CAIE A-Level Physics AS 4.3 Density and Pressure Questions

Practise calculating density, pressure and hydrostatic changes, explaining upthrust and applying Archimedes’ principle to floating or suspended bodies.

Exam points

use ρ = m/V to calculate or compare density, mass and volume for solids, liquids or composite materials
calculate pressure from force and area and use Δp = ρgΔh for a fluid column or pressure difference
explain hydrostatic pressure and derive or apply its dependence on density, gravitational field strength and depth
explain the origin of upthrust from the pressure difference across an immersed object
calculate upthrust with F = ρgV and apply Archimedes’ principle with force balance to floating or suspended bodies

Covered syllabus points

4.3.1—Density 4.3.2—Pressure 4.3.3—Derive, from the definitions of pressure and density, the equation 4.3.4—The equation ∆p = ρg∆h 4.3.5—The upthrust acting on an object in a fluid is due to a difference in 4.3.6—The upthrust acting on an object in a fluid using the equation F = ρgV

[Maximum number: 4]

(a)

A cylinder is suspended from the end of a string. The cylinder is stationary in water with the axis of the cylinder vertical, as shown in Fig. 2.1.

Fig. 2.1 (not to scale)

The cylinder has weight 0.84 N , height h and a circular cross-section of diameter 0.031 m . The density of the water is $1.0 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ . The difference between the pressures on the top and bottom faces of the cylinder is 520 Pa .

[ 4 ]

(i)

Calculate the height h of the cylinder.

h =m [2

[ 2 ]

$\begin{aligned} (\Delta) p=\rho g(\Delta) h 520=1000 \times 9.81 \times h \end{aligned}$ $h=0.053 \mathrm{~m}$

(ii)

Show that the upthrust acting on the cylinder is 0.39 N .

[ 2 ]

$\begin{aligned} (\text { upthrust } =)(\Delta) p \times A =(\Delta) p \times \pi(d / 2)^{2} \text { or }(\Delta) p \times \pi r^{2} \end{aligned}$ $=520 \times \pi(0.031 / 2)^{2}=0.39(\mathrm{~N})$

[Maximum number: 10]

A beaker in air contains a liquid. The base of the beaker is in contact with the liquid and has area A, as shown in Fig. 4.1.

Fig. 4.1

The liquid has density $\rho$ and fills the beaker to a depth h.

(a)

By using the definitions of pressure and density, show that

p=\rho g h

where p is the pressure due to the liquid that is exerted on the base of the beaker and g is the acceleration of free fall.

[ 3 ]

$\rho=m / V$ or $\rho=m / A h$
p=F / A or p=W / A
appropriate algebra leading to $p=\rho g h$

e.g. $p=\rho A h g / A$ or $\rho V g / A$ or $\rho V g /(V / h)$ and (so) $p=\rho g h$

(b)

Suggest why the equation in (a) does not give the total pressure on the base of the beaker.

[ 1 ]

(c)

Fig. 4.2 shows the variation of the total pressure inside the liquid with depth x below the surface.

Fig. 4.2

Determine the density of the liquid.
density = $\mathrm{kg} \mathrm{m}^{-3}$

[ 3 ]

$\Delta p=\rho g \Delta h$

e.g. $(9.66-9.60) \times 10^{4} / 8.0 \times 10^{-2}=\rho \times 9.81$ $\rho=760-770 \mathrm{~kg} \mathrm{~m}^{-3}$

(d)

A solid cylinder is held stationary by a wire so that the base of the cylinder is level with the surface of the liquid, as shown in Fig. 4.3.

Fig. 4.3 (not to scale)

The cylinder has length $4.0 \times 10^{-2} \mathrm{~m}$ and cross-sectional area $3.7 \times 10^{-4} \mathrm{~m}^{2}$ . The tension in the wire is 0.53 N .

The cylinder is now lowered and then held stationary by the wire so that the top of the cylinder is level with the surface of the liquid.

Calculate the new tension in the wire.
tension = N

[ 3 ]

$\begin{aligned} F =\rho g V =760 \times 9.81 \times 3.7 \times 10^{-4} \times 4.0 \times 10^{-2}(=0.11 \mathrm{~N}) \end{aligned}$
tension =0.53-0.11
$\text { = } 0.42 \mathrm{~N}$
or

$\begin{aligned} F =(\Delta) p \times A =(9.63-9.60) \times 10^{4} \times 3.7 \times 10^{-4}(=0.11 \mathrm{~N}) \end{aligned}$
tension =0.53-0.11
$=0.42 \mathrm{~N}$

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Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)