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 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?
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 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 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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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
8.66 Nm
D.
17.32 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer:
C
— 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
17.32 Nm
D.
34.64 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer:
C
— 17.32 Nm
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Q. A force of 25 N is applied at an angle of 30 degrees to the horizontal while moving an object 10 m. What is the work done?
A.
100 J
B.
125 J
C.
150 J
D.
175 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 25 N × 10 m × cos(30°) = 216.5 J.
Correct Answer:
B
— 125 J
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Q. A force of 25 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m. What is the work done?
A.
37.5 J
B.
50 J
C.
75 J
D.
100 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 25 N × 3 m × cos(60°) = 37.5 J.
Correct Answer:
A
— 37.5 J
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
A.
15 Nm
B.
30 Nm
C.
60 Nm
D.
52 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer:
D
— 52 Nm
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Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?
A.
3 m/s²
B.
6 m/s²
C.
9 m/s²
D.
12 m/s²
Show solution
Solution
Using F = ma, a = F/m = 30 N / 5 kg = 6 m/s².
Correct Answer:
B
— 6 m/s²
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
A.
10 Nm
B.
15 Nm
C.
20 Nm
D.
25 Nm
Show solution
Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer:
C
— 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
20 Nm
B.
40 Nm
C.
34.64 Nm
D.
69.28 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer:
C
— 34.64 Nm
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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
86.6 Nm
Show solution
Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 N·m
B.
50 N·m
C.
86.6 N·m
D.
100 N·m
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer:
C
— 86.6 N·m
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