Q. A fluid with a viscosity of 0.1 Pa·s flows through a pipe of radius 0.05 m. What is the shear stress if the flow velocity is 1 m/s?
A.
0.1 Pa
B.
0.2 Pa
C.
0.4 Pa
D.
0.5 Pa
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Solution
Shear stress = viscosity × (velocity/radius) = 0.1 × (1/0.05) = 2 Pa.
Correct Answer: B — 0.2 Pa
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Q. A fluid with a viscosity of 0.1 Pa·s flows through a pipe of radius 0.05 m. What is the shear stress if the flow rate is 0.01 m³/s?
A.
0.4 Pa
B.
0.2 Pa
C.
0.1 Pa
D.
0.5 Pa
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Solution
Using the formula for shear stress, τ = (4 * η * Q) / (π * r^3), we find τ = 0.4 Pa.
Correct Answer: A — 0.4 Pa
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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 for time t, what is the final angular velocity?
A.
ω + (τ/I)t
B.
ω - (τ/I)t
C.
ω + (I/τ)t
D.
ω - (I/τ)t
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Solution
Using the equation ω_f = ω + αt, where α = τ/I, we get ω_f = ω + (τ/I)t.
Correct Answer: A — ω + (τ/I)t
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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
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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 flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied to it, what is the angular acceleration α?
A.
τ/I
B.
I/τ
C.
Iω/τ
D.
τω/I
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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 flywheel is a device used to store rotational energy. If the moment of inertia of the flywheel is I and it rotates with an angular velocity ω, what is its rotational kinetic energy? (2020)
A.
(1/2)Iω^2
B.
(1/4)Iω^2
C.
(1/3)Iω^2
D.
(1/5)Iω^2
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Solution
The rotational kinetic energy is given by K.E. = (1/2)Iω^2.
Correct Answer: A — (1/2)Iω^2
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Q. A flywheel is rotating at 1000 rpm. If it is brought to rest in 10 seconds, what is the average angular deceleration?
A.
100 rad/s²
B.
10 rad/s²
C.
20 rad/s²
D.
50 rad/s²
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Solution
First convert rpm to rad/s: 1000 rpm = (1000 * 2π)/60 rad/s. Then use α = (ωf - ωi)/t = (0 - (1000 * 2π)/60) / 10.
Correct Answer: C — 20 rad/s²
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Q. A flywheel is rotating with an angular speed of 10 rad/s. If it is brought to rest in 5 seconds, what is the angular deceleration? (2020)
A.
2 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
20 rad/s²
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Solution
Using the formula α = (ω - ω₀)/t, we have α = (0 - 10)/5 = -2 rad/s², so the deceleration is 2 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A flywheel is rotating with an angular speed of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
A.
4 rad/s²
B.
5 rad/s²
C.
20 rad/s²
D.
0 rad/s²
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Solution
Angular deceleration α = (final angular speed - initial angular speed) / time = (0 - 20 rad/s) / 5 s = -4 rad/s².
Correct Answer: A — 4 rad/s²
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Q. A flywheel is rotating with an angular speed of 20 rad/s. If it experiences a torque of 5 Nm, what is the time taken to stop it?
A.
8 s
B.
4 s
C.
10 s
D.
5 s
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Solution
Using τ = Iα, we find α = τ/I. Assuming I = 1 kg·m², α = 5 rad/s². Time to stop = ω/α = 20 rad/s / 5 rad/s² = 4 s.
Correct Answer: B — 4 s
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Q. A flywheel is rotating with an angular speed of 30 rad/s. If it comes to rest in 10 seconds, what is the angular deceleration? (2020)
A.
3 rad/s²
B.
6 rad/s²
C.
2 rad/s²
D.
1 rad/s²
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Solution
Using the formula α = (ω - ω₀)/t, we have α = (0 - 30 rad/s) / 10 s = -3 rad/s².
Correct Answer: A — 3 rad/s²
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Q. A flywheel is rotating with an angular velocity of 10 rad/s. If it is brought to rest in 5 seconds, what is the angular deceleration? (2020)
A.
2 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
20 rad/s²
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Solution
Angular deceleration α = (final angular velocity - initial angular velocity) / time = (0 - 10) / 5 = -2 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 10 rad/s. If it is subjected to a torque of 5 Nm, what is the angular acceleration?
A.
0.5 rad/s²
B.
2 rad/s²
C.
0.2 rad/s²
D.
1 rad/s²
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Solution
Using τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration. Assuming I = 25 kg·m², α = τ/I = 5/25 = 0.2 rad/s².
Correct Answer: B — 2 rad/s²
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Q. A flywheel is rotating with an angular velocity of 10 rad/s. If the moment of inertia of the flywheel is 2 kg·m², what is its rotational kinetic energy? (2020)
A.
100 J
B.
50 J
C.
20 J
D.
10 J
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Solution
Rotational kinetic energy K.E. = (1/2)Iω² = (1/2)(2)(10)² = 100 J.
Correct Answer: B — 50 J
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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it comes to rest in 3 seconds, what is the angular deceleration?
A.
5 rad/s²
B.
10 rad/s²
C.
15 rad/s²
D.
20 rad/s²
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Solution
Angular deceleration = (final angular velocity - initial angular velocity) / time = (0 - 15) / 3 = -5 rad/s², so the magnitude is 5 rad/s².
Correct Answer: B — 10 rad/s²
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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it experiences a torque of 3 N·m, what is the angular acceleration?
A.
0.2 rad/s²
B.
0.5 rad/s²
C.
1 rad/s²
D.
5 rad/s²
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Solution
Angular acceleration α = Torque/I. Assuming I = 6 kg·m², α = 3 N·m / 6 kg·m² = 0.5 rad/s².
Correct Answer: B — 0.5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
A.
4 rad/s²
B.
5 rad/s²
C.
20 rad/s²
D.
0 rad/s²
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Solution
Angular deceleration α = (ω_f - ω_i) / t = (0 - 20 rad/s) / 5 s = -4 rad/s², so the magnitude is 4 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it experiences a constant torque that reduces its angular velocity to 10 rad/s in 5 seconds, what is the magnitude of the torque if the moment of inertia is 4 kg·m²?
A.
8 N·m
B.
4 N·m
C.
2 N·m
D.
10 N·m
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Solution
The angular deceleration α = (ω_final - ω_initial) / time = (10 - 20) / 5 = -2 rad/s². Torque τ = Iα = 4 kg·m² * (-2 rad/s²) = -8 N·m, so the magnitude is 8 N·m.
Correct Answer: B — 4 N·m
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Q. A food item contains 10 grams of carbohydrates, 5 grams of protein, and 2 grams of fat. How many total calories does this food item provide? (Carbohydrates = 4 cal/g, Protein = 4 cal/g, Fat = 9 cal/g) (2022)
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Solution
Total calories = (10g * 4) + (5g * 4) + (2g * 9) = 40 + 20 + 18 = 78 calories.
Correct Answer: C — 62
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Q. A food label states that a serving contains 5 grams of sugar. If a person consumes 3 servings, how many grams of sugar do they consume?
A.
10 grams
B.
12 grams
C.
15 grams
D.
18 grams
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Solution
5 grams/serving * 3 servings = 15 grams.
Correct Answer: C — 15 grams
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Q. A football field is 100 meters long and 64 meters wide. What is the area of the field in square meters? (2019)
A.
6400
B.
10000
C.
8000
D.
6000
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Solution
Area = length × width = 100 m × 64 m = 6400 square meters.
Correct Answer: A — 6400
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Q. A football match lasts for 90 minutes. If a player runs at an average speed of 8 km/h, how far does he run in the entire match? (2019)
A.
10 km
B.
12 km
C.
14 km
D.
15 km
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Solution
Distance = Speed × Time. Time in hours = 90/60 = 1.5 hours. Distance = 8 km/h × 1.5 h = 12 km.
Correct Answer: B — 12 km
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Q. A force is measured as 100 N with an uncertainty of ±2 N. What is the maximum possible value of the force?
A.
102 N
B.
98 N
C.
100 N
D.
104 N
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Solution
Maximum possible value = measured value + uncertainty = 100 + 2 = 102 N.
Correct Answer: A — 102 N
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Q. A force is measured as 50 N with an uncertainty of ±1 N. What is the percentage uncertainty in the force measurement?
A.
2%
B.
1%
C.
0.5%
D.
0.1%
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Solution
Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (1 / 50) * 100 = 2%.
Correct Answer: B — 1%
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Q. A force is measured as 50 N with an uncertainty of ±2 N. What is the percentage uncertainty in the force measurement?
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Solution
Percentage uncertainty = (absolute uncertainty / measured value) * 100 = (2 / 50) * 100 = 4%.
Correct Answer: A — 4%
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Q. A force is measured as 50 N with an uncertainty of ±2 N. What is the relative uncertainty in this force measurement?
A.
0.04
B.
0.02
C.
0.01
D.
0.05
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Solution
Relative uncertainty = (absolute uncertainty / measured value) = 2 / 50 = 0.04 or 4%.
Correct Answer: B — 0.02
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Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?
A.
2.0 Nm
B.
5.0 Nm
C.
10.0 Nm
D.
20.0 Nm
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Solution
Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.
Correct Answer: A — 2.0 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
20 Nm
D.
15 Nm
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Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot point?
A.
5 Nm
B.
10 Nm
C.
20 Nm
D.
15 Nm
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Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
A.
10 Nm
B.
17.32 Nm
C.
20 Nm
D.
5 Nm
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Solution
Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.
Correct Answer: B — 17.32 Nm
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