Physics Syllabus (JEE Main)
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 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 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 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 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 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 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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Q. A force of 10 N is applied to a 2 kg object at rest. What is the final velocity of the object after 5 seconds?
A.
2 m/s
B.
5 m/s
C.
10 m/s
D.
15 m/s
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Solution
Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s². Final velocity v = u + at = 0 + 5 * 5 = 25 m/s.
Correct Answer: B — 5 m/s
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Q. A force of 10 N is applied to a 2 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 4 N?
A.
6 N
B.
10 N
C.
14 N
D.
4 N
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Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
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Q. A force of 10 N is applied to a 2 kg object. What is the object's acceleration?
A.
5 m/s²
B.
10 m/s²
C.
2 m/s²
D.
20 m/s²
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Solution
Using F = ma, acceleration a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: A — 5 m/s²
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Q. A force of 10 N is applied to a 2 kg object. What is the resulting acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, we have a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: A — 2 m/s²
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Q. A force of 15 N is applied at an angle of 30° to the horizontal while moving an object 4 m. What is the work done by the force?
A.
30 J
B.
60 J
C.
90 J
D.
120 J
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Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(30°) = 15 N × 4 m × (√3/2) = 60 J.
Correct Answer: C — 90 J
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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?
A.
7.5 Nm
B.
12.5 Nm
C.
15 Nm
D.
25 Nm
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Solution
Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.
Correct Answer: B — 12.5 Nm
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Q. A force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m. What is the work done?
A.
30 J
B.
60 J
C.
45 J
D.
75 J
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Solution
Work Done = F * d * cos(θ) = 15 N * 4 m * cos(60°) = 30 J
Correct Answer: C — 45 J
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Q. A force of 15 N is applied to a 3 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 5 N?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
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Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer: A — 10 N
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Q. A force of 15 N is applied to a 3 kg object. What is the object's acceleration?
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
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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 15 N is applied to a 3 kg object. What is the resulting acceleration?
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
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Solution
Using F = ma, we find a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. A force of 15 N is applied to a 5 kg object. What is the object's acceleration?
A.
3 m/s²
B.
2 m/s²
C.
1 m/s²
D.
4 m/s²
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Solution
Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer: A — 3 m/s²
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Q. A force of 15 N is applied to move an object 3 m at an angle of 60 degrees to the horizontal. What is the work done by the force?
A.
15 J
B.
20 J
C.
30 J
D.
45 J
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Solution
Work done = F × d × cos(θ) = 15 N × 3 m × cos(60°) = 15 N × 3 m × 0.5 = 22.5 J.
Correct Answer: C — 30 J
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Q. A force of 15 N is applied to move an object 4 m in the direction of the force. What is the work done?
A.
30 J
B.
60 J
C.
75 J
D.
90 J
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Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer: D — 90 J
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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 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(30°) = 20 N × 1 m × 0.5 = 10 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 20 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.
20 Nm
C.
30 Nm
D.
40 Nm
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Solution
Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.
Correct Answer: B — 20 Nm
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