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FY13CE Physics study materials
FY13CE Physics study resources

FY13CE Physics practice questions

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Question 1Mechanics

A rugby ball is kicked from ground level across a level playing field with an initial velocity of 20.0 m/s at an angle of 30.0° above the horizontal. Taking g = 9.80 m/s² and neglecting air resistance, what is the total time of flight of the ball before it strikes the ground?

  1. 2.04 s Correct answer.
  2. 1.02 s
  3. 3.53 s
  4. 4.08 s

Explanation

The initial vertical component of velocity is v_0y = v_0 sin θ = 20.0 sin(30.0°) = 10.0 m/s. The time of flight for a projectile landing at the same vertical level is given by T = (2 v_0y) / g = (2 × 10.0 m/s) / 9.80 m/s² = 2.04 s.

Question 2Mechanics

A projectile is launched from ground level over flat terrain with an initial speed of 28.0 m/s at an angle of 45.0° to the horizontal. Assuming air resistance is negligible and g = 9.80 m/s², what is the maximum vertical height reached by the projectile above the launch level?

  1. 10.0 m
  2. 20.0 m Correct answer.
  3. 28.3 m
  4. 40.0 m

Explanation

The vertical component of initial velocity is v_0y = 28.0 sin(45.0°) = 19.80 m/s. Using v_y² = v_0y² - 2gh at maximum height (where v_y = 0), h_max = v_0y² / (2g) = (19.80 m/s)² / (2 × 9.80 m/s²) = 392.0 / 19.60 = 20.0 m.

Question 3Mechanics

A rescue package is thrown horizontally from the edge of a vertical sea cliff 78.4 m above sea level with an initial horizontal speed of 15.0 m/s. Assuming g = 9.80 m/s² and negligible air drag, what is the speed of the package immediately before striking the ocean surface?

  1. 39.2 m/s
  2. 15.0 m/s
  3. 42.0 m/s Correct answer.
  4. 54.2 m/s

Explanation

The vertical velocity upon landing satisfies v_y² = 2gh = 2(9.80 m/s²)(78.4 m) = 1536.64 m²/s², so v_y = 39.20 m/s. The horizontal velocity remains constant at v_x = 15.0 m/s. The final speed is v = √(v_x² + v_y²) = √(15.0² + 39.20²) = √(225.0 + 1536.64) = √1761.64 = 42.0 m/s.

Question 4Mechanics

A 5.00 kg block slides down an inclined plane that makes an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the block and the surface is 0.200. Taking g = 9.80 m/s², what is the linear acceleration of the block down the incline?

  1. 3.20 m/s² Correct answer.
  2. 4.90 m/s²
  3. 1.70 m/s²
  4. 6.60 m/s²

Explanation

The net acceleration down the incline is a = g(sin θ - μ_k cos θ). Substituting the known values: a = 9.80 [sin(30.0°) - 0.200 cos(30.0°)] = 9.80 [0.500 - 0.200(0.8660)] = 9.80 [0.500 - 0.1732] = 9.80(0.3268) = 3.20 m/s².

Question 5Mechanics

In an Atwood machine, two masses m_1 = 3.00 kg and m_2 = 5.00 kg are connected by a light inextensible cord passing over a frictionless, massless pulley. When the system is released from rest, what is the tension in the connecting cord? (Take g = 9.80 m/s²)

  1. 19.6 N
  2. 29.4 N
  3. 36.8 N Correct answer.
  4. 49.0 N

Explanation

The acceleration of the system is a = (m_2 - m_1)g / (m_1 + m_2) = (5.00 - 3.00)(9.80) / 8.00 = 2.45 m/s². The tension in the cord is T = m_1(g + a) = 3.00(9.80 + 2.45) = 3.00(12.25) = 36.75 N ≈ 36.8 N (or T = 2 m_1 m_2 g / (m_1 + m_2) = 2(3)(5)(9.8) / 8 = 36.75 N).

Question 6Mechanics

A 12.0 kg wooden crate rests on a horizontal concrete floor. A minimum horizontal pulling force of 47.0 N is required to just start the crate moving. Taking g = 9.80 m/s², what is the coefficient of static friction between the crate and the floor?

  1. 0.255
  2. 0.400 Correct answer.
  3. 0.480
  4. 0.520

Explanation

At the verge of slipping, the maximum static friction equals the applied horizontal force: f_s,max = μ_s N = μ_s mg. Therefore, μ_s = F / (mg) = 47.0 N / (12.0 kg × 9.80 m/s²) = 47.0 / 117.6 = 0.3997 ≈ 0.400.

Question 7Mechanics

A 1200 kg car travelling north at 20.0 m/s collides with and couples to a stationary 800 kg trailer. Assuming external resistive forces are negligible during the collision, what is the velocity of the combined car and trailer immediately after coupling?

  1. 8.00 m/s north
  2. 10.0 m/s north
  3. 12.0 m/s north Correct answer.
  4. 15.0 m/s north

Explanation

By conservation of linear momentum: m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'. Here (1200 kg)(20.0 m/s) + 0 = (1200 kg + 800 kg) v'. Thus 24000 kg·m/s = (2000 kg) v', giving v' = 12.0 m/s north.

Question 8Mechanics

A 0.400 kg puck sliding east at 6.00 m/s on a frictionless horizontal air table strikes an identical 0.400 kg stationary puck. After collision, the first puck moves at 30.0° north of east with a speed of 5.20 m/s. What is the magnitude of the velocity of the second puck?

  1. 3.00 m/s Correct answer.
  2. 0.80 m/s
  3. 2.60 m/s
  4. 4.50 m/s

Explanation

By conservation of momentum in component form (masses cancel out): along x, 6.00 = 5.20 cos(30.0°) + v_2x, so v_2x = 6.00 - 4.503 = 1.497 m/s. Along y, 0 = 5.20 sin(30.0°) + v_2y, so v_2y = -2.600 m/s. The resultant speed is v_2 = √(1.497² + (-2.600)²) = √(2.241 + 6.760) = √9.001 = 3.00 m/s.

Question 9Mechanics

A 0.150 kg cricket ball travelling horizontally at 32.0 m/s is struck straight back toward the bowler by a batsman. The ball leaves the bat in the opposite direction at 40.0 m/s. If the impact lasts for 2.50 ms (0.00250 s), what is the magnitude of the average force exerted on the ball by the bat?

  1. 480 N
  2. 2160 N
  3. 4320 N Correct answer.
  4. 7200 N

Explanation

Taking the outgoing direction as positive: Δp = m[v_f - (-v_i)] = 0.150[40.0 - (-32.0)] = 0.150(72.0) = 10.80 kg·m/s. The average force is F_avg = Δp / Δt = 10.80 kg·m/s / 0.00250 s = 4320 N.

Question 10Mechanics

A 2.00 kg block slides along a frictionless horizontal track at 8.00 m/s before entering a rough section of length 5.00 m where the coefficient of kinetic friction is 0.350. Taking g = 9.80 m/s², what is the speed of the block as it exits the rough section?

  1. 2.55 m/s
  2. 4.12 m/s
  3. 5.45 m/s Correct answer.
  4. 6.85 m/s

Explanation

Initial kinetic energy is KE_i = (1/2)m v_i² = (1/2)(2.00)(8.00²) = 64.00 J. Work done by friction is W_f = -f_k d = -(μ_k m g)d = -(0.350 × 2.00 × 9.80 × 5.00) = -34.30 J. Final kinetic energy is KE_f = 64.00 - 34.30 = 29.70 J. Therefore, (1/2)(2.00)v_f² = 29.70 J, so v_f = √29.70 = 5.45 m/s.

Question 11Mechanics

A highway curve of radius 150 m is to be banked so that vehicles travelling at a design speed of 25.0 m/s can safely negotiate the turn without relying on tyre friction. What is the required banking angle of the road? (Take g = 9.80 m/s²)

  1. 12.4°
  2. 23.0° Correct answer.
  3. 35.8°
  4. 45.0°

Explanation

For an ideally banked turn without friction: tan θ = v² / (rg). Substituting the given values: tan θ = (25.0 m/s)² / (150 m × 9.80 m/s²) = 625.0 / 1470.0 = 0.4252. Therefore, θ = arctan(0.4252) = 23.03° ≈ 23.0°.

Question 12Mechanics

A 1000 kg car travels around an unbanked, level circular track of radius 80.0 m. If the coefficient of static friction between the tyres and the road surface is 0.700, what is the maximum speed at which the car can round the bend without skidding? (Take g = 9.80 m/s²)

  1. 16.6 m/s
  2. 23.4 m/s Correct answer.
  3. 27.8 m/s
  4. 54.9 m/s

Explanation

The required centripetal force is provided entirely by static friction: f_s,max = m v_max² / r => μ_s m g = m v_max² / r. Therefore, v_max = √(μ_s g r) = √(0.700 × 9.80 × 80.0) = √548.8 = 23.43 m/s ≈ 23.4 m/s.

Question 13Mechanics

A heavy flywheel starts from rest and accelerates with a uniform angular acceleration of 15.0 rad/s² for 8.00 s. What is the total angular displacement of the flywheel during this acceleration phase?

  1. 240 rad
  2. 480 rad Correct answer.
  3. 960 rad
  4. 1920 rad

Explanation

Using the rotational kinematic equation θ = ω_0 t + (1/2)α t² with ω_0 = 0: θ = (1/2)(15.0 rad/s²)(8.00 s)² = (1/2)(15.0)(64.0) = 480 rad.

Question 14Mechanics

A circular saw blade spinning at 3000 rev/min is switched off and decelerates uniformly to rest in a time span of 20.0 s. How many complete revolutions does the blade make while coming to a halt?

  1. 250 revolutions
  2. 500 revolutions Correct answer.
  3. 1000 revolutions
  4. 3000 revolutions

Explanation

Initial angular velocity in revolutions per second is ω_0 = 3000 rev / 60 s = 50.0 rev/s. Since deceleration is uniform, the average angular speed is ω_avg = (50.0 + 0) / 2 = 25.0 rev/s. The total number of revolutions is N = ω_avg × t = (25.0 rev/s)(20.0 s) = 500 revolutions.

Question 15Mechanics

A uniform wooden plank of length 4.00 m and mass 30.0 kg is supported horizontally on two trestles placed 0.50 m from each end. A painter of mass 70.0 kg stands 1.20 m from the left end of the plank. Taking g = 9.80 m/s², what vertical upward force is exerted on the plank by the left support?

  1. 490 N
  2. 539 N
  3. 673 N Correct answer.
  4. 784 N

Explanation

The distance between the two trestles (supports A and B) is 4.00 - 2(0.50) = 3.00 m. Support A is at x = 0.50 m, Support B is at x = 3.50 m. The plank's centre of mass is at x = 2.00 m (weight W_p = 30.0 × 9.80 = 294 N). The painter is at x = 1.20 m (weight W_m = 70.0 × 9.80 = 686 N). Taking torques about B: R_A(3.00) = W_m(3.50 - 1.20) + W_p(3.50 - 2.00) = 686(2.30) + 294(1.50) = 1577.8 + 441.0 = 2018.8 N·m. Thus R_A = 2018.8 / 3.00 = 672.9 N ≈ 673 N.

Question 16Mechanics

A thin-walled hollow hoop and a solid cylinder of identical mass M and outer radius R roll down a rough inclined plane from rest without slipping. Which object reaches the bottom of the incline first, and why?

  1. The solid cylinder reaches the bottom first because it has a smaller moment of inertia, allocating a greater fraction of potential energy to translational kinetic energy. Correct answer.
  2. The hollow hoop reaches the bottom first because its mass is concentrated at the perimeter, generating greater downward momentum.
  3. Both reach the bottom simultaneously because acceleration down an incline is independent of body shape.
  4. The hollow hoop reaches the bottom first because it experiences larger static friction which accelerates it faster.

Explanation

Linear acceleration down the incline is a = g sin θ / [1 + I / (MR²)]. For a solid cylinder, I = (1/2)MR², giving a = (2/3)g sin θ = 0.67g sin θ. For a hoop, I = MR², giving a = (1/2)g sin θ = 0.50g sin θ. Because the solid cylinder has a smaller moment of inertia, less energy is stored in rotation and more in translation, resulting in greater acceleration.

Question 17Mechanics

A solid uniform sphere of mass 4.00 kg and radius 0.150 m rolls without slipping down an incline through a vertical height of 2.80 m, starting from rest. What is the linear speed of its centre of mass at the base of the incline? (Take g = 9.80 m/s²)

  1. 5.24 m/s
  2. 6.26 m/s Correct answer.
  3. 7.41 m/s
  4. 8.10 m/s

Explanation

By conservation of energy: Mgh = (1/2)M v² + (1/2)I ω². For a solid sphere, I = (2/5)M R² and ω = v / R, so rotational KE is (1/2)(2/5 M R²)(v / R)² = (1/5)M v². Total KE is (1/2 + 1/5)M v² = (7/10)M v². Equating Mgh = (7/10)M v² yields v² = (10/7)gh = (10/7)(9.80)(2.80) = 39.20 m²/s², so v = √39.20 = 6.26 m/s.

Question 18Mechanics

A figure skater spins on frictionless ice with outstretched arms at an angular speed of 2.50 rad/s. Her initial rotational inertia is 3.60 kg·m². When she draws her arms tightly in toward her chest, her moment of inertia reduces to 1.20 kg·m². What is her new angular speed?

  1. 0.83 rad/s
  2. 2.50 rad/s
  3. 5.00 rad/s
  4. 7.50 rad/s Correct answer.

Explanation

In the absence of external net torque, angular momentum is conserved: L = I_1 ω_1 = I_2 ω_2. Substituting values: (3.60 kg·m²)(2.50 rad/s) = (1.20 kg·m²) ω_2, so ω_2 = (3.60 / 1.20)(2.50) = 3 × 2.50 = 7.50 rad/s.

Question 19Mechanics

A uniform thin rod of length 1.20 m and mass 1.80 kg is pivoted at its top end about a frictionless horizontal axis. A lump of clay of mass 0.200 kg moving horizontally at 15.0 m/s strikes the bottom tip of the hanging rod and sticks to it. What is the angular speed of the rod-clay system immediately following the collision?

  1. 2.34 rad/s
  2. 3.13 rad/s Correct answer.
  3. 4.17 rad/s
  4. 6.25 rad/s

Explanation

Angular momentum about the pivot is conserved: L_i = m_c v L = (0.200 kg)(15.0 m/s)(1.20 m) = 3.60 kg·m²/s. The moment of inertia of the rod pivoted at one end is I_rod = (1/3)M L² = (1/3)(1.80)(1.20²) = 0.864 kg·m². The stuck clay has I_c = m_c L² = (0.200)(1.20²) = 0.288 kg·m². Total inertia is I_tot = 0.864 + 0.288 = 1.152 kg·m². Hence, ω = L_i / I_tot = 3.60 / 1.152 = 3.125 rad/s ≈ 3.13 rad/s.

Question 20Mechanics

The acceleration due to gravity at Earth's surface is 9.80 m/s². Assuming Earth is a uniform sphere of radius R, what is the acceleration due to gravity at an altitude h = R above Earth's surface?

  1. 1.23 m/s²
  2. 2.45 m/s² Correct answer.
  3. 4.90 m/s²
  4. 9.80 m/s²

Explanation

Gravitational field strength is inversely proportional to the square of the distance from the centre of the Earth: g(r) = GM / r². At altitude h = R, the distance from Earth's centre is r = R + h = 2R. Thus, g = g_surface / (2)² = 9.80 / 4 = 2.45 m/s².

Question 21Mechanics

A weather satellite orbits Earth in a circular orbit at an orbital radius of 7.20 × 10⁶ m from Earth's centre. Given that Earth's mass is 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N·m²/kg², what is the orbital speed of the satellite?

  1. 5.26 × 10³ m/s
  2. 7.44 × 10³ m/s Correct answer.
  3. 1.05 × 10⁴ m/s
  4. 5.53 × 10⁷ m/s

Explanation

Equating gravitational force to centripetal force: GMm / r² = m v² / r, so v = √(GM / r). Substituting values: v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (7.20 × 10⁶)] = √[3.982 × 10¹⁴ / 7.20 × 10⁶] = √5.531 × 10⁷ = 7437 m/s ≈ 7.44 × 10³ m/s.

Question 22Mechanics

Satellite A moves in a circular orbit around a planet with an orbital radius r and has an orbital period of 8.00 days. Satellite B moves in a circular orbit around the same planet with an orbital radius of 4r. What is the orbital period of Satellite B?

  1. 16.0 days
  2. 32.0 days
  3. 64.0 days Correct answer.
  4. 128 days

Explanation

According to Kepler's Third Law: T² ∝ r³, which implies (T_B / T_A)² = (r_B / r_A)³ = (4)³ = 64. Taking the square root gives T_B / T_A = √64 = 8. Therefore, T_B = 8 × 8.00 days = 64.0 days.

Question 23Mechanics

The escape speed from the surface of a spherical planet of mass M and radius R is v_esc. A second planet has twice the mass (2M) and half the radius (R/2). What is the escape speed from the surface of the second planet?

  1. v_esc
  2. √2 v_esc
  3. 2 v_esc Correct answer.
  4. 4 v_esc

Explanation

The escape speed from the surface of a spherical body is given by v_esc = √(2GM / R). For the second planet: v_esc' = √[2G(2M) / (R/2)] = √[4 × (2GM / R)] = 2 √(2GM / R) = 2 v_esc.

Question 24Mechanics

Which statement correctly describes the necessary orbital conditions for an artificial satellite to be geostationary?

  1. It must orbit eastward directly above the equator with an orbital period of exactly 24 hours. Correct answer.
  2. It can orbit at any arbitrary latitude provided its orbital period is 12 hours.
  3. It must orbit westward along the equator so its motion opposes the rotation of the Earth.
  4. It must orbit in a polar trajectory passing over both geographic poles every 24 hours.

Explanation

To remain permanently above the same point on Earth's surface, a geostationary satellite must orbit in the equatorial plane from west to east (eastward) with an orbital period that matches Earth's sidereal rotational period (24 hours).

Question 25Mechanics

A student of mass 60.0 kg stands on a bathroom spring scale calibrated in newtons inside an elevator. If the elevator accelerates downwards at a constant rate of 2.20 m/s², what reading does the scale display? (Take g = 9.80 m/s²)

  1. 132 N
  2. 456 N Correct answer.
  3. 588 N
  4. 720 N

Explanation

The scale reading equals the normal contact force N. From Newton's second law for downward acceleration: mg - N = ma, so N = m(g - a). Substituting the numbers: N = 60.0(9.80 - 2.20) = 60.0(7.60) = 456 N.

Question 26Electricity Magnetism

A proton (charge q = +1.60 × 10⁻¹⁹ C) enters a uniform magnetic field of 0.500 T with a velocity of 4.00 × 10⁶ m/s at an angle of 30.0° to the direction of the magnetic field. What is the magnitude of the magnetic force acting on the proton?

  1. 0.80 × 10⁻¹³ N
  2. 1.60 × 10⁻¹³ N Correct answer.
  3. 2.77 × 10⁻¹³ N
  4. 3.20 × 10⁻¹³ N

Explanation

The magnetic Lorentz force on a moving charge is F = q v B sin θ = (1.60 × 10⁻¹⁹ C)(4.00 × 10⁶ m/s)(0.500 T) sin(30.0°) = (3.20 × 10⁻¹³ N)(0.500) = 1.60 × 10⁻¹³ N.

Question 27Electricity Magnetism

An electron (mass m = 9.11 × 10⁻³¹ kg, charge e = 1.60 × 10⁻¹⁹ C) travels at 6.00 × 10⁶ m/s perpendicular to a uniform magnetic field of 2.00 × 10⁻³ T. What is the radius of the resulting circular trajectory?

  1. 0.85 cm
  2. 1.71 cm Correct answer.
  3. 3.42 cm
  4. 5.85 cm

Explanation

Centripetal force is provided by the magnetic force: m v² / r = q v B, so r = m v / (q B). Substituting the given values: r = [(9.11 × 10⁻³¹ kg)(6.00 × 10⁶ m/s)] / [(1.60 × 10⁻¹⁹ C)(2.00 × 10⁻³ T)] = (5.466 × 10⁻²⁴) / (3.20 × 10⁻²²) = 0.01708 m = 1.71 cm.

Question 28Electricity Magnetism

In a velocity selector, mutually perpendicular uniform electric and magnetic fields are established. If the electric field strength is 8.00 × 10⁴ V/m and the magnetic field strength is 0.200 T, what speed must a charged ion possess to traverse the selector without undergoing any deflection?

  1. 1.60 × 10⁴ m/s
  2. 2.50 × 10⁵ m/s
  3. 4.00 × 10⁵ m/s Correct answer.
  4. 8.00 × 10⁵ m/s

Explanation

Zero deflection occurs when the electric force balances the magnetic Lorentz force: q E = q v B. Cancelling q gives v = E / B = (8.00 × 10⁴ V/m) / (0.200 T) = 4.00 × 10⁵ m/s.

Question 29Electricity Magnetism

A straight conducting wire of length 0.400 m carries a current of 5.00 A perpendicular to a uniform magnetic field of 0.600 T. What is the magnitude of the magnetic force exerted on the wire?

  1. 0.48 N
  2. 0.60 N
  3. 1.20 N Correct answer.
  4. 2.40 N

Explanation

The magnetic force on a straight current-carrying wire is F = B I L sin θ. For a perpendicular field (θ = 90°): F = (0.600 T)(5.00 A)(0.400 m)(1) = 1.20 N.

Question 30Electricity Magnetism

Two long, straight parallel conductors separated by 0.0500 m in air carry currents of 4.00 A and 6.00 A in opposite directions. Taking μ_0 = 4π × 10⁻⁷ T·m/A, what is the magnetic force per unit length between the conductors, and what is its nature?

  1. 4.80 × 10⁻⁵ N/m, attractive
  2. 9.60 × 10⁻⁵ N/m, attractive
  3. 9.60 × 10⁻⁵ N/m, repulsive Correct answer.
  4. 1.92 × 10⁻⁴ N/m, repulsive

Explanation

The force per unit length is F / L = (μ_0 I_1 I_2) / (2π d) = [(4π × 10⁻⁷ T·m/A)(4.00 A)(6.00 A)] / [2π(0.0500 m)] = (2 × 10⁻⁷ × 24.0) / 0.0500 = 4.80 × 10⁻⁶ / 0.0500 = 9.60 × 10⁻⁵ N/m. Because the currents flow in opposite directions (antiparallel), the magnetic interaction is repulsive.

Question 31Electricity Magnetism

A rectangular coil consisting of 50 turns with dimensions 0.100 m by 0.200 m carries a current of 2.50 A in a uniform magnetic field of 0.400 T. What is the maximum torque that can be exerted on the coil?

  1. 0.20 N·m
  2. 0.50 N·m
  3. 1.00 N·m Correct answer.
  4. 2.00 N·m

Explanation

The area of the coil is A = 0.100 m × 0.200 m = 0.0200 m². The maximum torque occurs when the magnetic field lies in the plane of the coil (θ = 90° to normal): τ_max = N I A B = (50)(2.50 A)(0.0200 m²)(0.400 T) = 1.00 N·m.

Question 32Electricity Magnetism

What is the magnetic field strength at a perpendicular distance of 0.0400 m from a long, straight wire carrying a direct current of 8.00 A? (Take μ_0 = 4π × 10⁻⁷ T·m/A)

  1. 1.00 × 10⁻⁵ T
  2. 2.00 × 10⁻⁵ T
  3. 4.00 × 10⁻⁵ T Correct answer.
  4. 8.00 × 10⁻⁵ T

Explanation

From Ampère's law for a long straight conductor: B = μ_0 I / (2π r) = [(4π × 10⁻⁷ T·m/A)(8.00 A)] / [2π(0.0400 m)] = (2 × 10⁻⁷ × 8.00) / 0.0400 = 1.60 × 10⁻⁶ / 0.0400 = 4.00 × 10⁻⁵ T = 40.0 μT.

Question 33Electricity Magnetism

A long solenoid of length 0.500 m is tightly wound with 1500 turns of copper wire. When carrying a steady current of 2.00 A, what is the magnetic field strength inside the central core of the solenoid? (Take μ_0 = 4π × 10⁻⁷ T·m/A ≈ 1.257 × 10⁻⁶ T·m/A)

  1. 1.88 mT
  2. 3.77 mT
  3. 7.54 mT Correct answer.
  4. 15.1 mT

Explanation

The turn density is n = N / L = 1500 turns / 0.500 m = 3000 turns/m. The central magnetic field in an ideal solenoid is B = μ_0 n I = (4π × 10⁻⁷ T·m/A)(3000 m⁻¹)(2.00 A) = 2.4π × 10⁻³ T ≈ 7.54 × 10⁻³ T = 7.54 mT.

Question 34Electricity Magnetism

A flat circular loop of wire of radius 0.100 m is situated in a uniform magnetic field of 0.500 T. If the plane of the loop makes an angle of 30.0° with the magnetic field lines, what is the magnetic flux passing through the loop?

  1. 3.93 × 10⁻³ Wb
  2. 7.85 × 10⁻³ Wb Correct answer.
  3. 1.36 × 10⁻² Wb
  4. 1.57 × 10⁻² Wb

Explanation

The loop area is A = π r² = π(0.100 m)² = 0.0100π m² ≈ 0.03142 m². The angle between the normal to the loop and the magnetic field vector is θ = 90.0° - 30.0° = 60.0°. Therefore, magnetic flux is Φ = B A cos θ = (0.500 T)(0.03142 m²) cos(60.0°) = (0.01571)(0.500) = 7.85 × 10⁻³ Wb.

Question 35Electricity Magnetism

A flat circular coil consisting of 200 turns of wire, each of area 0.0250 m², is placed perpendicular to a uniform magnetic field. The magnetic field decreases steadily from 0.800 T to 0.200 T in an interval of 0.300 s. What is the magnitude of the average induced EMF in the coil?

  1. 2.50 V
  2. 5.00 V
  3. 10.0 V Correct answer.
  4. 20.0 V

Explanation

Change in magnetic field is ΔB = 0.800 T - 0.200 T = 0.600 T. Change in flux through one turn is ΔΦ = (ΔB) A = (0.600 T)(0.0250 m²) = 0.0150 Wb. By Faraday's law of induction: |ε| = N (ΔΦ / Δt) = 200 × (0.0150 Wb / 0.300 s) = 200 × 0.0500 V = 10.0 V.

Question 36Electricity Magnetism

A bar magnet is dropped vertically, north pole pointing downward, through a horizontal stationary copper loop. According to Lenz's law, as viewed from above the loop:

  1. As the magnet approaches, the induced current flows counter-clockwise to repel the magnet, and after the magnet passes through and recedes below the loop, the current flows clockwise. Correct answer.
  2. The induced current flows clockwise as the magnet approaches and remains clockwise as it falls away below the loop.
  3. No current is induced because copper is a non-ferromagnetic conductor.
  4. As the magnet approaches, the induced current flows clockwise to attract the incoming north pole downward.

Explanation

According to Lenz's law, the induced current creates a magnetic field that opposes the change in magnetic flux. As the north pole approaches downward, downward flux increases, inducing a counter-clockwise current (viewed from above) that creates an upward repulsive magnetic field. When leaving below the loop, downward flux decreases, inducing a clockwise current that creates a downward magnetic field to oppose the magnet's departure.

Question 37Electricity Magnetism

A conducting rod of length 0.500 m moves at a constant speed of 8.00 m/s perpendicular to a uniform magnetic field of 0.750 T along two frictionless, zero-resistance parallel rails connected to a 3.00 Ω resistor. What is the magnitude of the motional EMF induced across the ends of the rod?

  1. 1.50 V
  2. 3.00 V Correct answer.
  3. 4.50 V
  4. 6.00 V

Explanation

The motional electromotive force induced across a conductor moving perpendicular to a magnetic field is ε = B L v = (0.750 T)(0.500 m)(8.00 m/s) = 3.00 V.

Question 38Electricity Magnetism

An ideal conducting rod of length 0.500 m slides on conducting rails at 8.00 m/s through a uniform 0.750 T magnetic field. The rod, velocity and field are mutually perpendicular, total circuit resistance is 3.00 ohms, and friction is neglected. What external mechanical power maintains constant speed?

  1. 0.75 W
  2. 1.50 W
  3. 3.00 W Correct answer.
  4. 9.00 W

Explanation

The induced EMF is ε = 3.00 V. Induced current is I = ε / R = 3.00 V / 3.00 Ω = 1.00 A. The magnetic retarding force opposing motion is F_mag = I L B = (1.00 A)(0.500 m)(0.750 T) = 0.375 N. To maintain constant velocity, applied force equals F_mag. The mechanical power input is P = F v = (0.375 N)(8.00 m/s) = 3.00 W (which perfectly matches the electrical power dissipated P_elec = I² R = 1.00² × 3.00 = 3.00 W).

Question 39Electricity Magnetism

Why are the soft iron cores of power transformers manufactured from stacks of thin, insulated iron laminations rather than from a single solid block of iron?

  1. To restrict the circulation of eddy currents, thereby minimizing thermal energy dissipation and power losses. Correct answer.
  2. To prevent the magnetic field from alternating in direction during alternating current cycles.
  3. To eliminate magnetic hysteresis within the ferromagnetic core material completely.
  4. To allow high-voltage electrical insulation oil to circulate directly between individual coil windings.

Explanation

Alternating magnetic flux induces circulating currents (eddy currents) in the conducting core, causing significant I²R resistive heating losses. Laminating the core and insulating each sheet breaks up the large closed conductive loops into very narrow pathways of high resistance, drastically reducing eddy current power loss.

Question 40Electricity Magnetism

An inductor of self-inductance 0.250 H carries a steady direct current of 4.00 A. How much energy is stored in the magnetic field within the inductor?

  1. 0.50 J
  2. 1.00 J
  3. 2.00 J Correct answer.
  4. 4.00 J

Explanation

The magnetic potential energy stored in an inductor is U_B = (1/2) L I² = (1/2)(0.250 H)(4.00 A)² = (1/2)(0.250)(16.0) = 2.00 J.

Question 41Electricity Magnetism

An ideal transformer connected to a 240 V AC mains supply has 1200 turns in its primary winding and 50 turns in its secondary winding. What is the output voltage delivered by the secondary coil?

  1. 5.00 V
  2. 10.0 V Correct answer.
  3. 20.0 V
  4. 5760 V

Explanation

For an ideal transformer, the turns ratio equals the voltage ratio: V_s / V_p = N_s / N_p. Therefore, V_s = V_p (N_s / N_p) = 240 V × (50 / 1200) = 240 V × (1 / 24) = 10.0 V.

Question 42Electricity Magnetism

A step-down transformer operates with an efficiency of 90.0%. The primary winding draws 2.50 A from a 240 V AC supply. If the secondary voltage is 24.0 V, what current is delivered to the load on the secondary side?

  1. 20.0 A
  2. 22.5 A Correct answer.
  3. 25.0 A
  4. 27.8 A

Explanation

Primary input power is P_in = V_p I_p = (240 V)(2.50 A) = 600 W. With an efficiency of 90.0%, secondary output power is P_out = η P_in = 0.900 × 600 W = 540 W. The secondary current is I_s = P_out / V_s = 540 W / 24.0 V = 22.5 A.

Question 43Electricity Magnetism

A sinusoidal alternating current voltage source has an RMS (root-mean-square) value of 240 V. What is the peak (amplitude) voltage of this AC supply?

  1. 120 V
  2. 170 V
  3. 240 V
  4. 339 V Correct answer.

Explanation

For a sinusoidal alternating voltage waveform, the relationship between peak voltage and RMS voltage is V_peak = V_rms × √2 = 240 V × 1.4142 = 339.4 V ≈ 339 V.

Question 44Electricity Magnetism

An ideal inductor of inductance 50.0 mH (0.0500 H) is connected across a 50.0 Hz AC source. What is the inductive reactance of the inductor?

  1. 2.50 Ω
  2. 7.85 Ω
  3. 15.7 Ω Correct answer.
  4. 31.4 Ω

Explanation

Inductive reactance is given by X_L = 2π f L. Substituting values: X_L = 2π(50.0 Hz)(0.0500 H) = 5.00π Ω ≈ 15.71 Ω = 15.7 Ω.

Question 45Electricity Magnetism

A 20.0 μF capacitor is connected to an audio signal generator delivering a sinusoidal voltage at a frequency of 500 Hz. What is the capacitive reactance of the capacitor?

  1. 7.96 Ω
  2. 15.9 Ω Correct answer.
  3. 31.8 Ω
  4. 62.8 Ω

Explanation

Capacitive reactance is given by X_C = 1 / (2π f C). Substituting the values: X_C = 1 / [2π(500 s⁻¹)(20.0 × 10⁻⁶ F)] = 1 / (0.0200π) = 1 / 0.06283 = 15.915 Ω ≈ 15.9 Ω.

Question 46Electricity Magnetism

A series RLC circuit comprises a resistor of resistance R = 40.0 Ω, an inductor with inductive reactance X_L = 80.0 Ω, and a capacitor with capacitive reactance X_C = 50.0 Ω connected across an AC voltage source. What is the total impedance Z of the circuit?

  1. 40.0 Ω
  2. 50.0 Ω Correct answer.
  3. 70.0 Ω
  4. 170 Ω

Explanation

In a series RLC circuit, the net reactance is X = X_L - X_C = 80.0 Ω - 50.0 Ω = 30.0 Ω. The impedance is given by the phasor magnitude Z = √[R² + (X_L - X_C)²] = √[40.0² + 30.0²] = √[1600 + 900] = √2500 = 50.0 Ω.

Question 47Electricity Magnetism

A series RLC circuit contains an inductor of inductance 200 mH (0.200 H) and a capacitor of capacitance 5.00 μF (5.00 × 10⁻⁶ F). What is the resonant frequency f_0 of this circuit?

  1. 79.6 Hz
  2. 159 Hz Correct answer.
  3. 318 Hz
  4. 1000 Hz

Explanation

Resonant frequency occurs when X_L = X_C, yielding f_0 = 1 / [2π √(L C)]. Substituting the given values: √(L C) = √[(0.200 H)(5.00 × 10⁻⁶ F)] = √(1.00 × 10⁻⁶ s²) = 1.00 × 10⁻³ s. Thus, f_0 = 1 / [2π × 1.00 × 10⁻³] = 1000 / (2π) = 500 / π = 159.15 Hz ≈ 159 Hz.

Question 48Electricity Magnetism

An electrical load connected across a 230 V RMS AC supply has a total impedance of Z = 25.0 Ω and a pure resistance of R = 15.0 Ω. What is the power factor of the load and the true (active) average power consumed?

  1. Power factor = 0.600; Power = 1.27 kW Correct answer.
  2. Power factor = 0.600; Power = 2.12 kW
  3. Power factor = 0.800; Power = 1.69 kW
  4. Power factor = 1.00; Power = 3.53 kW

Explanation

The power factor is cos φ = R / Z = 15.0 Ω / 25.0 Ω = 0.600. The RMS current is I_rms = V_rms / Z = 230 V / 25.0 Ω = 9.20 A. The true power dissipated is P = V_rms I_rms cos φ = (230 V)(9.20 A)(0.600) = 1269.6 W ≈ 1.27 kW (or P = I_rms² R = 9.20² × 15.0 = 1269.6 W).

Question 49Electricity Magnetism

In an alternating current circuit containing a pure inductor with zero electrical resistance, what is the phase relationship between the alternating voltage across the inductor and the alternating current through it?

  1. The voltage leads the current by a phase angle of 90° (π/2 radians). Correct answer.
  2. The current leads the voltage by a phase angle of 90° (π/2 radians).
  3. The voltage and current are in phase with zero phase difference.
  4. The voltage and current are in anti-phase with a phase difference of 180° (π radians).

Explanation

Because the back-EMF of an inductor is proportional to the rate of change of current (v_L = L di/dt), current cannot change instantaneously. When current is sinusoidal, voltage reaches its positive peak a quarter-cycle (90° or π/2 rad) ahead of the current; that is, voltage leads current by 90°.

Question 50Electricity Magnetism

An ideal LC oscillation circuit consists of a 1.00 μF capacitor initially charged to a potential difference of 20.0 V, connected across a 0.400 mH (4.00 × 10⁻⁴ H) inductor. Assuming zero resistance, what is the maximum current that flows through the inductor during the oscillations?

  1. 0.50 A
  2. 1.00 A Correct answer.
  3. 2.00 A
  4. 4.00 A

Explanation

By conservation of electromagnetic energy: (1/2) C V_max² = (1/2) L I_max². Substituting values: (1/2)(1.00 × 10⁻⁶ F)(20.0 V)² = (1/2)(4.00 × 10⁻⁴ H) I_max². This simplifies to (1.00 × 10⁻⁶)(400) = (4.00 × 10⁻⁴) I_max², so 4.00 × 10⁻⁴ = (4.00 × 10⁻⁴) I_max², giving I_max² = 1.00 A², hence I_max = 1.00 A.