DSE Geography Practice Paper: Fluvial Processes And Coastal Landforms (12 Questions, 50 Marks, 75 Minutes)

DSE Geography Practice Paper: Fluvial Processes and Coastal Landforms (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Geography: Rivers and Coasts

DSE Geography: 12 original questions, 50 marks, 75 minutes on river gradient, discharge, sediment load, waves, longshore drift and coastal management.

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Paper details　Level: S6 | Subject: Geography | Time: 75 minutes | Marks: 50 | Questions: 12

Section A: Fluvial processes and hydrology (Questions 1-6, 24 marks)

Source A (illustrative gauging record written for this paper, not real measurements)

A gauging station in the middle course drains a catchment of 480 square kilometres. Records show a dry-season discharge of about 18 cubic metres per second, rising to a wet-season peak of 210 cubic metres per second, while mean suspended sediment concentration rises from 0.2 to 1.6 kilograms per cubic metre. A reservoir built in 1985 lies 12 kilometres upstream, and both banks 3 kilometres downstream have been lined with concrete. Over the last ten years the highest recorded stage has risen by 0.4 metres while the bed level has fallen by 0.3 metres.

Show the formula, the substitution and the units in calculation questions.

#	Question	Answer and marking notes
1	A river section falls from 500 m to 100 m above sea level over a channel length of 25 km. Calculate the river gradient in m per km and show your working. (4 marks)	Marking points: drop = 500 – 100 = 400 m (1 mark); gradient = drop / channel length (1 mark); 400 / 25 = 16 m per km (2 marks). Deduct 1 mark for a missing unit.
2	A channel cross-section is 20 m wide with a mean depth of 3 m, and the mean velocity is 1.2 m/s. Calculate the cross-sectional area and the discharge, stating the units. (4 marks)	Marking points: area = width x depth = 20 x 3 = 60 m2 (2 marks); discharge Q = area x velocity = 60 x 1.2 = 72 m3/s (2 marks). Deduct 1 mark if the unit is given as m3 without the time base.
3	For the same section the wetted perimeter is 26 m. Calculate the hydraulic radius and explain how a larger hydraulic radius affects velocity. (4 marks)	Marking points: hydraulic radius R = area / wetted perimeter = 60 / 26 = 2.31 m (2 marks). A larger R means a smaller proportion of the water is in contact with the bed and banks, so friction is relatively less and, for the same slope, velocity and sediment transport are higher (2 marks).
4	The river carries a mean suspended sediment concentration of 0.8 kg/m3. Using the discharge from Question 2, calculate the sediment load per second and per day in tonnes (to the nearest tonne). (4 marks)	Marking points: load per second = 72 x 0.8 = 57.6 kg/s (2 marks); per day = 57.6 x 86400 = 4,976,640 kg, about 4,977 tonnes (2 marks). A candidate who forgets the 86400 s conversion or the kg to tonne conversion scores at most 2.
5	The bed slope is 0.002 and the channel is 30 m wide. Using total stream power P = rhogQ*S (rho = 1000 kg/m3, g = 9.81 m/s2), calculate the total stream power and the stream power per unit width. (4 marks)	Marking points: P = 1000 x 9.81 x 72 x 0.002 = about 1413 W (2 marks); per unit width = P / channel width = 1413 / 30 = about 47.1 W/m (2 marks). A note that greater stream power allows coarser material to be moved earns credit within the second pair of marks.
6	For deep water waves, celerity c = 1.56T m/s and wavelength L = 1.56T^2 m. For a wave period of 10 s, calculate c and L, then explain how celerity, wavelength and wave height change as the wave enters shallow water. (4 marks)	Marking points: c = 1.56 x 10 = 15.6 m/s (1 mark); L = 1.56 x 10^2 = 156 m (1 mark); in shallow water, where depth is less than half the deep-water wavelength, both celerity and wavelength decrease as the wave shoals, wave height increases, the crest steepens and the wave breaks where depth is about 1.3 times the wave height (2 marks).

Section B: Coastal landforms and sediment (Questions 7-9, 12 marks)

Answer in complete sentences.

#	Question	Answer and marking notes
7	A beach is 1000 m long with a beach scarp 3 m high. Its annual sediment budget shows an input of 15000 cubic metres and an output of 18000 cubic metres. Calculate the net sediment change and the mean rate of beach retreat in metres per year. (4 marks)	Marking points: net change = 15000 – 18000 = minus 3000 m3, a net loss of 3000 m3 per year (2 marks); retreat = volume loss / (beach length x scarp height) = 3000 / (1000 x 3) = 1.0 m per year (2 marks). Only the net loss without converting to a retreat rate scores at most 2.
8	A cliff section is 40 m high and 500 m long and retreats on average 0.2 m per year. Calculate the volume of rock eroded each year and state two factors that would speed up the retreat. (4 marks)	Marking points: volume = height x length x retreat rate = 40 x 500 x 0.2 = 4000 m3 per year (2 marks). Speeding factors, any two at 1 mark each: high wave energy or frequent storm surges; well-developed joints and bedding with weak rock resistance; collapse of the cliff above an enlarged notch; removal of talus by longshore drift leaving the cliff foot unprotected; reduced sediment supply from human activity such as dams or sand mining.
9	For a flood with a return period of 50 years, what is the annual exceedance probability, and what is the probability of at least one such flood in the next 30 years (percentage, one decimal place)? (4 marks)	Marking points: annual exceedance probability = 1 / 50 = 2.0% (1 mark); 30-year probability = 1 – (1 – 0.02)^30 = 45.5% (2 marks); a further mark for noting that a longer return period means a lower annual chance but that cumulative risk over a long period cannot be ignored (1 mark).

Section C: Data application and evaluation (Questions 10-12, 14 marks)

Evaluation questions require both sides of the argument and a clear conclusion.

#	Question	Answer and marking notes
10	Sample A has a grain diameter of 0.25 mm and sample B 0.0625 mm. Using phi = -log2(d), calculate the phi value of each and state which sample is coarser. (4 marks)	Marking points: sample A phi = -log2(0.25) = 2.0 (1 mark); sample B phi = -log2(0.0625) = 4.0 (1 mark); a larger phi means a finer grain, so sample A at phi 2.0 is coarser (2 marks). Merely stating that A is coarser without calculation earns 1 mark.
11	Explain two reasons why the long profile is gentle and the sediment is finer in the lower course, and state how a fall in base level affects river erosion. (4 marks)	Marking points, any two at 1 mark each: the lower course lies close to base level so the drop and gradient are small, reducing velocity and stream power; transport competence falls with velocity so coarse material is deposited upstream while silt and clay travel further; the channel is wide and shallow, so friction is relatively greater; discharge is larger but the gradient effect dominates. Base level effect, 2 marks: a fall in base level, from a sea-level fall or uplift, causes rejuvenation, renewed downcutting, and the formation of terraces and incised meanders.
12	A coast affected by longshore drift is considering both a groyne and beach nourishment. Compare the effectiveness and limitations of the two approaches and give your recommendation. (6 marks)	Marking points: a groyne traps sediment, widening the beach immediately updrift (2 marks), but it cuts off supply downdrift so erosion there accelerates, and capital plus maintenance costs are high while the structure disrupts the shoreline and littoral exchange (2 marks); nourishment replaces the sediment directly, keeps the amenity and ecology of the beach, but must be repeated, so the cost accrues over time, and unsuitable source material or grain size is quickly lost (2 marks); a judgement is required, for example nourishment supported by vegetation first, with short groynes only at erosion hotspots and compensation for downdrift losses (2 marks).

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A TutorZone practice paper; original questions and passages, not publisher material.