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2026 Statistics
Key Facts: La Rioja PAU Physics Exam
90 min
Time Limit
Universidad de La Rioja
0–10
Grading Scale
Comisión PAU La Rioja
4.0
Min. Access Phase Score
Spanish PAU Regulations
EUR 55.43
Base Registration Fee
Universidad de La Rioja
100
Practice Questions
OpenExamPrep English Bank
The La Rioja PAU Physics exam is a 90-minute examination set by the Universidad de La Rioja (UR) / Comisión Organizadora de la PAU de La Rioja with a base registration fee of EUR 55.43. Graded on a 0–10 scale, a minimum mark of 4.0 in the Access Phase is required to combine with high school GPA.
Sample La Rioja PAU Physics Practice Questions
Try these sample questions to test your La Rioja PAU Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Two point masses m1 = 1000 kg and m2 = 4000 kg are separated by a distance of r = 2.0 m in a vacuum. Using Newton's Law of Universal Gravitation (G = 6.674 x 10^-11 N m^2/kg^2), what is the magnitude of the gravitational force between them?
A.6.67 x 10^-5 N
B.1.33 x 10^-4 N
C.3.34 x 10^-5 N
D.2.67 x 10^-7 N
Explanation: The gravitational force is 6.67 x 10^-5 N. Applying Newton's law of universal gravitation: F = G * m1 * m2 / r^2 = (6.674 x 10^-11) * 1000 * 4000 / (2.0)^2 = (6.674 x 10^-11) * (4.0 x 10^6) / 4.0 = 6.674 x 10^-5 N.
2A spherical planet has mass M = 6.00 x 10^24 kg and radius R = 6.00 x 10^6 m. Using G = 6.674 x 10^-11 N m^2/kg^2, what is the surface gravitational field strength g?
A.6.67 N/kg
B.11.1 N/kg
C.9.80 N/kg
D.66.7 N/kg
Explanation: The surface gravitational field strength is 11.1 N/kg. Gravitational field strength is g = G * M / R^2 = (6.674 x 10^-11) * (6.00 x 10^24) / (6.00 x 10^6)^2 = (4.0044 x 10^14) / (3.60 x 10^13) = 11.12 N/kg.
3What is the weight of an astronaut with mass m = 70.0 kg on the Moon's surface, where local gravitational acceleration is g_moon = 1.62 m/s^2?
A.686 N
B.113 N
C.43.2 N
D.162 N
Explanation: The astronaut's weight on the Moon is 113 N. Weight is given by W = m * g_moon = 70.0 kg * 1.62 m/s^2 = 113.4 N.
4Two point charges q1 = +2.00 μC and q2 = +5.00 μC are separated by d = 0.100 m in air (k_e = 8.99 x 10^9 N m^2/C^2). What is the magnitude of the electrostatic repulsive force between them?
A.0.899 N
B.8.99 N
C.89.9 N
D.4.49 N
Explanation: The electrostatic repulsive force is 8.99 N. By Coulomb's Law: F = k_e * |q1 * q2| / d^2 = (8.99 x 10^9) * (2.00 x 10^-6) * (5.00 x 10^-6) / (0.100)^2 = (8.99 x 10^-2) / 0.0100 = 8.99 N.
5What is the electric field magnitude E at distance r = 0.200 m from a point charge q = +4.00 nC in a vacuum (k_e = 8.99 x 10^9 N m^2/C^2)?
A.179.8 N/C
B.899 N/C
C.449.5 N/C
D.3596 N/C
Explanation: The electric field magnitude is 899 N/C. Electric field intensity is E = k_e * q / r^2 = (8.99 x 10^9) * (4.00 x 10^-9) / (0.200)^2 = 35.96 / 0.0400 = 899 N/C.
6What is the electric potential V at distance r = 0.500 m from a isolated point charge q = +10.0 nC in a vacuum (k_e = 8.99 x 10^9 N m^2/C^2)?
A.179.8 V
B.359.6 V
C.89.9 V
D.17.98 V
Explanation: The electric potential is 179.8 V. Electric potential is V = k_e * q / r = (8.99 x 10^9) * (10.0 x 10^-9) / 0.500 = 89.9 / 0.500 = 179.8 V.
7A charge q = +3.00 μC moves with speed v = 200 m/s perpendicular to a uniform magnetic field B = 0.500 T. What is the magnitude of the magnetic Lorentz force acting on the charge?
A.3.00 x 10^-4 N
B.6.00 x 10^-4 N
C.1.50 x 10^-4 N
D.3.00 x 10^-1 N
Explanation: The magnetic Lorentz force is 3.00 x 10^-4 N. Lorentz force for perpendicular motion is F = |q| * v * B * sin(90°) = (3.00 x 10^-6 C) * (200 m/s) * (0.500 T) * 1.0 = 3.00 x 10^-4 N.
8What is the magnetic field strength B at distance r = 0.050 m from an infinitely long straight wire carrying electric current I = 10.0 A in a vacuum (μ0 = 4π x 10^-7 T m/A)?
A.4.00 x 10^-5 T
B.2.00 x 10^-5 T
C.8.00 x 10^-5 T
D.1.26 x 10^-4 T
Explanation: The magnetic field strength is 4.00 x 10^-5 T. By Ampère's Law / Biot-Savart Law: B = μ0 * I / (2π * r) = (4π x 10^-7 * 10.0) / (2π * 0.050) = (2 x 10^-6) / 0.050 = 4.00 x 10^-5 T.
9A planar loop of area A = 0.020 m^2 is placed perpendicular to a uniform magnetic field B = 0.400 T (angle θ = 0° between field and surface normal). What is the magnetic flux Φ through the loop?
A.8.00 x 10^-3 Wb
B.1.60 x 10^-2 Wb
C.4.00 x 10^-3 Wb
D.0 Wb
Explanation: The magnetic flux through the loop is 8.00 x 10^-3 Wb. Magnetic flux is Φ = B * A * cos(θ) = 0.400 T * 0.020 m^2 * cos(0°) = 8.00 x 10^-3 Wb.
10A mass undergoes simple harmonic motion with amplitude A = 0.050 m and angular frequency ω = 10.0 rad/s. What is its maximum speed v_max?
A.0.500 m/s
B.5.00 m/s
C.0.050 m/s
D.0.250 m/s
Explanation: The maximum speed is 0.500 m/s. In simple harmonic motion, maximum speed occurs at equilibrium (x = 0) and is v_max = ω * A = 10.0 rad/s * 0.050 m = 0.500 m/s.
About the La Rioja PAU Physics Practice Questions
Verified exam format metadata for La Rioja PAU Physics Exam / Física 2º Bachillerato (Universidad de La Rioja) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.