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Q. A cyclist is moving in a circular path of radius 10 m at a speed of 5 m/s. What is the angle of banking required to prevent slipping?

A. 30°
B. 45°
C. 60°
D. 90°

Solution

tan(θ) = v²/(rg) = (5²)/(10*9.8) = 0.25. θ = tan⁻¹(0.25) ≈ 14°.

Correct Answer: A — 30°

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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 5 m/s. What is the angular velocity of the cyclist?

A. 0.2 rad/s
B. 0.5 rad/s
C. 1 rad/s
D. 2 rad/s

Solution

Angular velocity (ω) = v/r = 5 m/s / 15 m = 1/3 rad/s.

Correct Answer: B — 0.5 rad/s

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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 6 m/s. What is the angular velocity of the cyclist?

A. 0.4 rad/s
B. 0.6 rad/s
C. 0.8 rad/s
D. 1.0 rad/s

Solution

Angular velocity (ω) = v/r = 6 m/s / 15 m = 0.4 rad/s.

Correct Answer: B — 0.6 rad/s

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Q. A cyclist is moving in a circular track of radius 30 m with a speed of 15 m/s. What is the net force acting on the cyclist if the mass of the cyclist is 60 kg?

A. 180 N
B. 120 N
C. 90 N
D. 60 N

Solution

Centripetal force F = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.

Correct Answer: A — 180 N

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Q. A cyclist is moving in a circular track of radius 30 m. If he completes one round in 12 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(30) = 60π m. Average speed = distance/time = 60π m / 12 s = 5 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the angular velocity of the cyclist?

A. π/6 rad/s
B. π/3 rad/s
C. 2π/6 rad/s
D. 2π/3 rad/s

Solution

Angular velocity (ω) = 2π/T = 2π/12 = π/6 rad/s.

Correct Answer: B — π/3 rad/s

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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the average speed of the cyclist?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(30 m). Average speed = distance/time = (2π(30 m))/12 s ≈ 5 m/s.

Correct Answer: B — 10 m/s

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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 is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist towards the center of the track?

A. 50 N
B. 75 N
C. 100 N
D. 125 N

Solution

Centripetal force (F_c) = mv²/r. Assuming mass m = 100 kg, F_c = (100 kg)(15 m/s)² / (30 m) = 75 N.

Correct Answer: C — 100 N

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Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist if the mass of the cyclist is 60 kg?

A. 180 N
B. 120 N
C. 90 N
D. 60 N

Solution

Centripetal force F_c = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.

Correct Answer: A — 180 N

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Q. A cyclist is negotiating a circular turn of radius 30 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent slipping?

A. 0.25
B. 0.5
C. 0.75
D. 1

Solution

Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15²)/(30*9.8) = 0.25.

Correct Answer: B — 0.5

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Q. A cyclist is pedaling at a constant speed and exerts a power of 200 W. If the cyclist increases their power output to 400 W, what happens to their speed assuming no other forces act?

A. Speed remains the same
B. Speed doubles
C. Speed increases by 41%
D. Speed increases by 100%

Solution

Power is proportional to the cube of the speed in cycling. If power doubles, speed increases by a factor of (2)^(1/3) which is approximately 1.26, or about a 41% increase.

Correct Answer: C — Speed increases by 41%

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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 100 m in 5 seconds. What is the average speed of the cyclist?

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 25 m/s

Solution

Average speed = total distance / total time = 100 m / 5 s = 20 m/s.

Correct Answer: A — 10 m/s

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Q. A cyclist travels 100 m north and then 100 m east. What is the magnitude of the displacement from the starting point?

A. 100 m
B. 141.4 m
C. 200 m
D. 50 m

Solution

Displacement = √(100² + 100²) = √20000 = 141.4 m.

Correct Answer: B — 141.4 m

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Q. A cyclist travels 100 m north and then 100 m east. What is the magnitude of the displacement? (2000)

A. 100 m
B. 141.42 m
C. 200 m
D. 250 m

Solution

Displacement = √(100² + 100²) = √20000 = 141.42 m.

Correct Answer: B — 141.42 m

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Q. A cyclist travels 15 km in 30 minutes. What is his speed in km/h? (2023)

A. 25 km/h
B. 30 km/h
C. 20 km/h
D. 15 km/h

Solution

Speed = Distance/Time = 15 km / 0.5 h = 30 km/h.

Correct Answer: A — 25 km/h

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Q. A cyclist travels 15 km in 30 minutes. What is the speed of the cyclist in km/h? (2019)

A. 25 km/h
B. 30 km/h
C. 35 km/h
D. 40 km/h

Solution

Speed = Distance / Time = 15 km / 0.5 hours = 30 km/h.

Correct Answer: A — 25 km/h

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Q. A cyclist travels 30 km in 1.5 hours. What is his average speed in km/h? (2023)

A. 20 km/h
B. 25 km/h
C. 30 km/h
D. 15 km/h

Solution

Average speed = Total distance / Total time = 30 km / 1.5 h = 20 km/h.

Correct Answer: B — 25 km/h

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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 15 seconds. What distance does the cyclist cover?

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 30 seconds. How far does the cyclist travel?

A. 300 m
B. 200 m
C. 150 m
D. 250 m

Solution

Distance = speed × time = 10 m/s × 30 s = 300 m.

Correct Answer: A — 300 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 km/h. If he encounters a headwind of 5 km/h, what is his effective speed?

A. 10 km/h
B. 15 km/h
C. 20 km/h
D. 25 km/h

Solution

Effective speed = speed of cyclist - speed of wind = 15 - 5 = 10 km/h.

Correct Answer: A — 10 km/h

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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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Major Competitive Exams

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