JEE Main Preparation Resources for Competitive Exams

My Learning Cart

JEE Main

Download Q&A

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. If a simple harmonic oscillator has a total energy E, what is the kinetic energy when the displacement is half of the amplitude?

A. E/4
B. E/2
C. 3E/4
D. E

Solution

The total energy E is conserved. When the displacement is half the amplitude, the potential energy is (1/2)E, so the kinetic energy is E - (1/2)E = (1/2)E.

Correct Answer: C — 3E/4

Q. If a soap film is formed on a wire frame, what is the effect of adding more soap to the film?

A. Surface tension increases
B. Surface tension decreases
C. Surface tension remains the same
D. Surface tension becomes zero

Solution

Adding more soap decreases the surface tension because soap molecules disrupt the cohesive forces between water molecules.

Correct Answer: B — Surface tension decreases

Q. If a solenoid has n turns per unit length and carries a current I, what is the magnetic field inside the solenoid?

A. 0
B. μ₀nI
C. μ₀I/n
D. μ₀I

Solution

The magnetic field inside a solenoid is given by B = μ₀nI.

Correct Answer: B — μ₀nI

Q. If a solid cylinder is rotated about its diameter, what is its moment of inertia?

A. 1/2 MR^2
B. 1/4 MR^2
C. 1/3 MR^2
D. MR^2

Solution

The moment of inertia of a solid cylinder about its diameter is I = 1/4 MR^2.

Correct Answer: B — 1/4 MR^2

Q. If a solid cylinder rolls without slipping, what fraction of its total kinetic energy is translational?

A. 1/3
B. 1/2
C. 2/3
D. 1

Solution

For a solid cylinder, the total kinetic energy is KE_total = KE_translational + KE_rotational = (1/2)mv^2 + (1/2)(1/2)mR^2(ω^2). Since ω = v/R, the translational part is 2/3 of the total.

Correct Answer: C — 2/3

Q. If a solid cylinder rolls without slipping, what is the ratio of its translational kinetic energy to its rotational kinetic energy?

A. 1:1
B. 2:1
C. 1:2
D. 3:1

Solution

For a solid cylinder, the translational kinetic energy (KE_trans) is (1/2)mv² and the rotational kinetic energy (KE_rot) is (1/2)(Iω²). The ratio KE_trans:KE_rot is 1:2.

Correct Answer: C — 1:2

Q. If a solid disk rolls without slipping, what fraction of its total energy is translational at the bottom of an incline?

A. 1/4
B. 1/3
C. 1/2
D. 2/3

Solution

For a solid disk, the translational kinetic energy is 1/3 of the total kinetic energy when rolling without slipping.

Correct Answer: B — 1/3

Q. If a solid sphere and a hollow sphere have the same mass and radius, which one will roll down an incline faster?

A. Solid sphere
B. Hollow sphere
C. Both will roll at the same speed
D. Depends on the angle of incline

Solution

The solid sphere will roll down the incline faster because it has a smaller moment of inertia compared to the hollow sphere.

Correct Answer: A — Solid sphere

Q. If a solid sphere and a solid cylinder of the same mass and radius are released from rest at the same height, which will have a greater speed at the bottom?

A. Solid sphere
B. Solid cylinder
C. Both have the same speed
D. Depends on the mass

Solution

Both will have the same speed at the bottom due to conservation of energy, as they start from the same height.

Correct Answer: C — Both have the same speed

Q. If a solid sphere of mass M and radius R is rotating about an axis through its center, what is its moment of inertia?

A. 2/5 MR^2
B. 3/5 MR^2
C. 1/2 MR^2
D. 1/3 MR^2

Solution

The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.

Correct Answer: A — 2/5 MR^2

Q. If a solid sphere of radius R and mass M is rotating about an axis through its center, what is its moment of inertia?

A. 2/5 MR^2
B. 3/5 MR^2
C. 1/2 MR^2
D. 1/3 MR^2

Solution

The moment of inertia of a solid sphere about its center is I = 2/5 MR^2.

Correct Answer: A — 2/5 MR^2

Q. If a solution has a vapor pressure of 80 mmHg and the pure solvent has a vapor pressure of 100 mmHg, what is the mole fraction of the solvent?

A. 0.8
B. 0.75
C. 0.6
D. 0.5

Solution

Using Raoult's Law, the mole fraction of the solvent can be calculated as 80 mmHg / 100 mmHg = 0.8, which corresponds to a mole fraction of 0.75 for the solvent.

Correct Answer: B — 0.75

Q. If a sound wave has a frequency of 440 Hz, what is its period?

A. 0.00227 s
B. 0.0045 s
C. 0.01 s
D. 0.1 s

Solution

The period (T) is the inverse of frequency (f); T = 1/f = 1/440 ≈ 0.00227 s.

Correct Answer: A — 0.00227 s

Q. If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string if the mass of the stone is 1 kg?

A. 2 N
B. 4 N
C. 8 N
D. 10 N

Solution

Centripetal force = m(v²/r) = 1(4²/2) = 8 N. Tension in the string = 8 N.

Correct Answer: C — 8 N

Q. If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string?

A. 2 N
B. 4 N
C. 8 N
D. 16 N

Solution

Centripetal force (F_c) = mv²/r. Assuming mass (m) = 1 kg, F_c = 1(4²)/2 = 8 N.

Correct Answer: C — 8 N

Q. If a student guesses on a multiple-choice test with 4 options for each question, what is the probability of guessing the correct answer?

A. 1/4
B. 1/2
C. 1/3
D. 1/5

Solution

Probability of guessing correctly = 1/4.

Correct Answer: A — 1/4

Q. If a student guesses on a multiple-choice test with 4 options per question, what is the probability of guessing the correct answer?

A. 1/4
B. 1/2
C. 1/3
D. 1/5

Solution

Probability of guessing correctly = 1/4.

Correct Answer: A — 1/4

Q. If a student is selected at random from a group of 20 students, 12 of whom are boys, what is the probability that the student is a boy?

A. 3/5
B. 2/5
C. 1/2
D. 1/3

Solution

Probability = Number of boys / Total students = 12/20 = 3/5.

Correct Answer: A — 3/5

Q. If a student is selected at random from a group of 20 students, 12 of whom are boys, what is the probability that the student is a girl?

A. 1/5
B. 2/5
C. 3/5
D. 1/2

Solution

Number of girls = 20 - 12 = 8. Probability = 8/20 = 2/5.

Correct Answer: C — 3/5

Q. If a student is selected at random from a group of 20 students, 8 of whom are girls, what is the probability that the selected student is a boy?

A. 1/2
B. 3/5
C. 2/5
D. 1/5

Solution

There are 20 - 8 = 12 boys in the group. The probability of selecting a boy is the number of boys divided by the total number of students, which is 12/20 = 3/5.

Correct Answer: B — 3/5

Q. If a student is selected at random from a group of 20 students, where 12 are girls and 8 are boys, what is the probability that the selected student is a boy?

A. 2/5
B. 1/2
C. 3/5
D. 1/4

Solution

The probability of selecting a boy is the number of boys divided by the total number of students, which is 8/20 = 2/5.

Correct Answer: B — 1/2

Q. If a student is selected at random from a group of 20 students, where 8 are girls, what is the probability that the selected student is a boy?

A. 3/5
B. 1/4
C. 1/2
D. 2/5

Solution

The number of boys is 20 - 8 = 12. The probability of selecting a boy is 12/20 = 3/5.

Correct Answer: A — 3/5

Q. If a system absorbs 100 J of heat and does 40 J of work, what is the change in internal energy?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

Using the first law of thermodynamics, ΔU = Q - W = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

Q. If a system absorbs 200 J of heat and does 50 J of work, what is the change in internal energy?

A. 150 J
B. 250 J
C. 200 J
D. 50 J

Solution

Using the First Law of Thermodynamics, ΔU = Q - W = 200 J - 50 J = 150 J.

Correct Answer: A — 150 J

Q. If a system does 150 J of work on the surroundings and absorbs 100 J of heat, what is the change in internal energy?

A. -50 J
B. 50 J
C. 100 J
D. 250 J

Solution

Using the first law, ΔU = Q - W = 100 J - 150 J = -50 J.

Correct Answer: A — -50 J

Q. If a temperature is recorded as 25°C with an error of ±0.5°C, what is the maximum possible temperature?

A. 25.5°C
B. 24.5°C
C. 26°C
D. 25°C

Solution

Maximum possible temperature = Measured value + Error = 25 + 0.5 = 25.5°C.

Correct Answer: A — 25.5°C

Q. If a temperature is recorded as 25°C with an error of ±0.5°C, what is the minimum temperature?

A. 24.5°C
B. 25°C
C. 25.5°C
D. 26°C

Solution

Minimum temperature = Recorded temperature - Error = 25 - 0.5 = 24.5°C.

Correct Answer: A — 24.5°C

Q. If a thermometer reads 100°C with an error of ±0.5°C, what is the maximum possible temperature?

A. 100.5°C
B. 99.5°C
C. 101°C
D. 100°C

Solution

Maximum possible temperature = Measured value + Error = 100 + 0.5 = 100.5°C.

Correct Answer: A — 100.5°C

Q. If a torque of 15 Nm is produced by a force acting at a distance of 0.3 m from the pivot, what is the magnitude of the force?

A. 50 N
B. 45 N
C. 40 N
D. 30 N

Solution

Force = Torque / Distance = 15 Nm / 0.3 m = 50 N.

Correct Answer: A — 50 N

Q. If a torque of 25 Nm is applied and the lever arm is 5 m, what is the angle at which the force is applied if the force is 10 N?

A. 0 degrees
B. 30 degrees
C. 60 degrees
D. 90 degrees

Solution

Torque = Force × Distance × sin(θ) => 25 Nm = 10 N × 5 m × sin(θ) => sin(θ) = 0.5 => θ = 30 degrees.

Correct Answer: C — 60 degrees

Showing 3151 to 3180 of 10700 (357 Pages)

School

College

Degree

Competitve

Skills

Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely