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100+ Free Castilla-La Mancha PAU Physics (Física - UCLM 2026) Practice Questions

Castilla-La Mancha PAU Physics Examination — Universidad de Castilla-La Mancha (UCLM 2026) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Castilla-La Mancha PAU Physics (Física - UCLM 2026) Exam

UCLM 2026

University of Castilla-La Mancha PAU Tribunal authority

UCLM PAU Regulations

90 Mins

Official examination time limit

UCLM PAU Schedule

5 Blocks

Gravitation, Electromagnetism, Waves/Optics, Relativity/Quantum, Nuclear Physics

2º Bachillerato Physics Syllabus

Min 4.0

Minimum Access Phase score required for university admission mark calculation

UCLM Admission Regulations

100

Practice questions with worked solutions in OpenExamPrep

OpenExamPrep

The UCLM PAU 2026 Physics examination evaluates conceptual understanding and worked quantitative calculation skills across 5 core physics domains in a 90-minute paper. This English-language multiple-choice bank is a study adaptation of the official written PAU paper, not a replica of it.

Sample Castilla-La Mancha PAU Physics (Física - UCLM 2026) Practice Questions

Try these sample questions to test your Castilla-La Mancha PAU Physics (Física - UCLM 2026) 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 of 500 kg and 800 kg are separated by a distance of 2.0 m. What is the magnitude of the gravitational attraction force between them? (Data: G = 6.67×10⁻¹¹ N·m²/kg²)
A.1.33×10⁻5 N
B.6.67×10⁻⁶ N
C.3.34×10⁻6 N
D.2.67×10⁻5 N
Explanation: Applying Newton's law of universal gravitation: F = G · (m₁ · m₂) / r² = 6.67×10⁻¹¹ · (500 · 800) / (2.0)² = 6.67×10⁻¹¹ · 400 000 / 4 = 6.67×10⁻⁶ N.
2A satellite of mass m describes a circular orbit of radius r = 2 R_T around the Earth (mass M_T). What is the expression for its orbital velocity v_o?
A.v_o = √(G · M_T / R_T)
B.v_o = √(G · M_T / (2 R_T))
C.v_o = √(2 G · M_T / R_T)
D.v_o = G · M_T / (2 R_T²)
Explanation: Equating the gravitational force to the centripetal force: G · M_T · m / r² = m · v_o² / r ⇒ v_o = √(G · M_T / r). Substituting r = 2 R_T gives v_o = √(G · M_T / (2 R_T)).
3Calculate the escape velocity from the surface of a planet of mass M = 4.0×10²⁴ kg and radius R = 5.0×10⁶ m. (Data: G = 6.67×10⁻¹¹ N·m²/kg²)
A.7.31 km/s
B.5.17 km/s
C.14.62 km/s
D.10.33 km/s
Explanation: The escape velocity is v_e = √(2 G M / R) = √(2 · 6.67×10⁻¹¹ · 4.0×10²⁴ / 5.0×10⁶) = √(1.0672×10⁸) ≈ 10 330 m/s = 10.33 km/s.
4According to Kepler's third law, if the period of revolution of a planet A at a distance r from the Sun is 1.0 year, what will be the period of a planet B located at a distance 4r from the Sun?
A.8.0 years
B.16.0 years
C.2.0 years
D.4.0 years
Explanation: According to Kepler's 3rd law, T² / r³ = constant. For planet B: T_B² / (4r)³ = T_A² / r³ ⇒ T_B² = 64 · T_A² ⇒ T_B = √64 · T_A = 8.0 years.
5A body of 100 kg mass is moved from a point A where the gravitational potential is V_A = -5.0×10⁷ J/kg to a point B where V_B = -2.0×10⁷ J/kg. What work does the gravitational field do?
A.-7.0×10⁹ J
B.+3.0×10⁹ J
C.-3.0×10⁹ J
D.+7.0×10⁹ J
Explanation: The work done by a conservative force such as the gravitational field is W_field = -ΔE_p = -m · (V_B - V_A) = -100 · (-2.0×10⁷ - (-5.0×10⁷)) = -100 · (3.0×10⁷) = -3.0×10⁹ J.
6What is the relationship of the work done by the gravitational force when displacing a mass between two points on an equipotential surface?
A.It is maximum and positive
B.It is maximum and negative
C.It is zero, since the potential difference between the two points is null
D.It depends on the path taken between the two points
Explanation: An equipotential surface is defined as the locus of points where the potential V is constant. Since W = -m ΔV and ΔV = 0, the work done by the gravitational field is zero.
7If the Earth's radius is R_T and the acceleration of gravity at the surface is g₀ = 9.80 m/s², at what height h above the Earth's surface is the acceleration of gravity reduced to g₀ / 4?
A.h = R_T
B.h = 2 R_T
C.h = 0.5 R_T
D.h = 3 R_T
Explanation: The field intensity at height h is g(h) = G M_T / (R_T + h)² = g₀ · [R_T / (R_T + h)]². For g(h) = g₀ / 4: [R_T / (R_T + h)]² = 1/4 ⇒ R_T / (R_T + h) = 1/2 ⇒ R_T + h = 2 R_T ⇒ h = R_T.
8A satellite of mass m = 200 kg orbits at a height h = R_T above the Earth's surface (where R_T = 6.37×10⁶ m and M_T = 5.97×10²⁴ kg). What is the total mechanical energy E_m of the satellite? (G = 6.67×10⁻¹¹ N·m²/kg²)
A.-6.25×10⁹ J
B.+3.13×10⁹ J
C.-3.13×10⁹ J
D.-1.25×10¹⁰ J
Explanation: The orbital radius is r = R_T + h = 2 R_T = 1.274×10⁷ m. The mechanical energy of a satellite in circular orbit is E_m = -G M_T m / (2r) = - (6.67×10⁻¹¹ · 5.97×10²⁴ · 200) / (2 · 1.274×10⁷) = - (7.964×10¹⁶) / (2.548×10⁷) ≈ -3.13×10⁹ J.
9Kepler's second law (law of areas) is a direct consequence of the conservation of a fundamental physical quantity. Which quantity is it?
A.The angular momentum of the planet about the center of the Sun
B.The linear momentum of the planet
C.The kinetic energy of the planet
D.The gravitational potential of the system
Explanation: Since the gravitational force is a central force (its moment M = r × F = 0), the areal velocity is constant, which directly expresses the conservation of angular momentum L = r × p = constant.
10What height above the Earth's surface must the orbit of a geostationary satellite have so that its period is T = 24 hours (86 400 s)? (Data: M_T = 5.97×10²⁴ kg, R_T = 6370 km, G = 6.67×10⁻¹¹ N·m²/kg²)
A.42 200 km
B.20 200 km
C.6 370 km
D.35 800 km
Explanation: From T² = (4π² / (G M_T)) · r³: r³ = G M_T T² / (4π²) = (6.67×10⁻¹¹ · 5.97×10²⁴ · 86400²) / (39.478) ≈ 7.54×10²² m³ ⇒ r ≈ 42 200 km. The height h = r - R_T = 42 200 - 6 370 = 35 830 km ≈ 35 800 km.

About the Castilla-La Mancha PAU Physics (Física - UCLM 2026) Practice Questions

Verified exam format metadata for Castilla-La Mancha PAU Physics Examination — Universidad de Castilla-La Mancha (UCLM 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.