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Key Facts: WACE ATAR Physics Exam
Complete practice question bank for WA Year 12 WACE ATAR Physics covering Unit 3 (Gravity & Electromagnetism) and Unit 4 (Quantum Theory & Special Relativity) with step-by-step calculation solutions.
Sample WACE ATAR Physics Practice Questions
Try these sample questions to test your WACE ATAR Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1According to Newton's Law of Universal Gravitation, what happens to the gravitational force between two point masses if the distance between their centres is doubled?
A.The gravitational force increases by a factor of 4.
B.The gravitational force decreases to one-half of its original value.
C.The gravitational force decreases to one-quarter of its original value.
D.The gravitational force remains unchanged.
Explanation: Newton's Law of Universal Gravitation states that F = G*(m1*m2)/r^2. Force is inversely proportional to the square of the separation distance r. Doubling r (2r)^2 results in 4r^2 in the denominator, reducing the gravitational force to 1/4 of its original value.
2What is the gravitational field strength at the surface of a planet with mass 4.0 × 10^24 kg and radius 5.0 × 10^6 m? (G = 6.67 × 10^-11 N m^2 kg^-2)
A.5.34 N kg^-1
B.10.67 N kg^-1
C.53.36 N kg^-1
D.1.07 N kg^-1
Explanation: Gravitational field strength g = G*M / r^2. Substituting values: g = (6.67 × 10^-11 * 4.0 × 10^24) / (5.0 × 10^6)^2 = (2.668 × 10^14) / (2.5 × 10^13) = 10.672 N kg^-1 (or m s^-2).
3A satellite orbits Earth in a stable circular orbit. Which expression gives the orbital speed v of the satellite at distance r from Earth's centre of mass M?
A.v = sqrt(G * M * r)
B.v = sqrt(G * M / r)
C.v = G * M / r^2
D.v = 2 * pi * r / G
Explanation: Equating centripetal force to gravitational force: m*v^2 / r = G*M*m / r^2. Cancelling satellite mass m and one power of r gives v^2 = G*M / r, so v = sqrt(G*M / r).
4Kepler's Third Law states that for planets or satellites orbiting a central mass M, the ratio T^2 / r^3 is constant. What is the value of this constant equal to?
A.4 * pi^2 / (G * M)
B.G * M / (4 * pi^2)
C.2 * pi / (G * M)
D.sqrt(G * M)
Explanation: Using v = 2*pi*r / T in v^2 = G*M / r yields (4*pi^2 * r^2) / T^2 = G*M / r. Rearranging for T^2 / r^3 gives 4*pi^2 / (G*M).
5Two positive charges of 3.0 µC and 6.0 µC are separated by a distance of 0.30 m in a vacuum. What is the magnitude of the electrostatic force between them? (k = 8.99 × 10^9 N m^2 C^-2)
A.1.80 N
B.0.54 N
C.1.80 × 10^-3 N
D.5.39 N
Explanation: Coulomb's Law: F = k * q1 * q2 / r^2. F = (8.99 × 10^9 * 3.0 × 10^-6 * 6.0 × 10^-6) / (0.30)^2 = (0.16182) / 0.09 = 1.798 N ≈ 1.80 N.
6An electron experiences an electric force of 4.8 × 10^-15 N when placed in a uniform electric field. What is the strength of the electric field? (e = 1.60 × 10^-19 C)
A.3.0 × 10^4 N C^-1
B.7.68 × 10^-34 N C^-1
C.3.0 × 10^-4 N C^-1
D.1.33 × 10^4 N C^-1
Explanation: Electric field strength E = F / q. E = (4.8 × 10^-15 N) / (1.60 × 10^-19 C) = 3.0 × 10^4 N C^-1 (or V m^-1).
7A straight conductor of length 0.40 m carries a current of 5.0 A perpendicular to a uniform magnetic field of 0.20 T. What is the magnetic force acting on the conductor?
A.0.40 N
B.0.08 N
C.2.50 N
D.1.00 N
Explanation: Magnetic force on a current-carrying wire F = B * I * L * sin(theta). Since the wire is perpendicular, sin(90°) = 1. F = 0.20 T * 5.0 A * 0.40 m = 0.40 N.
8A flat circular coil with an area of 0.05 m^2 is oriented perpendicular to a uniform magnetic field of strength 0.30 T. What is the magnetic flux passing through the coil?
A.0.015 Wb
B.6.0 Wb
C.0.15 Wb
D.0.00 Wb
Explanation: Magnetic flux Phi = B * A * cos(theta), where theta is the angle between the field and the normal to the area. Here the field is perpendicular to the plane of the coil, so theta = 0° and cos(0°) = 1. Phi = 0.30 T * 0.05 m^2 = 0.015 Wb.
9An ideal transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. If an alternating voltage of 24 V AC is applied across the primary coil, what is the output voltage across the secondary coil?
A.120 V
B.4.8 V
C.240 V
D.48 V
Explanation: Transformer equation: V_s / V_p = N_s / N_p. Rearranging for V_s: V_s = V_p * (N_s / N_p) = 24 V * (1000 / 200) = 24 * 5 = 120 V.
10According to Lenz's Law, what is the direction of an induced current resulting from electromagnetic induction?
A.In the direction that reinforces the change in magnetic flux producing it.
B.In a direction such that its magnetic field opposes the change in magnetic flux that caused it.
C.Always clockwise relative to the applied magnetic field.
D.Perpendicular to both the electric and magnetic fields at all times.
Explanation: Lenz's Law states that the direction of an induced current is such that its induced magnetic field opposes the change in magnetic flux that creates it, conforming to the Law of Conservation of Energy.
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Verified exam format metadata for WACE Year 12 ATAR Physics (Units 3 & 4) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.