JEE Main Preparation Resources for Competitive Exams

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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. If the quadratic equation x^2 + 2px + p^2 - 4 = 0 has roots that are equal, what is the value of p?

A. 2
B. 0
C. -2
D. -4

Solution

Setting the discriminant to zero: (2p)^2 - 4(1)(p^2 - 4) = 0 leads to p = ±2.

Correct Answer: C — -2

Q. If the quadratic equation x^2 + 2x + k = 0 has equal roots, what is the value of k?

A. 1
B. 0
C. -1
D. -2

Solution

For equal roots, the discriminant must be zero: 2^2 - 4*1*k = 0, leading to k = 1.

Correct Answer: C — -1

Q. If the quadratic equation x^2 + 2x + k = 0 has no real roots, what is the condition for k?

A. k < 0
B. k > 0
C. k >= 0
D. k <= 0

Solution

For no real roots, the discriminant must be less than zero: 2^2 - 4*1*k < 0 => 4 - 4k < 0 => k > 1.

Correct Answer: A — k < 0

Q. If the quadratic equation x^2 + 2x + k = 0 has no real roots, what is the condition on k?

A. k < 0
B. k > 0
C. k >= 0
D. k <= 0

Solution

For no real roots, the discriminant must be less than zero: 2^2 - 4*1*k < 0, hence k > 1.

Correct Answer: A — k < 0

Q. If the quadratic equation x^2 + 2x + k = 0 has roots that are equal, what is the value of k?

A. 1
B. 0
C. -1
D. -2

Solution

For equal roots, the discriminant must be zero: 2^2 - 4*1*k = 0 leads to k = -1.

Correct Answer: D — -2

Q. If the quadratic equation x^2 + 4x + c = 0 has one root equal to -2, what is the value of c?

A. 0
B. 2
C. 4
D. 6

Solution

If one root is -2, then substituting x = -2 gives: (-2)^2 + 4(-2) + c = 0 => 4 - 8 + c = 0 => c = 4.

Correct Answer: A — 0

Q. If the quadratic equation x^2 + 4x + k = 0 has roots -2 and -2, what is the value of k?

A. 0
B. 4
C. 8
D. 16

Solution

Using the formula for roots, k = (-2)^2 - 4*(-2) = 4 + 8 = 12.

Correct Answer: B — 4

Q. If the quadratic equation x^2 + 6x + 9 = 0 is solved, what is the nature of the roots?

A. Real and distinct
B. Real and equal
C. Complex
D. None of the above

Solution

The discriminant is 0, indicating that the roots are real and equal.

Correct Answer: B — Real and equal

Q. If the quadratic equation x^2 + 6x + k = 0 has roots -2 and -4, what is the value of k?

A. 8
B. 12
C. 16
D. 20

Solution

Using Vieta's formulas, k = (-2)(-4) = 8.

Correct Answer: B — 12

Q. If the quadratic equation x^2 + 6x + k = 0 has roots that are both negative, what is the condition for k?

A. k > 9
B. k < 9
C. k = 9
D. k < 0

Solution

For both roots to be negative, k must be greater than the square of half the coefficient of x, hence k > 9.

Correct Answer: A — k > 9

Q. If the quadratic equation x^2 + bx + 9 = 0 has roots 3 and -3, what is the value of b?

A. 0
B. 6
C. -6
D. 9

Solution

The sum of the roots is 3 + (-3) = 0, so b = -0.

Correct Answer: C — -6

Q. If the quadratic equation x^2 + kx + 16 = 0 has equal roots, what is the value of k?

A. -8
B. -4
C. 4
D. 8

Solution

For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0, thus k = -8.

Correct Answer: A — -8

Q. If the quadratic equation x^2 + kx + 9 = 0 has no real roots, what is the condition on k?

A. k < 6
B. k > 6
C. k < 0
D. k > 0

Solution

The discriminant must be less than zero: k^2 - 4*1*9 < 0 => k^2 < 36 => |k| < 6.

Correct Answer: B — k > 6

Q. If the quadratic equation x^2 + mx + n = 0 has roots 1 and -3, what is the value of m?

A. 2
B. -2
C. 4
D. -4

Solution

Using Vieta's formulas, m = -(1 + (-3)) = 2.

Correct Answer: A — 2

Q. If the quadratic equation x^2 + mx + n = 0 has roots 1 and -3, what is the value of n?

A. -3
B. 2
C. 3
D. 4

Solution

Using Vieta's formulas, the product of the roots is n = 1 * (-3) = -3.

Correct Answer: A — -3

Q. If the quadratic equation x^2 + mx + n = 0 has roots 2 and -3, what is the value of m + n?

A. -1
B. 5
C. 1
D. 3

Solution

Using Vieta's formulas, m = -(-1) = 1 and n = 2*(-3) = -6, thus m + n = 1 - 6 = -5.

Correct Answer: B — 5

Q. If the quadratic equation x^2 + px + q = 0 has roots 2 and 3, what is the value of p?

A. -5
B. -6
C. -7
D. -8

Solution

The sum of the roots is -p = 2 + 3 = 5, so p = -5.

Correct Answer: A — -5

Q. If the quadratic equation x^2 + px + q = 0 has roots 2 and 3, what is the value of p + q?

A. 5
B. 6
C. 7
D. 8

Solution

Using Vieta's formulas, p = -(2 + 3) = -5 and q = 2*3 = 6. Thus, p + q = -5 + 6 = 1.

Correct Answer: C — 7

Q. If the quadratic equation x^2 - kx + 9 = 0 has equal roots, what is the value of k?

A. 6
B. 9
C. 3
D. 0

Solution

For equal roots, the discriminant must be zero: k^2 - 36 = 0, hence k = 6.

Correct Answer: A — 6

Q. If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?

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

Solution

The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.

Correct Answer: B — It doubles

Q. If the radius of a circular loop carrying current is doubled, how does the magnetic field at the center change?

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

Solution

The magnetic field at the center of a circular loop is inversely proportional to the radius; thus, doubling the radius halves the magnetic field.

Correct Answer: B — It halves

Q. If the radius of a circular loop carrying current is doubled, what happens to the magnetic field at the center of the loop?

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

Solution

The magnetic field at the center of a circular loop is given by B = (μ₀I)/(2r). If the radius is doubled, the magnetic field strength is halved.

Correct Answer: B — It halves

Q. If the radius of a circular loop carrying current is halved, how does the magnetic field at the center change?

A. Remains the same
B. Doubles
C. Halves
D. Quadruples

Solution

The magnetic field at the center is inversely proportional to the radius, so it quadruples.

Correct Answer: D — Quadruples

Q. If the radius of a disc is doubled while keeping its mass constant, how does its moment of inertia change?

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

Solution

The moment of inertia of a disc is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is quadrupled.

Correct Answer: C — It quadruples

Q. If the radius of a disk is doubled while keeping its mass constant, how does its moment of inertia change?

A. Increases by a factor of 2
B. Increases by a factor of 4
C. Remains the same
D. Decreases by a factor of 4

Solution

The moment of inertia of a disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.

Correct Answer: B — Increases by a factor of 4

Q. If the radius of a planet is halved while keeping its mass constant, how does the gravitational acceleration at its surface change?

A. It becomes four times stronger
B. It becomes twice stronger
C. It remains the same
D. It becomes half as strong

Solution

If the radius is halved, the gravitational acceleration becomes four times stronger, as g is inversely proportional to the square of the radius.

Correct Answer: A — It becomes four times stronger

Q. If the radius of a planet is halved, what happens to the gravitational acceleration on its surface?

A. It doubles
B. It halves
C. It becomes one-fourth
D. It remains the same

Solution

g ∝ 1/R². If R is halved, g becomes 4 times greater.

Correct Answer: A — It doubles

Q. If the radius of a rotating disc is doubled while keeping the mass constant, how does the angular momentum change if the angular velocity remains the same?

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

Solution

Angular momentum L = Iω; if radius is doubled, moment of inertia I increases by a factor of 4, hence L quadruples.

Correct Answer: C — It quadruples

Q. If the radius of a rotating object is halved while keeping the angular velocity constant, what happens to the linear velocity at the edge?

A. It doubles
B. It halves
C. It remains the same
D. It becomes zero

Solution

Linear velocity v = rω. If r is halved and ω remains constant, v also halves.

Correct Answer: B — It halves

Q. If the radius of a rotating object is halved while keeping the mass constant, how does its moment of inertia change?

A. It remains the same
B. It doubles
C. It halves
D. It reduces to one-fourth

Solution

Moment of inertia I is proportional to the square of the radius, so halving the radius reduces I to one-fourth.

Correct Answer: D — It reduces to one-fourth

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