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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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?
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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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Physics Syllabus (JEE Main) MCQ & Objective Questions
The Physics Syllabus for JEE Main is crucial for students aiming to excel in their exams. Understanding this syllabus not only helps in grasping fundamental concepts but also enhances problem-solving skills through practice. Engaging with MCQs and objective questions is essential for effective exam preparation, as it allows students to identify important questions and strengthen their knowledge base.
What You Will Practise Here
Mechanics: Laws of Motion, Work, Energy, and Power
Thermodynamics: Laws of Thermodynamics, Heat Transfer
Waves and Oscillations: Simple Harmonic Motion, Wave Properties
Electromagnetism: Electric Fields, Magnetic Fields, and Circuits
Optics: Reflection, Refraction, and Optical Instruments
Modern Physics: Quantum Theory, Atomic Models, and Nuclear Physics
Fluid Mechanics: Properties of Fluids, Bernoulli's Principle
Exam Relevance
The Physics Syllabus (JEE Main) is integral to various examinations, including CBSE, State Boards, and competitive exams like NEET and JEE. Questions often focus on conceptual understanding and application of theories. Common patterns include numerical problems, conceptual MCQs, and assertion-reason type questions, which test both knowledge and analytical skills.
Common Mistakes Students Make
Misinterpreting the question stem, leading to incorrect answers.
Neglecting units and dimensions in calculations.
Overlooking the significance of diagrams in understanding concepts.
Confusing similar concepts, such as velocity and acceleration.
Failing to apply formulas correctly in different contexts.
FAQs
Question: What are the key topics in the Physics Syllabus for JEE Main?Answer: Key topics include Mechanics, Thermodynamics, Waves, Electromagnetism, Optics, Modern Physics, and Fluid Mechanics.
Question: How can I improve my performance in Physics MCQs?Answer: Regular practice of MCQs, understanding concepts deeply, and revising important formulas can significantly enhance your performance.
Start solving practice MCQs today to test your understanding of the Physics Syllabus (JEE Main). This will not only boost your confidence but also prepare you effectively for your upcoming exams. Remember, consistent practice is the key to success!
