Fig. 10.1 shows a wire of length l and cross-sectional area A. [Figure 10.1]

1. (a)(i)State how the resistance of the wire in Fig. 10.1 depends on: l[1]
2. (a)(ii)A.[1]
3. (b)The cross-sectional area of a piece of metal wire is 7.5 × 10⁻⁴ cm². The resistance of a 1.0 m length of the same wire is 6.4 Ω. The wire is made from metal W. Fig. 10.2 shows a solid cube of side 1.0 cm. It is also made from metal W. [Figure 10.2] Calculate the resistance between the two opposite faces A and B of the cube.[2]
4. (c)(i)The wire in part (b) is taped to a metre rule. Fig. 10.3 shows that a 1.0m length of the wire (resistance 6.4Ω) is connected in series with a switch, a cell of electromotive force (e.m.f.) 1.2V and a resistor of resistance 9.6 Ω. [Figure 10.3] The switch is closed. Explain what is meant by electromotive force (e.m.f.).[2]
5. (c)(ii)Calculate the potential difference (p.d.) across the 1.0m length of the wire.[3]
6. (c)(iii)One input terminal of an oscilloscope is connected to the wire at point P, the 0cm mark of the metre rule. The other terminal of the oscilloscope is connected to a sliding contact. Initially, this contact touches the wire at point P. The Y-gain setting on the oscilloscope is 0.20V/cm. Fig. 10.4 shows the screen of the oscilloscope with a horizontal trace across the middle of the screen. [Figure 10.4] The sliding contact is slowly moved along the wire until it reaches the other end of the metre rule. Describe and explain what happens to the trace on the screen.[3]
7. (d)(i)A second, identical 1.2V cell is connected in parallel with the cell in the circuit in Fig. 10.3. State one advantage of using two cells in parallel rather than a single cell.[1]
8. (d)(ii)State and explain the effect on the trace in (c)(iii) of adding the second cell in parallel.[2]

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