CIE iGCSE Co-Ordinated Science P4.1 Simple phenomena of magnetism Exam Style Questions Paper 4
Question
(a) An iron magnet picks up two iron nails as shown in Fig. 6.1.
Explain why the nails do not hang vertically.
▶️Answer/Explanation
Answer: ref. to induced magnetism (in) nails ; two nail heads / north poles / like poles, will repel each other
Detailed Explanation: 1. The magnet induces temporary magnetism in the iron nails 2. Each nail becomes a magnet with: – North pole at end nearest the magnet’s south pole – South pole at end nearest the magnet’s north pole 3. The two adjacent nail heads become like poles (both north or both south) 4. Like magnetic poles repel each other (N-N or S-S repulsion) 5. This causes the nails to splay outward rather than hang straight down
(b) An isotope of iron has a nuclide notation 6026Fe and decays by beta particle emission to an isotope of cobalt.
(i) State what is meant by the term isotope.
▶️Answer/Explanation
Answer: atoms having same atomic number / proton number and different mass number / neutron number
Key Characteristics: 1. Same element (same number of protons) 2. Different numbers of neutrons 3. Same chemical properties (same electron configuration) 4. Different physical properties (density, radioactivity, etc.) 5. Example: 56Fe, 57Fe, and 58Fe are all stable iron isotopes
(ii) Use nuclide notation to complete the symbol equation for this β-decay process.
▶️Answer/Explanation
Answer: 6026Fe → 6027Co + 0-1e
Beta Decay Process: 1. A neutron converts to a proton + electron (β⁻ particle) 2. Atomic number increases by 1 (Fe 26 → Co 27) 3. Mass number remains same (60) 4. The emitted electron is the beta particle (0-1e) 5. Typical half-life for 60Fe is 2.6 million years
(c) An iron wire of length 0.50 m has a cross sectional area of 4.0 × 10-5 m2 and a resistance of 1.21 × 10-3 Ω.
Calculate the resistance of an iron wire of length 0.25 m that has a cross sectional area of 8.0 × 10-5 m2.
▶️Answer/Explanation
Answer: 3.0 × 10-4 Ω
Calculation Steps: Using R = ρL/A (where ρ is resistivity, constant for iron): 1. Original wire: R₁ = ρ(0.50)/(4.0×10-5) = 1.21×10-3 Ω 2. New wire has: – Half the length (0.25 m → decreases resistance) – Double the area (8.0×10-5 m² → decreases resistance) 3. Resistance changes by factor of (½)/(2) = ¼ 4. New resistance R₂ = ¼ × 1.21×10-3 = 3.025×10-4 Ω 5. Round to 2 significant figures: 3.0 × 10-4 Ω
(d) A block of iron is on a bench. The surface of the block of iron in contact with the bench has an area of 144 cm2. The mass of the block of iron is 13.6 kg. Calculate the pressure exerted by the block of iron on the bench in N/cm2. gravitational field strength = 10 N / kg
▶️Answer/Explanation
Answer: 0.94 N/cm2
Calculation: 1. Weight = mass × g = 13.6 kg × 10 N/kg = 136 N 2. Contact area = 144 cm2 3. Pressure = Force/Area = 136 N / 144 cm2 4. = 0.944… N/cm2 5. Round to 2 significant figures: 0.94 N/cm2 Key Points: – Pressure is force per unit area – 1 N/cm2 = 10,000 Pa (SI units) – Iron’s density (7.87 g/cm3) means this block has volume ≈ 1727 cm3