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 electric potential due to a dipole at a point along the axial line at a distance 'r' from the center of the dipole?
A.
0
B.
k * p / r²
C.
k * p / r
D.
k * p / 2r
Show solution
Solution
The potential V = k * p / r, where p is the dipole moment.
Correct Answer:
C
— k * p / r
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Q. What is the electric potential due to a point charge at a distance r?
A.
k * q / r
B.
k * q / r^2
C.
k * q * r
D.
k * q * r^2
Show solution
Solution
The electric potential V due to a point charge q at a distance r is given by V = k * q / r.
Correct Answer:
A
— k * q / r
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Q. What is the electric potential due to a point charge of 5 μC at a distance of 2 m?
A.
0 V
B.
2250 V
C.
1125 V
D.
4500 V
Show solution
Solution
Electric potential V = k * q / r = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m) = 2250 V.
Correct Answer:
B
— 2250 V
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Q. What is the electric potential energy of a charge of 1 C at a point where the electric potential is 10 V?
A.
10 J
B.
1 J
C.
0 J
D.
100 J
Show solution
Solution
Electric potential energy U = q * V = 1 C * 10 V = 10 J.
Correct Answer:
A
— 10 J
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Q. What is the electric potential energy of a charge of 1 C placed in an electric field of 10 N/C at a distance of 2 m?
A.
20 J
B.
10 J
C.
5 J
D.
2 J
Show solution
Solution
Electric potential energy U = q * V = q * E * d = 1 C * 10 N/C * 2 m = 20 J.
Correct Answer:
A
— 20 J
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Q. What is the electric potential energy of a charge of 1 μC placed in an electric potential of 200 V?
A.
0.2 mJ
B.
0.1 mJ
C.
0.4 mJ
D.
0.5 mJ
Show solution
Solution
Electric potential energy U = q * V = 1 × 10^-6 C * 200 V = 0.2 mJ.
Correct Answer:
A
— 0.2 mJ
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Q. What is the electric potential energy of a system of two charges +q and -q separated by a distance r?
A.
0
B.
kq²/r
C.
-kq²/r
D.
kq/r
Show solution
Solution
The electric potential energy U = k(q1*q2)/r = k(+q*-q)/r = -kq²/r.
Correct Answer:
C
— -kq²/r
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Q. What is the electric potential energy of a system of two charges of +2 μC and -2 μC separated by 0.5 m?
A.
-72 J
B.
72 J
C.
0 J
D.
-36 J
Show solution
Solution
U = k * (q1 * q2) / r = (9 × 10^9 N m²/C²) * (2 × 10^-6 C * -2 × 10^-6 C) / 0.5 m = -72 J.
Correct Answer:
D
— -36 J
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Q. What is the electric potential energy of a system of two point charges Q1 and Q2 separated by a distance r?
A.
kQ1Q2/r
B.
kQ1Q2/2r
C.
kQ1Q2/r²
D.
kQ1Q2
Show solution
Solution
The electric potential energy U of a system of two point charges is given by U = kQ1Q2/r.
Correct Answer:
A
— kQ1Q2/r
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Q. What is the electron configuration of Chromium (Cr)?
A.
[Ar] 4s2 3d4
B.
[Ar] 4s1 3d5
C.
[Ar] 4s2 3d5
D.
[Ar] 4s2 3d3
Show solution
Solution
Chromium has an electron configuration of [Ar] 4s1 3d5 due to stability from half-filled d-orbitals.
Correct Answer:
B
— [Ar] 4s1 3d5
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Q. What is the electron configuration of Magnesium?
A.
1s² 2s² 2p⁶ 3s²
B.
1s² 2s² 2p⁶ 3s² 3p¹
C.
1s² 2s² 2p⁶ 3s² 3p²
D.
1s² 2s² 2p⁶ 3s² 3d¹
Show solution
Solution
The electron configuration of Magnesium (Mg) is 1s² 2s² 2p⁶ 3s².
Correct Answer:
A
— 1s² 2s² 2p⁶ 3s²
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Q. What is the electron configuration of oxygen (O)?
A.
1s2 2s2 2p4
B.
1s2 2s2 2p6
C.
1s2 2s1 2p5
D.
1s2 2s2 2p3
Show solution
Solution
The electron configuration of oxygen (atomic number 8) is 1s2 2s2 2p4.
Correct Answer:
A
— 1s2 2s2 2p4
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Q. What is the electron configuration of potassium (K)?
A.
1s² 2s² 2p⁶ 3s¹
B.
1s² 2s² 2p⁶ 3s²
C.
1s² 2s² 2p⁶ 3s² 3p¹
D.
1s² 2s² 2p⁶ 3s² 3p²
Show solution
Solution
The electron configuration of potassium (K) is 1s² 2s² 2p⁶ 3s¹.
Correct Answer:
A
— 1s² 2s² 2p⁶ 3s¹
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Q. What is the electron configuration of the alkali metal potassium (K)?
A.
1s² 2s² 2p⁶ 3s¹
B.
1s² 2s² 2p⁶ 3s²
C.
1s² 2s² 2p⁶ 3s² 3p¹
D.
1s² 2s² 2p⁶ 3s² 3p²
Show solution
Solution
The electron configuration of potassium (K) is 1s² 2s² 2p⁶ 3s¹.
Correct Answer:
A
— 1s² 2s² 2p⁶ 3s¹
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Q. What is the electron configuration of the alkaline earth metal magnesium?
A.
1s² 2s² 2p⁶ 3s²
B.
1s² 2s² 2p⁶ 3s² 3p²
C.
1s² 2s² 2p⁶ 3s² 3p⁶
D.
1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Show solution
Solution
The electron configuration of magnesium (Mg) is 1s² 2s² 2p⁶ 3s².
Correct Answer:
A
— 1s² 2s² 2p⁶ 3s²
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Q. What is the electron configuration of the element with atomic number 12?
A.
1s2 2s2 2p6 3s2
B.
1s2 2s2 2p6 3s2 3p1
C.
1s2 2s2 2p6 3s2 3p6
D.
1s2 2s2 2p6 3s2 3d10
Show solution
Solution
The element with atomic number 12 is Magnesium (Mg) with the configuration 1s2 2s2 2p6 3s2.
Correct Answer:
A
— 1s2 2s2 2p6 3s2
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Q. What is the electron configuration of the element with atomic number 26?
A.
1s2 2s2 2p6 3s2 3p6 4s2 3d6
B.
1s2 2s2 2p6 3s2 3p6 4s2 3d5
C.
1s2 2s2 2p6 3s2 3p6 4s2 3d7
D.
1s2 2s2 2p6 3s2 3p6 4s2 3d8
Show solution
Solution
The electron configuration of iron (Fe), which has an atomic number of 26, is 1s2 2s2 2p6 3s2 3p6 4s2 3d6.
Correct Answer:
A
— 1s2 2s2 2p6 3s2 3p6 4s2 3d6
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Q. What is the electron configuration of the s-block element magnesium?
A.
1s² 2s² 2p⁶ 3s²
B.
1s² 2s² 2p⁶ 3s² 3p¹
C.
1s² 2s² 2p⁶ 3s² 3p²
D.
1s² 2s² 2p⁶ 3s² 3d¹
Show solution
Solution
The electron configuration of magnesium (Mg) is 1s² 2s² 2p⁶ 3s².
Correct Answer:
A
— 1s² 2s² 2p⁶ 3s²
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Q. What is the electron pair geometry of NH3?
A.
Tetrahedral
B.
Trigonal planar
C.
Octahedral
D.
Linear
Show solution
Solution
NH3 has four electron pairs (three bonding pairs and one lone pair), giving it a tetrahedral electron pair geometry.
Correct Answer:
A
— Tetrahedral
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Q. What is the electron pair geometry of SF6?
A.
Tetrahedral
B.
Trigonal bipyramidal
C.
Octahedral
D.
Square planar
Show solution
Solution
SF6 has six bonding pairs and no lone pairs, resulting in an octahedral electron pair geometry.
Correct Answer:
C
— Octahedral
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Q. What is the electronic configuration of a chlorine ion (Cl-)?
A.
1s2 2s2 2p6 3s2 3p5
B.
1s2 2s2 2p6
C.
1s2 2s2 2p6 3s2 3p6
D.
1s2 2s2 2p6 3s2 3p4
Show solution
Solution
Chlorine ion (Cl-) has gained an electron, resulting in the configuration 1s2 2s2 2p6.
Correct Answer:
B
— 1s2 2s2 2p6
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Q. What is the electronic configuration of a neutral oxygen atom?
A.
1s2 2s2 2p4
B.
1s2 2s2 2p6
C.
1s2 2s2 2p2
D.
1s2 2s2 2p5
Show solution
Solution
Oxygen has 8 electrons. The electronic configuration is 1s2 2s2 2p4.
Correct Answer:
A
— 1s2 2s2 2p4
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Q. What is the electronic configuration of a neutral sodium atom?
A.
1s2 2s2 2p6 3s1
B.
1s2 2s2 2p6
C.
1s2 2s2 2p5
D.
1s2 2s2 2p6 3s2
Show solution
Solution
Sodium has 11 electrons, and its configuration is 1s2 2s2 2p6 3s1.
Correct Answer:
A
— 1s2 2s2 2p6 3s1
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Q. What is the electronic configuration of a sodium ion (Na+)?
A.
1s2 2s2 2p6
B.
1s2 2s2 2p6 3s1
C.
1s2 2s2 2p5
D.
1s2 2s2 2p6 3s2
Show solution
Solution
Sodium ion (Na+) has lost one electron, resulting in the configuration 1s2 2s2 2p6.
Correct Answer:
A
— 1s2 2s2 2p6
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Q. What is the electronic configuration of Chlorine?
A.
1s2 2s2 2p6 3s2 3p5
B.
1s2 2s2 2p6 3s2 3p4
C.
1s2 2s2 2p6 3s2 3p6
D.
1s2 2s2 2p6 3s2 3p3
Show solution
Solution
Chlorine (Cl) has 17 electrons, so its configuration is 1s2 2s2 2p6 3s2 3p5.
Correct Answer:
A
— 1s2 2s2 2p6 3s2 3p5
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Q. What is the electronic configuration of Iron?
A.
[Ar] 4s2 3d6
B.
[Ar] 4s2 3d8
C.
[Ar] 4s2 3d5
D.
[Ar] 4s2 3d7
Show solution
Solution
Iron (Fe) has the configuration [Ar] 4s2 3d6.
Correct Answer:
A
— [Ar] 4s2 3d6
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Q. What is the electronic configuration of oxygen?
A.
1s2 2s2 2p4
B.
1s2 2s2 2p6
C.
1s2 2s2 2p2
D.
1s2 2s2 2p5
Show solution
Solution
Oxygen has 8 electrons, and its electronic configuration is 1s2 2s2 2p4.
Correct Answer:
A
— 1s2 2s2 2p4
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Q. What is the electronic configuration of the alkali metal potassium (K)?
A.
1s² 2s² 2p⁶ 3s¹
B.
1s² 2s² 2p⁶ 3s²
C.
1s² 2s² 2p⁶ 3s² 3p¹
D.
1s² 2s² 2p⁶ 3s² 3p⁶
Show solution
Solution
The electronic configuration of potassium (K) is 1s² 2s² 2p⁶ 3s¹.
Correct Answer:
A
— 1s² 2s² 2p⁶ 3s¹
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Q. What is the electronic configuration of the chloride ion (Cl-)?
A.
1s2 2s2 2p6 3s2 3p5
B.
1s2 2s2 2p6 3s2 3p6
C.
1s2 2s2 2p6 3s2 3p4
D.
1s2 2s2 2p6 3s2 3p3
Show solution
Solution
Chloride ion has gained one electron, making its configuration 1s2 2s2 2p6 3s2 3p6.
Correct Answer:
B
— 1s2 2s2 2p6 3s2 3p6
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Q. What is the electronic configuration of the element with atomic number 104?
A.
[Rn] 5f14 6d2
B.
[Rn] 5f14 6d4
C.
[Rn] 5f14 6d8
D.
[Rn] 5f14 7s2
Show solution
Solution
The electronic configuration of Rutherfordium (atomic number 104) is [Rn] 5f14 6d4.
Correct Answer:
B
— [Rn] 5f14 6d4
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