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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 projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 20 m/s
B. 28.28 m/s
C. 30 m/s
D. 40 m/s

Solution

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Correct Answer: B — 28.28 m/s

Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the time of flight?

A. 4 s
B. 5 s
C. 6 s
D. 7 s

Solution

Time of flight (T) = (2u * sin(θ)) / g = (2*40*sin(45))/9.8 = 5.14 s.

Correct Answer: B — 5 s

Q. A projectile is thrown with a speed of 20 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 10 m/s
B. 14.14 m/s
C. 20 m/s
D. 28.28 m/s

Solution

Vertical component (u_y) = u * sin(θ) = 20 * (√2/2) = 14.14 m/s.

Correct Answer: B — 14.14 m/s

Q. A proton moves in a magnetic field and experiences a force. If the velocity of the proton is doubled, what happens to the magnetic force?

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

Solution

The magnetic force is proportional to the velocity of the charge, so if the velocity is doubled, the magnetic force also doubles.

Correct Answer: A — It doubles

Q. A ray of light in a medium with n=1.33 strikes the boundary with air at an angle of 45°. What is the behavior of the ray?

A. Total internal reflection
B. Partial reflection and refraction
C. Complete absorption
D. Total refraction

Solution

Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.

Correct Answer: B — Partial reflection and refraction

Q. A ray of light in glass (n=1.5) strikes the glass-air interface at an angle of 30°. What will be the angle of refraction in air?

A. 60°
B. 30°
C. 45°
D. 90°

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.

Correct Answer: A — 60°

Q. A ray of light passes from air into glass at an angle of incidence of 30 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?

A. 20 degrees
B. 30 degrees
C. 18.4 degrees
D. 22 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), i = 30 degrees, n2 = 1.5 (glass). Thus, sin(r) = (1 * sin(30))/1.5 = 0.333, giving r ≈ 18.4 degrees.

Correct Answer: C — 18.4 degrees

Q. A ray of light passes from air into glass at an angle of incidence of 45 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 45 degrees. Solving gives r = 30 degrees.

Correct Answer: A — 30 degrees

Q. A ray of light passes from air into glass with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?

A. 20 degrees
B. 30 degrees
C. 18.4 degrees
D. 22 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1, n2 = 1.5, i = 30 degrees. Solving gives r ≈ 18.4 degrees.

Correct Answer: C — 18.4 degrees

Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction?

A. 22 degrees
B. 30 degrees
C. 45 degrees
D. 18 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water), θ1 = 30 degrees. Thus, sin(θ2) = (1 * sin(30))/1.33 = 0.375. Therefore, θ2 = sin^(-1)(0.375) which is approximately 22 degrees.

Correct Answer: A — 22 degrees

Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction in water (n = 1.33)?

A. 22.5 degrees
B. 30 degrees
C. 40 degrees
D. 45 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so 1 * 0.5 = 1.33 * sin(θ2). Solving gives θ2 ≈ 22.5 degrees.

Correct Answer: A — 22.5 degrees

Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction in water? (Refractive index of water = 1.33)

A. 22.5 degrees
B. 30 degrees
C. 40 degrees
D. 20 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2), we find θ2 = sin^(-1)(sin(30 degrees)/1.33) = 22.5 degrees.

Correct Answer: A — 22.5 degrees

Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction?

A. 22 degrees
B. 30 degrees
C. 45 degrees
D. 18 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water). Solving gives θ2 ≈ 22 degrees.

Correct Answer: A — 22 degrees

Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water? (Refractive index of water = 1.33)

A. 22.1 degrees
B. 30 degrees
C. 36.9 degrees
D. 45 degrees

Solution

Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.

Correct Answer: A — 22.1 degrees

Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water (n = 1.33)?

A. 22.5 degrees
B. 30 degrees
C. 22 degrees
D. 40 degrees

Solution

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so sin(θ2) = (1 * 0.5) / 1.33 = 0.375. Thus, θ2 = sin^(-1)(0.375) ≈ 22.5 degrees.

Correct Answer: A — 22.5 degrees

Q. A ray of light passes through the center of curvature of a concave mirror. What will be the angle of reflection?

A. 0 degrees
B. 30 degrees
C. 45 degrees
D. 90 degrees

Solution

When a ray of light passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.

Correct Answer: A — 0 degrees

Q. A ray of light passes through the center of curvature of a concave mirror. What is the angle of reflection?

A. 0 degrees
B. 30 degrees
C. 45 degrees
D. 90 degrees

Solution

When a ray passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.

Correct Answer: A — 0 degrees

Q. A ray of light passes through the optical center of a lens. What happens to the ray?

A. It bends towards the normal
B. It bends away from the normal
C. It continues in a straight line
D. It reflects back

Solution

A ray of light passing through the optical center of a lens continues in a straight line without bending.

Correct Answer: C — It continues in a straight line

Q. A ray of light strikes a plane mirror at an angle of 30 degrees. What is the angle of reflection?

A. 30 degrees
B. 60 degrees
C. 90 degrees
D. 45 degrees

Solution

According to the law of reflection, the angle of reflection is equal to the angle of incidence, which is 30 degrees.

Correct Answer: A — 30 degrees

Q. A ray of light strikes a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection?

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is 30 degrees.

Correct Answer: A — 30 degrees

Q. A ray of light traveling in air strikes the surface of water at an angle of incidence of 45 degrees. What is the angle of refraction?

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). For air (n1=1) and water (n2=1.33), we find r = 45 degrees.

Correct Answer: B — 45 degrees

Q. A rectangular plate of mass M and dimensions a x b is rotated about an axis along one of its edges. What is its moment of inertia?

A. 1/3 Ma^2
B. 1/12 Ma^2 + 1/3 Mb^2
C. 1/2 Ma^2
D. 1/4 Ma^2 + 1/3 Mb^2

Solution

The moment of inertia of a rectangular plate about an edge is I = 1/12 Ma^2 + 1/3 Mb^2.

Correct Answer: B — 1/12 Ma^2 + 1/3 Mb^2

Q. A resistor of 10 ohms is connected to a 20V battery. What is the current flowing through the resistor?

A. 0.5 A
B. 1 A
C. 2 A
D. 5 A

Solution

Using Ohm's Law, I = V/R = 20V / 10Ω = 2 A.

Correct Answer: C — 2 A

Q. A rigid body is rotating about a fixed axis with an angular velocity ω. If its moment of inertia is I, what is its angular momentum?

A. Iω
B. ω/I
C. I/ω
D. Iω^2

Solution

Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.

Correct Answer: A — Iω

Q. A rigid body is rotating about a fixed axis with an angular velocity ω. If the moment of inertia of the body is I, what is the angular momentum of the body?

A. Iω
B. ω/I
C. I/ω
D. Iω^2

Solution

Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.

Correct Answer: A — Iω

Q. A rigid body is rotating about a fixed axis with an angular velocity ω. If the moment of inertia of the body is I, what is its angular momentum?

A. Iω
B. ω/I
C. I/ω
D. Iω^2

Solution

Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.

Correct Answer: A — Iω

Q. A rigid body is rotating about a fixed axis. If the moment of inertia of the body is I and it is rotating with an angular velocity ω, what is its angular momentum?

A. Iω
B. I/ω
C. Iω^2
D. ω/I

Solution

Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.

Correct Answer: A — Iω

Q. A rigid body rotates about a fixed axis with an angular velocity ω. If the moment of inertia of the body is I, what is the angular momentum of the body?

A. Iω
B. ω/I
C. I/ω
D. Iω^2

Solution

Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.

Correct Answer: A — Iω

Q. A roller coaster at the top of a hill has a height of 30 m. If it has a mass of 500 kg, what is its potential energy at the top?

A. 1500 J
B. 3000 J
C. 7500 J
D. 15000 J

Solution

Potential energy = mgh = 500 * 9.8 * 30 = 147000 J.

Correct Answer: D — 15000 J

Q. A roller coaster at the top of a hill has a height of 30 m. If it has a mass of 500 kg, what is its potential energy? (g = 9.8 m/s²)

A. 14700 J
B. 9800 J
C. 15000 J
D. 5000 J

Solution

Potential energy is given by PE = mgh. Here, PE = 500 * 9.8 * 30 = 14700 J.

Correct Answer: A — 14700 J

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