MYP Physics · unit 19
Physics in society
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
An electrostatic sprayer and an unintended spark
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A conducting sprayer nozzle behaves as a 100 pF capacitor relative to its surroundings. The supplied charge–potential data describe it. It operates at +20 kV and transfers 2.0 nC to each paint droplet. Assume identical droplets. A grounding connection can discharge the nozzle; no calculation here establishes a safe ignition threshold. Supplied model: Use Q = CV and stored electrostatic energy U = ½QV. Supplied data (Potential / kV; Charge / μC): (0, 0), (10, 1), (20, 2). (a) Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge. [3 marks] (b) Explain why charged paint is attracted to a grounded conductor and how grounding differs from insulation as risk control. [3 marks]
Nozzle charge versus potential
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Charge / μC
Horizontal axis (x): Potential / kV
Read the exact data as a table
| Potential / kV | Charge / μC |
|---|---|
| 0 | 0 |
| 10 | 1 |
| 20 | 2 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge.
Reveal the worked solution
- The graph has C = Q/V = (1.0×10⁻⁶)/(1.0×10⁴) = 100 pF; at 20 kV charge is 2.0×10⁻⁶ C.
- Stored energy is ½QV = ½(2.0×10⁻⁶)(20000) = 0.020 J. The factor ½ accounts for the potential rising during charging.
- Q/qdrop = (2.0×10⁻⁶)/(2.0×10⁻⁹) = 1000 droplets if that charge were transferred with no replenishing supply.
- A positive droplet induces negative surface charge on the nearby conducting workpiece. Grounding lets electrons flow from Earth, maintaining a suitable potential and sustained attraction.
- Conductors allow mobile charge to redistribute; an insulator can retain local deposited charge. Static charge is not necessarily a continuous electric current.
- Bonding and grounding reduce charge accumulation and potential differences. Insulation can prevent contact current but may retain static charge, so its role must match the actual discharge or ignition risk.
Answer (a) Q = 2.0 μC; stored energy = 0.020 J; equivalent droplet count = 1000. (b) Charge redistribution produces attraction; grounding provides a discharge path while insulation restricts charge flow.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A household fault that a fuse may not detect
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 230 V appliance normally uses 1150 W and has a metal case connected to earth. In a hypothetical insulation fault, the effective live-to-earth path resistance becomes 46 Ω. In a separate scenario with direct contact with live or a failed protective-earth connection, model the person-to-earth path as 2300 Ω with 230 V across it. This is not the voltage across a person touching a correctly earthed case. A fitted fuse is rated 13 A, and an RCD trips when live–neutral current imbalance exceeds 30 mA. These are simplified fault calculations, not instructions for live testing. Supplied data (Path resistance / Ω; Current / A): (46, 5), (230, 1), (2300, 0.1). (a) Find normal current, the 46 Ω fault current and the modelled person-path current. Explain why the fuse rating alone is insufficient protection. [3 marks] (b) Explain how earthing, double insulation, a fuse and an RCD address different failure conditions. [3 marks]
Hypothetical fault current versus path resistance
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Current / A
Horizontal axis (x): Path resistance / Ω
Read the exact data as a table
| Path resistance / Ω | Current / A |
|---|---|
| 46 | 5 |
| 230 | 1 |
| 2300 | 0.1 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find normal current, the 46 Ω fault current and the modelled person-path current. Explain why the fuse rating alone is insufficient protection.
Reveal the worked solution
- Normal operating current is P/V = 1150/230 = 5.0 A.
- The fault path would carry 230/46 = 5.0 A, while the person-path model gives 230/2300 = 0.10 A = 100 mA.
- Neither additional path alone exceeds a 13 A fuse rating, and fuses have time–current characteristics. A potentially harmful earth leakage can therefore exist without prompt fuse operation.
- A protective earth conductor links accessible metal to earth and can carry fault current. Double-insulated equipment uses separate insulation barriers and suitable construction instead of an exposed earthed case.
- A fuse or overcurrent breaker protects the circuit against excessive current, with an operating time that depends on current; it does not directly measure a small body current.
- An RCD compares live and neutral currents. A 100 mA earth leakage exceeds the stated 30 mA threshold, but trip timing and the complete installation still matter. Classroom calculations cannot establish that contact is safe.
Answer (a) Normal current = 5.0 A; earth-fault current = 5.0 A; person-path current = 0.10 A. (b) Earthing provides a fault path; insulation restricts contact; a fuse interrupts excessive current; an RCD detects imbalance to earth.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A motor coil that must keep turning
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A rectangular coil has 20 turns and area 0.010 m² in a uniform 0.20 T field. It carries 2.0 A. The supplied graph gives torque versus sinθ, where θ is the angle between the coil's area normal and the magnetic field. Use τ = NBIA sinθ. Supplied data (sin θ; Torque / N m): (0, 0), (0.5, 0.04), (1, 0.08). (a) Find the maximum torque, torque at θ = 30°, and mechanical power when the instantaneous torque is 0.060 N m at 50 rad s⁻¹. [3 marks] (b) Explain why a DC motor needs current reversal and assess one energy-loss mechanism and one useful application trade-off. [3 marks]
Coil torque calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Torque / N m
Horizontal axis (x): sin θ
Read the exact data as a table
| sin θ | Torque / N m |
|---|---|
| 0 | 0 |
| 0.5 | 0.04 |
| 1 | 0.08 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find the maximum torque, torque at θ = 30°, and mechanical power when the instantaneous torque is 0.060 N m at 50 rad s⁻¹.
Reveal the worked solution
- The torque factor is NBIA = 20(0.20)(2.0)(0.010) = 0.080 N m.
- At θ = 30°, multiply by sin30° = 0.50 to obtain 0.040 N m. Torque is greatest when the normal is perpendicular to the field.
- Instantaneous mechanical power is τω = 0.060(50) = 3.0 W. Opposite forces on opposite sides form a couple, even if the net force on the whole coil is zero.
- Without reversal, the magnetic couple would reverse after half a turn and oppose continued rotation. Inertia carries the coil through positions of zero torque.
- Electrical input is divided between mechanical output and losses such as I²R heating, bearing friction and magnetic losses; power output cannot exceed input.
- Increasing current increases torque but raises resistive heating quadratically. A lifting motor needs sufficient torque while meeting temperature and efficiency constraints.
Answer (a) Maximum torque = 0.080 N m; at 30° = 0.040 N m; mechanical power = 3.0 W. (b) A split-ring commutator reverses coil current each half-turn, maintaining torque in the rotation direction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Choosing a tracer with physical and biological clearance
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A diagnostic tracer has physical half-life 6 h and biological half-life 12 h in the relevant organ. Assume the two removal processes are independent exponentials and initial organ activity is 80 MBq. The supplied graph gives the expected organ activity for the combined removal. Use 1/Teffective = 1/Tphysical + 1/Tbiological. Supplied data (Time / h; Activity / MBq): (0, 80), (4, 40), (8, 20), (12, 10). (a) Find effective half-life and expected organ activity after 12 h. Explain why biological removal does not change a nucleus's physical decay law. [3 marks] (b) Evaluate replacing this tracer by one with the same biological behaviour but physical half-life 30 years. Name the evidence needed for a patient-benefit decision. [3 marks]
Expected organ activity
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Activity / MBq
Horizontal axis (x): Time / h
Read the exact data as a table
| Time / h | Activity / MBq |
|---|---|
| 0 | 80 |
| 4 | 40 |
| 8 | 20 |
| 12 | 10 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find effective half-life and expected organ activity after 12 h. Explain why biological removal does not change a nucleus's physical decay law.
Reveal the worked solution
- 1/Teff = 1/6+1/12 = 1/4 h⁻¹, so Teff = 4 h.
- Twelve hours is three effective half-lives, giving 80/2³ = 10 MBq remaining in the organ.
- Biological clearance moves radionuclides out of the measured organ or body; it does not alter their intrinsic decay constant. Cleared material can still be radioactive elsewhere.
- A useful diagnostic tracer should remain measurable for preparation and imaging, then have limited unnecessary retention and exposure; 'shortest possible' is not automatically best if it decays before use.
- The emitted radiation must escape sufficiently for detection, while energy, branching ratio, chemistry and organ uptake determine image quality and dose.
- Compare expected diagnostic benefit, administered activity, absorbed and effective dose estimates, alternatives, logistics and patient-specific circumstances. Activity in MBq alone is not a complete measure of risk or image quality.
Answer (a) Effective half-life = 4 h; organ activity after 12 h = 10 MBq. (b) A very long physical half-life may leave unnecessary prolonged activity; suitability also depends on emissions, uptake and diagnostic performance.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.