CIE iGCSE Co-Ordinated Science P2.2.1 Thermal expansion of solids, liquids and gases Exam Style Questions Paper 4

(a)(i) Volume = \(6.50 \times 0.300 = 1.95 \, \text{cm}^3\)

Volume of a cylinder/cuboid-style solid here equals cross-sectional area multiplied by length.
\(1.95 \, \text{cm}^3\) confirms the value given in the question, showing the working clearly.

(a)(ii) 2.26 g/cm\(^3\)

Density is calculated using \(\rho = \dfrac{m}{V}\).
\(\rho = \dfrac{4.40}{1.95} = 2.256 \approx 2.26 \, \text{g/cm}^3\).

(b)(i) 1.5 V

In this single-loop circuit the voltmeter reads the same potential difference as the battery, since the graphite cylinder is the only component across it.

(b)(ii) 2.5 Ω

Resistance is found from \(R = \dfrac{V}{I}\).
\(R = \dfrac{1.5}{0.60} = 2.5 \, \Omega\).

(b)(iii) The two changes cancel out

Doubling the length doubles the resistance (longer path for charge carriers).
Doubling the cross-sectional area halves the resistance (more parallel pathways for current).
Since one effect doubles R while the other halves it, the overall resistance stays the same.

(c) Conduction by particle vibration and free electrons

In a solid, atoms vibrate about fixed positions, and these vibrations are passed on to neighbouring atoms, transferring thermal energy through the lattice.
In conductors such as graphite, free (delocalised) electrons also move through the structure, carrying thermal energy rapidly from hotter to cooler regions.