S4 Geography Paper: Rivers, Fluvial Landforms And The Water Cycle (100 Marks)

S4 Geography Paper: Rivers, Fluvial Landforms and the Water Cycle (100 marks)

2026-11-12

Geography – Rivers and Fluvial Landforms

S4 Geography: 6 questions, 24 parts, 100 marks, 75 min. Water cycle runoff, discharge, gradients, meanders, erosion and deposition, flood control.

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Paper details　

Section A: The Water Cycle and Catchment Runoff (16 marks)

A catchment of area 45 km2 receives 2200 mm of rain a year. The runoff ratio, that is runoff divided by rainfall, is 0.35. Take 1 km2 = 1 000 000 m2.

#	Question	Answer and marking notes
1	(a) Find the total volume of rain falling on the catchment in one year, in millions of cubic metres. (4 marks)	Answer: Area = 45 x 1 000 000 = 45 000 000 m2; rainfall depth = 2200 mm = 2.2 m; volume = 45 000 000 x 2.2 = 99.0 million m3. Marking: 1 mark for the area, 1 for the depth, 1 for the multiplication, 1 for the unit.
2	(b) Find the annual runoff and the combined loss through evapotranspiration and infiltration. (4 marks)	Answer: Runoff = 0.35 x 99.0 = 34.65 million m3; evapotranspiration plus infiltration = 99.0 – 34.65 = 64.35 million m3. Marking: 2 marks for runoff, 2 for the remainder.
3	(c) Calculate this loss as a percentage of rainfall, and explain why the percentage falls after vegetation is cleared. (4 marks)	Answer: 64.35 / 99.0 x 100% = 65.0 per cent. After clearance there is little interception and transpiration, rain reaches the soil directly and compacts it, so infiltration slows and surface runoff is rapid; the loss percentage falls. Marking: 2 marks for the percentage, 2 for the explanation.
4	(d) The upper catchment is urbanised. Explain the effects on peak discharge, on the timing of the peak and on downstream water quality. (4 marks)	Answer: Impermeable surfaces cut infiltration and raise the runoff ratio, so peak discharge increases; water reaches the channel quickly through drains, so the peak arrives earlier and lasts for a shorter time; surface wash carries sediment, oil and pollutants, raising turbidity and pollutant concentrations downstream. Marking: 1 mark for the higher peak, 1 for the earlier timing, 2 for water quality.

Section B: Channel Cross-section, Discharge and Velocity (17 marks)

A channel is 12 m wide with a mean depth of 1.5 m and a mean velocity of 0.80 m/s. Discharge equals cross-sectional area multiplied by velocity.

#	Question	Answer and marking notes
5	(a) Find the cross-sectional area of the channel. (3 marks)	Answer: Cross-sectional area = 12 x 1.5 = 18.0 m2. Marking: 1 mark for the formula, 1 for the value, 1 for the unit.
6	(b) Find the discharge of the channel. (4 marks)	Answer: Discharge = 18.0 x 0.80 = 14.40 m3/s. Marking: 1 mark for the formula, 1 for substitution, 1 for the value, 1 for the unit.
7	(c) Find the volume passing in one day in millions of cubic metres, taking one day as 86 400 s. (4 marks)	Answer: Daily volume = 14.40 x 86 400 = 1 244 160 m3 = 1.24 million m3. Marking: 2 marks for multiplying by the seconds, 1 for the unit conversion, 1 for the answer.
8	(d) After a storm the discharge rises to 21.6 m3/s with the same cross-section. Find the new mean velocity and explain why crossing the river is then dangerous. (6 marks)	Answer: Velocity = 21.60 / 18.0 = 1.2 m/s, 50 per cent higher than before. Depth and velocity both rise, so the kinetic energy and drag of the flow increase greatly; a person cannot stand steadily once the velocity exceeds about 1 m/s, and scour plus floating debris add further danger. Marking: 3 marks for the new velocity, 1 for the greater energy, 2 for the loss of footing.

Section C: Long Profile and Channel Gradient (17 marks)

A river is 60 km long, rising at 480 m above sea level and reaching the sea at 0 m. The first 10 km fall 320 m and the remaining 50 km fall to sea level.

#	Question	Answer and marking notes
9	(a) Find the average gradient of the whole river in m/km and as a percentage. (4 marks)	Answer: Average gradient = 480/60 = 8.0 m/km, that is 0.80 per cent. Marking: 2 marks for m/km, 2 for the percentage.
10	(b) Find the gradient of the upper and the lower section, and state how many times steeper the upper one is. (4 marks)	Answer: Upper = 320/10 = 32.0 m/km; lower = 160.0/50.0 = 3.2 m/km; the upper section is 10.0 times steeper. Marking: 1 mark for the upper, 2 for the lower, 1 for the ratio.
11	(c) Explain why the gradient decreases downstream. (4 marks)	Answer: Vertical erosion lowers the bed towards base level, wearing down the steep upper course; lower down, lateral erosion and deposition widen the valley and build up alluvium, flattening the bed, so the gradient falls. Marking: 2 marks for base level, 2 for lateral erosion and deposition.
12	(d) Explain how gradient affects the type of erosion, the energy of the river and the shape of the valley cross-profile. (5 marks)	Answer: A steep gradient gives high velocity and energy, so vertical incision and abrasion dominate, producing a narrow deep V-shaped valley with waterfalls and rapids. A gentle gradient gives low energy, so lateral erosion and deposition dominate, producing a wide shallow valley with a floodplain. Marking: 2 marks for the upper course, 2 for the lower course, 1 for linking energy to the type of erosion.

Section D: Meanders, Sinuosity and Ox-bow Lakes (17 marks)

A valley reach has a straight-line distance of 8.0 km while the channel itself is 11.2 km long. In one meander the concave bank has a velocity of 1.5 m/s and the convex bank 0.4 m/s.

#	Question	Answer and marking notes
13	(a) Calculate the sinuosity index of this reach. (4 marks)	Answer: Sinuosity = channel length / valley length = 11.2/8.0 = 1.40. Marking: 2 marks for the formula, 1 for the value, 1 for stating that it has no unit.
14	(b) Classify the degree of meandering and explain what the index implies for flooding and land use. (4 marks)	Answer: An index of 1.0 to 1.2 is straight, 1.2 to 1.5 is meandering and above 1.5 is highly sinuous, so this reach is moderately meandering. A sinuous channel has more friction and slower flow, so the flood peak moves downstream more slowly, but erosion on the outer bends threatens embankments and farmland and the longer channel takes up more land. Marking: 2 marks for the classification, 1 for flooding, 1 for land use.
15	(c) Find how many times faster the concave bank flows, and explain why the two banks behave differently. (4 marks)	Answer: 1.5/0.4 = 3.75 times. On a bend centrifugal force drives water to the outer bank, tilting the surface and setting up helical flow: surface water attacks the concave bank and cuts a river cliff, while water returns along the bed towards the convex bank, where slower flow deposits sand and gravel as a slip-off slope. Marking: 1 mark for the ratio, 2 for the centrifugal and helical flow, 1 for the two banks.
16	(d) Describe how an ox-bow lake forms and state its features on a floodplain. (5 marks)	Answer: A meander erodes its outer banks and deposits on its inner banks so the neck narrows; a flood cuts across the neck and leaves a straighter channel, and the old bend is sealed at both ends by deposition, so it stands as an ox-bow lake that slowly silts up into a marsh or hollow. On the floodplain it appears as a curved water body or curved hollow with abandoned meander scrolls and levees. Marking: 1 mark for the narrowing neck, 2 for the flood cut-off, 1 for the isolated lake, 1 for the landform features.

Section E: Erosion, Transport and Deposition (16 marks)

A river has a discharge of 15 m3/s and a suspended sediment concentration of 250 mg/L. Take 1 m3 = 1000 L and 1 tonne = 1000 kg.

#	Question	Answer and marking notes
17	(a) Find the mass of suspended sediment in one cubic metre of water in kg. (3 marks)	Answer: 250 mg/L x 1000 L = 250 000 mg = 0.25 kg/m3. Marking: 1 mark for litres to cubic metres, 1 for mg to kg, 1 for the answer.
18	(b) Find the mass of suspended sediment transported each second. (3 marks)	Answer: 0.25 x 15 = 3.75 kg/s. Marking: 2 marks for the multiplication, 1 for the unit.
19	(c) Find the total mass of suspended sediment carried in one day in tonnes. (5 marks)	Answer: 3.75 x 86 400 = 324 000 kg = 324.0 tonnes. A reservoir at the mouth would trap about 118 000 tonnes a year, enough to cut its storage substantially within decades. Marking: 2 marks for the seconds, 1 for tonnes, 1 for the value, 1 for the inference.
20	(d) Explain why sediment settles in order of grain size when velocity falls, and how a delta forms at a river mouth. (5 marks)	Answer: Coarse material such as pebbles and coarse sand needs a high velocity to be moved or held in suspension, so it settles first when velocity drops, while fine sand, silt and clay stay in suspension longer and settle further away, giving size sorting. Where a river enters a sea or lake its velocity falls sharply, so the load is dumped at one place; a subaqueous shoal builds up, emerges, splits the channel into distributaries and progrades seawards to form a delta. Marking: 3 marks for sorting, 2 for the delta.

Section F: Storm Runoff and Flood Management (17 marks)

A catchment of 25 km2 receives 100 mm of rain in 6 hours with a runoff ratio of 0.60. A detention basin of 200 000 m3 capacity is proposed.

#	Question	Answer and marking notes
21	(a) Find the total volume of rain falling on the catchment in millions of cubic metres. (3 marks)	Answer: Area = 25 x 1 000 000 = 25 000 000 m2; depth = 100 mm = 0.1 m; volume = 2 500 000 m3 = 2.5 million m3. Marking: 2 marks for the conversions, 1 for the answer.
22	(b) Find the total runoff generated by the storm. (4 marks)	Answer: Runoff = 0.60 x 2.5 = 1.5 million m3. Marking: 3 marks for applying the ratio, 1 for the unit.
23	(c) Assuming the runoff leaves evenly over 6 hours, find the mean discharge, and explain why the real flood peak is much higher. (5 marks)	Answer: 6 hours = 21 600 s; discharge = 1 500 000/21 600 = 69.4 m3/s. In a real storm the rain intensity is concentrated in a short burst, so the weighted peak can be several times the mean, and steep slopes with narrow channels deliver water to the trunk stream quickly. Marking: 3 marks for the mean discharge, 1 for the burst intensity, 1 for rapid delivery.
24	(d) Calculate the percentage of the runoff the basin can hold, judge whether it is adequate, and suggest two supporting measures. (5 marks)	Answer: 200 000/1 500 000 x 100% = 13.3 per cent, only about one eighth of the runoff, so it is clearly inadequate and can trim only a small part of the peak. Supporting measures include tree planting and terracing upstream to raise infiltration, restoring flood storage fields and wetlands, widening and clearing the channel, raising embankments with development control on the floodplain, and better drains with rain gardens. Marking: 2 marks for the percentage, 1 for the judgement, 2 for two measures.

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