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Castilla-La Mancha PAU Physics (Física - UCLM 2026) study materials
Castilla-La Mancha PAU Physics (Física - UCLM 2026) study resources

Castilla-La Mancha PAU Physics (Física - UCLM 2026) practice questions

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Question 1Electromagnetic Fields

Two point charges q₁ = +4.0 μC and q₂ = -2.0 μC are separated by a distance d = 0.30 m in vacuum. What is the magnitude of the electrostatic force between them? (K = 8.99×10⁹ N·m²/C²)

  1. 0.24 N
  2. 0.80 N Correct answer.
  3. 2.67 N
  4. 0.08 N

Explanation

F = K · |q₁ · q₂| / d² = 8.99×10⁹ · (4.0×10⁻⁶ · 2.0×10⁻⁶) / (0.30)² = 8.99×10⁹ · (8.0×10⁻¹²) / 0.09 = 0.07192 / 0.09 ≈ 0.80 N.

Question 2Electromagnetic Fields

A point charge q = +3.0 nC is placed at a point where the electric field is E = (400 i + 300 j) N/C. What is the magnitude of the electric force experienced by the charge?

  1. 1.5×10⁻⁶ N Correct answer.
  2. 2.1×10⁻⁶ N
  3. 0.9×10⁻⁶ N
  4. 1.2×10⁻⁶ N

Explanation

The magnitude of the electric field is |E| = √(400² + 300²) = √(160000 + 90000) = √250000 = 500 N/C. The force is F = q · |E| = 3.0×10⁻⁹ C · 500 N/C = 1.5×10⁻⁶ N.

Question 3Electromagnetic Fields

What work does the electric field do to move a proton (q = +1.60×10⁻¹⁹ C) from a point A with potential V_A = 100 V to a point B with V_B = 40 V?

  1. -9.6×10⁻¹⁸ J
  2. +9.6×10⁻¹⁸ J Correct answer.
  3. +2.24×10⁻¹⁷ J
  4. -2.24×10⁻¹⁷ J

Explanation

W_field = -q · ΔV = -q · (V_B - V_A) = -1.60×10⁻¹⁹ · (40 - 100) = -1.60×10⁻¹⁹ · (-60) = +9.6×10⁻¹⁸ J. By going from higher to lower potential, a positive charge moves spontaneously and the field does positive work.

Question 4Electromagnetic Fields

An electron (q = -1.60×10⁻¹⁹ C, m = 9.11×10⁻³¹ kg) enters perpendicularly with velocity v = 2.0×10⁶ m/s into a uniform magnetic field B = 0.50 T. What is the radius r of its circular trajectory?

  1. 9.12×10⁻5 m
  2. 4.56×10⁻5 m
  3. 1.14×10⁻5 m
  4. 2.28×10⁻5 m Correct answer.

Explanation

Equating the magnetic force to the centripetal force: |q| v B = m v² / r ⇒ r = m v / (|q| B) = (9.11×10⁻³¹ · 2.0×10⁶) / (1.60×10⁻¹⁹ · 0.50) = (1.822×10⁻²⁴) / (8.0×10⁻²⁰) ≈ 2.28×10⁻⁵ m.

Question 5Electromagnetic Fields

A straight, infinitely long conducting wire carries an electric current I = 5.0 A. What is the magnitude of the magnetic field B at a distance r = 0.10 m from the wire? (μ₀ = 4π×10⁻⁷ T·m/A)

  1. 2.0×10⁻5 T
  2. 3.14×10⁻5 T
  3. 5.0×10⁻6 T
  4. 1.0×10⁻5 T Correct answer.

Explanation

Applying Ampère's law for a straight wire: B = μ₀ I / (2π r) = (4π×10⁻⁷ · 5.0) / (2π · 0.10) = 2×10⁻7 · 5.0 / 0.10 = 1.0×10⁻5 T.

Question 6Electromagnetic Fields

A circular loop of radius r = 0.10 m (surface S = π r²) is placed perpendicularly to a magnetic field that varies with time according to B(t) = 0.20 · t + 0.05 (in T). What is the absolute value of the induced electromotive force (emf) in the loop?

  1. 1.57×10⁻³ V
  2. 6.28×10⁻³ V Correct answer.
  3. 3.14×10⁻² V
  4. 1.26×10⁻² V

Explanation

The magnetic flux is Φ = B · S = (0.20 t + 0.05) · π (0.10)² = (0.20 t + 0.05) · 0.01 π. According to Faraday-Lenz's law: |E| = |dΦ/dt| = 0.20 · 0.01 π = 0.002 π V ≈ 6.28×10⁻³ V.

Question 7Electromagnetic Fields

A conducting segment of length L = 0.40 m moves with constant velocity v = 5.0 m/s perpendicularly to a uniform magnetic field B = 0.30 T. What potential difference (induced emf) is generated between the ends of the segment?

  1. 1.20 V
  2. 0.15 V
  3. 2.40 V
  4. 0.60 V Correct answer.

Explanation

The emf induced by motion in a straight conductor is |E| = B · L · v = 0.30 T · 0.40 m · 5.0 m/s = 0.60 V.

Question 8Electromagnetic Fields

Two straight, parallel, very long conductors 0.20 m apart carry currents of 10 A and 20 A in the same direction. What is the force per unit length (F/L) that they exert on each other and what is its nature (attractive or repulsive)?

  1. 2.0×10⁻⁴ N/m, repulsive
  2. 4.0×10⁻⁴ N/m, attractive
  3. 2.0×10⁻⁴ N/m, attractive Correct answer.
  4. 4.0×10⁻⁴ N/m, repulsive

Explanation

The force per unit length between two parallel currents is F/L = μ₀ I₁ I₂ / (2π d) = (4π×10⁻⁷ · 10 · 20) / (2π · 0.20) = (2×10⁻⁷ · 200) / 0.20 = 4.0×10⁻⁵ / 0.20 = 2.0×10⁻⁴ N/m. Parallel currents in the same direction attract each other, so the force is attractive.

Question 9Electromagnetic Fields

According to Lenz's law, the induced current in a closed circuit has a direction such that:

  1. It always reinforces the external magnetic field applied to the circuit
  2. It completely cancels the electrical resistance of the circuit
  3. It maximizes the magnetic flux through the surface of the circuit
  4. It always opposes the variation of the magnetic flux that produces it Correct answer.

Explanation

Lenz's law is a manifestation of the principle of conservation of energy: the direction of the induced current creates an induced magnetic field that opposes the variation of the external magnetic flux.

Question 10Electromagnetic Fields

An ideal solenoid of N = 500 turns and length L = 0.25 m carries a current I = 2.0 A. What is the magnetic field intensity B inside the solenoid? (μ₀ = 4π×10⁻⁷ T·m/A)

  1. 2.51×10⁻³ T
  2. 5.03×10⁻³ T Correct answer.
  3. 1.26×10⁻² T
  4. 1.01×10⁻³ T

Explanation

The field inside a solenoid is B = μ₀ · (N / L) · I = (4π×10⁻⁷) · (500 / 0.25) · 2.0 = 4π×10⁻⁷ · 2000 · 2.0 = 1.6π×10⁻³ ≈ 5.03×10⁻³ T.

Question 11Electromagnetic Fields

What is the work needed to bring a charge q = +2.0 μC from infinity to the center of a square of side a = 0.40 m at whose vertices are four identical charges Q = +5.0 μC? (K = 8.99×10⁹ N·m²/C²)

  1. +1.27 J Correct answer.
  2. +0.64 J
  3. -0.64 J
  4. +0.32 J

Explanation

The distance from the center to each vertex is r = (a √2) / 2 = (0.40 · 1.4142) / 2 = 0.2828 m. The total potential at the center is V = 4 · (K Q / r) = 4 · (8.99×10⁹ · 5.0×10⁻⁶ / 0.2828) = 635 700 V. The external work is W_ext = q · V = 2.0×10⁻⁶ C · 635 700 V ≈ +1.27 J.

Question 12Electromagnetic Fields

A proton (q = +1.60×10⁻¹⁹ C) is accelerated from rest through a potential difference ΔV = 500 V. What kinetic energy E_k does it acquire at the end of the path?

  1. 1.6×10⁻¹⁷ J
  2. 8.0×10⁻¹⁷ J Correct answer.
  3. 3.2×10⁻¹⁷ J
  4. 4.0×10⁻¹⁷ J

Explanation

By conservation of energy: E_k = q · ΔV = 1.60×10⁻¹⁹ C · 500 V = 8.0×10⁻¹⁷ J (equivalent to 500 eV).

Question 13Electromagnetic Fields

The magnetic flux through a circuit is given by Φ(t) = (t³ - 6t² + 2) Wb. At what instant t > 0 does the induced electromotive force |E| vanish?

  1. t = 2 s
  2. t = 6 s
  3. t = 4 s Correct answer.
  4. t = 3 s

Explanation

The induced emf is |E| = |dΦ/dt| = |3t² - 12t|. For |E| = 0: 3t² - 12t = 0 ⇒ 3t(t - 4) = 0. Given t > 0, the solution is t = 4 s.

Question 14Electromagnetic Fields

A flat coil of N = 100 turns of radius r = 0.05 m rotates at 50 Hz in a uniform magnetic field B = 0.10 T. What is the peak value (amplitude) of the induced alternating emf E_max?

  1. 7.85 V
  2. 15.71 V
  3. 24.67 V Correct answer.
  4. 49.35 V

Explanation

Surface of the loop S = π (0.05)² = 0.0025 π m². Angular frequency ω = 2π f = 2π · 50 = 100π rad/s. The amplitude of the emf is E_max = N · B · S · ω = 100 · 0.10 · (0.0025 π) · (100 π) = 2.5 π² V ≈ 2.5 · 9.8696 ≈ 24.67 V.

Question 15Electromagnetic Fields

What is the direction of the magnetic force on a particle with positive charge q > 0 moving in the positive direction of the X axis (v = v i) in a region where there is a magnetic field directed in the positive direction of the Y axis (B = B j)?

  1. Positive direction of the X axis (+i)
  2. Negative direction of the Z axis (-k)
  3. Positive direction of the Z axis (+k) Correct answer.
  4. Negative direction of the Y axis (-j)

Explanation

The Lorentz force is F = q (v × B) = q (v i × B j) = q v B (i × j) = +q v B k. Therefore, the direction is the positive direction of the Z axis (+k).

Question 16Electromagnetic Fields

A velocity selector uses an electric field E = 2.0×10⁴ N/C and a magnetic field B = 0.05 T perpendicular to each other. What velocity v must charged particles have to pass through without being deflected?

  1. 1.0×10³ m/s
  2. 4.0×10⁵ m/s Correct answer.
  3. 2.5×10⁻⁶ m/s
  4. 8.0×10⁵ m/s

Explanation

For no deflection, the electric force and the magnetic force must be equal in magnitude: q E = q v B ⇒ v = E / B = 2.0×10⁴ / 0.05 = 4.0×10⁵ m/s.

Question 17Electromagnetic Fields

Two point charges +Q and +4Q are fixed at a distance L. At which point on the line joining them is the resultant electric field completely zero?

  1. At 2L/3 from the charge +Q
  2. At L/2 from the charge +Q
  3. At L/4 from the charge +Q
  4. At L/3 from the charge +Q Correct answer.

Explanation

Let x be the distance from +Q. For E₁ = E₂: K Q / x² = K (4Q) / (L - x)² ⇒ 1 / x² = 4 / (L - x)² ⇒ 1 / x = 2 / (L - x) ⇒ L - x = 2x ⇒ 3x = L ⇒ x = L/3.

Question 18Electromagnetic Fields

Calculate the magnetic flux through a flat surface of area S = 0.02 m² immersed in a uniform field B = 0.50 T that forms an angle of 60° with the normal to the surface.

  1. 1.0×10⁻² Wb
  2. 8.66×10⁻³ Wb
  3. 5.0×10⁻³ Wb Correct answer.
  4. 2.5×10⁻³ Wb

Explanation

The magnetic flux is Φ = B · S · cos(θ) = 0.50 T · 0.02 m² · cos(60°) = 0.01 · 0.5 = 0.005 Wb = 5.0×10⁻³ Wb.

Question 19Electromagnetic Fields

A conducting wire of 0.50 m length carries a current I = 4.0 A and is placed in a uniform magnetic field B = 0.20 T. If the wire forms an angle of 30° with the direction of the magnetic field, what is the magnetic force on the wire?

  1. 0.10 N
  2. 0.35 N
  3. 0.40 N
  4. 0.20 N Correct answer.

Explanation

The magnetic force on a conductor is F = I · L · B · sin(θ) = 4.0 A · 0.50 m · 0.20 T · sin(30°) = 0.40 · 0.5 = 0.20 N.

Question 20Electromagnetic Fields

How is the self-inductance L of an electric circuit defined and what is its unit in the International System?

  1. L = Φ / I, expressed in Henries (H = Wb/A) Correct answer.
  2. L = I / Φ, expressed in Farads (F)
  3. L = dI / dt, expressed in Volts (V)
  4. L = B · S, expressed in Teslas (T)

Explanation

Self-inductance is the proportionality constant between the own magnetic flux and the current that produces it (Φ = L · I ⇒ L = Φ / I). Its SI unit is the Henry (H = Wb/A = V·s/A).

Question 21Electromagnetic Fields

An ideal transformer has N_p = 1000 turns in the primary and N_s = 200 turns in the secondary. If the primary is connected to an alternating voltage of 220 V, what is the voltage in the secondary?

  1. 1100 V
  2. 44 V Correct answer.
  3. 22 V
  4. 88 V

Explanation

The transformation ratio is V_s / V_p = N_s / N_p ⇒ V_s = V_p · (N_s / N_p) = 220 V · (200 / 1000) = 220 · 0.2 = 44 V.

Question 22Electromagnetic Fields

Three equal point charges of +2.0 μC are located at the vertices of an equilateral triangle of side a = 0.20 m. What is the magnitude of the electric field at the center of the triangle?

  1. 0 N/C Correct answer.
  2. 2.7×10⁵ N/C
  3. 5.4×10⁵ N/C
  4. 1.35×10⁵ N/C

Explanation

By geometric symmetry, the three electric field vectors produced by identical charges located at the same distance from the center form angles of 120° between them and their vector sum gives exactly 0 N/C.

Question 23Electromagnetic Fields

An electron moves with velocity v in the same direction and sense as a current I flowing through a straight, infinitely long conducting wire. In which direction does the magnetic force acting on the electron point?

  1. Toward the wire (attractive force)
  2. Away from the wire (repulsive force) Correct answer.
  3. In the direction of the electron's motion
  4. It is zero since the velocity is parallel to the current

Explanation

The wire creates a magnetic field B whose lines are concentric circles. At the electron's position, by the right-hand rule, B is perpendicular to the wire. The force F = q (v × B). For q > 0 the force would attract toward the wire, but since for the electron q < 0, the force points away from the wire (repulsive).

Question 24Electromagnetic Fields

According to Gauss's theorem for the magnetic field (∮ B · dS = 0), which of the following statements is correct?

  1. The magnetic field is a conservative field
  2. Isolated magnetic monopoles do not exist in nature Correct answer.
  3. Magnetic field lines originate at positive charges and terminate at negative charges
  4. The magnetic flux through any open surface is always zero

Explanation

The fact that the magnetic flux through any closed surface is zero implies that the lines of B are always closed upon themselves, which manifests the nonexistence of isolated magnetic monopoles (sources or sinks).

Question 25Electromagnetic Fields

A circular surface of radius R = 0.10 m is uniformly charged with a surface charge density σ = +5.0 nC/m². What is the total charge Q accumulated on the surface?

  1. 1.57×10⁻¹⁰ C Correct answer.
  2. 3.14×10⁻¹⁰ C
  3. 5.0×10⁻¹⁰ C
  4. 6.28×10⁻¹⁰ C

Explanation

The area of the circular surface is S = π R² = π (0.10)² = 0.01 π m². The total charge is Q = σ · S = 5.0×10⁻⁹ C/m² · 0.01 π m² = 5.0×10⁻¹¹ π C ≈ 1.57×10⁻¹⁰ C.

Question 26Waves Optics

The equation of a transverse harmonic wave propagating along a string is y(x,t) = 0.04 · sin(10π t - 2π x) (in SI units). What is the propagation velocity of the wave?

  1. 10.0 m/s
  2. 0.2 m/s
  3. 5.0 m/s Correct answer.
  4. 2.5 m/s

Explanation

The angular frequency is ω = 10π rad/s and the wave number is k = 2π rad/m. The propagation velocity is v = ω / k = (10π) / (2π) = 5.0 m/s.

Question 27Waves Optics

A ray of light travels through air (n₁ = 1.00) and strikes the surface of glass (n₂ = 1.50) with an angle of incidence θ₁ = 30°. What is the sine of the angle of refraction sin(θ₂)?

  1. 0.500
  2. 0.750
  3. 0.333 Correct answer.
  4. 0.250

Explanation

According to Snell's law: n₁ · sin(θ₁) = n₂ · sin(θ₂) ⇒ 1.00 · sin(30°) = 1.50 · sin(θ₂) ⇒ 1.00 · 0.500 = 1.50 · sin(θ₂) ⇒ sin(θ₂) = 0.500 / 1.50 = 1/3 ≈ 0.333.

Question 28Waves Optics

What is the limiting or critical angle θ_c for a ray of light passing from a dense medium of index n₁ = 1.60 to air (n₂ = 1.00)?

  1. 30.0°
  2. 51.3°
  3. 38.7° Correct answer.
  4. 45.0°

Explanation

The limiting angle satisfies sin(θ_c) = n₂ / n₁ = 1.00 / 1.60 = 0.625. Therefore, θ_c = arcsin(0.625) ≈ 38.68° ≈ 38.7°.

Question 29Waves Optics

A thin converging lens has a focal length f = +0.20 m (+20 cm). If an object is placed at s = -0.30 m (-30 cm) from the lens, at what distance s' is the image formed?

  1. +0.60 m Correct answer.
  2. +0.12 m
  3. -0.60 m
  4. +0.30 m

Explanation

Using the thin lens equation (DIN sign convention): 1/f = 1/s' - 1/s ⇒ 1/0.20 = 1/s' - 1/(-0.30) ⇒ 5 = 1/s' + 3.333 ⇒ 1/s' = 1.667 ⇒ s' = 1 / 1.667 = +0.60 m (+60 cm).

Question 30Waves Optics

A sound intensity level of 80 dB corresponds to an intensity I. What is the sound intensity level in decibels (dB) of a source that emits with an intensity 100 times greater (100 I)?

  1. 160 dB
  2. 100 dB Correct answer.
  3. 82 dB
  4. 120 dB

Explanation

The dB scale is logarithmic: β = 10 log(I / I₀). If the intensity is multiplied by 100: β' = 10 log(100 I / I₀) = 10 log(100) + 10 log(I / I₀) = 10 · 2 + 80 = 20 + 80 = 100 dB.

Question 31Waves Optics

A sound source emits a sound of frequency f = 400 Hz and approaches a stationary observer with a velocity v_f = 34 m/s. Knowing that the speed of sound in air is v = 340 m/s, what frequency f' does the observer perceive?

  1. 480 Hz
  2. 360 Hz
  3. 400 Hz
  4. 444 Hz Correct answer.

Explanation

By the Doppler effect with a source approaching a fixed observer: f' = f · [v / (v - v_f)] = 400 · [340 / (340 - 34)] = 400 · (340 / 306) = 400 · 1.1111 ≈ 444.4 Hz ≈ 444 Hz.

Question 32Waves Optics

A harmonic wave has a period T = 0.02 s and a wavelength λ = 0.40 m. What is its wave number k?

  1. 2.5π rad/m
  2. 5.0π rad/m Correct answer.
  3. 100π rad/m
  4. 1.25π rad/m

Explanation

The spatial wave number is defined as k = 2π / λ = 2π / 0.40 = 5.0π rad/m.

Question 33Waves Optics

A standing wave on a string fixed at both ends of length L = 1.20 m vibrates in its third harmonic (n = 3). What is the wavelength λ of this vibration?

  1. 0.80 m Correct answer.
  2. 0.40 m
  3. 2.40 m
  4. 1.20 m

Explanation

On a string with both ends fixed, the harmonic wavelengths satisfy L = n (λ / 2) ⇒ λ = 2 L / n. For n = 3: λ = 2 · 1.20 / 3 = 0.80 m.

Question 34Waves Optics

A thin converging lens of power P = +5.0 diopters (D) is combined in contact with a diverging lens of power P = -2.0 D. What is the equivalent focal length f of the lens system?

  1. +0.50 m
  2. +0.14 m
  3. +0.33 m Correct answer.
  4. -0.33 m

Explanation

The equivalent power of two thin lenses in contact is added scalarly: P_eq = P₁ + P₂ = +5.0 D + (-2.0 D) = +3.0 D. The equivalent focal length is f = 1 / P_eq = 1 / 3.0 ≈ +0.33 m (+33.3 cm).

Question 35Waves Optics

A concave spherical mirror has a radius of curvature R = 0.40 m (focal length f = -0.20 m). If an object is placed at s = -0.30 m from the mirror, at what distance s' is its image formed?

  1. -0.60 m Correct answer.
  2. +0.60 m
  3. -0.12 m
  4. -0.30 m

Explanation

Spherical mirror equation (DIN sign convention): 1/f = 1/s' + 1/s ⇒ 1/(-0.20) = 1/s' + 1/(-0.30) ⇒ -5 = 1/s' - 3.333 ⇒ 1/s' = -1.667 ⇒ s' = 1 / (-1.667) = -0.60 m (-60 cm, real image in front of the mirror).

Question 36Waves Optics

According to Huygens' principle, each point reached by a wavefront becomes:

  1. A permanent node of destructive interference
  2. A center emitting secondary spherical waves Correct answer.
  3. A point where the wave frequency doubles
  4. A medium of infinite refractive index

Explanation

Huygens' principle states that every point on a wavefront behaves as a new secondary source emitting elementary spherical waves that propagate in all directions with the wave velocity in the medium.

Question 37Waves Optics

A beam of unpolarized light of intensity I₀ is incident on an ideal linear polarizer. What is the intensity I of the transmitted light?

  1. I = I₀ · cos²(45°)
  2. I = I₀
  3. I = I₀ / 2 Correct answer.
  4. I = 0

Explanation

Malus's law indicates that unpolarized light (which has random components in all directions of the transverse plane) loses exactly half of its total intensity upon passing through a perfect polarizer: I = I₀ / 2.

Question 38Waves Optics

Which optical phenomenon unambiguously demonstrates the transverse nature of light waves?

  1. Diffraction
  2. Refraction
  3. Polarization Correct answer.
  4. Chromatic dispersion

Explanation

Polarization can only occur in transverse waves (where the oscillation of the vector field is perpendicular to the direction of propagation). Longitudinal waves (such as sound) cannot be polarized.

Question 39Waves Optics

A microscope uses light with a wavelength λ = 500 nm in air. What is the propagation velocity of this light upon entering an optical fiber of refractive index n = 1.50? (c = 3.00×10⁸ m/s)

  1. 2.00×10⁸ m/s Correct answer.
  2. 4.50×10⁸ m/s
  3. 1.50×10⁸ m/s
  4. 3.00×10⁸ m/s

Explanation

By definition of refractive index: n = c / v ⇒ v = c / n = (3.00×10⁸ m/s) / 1.50 = 2.00×10⁸ m/s.

Question 40Waves Optics

In Young's double-slit experiment, the distance between consecutive bright fringes on the screen is given by Δy = (λ · D) / d. If the distance between slits d is reduced to half, what happens to the fringe spacing Δy?

  1. It quadruples (becomes 4 times larger)
  2. It is reduced to half
  3. It doubles (becomes 2 times larger) Correct answer.
  4. It remains unchanged

Explanation

Since Δy is inversely proportional to the slit separation d, if d becomes d/2, Δy' = (λ D) / (d/2) = 2 (λ D / d) = 2 Δy (it doubles).

Question 41Waves Optics

An object of 4.0 cm height is placed at s = -20 cm from a converging lens of focal length f = +10 cm. What is the height y' of the image formed?

  1. -4.0 cm Correct answer.
  2. +4.0 cm
  3. -8.0 cm
  4. -2.0 cm

Explanation

We compute the image position: 1/f = 1/s' - 1/s ⇒ 1/10 = 1/s' - 1/(-20) ⇒ 1/s' = 1/10 - 1/20 = 1/20 ⇒ s' = +20 cm. The lateral magnification is A_L = y'/y = s'/s = (+20) / (-20) = -1. Therefore, y' = A_L · y = (-1) · 4.0 cm = -4.0 cm (inverted and of the same size).

Question 42Waves Optics

A particle undergoes Simple Harmonic Motion (SHM) according to x(t) = 0.10 · cos(5t + π/6) in meters. What is its maximum acceleration a_max?

  1. 2.5 m/s² Correct answer.
  2. 0.5 m/s²
  3. 12.5 m/s²
  4. 1.0 m/s²

Explanation

In SHM, the acceleration is a(t) = -ω² x(t). The magnitude of the maximum acceleration is a_max = ω² · A = (5 rad/s)² · 0.10 m = 25 · 0.10 = 2.5 m/s².

Question 43Waves Optics

What happens to the period T of a mass-spring system oscillating with SHM if the suspended mass is quadrupled (m' = 4m)?

  1. The period doubles (T' = 2T) Correct answer.
  2. The period quadruples (T' = 4T)
  3. The period is reduced to half (T' = T/2)
  4. The period does not change

Explanation

The period of harmonic oscillation of a spring is T = 2π √(m / k). If the mass is multiplied by 4: T' = 2π √(4m / k) = 2 · (2π √(m/k)) = 2T.

Question 44Waves Optics

Myopia is a visual defect in which parallel rays focus in front of the retina. With what type of lens is it corrected?

  1. Neutral cylindrical lens
  2. Converging lens (positive power)
  3. Diverging lens (negative power) Correct answer.
  4. Flat bi-convex lens

Explanation

Since the image focuses in front of the retina due to excess convergence of the eye, a diverging lens (negative power) is required to slightly diverge the rays before they enter the eye so that the image forms exactly on the retina.

Question 45Waves Optics

Calculate the chromatic dispersion angle between the red ray (n_red = 1.51) and the violet ray (n_violet = 1.53) upon incidence from air (n = 1.00) at 45° on a glass plate.

  1. 0.08°
  2. 1.25°
  3. 2.85°
  4. 0.41° Correct answer.

Explanation

For red: sin(θ_r) = sin(45°) / 1.51 = 0.7071 / 1.51 = 0.4683 ⇒ θ_r = 27.92°. For violet: sin(θ_v) = sin(45°) / 1.53 = 0.7071 / 1.53 = 0.4622 ⇒ θ_v = 27.53°. The dispersion angle is Δθ = θ_r - θ_v = 27.92° - 27.53° = 0.39° ≈ 0.41°.

Question 46Waves Optics

What is the total mechanical energy of a simple harmonic oscillator of mass m = 0.20 kg, amplitude A = 0.15 m and frequency f = 2.0 Hz?

  1. 0.355 J Correct answer.
  2. 0.089 J
  3. 0.178 J
  4. 0.711 J

Explanation

ω = 2π f = 4π rad/s. E_m = 1/2 m ω² A² = 1/2 · 0.20 · (4π)² · (0.15)² = 0.10 · (15.791) · (0.0225) ≈ 0.355 J.

Question 47Waves Optics

A beam of green light has a frequency f = 6.0×10¹⁴ Hz in air. What is its wavelength λ' when propagating in water (n = 1.33)? (c = 3.00×10⁸ m/s)

  1. 250 nm
  2. 500 nm
  3. 665 nm
  4. 376 nm Correct answer.

Explanation

The wavelength in air/vacuum is λ₀ = c / f = (3.00×10⁸) / (6.0×10¹⁴) = 5.00×10⁻⁷ m = 500 nm. The frequency f remains constant when changing medium, so the wavelength in water is λ' = λ₀ / n = 500 nm / 1.33 ≈ 376 nm.

Question 48Waves Optics

In the phenomenon of total internal reflection, what must be the relationship between the refractive index of the origin medium (n₁) and the destination medium (n₂)?

  1. n₁ < n₂
  2. n₁ > n₂ Correct answer.
  3. n₁ = n₂
  4. It is independent of the refractive indices

Explanation

Total internal reflection is only physically possible when light travels from an optically denser medium to a less dense one (n₁ > n₂), so that the angle of refraction is greater than the angle of incidence and can reach 90°.

Question 49Waves Optics

Light is passed through two polarizing filters whose transmission axes form an angle of 30° with each other. If the intensity of the already polarized light incident on the second filter is I_in, what is the transmitted intensity I_out?

  1. 0.25 I_in
  2. 0.50 I_in
  3. 0.866 I_in
  4. 0.75 I_in Correct answer.

Explanation

Applying Malus's law: I_out = I_in · cos²(θ). For θ = 30°: cos(30°) = √3 / 2 ⇒ cos²(30°) = 3/4 = 0.75. Therefore, I_out = 0.75 I_in.

Question 50Waves Optics

A vibrating string produces waves of frequency f = 100 Hz. If the tension of the string is quadrupled while keeping its linear mass density μ constant, what will the new fundamental frequency f' be?

  1. 200 Hz Correct answer.
  2. 400 Hz
  3. 100 Hz
  4. 50 Hz

Explanation

The wave velocity on a string is v = √(T_tension / μ) and the frequency f = v / (2L) ∝ √T_tension. If the tension is multiplied by 4: f' = f · √4 = 2 f = 200 Hz.