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

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

Solution

The force acting on the object is its weight, F = mg = 10 kg * 10 m/s² = 100 N.

Correct Answer: B — 20 N

Q. A 10 kg object is hanging at rest from a rope. What is the tension in the rope?

A. 0 N
B. 10 N
C. 100 N
D. 50 N

Solution

The tension in the rope must balance the weight of the object. T = mg = 10 kg * 10 m/s² = 100 N.

Correct Answer: C — 100 N

Q. A 10 kg object is hanging from a rope. What is the tension in the rope when the object is at rest?

A. 0 N
B. 10 N
C. 100 N
D. 50 N

Solution

At rest, the tension in the rope equals the weight of the object: T = mg = 10 kg * 10 m/s² = 100 N.

Correct Answer: C — 100 N

Q. A 10 kg object is lifted to a height of 10 m. How much work is done against gravity?

A. 0 J
B. 100 J
C. 200 J
D. 1000 J

Solution

Work done against gravity is equal to the change in potential energy, W = mgh = 10 * 9.8 * 10 = 980 J.

Correct Answer: D — 1000 J

Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity?

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

Solution

Work done = mass × g × height = 10 kg × 9.8 m/s² × 2 m = 196 J.

Correct Answer: B — 40 J

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

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

Solution

Work done = mgh = 10 * 9.8 * 5 = 490 J.

Correct Answer: B — 100 J

Q. A 10 kg object is lifted to a height of 5 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 = 10 kg * 9.8 m/s² * 5 m = 490 J.

Correct Answer: B — 100 J

Q. A 10 kg object is lifted to a height of 5 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 = mgh = 10 * 9.8 * 5 = 490 J.

Correct Answer: B — 100 J

Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 10 m/s. What is the centripetal force acting on the object?

A. 20 N
B. 10 N
C. 5 N
D. 50 N

Solution

Centripetal force F = mv²/r = 10 kg * (10 m/s)² / 5 m = 200 N.

Correct Answer: A — 20 N

Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 4 m/s. What is the centripetal force acting on the object?

A. 8 N
B. 16 N
C. 32 N
D. 40 N

Solution

Centripetal force F = mv²/r = 10 kg * (4 m/s)² / 5 m = 32 N.

Correct Answer: B — 16 N

Q. A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?

A. 45 J
B. 90 J
C. 135 J
D. 180 J

Solution

Work done = Change in Kinetic Energy = 0 - (0.5 × 10 kg × (3 m/s)²) = -45 J.

Correct Answer: B — 90 J

Q. A 10 kg object is moving with a speed of 3 m/s. What is its kinetic energy?

A. 45 J
B. 30 J
C. 60 J
D. 90 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 10 kg × (3 m/s)² = 45 J.

Correct Answer: A — 45 J

Q. A 10 kg object is moving with a speed of 5 m/s. What is its total mechanical energy?

A. 125 J
B. 250 J
C. 500 J
D. 1000 J

Solution

Total mechanical energy = KE + PE. KE = 0.5 * 10 * (5)² = 125 J. Assuming PE = 0, total energy = 125 J.

Correct Answer: B — 250 J

Q. A 10 kg object is moving with a velocity of 2 m/s. If it comes to rest, what is the change in momentum?

A. 20 kg·m/s
B. 10 kg·m/s
C. 5 kg·m/s
D. 0 kg·m/s

Solution

Initial momentum = mv = 10 kg * 2 m/s = 20 kg·m/s. Final momentum = 0. Change in momentum = 20 kg·m/s - 0 = 20 kg·m/s.

Correct Answer: A — 20 kg·m/s

Q. A 10 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

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

Solution

Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.

Correct Answer: C — 45 J

Q. A 10 kg object is moving with a velocity of 3 m/s. What is its momentum?

A. 10 kg·m/s
B. 30 kg·m/s
C. 3 kg·m/s
D. 0.3 kg·m/s

Solution

Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.

Correct Answer: B — 30 kg·m/s

Q. A 10 kg object is moving with a velocity of 3 m/s. What is the total mechanical energy if it is at a height of 5 m?

A. 150 J
B. 180 J
C. 300 J
D. 450 J

Solution

Total Mechanical Energy = Kinetic Energy + Potential Energy = 0.5 × 10 kg × (3 m/s)² + 10 kg × 9.8 m/s² × 5 m = 45 J + 490 J = 535 J.

Correct Answer: C — 300 J

Q. A 10 kg object is moving with a velocity of 4 m/s. What is its momentum?

A. 40 kg m/s
B. 20 kg m/s
C. 10 kg m/s
D. 30 kg m/s

Solution

Momentum (p) = m * v = 10 kg * 4 m/s = 40 kg m/s

Correct Answer: A — 40 kg m/s

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

A. 10 kg·m/s
B. 25 kg·m/s
C. 50 kg·m/s
D. 75 kg·m/s

Solution

Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.

Correct Answer: C — 50 kg·m/s

Q. A 10 kg object is pushed with a force of 30 N. If the frictional force is 10 N, what is the net force?

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

Solution

Net force = applied force - friction = 30 N - 10 N = 20 N.

Correct Answer: B — 20 N

Q. A 10 kg object is thrown upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 11.5 m
B. 22.5 m
C. 15.3 m
D. 10.0 m

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (15)²/(2*9.8) = 11.5 m.

Correct Answer: A — 11.5 m

Q. A 10 kg object is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 10 m/s²)

A. 11.25 m
B. 22.5 m
C. 15 m
D. 7.5 m

Solution

Using energy conservation: KE_initial = PE_max; 1/2 * m * v^2 = m * g * h; h = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m

Correct Answer: A — 11.25 m

Q. A 10 Ω resistor and a 20 Ω resistor are connected in series. What is the total resistance?

A. 10 Ω
B. 20 Ω
C. 30 Ω
D. 5 Ω

Solution

In series, the total resistance R = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.

Correct Answer: C — 30 Ω

Q. A 10-ohm resistor and a 20-ohm resistor are connected in series. What is the total resistance?

A. 10 ohms
B. 20 ohms
C. 30 ohms
D. 5 ohms

Solution

In series, the total resistance R = R1 + R2 = 10 + 20 = 30 ohms.

Correct Answer: C — 30 ohms

Q. A 10-ohm resistor has a voltage of 20 volts across it. What is the current flowing through the resistor?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Using Ohm's Law, I = V / R = 20 V / 10 Ω = 2 A.

Correct Answer: B — 2 A

Q. A 10-ohm resistor has a voltage of 50 volts across it. What is the current flowing through the resistor?

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

Solution

Using Ohm's Law, I = V / R = 50 V / 10 Ω = 5 A.

Correct Answer: B — 10 A

Q. A 10-ohm resistor is connected to a 20-volt battery. What is the power dissipated by the resistor?

A. 40 W
B. 20 W
C. 10 W
D. 2 W

Solution

Power (P) can be calculated using P = V^2 / R = 20^2 / 10 = 400 / 10 = 40 W.

Correct Answer: A — 40 W

Q. A 1000 kg car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the average power exerted by the engine?

A. 2000 W
B. 4000 W
C. 5000 W
D. 6000 W

Solution

First, calculate the work done: W = (1/2)mv^2 = (1/2)(1000 kg)(20 m/s)^2 = 200000 J. Then, power is P = W/t = 200000 J / 10 s = 20000 W.

Correct Answer: C — 5000 W

Q. A 1000 W heater operates for 1 hour. How much energy does it consume?

A. 3600 J
B. 1000 J
C. 3600000 J
D. 100000 J

Solution

Energy consumed is E = P * t. Here, P = 1000 W and t = 1 hour = 3600 seconds. Thus, E = 1000 W * 3600 s = 3600000 J.

Correct Answer: C — 3600000 J

Q. A 1000 W heater operates for 2 hours. How much energy does it consume?

A. 7200000 J
B. 2000000 J
C. 3600000 J
D. 1000000 J

Solution

Energy consumed can be calculated using E = P * t. Here, P = 1000 W and t = 2 hours = 7200 seconds. Thus, E = 1000 W * 7200 s = 7200000 J.

Correct Answer: C — 3600000 J

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