A-LevelPhysicsForces, density and pressure9702/52 · Oct/Nov 2020 · Q215 Marks
A student investigates the behaviour of a liquid inside a narrow tube, as shown in Fig. 2.1. d h liquid [Figure 2.1] (not to scale) When the tube is placed in the liquid, the liquid rises in the tube. The student measures the internal diameter d of the tube and the maximum height h that the liquid rises in the tube. The experiment is repeated for tubes of different diameter. It is suggested that h and d are related by the equation h = 4σ / (dρg) where ρ is the density of the liquid, g is the acceleration of free fall and σ is a constant.
- (a)A graph is plotted of h on the y-axis against 1/d on the x-axis. Determine an expression for the gradient.[1]
- (b)Values of d and h are given in Table 2.1. Table 2.1 | d/mm | 1/d /mm⁻¹ | h/mm | |-----------|-----------|------| | 1.1 ± 0.1 | | 18.3 | | 1.3 ± 0.1 | | 15.1 | | 1.5 ± 0.1 | | 13.1 | | 1.7 ± 0.1 | | 11.6 | | 2.0 ± 0.1 | | 9.9 | | 2.3 ± 0.1 | | 8.6 | Calculate and record values of 1/d /mm⁻¹ in Table 2.1. Include the absolute uncertainties in 1/d[2]
- (c)(i)Plot a graph of h/mm against 1/d /mm⁻¹. Include error bars for 1/d[2]
- (c)(ii)Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled.[2]
- (c)(iii)Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.[2]
- (d)(i)The student determines the density of the liquid by weighing the empty beaker, adding a measured volume of the liquid and then reweighing the beaker. The student then uses the relationship density = mass / volume The student's results are: mass of empty beaker = (422 ± 1)g mass of beaker and liquid = (606 ± 1)g volume of liquid = (146 ± 2)cm³. Determine the density ρ of the liquid.[1]
- (d)(ii)Determine the percentage uncertainty in ρ.[1]
- (e)Using your answers to (a), (c)(iii) and (d), determine the value of σ. Include an appropriate unit and the absolute uncertainty in your answer. Data: g = 9.81 ms⁻²[3]
- (f)The experiment is repeated with a different tube. The height h is measured to be 22.0mm. Determine the diameter d of this tube.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] gradient = 4σ / ρg
- [1] (1/d)/mm⁻¹ values: 0.91 or 0.909, 0.77 or 0.769, 0.67 or 0.667, 0.59 or 0.588, 0.50 or 0.500, 0.43 or 0.435
- [1] Absolute uncertainties in 1/d from ± (0.08 or 0.09) to ± (0.01 or 0.02).
- [1] Six points plotted correctly. Must be accurate to nearest half a small square. Diameter of points must be less than half a small square.
- [1] Error bars in 1/d plotted correctly. All error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
- [1] Line of best fit drawn. Points must be balanced. Do not accept line from top to bottom point. Line must pass between (0.52, 10.5) and (0.55, 10.5) and between (0.84, 17.0) and (0.87, 17.0).
- [1] Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted.
- [1] Gradient determined with clear substitution of data points into Δy / Δx. Distance between data points must be at least half the length of the drawn line.
- [1] Gradient of worst acceptable line determined.
- [1] uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
- [1] ρ = (0.606 – 0.422) / (146×10⁻⁶) = 0.184 / (146×10⁻⁶) = 1260 kg m⁻³ and given to three or four significant figures.
- [1] % uncertainty in ρ = % uncertainty in m + % uncertainty in V ((2/184) + (2/146)) ×100 = 1.087 + 1.37 = 2.5% or using max ρ = 186 / 144 = 1292 and/or min ρ = 182 / 146 = 1230
- [1] σ determined using gradient with correct substitution shown. σ = (ρg × gradient) / 4 or (d)(i)×9.81×(c)(iii) / 4
- [1] σ determined using gradient and correct SI unit given (N m⁻¹ or kg s⁻²).
- [1] Absolute uncertainty in σ determined with correct substitution shown. uncertainty = ((d)(ii)/100) + (Δgradient/gradient) × σ or uncertainty = ((2/184) + (2/146) + (Δgradient/gradient)) × σ or using max σ = max(d)(i)×9.81×max(c)(iii) / 4 or using min σ = min (d)(i)×9.81×min (c)(iii) / 4
- [1] d determined to a minimum of two significant figures. d = gradient / h or d = (4 x (e)(i)) / (h x (d)(i) x 9.81)
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