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. What is the oxidation state of phosphorus in H3PO4?

A. +5
B. +3
C. 0
D. -3

Solution

In H3PO4, phosphorus has an oxidation state of +5.

Correct Answer: A — +5

Q. What is the oxidation state of phosphorus in phosphoric acid (H3PO4)?

A. +3
B. +5
C. 0
D. -3

Solution

In phosphoric acid, phosphorus has an oxidation state of +5.

Correct Answer: B — +5

Q. What is the oxidation state of sulfur in H2SO4?

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

Solution

In H2SO4, the oxidation state of sulfur is +6.

Correct Answer: C — +6

Q. What is the oxidation state of Uranium in Uranium Hexafluoride (UF6)?

A. +4
B. +6
C. +5
D. 0

Solution

In UF6, Uranium is in the +6 oxidation state as it forms six bonds with fluorine.

Correct Answer: B — +6

Q. What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?

A. y = 2x^2 + 1
B. y = 4x^2 + 1
C. y = 2x^2
D. y = 4x^2 + C

Solution

Integrating gives y = 2x^2 + C. Using the initial condition, C = 1.

Correct Answer: A — y = 2x^2 + 1

Q. What is the particular solution of the equation dy/dx = 2y with the initial condition y(0) = 1?

A. y = e^(2x)
B. y = e^(2x) + 1
C. y = 1 + e^(2x)
D. y = 1 + 2x

Solution

The general solution is y = Ce^(2x). Using the initial condition y(0) = 1, we find C = 1.

Correct Answer: A — y = e^(2x)

Q. What is the particular solution of the equation dy/dx = 2y with y(0) = 5?

A. y = 5e^(2x)
B. y = 2e^(2x)
C. y = 5e^(-2x)
D. y = 5 + 2x

Solution

The general solution is y = Ce^(2x). Using y(0) = 5 gives C = 5.

Correct Answer: A — y = 5e^(2x)

Q. What is the path difference for light waves from two coherent sources at an angle of 45° to the line joining the sources at a distance of 1 m?

A. 0.5 m
B. 0.707 m
C. 1 m
D. 0.25 m

Solution

Path difference = d sin θ = 1 m * sin(45°) = 1 m * √2/2 = 0.707 m.

Correct Answer: B — 0.707 m

Q. What is the path difference for light waves from two coherent sources at an angle of 45° to the line joining them?

A. λ/2
B. λ
C. √2λ
D. 0

Solution

Path difference = d sin θ = d sin(45°) = d(√2/2). For d = λ, path difference = √2λ.

Correct Answer: C — √2λ

Q. What is the path difference for the first minimum in a double-slit experiment?

A. λ/2
B. λ
C. 3λ/2
D. 2λ

Solution

The path difference for the first minimum is λ/2, which leads to destructive interference.

Correct Answer: A — λ/2

Q. What is the percentage by mass of a solution containing 20 grams of solute in 180 grams of solution?

A. 10%
B. 20%
C. 15%
D. 25%

Solution

Percentage by mass = (mass of solute / mass of solution) * 100 = (20 g / (20 g + 180 g)) * 100 = 20%.

Correct Answer: B — 20%

Q. What is the percentage by mass of a solution containing 5 g of NaCl in 95 g of water?

A. 5%
B. 10%
C. 15%
D. 20%

Solution

Percentage by mass = (mass of solute / total mass) × 100 = (5 g / (5 g + 95 g)) × 100 = 5%.

Correct Answer: A — 5%

Q. What is the percentage composition of carbon in glucose (C6H12O6)?

A. 40%
B. 50%
C. 60%
D. 30%

Solution

Molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 g. Percentage of carbon = (72/180) x 100 = 40%.

Correct Answer: A — 40%

Q. What is the period number of the element with atomic number 20?

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

Solution

The element with atomic number 20 is Calcium, which is in period 4.

Correct Answer: C — 4

Q. What is the period of a pendulum that is 1 meter long?

A. 1 s
B. 2 s
C. 0.5 s
D. 3 s

Solution

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

Correct Answer: B — 2 s

Q. What is the period of a satellite in a circular orbit at a height of 300 km above the Earth's surface?

A. 90 minutes
B. 60 minutes
C. 120 minutes
D. 30 minutes

Solution

The period of a satellite in a circular orbit at a height of 300 km is approximately 90 minutes.

Correct Answer: A — 90 minutes

Q. What is the period of a satellite in a low Earth orbit (LEO) compared to a satellite in a geostationary orbit?

A. Longer than a geostationary orbit
B. Shorter than a geostationary orbit
C. Equal to a geostationary orbit
D. Depends on the mass of the satellite

Solution

Satellites in low Earth orbit have a much shorter orbital period compared to geostationary satellites due to their proximity to Earth.

Correct Answer: B — Shorter than a geostationary orbit

Q. What is the periodic trend for electronegativity across a period?

A. Increases
B. Decreases
C. Remains the same
D. Varies irregularly

Solution

Electronegativity increases across a period from left to right.

Correct Answer: A — Increases

Q. What is the pH of a 0.01 M HCl solution?

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

Solution

The pH of a 0.01 M HCl solution is 2, as HCl is a strong acid and fully dissociates.

Correct Answer: B — 2

Q. What is the pH of a 0.01 M NaOH solution?

A. 12
B. 2
C. 10
D. 1

Solution

For NaOH, which is a strong base, pOH = -log[OH-]. [OH-] = 0.01 M, so pOH = 2. Therefore, pH = 14 - pOH = 14 - 2 = 12.

Correct Answer: A — 12

Q. What is the pH of a 0.01 M solution of sodium acetate (CH3COONa)?

A. 4.76
B. 7.00
C. 9.24
D. 10.00

Solution

Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]), where pKa of acetic acid is 4.76.

Correct Answer: C — 9.24

Q. What is the pH of a 0.01 M solution of sodium hydroxide (NaOH)?

A. 12
B. 2
C. 10
D. 14

Solution

pH = 14 - pOH; pOH = -log[OH-] = -log(0.01) = 2; pH = 14 - 2 = 12.

Correct Answer: A — 12

Q. What is the pH of a 0.01 M solution of sulfuric acid (H2SO4)?

A. 1
B. 2
C. 3
D. 0.5

Solution

H2SO4 is a strong acid and dissociates completely, so pH = -log(0.01) = 2.

Correct Answer: A — 1

Q. What is the pH of a 0.05 M solution of sulfuric acid (H2SO4)?

A. 1
B. 1.3
C. 1.7
D. 2

Solution

H2SO4 is a strong acid and dissociates completely; [H+] = 0.05 M, so pH = -log(0.05) ≈ 1.3

Correct Answer: B — 1.3

Q. What is the pH of a 0.1 M NaOH solution?

A. 10
B. 11
C. 12
D. 13

Solution

pOH = -log[OH-] = -log(0.1) = 1; pH = 14 - pOH = 14 - 1 = 13

Correct Answer: C — 12

Q. What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8 x 10^-5?

A. 2.87
B. 4.76
C. 3.87
D. 5.00

Solution

Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.

Correct Answer: B — 4.76

Q. What is the pH of a 0.1 M solution of ammonium chloride (NH4Cl)?

A. 5.1
B. 5.5
C. 6.1
D. 6.5

Solution

NH4Cl is a salt of a weak base and a strong acid; it hydrolyzes to give H+, resulting in a pH < 7.

Correct Answer: C — 6.1

Q. What is the pH of a 0.1 M solution of hydrochloric acid (HCl)?

A. 1
B. 0.1
C. 10
D. 0

Solution

The pH of a strong acid like HCl is calculated using the formula pH = -log[H+]. For 0.1 M HCl, [H+] = 0.1 M, so pH = -log(0.1) = 1.

Correct Answer: A — 1

Q. What is the pH of a 0.1 M solution of potassium hydrogen phthalate (KHP)?

A. 4.0
B. 5.0
C. 6.0
D. 7.0

Solution

KHP is a weak acid; its pKa is approximately 5.0, so the pH of a 0.1 M solution is around 5.0.

Correct Answer: B — 5.0

Q. What is the pH of a 0.1 M solution of sodium acetate (CH3COONa)?

A. 4.76
B. 7
C. 9.24
D. 10

Solution

pH = 7 + 0.5(pKa + log[C]) = 7 + 0.5(4.76 + log(0.1)) = 9.24

Correct Answer: C — 9.24

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