Physics Syllabus (JEE Main)

Current Electricity Electrostatics Gravitation Kinematics Kinetic Theory of Gases Laws of Motion Magnetism & EMI Modern Physics Optics Oscillations & Waves Properties of Matter Rotational Motion Thermodynamics Units & Measurement Work, Energy & Power

Q. A 2 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 7 m/s
C. 5 m/s
D. 15 m/s

Solution

Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 5 m) = 10 m/s.

Correct Answer: A — 10 m/s

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

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

Solution

Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Correct Answer: B — 20 J

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

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

Solution

Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Correct Answer: B — 20 J

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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 19.6 J
B. 39.2 J
C. 49 J
D. 98 J

Solution

Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Correct Answer: B — 39.2 J

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Q. A 2 kg object is dropped from a height of 5 m. What is the work done by gravity on the object just before it hits the ground?

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

Solution

Work done by gravity = m × g × h = 2 kg × 9.8 m/s² × 5 m = 98 J.

Correct Answer: B — 20 J

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Q. A 2 kg object is in free fall. What is the net force acting on it?

A. 0 N
B. 2 N
C. 20 N
D. 19.6 N

Solution

The net force is equal to the weight of the object: F = mg = 2 kg * 9.8 m/s² = 19.6 N.

Correct Answer: C — 20 N

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Q. A 2 kg object is moving with a speed of 3 m/s. What is its kinetic energy?

A. 6 J
B. 9 J
C. 12 J
D. 18 J

Solution

Kinetic energy = 0.5 × m × v² = 0.5 × 2 kg × (3 m/s)² = 9 J.

Correct Answer: B — 9 J

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Q. A 2 kg object is moving with a speed of 3 m/s. What is its total mechanical energy?

A. 9 J
B. 12 J
C. 15 J
D. 18 J

Solution

Total mechanical energy = kinetic energy + potential energy. KE = 0.5mv² = 0.5 * 2 * (3)² = 9 J.

Correct Answer: B — 12 J

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Q. A 2 kg object is moving with a velocity of 3 m/s. What is its momentum?

A. 3 kg·m/s
B. 6 kg·m/s
C. 9 kg·m/s
D. 12 kg·m/s

Solution

Momentum = mass × velocity = 2 kg × 3 m/s = 6 kg·m/s.

Correct Answer: B — 6 kg·m/s

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Q. A 2 kg object is moving with a velocity of 5 m/s. If it collides elastically with a stationary object of mass 3 kg, what will be the final velocity of the 2 kg object?

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

Solution

Using conservation of momentum and kinetic energy, the final velocity of the 2 kg object after an elastic collision is 1 m/s.

Correct Answer: A — 1 m/s

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Q. A 2 kg object is pulled with a force of 8 N. What is the acceleration of the object?

A. 2 m/s²
B. 4 m/s²
C. 6 m/s²
D. 8 m/s²

Solution

Using F = ma, we find a = F/m = 8 N / 2 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A 2 kg object is raised to a height of 5 m. What is the potential energy gained?

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

Solution

Potential Energy (PE) = m * g * h = 2 kg * 9.8 m/s^2 * 5 m = 98 J

Correct Answer: B — 20 J

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Q. A 2 kg object is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the object?

A. 8 kg·m/s
B. 2 kg·m/s
C. 4 kg·m/s
D. 16 kg·m/s

Solution

Momentum p = mv = 2 kg * 4 m/s = 8 kg·m/s.

Correct Answer: A — 8 kg·m/s

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Q. A 2 kg object is sliding on a frictionless surface with a velocity of 5 m/s. What is the momentum of the object?

A. 10 kg·m/s
B. 5 kg·m/s
C. 2 kg·m/s
D. 1 kg·m/s

Solution

Momentum p = mv = 2 kg * 5 m/s = 10 kg·m/s.

Correct Answer: A — 10 kg·m/s

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Q. A 2 kg object is thrown 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: KE_initial = PE_max; 0.5 × 2 kg × (10 m/s)² = 2 kg × 9.8 m/s² × h; h = 5.1 m.

Correct Answer: A — 5 m

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Q. A 2 kg object is thrown upwards with a velocity of 10 m/s. What is its maximum height? (g = 9.8 m/s²)

A. 5.1 m
B. 10.2 m
C. 20.4 m
D. 15.3 m

Solution

Using the formula v² = u² - 2gh, where v = 0 at max height: 0 = (10)² - 2 * 9.8 * h, h = 5.1 m.

Correct Answer: A — 5.1 m

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Q. A 2 kg object is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

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

Solution

Using energy conservation: KE at bottom = PE at top; 1/2 mv^2 = mgh; h = v^2/(2g) = (10^2)/(2*10) = 5 m

Correct Answer: B — 10 m

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Q. A 20 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

A. 0 N
B. 20 N
C. 200 N
D. 10 N

Solution

The force acting on the object is its weight: F = mg = 20 kg * 10 m/s² = 200 N.

Correct Answer: C — 200 N

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Q. A 20 kg object is pulled with a force of 100 N. If the frictional force is 40 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

Net force = applied force - friction = 100 N - 40 N = 60 N. Acceleration a = F/m = 60 N / 20 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 2000 W heater operates for 3 hours. How much energy does it consume in kilowatt-hours?

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

Solution

Energy consumed in kilowatt-hours is calculated as E = P * t. Here, P = 2000 W = 2 kW and t = 3 hours. Thus, E = 2 kW * 3 h = 6 kWh.

Correct Answer: A — 6 kWh

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Q. A 25 kg object is pulled with a force of 100 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 = 100 N / 25 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 25 kg object is pushed with a force of 100 N. If the frictional force is 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

Net force = 100 N - 25 N = 75 N. Acceleration a = F/m = 75 N / 25 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 25 kg object is subjected to a force of 50 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, we find a = F/m = 50 N / 25 kg = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 5.1 m
B. 10.2 m
C. 15.3 m
D. 20.0 m

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (10)²/(2*9.8) = 5.1 m.

Correct Answer: A — 5.1 m

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Q. A 3 kg block is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the block's acceleration assuming no friction?

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

Solution

Acceleration a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: A — 2 m/s²

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Q. A 3 kg block is on a table and a horizontal force of 15 N is applied. If the frictional force is 5 N, what is the acceleration of the block?

A. 2 m/s²
B. 3 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration a = F/m = 10 N / 3 kg = 3.33 m/s².

Correct Answer: A — 2 m/s²

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Q. A 3 kg block is pulled with a force of 12 N. If the frictional force is 3 N, what is the acceleration of the block?

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

Solution

Net force = applied force - friction = 12 N - 3 N = 9 N. Acceleration a = F/m = 9 N / 3 kg = 3 m/s².

Correct Answer: B — 4 m/s²

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Q. A 3 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?

A. 3.9 m/s²
B. 4.9 m/s²
C. 9.8 m/s²
D. 1.5 m/s²

Solution

The acceleration down the incline is given by a = g * sin(θ) = 9.8 m/s² * sin(30°) = 4.9 m/s².

Correct Answer: A — 3.9 m/s²

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Q. A 3 kg block is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the block?

A. 12 kg·m/s
B. 8 kg·m/s
C. 6 kg·m/s
D. 4 kg·m/s

Solution

Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.

Correct Answer: A — 12 kg·m/s

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Q. A 3 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum force exerted by the spring when stretched?

A. 100 N
B. 200 N
C. 300 N
D. 600 N

Solution

The maximum force exerted by the spring is F = kx. Assuming maximum stretch x = 1 m, F = 200 N/m * 1 m = 200 N.

Correct Answer: B — 200 N

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Physics Syllabus (JEE Main) MCQ & Objective Questions

The Physics Syllabus for JEE Main is crucial for students aiming to excel in their exams. Understanding this syllabus not only helps in grasping fundamental concepts but also enhances problem-solving skills through practice. Engaging with MCQs and objective questions is essential for effective exam preparation, as it allows students to identify important questions and strengthen their knowledge base.

What You Will Practise Here

Mechanics: Laws of Motion, Work, Energy, and Power
Thermodynamics: Laws of Thermodynamics, Heat Transfer
Waves and Oscillations: Simple Harmonic Motion, Wave Properties
Electromagnetism: Electric Fields, Magnetic Fields, and Circuits
Optics: Reflection, Refraction, and Optical Instruments
Modern Physics: Quantum Theory, Atomic Models, and Nuclear Physics
Fluid Mechanics: Properties of Fluids, Bernoulli's Principle

Exam Relevance

The Physics Syllabus (JEE Main) is integral to various examinations, including CBSE, State Boards, and competitive exams like NEET and JEE. Questions often focus on conceptual understanding and application of theories. Common patterns include numerical problems, conceptual MCQs, and assertion-reason type questions, which test both knowledge and analytical skills.

Common Mistakes Students Make

Misinterpreting the question stem, leading to incorrect answers.
Neglecting units and dimensions in calculations.
Overlooking the significance of diagrams in understanding concepts.
Confusing similar concepts, such as velocity and acceleration.
Failing to apply formulas correctly in different contexts.

FAQs

Question: What are the key topics in the Physics Syllabus for JEE Main?
Answer: Key topics include Mechanics, Thermodynamics, Waves, Electromagnetism, Optics, Modern Physics, and Fluid Mechanics.

Question: How can I improve my performance in Physics MCQs?
Answer: Regular practice of MCQs, understanding concepts deeply, and revising important formulas can significantly enhance your performance.

Start solving practice MCQs today to test your understanding of the Physics Syllabus (JEE Main). This will not only boost your confidence but also prepare you effectively for your upcoming exams. Remember, consistent practice is the key to success!

Physics Syllabus (JEE Main)

Physics Syllabus (JEE Main) MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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