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Topic 5 Practice Paper: Pressure — mark scheme

QuizQuipPractice Series
QQ/IGCSE-PHYS/T5

IGCSE · Cambridge IGCSE Physics (Extended (Core + Supplement))

Mark scheme

Pressure

Topic 5 Practice Paper · Maximum mark: 30

Notes for markers

  • Award one mark for each point marked [1], up to the marks shown for the part.
  • Accept answers worded differently that show the same understanding.
  • Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

QuestionAnswer and marking pointsMarks
Total for Question 11
Total for Question 21
Total for Question 31
Total for Question 41
Total for Question 51
Total for Question 61
Total for Question 71
Total for Question 81
Total for Question 91
Total for Question 101

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
11(a)

2 × 0.015 = 0.030 m². [1]

1
11(b)

p = 480 ÷ 0.030 = 16 000 Pa.

  • division 480 ÷ 0.030 [1]
  • 16 000 Pa [1]
2
11(c)

It doubles (to about 32 000 Pa): the same weight now acts through half the contact area, and p = F/A with A halved gives p doubled.

  • doubles [1]
  • half the area at the same force [1]
2
Total for Question 115
12(a)

p = 3600 ÷ 0.090 = 40 000 Pa.

  • substitution 3600 ÷ 0.090 [1]
  • 40 000 Pa [1]
2
12(b)

The force doubles to 7200 N at the same area, so the pressure doubles to 80 000 Pa.

  • force doubled [1]
  • 80 000 Pa [1]
2
12(c)

Area doubled with the same force halves the pressure to 20 000 Pa.

  • area doubled [1]
  • 20 000 Pa [1]
2
Total for Question 126
13(a)

A greater depth of water sits above the valve, so the column of water above it weighs more and pushes harder, increasing the pressure at the valve.

  • more water above the valve [1]
  • greater weight of the column gives greater pressure [1]
2
13(b)

Δp = ρgΔh = 1000 × 9.8 × 8.0 = 78 400 Pa ≈ 78 kPa.

  • substitution 1000 × 9.8 × 8.0 [1]
  • 78 400 Pa [1]
2
13(c)

1000 × 9.8 × 2.0 = 19 600 Pa ≈ 20 kPa — one quarter of the earlier value, in proportion to the depth.

  • 19 600 Pa [1]
1
Total for Question 135
14(a)

p = 2400 ÷ 0.0020 = 1 200 000 Pa = 1.2 MPa.

  • division 2400 ÷ 0.0020 [1]
  • 1 200 000 Pa [1]
2
14(b)

The small pad’s 1.2 MPa exceeds the 800 kPa threshold and would crack the plinth. The wide pad gives 2400 ÷ 0.60 = 4000 Pa (4 kPa), far below the threshold — choose the wide pad.

  • wide pad pressure 4000 Pa [1]
  • choice justified against the 800 kPa threshold [1]
2
Total for Question 144

— End of mark scheme —