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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.

Chemistry Syllabus (JEE Main) Mathematics Syllabus (JEE Main) Physics Syllabus (JEE Main)

Q. If the area of a rectangle is calculated as 50 m² with a length of 10 m and an uncertainty of ±0.1 m in length, what is the uncertainty in the area?

A. 1 m²
B. 0.5 m²
C. 0.2 m²
D. 0.1 m²

Solution

Uncertainty in area = 2 * length * uncertainty in length = 2 * 10 * 0.1 = 2 m².

Correct Answer: B — 0.5 m²

Q. If the area of a triangle is 30 square units and the base is 10 units, what is the height?

A. 3
B. 5
C. 6
D. 10

Solution

Area = 1/2 * base * height => height = 30/(1/2 * 10) = 6.

Correct Answer: B — 5

Q. If the area of triangle ABC is 30 cm² and the base BC = 10 cm, what is the height from A to BC?

A. 5 cm
B. 6 cm
C. 7 cm
D. 8 cm

Solution

Area = 1/2 * base * height. Therefore, 30 = 1/2 * 10 * height, height = 6 cm.

Correct Answer: B — 6 cm

Q. If the area of triangle ABC is 30 square units and the base BC = 10 units, what is the height from A to BC?

A. 5
B. 6
C. 7
D. 8

Solution

Area = 1/2 * base * height => 30 = 1/2 * 10 * height => height = 30 / 5 = 6.

Correct Answer: A — 5

Q. If the area of triangle ABC is 30 square units and the base BC is 10 units, what is the height from A to BC?

A. 3
B. 5
C. 6
D. 7

Solution

Area = 1/2 * base * height. Thus, 30 = 1/2 * 10 * height. Height = 6.

Correct Answer: B — 5

Q. If the area of triangle ABC is 60 cm² and the base BC = 12 cm, what is the height from A to BC?

A. 5 cm
B. 10 cm
C. 12 cm
D. 15 cm

Solution

Area = (1/2) * base * height. Therefore, 60 = (1/2) * 12 * height, height = 10 cm.

Correct Answer: B — 10 cm

Q. If the area of triangle JKL is 30 cm² and the base JK is 10 cm, what is the height from point L?

A. 3 cm
B. 6 cm
C. 5 cm
D. 4 cm

Solution

Area = 1/2 * base * height. 30 = 1/2 * 10 * height. Therefore, height = 6 cm.

Correct Answer: C — 5 cm

Q. If the azimuthal quantum number l = 1, what is the shape of the orbital?

A. Spherical
B. Dumbbell
C. Double dumbbell
D. None of the above

Solution

The azimuthal quantum number l = 1 corresponds to p orbitals, which have a dumbbell shape.

Correct Answer: B — Dumbbell

Q. If the balance length of a potentiometer is 50cm for a cell of unknown emf, and the potential gradient is 4 V/m, what is the emf of the cell?

A. 2V
B. 4V
C. 6V
D. 8V

Solution

The emf is calculated as V = potential gradient × length = 4 V/m × 0.5 m = 2 V.

Correct Answer: B — 4V

Q. If the balancing length of a potentiometer is found to be 40 cm for a cell of emf 2V, what is the potential gradient if the total length of the wire is 100 cm?

A. 5 V/m
B. 2 V/m
C. 4 V/m
D. 3 V/m

Solution

The potential gradient is V/L = 2V/0.4m = 5 V/m.

Correct Answer: A — 5 V/m

Q. If the center of a circle is at (0, 0) and it passes through the point (3, 4), what is the equation of the circle?

A. x² + y² = 25
B. x² + y² = 12
C. x² + y² = 7
D. x² + y² = 16

Solution

The radius is 5 (distance from (0,0) to (3,4)), so the equation is x² + y² = 5² = 25.

Correct Answer: A — x² + y² = 25

Q. If the center of a circle is at (0, 0) and it passes through the point (3, 4), what is the radius of the circle?

A. 5
B. 7
C. 6
D. 4

Solution

The radius is the distance from the center to the point, which is √(3² + 4²) = √25 = 5.

Correct Answer: A — 5

Q. If the charge density of a non-conducting sphere increases linearly with radius, how does the electric field vary inside the sphere?

A. Linearly with radius
B. Quadratically with radius
C. Constant
D. Inversely with radius

Solution

The electric field inside a non-conducting sphere with linearly increasing charge density varies linearly with radius.

Correct Answer: A — Linearly with radius

Q. If the charge density of a non-uniform spherical charge distribution varies as ρ(r) = kr², what is the electric field at the center of the sphere?

A. 0
B. k/3ε₀
C. k/4ε₀
D. k/2ε₀

Solution

At the center of a non-uniform spherical charge distribution, the electric field is zero due to symmetry.

Correct Answer: A — 0

Q. If the charge density of a spherical charge distribution increases linearly with radius, how does the electric field vary inside the sphere?

A. Linearly with radius
B. Quadratically with radius
C. Inversely with radius
D. Constant

Solution

The electric field inside a non-uniform charge distribution varies quadratically with radius due to the increasing charge density.

Correct Answer: B — Quadratically with radius

Q. If the charge inside a closed surface is doubled, what happens to the electric flux through the surface?

A. It doubles
B. It halves
C. It remains the same
D. It becomes zero

Solution

According to Gauss's law, the electric flux is directly proportional to the enclosed charge, so it doubles.

Correct Answer: A — It doubles

Q. If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

Capacitance C = Q/V; if Q is doubled and V remains constant, C must halve.

Correct Answer: B — It halves

Q. If the charge on a capacitor is doubled, what happens to the energy stored in it?

A. It doubles
B. It quadruples
C. It remains the same
D. It halves

Solution

The energy stored in a capacitor is given by U = 1/2 * C * V^2. If the charge is doubled, the energy stored quadruples.

Correct Answer: B — It quadruples

Q. If the circumference of a circle is measured as 31.4 cm with an error of 0.2 cm, what is the percentage error?

A. 0.64%
B. 0.5%
C. 1%
D. 0.2%

Solution

Percentage error = (Absolute error / Measured value) * 100 = (0.2 / 31.4) * 100 ≈ 0.64%.

Correct Answer: A — 0.64%

Q. If the circumference of a circle is measured as 31.4 cm with an error of 0.2 cm, what is the radius?

A. 5 cm
B. 4.5 cm
C. 5.0 cm
D. 6 cm

Solution

Circumference = 2πr. Measured circumference = 31.4 cm. Thus, r = 31.4 / (2π) ≈ 5.0 cm.

Correct Answer: C — 5.0 cm

Q. If the circumradius of triangle ABC is 10 cm and the area is 48 cm², what is the length of the side opposite to angle A?

A. 12 cm
B. 14 cm
C. 16 cm
D. 18 cm

Solution

Using the formula R = (abc)/(4 * Area), we can find the side opposite to angle A. Let a = side opposite to A. Then, a = (4 * Area * R) / (bc) = (4 * 48 * 10) / (b * c).

Correct Answer: B — 14 cm

Q. If the circumradius R of triangle ABC is 5 cm, what is the maximum area of the triangle?

A. 12.5 cm²
B. 15 cm²
C. 20 cm²
D. 25 cm²

Solution

The maximum area of a triangle with circumradius R is given by the formula Area = (abc)/(4R). For maximum area, the triangle should be equilateral, thus Area = (3√3/4) * (R^2) = (3√3/4) * (5^2) = 25√3/4 cm².

Correct Answer: C — 20 cm²

Q. If the concentration of reactants is increased in a system at equilibrium, what will happen to the position of equilibrium?

A. Shift to the right
B. Shift to the left
C. No change
D. Depends on temperature

Solution

According to Le Chatelier's principle, increasing the concentration of reactants will shift the equilibrium position to the right, favoring the formation of products.

Correct Answer: A — Shift to the right

Q. If the coordinates of points A and B are (1, 2) and (3, 4) respectively, what is the distance AB?

A. 2√2
B. √8
C. 2
D. 4

Solution

Distance AB = √((3-1)² + (4-2)²) = √(4) = 2.

Correct Answer: A — 2√2

Q. If the coordinates of points A and B are (1, 2) and (4, 6) respectively, what is the distance AB?

A. 5
B. 4
C. 3
D. 2

Solution

Distance AB = √((4-1)² + (6-2)²) = √(9 + 16) = √25 = 5.

Correct Answer: A — 5

Q. If the coordinates of the vertices of a triangle are (1, 1), (4, 5), and (7, 2), what is the perimeter of the triangle?

A. 12
B. 14
C. 16
D. 18

Solution

Calculate distances AB, BC, CA and sum them: AB = 5, BC = 5, CA = 7. Perimeter = 5 + 5 + 7 = 17.

Correct Answer: B — 14

Q. If the coordinates of the vertices of a triangle are (1, 2), (3, 4), and (5, 2), what is the perimeter of the triangle?

A. 8
B. 10
C. 12
D. 14

Solution

Perimeter = AB + BC + CA = √[(3-1)² + (4-2)²] + √[(5-3)² + (2-4)²] + √[(1-5)² + (2-2)²] = 2.83 + 2.83 + 4 = 10.

Correct Answer: B — 10

Q. If the coordinates of the vertices of a triangle are (1, 2), (4, 6), and (7, 2), what is the perimeter of the triangle?

A. 12
B. 14
C. 16
D. 18

Solution

Perimeter = AB + BC + CA = √[(4-1)² + (6-2)²] + √[(7-4)² + (2-6)²] + √[(1-7)² + (2-2)²] = 3 + √(9 + 16) + 6 = 3 + 5 + 6 = 14.

Correct Answer: B — 14

Q. If the coordinates of the vertices of a triangle are A(1, 1), B(4, 5), and C(7, 2), what is the area of the triangle?

A. 10
B. 12
C. 14
D. 16

Solution

Area = 1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | = 1/2 | 1(5-2) + 4(2-1) + 7(1-5) | = 1/2 | 3 + 4 - 28 | = 1/2 * 21 = 10.5.

Correct Answer: B — 12

Q. If the critical angle for a certain material is 30°, what is the refractive index of that material?

A. 1.00
B. 1.15
C. 1.73
D. 2.00

Solution

The refractive index n can be calculated using n = 1/sin(θc). For θc = 30°, n = 1/sin(30°) = 1/0.5 = 2.00.

Correct Answer: C — 1.73

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