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. At what temperature does a reaction become spontaneous if ΔH = 50 kJ and ΔS = 0.1 kJ/K?
A.
500 K
B.
250 K
C.
1000 K
D.
200 K
Show solution
Solution
Set ΔG = 0: 0 = ΔH - TΔS; T = ΔH/ΔS = 50 kJ / 0.1 kJ/K = 500 K.
Correct Answer:
A
— 500 K
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Q. At what temperature does the Gibbs Free Energy change from negative to positive?
A.
At absolute zero
B.
At the melting point
C.
At the boiling point
D.
At the transition temperature
Show solution
Solution
The Gibbs Free Energy changes from negative to positive at the transition temperature, where the system shifts from one phase to another.
Correct Answer:
D
— At the transition temperature
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Q. At what temperature does the volume of a gas become zero according to Charles's Law?
A.
0 K
B.
-273.15 °C
C.
273.15 K
D.
None of the above
Show solution
Solution
According to Charles's Law, the volume of a gas approaches zero at absolute zero, which is -273.15 °C.
Correct Answer:
B
— -273.15 °C
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Q. At what temperature does the volume of a gas theoretically become zero?
A.
0°C
B.
0 K
C.
273 K
D.
100 K
Show solution
Solution
According to Charles's Law, the volume of a gas approaches zero at absolute zero, which is 0 K.
Correct Answer:
B
— 0 K
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Q. At what temperature will the RMS speed of a gas be 1000 m/s if its molar mass is 0.044 kg/mol? (R = 8.314 J/(mol K))
A.
500 K
B.
600 K
C.
700 K
D.
800 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), we solve for T: T = (v_rms^2 * M) / (3R) = (1000^2 * 0.044) / (3 * 8.314) = 700 K.
Correct Answer:
C
— 700 K
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Q. At what temperature will the RMS speed of a gas be 1000 m/s if its molar mass is 0.044 kg/mol?
A.
300 K
B.
400 K
C.
500 K
D.
600 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), we rearrange to find T = (v_rms^2 * M) / (3R). Plugging in values gives T approximately 500 K.
Correct Answer:
C
— 500 K
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Q. At what temperature will the RMS speed of a gas be 300 m/s if its molar mass is 28 g/mol?
A.
300 K
B.
600 K
C.
900 K
D.
1200 K
Show solution
Solution
Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T. Setting v_rms = 300 m/s and M = 28 g/mol, we find T = (M * v_rms^2)/(3R) = 600 K.
Correct Answer:
B
— 600 K
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Q. At what temperature will the RMS speed of a gas be 500 m/s if its molar mass is 0.02 kg/mol? (2000)
A.
250 K
B.
500 K
C.
1000 K
D.
2000 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), rearranging gives T = (v_rms^2 * M) / (3R). Substituting values gives T = 500 K.
Correct Answer:
B
— 500 K
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Q. At what temperature will the RMS speed of a gas be 600 m/s if its molar mass is 0.02 kg/mol?
A.
300 K
B.
600 K
C.
900 K
D.
1200 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Plugging in values gives T = (600^2 * 0.02) / (3 * 8.314) = 900 K.
Correct Answer:
C
— 900 K
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Q. Calculate the area between the curves y = x and y = x^2 from x = 0 to x = 1.
A.
0.25
B.
0.5
C.
0.75
D.
1
Show solution
Solution
The area is given by the integral from 0 to 1 of (x - x^2) dx. This evaluates to [x^2/2 - x^3/3] from 0 to 1 = (1/2 - 1/3) = 1/6 = 0.5.
Correct Answer:
B
— 0.5
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Q. Calculate the area between the curves y = x^2 and y = 2x from x = 0 to x = 2.
Show solution
Solution
The area is given by the integral from 0 to 2 of (2x - x^2) dx. This evaluates to [x^2 - x^3/3] from 0 to 2 = (4 - 8/3) = 4/3.
Correct Answer:
A
— 2
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Q. Calculate the area between the curves y = x^2 and y = 4 from x = 0 to x = 2.
Show solution
Solution
The area is given by the integral from 0 to 2 of (4 - x^2) dx. This evaluates to [4x - x^3/3] from 0 to 2 = (8 - 8/3) = 16/3.
Correct Answer:
A
— 4
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Q. Calculate the area under the curve y = cos(x) from x = 0 to x = π/2.
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Solution
The area under the curve y = cos(x) from x = 0 to x = π/2 is given by ∫(from 0 to π/2) cos(x) dx = [sin(x)] from 0 to π/2 = 1 - 0 = 1.
Correct Answer:
A
— 1
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Q. Calculate the area under the curve y = x^2 + 2x from x = 0 to x = 2.
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Solution
The area under the curve is given by ∫(from 0 to 2) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 2 = (8/3 + 4) = 20/3.
Correct Answer:
B
— 6
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Q. Calculate the area under the curve y = x^4 from x = 0 to x = 2.
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Solution
The area under the curve y = x^4 from x = 0 to x = 2 is given by ∫(from 0 to 2) x^4 dx = [x^5/5] from 0 to 2 = (32/5) - 0 = 32/5.
Correct Answer:
B
— 8
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Q. Calculate the derivative of f(x) = e^(2x).
A.
2e^(2x)
B.
e^(2x)
C.
2xe^(2x)
D.
e^(x)
Show solution
Solution
Using the chain rule, f'(x) = d/dx(e^(2x)) = 2e^(2x).
Correct Answer:
A
— 2e^(2x)
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Q. Calculate the derivative of f(x) = x^2 * e^x.
A.
(2x + x^2)e^x
B.
2xe^x
C.
x^2e^x
D.
(x^2 + 2x)e^x
Show solution
Solution
Using the product rule, f'(x) = d/dx(x^2 * e^x) = (x^2 + 2x)e^x.
Correct Answer:
D
— (x^2 + 2x)e^x
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Q. Calculate the determinant of the matrix [[1, 2], [3, 4]].
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Solution
Determinant = (1*4) - (2*3) = 4 - 6 = -2.
Correct Answer:
A
— -2
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Q. Calculate the determinant of the matrix \( B = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \).
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Solution
The determinant is calculated as \( 2*7 - 3*5 = 14 - 15 = -1 \).
Correct Answer:
D
— 10
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Q. Calculate the determinant of the matrix \( G = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 3 & 1 & 2 \end{pmatrix} \).
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Solution
The determinant is calculated as \( 2(0*2 - 1*1) - 1(1*2 - 3*1) + 3(1*1 - 3*0) = 0 \).
Correct Answer:
A
— -1
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Q. Calculate the determinant of the matrix \( \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \).
Show solution
Solution
The determinant is calculated as \( 2*4 - 3*1 = 8 - 3 = 5 \).
Correct Answer:
A
— 5
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Q. Calculate the determinant of the matrix \( \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \).
Show solution
Solution
The determinant is calculated as (2*7) - (3*5) = 14 - 15 = -1.
Correct Answer:
A
— 1
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Q. Calculate the determinant of the matrix: | 1 1 1 | | 2 2 2 | | 3 3 3 |
Show solution
Solution
The rows are linearly dependent, hence the determinant is 0.
Correct Answer:
A
— 0
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Q. Calculate the determinant \( \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 4 & 1 \end{vmatrix} \).
Show solution
Solution
The determinant is 0 because the rows are linearly dependent.
Correct Answer:
A
— 0
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Q. Calculate the determinant \( \begin{vmatrix} 2 & 3 \\ 5 & 7 \end{vmatrix} \)
Show solution
Solution
The determinant is \( 2*7 - 3*5 = 14 - 15 = -1 \).
Correct Answer:
A
— 1
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Q. Calculate the determinant \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \)
A.
ad - bc
B.
ab + cd
C.
ac - bd
D.
bc - ad
Show solution
Solution
The determinant is calculated as \( ad - bc \).
Correct Answer:
A
— ad - bc
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Q. Calculate the determinant | 1 0 0 | | 0 1 0 | | 0 0 1 |.
Show solution
Solution
The determinant of the identity matrix is 1.
Correct Answer:
B
— 1
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Q. Calculate the determinant | 2 3 | | 4 5 | + | 1 1 | | 1 1 |.
Show solution
Solution
The first determinant is -2 and the second is 0, so the total is -2 + 0 = -2.
Correct Answer:
B
— 1
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Q. Calculate the determinant: | 2 3 1 | | 1 0 2 | | 0 1 3 |.
Show solution
Solution
The determinant evaluates to 0 as the rows are linearly dependent.
Correct Answer:
A
— -1
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Q. Calculate the determinant: | 2 3 1 | | 1 0 4 | | 0 5 2 |.
Show solution
Solution
Using the determinant formula, we find that the determinant evaluates to 0.
Correct Answer:
A
— -1
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