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Q. A torque of 10 N·m is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?

A. 5 rad/s²
B. 10 rad/s²
C. 2 rad/s²
D. 20 rad/s²

Solution

Using τ = Iα, we have α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².

Correct Answer: A — 5 rad/s²

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Q. A torque of 12 Nm is applied to a lever arm of 0.4 m. What is the force applied?

A. 30 N
B. 25 N
C. 20 N
D. 15 N

Solution

Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.

Correct Answer: C — 20 N

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Q. A torque of 12 Nm is applied to a wheel of radius 0.4 m. What is the force applied at the edge of the wheel?

A. 30 N
B. 20 N
C. 15 N
D. 10 N

Solution

Torque (τ) = r × F, thus F = τ / r = 12 Nm / 0.4 m = 30 N.

Correct Answer: B — 20 N

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Q. A torque of 12 Nm is applied to a wheel with a radius of 0.4 m. What is the force applied tangentially to the wheel?

A. 15 N
B. 30 N
C. 40 N
D. 50 N

Solution

Using τ = F × r, we have F = τ / r = 12 Nm / 0.4 m = 30 N.

Correct Answer: B — 30 N

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Q. A torque of 12 Nm is produced by a force acting at a distance of 0.4 m from the pivot. What is the magnitude of the force?

A. 20 N
B. 30 N
C. 40 N
D. 50 N

Solution

Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.

Correct Answer: A — 20 N

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Q. A torque of 12 Nm is produced by a force acting at a distance of 4 m from the pivot. What is the magnitude of the force?

A. 2 N
B. 3 N
C. 4 N
D. 5 N

Solution

Force = Torque / Distance = 12 Nm / 4 m = 3 N.

Correct Answer: B — 3 N

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Q. A torque of 15 N·m is applied to a wheel with a radius of 0.3 m. What is the force applied tangentially to the wheel?

A. 25 N
B. 50 N
C. 45 N
D. 30 N

Solution

Torque (τ) = Force (F) × Radius (r) => F = τ / r = 15 N·m / 0.3 m = 50 N.

Correct Answer: B — 50 N

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Q. A torque of 25 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?

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

Solution

Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.

Correct Answer: A — 50 N

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Q. A torque of 25 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?

A. 10 N
B. 25 N
C. 50 N
D. 5 N

Solution

Torque (τ) = Force (F) × Radius (r) => F = τ / r = 25 Nm / 0.5 m = 50 N.

Correct Answer: A — 10 N

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Q. A torque of 30 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?

A. 60 N
B. 30 N
C. 15 N
D. 75 N

Solution

Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.

Correct Answer: A — 60 N

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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially?

A. 15 N
B. 30 N
C. 60 N
D. 75 N

Solution

Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.

Correct Answer: C — 60 N

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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?

A. 15 N
B. 30 N
C. 60 N
D. 75 N

Solution

Torque (τ) = F × r, thus F = τ / r = 30 Nm / 0.5 m = 60 N.

Correct Answer: C — 60 N

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Q. A torque of 40 Nm is required to rotate a wheel. If the radius of the wheel is 0.4 m, what is the force applied tangentially?

A. 100 N
B. 80 N
C. 60 N
D. 40 N

Solution

Force = Torque / Radius = 40 Nm / 0.4 m = 100 N.

Correct Answer: B — 80 N

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Q. A torque of 5 Nm is applied to a wheel. If the radius of the wheel is 0.25 m, what is the force applied tangentially?

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

Solution

Torque (τ) = F × r, thus F = τ / r = 5 Nm / 0.25 m = 20 N.

Correct Answer: A — 10 N

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Q. A torque of 5 N·m is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?

A. 2.5 rad/s²
B. 5 rad/s²
C. 10 rad/s²
D. 1 rad/s²

Solution

Angular acceleration α = Torque/I = 5 N·m / 2 kg·m² = 2.5 rad/s².

Correct Answer: A — 2.5 rad/s²

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Q. A torque of 50 Nm is applied to a wheel with a radius of 0.25 m. What is the force applied at the edge of the wheel?

A. 100 N
B. 200 N
C. 250 N
D. 300 N

Solution

Force = Torque / Radius = 50 Nm / 0.25 m = 200 N.

Correct Answer: B — 200 N

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Q. A torque of 50 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied tangentially to the wheel?

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

Solution

Torque (τ) = Force (F) × Radius (r) => F = τ / r = 50 Nm / 0.5 m = 100 N.

Correct Answer: A — 100 N

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Q. A torque of 50 Nm is created by a force acting at a distance of 2 m. What is the force applied?

A. 20 N
B. 25 N
C. 30 N
D. 35 N

Solution

Force = Torque / Distance = 50 Nm / 2 m = 25 N.

Correct Answer: B — 25 N

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Q. A torque τ is applied to a rigid body with moment of inertia I. If the body starts from rest, what is the angular displacement θ after time t?

A. (1/2)(τ/I)t^2
B. (τ/I)t^2
C. (1/2)(I/τ)t^2
D. (I/τ)t^2

Solution

Using the equation of motion for rotation, θ = (1/2)αt^2, where α = τ/I, thus θ = (1/2)(τ/I)t^2.

Correct Answer: A — (1/2)(τ/I)t^2

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Q. A torque τ is applied to a rotating object with moment of inertia I. If the object starts from rest, what is its angular acceleration α?

A. τ/I
B. I/τ
C. Iτ
D. τI

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 tower is 100 meters high. From a point on the ground, the angle of elevation to the top of the tower is 30 degrees. How far is the point from the base of the tower?

A. 100√3 meters
B. 50√3 meters
C. 200 meters
D. 150 meters

Solution

Distance = height / tan(angle) = 100 / (1/√3) = 100√3 meters.

Correct Answer: A — 100√3 meters

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Q. A tower is 40 meters high. From a point on the ground, the angle of elevation to the top of the tower is 60 degrees. How far is the point from the base of the tower?

A. 20√3 meters
B. 40 meters
C. 20 meters
D. 30 meters

Solution

Distance = height / tan(angle) = 40 / tan(60°) = 40 / √3 = 20√3 meters.

Correct Answer: A — 20√3 meters

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Q. A tower is 50 meters high. From a point on the ground, the angle of elevation to the top of the tower is 30 degrees. What is the distance from the point to the base of the tower?

A. 25√3 m
B. 50 m
C. 25 m
D. 50√3 m

Solution

Using tan(30°) = height/distance, we have distance = height/tan(30°) = 50/√3 = 25√3 m.

Correct Answer: A — 25√3 m

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Q. A tower is 60 meters high. From a point on the ground, the angle of elevation to the top of the tower is 45 degrees. How far is the point from the base of the tower?

A. 30 meters
B. 60 meters
C. 45 meters
D. 75 meters

Solution

Distance = height / tan(angle) = 60 / 1 = 60 meters.

Correct Answer: B — 60 meters

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Q. A tower is 80 meters high. From a point on the ground, the angle of elevation to the top of the tower is 60 degrees. How far is the point from the base of the tower?

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

Solution

Distance = height / tan(angle) = 80 / √3 = 40 m.

Correct Answer: A — 40 m

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Q. A tower is standing on a horizontal ground. The angle of elevation of the top of the tower from a point on the ground is 30 degrees. If the height of the tower is 50 meters, how far is the point from the base of the tower?

A. 50√3 m
B. 100 m
C. 50 m
D. 100√3 m

Solution

Using tan(30°) = height/distance, we have distance = height/tan(30°) = 50/(1/√3) = 50√3 m.

Correct Answer: A — 50√3 m

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Q. A tower is standing on a horizontal ground. The angle of elevation of the top of the tower from a point on the ground is 30 degrees. If the height of the tower is 10√3 meters, how far is the point from the base of the tower?

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

Solution

Using tan(30°) = height/distance, we have 1/√3 = 10√3/distance. Therefore, distance = 10√3 * √3 = 30 m.

Correct Answer: B — 20 m

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Q. A tower is standing on a horizontal ground. The angle of elevation of the top of the tower from a point on the ground is 30 degrees. If the height of the tower is 10√3 m, how far is the point from the base of the tower?

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

Solution

Using tan(30°) = height/distance, we have 1/√3 = 10√3/distance. Therefore, distance = 10√3 * √3 = 30 m.

Correct Answer: A — 10 m

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Q. A train is moving at 72 km/h and a bird flies in the opposite direction at 18 km/h. What is the speed of the bird relative to the train?

A. 54 km/h
B. 72 km/h
C. 90 km/h
D. 18 km/h

Solution

Relative speed = Speed of train + Speed of bird = 72 km/h + 18 km/h = 90 km/h.

Correct Answer: A — 54 km/h

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Q. A train is moving at 72 km/h and passes a platform in 20 seconds. What is the length of the train?

A. 100 m
B. 200 m
C. 300 m
D. 400 m

Solution

Length of train = Speed × Time = (72 km/h × (1000 m/1 km) / (3600 s/1 h)) × 20 s = 400 m.

Correct Answer: B — 200 m

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Engineering & Architecture Admissions

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