Geomorphology: Concepts, Processes and Agents, Landform Evolution, Tectonic, Climatic and Base-Level Controls, and the Quaternary

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. Which statement best expresses Hack's idea of dynamic equilibrium in landscapes?

   1. Landforms pass through youth, maturity and old age after a single rapid uplift followed by a long stillstand, so the form of a landscape records how long ago it was uplifted
   2. Slopes retreat parallel to themselves, leaving a gently inclined pediment at their foot with scattered residual hills, and this pediplain is taken as the final state of the landscape
   3. Form adjusts to rock resistance and to the rate of the processes acting on it, so differences in form between areas can reflect material and process rather than elapsed time
   4. Relief is made only while uplift is acting and is destroyed as soon as uplift stops, so every landscape tends towards the same flat end form whatever its rock type
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   Answer: C — Form adjusts to rock resistance and to the rate of the processes acting on it, so differences in form between areas can reflect material and process rather than elapsed time

   Dynamic equilibrium ties landscape form to rock resistance and process rate, so form need not carry any message about elapsed time. The tempting alternative, passage through youth, maturity and old age, is Davis's cycle of erosion, which reads form as a stage in time. Parallel retreat of slopes towards a pediplain is King's model, and the idea that every landscape converges on one flat end form is not a recognised model at all.

2. In Lester King's model of pediplanation, a landscape evolves mainly by:

   1. parallel retreat of steep scarps, which leaves a gently sloping pediment at their foot and isolated residual hills (inselbergs) standing in front of them
   2. steady decline of slope angle everywhere as the whole profile is lowered towards base level, ending in a peneplain of very low relief and gentle convex hills
   3. vertical downcutting by rivers alone, with the interfluve slopes keeping the same angle and the same position while the valleys deepen beneath them
   4. retreat of the coastline while the inland relief stays constant, so that landforms develop only in a narrow belt along the shoreline as sea level changes
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   Answer: A — parallel retreat of steep scarps, which leaves a gently sloping pediment at their foot and isolated residual hills (inselbergs) standing in front of them

   King's pediplanation is driven by parallel retreat of scarps, which widens the pediment and leaves inselbergs as remnants. Steady decline of slope angle ending in a peneplain is the tempting alternative but is Davis's slope decline. Rivers cutting down while the interfluves stay fixed describes downcutting without slope retreat, which is no named model of general landscape evolution, and a coastline-only control leaves the interior unexplained.

3. According to the Hjulström curve for the erosion of loose, uniform grains by flowing water, which grains are entrained at the lowest flow velocity?

   1. Clay finer than 0.004 mm, because the smallest particles are the lightest and are therefore the easiest to lift from the bed
   2. Fine silt of about 0.01 mm, because it lies between clay and sand in size and so has the smallest critical erosion velocity of all
   3. Gravel of about 10 mm, because large grains project into the fastest part of the flow and are the easiest to roll along the bed
   4. Sand of roughly 0.2 to 0.5 mm, because it is light enough to lift yet lacks the cohesion that binds the finer grains to the bed
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   Answer: D — Sand of roughly 0.2 to 0.5 mm, because it is light enough to lift yet lacks the cohesion that binds the finer grains to the bed

   The critical erosion velocity is lowest for sand of roughly 0.2 to 0.5 mm. Clay and fine silt are held to the bed by cohesion and need higher velocities, so smaller is not easier, which is why clay is the tempting wrong choice. Gravel is heavy and needs a higher velocity again. Once moving, fine particles stay in suspension at velocities well below those that eroded them.

4. A drainage basin has a hypsometric integral of 0.72 and a convex-up hypsometric curve. Using Strahler's commonly quoted divisions (above 0.6 youthful, 0.35 to 0.6 mature, below 0.35 old), what does this indicate?

   1. An old, monadnock-stage basin, because most of its original volume has already been removed and only low residual hills remain
   2. A youthful basin, because most of its volume still stands high above its lowest point and comparatively little has yet been removed
   3. A mature basin in equilibrium, because an S-shaped curve with an integral near 0.5 is the signature of balance between uplift and erosion
   4. No inference is possible, because the integral measures only the gradient of the main channel and not how the area is spread with height
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   Answer: B — A youthful basin, because most of its volume still stands high above its lowest point and comparatively little has yet been removed

   HI = (mean − minimum) / (maximum − minimum) is the share of the basin's relief volume above the lowest point, so 0.72 with a convex curve means a high-standing, little-dissected basin: youthful in Strahler's scheme. An S-shaped curve with HI near 0.5 would be mature, which is the tempting choice only if the integral is misread. The integral does describe how area is distributed with height, not channel gradient.

5. Which of the following statements about geomorphic systems are correct?

   1. An intrinsic threshold can be crossed with no change in any external control, for example when slowly thickening regolith on a slope finally fails
   2. In steady state the rate of rock uplift equals the rate of erosion, so the mean elevation of the landscape stays constant
   3. Complex response means that a landscape reacts to an external change in a single step that is completed at once, leaving no later internal adjustments
   4. Over long periods the largest share of sediment is always moved by the rarest and largest flood in the record, since only extreme floods exceed the threshold
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   Answer: A — An intrinsic threshold can be crossed with no change in any external control, for example when slowly thickening regolith on a slope finally fails; B — In steady state the rate of rock uplift equals the rate of erosion, so the mean elevation of the landscape stays constant

   An intrinsic threshold is crossed by internal change alone, and steady state means uplift equals erosion, so those two statements are correct. Complex response is the opposite of a single immediate step: one external change triggers a chain of internal adjustments such as incision, a sediment pulse and aggradation. The Wolman and Miller result is that moderate, fairly frequent flows usually move most sediment, so the claim of the rarest flood is wrong.

6. A drainage basin has a minimum elevation of 150 m, a mean elevation of 640 m and a maximum elevation of 950 m. Its hypsometric integral, to two decimal places, is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 0.61

   HI = (h_mean − h_min) / (h_max − h_min) = (640 − 150) / (950 − 150) = 490 / 800 = 0.6125, which rounds to 0.61. Using the mean elevation divided by the maximum elevation (0.67) instead of measuring from the minimum is the usual slip.

7. A river reach in equilibrium suddenly receives much more coarse sediment from an upstream landslide, while its discharge and the grain size of its bed stay the same. According to Lane's balance (Q S proportional to Q_s D_50, with Q the discharge, S the slope, Q_s the sediment supply and D_50 the median grain size), the reach will tend to:

   1. incise, because the extra sediment supplies more tools for abrasion and so lowers the slope needed to carry the load
   2. aggrade and steepen (or braid) until the slope has risen enough to carry the larger load that now arrives
   3. keep the same slope, because with discharge and bed grain size unchanged the right-hand side of the balance is unaffected
   4. narrow into a single meandering channel with a lower slope, because the balance favours a smaller channel when the load increases
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   Answer: B — aggrade and steepen (or braid) until the slope has risen enough to carry the larger load that now arrives

   In Lane's balance the right-hand side Q_s D_50 has risen while Q is unchanged, so S on the left must rise to restore balance: the reach aggrades and steepens (or braids). The tempting alternative, incision from extra tools, ignores that a supply beyond the transport capacity buries the bed. Keeping the slope unchanged would leave the balance violated, and a lower slope would reduce capacity further.

8. A wide river channel has flow depth h = 2.0 m and water-surface slope S = 0.0015. Take water density 1000 kg/m³, grain density 2650 kg/m³, g = 9.8 m/s² and a critical Shields parameter of 0.05. Using bed shear stress τ = ρ g h S, the diameter of the largest grain the flow can just entrain is ____ mm (to the nearest whole millimetre).

   Numerical answer — type the value.

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   Answer: 36

   τ = ρ g h S = 1000 × 9.8 × 2.0 × 0.0015 = 29.4 Pa. From θc = τ / [(ρs − ρ) g D], D = τ / [θc (ρs − ρ) g] = 29.4 / (0.05 × 1650 × 9.8) = 29.4 / 808.5 = 0.03636 m, about 36 mm. Using ρs instead of (ρs − ρ) in the denominator, which ignores buoyancy, gives about 23 mm and is the usual slip.

9. Two neighbouring catchments lie in the same rock type and climate. Both have reached steady state under detachment-limited incision E = K A^m S^n with n = 1 and a common K. The steepness coefficient k_s in S = k_s A^(−θ) of one catchment is twice that of the other. What follows?

   1. Its concavity θ is also twice as large, because steeper channels always have more strongly concave long profiles
   2. It is twice as old, because steepness in a steady-state catchment grows in proportion to the time since uplift began
   3. Its drainage area is twice as large, because k_s is the value of the slope at a drainage area of one unit and so scales with area
   4. Its rock uplift rate is about twice as large, because at steady state with n = 1 the steepness is k_s = U / K and K is shared
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   Answer: D — Its rock uplift rate is about twice as large, because at steady state with n = 1 the steepness is k_s = U / K and K is shared

   Setting U = E gives S = (U/K)^(1/n) A^(−m/n), so for n = 1, k_s = U / K and a shared K means k_s is proportional to U. The concavity θ = m/n does not depend on U, which rules out the concavity claim. Area is the variable on the x-axis, not part of k_s, and in steady state the form no longer evolves with time, so age plays no part.

10. A channel in steady state follows detachment-limited incision E = K A^m S^n with m = 1, n = 2 and K = 1 × 10⁻⁹ m⁻¹ yr⁻¹ (E in m/yr, A in m²). The rock uplift rate is 0.16 mm/yr and the drainage area at the point considered is 1 km². The channel slope there, to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 0.4

    At steady state U = E, so S^n = U / (K A^m). Here U = 1.6 × 10⁻⁴ m/yr and K A^m = 10⁻⁹ × 10⁶ = 10⁻³, so S² = 0.16 and S = 0.4. Taking S = U / (K A), as for n = 1, would give 0.16, and forgetting to convert 1 km² to 10⁶ m² gives a wildly different result.

11. Which sequence correctly describes Darwin's model for the development of an atoll, and its cause?

    1. Atoll, then barrier reef, then fringing reef, as a rising volcanic island steadily loses the reef that grew around it
    2. Barrier reef, then fringing reef, then atoll, as the island rises and strands the reef above sea level in a ring around a lagoon
    3. Fringing reef, then barrier reef, then atoll, as a volcanic island subsides and the coral grows upward to stay near the sea surface
    4. Fringing reef, then atoll, then barrier reef, as the lagoon fills with sediment and the island is slowly uplifted by isostatic rebound
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    Answer: C — Fringing reef, then barrier reef, then atoll, as a volcanic island subsides and the coral grows upward to stay near the sea surface

    Darwin's sequence is fringing reef, barrier reef and atoll, driven by subsidence of the volcanic island while reef-building coral keeps growing upward. The reversed order, atoll first, would need the island to rise and has no reef-growth logic. Placing the atoll before the barrier reef contradicts the progressive widening of the lagoon as the island sinks.

12. The depth of ocean floor younger than about 70 Ma is given empirically by d = 2500 + 350 √t, with d in metres and t in Ma. The depth of seafloor aged 49 Ma, in metres, is ____.

    Numerical answer — type the value.

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    Answer: 4950

    d = 2500 + 350 × √49 = 2500 + 350 × 7 = 2500 + 2450 = 4950 m. The depth increases as the square root of age because the lithosphere cools and thickens, and so subsides, in proportion to √t. Using t instead of √t would give an absurd depth of over 19 km.

13. At a stretch of coast the land is rising at 2.0 mm/yr while eustatic (global) sea level is rising at 1.5 mm/yr. Which statement is correct?

    1. Relative sea level is falling at 0.5 mm/yr, so rivers reaching the coast tend to incise and emerged shorelines can form
    2. Relative sea level is rising at 3.5 mm/yr, so valleys are drowned and estuaries form along the coast
    3. Relative sea level is rising at 0.5 mm/yr, so rivers aggrade their lower reaches and the coastline retreats inland
    4. Relative sea level is unchanged, because tectonic and eustatic changes always cancel each other at a passive coast
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    Answer: A — Relative sea level is falling at 0.5 mm/yr, so rivers reaching the coast tend to incise and emerged shorelines can form

    Relative sea-level change = eustatic change − land uplift = 1.5 − 2.0 = −0.5 mm/yr, a relative fall. A relative fall lowers the base level, so rivers incise and shorelines are left emerged. Adding the two rates (3.5 mm/yr) is the tempting error and is wrong because uplift of the land works against a rising sea. The two rates cancel only if they happen to be equal, which they are not here.

14. Which of the following statements about river terraces and incision are correct?

    1. A strath terrace is a bedrock surface planed by lateral erosion, which is left abandoned when the river cuts down below it
    2. A fill terrace records a phase of aggradation followed by incision into the alluvium that had been deposited
    3. Matching terrace heights above the channel at two places prove by themselves that the terraces were abandoned at the same time
    4. A fall of base level starts incision at the river mouth, and the response travels upstream as knickpoints
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    Answer: A — A strath terrace is a bedrock surface planed by lateral erosion, which is left abandoned when the river cuts down below it; B — A fill terrace records a phase of aggradation followed by incision into the alluvium that had been deposited; D — A fall of base level starts incision at the river mouth, and the response travels upstream as knickpoints

    A strath terrace is an abandoned bedrock bevel, a fill terrace is aggradation followed by incision, and a base-level fall begins incision at the mouth with the signal migrating upstream as knickpoints, so those three statements are correct. Similar heights do not prove the same age: terraces can be diachronous, and correlation needs dating, so that statement is wrong.

15. A bedrock strath terrace now lies 14 m above the present channel bed and was abandoned 35 ka ago. Assuming the river has kept pace with rock uplift at a steady rate since then, the rock uplift rate, in mm/yr to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 0.4

    Incision rate = height / age = 14 m / 35 000 yr = 4 × 10⁻⁴ m/yr = 0.4 mm/yr. Because the river kept pace with uplift, the rock uplift rate equals this incision rate. Dividing 14 by 35 gives 0.4 m per ka, which is the same number in mm/yr.

16. A mountain belt in local (Airy-type) isostatic equilibrium has 5.5 km of rock removed by erosion. Take crustal density 2700 kg/m³ and mantle density 3300 kg/m³, and ignore flexural strength. After isostatic readjustment, the net lowering of the land surface is ____ m.

    Numerical answer — type the value.

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    Answer: 1000

    Rock uplift compensates a fraction ρc/ρm = 2700/3300 = 0.818 of the eroded thickness, so the surface lowers by e (1 − ρc/ρm) = 5500 × (600/3300) = 1000 m. The rock uplift is 4500 m. Giving the answer as 5500 m, which ignores isostasy, or as 4500 m, which is the rebound rather than the lowering, are the two usual slips.

17. In a deep-sea oxygen-isotope record, an even-numbered marine isotope stage such as MIS 2 corresponds to a:

    1. glacial period, with foraminifera enriched in ¹⁸O because the lighter ¹⁶O is stored in the growing ice sheets
    2. glacial period, with foraminifera depleted in ¹⁸O because the growing ice sheets preferentially take up the heavier isotope
    3. interglacial period, with foraminifera enriched in ¹⁸O because warm surface water evaporates less and leaves heavier oxygen behind
    4. interglacial period, with foraminifera depleted in ¹⁸O because melting ice returns the lighter oxygen to the ocean
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    Answer: A — glacial period, with foraminifera enriched in ¹⁸O because the lighter ¹⁶O is stored in the growing ice sheets

    Even-numbered marine isotope stages are glacial. Evaporation favours the light isotope ¹⁶O, which is locked in ice sheets, so the ocean, and the foraminifera that grow in it, become enriched in ¹⁸O. The mirror-image statement that ice takes up the heavier isotope is wrong, and MIS 1, the Holocene, is the odd-numbered interglacial.

18. Which of the following statements about the Quaternary record and its dating are correct?

    1. Axial tilt (obliquity) varies with a period of about 41 ka and paced the glacial cycles of the early Pleistocene
    2. Radiocarbon dating is routinely applied to charcoal, wood and shell material as old as about 500 ka
    3. Luminescence (OSL) dating gives the time since mineral grains were last exposed to daylight, that is, since they were buried
    4. At the last glacial maximum global sea level stood roughly 120 m below its present level
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    Answer: A — Axial tilt (obliquity) varies with a period of about 41 ka and paced the glacial cycles of the early Pleistocene; C — Luminescence (OSL) dating gives the time since mineral grains were last exposed to daylight, that is, since they were buried; D — At the last glacial maximum global sea level stood roughly 120 m below its present level

    Obliquity has a period of about 41 ka, OSL dates the last exposure to light, which is burial, and the last glacial maximum sea level was roughly 120 m lower, so those three are correct. Radiocarbon is limited to about 50 ka because too little carbon-14 remains beyond that, so the claim of 500 ka is wrong.