same day. What is the longitude of the research vessel in degrees East?

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Answer: 86

Answer

The research vessel is located at 86E86^\circ\text{E}.
The time difference between 12:00 PM and 7:16 PM is 7 hours and 16 minutes (436 minutes). Since 4 minutes correspond to 1° of longitude, the angular difference is 436 ÷ 4 = 109°. Because the vessel's solar time is later than the station's time, the vessel lies to the East. Moving 109° East from 23°W requires 23° to reach the 0° Greenwich Meridian and an additional 86° into the Eastern Hemisphere, placing the vessel at 86°E.

Step-by-Step Solution

1
Determine the time difference between the solar observation station and the research vessel.
Time difference = 19:16 - 12:00 = 7 hours and 16 minutes = 436 minutes.
Difference in local solar time corresponds to longitudinal separation.
2
Convert the time difference into longitudinal degrees using the rate of Earth's rotation (1=4 minutes1^\circ = 4\text{ minutes}).
Longitudinal difference = 436 ÷ 4 = 109°.
The Earth rotates 360° in 24 hours, which equals 1° for every 4 minutes of time difference.
3
Determine the direction of the vessel relative to the station.
The vessel is East of the station.
Local solar time at the vessel (7:16 PM) is ahead of the station (12:00 PM), meaning the vessel lies further East.
4
Calculate the absolute longitude in the Eastern Hemisphere.
Vessel longitude = 109° - 23° = 86°E.
Traversing 109° East starting from 23°W uses 23° to reach the Greenwich Meridian (0°) and the remaining 86° extends into the Eastern Hemisphere.

Key Concept

Calculating longitude from local solar time difference across meridians
Question 83Question

Which of the following landforms is formed by glacial deposition and is characterized as an elongated, streamlined, teardrop-shaped hill of unsorted till with its blunt end pointing towards the direction of ice advance?

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Answer: Drumlin

Answer

Drumlin
The correct answer is the drumlin. Drumlins are smooth, elongated, teardrop-shaped hills of unstratified glacial till deposited and shaped under moving ice. The steep, blunt end (stoss side) faces the direction from which the glacier advanced, while the gently sloping tail (lee side) points in the direction of ice movement.

Step-by-Step Solution

1
Identify the primary agent and process specified in the stem
The landform is formed by glacial deposition (ice carrying and dropping sediment).
Differentiating between erosional and depositional features narrows down glacial landform classifications.
2
Analyze the physical shape and orientation of the landform described
An elongated, teardrop-shaped hill composed of unsorted till with a steep blunt stoss side facing ice advance and a tapered lee side.
This specific morphology uniquely identifies a drumlin.
3
Distinguish from non-glacial and erosional distractors
Roche moutonnée is erosional, oxbow lake is fluvial, and scree slope is caused by weathering and mass wasting.
Eliminating features of other geological agents confirms the correct answer.

Key Concept

Glacial Depositional Landforms (Drumlins)
Estimated Time:50s
Question 84Question

Match each coastal geomorphic process or mechanism on the left with its corresponding characteristic landform on the right.

Click a left item, then click its matching right item

Items

Continuous wave attack and undercutting at the base of a rocky sea cliff
Lateral sediment drift across an estuary mouth forming a linear depositional ridge
Wave refraction behind an offshore island causing sediment to bridge the gap to the mainland
Eustatic rise in sea level drowning a river valley system along a coastline

Matches

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Answer

Continuous undercutting at a cliff base matches with Wave-cut platform; lateral sediment drift across an estuary matches with Spit; sediment bridging to an offshore island matches with Tombolo; and sea-level submergence of a river valley matches with Ria.
Each process correctly generates its characteristic landform: cliff base erosion produces a wave-cut platform, longshore drift creates a spit across sheltered water, wave energy dissipation behind an island forms a tombolo, and coastal submergence of river valleys results in a ria.

Step-by-Step Solution

1
Analyze cliff erosion processes.
Marine erosion at the cliff base forms a notch which collapses repeatedly, leaving a flat wave-cut platform exposed at low tide.
Hydraulic action and abrasion concentrate mechanical wave force at high-water level.
2
Analyze coastal sediment transport along estuaries.
Oblique swash and direct backwash move material along the coast until open water causes deposition, building a spit.
Reductions in current velocity where coastlines change direction force waves to drop carried sediment.
3
Evaluate wave interaction behind offshore islands.
Wave refraction creates a low-energy zone behind the island where deposition forms a connecting ridge called a tombolo.
Converging wave crests lose velocity in sheltered water behind coastal obstacles.
4
Examine the effect of relative sea-level rise on coastal river valleys.
Submergence of river valleys produces deep, branching estuaries called rias.
Post-glacial eustatic sea-level rise floods low-lying dendritic river drainage patterns.

Key Concept

Coastal erosional, depositional, and submergent processes and landforms
Question 85Question

A weather observer at a meteorological station needs to organize daily readings and equipment records. Match each weather element listed in Column I with its corresponding measuring instrument and standard unit of measurement in Column II. Which pairings correctly align every weather element with its designated instrument?

Click a left item, then click its matching right item

Items

Atmospheric Pressure
Wind Speed
Sunshine Duration
Relative Humidity

Matches

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Answer

Atmospheric Pressure matches with Aneroid Barometer (Millibars); Wind Speed matches with Cup Anemometer (Knots); Sunshine Duration matches with Campbell-Stokes Recorder (Hours); Relative Humidity matches with Hygrometer (Percentage).
Each weather element is correctly paired with its dedicated meteorological instrument and standard unit: Atmospheric Pressure is measured with an Aneroid Barometer (Millibars), Wind Speed with a Cup Anemometer (Knots), Sunshine Duration with a Campbell-Stokes Recorder (Hours), and Relative Humidity with a Hygrometer (Percentage).

Step-by-Step Solution

1
Identify the instrument and unit for Atmospheric Pressure
Atmospheric Pressure pairs with Aneroid Barometer (Millibars).
Air pressure is measured with barometers in millibars or hectopascals.
2
Identify the instrument and unit for Wind Speed
Wind Speed pairs with Cup Anemometer (Knots).
The rotation rate of anemometer cups measures wind velocity in knots or m/s.
3
Identify the instrument and unit for Sunshine Duration
Sunshine Duration pairs with Campbell-Stokes Recorder (Hours).
A glass sphere focuses rays onto a sensitized card to record sunshine duration in hours per day.
4
Identify the instrument and unit for Relative Humidity
Relative Humidity pairs with Hygrometer (Percentage).
Moisture relative to saturation level is measured using hygrometers or psychrometers as a percentage.

Key Concept

Weather Elements, Instruments, and Standard Units
Estimated Time:1m 30s
Question 86Question

A weather observer at a meteorological station requires a continuous visual trace of fluctuations in atmospheric pressure over a 24-hour cycle. Which instrument is designed specifically to record these continuous pressure changes on a rotating drum chart?

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Answer: Barograph

Answer

Barograph
The barograph is a self-recording barometer. It contains a series of aneroid capsules that expand and contract with changes in atmospheric pressure. This movement is magnified through levers to a pen arm, which draws a continuous trace (barogram) on a paper chart wrapped around a clockwork-driven rotating cylinder.

Step-by-Step Solution

1
Identify the target atmospheric parameter and measurement format requested in the stem
The target parameter is atmospheric pressure, requiring a continuous graphical record (trace chart) rather than a single instantaneous reading.
Different meteorological instruments serve specific parameters and reading methods (instantaneous vs. self-recording).
2
Match the target parameter to the corresponding self-recording weather instrument
Atmospheric pressure recorded continuously on a rotating chart drum corresponds to a barograph.
The suffix '-graph' indicates a self-recording instrument, and the prefix 'baro-' relates to atmospheric pressure.

Key Concept

Continuous Recording Weather Instruments
Estimated Time:1m 0s
Question 87Question

Cold ocean currents flowing along the western margins of continents in tropical latitudes significantly alter coastal weather. Which mechanism explains how the Benguela Current causes hyper-arid climatic conditions in the Namib Desert?

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Answer: It chills the low-level air mass, producing a temperature inversion that suppresses vertical convection and condensation.

Answer

The cold Benguela Current chills the lower atmospheric layer, creating a temperature inversion that stabilizes the air mass and suppresses convective precipitation.
The correct option states that the current chills the low-level air mass, producing a temperature inversion that suppresses vertical convection. As the cold Benguela Current flows northward along the Namibian coast, it cools the lower layers of the atmosphere. This dense, chilled air sits beneath warmer air aloft, creating a stable temperature inversion. Because warm air above prevents the cool surface air from rising, convective clouds cannot form, leading to extreme desert conditions along the coast.

Step-by-Step Solution

1
Identify the primary climatic control acting on the coastal desert region.
The Benguela Current is a cold ocean current flowing northward along the southwestern coast of Africa.
Ocean currents regulate air temperature and moisture capacity of overlying air masses.
2
Analyze how cold ocean currents influence atmospheric stability.
Air directly in contact with cold ocean surface waters is cooled from below, while upper atmospheric air remains warmer.
Cool air is denser than warm air, creating a stable thermal inversion layer where air cannot rise easily.
3
Relate atmospheric stability to precipitation outcomes.
Without vertical convection, moist air cannot rise, expand, cool adiabatically, and condense into rain clouds, producing fog and extreme aridity.
Vertical convective uplift is essential for thunderstorm and cloud formation.

Key Concept

Climatic Controls - Ocean Currents and Aridity
Question 88Question

Match each major vegetation biome on the left with its corresponding plant structural adaptation and characteristic soil type on the right.

Click a left item, then click its matching right item

Items

Tropical Rainforest
Mediterranean Woodland
Coniferous Boreal Forest (Taiga)
Tropical Savanna

Matches

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Answer

Tropical Rainforest matches with multilayered canopy, drip-tip leaves, and leached oxisols. Mediterranean Woodland matches with sclerophyllous waxy leaves and terra rossa soils. Coniferous Boreal Forest matches with needle-like leaves, snow-shedding branches, and podzols. Tropical Savanna matches with drought-resistant umbrella crowns and fire-resistant bark.
Each biome exhibits structural morphology suited to its moisture and thermal regime: rainforest plants feature drip-tips and buttress roots on oxisols; Mediterranean plants feature sclerophyllous waxy leaves on terra rossa soils; boreal conifers feature needle-like leaves on podzols; savanna vegetation features fire-resistant bark and umbrella crowns adapted to wet-dry cycles.

Step-by-Step Solution

1
Analyze Tropical Rainforest adaptations and soils.
Identified high rainfall leading to intense leaching (oxisols/latosols) and plant features like buttress roots and drip tips.
Extreme moisture and rapid organic decomposition create nutrient-poor, highly leached soils.
2
Analyze Mediterranean Woodland adaptations and soils.
Identified summer drought conditions requiring sclerophyllous (small, waxy) foliage on terra rossa soils.
High evaporation during dry summers necessitates structures that prevent transpiration loss.
3
Analyze Coniferous Boreal Forest adaptations and soils.
Identified cold subarctic climate requiring needle leaves and conical tree form on acidic podzol soils.
Slow decomposition of resinous pine needles acidifies the soil profile.
4
Analyze Tropical Savanna adaptations and soils.
Identified seasonal moisture shifts requiring umbrella crowns and thick bark.
Trees must survive prolonged dry seasons and seasonal bushfires.

Key Concept

Ecological adaptations of vegetation biomes to climatic controls and soil profiles
Estimated Time:1m 30s
Question 89Question

City A is located at longitude 25W25^\circ\text{W} and City B is located at longitude 65E65^\circ\text{E}. Calculate the difference in local solar time between the two cities in hours.

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Answer: 6

Answer

The time difference between City A and City B is 6 hours.
Because City A (25W25^\circ\text{W}) and City B (65E65^\circ\text{E}) lie in opposite hemispheres relative to the Greenwich Meridian (00^\circ), the total longitudinal difference between them is the sum of their absolute longitudes (25+65=9025^\circ + 65^\circ = 90^\circ). Given that Earth rotates 1515^\circ of longitude every hour (360/24 hours360^\circ / 24\text{ hours}), dividing 9090^\circ by 15/hour15^\circ\text{/hour} gives a total time difference of 6 hours.

Step-by-Step Solution

1
Calculate the total angular distance between the two longitudes.
25W+65E=9025^\circ\text{W} + 65^\circ\text{E} = 90^\circ
Because the two locations are in different hemispheres (West and East), their longitudinal values must be added to find the total separation across the Prime Meridian.
2
Convert longitudinal degrees into time difference.
90/15 per hour=6 hours90^\circ / 15^\circ\text{ per hour} = 6\text{ hours}
Earth completes one full rotation of 360360^\circ in 24 hours, which corresponds to an angular velocity of 1515^\circ per hour.

Key Concept

Calculating time difference from longitudinal distance across hemispheres.
Question 90Question

A ship captain at sea observes local solar noon (12:00 PM12:00\text{ PM}) on Monday. At that precise moment, a radio time signal from Greenwich (00^\circ) indicates that Greenwich Mean Time (GMT) is 05:20 PM05:20\text{ PM} on Monday. What is the longitude of the ship in degrees West of the Prime Meridian?

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Answer: 80

Answer

The longitude of the ship is 80W80^\circ\text{W} (80 degrees West).
The time difference between local solar noon (12:00 PM12:00\text{ PM}) and GMT (05:20 PM05:20\text{ PM}) is 5 hours and 20 minutes, or 5135\frac{1}{3} hours. Multiplying 5.333 hours5.333\text{ hours} by 1515^\circ per hour yields 8080^\circ. Since local solar time is behind GMT, the ship is situated 8080^\circ West of the Prime Meridian.

Step-by-Step Solution

1
Calculate the time difference between local solar time and Greenwich Mean Time (GMT)
05:20 PM12:00 PM=5 hours and 20 minutes=5.333 hours05:20\text{ PM} - 12:00\text{ PM} = 5\text{ hours and } 20\text{ minutes} = 5.333\text{ hours}
The distance in longitude is directly proportional to the difference between local time and GMT.
2
Convert the time difference into angular distance in degrees of longitude
513 hours×15/hour=805\frac{1}{3}\text{ hours} \times 15^\circ/\text{hour} = 80^\circ
Earth rotates 360360^\circ in 24 hours, which corresponds to 1515^\circ of longitude for every 1 hour of time difference.
3
Determine hemisphere direction based on whether local time is ahead or behind GMT
Western Hemisphere (80W80^\circ\text{W})
Local time (12:00 PM12:00\text{ PM}) is earlier than GMT (05:20 PM05:20\text{ PM}), placing the ship west of the Prime Meridian.

Key Concept

Calculating longitude using local time and Greenwich Mean Time (GMT)
Question 91Question

An aircraft departs from Town X, located at longitude 120W120^\circ\text{W}, at 4:00 a.m. local time on Monday. The non-stop flight to Town Y, located at longitude 75E75^\circ\text{E}, takes exactly 14 hours. What is the local time and day at Town Y when the aircraft lands?

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Answer: 7:00 a.m. on Tuesday

Answer

7:00 a.m. on Tuesday
The correct answer is 7:00 a.m. on Tuesday. Town X (120W120^\circ\text{W}) and Town Y (75E75^\circ\text{E}) are separated by 195195^\circ of longitude (120+75120^\circ + 75^\circ). Dividing by 1515^\circ per hour yields a 13-hour time difference. Because Town Y is located east of Town X, local time at Town Y is 13 hours ahead of Town X. Thus, when the plane takes off at 4:00 a.m. Monday in Town X, the local time in Town Y is 5:00 p.m. Monday. Adding the 14-hour flight duration to 5:00 p.m. Monday results in 7:00 a.m. on Tuesday.

Step-by-Step Solution

1
Calculate total longitudinal difference between Town X and Town Y
Longitudinal difference = 120+75=195120^\circ + 75^\circ = 195^\circ
Since Town X is in the Western Hemisphere (120W120^\circ\text{W}) and Town Y is in the Eastern Hemisphere (75E75^\circ\text{E}), their angular distances from the Prime Meridian (00^\circ) must be added together.
2
Convert the longitudinal difference into time difference
Time difference = 19515/hour=13 hours\frac{195^\circ}{15^\circ/\text{hour}} = 13\text{ hours}
The Earth rotates 360360^\circ in 24 hours, which corresponds to 1515^\circ per hour.
3
Determine the local time at Town Y at the exact moment of departure from Town X
Departure time at Town Y = 4:00 a.m. Monday + 13 hours = 5:00 p.m. (17:00) Monday
Places to the east gain time relative to places to the west ('East gain, West lose'). Town Y is east of Town X.
4
Add the flight duration to find the arrival time at Town Y
Arrival time at Town Y = 5:00 p.m. Monday + 14 hours = 7:00 a.m. Tuesday
Adding 14 hours to 5:00 p.m. (17:00) yields 31:00 hours. Subtracting 24 hours for a full day rollover gives 7:00 a.m. on the following day (Tuesday).

Key Concept

Calculating local time differences across Eastern and Western hemispheres combined with flight elapsed time and calendar day rollover.
Estimated Time:3m 0s
Question 92Question

Southeastern Nigeria experiences severe environmental degradation due to catastrophic gully erosion triggered by human activities and physical vulnerability. Arrange the following geomorphic stages in the correct chronological sequence of gully morphogenesis from initial disturbance to advanced structural enlargement.

Drag items to arrange them in the correct order

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Answer

The correct chronological sequence begins with vegetation clearance causing raindrop splash erosion, followed by surface runoff concentration into rills, then channel deepening through headward scouring into gullies, and culminates in groundwater sapping and sidewall mass slumping.
Gully erosion develops systematically from surface destabilization to deep structural collapse. The process begins with the removal of plant cover, exposing soil to splash erosion. Next, unabsorbed runoff concentrates into rills. As flow energy increases, rills deepen into gullies via headward erosion. Finally, when the gully floor reaches groundwater levels, seepage (basal sapping) and bank undercutting induce slope instability and mass slumping.

Step-by-Step Solution

1
Identify the initiating trigger of soil degradation.
Deforestation or land clearing exposes bare soil to splash erosion, breaking down soil aggregates.
Vegetation removal is the primary anthropogenic antecedent condition.
2
Trace the initial hydrological response of surface runoff.
Infiltrative capacity is exceeded, leading to sheet wash and small micro-channel (rill) incision.
Runoff gathers momentum and concentrates into discrete paths.
3
Determine the phase where rills transition into active gullies.
Concentrated flow scours deep into weak, un-consolidated subsoil strata, expanding rills through headward erosion.
Hydraulic force increases bed scouring depth beyond normal tillage or agricultural recovery.
4
Identify the mature stage dominated by subsurface hydrology and mass wasting.
Groundwater seepage (basal sapping) undermines sidewalls, causing structural mass slumping and rapid gully expansion.
Deep gullies intersect the local water table, introducing geotechnical failure mechanisms.

Key Concept

Morphogenesis of Gully Erosion and Mass Wasting Feedback Loops
Question 93Question

In subtropical latitudes (3040 N/S30^\circ\text{--}40^\circ\text{ N/S}), western continental margins experience dry summers and wet winters, whereas eastern continental margins at the same latitudes experience hot, humid summers with peak rainfall. Which atmospheric mechanism primarily accounts for this contrasting seasonal precipitation regime?

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Answer: The poleward shift of subtropical high-pressure cells and cool ocean currents suppressing summer rainfall on western margins, compared to summer onshore monsoon/trade winds over warm currents on eastern margins.

Answer

The contrasting seasonal precipitation regimes between subtropical western continental margins (Mediterranean climate) and eastern margins (China/Humid Subtropical climate) are driven by the seasonal migration of subtropical high-pressure cells and offshore cool ocean currents on the west versus onshore moist winds over warm currents on the east.
The Mediterranean climate on western continental margins at 304030^\circ\text{--}40^\circ latitude experiences dry summers because the seasonal movement of the sun shifts subtropical anticyclones poleward, placing these regions under dry, subsiding air enhanced by cool offshore ocean currents. In contrast, eastern margins at identical latitudes experience wet summers because maritime air masses drawn along the western periphery of subtropical high-pressure cells pass over warm ocean currents, bringing abundant moisture and rainfall.

Step-by-Step Solution

1
Identify the climatic zones at 304030^\circ\text{--}40^\circ N/S latitudes on western and eastern continental margins.
Western margins host Mediterranean climates (CsCs) characterized by dry summers and wet winters. Eastern margins host Humid Subtropical/China-type climates (CfaCfa) characterized by hot, wet summers.
Establishing the specific climatic profiles identifies the controlling mechanisms.
2
Analyze atmospheric circulation controls acting on western margins in summer.
Subtropical high-pressure anticyclones shift poleward in summer. Subsiding, stable air combined with cool offshore currents inhibits convection and condensation, leading to summer drought.
Explains why summer precipitation is suppressed on western continental margins.
3
Analyze atmospheric circulation controls acting on eastern margins in summer.
Eastern margins receive moist air blowing around the western perimeter of subtropical high-pressure cells, augmented by warm ocean currents which destabilize the lower atmosphere and create heavy convectional and frontal rain.
Explains why eastern margins receive peak precipitation during summer.

Key Concept

Subtropical atmospheric circulation cells and ocean current controls on coastal climate regimes
Estimated Time:1m 30s
Question 94Question

A weather monitoring station situated at longitude 55W55^\circ\text{W} records a local solar time of 08:20 AM08:20\text{ AM}. At the exact same instant, a maritime research vessel logs a local solar time of 02:40 PM02:40\text{ PM}. What is the longitude of the maritime research vessel in degrees East (E^\circ\text{E})?

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Answer: 40

Answer

The longitude of the maritime research vessel is 40E40^\circ\text{E}.
The time difference between 08:20 AM08:20\text{ AM} (55W55^\circ\text{W}) and 02:40 PM02:40\text{ PM} (14:4014:40) is 6 hours and 20 minutes6\text{ hours and } 20\text{ minutes}, which equals 380 minutes380\text{ minutes}. At a rate of 4 minutes per degree4\text{ minutes per degree} of longitude, this represents a longitudinal difference of 9595^\circ. Since the vessel's local solar time is ahead, it is situated to the East. Measuring 9595^\circ East from 55W55^\circ\text{W} crosses the Greenwich Meridian (00^\circ) after 5555^\circ, placing the vessel at 40E40^\circ\text{E}.

Step-by-Step Solution

1
Calculate the time difference between the weather monitoring station and the maritime research vessel.
Time difference = 14:4008:20=6 hours 20 minutes=380 minutes14:40 - 08:20 = 6\text{ hours } 20\text{ minutes} = 380\text{ minutes}.
Converting 02:40 PM to 24-hour solar time (14:40) allows direct subtraction of the times.
2
Convert the time difference into angular degrees of longitude.
Angular distance = 380 minutes4 minutes per degree=95\frac{380\text{ minutes}}{4\text{ minutes per degree}} = 95^\circ.
The Earth rotates 360360^\circ in 24 hours, which corresponds to 11^\circ of longitude every 4 minutes.
3
Determine the direction and calculate the vessel's longitude.
95 East55 West to Prime Meridian=40E95^\circ\text{ East} - 55^\circ\text{ West to Prime Meridian} = 40^\circ\text{E}.
Because the local solar time at the vessel is ahead (later in the day), the vessel is located to the East. Traveling 9595^\circ East starting from 55W55^\circ\text{W} uses 5555^\circ to reach 00^\circ (Prime Meridian), leaving 4040^\circ in the Eastern Hemisphere.

Key Concept

Longitude calculation from local solar time difference across the Prime Meridian
Estimated Time:1m 30s
Question 95Question

In physical geography, weathering and mass wasting operate through distinct chemical reactions and mechanical failure modes governed by specific climatic and geological conditions. Which option correctly pairs each weathering or mass wasting process on the left with its corresponding environmental mechanism and landform result on the right?

Click a left item, then click its matching right item

Items

Carbonation-Solution
Hydrolysis
Solifluction
Slumping (Rotational Slide)

Matches

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Answer

Carbonation-Solution pairs with the dissolution of limestone by carbonic acid forming karst landforms; Hydrolysis pairs with hydrogen ion substitution altering feldspar into kaolinite clay; Solifluction pairs with slow flowage of thawed saturated soil over permafrost; Slumping pairs with downslope displacement along a curved concave shear plane forming backward-tilted terraces.
Carbonation dissolves limestone via carbonic acid to produce karst landforms; Hydrolysis alters silicate minerals like feldspar to clay through hydrogen ion substitution; Solifluction is the slow downslope flow of saturated soil over permafrost; Slumping is rotational displacement along a concave surface forming step-like terraces.

Step-by-Step Solution

1
Analyze the chemical mechanism of Carbonation-Solution.
Carbonation requires carbon dioxide dissolved in water (H2CO3H_2CO_3) dissolving calcium carbonate in limestone to form underground karst topography.
Calcium carbonate is insoluble in pure water but dissolves readily in carbonated groundwater.
2
Examine the process of Hydrolysis.
Hydrolysis involves reactive H+H^+ ions replacing cations in silicate minerals such as feldspar, converting them into kaolinite clay.
Hydrolysis changes the fundamental chemical composition of rock minerals to produce soft hydrous clays.
3
Differentiate between Solifluction and Slumping mass wasting movements.
Solifluction requires permafrost acting as an impermeable layer for saturated soil flow in tundra regions, whereas Slumping involves rotational slip along a concave failure plane creating backward-tilted step terraces.
Solifluction is thermal-climatic and flow-based, while slumping is structural shear failure along a curved surface.

Key Concept

Chemical weathering mechanisms and mass wasting failure modes
Estimated Time:2m 0s
Question 96Question

Due to the Earth's rotation from west to east, local solar time varies with longitude across the globe. Arrange the following longitudes in order of their local solar time at any given moment, from the EARLIEST time of day to the LATEST time of day.

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Answer

The correct sequence from earliest to latest local time is 90W90^\circ\text{W}, 30W30^\circ\text{W}, 45E45^\circ\text{E}, and 120E120^\circ\text{E}.
Because the Earth rotates on its axis from West to East at a rate of 1515^\circ per hour, local solar time increases as you move eastwards. Therefore, the longitude located furthest west (90W90^\circ\text{W}) has the earliest time of day, followed by 30W30^\circ\text{W}, then 45E45^\circ\text{E}, and finally the longitude furthest east (120E120^\circ\text{E}) which has the latest time.

Step-by-Step Solution

1
Identify the relationship between Earth's rotation direction and local time.
Earth rotates from West to East, meaning eastern longitudes experience sunrise earlier and are ahead in time, while western longitudes are behind in time.
The sun appears to rise in the east and set in the west due to Earth's axial rotation.
2
Compare longitudes relative to the Greenwich Meridian (00^\circ).
Longitudes in the Western Hemisphere (W\text{W}) are behind 00^\circ, while longitudes in the Eastern Hemisphere (E\text{E}) are ahead of 00^\circ.
Moving west reduces local solar time (1=4 minutes1^\circ = 4\text{ minutes} behind), whereas moving east increases local solar time (1=4 minutes1^\circ = 4\text{ minutes} ahead).
3
Order the given longitudes from furthest west to furthest east.
90W30W45E120E90^\circ\text{W} \rightarrow 30^\circ\text{W} \rightarrow 45^\circ\text{E} \rightarrow 120^\circ\text{E}.
This spatial progression from west to east directly corresponds to chronological progression from earliest hour of the day to latest hour.

Key Concept

Earth's West-to-East rotation causes places located further East to have a local time ahead of places located further West.
Question 97Question

A coastal meteorological station situated at latitude 10N10^\circ\text{N} records a high mean annual temperature exceeding 27C27^\circ\text{C} and an annual rainfall total of 2800 mm2{}800\text{ mm}. The station receives heavy precipitation for ten months of the year, but experiences a brief, distinct dry period of two months where precipitation falls slightly below 60 mm60\text{ mm} per month. Despite this dry season, soil moisture reserves remain high enough to support dense tropical rainforest growth. Which Köppen climate classification code and underlying atmospheric control mechanism account for these observed conditions?

Show answer & explanation

Answer: Climate code AmAm, where seasonal wind reversals driven by differential land-sea heating produce intense onshore monsoon rainfall that compensates for the short dry season.

Answer

The correct answer is Climate code AmAm, where seasonal wind reversals driven by differential land-sea heating produce intense onshore monsoon rainfall that compensates for the short dry season.
The Tropical Monsoon climate (AmAm) is defined by a high total annual precipitation that compensates for a short dry season where monthly rainfall drops below 60 mm60\text{ mm}. The major controlling mechanism is the seasonal reversal of winds (monsoon wind system) caused by differential heating between oceans and landmasses, bringing abundant maritime moisture onshore.

Step-by-Step Solution

1
Analyze the rainfall and temperature profile of the meteorological station.
High annual temperature (>27C>27^\circ\text{C}), very high annual precipitation (2800 mm2{}800\text{ mm}), and a brief 2-month dry season (<60 mm<60\text{ mm} per month).
Identifying temperature and monthly precipitation thresholds is the first step in Köppen climate classification.
2
Determine the Köppen climate code based on empirical criteria.
The criteria match Tropical Monsoon (AmAm), where high overall precipitation compensates for a short dry spell, allowing rainforest vegetation to persist.
Köppen defines AmAm by high total annual rainfall despite having one or more months with less than 60 mm60\text{ mm} of rain.
3
Identify the primary climatic control mechanism for this region.
Seasonal wind reversals resulting from thermal pressure differences between landmasses and oceans during high-sun periods bring moist maritime air inland.
Monsoon circulation (AmAm) is primarily driven by seasonal shifts in pressure gradient and wind direction.

Key Concept

Köppen Tropical Monsoon (AmAm) Climate Classification and Atmospheric Controls
Question 98Question

A meteorological station situated in a tropical latitude at 15N15^\circ\text{N} at an elevation of 2200 meters2{}200\text{ meters} records a mean annual temperature of 14C14^\circ\text{C}, with its warmest month averaging 17C17^\circ\text{C} and its coldest month averaging 11C11^\circ\text{C}. Precipitation is heavily concentrated during the high-sun summer season under the influence of the Intertropical Convergence Zone (ITCZ), whereas the low-sun winter season is markedly dry. Under the Köppen climate classification system, how is this highland climate designated, and which primary climatic control accounts for its thermal departure from surrounding lowlands?

Show answer & explanation

Answer: Designated as Cw (Subtropical Highland climate); modified primarily by altitude through the environmental lapse rate.

Answer

The climate is designated as Cw (Subtropical Highland climate), modified primarily by altitude through the environmental lapse rate.
In the Köppen climate classification, when a tropical location is situated at high altitude, temperature drops at the environmental lapse rate (~6.5°C per 1,000 m). Because its warmest month is 17°C (below 18°C) and its coolest is 11°C (above -3°C), it is placed in Group C rather than Group A. Combined with a dry winter season under high-pressure subsidence, the full code is Cw (Subtropical Highland).

Step-by-Step Solution

1
Analyze thermal thresholds according to Köppen climate rules
Warmest month (17C17^\circ\text{C}) is <18C< 18^\circ\text{C} and coldest month (11C11^\circ\text{C}) is >3C> -3^\circ\text{C}, excluding Group A (Tropical, all months 18C\ge 18^\circ\text{C}) and assigning the station to Group C (Warm Temperate/Mesothermal).
Köppen criteria classify any location where the warmest month drops below 18°C as non-tropical.
2
Evaluate seasonal precipitation distribution pattern
Precipitation occurs during high-sun (summer) with a dry low-sun (winter) period, giving the precipitation letter code 'w' (winter dry).
The station's wet high-sun and dry low-sun cycle matches the 'w' modifier.
3
Identify the primary climatic control responsible for cooling
Altitude reduces surface temperature via the normal environmental lapse rate (roughly 6.5C6.5^\circ\text{C} per 1000 m1{}000\text{ m}).
At 15N15^\circ\text{N} latitude, lowland temperatures exceed 27C27^\circ\text{C}, so an elevation of 2200 m2{}200\text{ m} lowers temperatures into the temperate CwCw range.

Key Concept

Köppen Highland Climate Classification and Environmental Lapse Rate
Question 99Question

City X is located at longitude 15E15^\circ\text{E}. When the local time at City X is 2:00 PM2:00\text{ PM}, what is the local time at City Y, located at longitude 45E45^\circ\text{E}?

Show answer & explanation

Answer: 4:00 PM4:00\text{ PM}

Answer

The local time at City Y is 4:00 PM4:00\text{ PM}.
The difference in longitude between City X (15E15^\circ\text{E}) and City Y (45E45^\circ\text{E}) is 3030^\circ. Since 1515^\circ corresponds to 1 hour, 3030^\circ equals 2 hours. Because City Y lies to the east of City X, local time is ahead, so 2 hours must be added to 2:00 PM2:00\text{ PM}, giving 4:00 PM4:00\text{ PM}.

Step-by-Step Solution

1
Calculate the angular longitudinal difference between City X and City Y.
Difference =45E15E=30= 45^\circ\text{E} - 15^\circ\text{E} = 30^\circ.
Since both cities are in the Eastern Hemisphere, subtract the smaller longitude from the larger longitude.
2
Convert the longitudinal difference into a time difference.
Time difference =30÷15/hour=2 hours= 30^\circ \div 15^\circ/\text{hour} = 2\text{ hours}.
The Earth rotates 1515^\circ per hour (360360^\circ in 24 hours).
3
Determine whether to add or subtract the time difference.
Local time =2:00 PM+2 hours=4:00 PM= 2:00\text{ PM} + 2\text{ hours} = 4:00\text{ PM}.
City Y (45E45^\circ\text{E}) is east of City X (15E15^\circ\text{E}), so time is ahead (gain time going east).

Key Concept

Calculation of local time difference using longitudinal intervals
Estimated Time:45s
Question 100Question

Arrange the following geographical locations in order of INCREASING daylight duration (from shortest day length to longest day length) during the June Solstice (June 21st).

Drag items to arrange them in the correct order

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Answer

The correct sequence of locations from shortest to longest daylight duration on June 21st is: Antarctic Circle (6612S66\frac{1}{2}^\circ\text{S}), Tropic of Capricorn (2312S23\frac{1}{2}^\circ\text{S}), Equator (00^\circ), Tropic of Cancer (2312N23\frac{1}{2}^\circ\text{N}), and Arctic Circle (6612N66\frac{1}{2}^\circ\text{N}).
On June 21st (the June Solstice), the North Pole is inclined toward the Sun. As a result, day length increases continuously from south to north across the globe. Locations south of the Antarctic Circle (6612S66\frac{1}{2}^\circ\text{S}) experience 0 hours of sunlight (24 hours of darkness). Mid-latitude southern regions such as the Tropic of Capricorn (2312S23\frac{1}{2}^\circ\text{S}) experience short winter days (~10.5 hours). The Equator (00^\circ) always maintains an equal 12-hour day and night. Mid-latitude northern regions such as the Tropic of Cancer (2312N23\frac{1}{2}^\circ\text{N}) experience long summer days (~13.5 hours), and the region within the Arctic Circle (6612N66\frac{1}{2}^\circ\text{N}) receives continuous 24-hour daylight. Arranging these from shortest to longest daylight duration gives: Antarctic Circle, Tropic of Capricorn, Equator, Tropic of Cancer, and Arctic Circle.

Step-by-Step Solution

1
Identify Earth's orientation relative to the Sun on June 21st (June Solstice).
The Northern Hemisphere is tilted towards the Sun at an angle of 23.523.5^\circ, making the Tropic of Cancer (2312N23\frac{1}{2}^\circ\text{N}) the subsolar point.
Earth's axial tilt causes daylight duration to increase progressively from the South Pole toward the North Pole during the June solstice.
2
Determine daylight hours at the polar circles.
The Antarctic Circle (6612S66\frac{1}{2}^\circ\text{S}) receives 0 hours of daylight, while the Arctic Circle (6612N66\frac{1}{2}^\circ\text{N}) receives 24 hours of daylight.
The entire area south of the Antarctic Circle is in Earth's shadow, whereas the area north of the Arctic Circle remains continuously illuminated.
3
Determine daylight hours at intermediate latitudes.
The Tropic of Capricorn (2312S23\frac{1}{2}^\circ\text{S}) has ~10.5 hours of daylight, the Equator (00^\circ) has exactly 12 hours, and the Tropic of Cancer (2312N23\frac{1}{2}^\circ\text{N}) has ~13.5 hours.
Day length increases smoothly along the latitudinal gradient from south to north.
4
Order the locations from shortest to longest daylight duration.
Antarctic Circle (0 hrs0\text{ hrs}) < Tropic of Capricorn (10.5 hrs\sim 10.5\text{ hrs}) < Equator (12 hrs12\text{ hrs}) < Tropic of Cancer (13.5 hrs\sim 13.5\text{ hrs}) < Arctic Circle (24 hrs24\text{ hrs}).
This sequence correctly reflects increasing daylight hours.

Key Concept

Latitudinal Variation of Daylight Hours during Solstices
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