All Practice Exams
100+ Free Balearic PAU Physics Practice Questions
Balearic Islands PAU Physics Examination — UIB (Física 2º Bachillerato UIB 2026) practice questions are available now; exam metadata is being verified.
✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
100+ Questions
100% Free
Loading practice questions...
Same family resources
Explore More Spain Balearic Islands PAU (Prueba de Acceso a la Universidad)
Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.
2026 Statistics
Key Facts: Balearic PAU Physics Exam
90 Minutes
Exam time duration
UIB PAU Commission
EUR 69.21
Ordinary registration fee for Access Phase
UIB 2026 Fees
0–10 Scale
Grading scale for Bachillerato and PAU exams
Spanish Ministry of Education
4 Core Blocks
Gravitational Field, Electromagnetism, Waves & Optics, Modern Physics
UIB Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Balearic Islands PAU Physics (UIB 2026) is a 90-minute university entrance exam assessing 2º Bachillerato Physics across 4 core syllabus domains.
Sample Balearic PAU Physics Practice Questions
Try these sample questions to test your Balearic 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 and m2 are initially separated by a distance r, exerting a gravitational force F on each other. If the separation distance is doubled to 2r while keeping the masses constant, what is the new gravitational force between them?
A.2 F
B.4 F
C.0.25 F
D.0.5 F
Explanation: According to Newton's law of universal gravitation, F = G * m1 * m2 / r^2. Because gravitational force is inversely proportional to the square of the distance (1/r^2), doubling the separation distance r to 2r results in F' = G * m1 * m2 / (2r)^2 = F / 4 = 0.25 F.
2According to Kepler's third law of planetary motion, how does the orbital period T of a planet orbiting the Sun relate to its semi-major axis r?
A.T is directly proportional to r.
B.T^2 is directly proportional to r^3.
C.T^3 is directly proportional to r^2.
D.T is inversely proportional to r^2.
Explanation: Kepler's third law states that the square of the orbital period T of any planet is directly proportional to the cube of the semi-major axis r of its orbit (T^2 / r^3 = constant = 4 pi^2 / (G M_sun)).
3A planet has twice the mass of Earth (M_p = 2 M_E) and twice the radius of Earth (R_p = 2 R_E). If the surface gravitational acceleration on Earth is g_E, what is the surface gravitational acceleration g_p on this planet?
A.0.5 g_E
B.g_E
C.2 g_E
D.4 g_E
Explanation: Gravitational field strength at the surface is given by g = G M / R^2. Substituting M_p = 2 M_E and R_p = 2 R_E yields g_p = G (2 M_E) / (2 R_E)^2 = 2/4 * (G M_E / R_E^2) = 0.5 g_E.
4Why is the gravitational potential energy U = -G M m / r defined with a negative sign when using the standard reference point at infinity?
A.Because gravity is a repulsive force at large distances.
B.Because work must be done against the attractive gravitational force to move a mass to infinity, where U = 0.
C.Because gravitational potential energy is a vector pointing toward the center of mass.
D.Because mass values are defined as negative in bound gravitational systems.
Explanation: By convention, gravitational potential energy is chosen to be zero at infinite separation (r = infinity). Since gravity is an attractive force, bringing a mass from infinity closer to the primary mass releases energy, making the potential energy negative. Positive work must be done by an external force to separate the masses to infinity.
5Kepler's second law (law of equal areas) is a direct consequence of which fundamental conservation law?
A.Conservation of linear momentum
B.Conservation of mechanical energy
C.Conservation of angular momentum
D.Conservation of electric charge
Explanation: Because the gravitational force exerted on a planet acts strictly along the line joining the planet to the central body (central force), the net torque about the central body is zero. Consequently, angular momentum L = m r v sin(theta) is conserved, which implies a constant areal velocity (dA/dt = L / (2 m) = constant).
6What is the gravitational field strength at an altitude h equal to the Earth's radius R_E above the surface of Earth? (Take surface gravitational acceleration as g = 9.8 m/s^2).
A.9.8 m/s^2
B.4.9 m/s^2
C.2.45 m/s^2
D.0 m/s^2
Explanation: The distance from the center of Earth is r = R_E + h = R_E + R_E = 2 R_E. The gravitational field strength is g(r) = G M / r^2 = G M / (2 R_E)^2 = g_surface / 4 = 9.8 / 4 = 2.45 m/s^2.
7A satellite orbits Earth in a circular orbit of radius r with speed v. If the satellite is moved to a circular orbit of radius 4r, what will its new orbital speed be?
A.0.25 v
B.0.5 v
C.2 v
D.4 v
Explanation: Equating centripetal force to gravitational force gives m v^2 / r = G M m / r^2, so orbital speed is v = sqrt(G M / r). For a new radius r' = 4r, the new orbital speed is v' = sqrt(G M / (4r)) = 0.5 * sqrt(G M / r) = 0.5 v.
8Calculate the escape velocity from the surface of a planet with mass M = 6.0 x 10^24 kg and radius R = 6.4 x 10^6 m. (G = 6.67 x 10^-11 N m^2/kg^2).
A.7.9 km/s
B.11.2 km/s
C.15.8 km/s
D.22.4 km/s
Explanation: Escape velocity is obtained by setting initial mechanical energy to zero: 1/2 m v_esc^2 - G M m / R = 0, giving v_esc = sqrt(2 G M / R). Plugging in values: v_esc = sqrt(2 * 6.67e-11 * 6.0e24 / 6.4e6) = sqrt(1.25e8) = 11180 m/s = 11.2 km/s.
9A satellite of mass m = 500 kg is in a circular orbit around Earth (M = 6.0 x 10^24 kg) at a radius r = 1.0 x 10^7 m. What is its total mechanical energy E?
A.-1.0 x 10^10 J
B.-2.0 x 10^10 J
C.1.0 x 10^10 J
D.2.0 x 10^10 J
Explanation: For a circular orbit, kinetic energy is K = 1/2 G M m / r and potential energy is U = -G M m / r. Total mechanical energy is E = K + U = -G M m / (2 r). Calculating: E = -(6.67e-11 * 6.0e24 * 500) / (2 * 1.0e7) = -2.001e12 / 2.0e7 = -1.0 x 10^10 J.
10An object weighs 600 N on Earth's surface. What would it weigh on the surface of Planet X, which has 3 times Earth's mass and 2 times Earth's radius?
A.450 N
B.600 N
C.900 N
D.1350 N
Explanation: Weight is W = m g = m (G M / R^2). On Planet X: W_X = m G (3 M_E) / (2 R_E)^2 = 3/4 * (m G M_E / R_E^2) = 0.75 * W_E = 0.75 * 600 N = 450 N.
About the Balearic PAU Physics Practice Questions
Verified exam format metadata for Balearic Islands PAU Physics Examination — UIB (Física 2º Bachillerato UIB 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.