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. Given vectors P = (4, 1, 0) and Q = (1, 2, 3), find the scalar product P · Q.
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Solution
P · Q = 4*1 + 1*2 + 0*3 = 4 + 2 + 0 = 6.
Correct Answer:
B
— 11
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Q. How does the addition of a surfactant affect the surface tension of water?
A.
Increases surface tension
B.
Decreases surface tension
C.
No effect on surface tension
D.
Surface tension becomes negative
Show solution
Solution
Surfactants lower the surface tension of water by disrupting the cohesive forces between water molecules.
Correct Answer:
B
— Decreases surface tension
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Q. How does the addition of salt to water affect its surface tension?
A.
Increases surface tension
B.
Decreases surface tension
C.
No effect on surface tension
D.
Salt has no effect on water
Show solution
Solution
The addition of salt to water can increase the surface tension due to the ionic interactions that enhance the cohesive forces among water molecules.
Correct Answer:
A
— Increases surface tension
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Q. How does the addition of soap to water affect its surface tension?
A.
Increases surface tension
B.
Decreases surface tension
C.
No effect on surface tension
D.
Changes surface tension unpredictably
Show solution
Solution
The addition of soap decreases the surface tension of water by disrupting the cohesive forces between water molecules.
Correct Answer:
B
— Decreases surface tension
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Q. How does the gravitational field strength vary with distance from a point mass?
A.
It increases linearly.
B.
It decreases with the square of the distance.
C.
It remains constant.
D.
It decreases linearly.
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Solution
The gravitational field strength decreases with the square of the distance from the point mass.
Correct Answer:
B
— It decreases with the square of the distance.
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Q. How does the gravitational force between two objects change if both masses are doubled?
A.
It becomes four times stronger
B.
It becomes twice as strong
C.
It remains the same
D.
It becomes half as strong
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Solution
Doubling both masses results in a fourfold increase in gravitational force, as it is directly proportional to the product of the masses.
Correct Answer:
A
— It becomes four times stronger
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Q. How does the gravitational force between two objects change if the distance between them is halved?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
According to Newton's law of gravitation, if the distance is halved, the force increases by a factor of 4 (F ∝ 1/r²).
Correct Answer:
B
— It quadruples
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Q. How does the gravitational force between two objects change if the mass of one object is tripled?
A.
It triples
B.
It doubles
C.
It remains the same
D.
It increases by a factor of nine
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Solution
Gravitational force is directly proportional to the product of the masses. Tripling one mass triples the force.
Correct Answer:
A
— It triples
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Q. How does the gravitational potential change as you move away from a planet?
A.
It increases.
B.
It decreases.
C.
It remains constant.
D.
It oscillates.
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Solution
The gravitational potential decreases as you move away from a planet.
Correct Answer:
B
— It decreases.
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Q. How does the gravitational potential energy of a system of two masses change as they move closer together?
A.
It increases.
B.
It decreases.
C.
It remains constant.
D.
It becomes zero.
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Solution
The gravitational potential energy decreases as the two masses move closer together.
Correct Answer:
B
— It decreases.
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Q. How does the gravitational potential energy of an object change when it is lifted to a height 'h' above the ground?
A.
It decreases.
B.
It increases.
C.
It remains the same.
D.
It becomes zero.
Show solution
Solution
The gravitational potential energy increases as the object is lifted to a height 'h' above the ground.
Correct Answer:
B
— It increases.
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Q. How does the presence of a +M group affect the stability of a carbocation?
A.
Increases stability
B.
Decreases stability
C.
No effect
D.
Depends on the solvent
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Solution
A +M group increases the stability of a carbocation by donating electron density through resonance.
Correct Answer:
A
— Increases stability
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Q. How does the presence of a -I group affect the stability of a carbocation?
A.
Increases stability
B.
Decreases stability
C.
No effect
D.
Depends on the size of the group
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Solution
A -I group decreases the stability of a carbocation by withdrawing electron density.
Correct Answer:
B
— Decreases stability
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Q. How many atoms are in 2 moles of oxygen gas (O2)?
A.
6.022 x 10^23
B.
1.2044 x 10^24
C.
3.011 x 10^23
D.
12.044 x 10^23
Show solution
Solution
Each O2 molecule has 2 oxygen atoms. Number of atoms = moles x Avogadro's number x 2 = 2 moles x 6.022 x 10^23 molecules/mole x 2 = 1.2044 x 10^24 atoms.
Correct Answer:
B
— 1.2044 x 10^24
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Q. How many centimeters are there in a meter?
A.
10
B.
100
C.
1000
D.
10000
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Solution
There are 100 centimeters in a meter.
Correct Answer:
B
— 100
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Q. How many different ways can 3 red, 2 blue, and 1 green balls be arranged in a line?
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Solution
The total arrangements = 6! / (3! * 2! * 1!) = 60.
Correct Answer:
B
— 120
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Q. How many different ways can 4 people be seated at a round table?
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Solution
The number of arrangements of n people at a round table is (n-1)!. For 4 people, it is 3! = 6.
Correct Answer:
B
— 12
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Q. How many different ways can 4 students be selected from a group of 10?
A.
210
B.
120
C.
100
D.
90
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Solution
The number of ways to choose 4 from 10 is given by 10C4 = 210.
Correct Answer:
A
— 210
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Q. How many different ways can 6 people be seated at a round table?
A.
720
B.
120
C.
600
D.
480
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Solution
The number of arrangements of 6 people at a round table is (6-1)! = 5! = 120.
Correct Answer:
A
— 720
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Q. How many elements are in the power set of the empty set?
A.
0
B.
1
C.
2
D.
Infinite
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Solution
The power set of the empty set contains only one subset, which is the empty set itself. Thus, it has 1 element.
Correct Answer:
B
— 1
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Q. How many grams are in 0.25 moles of glucose (C6H12O6)?
A.
30 g
B.
45 g
C.
75 g
D.
90 g
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Solution
Molar mass of C6H12O6 = 6*12 + 12*1 + 6*16 = 180 g/mol. Mass = 0.25 moles x 180 g/mol = 45 g.
Correct Answer:
D
— 90 g
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Q. How many grams are in 0.25 moles of KCl?
A.
35 g
B.
70 g
C.
17.5 g
D.
140 g
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Solution
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol. Mass = moles x molar mass = 0.25 moles x 74.5 g/mol = 18.625 g.
Correct Answer:
B
— 70 g
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Q. How many grams are in 0.25 moles of sulfuric acid (H2SO4)?
A.
49 g
B.
98 g
C.
24.5 g
D.
12.25 g
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Solution
Molar mass of H2SO4 = 98 g/mol. Mass = moles x molar mass = 0.25 moles x 98 g/mol = 24.5 g.
Correct Answer:
A
— 49 g
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Q. How many grams are in 0.5 moles of sodium (Na)?
A.
11 g
B.
22 g
C.
5.5 g
D.
44 g
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Solution
The molar mass of sodium is approximately 23 g/mol. Therefore, 0.5 moles of Na weigh 0.5 x 23 g = 11 g.
Correct Answer:
A
— 11 g
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Q. How many grams are in 2 moles of carbon dioxide (CO2)?
A.
44 g
B.
88 g
C.
22 g
D.
66 g
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Solution
The molar mass of CO2 is 44 g/mol. Thus, 2 moles of CO2 = 2 x 44 g = 88 g.
Correct Answer:
B
— 88 g
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Q. How many grams are in 3 moles of glucose (C6H12O6)?
A.
180 g
B.
360 g
C.
540 g
D.
90 g
Show solution
Solution
Molar mass of glucose = 6*12 + 12*1 + 6*16 = 180 g/mol. Mass = moles x molar mass = 3 moles x 180 g/mol = 540 g.
Correct Answer:
B
— 360 g
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Q. How many grams of CO2 are produced from the complete combustion of 1 mole of C3H8?
A.
44 g
B.
88 g
C.
132 g
D.
22 g
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Solution
C3H8 + 5O2 → 3CO2 + 4H2O. 1 mole of C3H8 produces 3 moles of CO2. Mass = 3 * 44 g = 132 g.
Correct Answer:
B
— 88 g
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Q. How many grams of KCl are needed to prepare 0.5 moles of KCl solution?
A.
37.5 g
B.
74.5 g
C.
50 g
D.
100 g
Show solution
Solution
The molar mass of KCl is 39 g/mol (K) + 35.5 g/mol (Cl) = 74.5 g/mol. Therefore, 0.5 moles of KCl will weigh 0.5 x 74.5 g = 37.25 g.
Correct Answer:
B
— 74.5 g
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Q. How many grams of KCl are needed to prepare 0.5 moles of KCl solution? (Molar mass of KCl = 74.5 g/mol)
A.
37.25 g
B.
74.5 g
C.
148.5 g
D.
18.5 g
Show solution
Solution
To find the mass, use the formula: mass = moles x molar mass. Thus, 0.5 moles x 74.5 g/mol = 37.25 g.
Correct Answer:
A
— 37.25 g
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Q. How many grams of KCl are needed to prepare 0.5 moles of KCl?
A.
37.5 g
B.
45 g
C.
50 g
D.
60 g
Show solution
Solution
The molar mass of KCl is approximately 74.5 g/mol. Therefore, 0.5 moles of KCl will weigh 0.5 x 74.5 g = 37.25 g.
Correct Answer:
B
— 45 g
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