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. The condition for the lines represented by the equation x^2 + 2xy + y^2 = 0 to be coincident is:
A.
Discriminant > 0
B.
Discriminant = 0
C.
Discriminant < 0
D.
None of the above
Show solution
Solution
For the lines to be coincident, the discriminant must be equal to zero.
Correct Answer:
B
— Discriminant = 0
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Q. The condition for the lines represented by the equation x^2 + y^2 + 2xy = 0 to be coincident is:
A.
Discriminant = 0
B.
Discriminant > 0
C.
Discriminant < 0
D.
None of the above
Show solution
Solution
For the lines to be coincident, the discriminant of the quadratic must be zero.
Correct Answer:
A
— Discriminant = 0
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Q. The condition for the lines represented by the equation x^2 + y^2 - 4x - 6y + 9 = 0 to be coincident is:
A.
Discriminant = 0
B.
Discriminant > 0
C.
Discriminant < 0
D.
None of the above
Show solution
Solution
For the lines to be coincident, the discriminant of the quadratic must equal zero.
Correct Answer:
A
— Discriminant = 0
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Q. The coordinates of the centroid of a triangle with vertices at (0, 0), (6, 0), and (3, 6) are:
A.
(3, 2)
B.
(3, 3)
C.
(2, 3)
D.
(0, 0)
Show solution
Solution
Centroid = ((x1+x2+x3)/3, (y1+y2+y3)/3) = (9/3, 6/3) = (3, 2).
Correct Answer:
B
— (3, 3)
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Q. The coordinates of the centroid of a triangle with vertices at (2, 3), (4, 5), and (6, 1) are:
A.
(4, 3)
B.
(4, 4)
C.
(3, 3)
D.
(5, 3)
Show solution
Solution
Centroid = ((2+4+6)/3, (3+5+1)/3) = (4, 3).
Correct Answer:
A
— (4, 3)
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Q. The coordinates of the centroid of the triangle with vertices (0, 0), (6, 0), and (3, 6) are:
A.
(3, 2)
B.
(2, 3)
C.
(3, 3)
D.
(0, 0)
Show solution
Solution
Centroid = ((0+6+3)/3, (0+0+6)/3) = (3, 2).
Correct Answer:
A
— (3, 2)
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Q. The coordinates of the centroid of the triangle with vertices (2, 3), (4, 5), and (6, 7) are:
A.
(4, 5)
B.
(3, 4)
C.
(5, 6)
D.
(6, 5)
Show solution
Solution
Centroid = ((2+4+6)/3, (3+5+7)/3) = (4, 5).
Correct Answer:
B
— (3, 4)
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Q. The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:
A.
x = 1, 3
B.
x = 0, 2
C.
x = 2, 4
D.
x = 1, 2
Show solution
Solution
Finding f'(x) = 3x^2 - 12x + 9 and solving gives critical points at x = 1 and x = 3.
Correct Answer:
A
— x = 1, 3
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Q. The dimensional formula for work is:
A.
[M^1 L^2 T^-2]
B.
[M^1 L^1 T^-1]
C.
[M^0 L^2 T^-1]
D.
[M^1 L^0 T^0]
Show solution
Solution
The dimensional formula for work is [M^1 L^2 T^-2].
Correct Answer:
A
— [M^1 L^2 T^-2]
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Q. The displacement of a simple harmonic oscillator is given by x(t) = A cos(ωt + φ). What is the maximum displacement?
Show solution
Solution
The maximum displacement in SHM is equal to the amplitude A.
Correct Answer:
A
— A
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Q. The distance from the point (1, 2) to the line 2x + 3y - 6 = 0 is:
Show solution
Solution
Distance = |2(1) + 3(2) - 6| / √(2² + 3²) = |2 + 6 - 6| / √13 = 2/√13.
Correct Answer:
B
— 2
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Q. The distance from the point (3, 4) to the line 2x + 3y - 6 = 0 is:
Show solution
Solution
Distance = |2(3) + 3(4) - 6| / √(2² + 3²) = |6 + 12 - 6| / √13 = 12/√13.
Correct Answer:
B
— 2
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Q. The eccentricity of an ellipse is defined as e = c/a. If a = 10 and c = 6, what is the eccentricity?
A.
0.6
B.
0.8
C.
0.4
D.
0.5
Show solution
Solution
Eccentricity e = c/a = 6/10 = 0.6.
Correct Answer:
B
— 0.8
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Q. The electric potential due to a uniformly charged sphere at a point outside the sphere is equivalent to that of?
A.
A point charge at the center
B.
A point charge at the surface
C.
A point charge at the edge
D.
A hollow sphere
Show solution
Solution
The electric potential outside a uniformly charged sphere is the same as that due to a point charge located at the center of the sphere.
Correct Answer:
A
— A point charge at the center
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Q. The energy of a simple harmonic oscillator is proportional to which of the following?
A.
Displacement
B.
Velocity
C.
Square of amplitude
D.
Frequency
Show solution
Solution
The total energy of a simple harmonic oscillator is proportional to the square of the amplitude.
Correct Answer:
C
— Square of amplitude
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Q. The enthalpy change for a reaction can be calculated using which of the following?
A.
Bond energies
B.
Standard enthalpies of formation
C.
Calorimetry
D.
All of the above
Show solution
Solution
The enthalpy change for a reaction can be calculated using bond energies, standard enthalpies of formation, and calorimetry.
Correct Answer:
D
— All of the above
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Q. The enthalpy change for the reaction A + B → C is +50 kJ/mol. What can be said about the reaction?
A.
It is exothermic
B.
It is endothermic
C.
It is spontaneous
D.
It is at equilibrium
Show solution
Solution
A positive enthalpy change indicates that the reaction absorbs heat, thus it is endothermic.
Correct Answer:
B
— It is endothermic
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Q. The enthalpy of vaporization of a substance is defined as:
A.
The heat required to melt the substance
B.
The heat required to convert a liquid into a gas
C.
The heat released during condensation
D.
The heat required to raise the temperature of a substance
Show solution
Solution
The enthalpy of vaporization is the heat required to convert a liquid into a gas at constant temperature and pressure.
Correct Answer:
B
— The heat required to convert a liquid into a gas
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Q. The enthalpy of vaporization of water is approximately ____ kJ/mol.
A.
40.79
B.
60.79
C.
80.79
D.
100.79
Show solution
Solution
The enthalpy of vaporization of water is approximately 40.79 kJ/mol.
Correct Answer:
A
— 40.79
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Q. The enthalpy of vaporization of water is approximately:
A.
40.79 kJ/mol
B.
2260 kJ/mol
C.
100 kJ/mol
D.
60 kJ/mol
Show solution
Solution
The enthalpy of vaporization of water is approximately 2260 kJ/mol.
Correct Answer:
B
— 2260 kJ/mol
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Q. The entropy change for a phase transition at constant temperature is given by:
A.
ΔS = ΔH/T
B.
ΔS = T/ΔH
C.
ΔS = ΔH*T
D.
ΔS = ΔH + T
Show solution
Solution
For a phase transition at constant temperature, the change in entropy is given by ΔS = ΔH/T, where ΔH is the enthalpy change.
Correct Answer:
A
— ΔS = ΔH/T
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Q. The entropy change for a reaction can be calculated using which of the following?
A.
ΔS = ΣS(products) - ΣS(reactants)
B.
ΔS = ΣS(reactants) - ΣS(products)
C.
ΔS = Q/T
D.
ΔS = W/T
Show solution
Solution
The change in entropy for a reaction is calculated using the formula ΔS = ΣS(products) - ΣS(reactants).
Correct Answer:
A
— ΔS = ΣS(products) - ΣS(reactants)
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Q. The entropy of a perfect crystal at absolute zero is given by:
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is zero.
Correct Answer:
B
— 0
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Q. The entropy of a perfect crystal at absolute zero is:
A.
Maximum
B.
Minimum
C.
Undefined
D.
Infinite
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is zero, which is the minimum value.
Correct Answer:
B
— Minimum
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Q. The entropy of a perfect crystal at absolute zero temperature is given by which law?
A.
Third law of thermodynamics
B.
First law of thermodynamics
C.
Second law of thermodynamics
D.
Zeroth law of thermodynamics
Show solution
Solution
The third law of thermodynamics states that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.
Correct Answer:
A
— Third law of thermodynamics
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Q. The entropy of a perfect crystal at absolute zero temperature is given by:
A.
0
B.
1
C.
Infinity
D.
Depends on the substance
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is exactly zero.
Correct Answer:
A
— 0
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Q. The entropy of a perfect crystalline substance at absolute zero is given by which law?
A.
Third law of thermodynamics
B.
First law of thermodynamics
C.
Second law of thermodynamics
D.
Zeroth law of thermodynamics
Show solution
Solution
The third law of thermodynamics states that the entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero.
Correct Answer:
A
— Third law of thermodynamics
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Q. The entropy of a perfect crystalline substance at absolute zero is:
A.
Zero
B.
Maximum
C.
Undefined
D.
Infinite
Show solution
Solution
According to the third law of thermodynamics, the entropy of a perfect crystalline substance at absolute zero is zero.
Correct Answer:
A
— Zero
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Q. The equation of a line parallel to y = 2x + 3 and passing through (1, 1) is?
A.
y = 2x - 1
B.
y = 2x + 1
C.
y = 2x + 3
D.
y = 2x - 3
Show solution
Solution
Parallel lines have the same slope. Using point-slope form: y - 1 = 2(x - 1) => y = 2x - 1.
Correct Answer:
A
— y = 2x - 1
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Q. The equation of a line passing through (1, 2) and (3, 6) is:
A.
y = 2x
B.
y = 3x - 1
C.
y = x + 1
D.
y = 4x - 2
Show solution
Solution
Slope = (6-2)/(3-1) = 2. Using point-slope form: y - 2 = 2(x - 1) => y = 2x.
Correct Answer:
A
— y = 2x
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