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100+ Free Valencian Community PAU Physics Practice Questions
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Key Facts: Valencian Community PAU Physics Exam
The Valencian Community PAU Physics exam is organized by the Generalitat Valenciana Managing Commission in collaboration with UPV, UV, UA, UJI, and UMH. The 90-minute exam is scored from 0 to 10. Local questions on this platform provide 100 multiple-choice practice items covering the complete 2nd Bachillerato Physics curriculum with worked calculations.
Sample Valencian Community PAU Physics Practice Questions
Try these sample questions to test your Valencian Community PAU 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, how does the gravitational force between two point masses change if the distance between their centers is doubled?
A.The force increases by a factor of 4.
B.The force decreases to one-half of its original value.
C.The force decreases to one-fourth of its original value.
D.The force remains unchanged because mass is constant.
Explanation: Newton's law of universal gravitation states that gravitational force is inversely proportional to the square of the distance (F = G*m1*m2/r^2). Doubling the distance r multiplies the denominator by 2^2 = 4, reducing the force to 1/4 of its initial magnitude.
2Calculate the gravitational field strength (g) at Earth's surface, assuming Earth is a homogeneous sphere of mass M = 5.97 x 10^24 kg and radius R = 6.37 x 10^6 m. (Use G = 6.674 x 10^-11 N m^2/kg^2).
A.9.81 m/s^2
B.8.54 m/s^2
C.11.20 m/s^2
D.6.67 m/s^2
Explanation: Gravitational field strength is calculated using g = G * M / R^2. Substituting the given values: g = (6.674 x 10^-11 * 5.97 x 10^24) / (6.37 x 10^6)^2 = 3.984 x 10^14 / 4.058 x 10^13 = 9.81 m/s^2.
3What is the gravitational potential V at a distance r from a point mass M, taking zero potential at infinity?
A.V = G * M / r
B.V = -G * M / r
C.V = -G * M / r^2
D.V = G * M^2 / r
Explanation: Gravitational potential is defined as the work done per unit mass by an external force bringing a mass from infinity to distance r. Since gravity is an attractive force, potential is negative relative to infinity: V = -G * M / r.
4A space probe orbits Earth at an altitude h equal to Earth's radius (h = R_E). What is the gravitational acceleration experienced by the probe at this altitude in terms of surface acceleration g_0 = 9.81 m/s^2?
A.4.91 m/s^2 (g_0 / 2)
B.2.45 m/s^2 (g_0 / 4)
C.1.23 m/s^2 (g_0 / 8)
D.9.81 m/s^2 (g_0)
Explanation: The radial distance from Earth's center is r = R_E + h = R_E + R_E = 2 R_E. The gravitational acceleration is g' = G * M / (2 R_E)^2 = (G * M / R_E^2) / 4 = g_0 / 4 = 9.81 / 4 = 2.45 m/s^2.
5Which expression gives the orbital speed v of a satellite in a stable circular orbit of radius r around a planet of mass M?
A.v = sqrt(G * M / r)
B.v = G * M / r^2
C.v = sqrt(2 * G * M / r)
D.v = G * M / r
Explanation: Equating centripetal force to gravitational force: m * v^2 / r = G * M * m / r^2. Solving for v gives orbital speed v = sqrt(G * M / r).
6Satellite A orbits a central star at radius r_A. Satellite B orbits the same star at radius r_B = 4 r_A. According to Kepler's Third Law, what is the ratio of their orbital periods T_B / T_A?
A.2
B.4
C.8
D.16
Explanation: Kepler's Third Law states T^2 is proportional to r^3. Therefore, (T_B / T_A)^2 = (r_B / r_A)^3 = (4)^3 = 64. Taking the square root gives T_B / T_A = sqrt(64) = 8.
7Calculate the orbital speed of a satellite orbiting Earth at an altitude h = R_E (r = 2 R_E = 1.274 x 10^7 m). Given Earth mass M = 5.97 x 10^24 kg and G = 6.674 x 10^-11 N m^2/kg^2.
A.5.59 km/s
B.7.91 km/s
C.11.2 km/s
D.3.96 km/s
Explanation: Orbital speed v = sqrt(G * M / r) = sqrt((6.674 x 10^-11 * 5.97 x 10^24) / 1.274 x 10^7) = sqrt(3.127 x 10^7) = 5.592 x 10^3 m/s = 5.59 km/s.
8How much work W is done by Earth's gravitational field on a satellite of mass m = 500 kg as it moves from an initial distance r_i = 2 R_E to a final distance r_f = R_E? (Express answer using g_0 = 9.81 m/s^2 and R_E = 6.37 x 10^6 m).
A.+1.56 x 10^10 J
B.-1.56 x 10^10 J
C.+3.12 x 10^10 J
D.-3.12 x 10^10 J
Explanation: Work done by conservative gravity is W = -Delta U = -(U_f - U_i) = G*M*m*(1/R_E - 1/(2 R_E)) = G*M*m / (2 R_E). Since G*M = g_0 * R_E^2, W = m * g_0 * R_E / 2 = 500 * 9.81 * 6.37 x 10^6 / 2 = +1.56 x 10^10 J. Gravity does positive work as the satellite moves closer.
9For a satellite of mass m in a circular orbit of radius r around mass M, what is the relationship between its kinetic energy K, potential energy U, and total mechanical energy E?
A.K = -U / 2 and E = U / 2
B.K = U / 2 and E = -U / 2
C.K = -U and E = 0
D.K = -2 U and E = -U
Explanation: Kinetic energy K = G*M*m / (2r). Potential energy U = -G*M*m / r. Thus K = -U / 2. Total energy E = K + U = G*M*m / (2r) - G*M*m / r = -G*M*m / (2r) = U / 2.
10What is the ratio of the escape velocity v_esc from the surface of a spherical planet to the circular orbital velocity v_orb near its surface?
A.sqrt(2) approx 1.414
B.2
C.1/sqrt(2) approx 0.707
D.sqrt(3) approx 1.732
Explanation: Escape velocity is v_esc = sqrt(2 * G * M / R) and circular orbital velocity is v_orb = sqrt(G * M / R). Dividing them gives v_esc / v_orb = sqrt(2) approx 1.414.
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