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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. A 1000 W heater operates for 3 hours. How much energy does it consume?

A. 3000 J
B. 1080000 J
C. 3600000 J
D. 1000 J

Solution

Energy consumed is E = P * t. Here, P = 1000 W and t = 3 hours = 10800 seconds. Thus, E = 1000 W * 10800 s = 10800000 J.

Correct Answer: C — 3600000 J

Q. A 12 kg box is pulled with a force of 60 N at an angle of 30 degrees to the horizontal. What is the horizontal component of the force?

A. 30 N
B. 60 N
C. 51.96 N
D. 20 N

Solution

The horizontal component is F_horizontal = F * cos(θ) = 60 N * cos(30°) = 60 N * (√3/2) = 51.96 N.

Correct Answer: C — 51.96 N

Q. A 12 kg object is at rest on a horizontal surface. What is the normal force acting on it?

A. 0 N
B. 12 N
C. 120 N
D. 100 N

Solution

The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.

Correct Answer: C — 120 N

Q. A 12 kg object is in free fall. What is the force acting on it due to gravity?

A. 12 N
B. 24 N
C. 36 N
D. 48 N

Solution

The force due to gravity is F = mg = 12 kg * 9.8 m/s² = 117.6 N (approximately 120 N).

Correct Answer: B — 24 N

Q. A 12 kg object is moving with a velocity of 5 m/s. What is its kinetic energy?

A. 150 J
B. 120 J
C. 100 J
D. 75 J

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 12 kg * (5 m/s)² = 150 J.

Correct Answer: B — 120 J

Q. A 12 kg object is moving with a velocity of 5 m/s. What is its momentum?

A. 30 kg·m/s
B. 60 kg·m/s
C. 90 kg·m/s
D. 120 kg·m/s

Solution

Momentum p = mv = 12 kg * 5 m/s = 60 kg·m/s.

Correct Answer: B — 60 kg·m/s

Q. A 12 kg object is pulled with a force of 24 N. If the object experiences a frictional force of 4 N, what is its acceleration?

A. 1.67 m/s²
B. 2 m/s²
C. 2.5 m/s²
D. 3 m/s²

Solution

Net force = 24 N - 4 N = 20 N. Acceleration a = F/m = 20 N / 12 kg = 1.67 m/s².

Correct Answer: B — 2 m/s²

Q. A 15 kg box is pushed with a force of 30 N. If the frictional force opposing the motion is 10 N, what is the acceleration of the box?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Net force = applied force - friction = 30 N - 10 N = 20 N. Acceleration a = F/m = 20 N / 15 kg = 1.33 m/s².

Correct Answer: B — 2 m/s²

Q. A 15 kg box is pushed with a force of 45 N. What is the acceleration of the box? (Assume no friction)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, we have a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 15 kg box is pushed with a force of 60 N on a rough surface with a frictional force of 15 N. What is the box's acceleration?

A. 3 m/s²
B. 4 m/s²
C. 5 m/s²
D. 6 m/s²

Solution

Net force = 60 N - 15 N = 45 N. Acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 4 m/s²

Q. A 15 kg box is pushed with a force of 60 N. If the frictional force is 20 N, what is the acceleration of the box?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Net force = applied force - friction = 60 N - 20 N = 40 N. Acceleration a = F/m = 40 N / 15 kg = 2.67 m/s².

Correct Answer: B — 3 m/s²

Q. A 15 kg cart is pushed with a force of 60 N. If the frictional force opposing the motion is 15 N, what is the acceleration of the cart?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Net force = 60 N - 15 N = 45 N. Using F = ma, a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 15 kg object is at rest on a horizontal surface. What is the normal force acting on it?

A. 0 N
B. 15 N
C. 150 N
D. 75 N

Solution

Normal force = weight = mg = 15 kg * 10 m/s² = 150 N.

Correct Answer: C — 150 N

Q. A 15 kg object is at rest on a table. What is the force exerted by the table on the object?

A. 0 N
B. 15 N
C. 150 N
D. 10 N

Solution

The normal force equals the weight of the object: 15 kg * 10 m/s² = 150 N.

Correct Answer: C — 150 N

Q. A 15 kg object is in equilibrium. What is the net force acting on it?

A. 0 N
B. 15 N
C. 30 N
D. 45 N

Solution

In equilibrium, the net force acting on the object is 0 N.

Correct Answer: A — 0 N

Q. A 15 kg object is pushed with a force of 45 N. What is the acceleration of the object?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 15 kg object is subjected to a net force of 45 N. What is its acceleration?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: C — 3 m/s²

Q. A 2 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Using energy conservation: mgh = 0.5mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.

Correct Answer: A — 5 m

Q. A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.

Correct Answer: B — 10 m

Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 20.4 m
B. 30.4 m
C. 40.4 m
D. 50.4 m

Solution

Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.

Correct Answer: B — 30.4 m

Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

A. 10 m
B. 20 m
C. 30 m
D. 40 m

Solution

Using conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.

Correct Answer: B — 20 m

Q. A 2 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?

A. 4.9 m/s²
B. 9.8 m/s²
C. 3.9 m/s²
D. 1.96 m/s²

Solution

The acceleration a = g sin(θ) = 9.8 m/s² * sin(30°) = 9.8 m/s² * 0.5 = 4.9 m/s².

Correct Answer: A — 4.9 m/s²

Q. A 2 kg block is sliding on a frictionless surface. If a force of 6 N is applied, what is its acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 6 m/s²

Solution

Using F = ma, acceleration a = F/m = 6 N / 2 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 2 kg object is dropped from a height of 10 m. What is its potential energy at the top?

A. 20 J
B. 40 J
C. 60 J
D. 80 J

Solution

Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 10 m = 196 J.

Correct Answer: B — 40 J

Q. A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 30 m/s

Solution

Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: B — 14 m/s

Q. A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 17.15 m/s
B. 12.25 m/s
C. 14.14 m/s
D. 10.0 m/s

Solution

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 15) = 17.15 m/s.

Correct Answer: A — 17.15 m/s

Q. A 2 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 39.2 J
B. 196 J
C. 78.4 J
D. 98.0 J

Solution

Potential Energy (PE) = m × g × h = 2 kg × 9.8 m/s² × 20 m = 392 J.

Correct Answer: B — 196 J

Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 14 m/s
B. 19.8 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.

Correct Answer: B — 19.8 m/s

Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (Assume g = 10 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 40 m/s

Solution

Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 10 m/s² × 20 m) = 20 m/s.

Correct Answer: C — 20 m/s

Q. A 2 kg object is dropped from a height of 5 m. What is its potential energy at the top?

A. 10 J
B. 20 J
C. 30 J
D. 40 J

Solution

Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 5 m = 98 J.

Correct Answer: B — 20 J

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