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.
Q. What is the molecular geometry of BF3?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Bent
Show solution
Solution
BF3 has a trigonal planar geometry due to the three bonding pairs and no lone pairs on the central atom.
Correct Answer:
B
— Trigonal planar
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Q. What is the molecular geometry of CH4 according to VSEPR theory?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
According to VSEPR theory, CH4 has four bonding pairs and no lone pairs, resulting in a tetrahedral geometry.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular geometry of CH4?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
CH4 has a tetrahedral geometry due to four bonding pairs around the central carbon atom.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular geometry of methane (CH4)?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
Methane has a tetrahedral molecular geometry due to sp3 hybridization of the carbon atom.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular geometry of NH3 according to VSEPR theory?
A.
Trigonal planar
B.
Tetrahedral
C.
Bent
D.
Trigonal pyramidal
Show solution
Solution
NH3 has three bonding pairs and one lone pair, resulting in a trigonal pyramidal geometry.
Correct Answer:
D
— Trigonal pyramidal
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Q. What is the molecular geometry of SF4?
A.
Tetrahedral
B.
Trigonal bipyramidal
C.
Seesaw
D.
Square planar
Show solution
Solution
SF4 has four bonding pairs and one lone pair, resulting in a seesaw molecular geometry.
Correct Answer:
C
— Seesaw
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Q. What is the molecular geometry of SO2?
A.
Linear
B.
Trigonal planar
C.
Bent
D.
Tetrahedral
Show solution
Solution
SO2 has two bonding pairs and one lone pair, resulting in a bent molecular geometry.
Correct Answer:
C
— Bent
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Q. What is the molecular geometry of the molecule with the electronic configuration of 1s2 2s2 2p2?
A.
Linear
B.
Trigonal Planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
The electronic configuration corresponds to C2, which has a tetrahedral geometry due to sp3 hybridization.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular orbital configuration of F2?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(π2p)⁴(π*2p)²
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³(π*2p)²
Show solution
Solution
The correct configuration for F2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)².
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
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Q. What is the molecular orbital configuration of O2?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)²
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)⁴
Show solution
Solution
The correct configuration for O2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹.
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
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Q. What is the molecular orbital configuration of the F2 molecule?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)¹
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³(π*2p)²
Show solution
Solution
The correct configuration for F2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)².
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
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Q. What is the molecular orbital configuration of the O2 molecule?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)²
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)¹
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)²
Show solution
Solution
The correct configuration for O2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹.
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
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Q. What is the molecular shape of a molecule with the formula AX3E?
A.
Trigonal planar
B.
Tetrahedral
C.
Trigonal pyramidal
D.
Bent
Show solution
Solution
AX3E indicates three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.
Correct Answer:
C
— Trigonal pyramidal
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Q. What is the molecular shape of BF3 according to VSEPR theory?
A.
Bent
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
BF3 has three bonding pairs and no lone pairs, resulting in a trigonal planar shape.
Correct Answer:
B
— Trigonal planar
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Q. What is the molecular shape of NH3 according to VSEPR theory?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Trigonal pyramidal
Show solution
Solution
NH3 has three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.
Correct Answer:
D
— Trigonal pyramidal
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Q. What is the molecular weight of water (H2O)?
A.
16 g/mol
B.
18 g/mol
C.
20 g/mol
D.
22 g/mol
Show solution
Solution
The molecular weight of water is calculated as (2*1) + (16) = 18 g/mol.
Correct Answer:
B
— 18 g/mol
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Q. What is the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to its plane?
A.
1/2 MR^2
B.
MR^2
C.
1/4 MR^2
D.
2/3 MR^2
Show solution
Solution
The moment of inertia of a disk about an axis through its center is I = 1/2 MR^2.
Correct Answer:
A
— 1/2 MR^2
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Q. What is the moment of inertia of a solid cylinder of mass M and radius R about its central axis?
A.
1/2 MR^2
B.
1/3 MR^2
C.
MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a solid cylinder about its central axis is given by I = 1/2 MR^2.
Correct Answer:
A
— 1/2 MR^2
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Q. What is the moment of inertia of a solid disk about its central axis?
A.
(1/2)MR^2
B.
(1/3)MR^2
C.
(1/4)MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a solid disk about its central axis is (1/2)MR^2.
Correct Answer:
A
— (1/2)MR^2
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Q. What is the moment of inertia of a solid sphere about an axis through its center?
A.
(2/5)mr^2
B.
(1/2)mr^2
C.
(1/3)mr^2
D.
(5/2)mr^2
Show solution
Solution
The moment of inertia of a solid sphere about an axis through its center is given by I = (2/5)mr^2.
Correct Answer:
A
— (2/5)mr^2
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Q. What is the moment of inertia of a solid sphere of mass M and radius R about an axis through its center?
A.
2/5 MR^2
B.
3/5 MR^2
C.
1/2 MR^2
D.
MR^2
Show solution
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
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Q. What is the moment of inertia of a thin circular hoop of mass M and radius R about an axis through its center?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a thin circular hoop about an axis through its center is I = MR^2.
Correct Answer:
A
— MR^2
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Q. What is the moment of inertia of a thin circular plate of mass M and radius R about an axis through its center and perpendicular to its plane?
A.
1/2 MR^2
B.
MR^2
C.
1/4 MR^2
D.
1/3 MR^2
Show solution
Solution
The moment of inertia of a thin circular plate about an axis through its center is I = 1/2 MR^2.
Correct Answer:
A
— 1/2 MR^2
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Q. What is the moment of inertia of a thin circular ring of mass M and radius R about an axis perpendicular to its plane through its center?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a thin circular ring about an axis through its center is I = MR^2.
Correct Answer:
A
— MR^2
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Q. What is the moment of inertia of a thin circular ring of mass M and radius R about an axis through its center?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a thin circular ring about an axis through its center is I = MR^2.
Correct Answer:
A
— MR^2
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Q. What is the moment of inertia of a thin circular ring of mass M and radius R about an axis perpendicular to its plane and passing through its center?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a thin circular ring about an axis through its center is I = MR^2.
Correct Answer:
A
— MR^2
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Q. What is the moment of inertia of a thin circular ring of mass M and radius R about an axis through its center and perpendicular to its plane?
A.
MR^2
B.
1/2 MR^2
C.
2/3 MR^2
D.
1/3 MR^2
Show solution
Solution
The moment of inertia of a thin circular ring about an axis through its center is I = MR^2.
Correct Answer:
A
— MR^2
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Q. What is the moment of inertia of a thin rod of length L about an axis perpendicular to it and passing through its center?
A.
(1/3)ML^2
B.
(1/12)ML^2
C.
(1/2)ML^2
D.
ML^2
Show solution
Solution
The moment of inertia of a thin rod about an axis through its center is given by I = (1/12)ML^2.
Correct Answer:
B
— (1/12)ML^2
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Q. What is the moment of inertia of a thin rod of length L about an axis perpendicular to it and passing through one end?
A.
(1/3)ML^2
B.
(1/12)ML^2
C.
ML^2
D.
(1/2)ML^2
Show solution
Solution
The moment of inertia of a thin rod about an end is given by I = (1/3)ML^2.
Correct Answer:
A
— (1/3)ML^2
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Q. What is the moment of inertia of a thin spherical shell of mass M and radius R about an axis through its center?
A.
2/3 MR^2
B.
1/2 MR^2
C.
MR^2
D.
2 MR^2
Show solution
Solution
The moment of inertia of a thin spherical shell about an axis through its center is I = 2/3 MR^2.
Correct Answer:
A
— 2/3 MR^2
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