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Q. A coil has an inductance of 0.5 H and is connected to a 50 Hz AC supply. What is the inductive reactance? (2020)

A. 25.13 Ω
B. 31.42 Ω
C. 50 Ω
D. 100 Ω

Solution

Inductive reactance Xl = 2πfL = 2π(50)(0.5) = 31.42 Ω.

Correct Answer: B — 31.42 Ω

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Q. A coil of wire has 100 turns and is placed in a magnetic field of strength 0.5 T. If the area of the coil is 0.1 m² and the magnetic field is perpendicular to the coil, what is the magnetic flux through the coil? (2022)

A. 5 Wb
B. 0.5 Wb
C. 10 Wb
D. 0.05 Wb

Solution

Magnetic flux (Φ) = B × A × cos(θ). Here, B = 0.5 T, A = 0.1 m², and θ = 0° (cos(0) = 1). Thus, Φ = 0.5 T × 0.1 m² × 1 = 0.05 Wb.

Correct Answer: A — 5 Wb

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Q. A coil of wire has 100 turns and is placed in a magnetic field of strength 0.5 T. If the area of the coil is 0.1 m², what is the maximum magnetic flux through the coil? (2022)

A. 5 Wb
B. 0.5 Wb
C. 10 Wb
D. 1 Wb

Solution

Magnetic flux (Φ) = B × A × N = 0.5 T × 0.1 m² × 100 = 5 Wb.

Correct Answer: A — 5 Wb

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Q. A coil with 100 turns has a magnetic flux of 0.02 Wb passing through it. If the flux changes to 0.01 Wb in 2 seconds, what is the induced EMF? (2022)

A. 0.5 V
B. 1 V
C. 2 V
D. 5 V

Solution

Induced EMF = -N * (ΔΦ/Δt) = -100 * ((0.01 - 0.02) / 2) = 0.5 V.

Correct Answer: B — 1 V

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Q. A coil with 100 turns is placed in a magnetic field that changes from 0.5 T to 1.5 T in 2 seconds. What is the induced EMF in the coil?

A. 50 V
B. 100 V
C. 200 V
D. 400 V

Solution

Using Faraday's law, EMF = -N * (ΔB/Δt) = -100 * ((1.5 - 0.5) / 2) = -100 * (1 / 2) = -50 V. The magnitude is 50 V.

Correct Answer: B — 100 V

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Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force given by Fleming's left-hand rule? (2020)

A. Parallel to the field
B. Opposite to the field
C. Perpendicular to the field
D. In the direction of current

Solution

According to Fleming's left-hand rule, the force is perpendicular to both the magnetic field and the direction of current.

Correct Answer: C — Perpendicular to the field

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Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force? (2020)

A. Parallel to the field
B. Opposite to the field
C. Perpendicular to both current and field
D. In the direction of current

Solution

According to Fleming's left-hand rule, the force on a current-carrying conductor in a magnetic field is perpendicular to both the current and the magnetic field.

Correct Answer: C — Perpendicular to both current and field

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Q. A cyclist accelerates at 3 m/s² for 4 seconds. What is the final velocity if the initial velocity is 2 m/s? (2023)

A. 10 m/s
B. 11 m/s
C. 14 m/s
D. 20 m/s

Solution

Using v = u + at, we have v = 2 m/s + (3 m/s² * 4 s) = 2 + 12 = 14 m/s.

Correct Answer: B — 11 m/s

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the acceleration? (2023)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Acceleration = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the cyclist's acceleration? (2021)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Acceleration a = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A cyclist does 1500 J of work to climb a hill. If he takes 5 minutes, what is his average power output?

A. 5 W
B. 10 W
C. 15 W
D. 20 W

Solution

Power = Work / Time = 1500 J / (5 * 60 s) = 5 W.

Correct Answer: B — 10 W

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Q. A cyclist does 300 J of work to climb a hill. If the cyclist's mass is 75 kg, what is the height of the hill? (2023)

A. 0.4 m
B. 0.5 m
C. 0.6 m
D. 0.8 m

Solution

Potential Energy = mass × g × height; 300 J = 75 kg × 10 m/s² × height; height = 300 J / (75 kg × 10 m/s²) = 0.4 m

Correct Answer: B — 0.5 m

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Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the mass of the cyclist? (g = 9.8 m/s²)

A. 6.12 kg
B. 10 kg
C. 15 kg
D. 20 kg

Solution

Using Work = mgh; m = Work / (gh) = 300 J / (9.8 m/s² * 5 m) = 6.12 kg.

Correct Answer: A — 6.12 kg

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Q. A cyclist does 300 J of work to climb a hill. If the hill is 5 m high, what is the effective weight of the cyclist? (2022)

A. 30 kg
B. 60 kg
C. 90 kg
D. 120 kg

Solution

Weight = Work done / height = 300 J / 5 m = 60 kg

Correct Answer: B — 60 kg

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Q. A cyclist does 300 J of work to climb a hill. If the hill is 5 m high, what is the weight of the cyclist? (2020)

A. 60 kg
B. 30 kg
C. 50 kg
D. 40 kg

Solution

Work = m × g × h. 300 J = m × 9.8 m/s² × 5 m. m = 300 J / (9.8 m/s² × 5 m) = 6.12 kg.

Correct Answer: A — 60 kg

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Q. A cyclist does 600 J of work to climb a hill. If he takes 30 seconds, what is his average power output? (2022)

A. 10 W
B. 20 W
C. 30 W
D. 40 W

Solution

Power = Work / Time = 600 J / 30 s = 20 W.

Correct Answer: B — 20 W

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Q. A cyclist is moving at a speed of 10 m/s. If he doubles his speed, how much more kinetic energy will he have? (2023)

A. 4 times
B. 2 times
C. 3 times
D. 1.5 times

Solution

Kinetic Energy is proportional to the square of the speed. If speed is doubled, KE increases by a factor of 2^2 = 4.

Correct Answer: A — 4 times

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Q. A cyclist is moving at a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the total kinetic energy?

A. 800 J
B. 900 J
C. 1000 J
D. 1200 J

Solution

KE = 0.5 × m × v² = 0.5 × 75 kg × (15 m/s)² = 0.5 × 75 × 225 = 8437.5 J.

Correct Answer: C — 1000 J

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Q. A cyclist is moving at a speed of 15 m/s. If the total mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy?

A. 843.75 J
B. 562.5 J
C. 1687.5 J
D. 1125 J

Solution

Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 75 kg × (15 m/s)² = 0.5 × 75 × 225 = 8437.5 J.

Correct Answer: A — 843.75 J

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Q. A cyclist is moving up a hill and gains 3000 J of potential energy. If the mass of the cyclist and the bicycle is 75 kg, what is the height of the hill? (g = 9.8 m/s²)

A. 4.08 m
B. 3.06 m
C. 2.04 m
D. 1.5 m

Solution

Height (h) = PE / (mg) = 3000 J / (75 kg * 9.8 m/s²) = 4.08 m.

Correct Answer: A — 4.08 m

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Q. A cyclist is moving with a speed of 10 m/s. If he doubles his speed, how much more kinetic energy will he have? (2023)

A. 4 times
B. 2 times
C. 3 times
D. 1.5 times

Solution

Kinetic Energy is proportional to the square of the speed. If speed is doubled, KE increases by a factor of 2^2 = 4.

Correct Answer: A — 4 times

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Q. A cyclist is moving with a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy? (2022)

A. 8437.5 J
B. 1687.5 J
C. 5625 J
D. 2250 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 75 kg * (15 m/s)^2 = 8437.5 J.

Correct Answer: A — 8437.5 J

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Q. A cyclist of mass 70 kg accelerates at 2 m/s². What is the force exerted by the cyclist on the ground? (2021)

A. 70 N
B. 140 N
C. 210 N
D. 280 N

Solution

Using F = ma, F = 70 kg * 2 m/s² = 140 N.

Correct Answer: B — 140 N

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Q. A cyclist travels at a speed of 10 m/s for 15 seconds. What distance does he cover? (2022)

A. 100 m
B. 150 m
C. 200 m
D. 250 m

Solution

Distance = speed * time = 10 m/s * 15 s = 150 m.

Correct Answer: B — 150 m

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Q. A cyclist travels at a speed of 10 m/s for 5 seconds. What distance does he cover? (2023)

A. 25 m
B. 50 m
C. 75 m
D. 100 m

Solution

Distance = speed * time = 10 m/s * 5 s = 50 m.

Correct Answer: B — 50 m

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Q. A cyclist travels at a speed of 15 m/s for 2 minutes. How far does he travel?

A. 1800 m
B. 1200 m
C. 900 m
D. 600 m

Solution

Distance = speed * time. Time in seconds = 2 * 60 = 120 s. Distance = 15 m/s * 120 s = 1800 m.

Correct Answer: A — 1800 m

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Q. A cyclist travels at a speed of 15 m/s for 2 minutes. How far does the cyclist travel?

A. 1800 m
B. 1200 m
C. 900 m
D. 600 m

Solution

Distance = speed * time. Time in seconds = 2 minutes = 120 seconds. Distance = 15 m/s * 120 s = 1800 m.

Correct Answer: A — 1800 m

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Q. A cyclist travels at a speed of 15 m/s for 20 seconds. What distance does he cover?

A. 200 m
B. 300 m
C. 400 m
D. 500 m

Solution

Distance = speed * time = 15 m/s * 20 s = 300 m.

Correct Answer: C — 400 m

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Q. A cyclist travels at a speed of 15 m/s for 20 seconds. What distance does the cyclist cover?

A. 150 m
B. 300 m
C. 400 m
D. 600 m

Solution

Distance = speed * time = 15 m/s * 20 s = 300 m.

Correct Answer: B — 300 m

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Q. A cylinder rolls down an incline without slipping. If its mass is M and radius is R, what is the acceleration of its center of mass? (2020)

A. g/2
B. g/3
C. g/4
D. g/5

Solution

For rolling without slipping, a_cm = (2/3)g sin(θ).

Correct Answer: B — g/3

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