JEE Main MCQ & Objective Questions
The JEE Main exam is a crucial step for students aspiring to enter prestigious engineering colleges in India. It tests not only knowledge but also the ability to apply concepts effectively. Practicing MCQs and objective questions is essential for scoring better, as it helps in familiarizing students with the exam pattern and enhances their problem-solving skills. Engaging with practice questions allows students to identify important questions and strengthen their exam preparation.
What You Will Practise Here
Fundamental concepts of Physics, Chemistry, and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theories relevant to JEE Main
Diagrams and graphical representations for better understanding
Numerical problems and their step-by-step solutions
Previous years' JEE Main questions for real exam experience
Time management strategies while solving MCQs
Exam Relevance
The topics covered in JEE Main are not only significant for the JEE exam but also appear in various CBSE and State Board examinations. Many concepts are shared with the NEET syllabus, making them relevant across multiple competitive exams. Common question patterns include conceptual applications, numerical problems, and theoretical questions that assess a student's understanding of core subjects.
Common Mistakes Students Make
Misinterpreting the question stem, leading to incorrect answers
Neglecting units in numerical problems, which can change the outcome
Overlooking negative marking and not managing time effectively
Relying too heavily on rote memorization instead of understanding concepts
Failing to review and analyze mistakes from practice tests
FAQs
Question: How can I improve my speed in solving JEE Main MCQ questions?Answer: Regular practice with timed quizzes and focusing on shortcuts can significantly enhance your speed.
Question: Are the JEE Main objective questions similar to previous years' papers?Answer: Yes, many questions are based on previous years' patterns, so practicing them can be beneficial.
Question: What is the best way to approach JEE Main practice questions?Answer: Start with understanding the concepts, then attempt practice questions, and finally review your answers to learn from mistakes.
Now is the time to take charge of your preparation! Dive into solving JEE Main MCQs and practice questions to test your understanding and boost your confidence for the exam.
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? (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 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 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 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 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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