AI-assisted English content review; independent teacher approval pending.
Practice scope and review
Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
69 original questions across 13 selected topic areas. No account or payment required to practise.
Cambridge International AS & A Level Physics (9702), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.
4 questions in this topic
This is a limited practice preview and is not submitted for search indexing.
All topics
Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Drift and sensor dividers · Hall voltage and velocity selection · Rectification and smoothing · First-law work convention
4 of 4 questions shown
Loading saved responses
Question 1A wire carries 1.60 A with carrier density 8.00 x 1028m−3, area 1.00 x 10−6m2 and charge magnitude 1.60 x 10−19 C per carrier. Find drift-speed magnitude. Separately, a 6.00 V ideal divider has a 2.00 kilohm upper resistor and a lower NTC thermistor that falls from 4.00 to 1.00 kilohm on warming. Find both output voltages measured across the thermistor.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A wire carries 1.60 A with carrier density 8.00 x 1028m−3, area 1.00 x 10−6m2 and charge magnitude 1.60 x 10−19 C per carrier. Find drift-speed magnitude. Separately, a 6.00 V ideal divider has a 2.00 kilohm upper resistor and a lower NTC thermistor that falls from 4.00 to 1.00 kilohm on warming. Find both output voltages measured across the thermistor.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked I=Anvq and both explicitly supported NTC-divider fractions; area, number density and charge units give speed, not mobility. This is conceptual placement, not completion of the entire referenced subsection.
Anq = (1.00e-6)(8.00e28)(1.60e-19) = 12800; divide 1.60 by this factor. Drift speed is not the speed at which a circuit signal propagates.
Use Vout = Vsupply Rlower/(Rupper+Rlower) for each thermistor resistance, with no load attached. The increasing current does not imply increasing voltage across a falling fraction of the total resistance.
Self-marking points
View marking points
v = I/(Anq) = 1.25 x 10−4 m s−1.
Initial output = 6.00 x 4/(2+4) = 4.00 V.
Warm output = 6.00 x 1/(2+1) = 2.00 V.
The output across this lower NTC decreases on warming; the location of the output terminals matters.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2A Hall specimen has thickness 1.00 x 10−4 m, carrier density 5.00 x 1028m−3 and carrier charge magnitude 1.60 x 10−19 C. With current 2.00 A and perpendicular B = 0.500 T, calculate Hall-voltage magnitude using VH = BI/(ntq). Charged particles travel perpendicular to both crossed fields E = 3.00 x 104 V m−1 and B = 0.500 T, with electric and magnetic forces in opposite directions, and pass undeflected. Find the selected speed and explain the balance.3 indicative marks
Calculate and explain3 indicative marksCalculator allowed
A Hall specimen has thickness 1.00 x 10−4 m, carrier density 5.00 x 1028m−3 and carrier charge magnitude 1.60 x 10−19 C. With current 2.00 A and perpendicular B = 0.500 T, calculate Hall-voltage magnitude using VH = BI/(ntq). Charged particles travel perpendicular to both crossed fields E = 3.00 x 104 V m−1 and B = 0.500 T, with electric and magnetic forces in opposite directions, and pass undeflected. Find the selected speed and explain the balance.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Verified Hall microvolt scale and velocity selection from force balance; both are explicit 20.3 requirements, unlike treating all magnetic-field questions as identical. This is conceptual placement, not completion of the entire referenced subsection.
The Hall denominator is (5.00e28)(1.00e-4)(1.60e-19) = 8.00e5, so VH = 1.00/(8.00e5) V. A sign would require carrier type and orientations not supplied.
For the selector choose directions such that the two transverse forces oppose. Their magnitudes cancel at v = E/B; cancellation of charge magnitude makes this speed independent of the particle's charge magnitude.
Self-marking points
View marking points
Hall voltage magnitude = 1.25 x 10−6 V.
Undeflected particles require electric and magnetic forces equal and opposite: qE = qvB.
Selected speed = E/B = 6.00 x 104 m s−1.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3A sinusoidal source of peak voltage 12.0 V drives a 24.0 ohm resistor directly. Calculate rms voltage and mean power. Describe qualitatively the load waveform after an ideal full-wave bridge rectifier, and the effect of increasing a parallel smoothing capacitor at fixed load resistance. Ignore diode drops; no practical construction is requested.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A sinusoidal source of peak voltage 12.0 V drives a 24.0 ohm resistor directly. Calculate rms voltage and mean power. Describe qualitatively the load waveform after an ideal full-wave bridge rectifier, and the effect of increasing a parallel smoothing capacitor at fixed load resistance. Ignore diode drops; no practical construction is requested.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked RMS/mean-square distinction and qualitative bridge/smoothing behaviour; actual waveform sketch and electrical handling remain unassessed. This is conceptual placement, not completion of the entire referenced subsection.
The unrectified average voltage is zero but mean squared voltage is 12.02/2 = 72.0 V2. Divide by 24.0 ohm for heating power.
An ideal bridge routes current through the load in the same direction in both source half-cycles. A capacitor charges near peaks and supplies the load between them; greater RC reduces the fractional drop, not an increase in source frequency.
Self-marking points
View marking points
Vrms = 12.0/sqrt(2) = 8.49 V.
Mean power = Vrms^2/R = 3.00 W.
Full-wave rectification reverses negative half-cycles so both halves have the same output polarity.
A larger parallel capacitor discharges more slowly between peaks, reducing ripple at fixed load.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 4A gas absorbs 500 J by heating and expands by 2.00 x 10−3m3 against constant pressure 1.00 x 105 Pa. Find work done by the gas, work done on it and its internal-energy increase. Use Cambridge's convention deltaU = q + W, where W is work on the gas.3 indicative marks
Calculate and explain3 indicative marksCalculator allowed
A gas absorbs 500 J by heating and expands by 2.00 x 10−3m3 against constant pressure 1.00 x 105 Pa. Find work done by the gas, work done on it and its internal-energy increase. Use Cambridge's convention deltaU = q + W, where W is work on the gas.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked the syllabus's work-on convention explicitly and verified energy balance 500=200+300; repairs the bank's lack of direct constant-pressure first-law practice. This is conceptual placement, not completion of the entire referenced subsection.
Positive expansion transfers mechanical energy out of the gas. Calculate 1.00e5 x 2.00e-3 = 200 J before assigning the work-on sign.
With heating into the gas positive, 500 J enters and 200 J leaves as work; the remaining 300 J increases its molecular internal energy. This is constant-pressure work, not an integrated isothermal-work requirement.
Self-marking points
View marking points
Work by gas = p deltaV = 200 J.
Work on gas W = -200 J.
deltaU = 500 + (-200) = +300 J.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
