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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.

Chemistry Syllabus (JEE Main) Mathematics Syllabus (JEE Main) Physics Syllabus (JEE Main)

Q. A particle moves in a circular path with a radius of 4 m and completes one revolution in 8 seconds. What is the centripetal acceleration?

A. 0.5 m/s²
B. 1 m/s²
C. 2 m/s²
D. 4 m/s²

Solution

Centripetal acceleration (a_c) = v²/r. First find v = 2πr/T = 2π(4)/8 = π m/s. Then a_c = (π)²/4 = 2.5 m/s².

Correct Answer: B — 1 m/s²

Q. A particle moves in a circular path with a radius r and a constant speed v. If the speed is doubled, what happens to the angular momentum of the particle?

A. It remains the same
B. It doubles
C. It quadruples
D. It halves

Solution

Angular momentum L = mvr; if v is doubled, L also doubles.

Correct Answer: B — It doubles

Q. A particle moves in a straight line under the influence of a constant force of 12 N. If the particle moves 3 m, what is the work done by the force?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

Work done = Force × Distance = 12 N × 3 m = 36 J.

Correct Answer: B — 24 J

Q. A particle moves in a straight line under the influence of a constant force of 3 N. If it moves 6 m, what is the work done by the force?

A. 9 J
B. 12 J
C. 15 J
D. 18 J

Solution

Work done = Force × Distance = 3 N × 6 m = 18 J.

Correct Answer: D — 18 J

Q. A particle moves in a straight line under the influence of a constant force. If the initial kinetic energy is 100 J and the work done by the force is 50 J, what is the final kinetic energy?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Final kinetic energy = Initial kinetic energy + Work done = 100 J + 50 J = 150 J.

Correct Answer: C — 150 J

Q. A particle moves in a straight line with a constant velocity. What is its angular momentum about a point not on the line of motion?

A. Zero
B. Constant
C. Varies with time
D. Depends on distance

Solution

Angular momentum is constant as the particle moves with constant velocity.

Correct Answer: B — Constant

Q. A particle moves in a straight line with a constant velocity. What is the angular momentum of the particle about a point not on the line of motion?

A. Zero
B. Depends on the distance from the point
C. Infinite
D. Constant

Solution

The angular momentum is non-zero and depends on the distance from the point to the line of motion.

Correct Answer: B — Depends on the distance from the point

Q. A particle moves in a straight line with a velocity v. What is its angular momentum about a point P located at a distance d from the line of motion?

A. mv
B. mvd
C. mdv
D. 0

Solution

Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point P.

Correct Answer: B — mvd

Q. A particle moves in a straight line with an acceleration of 2 m/s². If its initial velocity is 3 m/s, what will be its velocity after 5 seconds?

A. 10 m/s
B. 13 m/s
C. 15 m/s
D. 20 m/s

Solution

Final velocity (v) = u + at = 3 + (2 * 5) = 13 m/s.

Correct Answer: B — 13 m/s

Q. A particle moves in a straight line with an acceleration of 4 m/s². If its initial velocity is 2 m/s, what will be its velocity after 5 seconds?

A. 22 m/s
B. 20 m/s
C. 18 m/s
D. 16 m/s

Solution

Using the formula: final velocity = initial velocity + acceleration * time. Final velocity = 2 + 4 * 5 = 22 m/s.

Correct Answer: A — 22 m/s

Q. A particle moves in a straight line with an initial velocity of 10 m/s and accelerates at 2 m/s². What will be its velocity after 5 seconds?

A. 20 m/s
B. 30 m/s
C. 40 m/s
D. 50 m/s

Solution

Final velocity = initial velocity + (acceleration * time) = 10 m/s + (2 m/s² * 5 s) = 20 m/s.

Correct Answer: B — 30 m/s

Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle?

A. mr
B. mr^2
C. m/r
D. m/r^2

Solution

The moment of inertia of a point mass is given by I = mr^2.

Correct Answer: B — mr^2

Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle about the axis?

A. mr
B. mr^2
C. m/r
D. m/r^2

Solution

The moment of inertia of a point mass about an axis is given by I = mr^2.

Correct Answer: B — mr^2

Q. A particle of mass m is moving in a circular path of radius r with a constant speed v. What is the angular momentum of the particle about the center of the circle?

A. mv
B. mvr
C. mr^2
D. mv^2

Solution

Angular momentum L = mvr, where v is the linear speed and r is the radius.

Correct Answer: B — mvr

Q. A particle with charge q moves with velocity v in a magnetic field B. What is the expression for the magnetic force acting on the particle?

A. F = qvB
B. F = qvB sin(θ)
C. F = qB
D. F = qvB cos(θ)

Solution

The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvB sin(θ), where θ is the angle between the velocity vector and the magnetic field vector.

Correct Answer: B — F = qvB sin(θ)

Q. A pendulum is measured to have a length of 2.0 m with an uncertainty of ±0.1 m. What is the relative uncertainty in the length?

A. 5%
B. 10%
C. 2.5%
D. 1%

Solution

Relative uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.

Correct Answer: C — 2.5%

Q. A pendulum of length 2 m swings from a height of 1 m. What is the speed at the lowest point of the swing? (g = 9.8 m/s²)

A. 4.4 m/s
B. 3.1 m/s
C. 2.8 m/s
D. 5.0 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.4 m/s.

Correct Answer: A — 4.4 m/s

Q. A pendulum swings back and forth with a period of 1 second. If the length of the pendulum is doubled, what will be the new period?

A. 1 s
B. 1.41 s
C. 2 s
D. 4 s

Solution

The period of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T ≈ 1.41 s.

Correct Answer: B — 1.41 s

Q. A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?

A. 2 m/s
B. 4 m/s
C. 6 m/s
D. 8 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.

Correct Answer: B — 4 m/s

Q. A pendulum swings from a height of 5 m. What is the speed at the lowest point of the swing?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.

Correct Answer: B — 10 m/s

Q. A pendulum swings with a maximum angle of 30 degrees. What is the approximate period of the pendulum if its length is 1 m?

A. 1.0 s
B. 1.5 s
C. 2.0 s
D. 2.5 s

Solution

The period T of a simple pendulum is given by T = 2π√(L/g). Here, L = 1 m and g = 9.8 m/s². Thus, T = 2π√(1/9.8) ≈ 2.0 s.

Correct Answer: C — 2.0 s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is doubled, what will be the new period?

A. 1 s
B. 1.41 s
C. 2 s
D. 4 s

Solution

The period T of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T = 1.41 s.

Correct Answer: B — 1.41 s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased to four times its original length, what will be the new period?

A. 1 s
B. 2 s
C. 4 s
D. √4 s

Solution

The period of a pendulum is given by T = 2π√(L/g). If L is increased to 4L, T becomes 2π√(4L/g) = 2T = 2 seconds.

Correct Answer: B — 2 s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased by a factor of 4, what will be the new period?

A. 1 s
B. 2 s
C. 4 s
D. √4 s

Solution

The period T = 2π√(L/g). If L is increased by a factor of 4, T increases by a factor of √4 = 2.

Correct Answer: B — 2 s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is tripled, what will be the new period?

A. 1 s
B. 2 s
C. 3 s
D. √3 s

Solution

The period T = 2π√(L/g). If L is tripled, T becomes √3 times longer, so T = 2 s.

Correct Answer: B — 2 s

Q. A pendulum swings with a period of 1 second. What is the length of the pendulum?

A. 0.25 m
B. 0.5 m
C. 1 m
D. 2 m

Solution

Length L = (gT^2)/(4π^2) = (9.8 * 1^2)/(4 * π^2) ≈ 1 m.

Correct Answer: C — 1 m

Q. A pendulum swings with a period of 1.5 seconds. What is the angular frequency of the pendulum?

A. 2π/1.5 rad/s
B. 4π/3 rad/s
C. π/1.5 rad/s
D. 3π/2 rad/s

Solution

Angular frequency (ω) = 2π/T = 2π/1.5 = 4π/3 rad/s.

Correct Answer: B — 4π/3 rad/s

Q. A pendulum swings with a period of 2 seconds. What is the frequency of the pendulum?

A. 0.25 Hz
B. 0.5 Hz
C. 1 Hz
D. 2 Hz

Solution

Frequency (f) is the reciprocal of the period (T). Therefore, f = 1/T = 1/2 = 0.5 Hz.

Correct Answer: B — 0.5 Hz

Q. A pendulum swings with a period of 2 seconds. What is the length of the pendulum?

A. 0.5 m
B. 1 m
C. 2 m
D. 4 m

Solution

The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T²g)/(4π²). Using g = 9.8 m/s², L = (2² * 9.8)/(4π²) ≈ 1 m.

Correct Answer: B — 1 m

Q. A pendulum swings with a period T. What is the period of a pendulum of length 4L?

A. 2T
B. T/2
C. T√2
D. 2√2T

Solution

The period T of a simple pendulum is given by T = 2π√(L/g). For length 4L, T' = 2π√(4L/g) = 2T.

Correct Answer: A — 2T

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