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Madrid PAU Physics practice questions

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Question 1Electromagnetic Fields

Two point charges q_1 = +2.0 µC and q_2 = +4.0 µC are placed in vacuum separated by distance d = 0.30 m. Given Coulomb constant K = 9.0 × 10⁹ N m²/C², what is the magnitude of electrostatic force between them?

  1. 0.80 N Correct answer.
  2. 2.40 N
  3. 0.24 N
  4. 8.00 N

Explanation

By Coulomb's Law, F = K * |q_1 * q_2| / d^2. Substituting values: F = (9.0 × 10⁹ × 2.0 × 10⁻⁶ × 4.0 × 10⁻⁶) / (0.30)^2 = 0.072 / 0.09 = 0.80 N.

Question 2Electromagnetic Fields

A point charge q = +5.0 nC is located on the y-axis at (0, 3.0 m). What is the electric field magnitude at the origin (0,0)? (K = 9.0 × 10⁹ N m²/C²).

  1. 5.0 N/C Correct answer.
  2. 15.0 N/C
  3. 45.0 N/C
  4. 1.67 N/C

Explanation

Electric field magnitude is E = K * |q| / r^2. Substituting: E = (9.0 × 10⁹ × 5.0 × 10⁻⁹) / (3.0)^2 = 45.0 / 9.0 = 5.0 N/C.

Question 3Electromagnetic Fields

How much work is done by the electric field when a charge q = +2.0 µC moves from point A (V_A = 100 V) to point B (V_B = 40 V)?

  1. 1.2 × 10⁻⁴ J Correct answer.
  2. -1.2 × 10⁻⁴ J
  3. 2.0 × 10⁻⁴ J
  4. 8.0 × 10⁻5 J

Explanation

Work done by conservative electric field is W_field = -Delta U = q (V_A - V_B). Substituting: W = 2.0 × 10⁻⁶ C × (100 V - 40 V) = 2.0 × 10⁻⁶ × 60 = 1.2 × 10⁻⁴ J.

Question 4Electromagnetic Fields

An electron (charge -e) is released from rest in a uniform electric field pointing in the positive x-direction. How will the electron move?

  1. It accelerates in the negative x-direction with constant acceleration. Correct answer.
  2. It accelerates in the positive x-direction with constant acceleration.
  3. It moves with constant velocity in the negative x-direction.
  4. It remains stationary because magnetic forces are absent.

Explanation

Electric force on a charge is F = q * E. For a negative charge q = -e, F = -e * E. Thus force and acceleration are directed opposite to the electric field vector E, i.e., in the negative x-direction, with constant magnitude a = e*E/m_e.

Question 5Electromagnetic Fields

An electron (m_e = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C) is accelerated from rest through an electric potential difference V = 100 V. What is its final speed?

  1. 5.93 × 10⁶ m/s Correct answer.
  2. 3.51 × 10¹³ m/s
  3. 2.97 × 10⁶ m/s
  4. 1.19 × 10⁷ m/s

Explanation

By energy conservation: 1/2 m_e v^2 = e V => v = sqrt(2 e V / m_e). Substituting: v = sqrt((2 × 1.6 × 10⁻¹⁹ × 100) / 9.1 × 10⁻³¹) = sqrt(3.516 × 10¹³) = 5.93 × 10⁶ m/s.

Question 6Electromagnetic Fields

An electron (m_e = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C) enters a uniform magnetic field B = 0.010 T perpendicular to its velocity v = 2.0 × 10⁶ m/s. What is the radius of its circular orbit?

  1. 1.14 mm Correct answer.
  2. 2.28 mm
  3. 0.57 mm
  4. 11.4 cm

Explanation

Magnetic Lorentz force supplies centripetal force: e v B = m_e v^2 / r => r = m_e v / (e B). Substituting: r = (9.1 × 10⁻³¹ × 2.0 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.010) = 1.82 × 10⁻²⁴ / 1.6 × 10⁻²¹ = 1.1375 × 10⁻³ m = 1.14 mm.

Question 7Electromagnetic Fields

How does the cyclotron orbital period T of a charged particle moving in a uniform perpendicular magnetic field depend on its speed v?

  1. T is completely independent of speed v. Correct answer.
  2. T is directly proportional to v.
  3. T is inversely proportional to v.
  4. T is directly proportional to v².

Explanation

Radius of circular path is r = m v / (q B), and orbital period is T = 2 pi r / v = 2 pi (m v / q B) / v = 2 pi m / (q B). Speed v cancels out, making period T independent of particle speed.

Question 8Electromagnetic Fields

A long straight wire carries a steady current I = 10.0 A. What is the magnetic field magnitude B at distance r = 0.050 m from the wire? (µ_0 = 4π × 10⁻⁷ T m/A).

  1. 4.0 × 10⁻⁵ T (40 µT) Correct answer.
  2. 2.0 × 10⁻⁵ T (20 µT)
  3. 1.26 × 10⁻⁴ T
  4. 8.0 × 10⁻⁵ T

Explanation

By Ampère's Law, B = µ_0 I / (2 pi r). Substituting: B = (4 pi × 10⁻⁷ × 10.0) / (2 pi × 0.050) = (2 × 10⁻⁶) / 0.050 = 4.0 × 10⁻⁵ T = 40 µT.

Question 9Electromagnetic Fields

Two long parallel straight wires carry electric currents of equal magnitude in opposite directions. The magnetic force between the two wires is:

  1. Repulsive, pushing the wires apart. Correct answer.
  2. Attractive, pulling the wires together.
  3. Zero, because opposite currents cancel magnetically.
  4. Rotational, exerting only a net torque.

Explanation

Parallel currents flowing in opposite directions produce magnetic field vectors at each other's positions such that by the right-hand rule, magnetic force pushes the wires apart (repulsive force).

Question 10Electromagnetic Fields

A circular coil of radius R = 0.10 m containing N = 50 turns carries current I = 2.0 A. What is the magnetic field magnitude at the center of the coil?

  1. 6.28 × 10⁻⁴ T Correct answer.
  2. 3.14 × 10⁻⁴ T
  3. 1.26 × 10⁻³ T
  4. 2.00 × 10⁻⁴ T

Explanation

Magnetic field at center of N-turn circular loop is B = µ_0 N I / (2 R). Substituting: B = (4 pi × 10⁻⁷ × 50 × 2.0) / (2 × 0.10) = (400 pi × 10⁻⁷) / 0.20 = 2000 pi × 10⁻⁷ = 2.0 pi × 10⁻⁴ T = 6.28 × 10⁻⁴ T.

Question 11Electromagnetic Fields

A positive charge q moves along the positive x-axis (+i) in a region with uniform magnetic field B pointing along the positive y-axis (+j). What is the direction of the magnetic force F?

  1. Positive z-direction (+k) Correct answer.
  2. Negative z-direction (-k)
  3. Positive x-direction (+i)
  4. Negative y-direction (-j)

Explanation

Magnetic force vector is F = q (v × B). For velocity in +i and field in +j, cross product is (+i) × (+j) = +k (positive z-direction). Since q is positive, force direction remains +k.

Question 12Electromagnetic Fields

In a velocity selector, mutually perpendicular electric (E = 1.2 × 10⁴ V/m) and magnetic (B = 0.40 T) fields are applied. What speed v must a charged particle have to pass through undeviated?

  1. 3.0 × 10⁴ m/s Correct answer.
  2. 4.8 × 10³ m/s
  3. 3.3 × 10⁻⁵ m/s
  4. 1.2 × 10⁴ m/s

Explanation

Undeviated motion requires electric and magnetic forces to be equal and opposite: q E = q v B => v = E / B. Substituting: v = (1.2 × 10⁴) / 0.40 = 3.0 × 10⁴ m/s.

Question 13Electromagnetic Fields

Under what condition is the magnetic flux Φ through a flat surface of area A in a uniform magnetic field B maximum?

  1. When the surface normal vector is parallel to the magnetic field lines (angle θ = 0°). Correct answer.
  2. When the surface normal vector is perpendicular to the magnetic field lines (angle θ = 90°).
  3. When the surface is inclined at 45° to the magnetic field lines.
  4. Magnetic flux is constant regardless of surface orientation.

Explanation

Magnetic flux is defined as Phi = B * A * cos(theta), where theta is the angle between the magnetic field B and the surface normal vector n. Cosine achieves its maximum value of 1.0 when theta = 0° (field parallel to surface normal / perpendicular to surface plane).

Question 14Electromagnetic Fields

A rectangular loop of area A = 0.040 m² is placed in a uniform magnetic field B = 0.50 T. If the plane of the loop makes an angle of 60° with magnetic field lines, what is the magnetic flux through the loop?

  1. 0.0173 Wb Correct answer.
  2. 0.0100 Wb
  3. 0.0200 Wb
  4. 0.0087 Wb

Explanation

The angle theta in Phi = B A cos(theta) is between magnetic field and normal vector to loop plane. If loop plane makes 60° with field lines, normal vector makes theta = 90° - 60° = 30° with field lines. Thus Phi = 0.50 × 0.040 × cos(30°) = 0.020 × (sqrt(3)/2) = 0.0173 Wb.

Question 15Electromagnetic Fields

A coil with N = 100 turns and area A = 0.020 m² is oriented perpendicular to a magnetic field. The field decreases uniformly from 0.80 T to 0.20 T in a time interval of 0.10 s. What is the magnitude of induced electromotive force (emf)?

  1. 12.0 V Correct answer.
  2. 16.0 V
  3. 1.20 V
  4. 4.0 V

Explanation

By Faraday's Law, induced emf magnitude is |emf| = N * |Delta Phi / Delta t| = N * A * |Delta B / Delta t|. Substituting: |emf| = 100 × 0.020 × (0.80 - 0.20) / 0.10 = 2.0 × (0.60 / 0.10) = 12.0 V.

Question 16Electromagnetic Fields

A flat coil with N = 200 turns and area A = 0.050 m² rotates at constant angular speed ω = 100π rad/s in a uniform magnetic field B = 0.10 T. What is the peak induced emf (emf_max)?

  1. 314 V Correct answer.
  2. 100 V
  3. 628 V
  4. 157 V

Explanation

Magnetic flux is Phi(t) = N B A cos(omega t). By Faraday's Law, emf(t) = -dPhi/dt = N B A omega sin(omega t). Peak emf is emf_max = N B A omega = 200 × 0.10 × 0.050 × 100 pi = 1.0 × 100 pi = 100 pi V ≈ 314 V.

Question 17Electromagnetic Fields

A conducting rod of length L = 0.50 m slides at constant speed v = 4.0 m/s along frictionless parallel rails in a uniform magnetic field B = 0.80 T perpendicular to the rail plane. What is the motional emf induced in the rod?

  1. 1.60 V Correct answer.
  2. 0.80 V
  3. 3.20 V
  4. 0.40 V

Explanation

Motional emf in a straight conductor moving perpendicular to B field is emf = B * L * v. Substituting: emf = 0.80 T × 0.50 m × 4.0 m/s = 1.60 V.

Question 18Electromagnetic Fields

Lenz's Law states that the direction of an induced electric current always:

  1. Opposes the change in magnetic flux that causes it. Correct answer.
  2. Aligns with the external magnetic field direction.
  3. Maximizes the rate of change of magnetic flux.
  4. Performs negative work on electrostatic potential charges.

Explanation

Lenz's Law is a consequence of energy conservation: induced currents produce magnetic fields that oppose the change in magnetic flux driving the induction.

Question 19Electromagnetic Fields

How much work is required to move a point charge q between two points situated on the same electrostatic equipotential surface?

  1. Zero Correct answer.
  2. q * V
  3. 1/2 q * V²
  4. Depends on path length between points.

Explanation

Work done by electrostatic force is W = -q Delta V. Since all points on an equipotential surface have equal electric potential (Delta V = 0), work done moving a charge along the surface is zero.

Question 20Electromagnetic Fields

A charged particle enters a uniform magnetic field B with velocity v directed at angle θ = 45° to the field lines. What trajectory will the particle follow?

  1. A helical trajectory around the magnetic field lines. Correct answer.
  2. A circular path in a plane perpendicular to the field.
  3. A straight line parallel to the magnetic field lines.
  4. A parabolic path toward the magnetic field boundary.

Explanation

Velocity component parallel to field (v_parallel = v cos 45°) remains constant because magnetic force on it is zero. Perpendicular component (v_perp = v sin 45°) produces circular motion. Superposition of uniform motion along field and circular motion perpendicular to field results in a helical trajectory.

Question 21Electromagnetic Fields

Two long parallel straight wires separated by distance d = 0.20 m carry currents I_1 = 5.0 A and I_2 = 10.0 A. What is the magnitude of force per unit length (F/L) between them?

  1. 5.0 × 10⁻⁵ N/m Correct answer.
  2. 2.5 × 10⁻⁵ N/m
  3. 1.0 × 10⁻⁴ N/m
  4. 5.0 × 10⁻⁶ N/m

Explanation

Force per unit length between parallel wires is F/L = µ_0 I_1 I_2 / (2 pi d). Substituting: F/L = (4 pi × 10⁻⁷ × 5.0 × 10.0) / (2 pi × 0.20) = (2 × 10⁻⁶ × 50) / 0.20 = (1.0 × 10⁻⁴) / 0.20 = 5.0 × 10⁻⁵ N/m.

Question 22Electromagnetic Fields

In electrostatic equilibrium, what are the internal electric field E_in and electric potential V_in inside a hollow spherical conductor carrying net charge Q and radius R?

  1. E_in = 0, V_in = K Q / R (uniform throughout interior) Correct answer.
  2. E_in = K Q / R², V_in = 0
  3. E_in = 0, V_in = 0
  4. E_in = K Q / r², V_in = K Q / r

Explanation

By Gauss's Law, free charges reside entirely on conductor surface in electrostatic equilibrium, so E_in = 0 inside cavity. Since E = -grad V = 0, electric potential V is constant throughout conductor volume and equal to surface value V_surface = K Q / R.

Question 23Electromagnetic Fields

A closed conductive loop with total electrical resistance R = 4.0 Ω experiences an induced emf of magnitude ε = 12.0 V. What is the rate of thermal energy dissipation (Joule heating power P) in the loop?

  1. 36.0 W Correct answer.
  2. 48.0 W
  3. 3.0 W
  4. 144.0 W

Explanation

Induced current is I = emf / R = 12.0 V / 4.0 Ω = 3.0 A. Thermal power dissipated by Joule heating is P = I^2 * R = (3.0)^2 × 4.0 = 9.0 × 4.0 = 36.0 W (or P = emf^2 / R = 144 / 4 = 36.0 W).

Question 24Electromagnetic Fields

Which SI units correctly correspond to magnetic flux (Φ) and magnetic field induction (B), respectively?

  1. Weber (Wb) and Tesla (T) Correct answer.
  2. Tesla (T) and Weber (Wb)
  3. Henry (H) and Farad (F)
  4. Gauss (G) and Maxwell (Mx)

Explanation

In SI units, magnetic flux Phi is measured in Weber (Wb = T m²), and magnetic field induction B is measured in Tesla (T = Wb/m²).

Question 25Electromagnetic Fields

A proton moves with speed v directly parallel to uniform magnetic field lines B (angle θ = 0°). What magnetic force acts on the proton?

  1. Zero force Correct answer.
  2. q v B
  3. 1/2 q v B
  4. q v B / m

Explanation

Magnetic force magnitude is F = q v B sin(theta). When velocity vector is parallel to magnetic field lines, theta = 0°, so sin(0°) = 0, resulting in zero magnetic force.

Question 26Waves Optics

A particle undergoes Simple Harmonic Motion (SHM) according to equation x(t) = 0.10 sin(10π t + π/4) in SI units. What is its oscillation period T?

  1. 0.20 s Correct answer.
  2. 5.0 s
  3. 0.10 s
  4. 10.0 s

Explanation

In SHM equation x(t) = A sin(omega t + phi_0), angular frequency is omega = 10 pi rad/s. Oscillation period is T = 2 pi / omega = 2 pi / (10 pi) = 0.20 s.

Question 27Waves Optics

A harmonic oscillator has amplitude A = 0.050 m and frequency f = 5.0 Hz. What are its maximum speed v_max and maximum acceleration a_max?

  1. v_max = 1.57 m/s, a_max = 49.3 m/s² Correct answer.
  2. v_max = 0.25 m/s, a_max = 1.25 m/s²
  3. v_max = 3.14 m/s, a_max = 98.7 m/s²
  4. v_max = 1.57 m/s, a_max = 15.7 m/s²

Explanation

Angular frequency is omega = 2 pi f = 10 pi rad/s ≈ 31.416 rad/s. Maximum speed v_max = A omega = 0.050 × 10 pi = 0.50 pi ≈ 1.57 m/s. Maximum acceleration a_max = A omega^2 = 0.050 × (10 pi)^2 = 5.0 pi^2 ≈ 49.35 m/s².

Question 28Waves Optics

A mass m = 0.20 kg attached to a spring oscillates with amplitude A = 0.10 m at frequency f = 10.0 Hz. What is the total mechanical energy of the oscillator?

  1. 3.95 J Correct answer.
  2. 1.97 J
  3. 7.89 J
  4. 0.39 J

Explanation

Angular frequency omega = 2 pi f = 20 pi rad/s. Total mechanical energy is E = 1/2 m omega^2 A^2 = 0.5 × 0.20 × (20 pi)^2 × (0.10)^2 = 0.10 × (400 pi^2) × 0.01 = 0.40 pi^2 ≈ 3.948 J.

Question 29Waves Optics

At what displacement x from the equilibrium position in SHM is kinetic energy E_k equal to potential energy E_p?

  1. x = ± A / √2 (approx. ±0.707 A) Correct answer.
  2. x = ± A / 2
  3. x = ± A / 4
  4. x = 0

Explanation

Potential energy is E_p = 1/2 k x^2 and total energy is E = 1/2 k A^2. When E_k = E_p, E_p = 1/2 E => 1/2 k x^2 = 1/2 (1/2 k A^2) => x^2 = A^2 / 2 => x = ± A / sqrt(2) ≈ ±0.707 A.

Question 30Waves Optics

What fundamental relation links wave speed v, wavelength λ, frequency f, and period T for a harmonic wave?

  1. v = λ f = λ / T Correct answer.
  2. v = λ T = λ / f
  3. v = f / λ
  4. v = T / λ

Explanation

Wave speed is distance traveled per unit time. One wavelength lambda is traversed in one period T, so v = lambda / T. Since f = 1/T, v = lambda * f.

Question 31Waves Optics

A progressive transverse wave is described by y(x,t) = 0.020 sin(100π t - 4π x) in SI units. What are its propagation speed v and wavelength λ?

  1. v = 25 m/s, λ = 0.50 m Correct answer.
  2. v = 50 m/s, λ = 1.00 m
  3. v = 25 m/s, λ = 2.00 m
  4. v = 100 m/s, λ = 0.25 m

Explanation

From general wave equation y(x,t) = A sin(omega t - k x), angular frequency omega = 100 pi rad/s (f = 50 Hz) and wave number k = 4 pi rad/m. Wavelength is lambda = 2 pi / k = 2 pi / (4 pi) = 0.50 m. Wave speed is v = omega / k = (100 pi) / (4 pi) = 25 m/s.

Question 32Waves Optics

What is the phase difference Δφ between two points along a wave path separated by distance Δx = 0.25 m, if wavelength λ = 1.00 m?

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

Explanation

Phase difference is related to spatial path difference by Delta phi = (2 pi / lambda) * Delta x. Substituting values: Delta phi = (2 pi / 1.00) × 0.25 = 0.50 pi rad = pi/2 rad (90°).

Question 33Waves Optics

For a transverse wave y(x,t) = 0.040 sin(20π t - 2π x) in SI units, what is the maximum transverse velocity v_y,max of a medium particle?

  1. 2.51 m/s (0.80π m/s) Correct answer.
  2. 1.26 m/s (0.40π m/s)
  3. 10.0 m/s
  4. 0.08 m/s

Explanation

Transverse particle velocity is v_y = dy/dt = A omega cos(omega t - k x). Maximum transverse speed is v_y,max = A omega. Here A = 0.040 m and omega = 20 pi rad/s, so v_y,max = 0.040 × 20 pi = 0.80 pi ≈ 2.513 m/s.

Question 34Waves Optics

A point sound source emits spherical waves uniformly into non-absorbing air. If sound intensity at distance r_1 = 2.0 m is I_1 = 1.0 × 10⁻⁴ W/m², what is the intensity I_2 at distance r_2 = 4.0 m?

  1. 2.5 × 10⁻⁵ W/m² Correct answer.
  2. 5.0 × 10⁻⁵ W/m²
  3. 1.25 × 10⁻⁵ W/m²
  4. 4.0 × 10⁻⁴ W/m²

Explanation

For spherical waves, intensity follows inverse-square law: I_1 * r_1^2 = I_2 * r_2^2 => I_2 = I_1 * (r_1 / r_2)^2. Substituting: I_2 = 1.0 × 10⁻⁴ × (2.0 / 4.0)^2 = 1.0 × 10⁻⁴ × 0.25 = 2.5 × 10⁻⁵ W/m².

Question 35Waves Optics

What is the sound intensity level β in decibels (dB) for a sound wave of intensity I = 1.0 × 10⁻⁶ W/m²? (Reference threshold I_0 = 1.0 × 10⁻¹² W/m²).

  1. 60 dB Correct answer.
  2. 120 dB
  3. 30 dB
  4. 90 dB

Explanation

Sound intensity level in decibels is beta = 10 log_10(I / I_0). Substituting: beta = 10 log_10(1.0 × 10⁻⁶ / 1.0 × 10⁻¹²) = 10 log_10(10⁶) = 10 × 6 = 60 dB.

Question 36Waves Optics

If distance from a point sound source is quadrupled (r_2 = 4 r_1), by how many decibels does the sound intensity level decrease?

  1. 12.0 dB decrease Correct answer.
  2. 6.0 dB decrease
  3. 4.0 dB decrease
  4. 24.0 dB decrease

Explanation

By inverse-square law, I_2 / I_1 = (r_1 / r_2)^2 = (1/4)^2 = 1/16. Change in decibel level is Delta beta = 10 log_10(I_2 / I_1) = 10 log_10(1/16) = -10 log_10(16) ≈ -10 × 1.204 = -12.04 dB (decrease of ~12 dB).

Question 37Waves Optics

A ambulance siren emits sound at frequency f = 500 Hz. If the ambulance approaches a stationary observer at speed v_s = 34.0 m/s, what frequency f' does the observer hear? (Speed of sound in air v = 340 m/s).

  1. 555.6 Hz Correct answer.
  2. 453.3 Hz
  3. 550.0 Hz
  4. 450.0 Hz

Explanation

For a source approaching a stationary observer, Doppler formula is f' = f * [v / (v - v_s)]. Substituting: f' = 500 × [340 / (340 - 34)] = 500 × (340 / 306) = 500 × 1.1111 = 555.6 Hz.

Question 38Waves Optics

In a standing wave, what is the distance between two consecutive nodes?

  1. λ / 2 (half wavelength) Correct answer.
  2. λ (one full wavelength)
  3. λ / 4 (quarter wavelength)
  4. 2λ (two wavelengths)

Explanation

Nodes are points of zero displacement in standing waves. Distance between adjacent nodes (or adjacent antinodes) is exactly half a wavelength (lambda / 2). Distance between a node and adjacent antinode is lambda / 4.

Question 39Waves Optics

A guitar string of length L = 0.80 m fixed at both ends has transverse wave speed v = 320 m/s. What is its fundamental frequency f_1?

  1. 200 Hz Correct answer.
  2. 400 Hz
  3. 100 Hz
  4. 250 Hz

Explanation

For a string fixed at both ends, fundamental mode has L = lambda_1 / 2 => lambda_1 = 2 L = 1.60 m. Fundamental frequency is f_1 = v / (2 L) = 320 / (2 × 0.80) = 320 / 1.60 = 200 Hz.

Question 40Waves Optics

What distinguishes transverse waves from longitudinal waves?

  1. In transverse waves, medium particle oscillation is perpendicular to wave propagation direction; in longitudinal waves, oscillation is parallel to propagation direction. Correct answer.
  2. Transverse waves require a material medium to propagate, whereas longitudinal waves can propagate through vacuum.
  3. Transverse waves travel faster than longitudinal waves in all media.
  4. Longitudinal waves exhibit polarization, while transverse waves do not.

Explanation

By definition, transverse waves involve medium oscillations perpendicular to energy propagation (e.g., light, string waves), whereas longitudinal waves involve oscillations parallel to propagation direction (e.g., sound in fluid). Only transverse waves can be polarized.

Question 41Waves Optics

A simple pendulum of length L has period T. If the pendulum length is quadrupled (L' = 4L), how does its period change?

  1. The period doubles (T' = 2T). Correct answer.
  2. The period quadruples (T' = 4T).
  3. The period is halved (T' = 0.5T).
  4. The period increases by factor of 16.

Explanation

Period of a simple pendulum is T = 2 pi sqrt(L / g). Quadrupling length L' = 4L gives T' = 2 pi sqrt(4L / g) = 2 * (2 pi sqrt(L / g)) = 2 T. Thus period doubles.

Question 42Waves Optics

A mass m = 0.50 kg is connected to a spring of force constant k = 200 N/m. What is its oscillation period T?

  1. 0.314 s (0.10π s) Correct answer.
  2. 0.157 s
  3. 0.628 s
  4. 3.18 s

Explanation

Period of mass-spring system is T = 2 pi sqrt(m / k). Substituting: T = 2 pi sqrt(0.50 / 200) = 2 pi sqrt(0.0025) = 2 pi × 0.050 = 0.10 pi s ≈ 0.3142 s.

Question 43Waves Optics

Two coherent point wave sources S_1 and S_2 emit in-phase waves of wavelength λ = 0.40 m. At a point P, path distances are r_1 = 2.20 m and r_2 = 1.40 m. What type of interference occurs at P?

  1. Fully constructive interference (path difference is an integer multiple of λ) Correct answer.
  2. Fully destructive interference (path difference is a half-integer multiple of λ)
  3. Partial interference with zero amplitude
  4. No wave interaction occurs

Explanation

Path difference is Delta r = |r_1 - r_2| = |2.20 - 1.40| = 0.80 m. Since Delta r / lambda = 0.80 / 0.40 = 2 (an integer n = 2), waves arrive in phase, resulting in fully constructive interference.

Question 44Waves Optics

According to Huygens' Principle, how does a primary wave front propagate through space?

  1. Every point on a primary wave front acts as a point source of secondary spherical wavelets that advance at the wave speed. Correct answer.
  2. Wave fronts propagate only along rectilinear ray lines without spreading.
  3. Secondary wavelets propagate backward toward the original source.
  4. Wave energy accumulates only at discrete nodal boundaries.

Explanation

Huygens' Principle states that every point on a wavefront serves as a source of secondary spherical wavelets. The new wavefront at a later time is the envelope tangent to these secondary wavelets.

Question 45Waves Optics

If 10 identical independent sound sources each produce a sound intensity level of 60 dB at a given point, what is the combined sound intensity level?

  1. 70 dB Correct answer.
  2. 600 dB
  3. 63 dB
  4. 80 dB

Explanation

For independent incoherent sources, total intensity is sum of individual intensities: I_total = 10 * I_1. Combined decibel level is beta_total = 10 log_10(10 I_1 / I_0) = 10 log_10(I_1 / I_0) + 10 log_10(10) = 60 + 10 = 70 dB.

Question 46Waves Optics

An observer moves at speed v_o = 17.0 m/s toward a stationary sound source emitting frequency f = 400 Hz. What frequency f' does the observer perceive? (Sound speed v = 340 m/s).

  1. 420 Hz Correct answer.
  2. 380 Hz
  3. 440 Hz
  4. 400 Hz

Explanation

For observer moving toward a stationary source, Doppler formula is f' = f * [(v + v_o) / v]. Substituting: f' = 400 × [(340 + 17) / 340] = 400 × (357 / 340) = 400 × 1.05 = 420 Hz.

Question 47Waves Optics

A flexible string of linear mass density µ = 0.010 kg/m is under tension T_s = 100 N. What is the speed of transverse waves on this string?

  1. 100 m/s Correct answer.
  2. 10 m/s
  3. 10,000 m/s
  4. 1,000 m/s

Explanation

Wave speed on a stretched string is given by v = sqrt(T_s / µ). Substituting values: v = sqrt(100 / 0.010) = sqrt(10,000) = 100 m/s.

Question 48Waves Optics

Which harmonics can resonate inside an organ pipe of length L closed at one end and open at the other?

  1. Only odd harmonics (n = 1, 3, 5, ...) Correct answer.
  2. Only even harmonics (n = 2, 4, 6, ...)
  3. All integer harmonics (n = 1, 2, 3, 4, ...)
  4. No standing wave harmonics can form in closed pipes.

Explanation

A closed-open pipe has a displacement node at closed end and antinode at open end. Supported resonant frequencies are f_n = n v / (4L) for odd integers n = 1, 3, 5, ... Even harmonics are absent.

Question 49Waves Optics

What is the defining characteristic of underdamped harmonic motion?

  1. Oscillation amplitude decays exponentially with time while frequency remains nearly constant. Correct answer.
  2. Oscillation frequency increases exponentially over time.
  3. Total mechanical energy remains strictly conserved.
  4. The system returns to equilibrium without making any oscillations.

Explanation

In underdamped harmonic motion, dissipative resistive forces cause mechanical energy loss, resulting in an exponential amplitude decay A(t) = A_0 e^(-gamma t) while oscillating at damped frequency omega_d.

Question 50Waves Optics

Resonance in a driven mechanical oscillator occurs when:

  1. The external driving frequency matches the natural frequency of the oscillator, producing maximum steady-state amplitude. Correct answer.
  2. The driving frequency is much higher than the natural frequency.
  3. Damping force becomes zero.
  4. Potential energy becomes zero at all points.

Explanation

Resonance occurs when the frequency of periodic driving force equals natural frequency omega_0 of the system. At resonance, energy transfer to oscillator is maximized, producing peak oscillation amplitude.