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 vapor pressure of a solution containing 1 mole of solute in 3 moles of solvent, assuming ideal behavior?
A.
0.25 P0
B.
0.75 P0
C.
0.5 P0
D.
P0
Show solution
Solution
Using Raoult's law, the vapor pressure of the solution = (moles of solvent / total moles) * P0 = (3 / (3 + 1)) * P0 = 0.75 P0.
Correct Answer:
B
— 0.75 P0
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Q. What is the vapor pressure of a solution containing a non-volatile solute compared to the pure solvent?
A.
Higher than the pure solvent
B.
Lower than the pure solvent
C.
Equal to the pure solvent
D.
Unpredictable
Show solution
Solution
The vapor pressure of a solution containing a non-volatile solute is lower than that of the pure solvent due to the presence of solute particles.
Correct Answer:
B
— Lower than the pure solvent
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Q. What is the variance of the data set {4, 8, 6, 5, 3}?
Show solution
Solution
Mean = 5.2; Variance = [(4-5.2)² + (8-5.2)² + (6-5.2)² + (5-5.2)² + (3-5.2)²] / 5 = 2.56.
Correct Answer:
A
— 2.5
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Q. What is the variance of the data set: 1, 2, 3, 4, 5?
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Solution
Mean = 3. Variance = [(1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)²] / 5 = 2.
Correct Answer:
B
— 1.5
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Q. What is the variance of the data set: 2, 4, 4, 4, 5, 5, 7, 9?
A.
2.5
B.
3.5
C.
4.5
D.
5.5
Show solution
Solution
Mean = 5.0; Variance = [(2-5)^2 + (4-5)^2 + (4-5)^2 + (4-5)^2 + (5-5)^2 + (5-5)^2 + (7-5)^2 + (9-5)^2] / 8 = 2.5.
Correct Answer:
A
— 2.5
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Q. What is the variance of the following data set: {5, 7, 8, 9, 10}?
Show solution
Solution
Mean = 7.8, Variance = [(5-7.8)² + (7-7.8)² + (8-7.8)² + (9-7.8)² + (10-7.8)²] / 5 = 3.
Correct Answer:
B
— 3
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Q. What is the vertex of the parabola defined by the equation y = -2(x - 1)^2 + 4?
A.
(1, 4)
B.
(1, -4)
C.
(4, 1)
D.
(-1, 4)
Show solution
Solution
The vertex form of a parabola is y = a(x - h)^2 + k. Here, h = 1 and k = 4, so the vertex is (1, 4).
Correct Answer:
A
— (1, 4)
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Q. What is the vertex of the parabola given by the equation y = -2(x - 1)^2 + 4?
A.
(1, 4)
B.
(1, -4)
C.
(4, 1)
D.
(-1, 4)
Show solution
Solution
The vertex form of a parabola is y = a(x - h)^2 + k. Here, h = 1 and k = 4, so the vertex is (1, 4).
Correct Answer:
A
— (1, 4)
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Q. What is the vertex of the parabola represented by the equation y = -2(x - 1)^2 + 4?
A.
(1, 4)
B.
(1, -4)
C.
(-1, 4)
D.
(-1, -4)
Show solution
Solution
The vertex form of a parabola is y = a(x - h)^2 + k. Here, h = 1 and k = 4, so the vertex is (1, 4).
Correct Answer:
A
— (1, 4)
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Q. What is the vertex of the parabola represented by the equation y = 2x^2 - 8x + 5?
A.
(2, -3)
B.
(2, -7)
C.
(4, -3)
D.
(4, -7)
Show solution
Solution
The vertex can be found using x = -b/(2a) = 4. Substituting x = 4 into the equation gives y = -3.
Correct Answer:
A
— (2, -3)
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Q. What is the viscosity of water at 20°C?
A.
0.001 Pa·s
B.
0.01 Pa·s
C.
0.1 Pa·s
D.
1 Pa·s
Show solution
Solution
The viscosity of water at 20°C is approximately 0.001 Pa·s.
Correct Answer:
A
— 0.001 Pa·s
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Q. What is the voltage across a 10 ohm resistor carrying a current of 0.5 A?
A.
5 V
B.
10 V
C.
15 V
D.
20 V
Show solution
Solution
Using Ohm's law, V = I * R = 0.5 A * 10 Ω = 5 V.
Correct Answer:
A
— 5 V
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Q. What is the voltage drop across a 3 ohm resistor carrying a current of 2 A?
A.
3 V
B.
6 V
C.
9 V
D.
12 V
Show solution
Solution
Using Ohm's law, V = I * R = 2 A * 3 ohms = 6 V.
Correct Answer:
B
— 6 V
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Q. What is the voltage drop across a 5 ohm resistor carrying a current of 2 A?
A.
5 V
B.
10 V
C.
15 V
D.
20 V
Show solution
Solution
Using Ohm's law, V = I * R = 2 A * 5 ohms = 10 V.
Correct Answer:
B
— 10 V
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Q. What is the voltage drop across a 5 ohm resistor carrying a current of 3 A?
A.
15 V
B.
10 V
C.
5 V
D.
20 V
Show solution
Solution
Using Ohm's law, V = I * R = 3 A * 5Ω = 15 V.
Correct Answer:
A
— 15 V
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Q. What is the voltage drop across a 5Ω resistor carrying a current of 3A?
A.
5V
B.
10V
C.
15V
D.
20V
Show solution
Solution
Using Ohm's law, V = I * R = 3A * 5Ω = 15V.
Correct Answer:
C
— 15V
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Q. What is the volume occupied by 1 mole of an ideal gas at STP (Standard Temperature and Pressure)?
A.
22.4 L
B.
24.0 L
C.
18.0 L
D.
20.0 L
Show solution
Solution
At STP, 1 mole of an ideal gas occupies 22.4 liters.
Correct Answer:
A
— 22.4 L
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Q. What is the volume occupied by 1 mole of an ideal gas at STP?
A.
22.4 L
B.
24 L
C.
20 L
D.
18 L
Show solution
Solution
At STP, 1 mole of an ideal gas occupies 22.4 liters.
Correct Answer:
A
— 22.4 L
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Q. What is the volume occupied by 4 moles of an ideal gas at STP?
A.
22.4 L
B.
44.8 L
C.
67.2 L
D.
89.6 L
Show solution
Solution
At STP, 1 mole of gas occupies 22.4 L. Therefore, 4 moles occupy 4 x 22.4 L = 89.6 L.
Correct Answer:
B
— 44.8 L
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Q. What is the volume of 1 liter in cubic meters?
A.
0.001
B.
0.01
C.
1
D.
1000
Show solution
Solution
1 liter is equal to 0.001 cubic meters.
Correct Answer:
A
— 0.001
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Q. What is the volume of 1 M NaOH solution required to obtain 0.5 moles of NaOH?
A.
0.5 L
B.
1 L
C.
2 L
D.
0.25 L
Show solution
Solution
Using the formula M = moles/volume, Volume = moles/M = 0.5 moles / 1 M = 0.5 L.
Correct Answer:
A
— 0.5 L
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Q. What is the volume of 1 mole of an ideal gas at STP (Standard Temperature and Pressure)?
A.
22.4 L
B.
24.0 L
C.
18.0 L
D.
20.0 L
Show solution
Solution
At STP, 1 mole of an ideal gas occupies 22.4 liters.
Correct Answer:
A
— 22.4 L
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Q. What is the volume of 1 mole of an ideal gas at STP?
A.
22.4 L
B.
24 L
C.
10 L
D.
1 L
Show solution
Solution
At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 liters.
Correct Answer:
A
— 22.4 L
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Q. What is the volume of 1 mole of gas at STP?
A.
22.4 L
B.
24 L
C.
20 L
D.
18 L
Show solution
Solution
At STP, 1 mole of gas occupies 22.4 liters.
Correct Answer:
A
— 22.4 L
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Q. What is the volume of 4 moles of an ideal gas at STP?
A.
22.4 L
B.
44.8 L
C.
89.6 L
D.
112 L
Show solution
Solution
Volume = moles x volume per mole = 4 moles x 22.4 L/mole = 89.6 L.
Correct Answer:
C
— 89.6 L
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Q. What is the volume of a 2 M solution that contains 4 moles of solute?
A.
2 L
B.
4 L
C.
1 L
D.
0.5 L
Show solution
Solution
Volume = moles of solute / molarity = 4 moles / 2 M = 2 L.
Correct Answer:
B
— 4 L
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Q. What is the volume of a cube with a side length of 2 meters?
A.
4 m³
B.
6 m³
C.
8 m³
D.
10 m³
Show solution
Solution
Volume = side³ = 2³ = 8 m³.
Correct Answer:
C
— 8 m³
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Q. What is the volume of a cube with a side length of 3 cm?
A.
9 cm³
B.
18 cm³
C.
27 cm³
D.
36 cm³
Show solution
Solution
The volume of a cube is given by side³, so 3 cm³ = 27 cm³.
Correct Answer:
C
— 27 cm³
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Q. What is the volume of CO2 produced at STP when 2 moles of C2H5OH are completely combusted?
A.
22.4 L
B.
44.8 L
C.
67.2 L
D.
89.6 L
Show solution
Solution
C2H5OH + 3O2 → 2CO2 + 3H2O. 2 moles of C2H5OH produce 4 moles of CO2. Volume = 4 * 22.4 L = 89.6 L.
Correct Answer:
B
— 44.8 L
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Q. What is the volume percent concentration of a solution containing 30 mL of ethanol in 150 mL of solution?
A.
20%
B.
25%
C.
30%
D.
15%
Show solution
Solution
Volume percent = (volume of solute / total volume) x 100 = (30 mL / 150 mL) x 100 = 20%.
Correct Answer:
B
— 25%
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