Density and pressure: past paper questions
Real questions on density and pressure
The newest 20 of 50 written questions. Open one to see it on the real paper.
- 9702/21 · October/November 2025 · Q2a(i)[1]
Define pressure.
See it on the paper → - 9702/21 · October/November 2025 · Q2a(ii)[2]
Explain how hydrostatic pressure results in an upthrust force acting on a solid object immersed in a liquid.
See it on the paper → - 9702/21 · October/November 2025 · Q2b(i)[3]
On Fig. 2.1, draw labelled arrows from the ball to show the directions of the three forces that act on the ball as it falls. (i)
See it on the paper → - 9702/21 · October/November 2025 · Q2c(i)[1]
Show that the upthrust force acting on the ball is 2.8 × 10 –3 N.
See it on the paper → - 9702/21 · October/November 2025 · Q2c(ii)[3]
Determine the terminal speed v of the ball. v = m s –1
See it on the paper → - 9702/23 · October/November 2025 · Q2a(i)[1]
Define pressure.
See it on the paper → - 9702/23 · October/November 2025 · Q2a(ii)[2]
Explain how hydrostatic pressure results in an upthrust force acting on a solid object immersed in a liquid.
See it on the paper → - 9702/23 · October/November 2025 · Q2b(i)[3]
On Fig. 2.1, draw labelled arrows from the ball to show the directions of the three forces that act on the ball as it falls. (i)
See it on the paper → - 9702/23 · October/November 2025 · Q2c(i)[1]
Show that the upthrust force acting on the ball is 2.8 × 10 –3 N.
See it on the paper → - 9702/23 · October/November 2025 · Q2c(ii)[3]
Determine the terminal speed v of the ball. v = m s –1
See it on the paper → - 9702/21 · May/June 2025 · Q3c(i)[1]
the air resistance acting on the car
See it on the paper → - 9702/23 · May/June 2025 · Q3c(i)[1]
the air resistance acting on the car
See it on the paper → - 9702/22 · February/March 2025 · Q2b(ii)[1]
The bottom of the cylinder is submerged in the water to depth y as shown in Fig. 2.2. The beam is still attached to the cylinder but not shown. force from beam, 1300 N cylinder, mass 11 kg y 0.78 m water Fig. 2.2 (not to scale) The cylinder has mass 11 kg and diameter 0.78 m. The beam exerts a vertical force of 1300 N...
See it on the paper → - 9702/22 · February/March 2025 · Q2b(iii)[2]
Calculate the depth y. The water has density 990 kg m –3 . y = m
See it on the paper → - 9702/22 · February/March 2025 · Q2b(iv)[2]
The person can stand anywhere between A and B. On Fig. 2.3, sketch the variation of the depth of the bottom of the cylinder with the distance of the person from A, for distances between 0 and 6.0 m. Numerical values are not required. depth / m 0 0 distance from A / m 6.0
See it on the paper → - 9702/21 · October/November 2024 · Q1a[1]
Define density.
See it on the paper → - 9702/21 · October/November 2024 · Q1b(i)[2]
Determine the density of the glass. density = kg m –3
See it on the paper → - 9702/21 · October/November 2024 · Q1b(ii)[3]
Calculate the percentage uncertainty in the density. percentage uncertainty = %
See it on the paper → - 9702/21 · October/November 2024 · Q1c[1]
The true value of the density of the glass is different from the answer in (b)(i) because of a systematic error in the measurements. Suggest one possible cause of this systematic error.
See it on the paper → - 9702/22 · February/March 2024 · Q2c(ii)[5]
The diver has a volume of 7.5 × 10 –2 m 3 . The density of the water is 1.0 × 10 3 kg m –3 . Show that the upthrust acting on the diver when he is entirely underwater is 740 N.
See it on the paper →
What examiners said
The majority of candidates equated the upthrust with the weight of the cylinder and the force of the beam on the cylinder. Some candidates did not give the required full working for the weight, jumping directly to weight = 100 N without explanation. Weaker candidates tried to use the given equation for the upthrust (F = gV) without success.
— Examiner report, February/March 2025, 9702/22 Q2(b)(ii) (Cambridge)
The stronger candidates used the expression for the upthrust and a correct expression for the volume of a cylinder to determine the length y. A significant number used the force from the beam (1300 N) and not the upthrust given (1400 N). Some candidates confused pressure with force and so wrongly stated that 1400 = 990 9.81 y. Others misused the density equation by confusing the mass of the cylind...
— Examiner report, February/March 2025, 9702/22 Q2(b)(iii) (Cambridge)
It was generally understood that the graph line would have a positive gradient. However, weaker candidates often made the mistake of drawing their graph line from the origin and did not appreciate that the graph would have a non-zero value of depth when the distance from A is zero, as there needs to be an upthrust due to the weight of the beam. Although the question explicitly stated the required...
— Examiner report, February/March 2025, 9702/22 Q2(b)(iv) (Cambridge)
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