A liquid of density ρ fills a container to a depth h, as shown in Fig. 2.1. [Figure 2.1] The base of the container has area A. (c) The liquid in the container is liquid L. Liquid M is now added to the container. The two liquids do not mix. The total depth of the liquids is 0.17 m. Fig. 2.2 shows how the pressure p inside the liquids varies with height x above the base of the container. [Figure 2.2] Use Fig. 2.2 to
- (a)Derive, from the definitions of pressure and density, the equation p = ρgh where p is the pressure exerted by the liquid on the base of the container and g is the acceleration of free fall.[3]
- (b)(i)State 1. the names of the three forces acting on the sphere, 2. a word equation that relates the magnitudes of these forces.[2]
- (b)(ii)State and explain the changes in energy that occur as the sphere falls.[2]
- (c)(i)state the value of atmospheric pressure, atmospheric pressure = _______________ Pa[1]
- (c)(ii)determine the density of liquid M. density = _______________ kgm⁻³[2]
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Mark scheme
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- 2(a) p = m / V or p = m / Ah B1 p = F / A or p = W / A B1 p = [pAhg] / A or p = [pVg] / [V/ h] (so) p = pgh A1
- 2(b)(i) 1. weight/gravitational (force) B1 upthrust (force)/buoyancy (force) drag/viscous/frictional (force)/fluid resistance/resistance 2. weight = upthrust + viscous (force) B1
- 2(b)(ii) α decrease in (gravitational) potential energy (of sphere) due to decrease in height (since E_p = mgh) B2 ∞ increase in thermal energy due to work done against viscous force/drag ∞ loss/change of (total) E_p equal to gain/change of thermal energy Any 2 points.
- 2(c)(i) atmospheric pressure = 9.1(0) × 10⁴ Pa A1
- 2(c)(ii) (Δ)p = ρg(Δ)h C1 (9.15 – 9.10) × 10⁴ = ρ × 9.81 × (0.17 – 0.10) ρ = 730 (728) kg m⁻³ A1
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