CAIE IGCSE Physics 1.8 Pressure Questions

Covered syllabus points

1.8.1—Pressure as force per unit area 1.8.2—Pressure varies with force and area in 1.8.3—Qualitatively how the pressure beneath 1.8.4—Recall/use: for the change in pressure

[Maximum number: 2]

A car has a weight of 13000 N . The car is supported by 4 tyres. The area of each tyre in contact with the road is $0.016 \mathrm{~m}^{2}$ .

Calculate the pressure on the road due to the weight of the car.

pressure =

$2.0 \times 10^{5} \mathrm{~Pa}$ OR 200000 Pa OR 200 kPa
( P= ) F / A OR $13000 /(0.016 \times 4)$

[Maximum number: 7]

(a)

Fig. 3.1 shows a person moving across an ice-covered pond to reach a ball on the ice.

Fig. 3.1

Explain why this way of moving across the ice is safer than walking. Use your understanding of pressure in your answer.

[ 3 ]

(force of gravity / weight of person is spread over a much) greater area
$p=F / A \mathbf{O R} p \propto 1 / A$
(force is same so) pressure is lower (so ice is less likely to crack)

(b)

Fig. 3.2 shows a side view of the pond with a layer of ice floating freely on the water.

Fig. 3.2

The surface area of the pond is $5.0 \mathrm{~m}^{2}$ .
The mass of the ice is 690 kg.
The density of water is $1000 \mathrm{~kg} / \mathrm{m}^{3}$ .
Point X is 0.45 m below the ice.
Calculate the pressure at point X due to the ice and the water.
pressure =

[ 4 ]

$5.8 \times 10^{3} \mathrm{~Pa}$
p (due to water) $=\rho g h \mathbf{~} \mathbf{R}(p=) \rho g h \mathbf{~} \mathbf{R}(p=) 1000 \times 9.8 \times 0.45 \mathbf{~ OR}(p=) 4410$ $W=m g \mathbf{OR}(W=) m g \mathbf{OR}(W=)(690 \times 9.8) \mathbf{OR}(W=) 6762 \mathbf{OR}(p($ due to ice )=) 1352.4
(pressure =) candidate's calculated pressure due to water + candidate's calculated pressure due to ice

OR total pressure $=[1000 \times 9.8 \times 0.45]+[(690 \times 9.8) / 5.0]$

OR total pressure =4410+1352.4

[Maximum number: 1]

Which row describes how the pressure beneath the surface of a liquid depends on the depth below the surface and on the density of the liquid?

depth below surface

density of liquid

pressure decreases with increasing depth

pressure decreases with increasing density

pressure decreases with increasing depth

pressure increases with increasing density

pressure increases with increasing depth

pressure decreases with increasing density

pressure increases with increasing depth

pressure increases with increasing density

[Maximum number: 3]

Fig. 3.1 shows a person moving across an ice-covered pond to reach a ball on the ice.

Fig. 3.1

Explain why this way of moving across the ice is safer than walking. Use your understanding of pressure in your answer.

(force of gravity / weight of person is spread over a much) greater area
$p=F / A \mathbf{O R} p \propto 1 / A$
(force is same so) pressure is lower (so ice is less likely to crack)

[Maximum number: 2]

Fig. 4.1 shows a bottle part-filled with water. The air inside the bottle is at the same pressure as the air outside the bottle. The bottle and its contents are at room temperature.

Fig. 4.1

Another plastic bottle is filled to the top with water. The height of the bottle is 40.0 cm . The density of water is $1.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ .

Calculate the pressure difference between the top and bottom of the water.

pressure difference =

3900 Pa
$\begin{aligned} (\Delta p=) \rho \mathrm{g}(\Delta) h \text { OR }(\Delta p=) 1.0 \times 10^{3} \times 9.8 \times 0.4 \text { OR } (\Delta p=) 1.0 \times 10^{3} \times 9.8 \times 40 \text { OR }(\Delta p=) 3.9 \times 10^{\mathrm{N}} \end{aligned}$

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CAIE IGCSE Physics 1.8 Pressure Topic Practice

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question 5