JEE Main Preparation Resources for Competitive Exams

My Learning Cart

JEE Main

Download Q&A

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 5 kg object is at rest on a surface. If a horizontal force of 15 N is applied, what is the object's acceleration? (Assume friction is negligible)

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

Solution

Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 50 J
C. 100 J
D. 200 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.

Correct Answer: C — 100 J

Q. A 5 kg object is dropped from a height of 20 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 980 J
B. 490 J
C. 196 J
D. 9800 J

Solution

Potential energy is given by PE = mgh. Here, PE = 5 * 9.8 * 20 = 980 J.

Correct Answer: A — 980 J

Q. A 5 kg object is dropped from a height of 20 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 100 J
C. 200 J
D. 500 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 20 = 980 J. Just before hitting the ground, total energy = 980 J.

Correct Answer: C — 200 J

Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (Assume g = 9.8 m/s²)

A. 5 N
B. 9.8 N
C. 49 N
D. 0 N

Solution

The force acting on the object is its weight, F = mg = 5 kg * 9.8 m/s² = 49 N.

Correct Answer: C — 49 N

Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

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

Solution

The only force acting on it is its weight, W = mg = 5 kg * 10 m/s² = 50 N.

Correct Answer: B — 10 N

Q. A 5 kg object is lifted to a height of 10 m. What is the gravitational potential energy gained?

A. 50 J
B. 100 J
C. 200 J
D. 500 J

Solution

Potential Energy (PE) = m * g * h = 5 kg * 10 m/s^2 * 10 m = 500 J

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained?

A. 50 J
B. 100 J
C. 200 J
D. 500 J

Solution

Potential Energy (PE) = m * g * h = 5 kg * 9.8 m/s^2 * 10 m = 490 J (approx. 500 J)

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: B — 100 J

Q. A 5 kg object is lifted to a height of 10 m. What is the work done against gravity?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 10 = 490 J.

Correct Answer: D — 200 J

Q. A 5 kg object is lifted to a height of 3 m. What is the work done against gravity?

A. 15 J
B. 30 J
C. 45 J
D. 60 J

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 3 = 147 J.

Correct Answer: B — 30 J

Q. A 5 kg object is moving with a velocity of 10 m/s. What is its kinetic energy?

A. 25 J
B. 50 J
C. 100 J
D. 250 J

Solution

Kinetic energy (KE) = 0.5 * mass * velocity² = 0.5 * 5 * (10)² = 250 J.

Correct Answer: C — 100 J

Q. A 5 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

A. 22.5 J
B. 30 J
C. 45 J
D. 60 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 5 kg × (3 m/s)² = 22.5 J.

Correct Answer: A — 22.5 J

Q. A 5 kg object is pulled with a force of 25 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 = 25 N / 5 kg = 5 m/s².

Correct Answer: C — 4 m/s²

Q. A 5 kg object is pushed with a force of 15 N. If the frictional force is 5 N, what is the acceleration of the object?

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

Solution

Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 2 m/s²

Q. A 5 kg object is subjected to a net force of 15 N. What is its acceleration?

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

Solution

Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 5 kg object is subjected to a net force of 15 N. What is the object's acceleration?

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

Solution

Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 5 kg object is subjected to a net force of 25 N. What is its acceleration?

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²

Solution

Using F = ma, a = F/m = 25 N / 5 kg = 5 m/s².

Correct Answer: C — 10 m/s²

Q. A 5 kg object is thrown vertically upward with a speed of 20 m/s. What is the maximum height reached by the object? (g = 10 m/s²)

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Solution

Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.

Correct Answer: B — 30 m

Q. A 5 ohm resistor and a 10 ohm resistor are connected in series. What is the total resistance?

A. 15 ohms
B. 5 ohms
C. 10 ohms
D. 2 ohms

Solution

In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.

Correct Answer: A — 15 ohms

Q. A 50 kg crate is at rest on a flat surface. What is the normal force acting on the crate?

A. 0 N
B. 50 N
C. 500 N
D. 1000 N

Solution

The normal force equals the weight of the crate, which is N = mg = 50 kg * 10 m/s² = 500 N.

Correct Answer: C — 500 N

Q. A 50 kg object is at rest on a surface. What is the normal force acting on it?

A. 0 N
B. 50 N
C. 100 N
D. 500 N

Solution

The normal force equals the weight of the object, which is F = mg = 50 kg * 9.8 m/s² = 490 N.

Correct Answer: C — 100 N

Q. A 50 kg object is in free fall. What is the net force acting on it?

A. 50 N
B. 100 N
C. 200 N
D. 500 N

Solution

The net force is equal to the weight of the object, F = mg = 50 kg * 9.8 m/s² = 490 N (approximately 500 N).

Correct Answer: B — 100 N

Q. A 50 kg object is moving with a constant velocity. What can be said about the net force acting on it?

A. It is zero
B. It is equal to its weight
C. It is equal to the applied force
D. It is maximum

Solution

If the object is moving with constant velocity, the net force acting on it is zero.

Correct Answer: A — It is zero

Q. A 50 kg object is moving with a velocity of 10 m/s. What is its momentum?

A. 500 kg m/s
B. 1000 kg m/s
C. 1500 kg m/s
D. 2000 kg m/s

Solution

Momentum = mass × velocity = 50 kg × 10 m/s = 500 kg m/s.

Correct Answer: B — 1000 kg m/s

Q. A 50 kg object is pulled with a force of 200 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, we have a = F/m = 200 N / 50 kg = 4 m/s².

Correct Answer: D — 5 m/s²

Q. A 50 kg object is pushed with a force of 200 N over a distance of 10 m. What is the work done? (2000)

A. 2000 J
B. 1000 J
C. 500 J
D. 3000 J

Solution

Work Done = Force * Distance = 200 N * 10 m = 2000 J

Correct Answer: A — 2000 J

Q. A 50 kg person jumps off a diving board with a speed of 5 m/s. What is the momentum of the person just before hitting the water?

A. 100 kg·m/s
B. 200 kg·m/s
C. 250 kg·m/s
D. 300 kg·m/s

Solution

Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.

Correct Answer: A — 100 kg·m/s

Q. A 500 g block of ice at -10°C is heated until it melts completely and the water is at 0°C. How much heat is required? (Specific heat of ice = 2.1 J/g°C, Latent heat of fusion = 334 J/g) (2000)

A. 1050 J
B. 1340 J
C. 1500 J
D. 2000 J

Solution

Heat to raise ice to 0°C = 500g * 2.1 J/g°C * 10°C = 10500 J. Heat to melt = 500g * 334 J/g = 167000 J. Total = 1050 J + 167000 J = 1340 J.

Correct Answer: B — 1340 J

Q. A 6 kg object is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the acceleration assuming no friction?

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

Solution

Using F = ma, acceleration a = F/m = 12 N / 6 kg = 2 m/s².

Correct Answer: B — 2 m/s²

|<
<
3
4
5
6
7
8
9
10
11
>
>|

Showing 181 to 210 of 10700 (357 Pages)

School

College

Degree

Competitve

Skills

Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely