Q. A 3 kg object is subjected to a net force of 12 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 4 kg object is at rest on a table. What is the force of static friction acting on it?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
If the object is at rest and no external force is applied, the static friction force is 0 N.
Correct Answer:
A
— 0 N
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Q. A 4 kg object is at rest on a table. What is the force of static friction if no external force is applied?
A.
0 N
B.
4 N
C.
40 N
D.
None of the above
Show solution
Solution
If no external force is applied, the force of static friction is 0 N.
Correct Answer:
A
— 0 N
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Q. A 4 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
The force acting on the object is its weight, F = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer:
C
— 40 N
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Q. A 4 kg object is hanging from a rope. What is the tension in the rope if the object is at rest?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
Show solution
Solution
The tension in the rope equals the weight of the object: T = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer:
C
— 40 N
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Q. A 4 kg object is lifted to a height of 3 m. What is the change in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Change in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Increase in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy? (g = 9.8 m/s²)
A.
117.6 J
B.
117 J
C.
120 J
D.
150 J
Show solution
Solution
Potential energy increase = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on the object?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
Show solution
Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 4 * 36 / 2 = 72 N / 2 = 36 N.
Correct Answer:
B
— 12 N
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on it?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
Show solution
Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 72 N / 2 = 36 N.
Correct Answer:
B
— 12 N
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Q. A 4 kg object is moving in a straight line with a velocity of 5 m/s. What is the kinetic energy of the object?
A.
50 J
B.
40 J
C.
20 J
D.
10 J
Show solution
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (5 m/s)² = 50 J.
Correct Answer:
A
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. If it comes to rest, what is the work done by friction?
A.
50 J
B.
75 J
C.
100 J
D.
125 J
Show solution
Solution
Work done = change in kinetic energy = 0 - 0.5 × 4 kg × (5 m/s)² = -50 J.
Correct Answer:
C
— 100 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is its kinetic energy?
A.
10 J
B.
20 J
C.
50 J
D.
100 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 50 J.
Correct Answer:
C
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is the total mechanical energy if it is at a height of 2 m?
A.
50 J
B.
60 J
C.
70 J
D.
80 J
Show solution
Solution
Total mechanical energy = Kinetic energy + Potential energy = 0.5 × 4 kg × (5 m/s)² + 4 kg × 9.8 m/s² × 2 m = 50 J + 78.4 J = 128.4 J.
Correct Answer:
C
— 70 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is its kinetic energy?
A.
8 J
B.
4 J
C.
16 J
D.
2 J
Show solution
Solution
Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 4 * (2^2) = 8 J
Correct Answer:
A
— 8 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is the change in momentum if the object comes to a stop?
A.
0 kg·m/s
B.
4 kg·m/s
C.
8 kg·m/s
D.
2 kg·m/s
Show solution
Solution
Change in momentum = final momentum - initial momentum = 0 - (4 kg * 2 m/s) = -8 kg·m/s.
Correct Answer:
C
— 8 kg·m/s
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Q. A 4 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
18 J
B.
24 J
C.
36 J
D.
12 J
Show solution
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (3 m/s)² = 18 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
10 kg·m/s
B.
20 kg·m/s
C.
15 kg·m/s
D.
25 kg·m/s
Show solution
Solution
Momentum p = mv = 4 kg * 5 m/s = 20 kg·m/s.
Correct Answer:
B
— 20 kg·m/s
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Q. A 4 kg object is pulled with a force of 20 N. If the frictional force is 4 N, what is the net force acting on the object?
A.
16 N
B.
20 N
C.
24 N
D.
4 N
Show solution
Solution
Net force = applied force - frictional force = 20 N - 4 N = 16 N.
Correct Answer:
A
— 16 N
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Q. A 4 kg object is pushed with a force of 16 N. What is the acceleration of the object?
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 = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. Kinetic energy = 0.5 × m × v². 60 J = 0.5 × 4 kg × v². v² = 30, v = √30 ≈ 5.48 m/s.
Correct Answer:
C
— 4 m/s
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. K.E = 0.5 × m × v², 60 J = 0.5 × 4 kg × v², v = 3.87 m/s.
Correct Answer:
C
— 4 m/s
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Q. A 4 kg object is pushed with a force of 20 N. If the frictional force is 4 N, what is the acceleration of the object?
A.
4 m/s²
B.
5 m/s²
C.
6 m/s²
D.
7 m/s²
Show solution
Solution
Net force = applied force - friction = 20 N - 4 N = 16 N. Acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
B
— 5 m/s²
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Q. A 4 kg object is subjected to a net force of 12 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, a = F/m = 12 N / 4 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. A 4 kg object is subjected to a net force of 16 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. A 5 kg block is placed on a surface with a coefficient of static friction of 0.6. What is the maximum static friction force before the block starts to move?
A.
30 N
B.
20 N
C.
15 N
D.
25 N
Show solution
Solution
Maximum static friction force = μs * N = 0.6 * (5 kg * 9.8 m/s²) = 29.4 N, approximately 30 N.
Correct Answer:
A
— 30 N
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Q. A 5 kg block is resting on a frictionless surface. A force of 10 N is applied to it. What is the acceleration of the block?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Show solution
Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer:
B
— 2 m/s²
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Q. A 5 kg block is resting on a frictionless surface. If a force of 10 N is applied to it, what will be its acceleration?
A.
1 m/s²
B.
2 m/s²
C.
5 m/s²
D.
10 m/s²
Show solution
Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer:
B
— 2 m/s²
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Q. A 5 kg block is resting on a horizontal surface. What is the normal force acting on the block?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
Show solution
Solution
The normal force equals the weight of the block, which is mass times gravity: 5 kg * 9.8 m/s² = 49 N.
Correct Answer:
B
— 10 N
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Q. A 5 kg object is at rest on a surface with a coefficient of static friction of 0.4. What is the maximum static friction force?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
Show solution
Solution
Maximum static friction force = μs * N = 0.4 * (5 kg * 9.8 m/s²) = 0.4 * 49 N = 19.6 N, approximately 20 N.
Correct Answer:
B
— 20 N
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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!
