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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 satellite is in a circular orbit around the Earth. If it moves to a higher orbit, what happens to its potential energy?

A. It increases.
B. It decreases.
C. It remains constant.
D. It becomes zero.

Solution

As the satellite moves to a higher orbit, its gravitational potential energy increases due to the increase in distance from the Earth's center.

Correct Answer: A — It increases.

Q. A satellite is in a circular orbit around the Earth. If its orbital radius is 4R, what is the gravitational force acting on it compared to that at the surface of the Earth?

A. 1/4
B. 1/16
C. 1/8
D. 1/2

Solution

The gravitational force decreases with the square of the distance. At 4R, the force is 1/(4^2) = 1/16 of the force at the surface.

Correct Answer: B — 1/16

Q. A satellite is in a circular orbit around the Earth. If its orbital radius is tripled, how does the orbital speed change?

A. It triples
B. It doubles
C. It remains the same
D. It is reduced to one-third

Solution

Orbital speed v = √(GM/R). If R is tripled, v becomes √(GM/(3R)) = v/√3, which is reduced to one-third.

Correct Answer: D — It is reduced to one-third

Q. A satellite is in a circular orbit around the Earth. If its speed is doubled, what happens to the radius of its orbit?

A. It remains the same
B. It doubles
C. It increases by a factor of four
D. It decreases by a factor of four

Solution

If the speed of a satellite is doubled, the radius of its orbit decreases by a factor of four due to the relationship between speed and radius in circular motion.

Correct Answer: D — It decreases by a factor of four

Q. A satellite is in a circular orbit around the Earth. If its speed is doubled, what will happen to its orbital radius?

A. It will remain the same.
B. It will double.
C. It will increase by a factor of four.
D. It will decrease by a factor of four.

Solution

If the speed of a satellite is doubled, the orbital radius will decrease by a factor of four, as orbital speed is inversely proportional to the square root of the radius.

Correct Answer: D — It will decrease by a factor of four.

Q. A satellite is in a circular orbit around the Earth. If its speed is increased, what will happen to its orbit?

A. It will remain circular
B. It will become elliptical
C. It will crash into the Earth
D. It will escape Earth's gravity

Solution

Increasing the speed of a satellite in a circular orbit will cause its orbit to become elliptical.

Correct Answer: B — It will become elliptical

Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is 7000 km and the speed of the satellite is 7.9 km/s, what is the centripetal acceleration?

A. 7.9 m/s²
B. 9.8 m/s²
C. 11.2 m/s²
D. 14.0 m/s²

Solution

Centripetal acceleration a_c = v²/r = (7.9 km/s)² / (7000 km) = 0.00063 km/s² = 6.3 m/s².

Correct Answer: A — 7.9 m/s²

Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is increased, what happens to the orbital speed of the satellite?

A. Increases
B. Decreases
C. Remains the same
D. Depends on the mass of the satellite

Solution

Orbital speed (v) = √(GM/r). As r increases, v decreases.

Correct Answer: B — Decreases

Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is tripled, how does the orbital speed change?

A. It triples
B. It doubles
C. It remains the same
D. It decreases by a factor of √3

Solution

Orbital speed v = √(GM/r). If r is tripled, v decreases by a factor of √3.

Correct Answer: D — It decreases by a factor of √3

Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is halved, what happens to the gravitational force acting on the satellite?

A. It remains the same
B. It doubles
C. It quadruples
D. It decreases by half

Solution

The gravitational force is inversely proportional to the square of the distance; halving the radius increases the force by a factor of four.

Correct Answer: C — It quadruples

Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is doubled, what happens to the gravitational force acting on the satellite?

A. It doubles
B. It halves
C. It becomes four times
D. It becomes one-fourth

Solution

Gravitational force ∝ 1/r². If radius is doubled, force becomes 1/(2²) = 1/4.

Correct Answer: D — It becomes one-fourth

Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is 7000 km and the gravitational acceleration is 9.8 m/s², what is the speed of the satellite?

A. 5.5 km/s
B. 7.9 km/s
C. 9.8 km/s
D. 11.2 km/s

Solution

Speed (v) = √(g*r) = √(9.8 m/s² * 7000 * 1000 m) = 7.9 km/s.

Correct Answer: B — 7.9 km/s

Q. A satellite is in a circular orbit around the Earth. If the satellite's speed is increased, what will happen to its orbit?

A. It will remain circular
B. It will become elliptical
C. It will crash into the Earth
D. It will escape Earth's gravity

Solution

Increasing the speed of a satellite in a circular orbit will cause it to move into an elliptical orbit.

Correct Answer: B — It will become elliptical

Q. A satellite is in a circular orbit around the Earth. What happens to its gravitational potential energy as it moves to a higher orbit?

A. It increases.
B. It decreases.
C. It remains constant.
D. It becomes zero.

Solution

As the satellite moves to a higher orbit, its distance from the Earth increases, leading to an increase in gravitational potential energy.

Correct Answer: A — It increases.

Q. A satellite is in a circular orbit around the Earth. What is the angular momentum of the satellite if its mass is m, its orbital radius is r, and its orbital speed is v?

A. mv^2/r
B. mvr
C. mr^2
D. mv

Solution

Angular momentum L = mvr, where v is the orbital speed and r is the radius of the orbit.

Correct Answer: B — mvr

Q. A satellite is in a circular orbit around the Earth. What is the gravitational potential energy of the satellite?

A. It is positive.
B. It is negative.
C. It is zero.
D. It is constant.

Solution

The gravitational potential energy of a satellite in orbit is negative due to the attractive force of gravity.

Correct Answer: B — It is negative.

Q. A satellite is in a circular orbit around the Earth. What is the relationship between its orbital speed v and the radius r of the orbit?

A. v = sqrt(G * M / r)
B. v = G * M / r^2
C. v = r * sqrt(G / M)
D. v = G * r / M

Solution

The orbital speed of a satellite is given by v = sqrt(G * M / r), where M is the mass of the Earth.

Correct Answer: A — v = sqrt(G * M / r)

Q. A satellite is in a circular orbit around the Earth. What is the relationship between its orbital speed (v) and the radius of the orbit (r)?

A. v = sqrt(G * M / r)
B. v = G * M / r
C. v = r / G * M
D. v = G * r / M

Solution

The orbital speed of a satellite is given by v = sqrt(G * M / r), where M is the mass of the Earth.

Correct Answer: A — v = sqrt(G * M / r)

Q. A satellite is in a circular orbit at a height of 300 km above the Earth's surface. What is the approximate speed of the satellite?

A. 7.9 km/s
B. 5.0 km/s
C. 10.0 km/s
D. 3.5 km/s

Solution

The orbital speed can be calculated using the formula v = √(GM/r). For a height of 300 km, the speed is approximately 7.9 km/s.

Correct Answer: A — 7.9 km/s

Q. A satellite is in a polar orbit. What is the significance of this orbit?

A. It allows the satellite to cover the entire surface of the Earth.
B. It is the fastest orbit available.
C. It is used only for communication satellites.
D. It is the most stable orbit.

Solution

A polar orbit allows the satellite to pass over the entire surface of the Earth as the planet rotates beneath it.

Correct Answer: A — It allows the satellite to cover the entire surface of the Earth.

Q. A satellite orbits the Earth at a height of 500 km. What is the approximate gravitational acceleration experienced by the satellite?

A. 9.8 m/s²
B. 7.9 m/s²
C. 8.7 m/s²
D. 6.0 m/s²

Solution

The gravitational acceleration experienced by a satellite at a height of 500 km is approximately 8.7 m/s².

Correct Answer: C — 8.7 m/s²

Q. A simple harmonic oscillator has a frequency of 5 Hz. What is the time period of the oscillator?

A. 0.2 s
B. 0.5 s
C. 1 s
D. 2 s

Solution

The time period (T) is the reciprocal of frequency (f). T = 1/f = 1/5 = 0.2 s.

Correct Answer: A — 0.2 s

Q. A simple harmonic oscillator has a mass of 0.5 kg and a spring constant of 200 N/m. What is the angular frequency of the oscillator?

A. 10 rad/s
B. 20 rad/s
C. 5 rad/s
D. 15 rad/s

Solution

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 0.5 kg. Thus, ω = √(200/0.5) = √400 = 20 rad/s.

Correct Answer: A — 10 rad/s

Q. A simple harmonic oscillator has a mass of 2 kg and a spring constant of 200 N/m. What is the angular frequency of the oscillator?

A. 5 rad/s
B. 10 rad/s
C. 20 rad/s
D. 15 rad/s

Solution

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 2 kg. Thus, ω = √(200/2) = √100 = 10 rad/s.

Correct Answer: B — 10 rad/s

Q. A simple harmonic oscillator has a mass of 2 kg and a spring constant of 50 N/m. What is the angular frequency of the oscillator?

A. 5 rad/s
B. 10 rad/s
C. 15 rad/s
D. 20 rad/s

Solution

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 50 N/m and m = 2 kg. Thus, ω = √(50/2) = √25 = 5 rad/s.

Correct Answer: B — 10 rad/s

Q. A simple harmonic oscillator has a maximum displacement of 0.1 m and a maximum speed of 2 m/s. What is the angular frequency?

A. 10 rad/s
B. 20 rad/s
C. 5 rad/s
D. 15 rad/s

Solution

ω = V_max/A = 2/0.1 = 20 rad/s.

Correct Answer: A — 10 rad/s

Q. A simple harmonic oscillator has a maximum displacement of 0.1 m. What is the maximum potential energy if the spring constant is 200 N/m?

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

Solution

Maximum potential energy (PE) = (1/2)kA^2 = (1/2)(200)(0.1^2) = 1 J.

Correct Answer: B — 2 J

Q. A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg. What is its period?

A. 0.5 s
B. 1 s
C. 2 s
D. 4 s

Solution

T = 2π√(m/k) = 2π√(2/200) = 1 s.

Correct Answer: B — 1 s

Q. A simple harmonic oscillator has a total energy E. If the amplitude is halved, what will be the new total energy?

A. E/4
B. E/2
C. E
D. 2E

Solution

The total energy E is proportional to the square of the amplitude. If the amplitude is halved, the new energy will be E/4.

Correct Answer: A — E/4

Q. A simple harmonic oscillator has a total energy of 50 J and an amplitude of 10 cm. What is the spring constant?

A. 200 N/m
B. 500 N/m
C. 1000 N/m
D. 2000 N/m

Solution

Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.

Correct Answer: B — 500 N/m

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