A solid cylinder of weight 24N is made of material of density 850kgm⁻³. The cylinder has a length of 0.18m, as shown in Fig. 1.1. [Figure 1.1]
- (a)Show that the cross-sectional area A of the cylinder is 0.016 m².[3]
- (b)The cylinder in (a) is attached by a spring to the bottom of a rigid container of liquid, as shown in Fig. 1.2. [Figure 1.2] The cylinder is in equilibrium with its bottom face at a depth of 0.17m below the surface of the liquid. The tension in the spring is 8.0N.
- (b)(i)Show that the upthrust acting on the cylinder due to the liquid is 32 Ν.[1]
- (b)(ii)Calculate the density of the liquid.[3]
- (c)Fig. 1.3 shows the variation of the tension F with the length of the spring in (b). [Figure 1.3]
- (c)(i)The tap at the bottom of the container is opened so that a fixed amount of liquid flows out of the container. The cylinder moves downwards so that the tension in the spring changes from 8.0N to 4.0N. Determine the change in the elastic potential energy of the spring.[3]
- (c)(ii)More liquid is let out of the container until the upthrust on the cylinder becomes 24 N. For the upthrust of 24 N, determine the length of the spring.[1]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [C1] m = ρV or ρAL
- [C1] W = mg
- [A1] (A =) 24 / (9.81 × 850 × 0.18) = 0.016 (m²)
- or
- [(C1)] P = F/A
- [(C1)] P = ρgh
- [(A1)] (A =) 24 / (9.81 × 850 × 0.18) = 0.016 (m²)
- [A1] (upthrust =) 24 + 8(.0) = 32 (N)
- [C1] (A)ρ = 32 / 0.016 (= 2000)
- [C1] (A)ρ = ρg(A)h
- [A1] ρ = 2000 / (9.81 × 0.17) = 1200 kg m⁻³
- [C1] E = ½Fx_or_E = ½kx²_or_E = area under graph
- [C1] (Δ)E = (½ × 8.0 × 0.40) – (½ × 4.0 × 0.20) or (½ × 20 × 0.40²) – (½ × 20 × 0.20²) or ½ × (4.0 + 8.0) × 0.20
- [A1] = 1.2 J
- [A1] length = 30 cm
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