Physics Study Resources for Competitive Exams

Physics

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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 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 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 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 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 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 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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Q. A dipole consists of two equal and opposite charges separated by a distance 'd'. What is the dipole moment? (2023)

A. qd
B. q/d
C. q^2d
D. qd^2

Solution

The dipole moment (p) is defined as p = qd, where q is the charge and d is the separation distance.

Correct Answer: A — qd

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Q. A dipole is placed in a uniform electric field. What is the torque experienced by the dipole? (2023)

A. pE
B. pE sin(θ)
C. pE cos(θ)
D. Zero

Solution

The torque (τ) experienced by a dipole in a uniform electric field is given by τ = pE sin(θ), where p is the dipole moment and θ is the angle between p and E.

Correct Answer: B — pE sin(θ)

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Q. A disc of radius R and mass M is rotating about its axis with an angular velocity ω. What is its rotational kinetic energy? (2020)

A. (1/2)Iω²
B. (1/2)Mω²
C. Iω
D. Mω²

Solution

Rotational kinetic energy K.E. = (1/2)Iω². For a disc, I = (1/2)MR², thus K.E. = (1/4)MR²ω².

Correct Answer: A — (1/2)Iω²

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Q. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied, what is the angular acceleration? (2021)

A. τ/I
B. I/τ
C. ω/τ
D. Iω

Solution

From Newton's second law for rotation, τ = Iα, thus α = τ/I.

Correct Answer: A — τ/I

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Q. A force of 10 N is applied to an object, causing it to accelerate at 2 m/s². What is the mass of the object? (2020)

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

Solution

Using F = ma, mass m = F/a = 10 N / 2 m/s² = 5 kg.

Correct Answer: A — 5 kg

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Q. A force of 10 N is applied to move a box 5 m across a floor. What is the work done on the box?

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J

Correct Answer: B — 20 J

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Q. A force of 10 N is applied to move a box 5 m across a floor. What is the work done by the force?

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J

Correct Answer: B — 20 J

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Q. A force of 10 N is applied to move a box 5 m along a horizontal surface. What is the work done by the force? (2021)

A. 20 J
B. 50 J
C. 10 J
D. 5 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J

Correct Answer: B — 50 J

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Q. A force of 10 N is applied to move an object 5 m in the direction of the force. What is the work done by the force?

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done (W) = Force (F) × Distance (d) × cos(θ). Here, θ = 0° (force and displacement are in the same direction). W = 10 N × 5 m × cos(0°) = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A force of 10 N is applied to move an object 5 m. What is the work done? (2023)

A. 50 J
B. 10 J
C. 5 J
D. 20 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: A — 50 J

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Q. A force of 15 N is applied to a 3 kg mass. What is the resulting acceleration? (2023)

A. 3 m/s²
B. 5 m/s²
C. 7 m/s²
D. 10 m/s²

Solution

Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².

Correct Answer: B — 5 m/s²

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Q. A force of 20 N is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done?

A. 40 J
B. 80 J
C. 20 J
D. 60 J

Solution

Work done (W) = F × d × cos(θ) = 20 N × 4 m × cos(60°) = 20 × 4 × 0.5 = 40 J.

Correct Answer: B — 80 J

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Q. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)

A. 3 m/s²
B. 4 m/s²
C. 5 m/s²
D. 6 m/s²

Solution

Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².

Correct Answer: B — 4 m/s²

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Q. A force of 50 N is applied to move an object 4 m. How much work is done?

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

Solution

Work = Force * Distance = 50 N * 4 m = 200 J.

Correct Answer: C — 200 J

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Physics

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