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La Rioja PAU Physics study materials
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La Rioja PAU Physics practice questions

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Question 1Electromagnetic Fields

Two 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?

  1. 0.899 N
  2. 8.99 N Correct answer.
  3. 89.9 N
  4. 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.

Question 2Electromagnetic Fields

What 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)?

  1. 179.8 N/C
  2. 899 N/C Correct answer.
  3. 449.5 N/C
  4. 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.

Question 3Electromagnetic Fields

What 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)?

  1. 179.8 V Correct answer.
  2. 359.6 V
  3. 89.9 V
  4. 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.

Question 4Electromagnetic Fields

A 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?

  1. 3.00 x 10^-4 N Correct answer.
  2. 6.00 x 10^-4 N
  3. 1.50 x 10^-4 N
  4. 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.

Question 5Electromagnetic Fields

What 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)?

  1. 4.00 x 10^-5 T Correct answer.
  2. 2.00 x 10^-5 T
  3. 8.00 x 10^-5 T
  4. 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.

Question 6Electromagnetic Fields

A 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?

  1. 8.00 x 10^-3 Wb Correct answer.
  2. 1.60 x 10^-2 Wb
  3. 4.00 x 10^-3 Wb
  4. 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.

Question 7Electromagnetic Fields

Two point charges q1 = +4.00 nC and q2 = +4.00 nC are placed at coordinates (0, +0.300 m) and (0, -0.300 m) respectively. What is the net electric field vector E_net at the origin (0, 0)?

  1. 0 N/C Correct answer.
  2. 800 N/C pointing in +y direction
  3. 800 N/C pointing in -y direction
  4. 400 N/C pointing in +x direction

Explanation

The net electric field vector at the origin is 0 N/C. The electric field E1 from q1 points downwards (-y) with magnitude k_e * q1 / r^2, while E2 from q2 points upwards (+y) with identical magnitude. By vector superposition, E_net = E1 + E2 = 0 N/C.

Question 8Electromagnetic Fields

How much work W_elec is done by the electrostatic field when moving a test charge q = +2.00 μC from point A (V_A = +100 V) to point B (V_B = +40.0 V)?

  1. +1.20 x 10^-4 J Correct answer.
  2. -1.20 x 10^-4 J
  3. +2.80 x 10^-4 J
  4. +2.00 x 10^-4 J

Explanation

The work done by the electric force is +1.20 x 10^-4 J. Work done by a conservative electric field is W_elec = - q * ΔV = q * (V_A - V_B) = (2.00 x 10^-6 C) * (+100 V - 40.0 V) = (2.00 x 10^-6) * (+60.0 V) = +1.20 x 10^-4 J.

Question 9Electromagnetic Fields

A proton (m_p = 1.673 x 10^-27 kg, q = +1.602 x 10^-19 C) is released from rest in a uniform electric field E = 500 N/C. What is the magnitude of its acceleration a?

  1. 4.79 x 10^10 m/s^2 Correct answer.
  2. 8.01 x 10^-17 m/s^2
  3. 2.99 x 10^11 m/s^2
  4. 9.58 x 10^10 m/s^2

Explanation

The magnitude of acceleration is 4.79 x 10^10 m/s^2. By Newton's second law: F_e = q * E = m_p * a => a = q * E / m_p = (1.602 x 10^-19 C * 500 N/C) / (1.673 x 10^-27 kg) = (8.010 x 10^-17) / (1.673 x 10^-27) = 4.788 x 10^10 m/s^2 = 4.79 x 10^10 m/s^2.

Question 10Electromagnetic Fields

An electron (m_e = 9.11 x 10^-31 kg, q = 1.602 x 10^-19 C) enters perpendicularly into a uniform magnetic field B = 0.0200 T with velocity v = 5.00 x 10^6 m/s. What is the radius r of its circular orbit?

  1. 1.42 mm Correct answer.
  2. 2.84 mm
  3. 0.71 mm
  4. 1.42 cm

Explanation

The radius of orbit is 1.42 mm. Centripetal force is provided by Lorentz force: m_e * v^2 / r = q * v * B => r = m_e * v / (q * B) = (9.11 x 10^-31 * 5.00 x 10^6) / (1.602 x 10^-19 * 0.0200) = (4.555 x 10^-24) / (3.204 x 10^-21) = 1.422 x 10^-3 m = 1.42 mm.

Question 11Electromagnetic Fields

What is the cyclotron frequency f of a proton (m_p = 1.673 x 10^-27 kg, q = 1.602 x 10^-19 C) moving in a uniform magnetic field B = 0.500 T?

  1. 7.62 MHz Correct answer.
  2. 15.2 MHz
  3. 3.81 MHz
  4. 47.9 MHz

Explanation

The cyclotron frequency is 7.62 MHz. Cyclotron frequency is f = q * B / (2π * m_p) = (1.602 x 10^-19 * 0.500) / (2π * 1.673 x 10^-27) = (8.010 x 10^-20) / (1.0512 x 10^-26) = 7.619 x 10^6 Hz = 7.62 MHz.

Question 12Electromagnetic Fields

Two long parallel straight wires separated by d = 0.100 m carry currents I1 = 5.00 A and I2 = 8.00 A in the same direction. What is the magnetic force per unit length F/L acting between them?

  1. 8.00 x 10^-5 N/m (attractive) Correct answer.
  2. 8.00 x 10^-5 N/m (repulsive)
  3. 4.00 x 10^-5 N/m (attractive)
  4. 1.60 x 10^-4 N/m (repulsive)

Explanation

The magnetic force per unit length is 8.00 x 10^-5 N/m (attractive). Force per unit length is F/L = μ0 * I1 * I2 / (2π * d) = (4π x 10^-7 * 5.00 * 8.00) / (2π * 0.100) = (2 x 10^-7 * 40.0) / 0.100 = 8.00 x 10^-5 N/m. Parallel currents carrying charges in the same direction attract.

Question 13Electromagnetic Fields

A circular coil of N = 100 turns and radius r = 0.100 m is placed perpendicular to a magnetic field. The field increases uniformly from B1 = 0 T to B2 = 0.500 T in time interval Δt = 0.200 s. What is the magnitude of the induced EMF |ε|?

  1. 7.85 V Correct answer.
  2. 15.7 V
  3. 0.0785 V
  4. 3.93 V

Explanation

The magnitude of the induced EMF is 7.85 V. Coil area is A = π * r^2 = π * (0.100)^2 = 0.031416 m^2. By Faraday's law: |ε| = N * |ΔΦ / Δt| = N * A * (ΔB / Δt) = 100 * 0.031416 m^2 * (0.500 T / 0.200 s) = 100 * 0.031416 * 2.50 = 7.854 V = 7.85 V.

Question 14Electromagnetic Fields

A straight conducting rod of length L = 0.400 m moves at constant speed v = 5.00 m/s perpendicular to a uniform magnetic field B = 0.300 T. What is the motional EMF induced across the ends of the rod?

  1. 0.600 V Correct answer.
  2. 1.20 V
  3. 0.300 V
  4. 0.120 V

Explanation

The induced motional EMF is 0.600 V. Motional electromotive force is ε = B * L * v = 0.300 T * 0.400 m * 5.00 m/s = 0.600 V.

Question 15Electromagnetic Fields

A square coil of N = 200 turns and side length s = 0.150 m rotates at f = 50.0 Hz in a uniform magnetic field B = 0.200 T. What is the peak induced electromotive force ε_max?

  1. 283 V Correct answer.
  2. 141 V
  3. 45.0 V
  4. 565 V

Explanation

The peak induced EMF is 283 V. Area is A = s^2 = 0.150^2 = 0.0225 m^2. Angular frequency is ω = 2π f = 2π * 50.0 = 314.16 rad/s. Peak EMF is ε_max = N * B * A * ω = 200 * 0.200 T * 0.0225 m^2 * 314.16 rad/s = 282.74 V ≈ 283 V.

Question 16Electromagnetic Fields

A rectangular coil of N = 50 turns with dimensions 0.100 m x 0.200 m (area A = 0.0200 m^2) carries electric current I = 2.00 A. It is oriented in a uniform magnetic field B = 0.500 T such that its plane is parallel to the field lines. What is the magnetic torque τ acting on the coil?

  1. 0.500 N m Correct answer.
  2. 0.250 N m
  3. 1.00 N m
  4. 0 N m

Explanation

The magnetic torque is 0.500 N m. Magnetic torque is τ = N * I * A * B * sin(θ). When the coil plane is parallel to field lines, the angle between surface normal and B field is θ = 90°, so sin(90°) = 1.0. τ = 50 * 2.00 A * 0.0200 m^2 * 0.500 T * 1.0 = 0.500 N m.

Question 17Electromagnetic Fields

An inductor with self-inductance L = 0.200 H carries a current changing at a constant rate dI/dt = -50.0 A/s. What is the self-induced electromotive force ε_ind?

  1. +10.0 V Correct answer.
  2. -10.0 V
  3. +2.50 V
  4. +250 V

Explanation

The self-induced EMF is +10.0 V. Self-induced EMF is given by Faraday's law of induction for inductors: ε_ind = - L * (dI/dt) = - (0.200 H) * (-50.0 A/s) = +10.0 V.

Question 18Electromagnetic Fields

Two point charges q1 = +8.00 nC and q2 = -8.00 nC form an electric dipole separated by d = 0.200 m along the x-axis. What are the electric field magnitude E and potential V at the midpoint between them?

  1. E = 1.44 x 10^4 N/C, V = 0 V Correct answer.
  2. E = 0 N/C, V = 0 V
  3. E = 1.44 x 10^4 N/C, V = 1440 V
  4. E = 0 N/C, V = 1440 V

Explanation

The values at the midpoint are E = 1.44 x 10^4 N/C and V = 0 V. Distance to midpoint is r = 0.100 m. E1 from positive charge points away (towards negative charge) with magnitude k_e * q1 / r^2 = (8.99 x 10^9 * 8.00 x 10^-9) / (0.100)^2 = 7192 N/C. E2 from negative charge points towards negative charge with same magnitude 7192 N/C. Being in the same direction, E_net = 7192 + 7192 = 14384 N/C ≈ 1.44 x 10^4 N/C. Electric potential is scalar: V = k_e (q1/r + q2/r) = k_e / r (8.00 nC - 8.00 nC) = 0 V.

Question 19Electromagnetic Fields

A velocity selector uses perpendicular uniform electric field E = 4.00 x 10^4 V/m and magnetic field B = 0.200 T. What speed v must charged particles have to pass through undeflected?

  1. 2.00 x 10^5 m/s Correct answer.
  2. 8.00 x 10^3 m/s
  3. 5.00 x 10^-6 m/s
  4. 1.00 x 10^5 m/s

Explanation

The required velocity is 2.00 x 10^5 m/s. Undeflected motion requires electric force to balance magnetic force: q * E = q * v * B => v = E / B = (4.00 x 10^4 V/m) / (0.200 T) = 2.00 x 10^5 m/s.

Question 20Electromagnetic Fields

Singly-ionized isotopes (q = +1.602 x 10^-19 C) of masses m1 = 3.50 x 10^-25 kg and m2 = 3.68 x 10^-25 kg enter a mass spectrometer with speed v = 1.00 x 10^5 m/s perpendicular to magnetic field B = 0.500 T. What is the spatial separation Δr between their circular paths at 180° deflection?

  1. 4.49 cm Correct answer.
  2. 2.25 cm
  3. 8.99 cm
  4. 1.12 cm

Explanation

The spatial separation at 180° deflection is 4.49 cm. Radius for mass m is r = m * v / (q * B). Radius difference is Δr = (m2 - m1) * v / (q * B) = (0.18 x 10^-25 kg * 1.00 x 10^5 m/s) / (1.602 x 10^-19 C * 0.500 T) = (1.80 x 10^-21) / (8.01 x 10^-20) = 0.02247 m = 2.247 cm. At 180° deflection, detector position separation is Δx = 2 * Δr = 2 * 2.247 cm = 4.494 cm ≈ 4.49 cm.

Question 21Electromagnetic Fields

A metal rod of length L = 0.500 m rotates in a horizontal plane about a vertical axis through one end with constant angular speed ω = 20.0 rad/s in a uniform vertical magnetic field B = 0.400 T. What is the induced EMF ε between its ends?

  1. 1.00 V Correct answer.
  2. 2.00 V
  3. 0.500 V
  4. 4.00 V

Explanation

The induced EMF is 1.00 V. For a rotating rod, effective linear velocity of midpoint is v_mid = ω * (L / 2). Induced EMF is ε = B * L * v_mid = (1/2) * B * L^2 * ω = 0.5 * (0.400 T) * (0.500 m)^2 * (20.0 rad/s) = 0.5 * 0.400 * 0.250 * 20.0 = 1.00 V.

Question 22Electromagnetic Fields

A plane electromagnetic wave travelling in a vacuum has peak electric field amplitude E0 = 300 V/m. What is the average energy density u_avg of the wave (ε0 = 8.854 x 10^-12 F/m)?

  1. 3.98 x 10^-7 J/m^3 Correct answer.
  2. 7.97 x 10^-7 J/m^3
  3. 1.99 x 10^-7 J/m^3
  4. 1.19 x 10^-4 J/m^3

Explanation

The average energy density is 3.98 x 10^-7 J/m^3. Average EM wave energy density is u_avg = (1/2) * ε0 * E0^2 = 0.5 * (8.854 x 10^-12 F/m) * (300 V/m)^2 = 0.5 * (8.854 x 10^-12) * (90000) = 3.984 x 10^-7 J/m^3.

Question 23Waves Optics

A 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?

  1. 0.500 m/s Correct answer.
  2. 5.00 m/s
  3. 0.050 m/s
  4. 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.

Question 24Waves Optics

A body of mass m = 0.500 kg is attached to an ideal spring of stiffness k = 50.0 N/m. What is the period T of oscillation?

  1. 0.628 s Correct answer.
  2. 0.100 s
  3. 1.57 s
  4. 3.14 s

Explanation

The period of oscillation is 0.628 s. The period of a spring-mass oscillator is T = 2π * √(m / k) = 2π * √(0.500 / 50.0) = 2π * √(0.0100) = 2π * 0.100 = 0.200π = 0.628 s.

Question 25Waves Optics

A musical note has frequency f = 440 Hz in air where sound propagation speed is v = 343 m/s. What is its wavelength λ?

  1. 0.780 m Correct answer.
  2. 1.28 m
  3. 150.9 m
  4. 0.390 m

Explanation

The wavelength is 0.780 m. Wave propagation relation is v = f * λ, so λ = v / f = 343 m/s / 440 Hz = 0.7795 m = 0.780 m.

Question 26Waves Optics

What is the angular wave number k for a harmonic wave with wavelength λ = 0.500 m?

  1. 12.57 rad/m Correct answer.
  2. 6.28 rad/m
  3. 3.14 rad/m
  4. 2.00 rad/m

Explanation

The angular wave number is 12.57 rad/m. By definition, wave number k = 2π / λ = 2π / 0.500 m = 4π = 12.566 rad/m = 12.57 rad/m.

Question 27Waves Optics

What is the sound intensity level β in decibels for a sound wave with intensity I = 1.00 x 10^-6 W/m^2 (reference intensity I0 = 1.00 x 10^-12 W/m^2)?

  1. 60 dB Correct answer.
  2. 120 dB
  3. 6 dB
  4. 600 dB

Explanation

The sound intensity level is 60 dB. Sound level in decibels is β = 10 * log10(I / I0) = 10 * log10(1.00 x 10^-6 / 1.00 x 10^-12) = 10 * log10(10^6) = 10 * 6 = 60 dB.

Question 28Waves Optics

A body undergoes simple harmonic motion according to the position equation x(t) = 0.0800 cos(5.00 t + π/3) in SI units. What is its maximum acceleration magnitude a_max?

  1. 2.00 m/s^2 Correct answer.
  2. 0.400 m/s^2
  3. 1.00 m/s^2
  4. 10.0 m/s^2

Explanation

The maximum acceleration magnitude is 2.00 m/s^2. In SHM, maximum acceleration is a_max = ω^2 * A = (5.00 rad/s)^2 * (0.0800 m) = 25.0 * 0.0800 = 2.00 m/s^2.

Question 29Waves Optics

A spring-mass system (k = 200 N/m, amplitude A = 0.100 m) oscillates without friction. What are its potential energy U and kinetic energy K at displacement x = 0.0600 m?

  1. U = 0.360 J, K = 0.640 J Correct answer.
  2. U = 0.640 J, K = 0.360 J
  3. U = 0.500 J, K = 0.500 J
  4. U = 0.180 J, K = 0.320 J

Explanation

The energies are U = 0.360 J and K = 0.640 J. Total energy is E = (1/2) * k * A^2 = 0.5 * 200 * (0.100)^2 = 1.00 J. Potential energy at x is U = (1/2) * k * x^2 = 0.5 * 200 * (0.0600)^2 = 100 * 0.00360 = 0.360 J. Kinetic energy is K = E - U = 1.00 - 0.360 = 0.640 J.

Question 30Waves Optics

What is the required length L of a simple pendulum to achieve an oscillation period T = 2.00 s at a location where g = 9.80 m/s^2?

  1. 0.993 m Correct answer.
  2. 0.496 m
  3. 1.99 m
  4. 3.97 m

Explanation

The pendulum length is 0.993 m. Simple pendulum period formula T = 2π * √(L / g) => T^2 = 4π^2 * L / g => L = g * T^2 / (4π^2) = 9.80 * (2.00)^2 / (39.478) = 39.20 / 39.478 = 0.9929 m = 0.993 m.

Question 31Waves Optics

A transverse harmonic wave is described by y(x,t) = 0.0400 sin(20.0π t - 2.00π x) in SI units. What is the propagation speed v and direction of motion?

  1. 10.0 m/s in +x direction Correct answer.
  2. 10.0 m/s in -x direction
  3. 0.100 m/s in +x direction
  4. 20.0 m/s in +x direction

Explanation

The propagation speed is 10.0 m/s in the +x direction. Angular frequency is ω = 20.0π rad/s and wave number is k = 2.00π rad/m. Wave speed is v = ω / k = (20.0π) / (2.00π) = 10.0 m/s. The minus sign in (ω t - k x) indicates propagation in the +x direction.

Question 32Waves Optics

An isotropic point acoustic source emits a sound power P = 10.0 W uniformly in all directions. What is the sound intensity I at distance r = 5.00 m from the source?

  1. 0.0318 W/m^2 Correct answer.
  2. 0.127 W/m^2
  3. 0.400 W/m^2
  4. 0.0637 W/m^2

Explanation

The sound intensity is 0.0318 W/m^2. By the inverse square law for spherical wave expansion: I = P / (4π * r^2) = 10.0 W / (4π * (5.00 m)^2) = 10.0 / (4π * 25.0) = 10.0 / 314.16 = 0.03183 W/m^2 = 0.0318 W/m^2.

Question 33Waves Optics

The sound level from a point source is measured as β1 = 80.0 dB at distance r1 = 2.00 m. At what distance r2 will the sound level drop by 20.0 dB to β2 = 60.0 dB?

  1. 20.0 m Correct answer.
  2. 10.0 m
  3. 40.0 m
  4. 8.00 m

Explanation

The required distance is 20.0 m. A drop of Δβ = -20.0 dB means intensity decreases by factor of 10^2 = 100 (since 10 * log10(100) = 20.0 dB). By inverse square law I ∝ 1/r^2, so r2 / r1 = √100 = 10.0 => r2 = 10.0 * 2.00 m = 20.0 m.

Question 34Waves Optics

A guitar string of fixed length L = 0.650 m has transverse wave speed v = 416 m/s. What is the fundamental frequency f1 of the standing wave on the string?

  1. 320 Hz Correct answer.
  2. 640 Hz
  3. 160 Hz
  4. 480 Hz

Explanation

The fundamental frequency is 320 Hz. For a string fixed at both ends, the fundamental wavelength is λ1 = 2 L = 2 * 0.650 m = 1.30 m. Fundamental frequency is f1 = v / λ1 = 416 m/s / 1.30 m = 320 Hz.

Question 35Waves Optics

In a standing wave pattern, the distance between two consecutive nodes is Δx = 0.200 m. What is the wavelength λ of the interfering waves?

  1. 0.400 m Correct answer.
  2. 0.200 m
  3. 0.100 m
  4. 0.800 m

Explanation

The wavelength is 0.400 m. Node-to-node distance in a standing wave is equal to half a wavelength: Δx = λ / 2 => λ = 2 * Δx = 2 * 0.200 m = 0.400 m.

Question 36Waves Optics

An ambulance siren emits frequency fs = 400 Hz while approaching a stationary listener at speed vs = 30.0 m/s in calm air (sound speed v = 340 m/s). What frequency f' does the listener hear?

  1. 439 Hz Correct answer.
  2. 368 Hz
  3. 435 Hz
  4. 471 Hz

Explanation

The observed frequency is 439 Hz. By the Doppler effect for an approaching source: f' = fs * v / (v - vs) = 400 Hz * 340 / (340 - 30.0) = 400 * 340 / 310 = 438.7 Hz ≈ 439 Hz.

Question 37Waves Optics

A mass m = 0.200 kg on a spring (k = 80.0 N/m) oscillates with amplitude A = 0.100 m. What is its speed v when passing through position x = 0.0800 m?

  1. 1.20 m/s Correct answer.
  2. 2.00 m/s
  3. 1.60 m/s
  4. 0.800 m/s

Explanation

The speed is 1.20 m/s. Angular frequency is ω = √(k / m) = √(80.0 / 0.200) = √400 = 20.0 rad/s. Speed in SHM is v = ω * √(A^2 - x^2) = 20.0 * √((0.100)^2 - (0.0800)^2) = 20.0 * √(0.0100 - 0.00640) = 20.0 * √(0.00360) = 20.0 * 0.0600 = 1.20 m/s.

Question 38Waves Optics

What is the phase difference Δφ in radians between two points separated by spatial distance Δx = 0.150 m along a progressive wave of wavelength λ = 0.600 m?

  1. π/2 rad (90°) Correct answer.
  2. π rad (180°)
  3. π/4 rad (45°)
  4. 2π rad (360°)

Explanation

The phase difference is π/2 rad (90°). Phase difference for spatial separation is Δφ = 2π * Δx / λ = 2π * (0.150 m / 0.600 m) = 2π * 0.250 = π/2 rad (90°).

Question 39Waves Optics

A damped harmonic oscillator consists of a mass m = 0.250 kg, spring of constant k = 100 N/m, and damping constant b = 0.500 kg/s. What is its damped angular frequency ω_d?

  1. 19.98 rad/s Correct answer.
  2. 20.00 rad/s
  3. 19.90 rad/s
  4. 19.75 rad/s

Explanation

The damped angular frequency is 19.98 rad/s. Undamped angular frequency is ω0 = √(k / m) = √(100 / 0.250) = √400 = 20.00 rad/s. Damping term is γ = b / (2m) = 0.500 / (2 * 0.250) = 1.00 s^-1. Damped frequency is ω_d = √(ω0^2 - γ^2) = √(400 - 1.00) = √399 = 19.975 rad/s ≈ 19.98 rad/s.

Question 40Waves Optics

Two simple harmonic motions along the same axis are given by x1(t) = 0.0300 cos(10.0 t) and x2(t) = 0.0400 sin(10.0 t) in SI units. What is the amplitude A of the resultant oscillation x(t) = x1(t) + x2(t)?

  1. 0.0500 m (5.00 cm) Correct answer.
  2. 0.0700 m (7.00 cm)
  3. 0.0100 m (1.00 cm)
  4. 0.0350 m (3.50 cm)

Explanation

The resultant amplitude is 0.0500 m (5.00 cm). Since sin(10.0 t) = cos(10.0 t - π/2), the phase difference between x1 and x2 is Δφ = π/2 rad (90°). By phasor addition: A = √(A1^2 + A2^2) = √((0.0300)^2 + (0.0400)^2) = √(0.00090 + 0.00160) = √(0.00250) = 0.0500 m.

Question 41Modern Physics

What is the energy of a photon of green light with wavelength λ = 500 nm in vacuum (h = 6.626 x 10^-34 J s, c = 3.00 x 10^8 m/s)?

  1. 3.98 x 10^-19 J Correct answer.
  2. 1.33 x 10^-27 J
  3. 9.94 x 10^-19 J
  4. 2.48 x 10^-19 J

Explanation

The photon energy is 3.98 x 10^-19 J. Photon energy is E = h * c / λ = (6.626 x 10^-34 * 3.00 x 10^8) / (500 x 10^-9) = (1.9878 x 10^-25) / (5.00 x 10^-7) = 3.976 x 10^-19 J = 3.98 x 10^-19 J.

Question 42Modern Physics

Sodium metal has work function W0 = 2.28 eV (1 eV = 1.602 x 10^-19 J). What is its photoelectric threshold frequency f0 (h = 6.626 x 10^-34 J s)?

  1. 5.51 x 10^14 Hz Correct answer.
  2. 3.44 x 10^14 Hz
  3. 8.82 x 10^14 Hz
  4. 1.20 x 10^15 Hz

Explanation

The threshold frequency is 5.51 x 10^14 Hz. Threshold condition is W0 = h * f0 => f0 = W0 / h = (2.28 * 1.602 x 10^-19 J) / (6.626 x 10^-34 J s) = (3.6526 x 10^-19) / (6.626 x 10^-34) = 5.512 x 10^14 Hz.

Question 43Modern Physics

What is the de Broglie wavelength λ of an electron (m_e = 9.11 x 10^-31 kg) moving at non-relativistic speed v = 2.00 x 10^6 m/s (h = 6.626 x 10^-34 J s)?

  1. 0.364 nm Correct answer.
  2. 0.728 nm
  3. 0.182 nm
  4. 3.64 nm

Explanation

The de Broglie wavelength is 0.364 nm. By de Broglie relation: λ = h / p = h / (m_e * v) = (6.626 x 10^-34) / (9.11 x 10^-31 * 2.00 x 10^6) = (6.626 x 10^-34) / (1.822 x 10^-24) = 3.637 x 10^-10 m = 0.364 nm.

Question 44Modern Physics

Ultraviolet light of wavelength λ = 300 nm illuminates a potassium surface (work function W0 = 2.30 eV). What is the maximum kinetic energy Ek,max of emitted photoelectrons in eV (h c = 1240 eV nm)?

  1. 1.83 eV Correct answer.
  2. 4.13 eV
  3. 2.30 eV
  4. 6.43 eV

Explanation

The maximum kinetic energy is 1.83 eV. Incident photon energy is E = h c / λ = 1240 eV nm / 300 nm = 4.133 eV. By Einstein's photoelectric equation: Ek,max = E - W0 = 4.133 eV - 2.30 eV = 1.833 eV = 1.83 eV.

Question 45Modern Physics

In a photoelectric experiment, emitted electrons have maximum kinetic energy Ek,max = 1.84 eV. What is the stopping potential Vs required to reduce photocurrent to zero?

  1. 1.84 V Correct answer.
  2. 0.92 V
  3. 3.68 V
  4. 2.30 V

Explanation

The stopping potential is 1.84 V. Stopping potential relation is e * Vs = Ek,max => Vs = Ek,max / e = 1.84 eV / e = 1.84 V.

Question 46Modern Physics

An electron (m_e = 9.11 x 10^-31 kg, q = 1.602 x 10^-19 C) is accelerated from rest through potential difference V = 100 V. What is its de Broglie wavelength λ (h = 6.626 x 10^-34 J s)?

  1. 0.123 nm Correct answer.
  2. 0.246 nm
  3. 0.061 nm
  4. 1.23 nm

Explanation

The de Broglie wavelength is 0.123 nm. Kinetic energy gained is K = q * V = 1.602 x 10^-17 J. Electron momentum is p = √(2 * m_e * K) = √(2 * 9.11 x 10^-31 * 1.602 x 10^-17) = √(2.9188 x 10^-47) = 5.4025 x 10^-24 kg m/s. De Broglie wavelength λ = h / p = 6.626 x 10^-34 / 5.4025 x 10^-24 = 1.226 x 10^-10 m = 0.123 nm.

Question 47Modern Physics

Light of frequency f = 1.00 x 10^15 Hz strikes a metal plate with work function W0 = 3.00 eV. What is the maximum kinetic energy Ek,max of emitted photoelectrons (h = 4.136 x 10^-15 eV s)?

  1. 1.14 eV Correct answer.
  2. 4.14 eV
  3. 3.00 eV
  4. 7.14 eV

Explanation

The maximum kinetic energy is 1.14 eV. Photon energy is E = h * f = (4.136 x 10^-15 eV s) * (1.00 x 10^15 s^-1) = 4.136 eV. Photoelectric equation: Ek,max = E - W0 = 4.136 eV - 3.00 eV = 1.136 eV ≈ 1.14 eV.

Question 48Modern Physics

What is the de Broglie wavelength λ of a thermal neutron (m_n = 1.675 x 10^-27 kg) in thermal equilibrium at temperature T = 300 K (k_B = 1.381 x 10^-23 J/K, average kinetic energy K = 3/2 k_B T, h = 6.626 x 10^-34 J s)?

  1. 0.145 nm Correct answer.
  2. 0.290 nm
  3. 0.072 nm
  4. 1.45 nm

Explanation

The de Broglie wavelength is 0.145 nm. Average kinetic energy is K = (3/2) * (1.381 x 10^-23 J/K) * (300 K) = 6.2145 x 10^-21 J. Thermal momentum is p = √(2 * m_n * K) = √(2 * 1.675 x 10^-27 * 6.2145 x 10^-21) = √(2.0819 x 10^-47) = 4.563 x 10^-24 kg m/s. De Broglie wavelength λ = h / p = (6.626 x 10^-34) / (4.563 x 10^-24) = 1.452 x 10^-10 m = 0.145 nm.

Question 49Modern Physics

An electron (m_e = 9.11 x 10^-31 kg) is confined within an atomic region of width Δx = 0.100 nm. By Heisenberg's Uncertainty Principle (Δx * Δp >= h / (4π)), what is the minimum uncertainty in its velocity Δv?

  1. 5.79 x 10^5 m/s Correct answer.
  2. 1.16 x 10^6 m/s
  3. 2.89 x 10^5 m/s
  4. 7.28 x 10^6 m/s

Explanation

The minimum velocity uncertainty is 5.79 x 10^5 m/s. Heisenberg's relation Δx * (m_e * Δv) >= h / (4π) => Δv >= h / (4π * m_e * Δx) = (6.626 x 10^-34 J s) / (4π * 9.11 x 10^-31 kg * 1.00 x 10^-10 m) = (6.626 x 10^-34) / (1.1448 x 10^-39) = 5.788 x 10^5 m/s ≈ 5.79 x 10^5 m/s.

Question 50Waves Optics

Light travels through glass with refractive index n = 1.50. Taking speed of light in vacuum c = 3.00 x 10^8 m/s, what is the speed of light v in glass?

  1. 2.00 x 10^8 m/s Correct answer.
  2. 4.50 x 10^8 m/s
  3. 1.50 x 10^8 m/s
  4. 3.00 x 10^8 m/s

Explanation

The speed of light in glass is 2.00 x 10^8 m/s. Refractive index is n = c / v, so v = c / n = (3.00 x 10^8 m/s) / 1.50 = 2.00 x 10^8 m/s.