Torque
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?
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A.
100 N
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B.
80 N
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C.
60 N
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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?
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A.
10 N
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B.
20 N
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C.
5 N
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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 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?
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A.
100 N
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B.
200 N
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C.
250 N
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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?
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A.
100 N
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B.
50 N
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C.
25 N
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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?
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A.
20 N
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B.
25 N
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C.
30 N
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D.
35 N
Solution
Force = Torque / Distance = 50 Nm / 2 m = 25 N.
Correct Answer: B — 25 N
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Q. If a force of 12 N is applied at an angle of 30 degrees to a lever arm of 1 m, what is the torque about the pivot?
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A.
6 Nm
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B.
10 Nm
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C.
12 Nm
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D.
15 Nm
Solution
Torque = Force × Distance × sin(θ) = 12 N × 1 m × sin(30°) = 12 N × 1 m × 0.5 = 6 Nm.
Correct Answer: A — 6 Nm
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Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1.5 m, what is the torque about the pivot?
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A.
3.75 Nm
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B.
7.5 Nm
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C.
11.25 Nm
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D.
12.99 Nm
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1.5 m × sin(30°) = 15 × 1.5 × 0.5 = 11.25 Nm.
Correct Answer: B — 7.5 Nm
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Q. If a torque of 15 Nm is produced by a force acting at a distance of 0.3 m from the pivot, what is the magnitude of the force?
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A.
50 N
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B.
45 N
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C.
40 N
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D.
30 N
Solution
Force = Torque / Distance = 15 Nm / 0.3 m = 50 N.
Correct Answer: A — 50 N
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Q. If a torque of 25 Nm is applied and the lever arm is 5 m, what is the angle at which the force is applied if the force is 10 N?
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A.
0 degrees
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B.
30 degrees
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C.
60 degrees
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D.
90 degrees
Solution
Torque = Force × Distance × sin(θ) => 25 Nm = 10 N × 5 m × sin(θ) => sin(θ) = 0.5 => θ = 30 degrees.
Correct Answer: C — 60 degrees
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Q. If 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?
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A.
50 N
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B.
40 N
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C.
30 N
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D.
20 N
Solution
Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 50 N
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Q. If a torque of 25 Nm is generated by a force acting at a distance of 0.5 m, what is the force applied?
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A.
50 N
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B.
40 N
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C.
30 N
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D.
20 N
Solution
Force = Torque / Distance = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 50 N
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Q. If a torque of 30 Nm is applied to a lever arm of 3 m, what is the force applied?
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A.
5 N
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B.
10 N
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C.
15 N
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D.
20 N
Solution
Force = Torque / Distance = 30 Nm / 3 m = 10 N.
Correct Answer: B — 10 N
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Q. If a torque of 8 Nm is produced by a force acting at a distance of 0.2 m, what is the force?
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A.
20 N
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B.
30 N
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C.
40 N
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D.
50 N
Solution
Force = Torque / Distance = 8 Nm / 0.2 m = 40 N.
Correct Answer: A — 20 N
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Q. If the angle between the force and the lever arm is 90 degrees, how does it affect the torque?
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A.
Torque is zero
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B.
Torque is maximum
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C.
Torque is half
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D.
Torque is minimum
Solution
Torque is maximum when the angle is 90 degrees because τ = F × r × sin(θ) and sin(90°) = 1.
Correct Answer: B — Torque is maximum
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Q. If the angle between the force and the lever arm is 90 degrees, what is the torque produced by a 15 N force applied at a distance of 2 m?
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A.
0 Nm
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B.
15 Nm
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C.
30 Nm
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D.
45 Nm
Solution
Torque (τ) = F × d × sin(θ) = 15 N × 2 m × sin(90°) = 30 Nm.
Correct Answer: C — 30 Nm
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Q. If the lever arm is doubled while keeping the force constant, how does the torque change?
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A.
It doubles
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B.
It triples
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C.
It remains the same
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D.
It halves
Solution
Torque is directly proportional to the lever arm; if the lever arm is doubled, the torque also doubles.
Correct Answer: A — It doubles
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Q. If the torque is doubled while keeping the distance constant, what happens to the force applied?
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A.
It doubles
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B.
It halves
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C.
It remains the same
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D.
It quadruples
Solution
Torque = Force × Distance; if Torque is doubled and Distance is constant, Force must also double.
Correct Answer: A — It doubles
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Q. If the torque on an object is zero, what can be said about the forces acting on it?
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A.
The object is at rest.
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B.
The net force is zero.
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C.
The forces are balanced.
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D.
The forces are acting along the same line.
Solution
If the torque is zero, it means that the forces are acting along the same line, resulting in no rotational effect.
Correct Answer: D — The forces are acting along the same line.
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Q. If the torque on an object is zero, which of the following must be true?
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A.
The net force is zero.
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B.
The object is at rest.
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C.
The forces are balanced.
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D.
The line of action of the forces passes through the pivot.
Solution
If the torque is zero, it means that the line of action of the forces passes through the pivot point.
Correct Answer: D — The line of action of the forces passes through the pivot.
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Q. If two equal and opposite forces are applied at the ends of a lever arm of length 1 m, what is the net torque about the center?
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A.
0 Nm
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B.
1 Nm
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C.
2 Nm
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D.
4 Nm
Solution
The net torque is zero because the forces are equal and opposite, producing no rotational effect.
Correct Answer: A — 0 Nm
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Q. If two equal and opposite forces are applied at the ends of a lever arm of length 4 m, what is the net torque about the center?
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A.
0 Nm
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B.
8 Nm
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C.
4 Nm
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D.
16 Nm
Solution
The net torque is zero because the forces are equal and opposite, resulting in no rotational effect.
Correct Answer: A — 0 Nm
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Q. If two equal and opposite forces are applied at the ends of a lever arm, what is the net torque about the center?
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A.
Zero
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B.
Equal to the force
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C.
Depends on the distance
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D.
Twice the force
Solution
The net torque is zero because the forces are equal and opposite, producing no rotational effect.
Correct Answer: A — Zero
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Q. If two forces of 5 N and 10 N are applied at the same distance from a pivot, which force produces more torque?
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A.
5 N
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B.
10 N
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C.
Both produce the same torque
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D.
It depends on the angle of application
Solution
Torque is directly proportional to the force applied; hence, the 10 N force produces more torque.
Correct Answer: B — 10 N
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Q. If two forces of equal magnitude are applied at different distances from a pivot, which will produce a greater torque?
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A.
The force applied further away
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B.
The force applied closer
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C.
Both produce equal torque
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D.
It depends on the angle
Solution
Torque is greater when the force is applied further away from the pivot, given equal magnitudes.
Correct Answer: A — The force applied further away
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Q. If two forces of equal magnitude are applied at opposite ends of a lever arm, what is the net torque about the center?
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A.
Zero
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B.
Equal to the force
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C.
Twice the force
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D.
Depends on the distance
Solution
The net torque is zero because the torques produced by the two forces cancel each other out.
Correct Answer: A — Zero
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Q. In a seesaw, if one child exerts a force of 30 N at a distance of 2 m from the pivot, what is the torque exerted by that child?
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A.
15 Nm
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B.
30 Nm
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C.
60 Nm
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D.
0 Nm
Solution
Torque (τ) = F × r = 30 N × 2 m = 60 Nm.
Correct Answer: C — 60 Nm
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Q. In a seesaw, if one child exerts a torque of 30 N·m on one side, what torque must the other child exert to balance it?
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A.
15 N·m
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B.
30 N·m
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C.
45 N·m
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D.
60 N·m
Solution
To balance the seesaw, the other child must exert an equal torque of 30 N·m.
Correct Answer: B — 30 N·m
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Q. What is the torque about a pivot if a force of 8 N is applied perpendicular to a lever arm of 0.75 m?
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A.
4 Nm
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B.
6 Nm
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C.
8 Nm
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D.
10 Nm
Solution
Torque = Force × Distance = 8 N × 0.75 m = 6 Nm.
Correct Answer: C — 8 Nm
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Q. What is the torque produced by a 60 N force acting at a distance of 0.75 m from the pivot point?
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A.
45 Nm
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B.
60 Nm
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C.
75 Nm
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D.
90 Nm
Solution
Torque = Force × Distance = 60 N × 0.75 m = 45 Nm.
Correct Answer: A — 45 Nm
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Q. What is the torque produced by a 60 N force applied at a distance of 0.75 m from the pivot point?
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A.
45 Nm
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B.
60 Nm
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C.
75 Nm
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D.
90 Nm
Solution
Torque = Force × Distance = 60 N × 0.75 m = 45 Nm.
Correct Answer: A — 45 Nm
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