Home / iGCSE Physics (0625) 1.8 Pressure Paper 3 -Exam Style Questions- New Syllabus

iGCSE Physics (0625) 1.8 Pressure Paper 3 -Exam Style Questions- New Syllabus

Question

Fig. 5.1 shows a metal block at room temperature on a table.
(a) Describe the arrangement, separation and motion of the particles in the metal block.
(b) (i) The temperature of the metal block decreases. Describe any changes in the motion and separation of the particles in the metal block.
(ii) A scientist cools the metal block until its temperature is close to absolute zero. Describe the motion of the particles in the metal block.
(c) The weight of the metal block is 26N. The area of the metal block in contact with the table is \(42cm^2\). Calculate the pressure on the table due to the metal block.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

Topic 2.1.2 — Particle model (Parts (a), (b)(i), (b)(ii))
Topic 1.8 — Pressure (Part (c))

▶️ Answer/Explanation

(a)
For the correct answer:
(particles are) fixed in position / in lattice OR regular / fixed arrangement / pattern
can only vibrate / no translational KE
close / closer (than in liquids or gases)

In a solid metal block, the particles are arranged in a regular, repeating pattern known as a lattice. They are held closely together by strong forces of attraction, which restricts their movement to small vibrations about their fixed positions; they possess no translational kinetic energy.

(b)(i)
For the correct answer:
(particles move) closer (as temperature decreases)
particles vibrate slower / less OR have smaller vibrations

As the temperature of the metal block decreases, the average kinetic energy of its particles reduces. This causes the particles to vibrate with less speed and amplitude, and their average separation distance decreases slightly as the block undergoes thermal contraction.

(b)(ii)
For the correct answer:
(at absolute zero particles have) least / smallest vibrations

Absolute zero ($-273^{\circ}\text{C}$ or $0\text{ K}$) is the lowest possible theoretical temperature. At this limit, particles within the metal block would possess the minimum possible kinetic energy and would exhibit only their least, or smallest, vibrations.

(c)
For the correct answer:
(P =) \(0.62 (\text{N} / \text{cm}^2)\)
(P =) 26 ÷ 42
(P =) F ÷ A

Pressure is defined as the force acting per unit area. Using the formula $P = \frac{F}{A}$, substitute the weight of the block for the force ($F = 26\text{ N}$) and the given contact area ($A = 42\text{ cm}^2$). The calculation yields $P = \frac{26}{42} = 0.619\text{ N/cm}^2$, which rounds to $0.62\text{ N/cm}^2$.

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