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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 disk and a ring of the same mass and radius are rolling without slipping. Which one will reach the bottom of an incline first?

A. Disk
B. Ring
C. Both will reach at the same time
D. Depends on the angle of incline

Solution

The disk will reach the bottom first because it has a smaller moment of inertia compared to the ring.

Correct Answer: A — Disk

Q. A disk is rotating with an angular velocity of 10 rad/s. If it experiences a constant angular acceleration of 2 rad/s², what will be its angular velocity after 5 seconds?

A. 20 rad/s
B. 10 rad/s
C. 30 rad/s
D. 15 rad/s

Solution

Using the formula ω = ω₀ + αt, we have ω = 10 + 2*5 = 20 rad/s.

Correct Answer: A — 20 rad/s

Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is the angular momentum of the disk?

A. (1/2)MR^2ω
B. MR^2ω
C. (1/4)MR^2ω
D. (3/2)MR^2ω

Solution

Angular momentum L = Iω = (1/2)MR^2ω, where I = (1/2)MR^2 for a disk.

Correct Answer: A — (1/2)MR^2ω

Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is its angular momentum?

A. (1/2)MR^2ω
B. MR^2ω
C. Mω
D. (1/4)MR^2ω

Solution

Angular momentum L = Iω, where I = (1/2)MR^2 for a disk.

Correct Answer: A — (1/2)MR^2ω

Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is its kinetic energy?

A. (1/2)Mω^2R^2
B. (1/2)Iω^2
C. (1/2)Mω^2
D. Mω^2R

Solution

The moment of inertia I of a disk about its axis is (1/2)MR^2. Therefore, the kinetic energy K.E. = (1/2)Iω^2 = (1/2)(1/2)MR^2ω^2 = (1/4)MR^2ω^2.

Correct Answer: B — (1/2)Iω^2

Q. A disk rolls down a slope of height h. What fraction of its total energy is translational at the bottom?

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

Solution

At the bottom, the total energy is converted into translational and rotational energy. The translational energy is 2/3 of the total energy.

Correct Answer: C — 2/3

Q. A disk rolls down a slope of height h. What is the ratio of translational to rotational kinetic energy at the bottom?

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

Solution

At the bottom, the total kinetic energy is split equally between translational and rotational for a disk, hence the ratio is 1:1.

Correct Answer: A — 1:1

Q. A disk rolls down an incline. If the height of the incline is h, what is the speed of the disk at the bottom assuming no energy losses?

A. √(gh)
B. √(2gh)
C. √(3gh)
D. √(4gh)

Solution

Using conservation of energy, potential energy at height h converts to kinetic energy at the bottom. The speed is √(2gh).

Correct Answer: B — √(2gh)

Q. A disk rolls without slipping on a horizontal surface. If its radius is R and it rolls with a linear speed v, what is the angular speed of the disk?

A. v/R
B. R/v
C. vR
D. v^2/R

Solution

The relationship between linear speed and angular speed for rolling without slipping is given by ω = v/R.

Correct Answer: A — v/R

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled while keeping the mass constant, what will be the new angular momentum?

A. 2Iω
B. 4Iω
C. Iω
D. I(2ω)

Solution

The moment of inertia I of a disk is proportional to r^2, so if the radius is doubled, I becomes 4I. Thus, angular momentum L = Iω becomes 4Iω.

Correct Answer: B — 4Iω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled while keeping the mass constant, what will be the new moment of inertia?

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

Solution

The moment of inertia of a disk is I = (1/2)MR^2. If the radius is doubled, the new moment of inertia becomes I' = (1/2)M(2R)^2 = 4I.

Correct Answer: B — 4I

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular momentum if the mass remains the same?

A. 2ω
B. 4ω
C. ω
D. ω/2

Solution

Angular momentum L = Iω. If the radius is doubled, the moment of inertia increases by a factor of 4, thus L = 4Iω.

Correct Answer: B — 4ω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular momentum?

A. 2Iω
B. 4Iω
C. Iω
D. I(2ω)

Solution

Angular momentum L = Iω, and if the radius is doubled, the moment of inertia I becomes 4I, thus L = 4Iω.

Correct Answer: B — 4Iω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular velocity to maintain the same linear velocity at the edge?

A. ω/2
B. ω
C. 2ω
D. 4ω

Solution

The linear velocity v = rω. If the radius is doubled, to maintain the same v, the angular velocity must remain ω.

Correct Answer: B — ω

Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular velocity to conserve angular momentum?

A. ω
B. 2ω
C. ω/2
D. ω/4

Solution

To conserve angular momentum, if the radius is doubled, the angular velocity must be halved.

Correct Answer: C — ω/2

Q. A distance is measured as 100 m with an error of 2 m. What is the absolute error?

A. 2 m
B. 0.02 m
C. 0.2 m
D. 20 m

Solution

The absolute error is given as 2 m.

Correct Answer: A — 2 m

Q. A diverging lens has a focal length of -10 cm. What is the image distance when the object is placed at 5 cm?

A. -3.33 cm
B. -5 cm
C. 3.33 cm
D. 5 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we have 1/(-10) = 1/v - 1/5. Solving gives v = -3.33 cm.

Correct Answer: A — -3.33 cm

Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 0.8 m, what is the torque about the hinges?

A. 8 Nm
B. 10 Nm
C. 16 Nm
D. 20 Nm

Solution

Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.

Correct Answer: C — 16 Nm

Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 1 m, what is the torque about the hinges?

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

Solution

Torque (τ) = F × d = 20 N × 1 m = 20 Nm.

Correct Answer: B — 20 Nm

Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1.2 m, what is the torque about the hinges?

A. 60 Nm
B. 50 Nm
C. 70 Nm
D. 40 Nm

Solution

Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.

Correct Answer: A — 60 Nm

Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1 m, what is the torque about the hinges?

A. 25 Nm
B. 50 Nm
C. 75 Nm
D. 100 Nm

Solution

Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.

Correct Answer: B — 50 Nm

Q. A double convex lens has a focal length of 10 cm. If it is made of a material with a refractive index of 1.5, what is the radius of curvature of each surface assuming they are equal?

A. 15 cm
B. 20 cm
C. 25 cm
D. 30 cm

Solution

Using the lens maker's formula, R = 2f(n-1) = 2*10*(1.5-1) = 20 cm.

Correct Answer: B — 20 cm

Q. A double convex lens has a radius of curvature of 20 cm on both sides. What is its focal length if the refractive index is 1.5?

A. 10 cm
B. 15 cm
C. 20 cm
D. 25 cm

Solution

Using the lens maker's formula, f = R/(n-1), we find f = 20/(1.5-1) = 40 cm.

Correct Answer: A — 10 cm

Q. A drop of liquid is in equilibrium on a surface. What is the condition for the drop to remain in equilibrium?

A. Weight equals surface tension
B. Weight equals gravitational force
C. Surface tension equals gravitational force
D. Surface tension equals buoyant force

Solution

For a drop to remain in equilibrium, the upward force due to surface tension must balance the downward gravitational force.

Correct Answer: C — Surface tension equals gravitational force

Q. A drop of liquid is spherical in shape. This is due to which of the following?

A. Gravity
B. Surface tension
C. Viscosity
D. Pressure

Solution

The spherical shape of a liquid drop is due to surface tension, which minimizes the surface area for a given volume.

Correct Answer: B — Surface tension

Q. A drop of oil spreads on water instead of sinking. This is due to:

A. Higher density of oil
B. Lower surface tension of oil
C. Higher surface tension of water
D. Lower viscosity of oil

Solution

The drop of oil spreads on water because the surface tension of water is higher than that of oil, causing the oil to spread out.

Correct Answer: C — Higher surface tension of water

Q. A drop of water is spherical in shape due to which property?

A. Viscosity
B. Surface tension
C. Capillarity
D. Density

Solution

A drop of water is spherical due to surface tension, which minimizes the surface area.

Correct Answer: B — Surface tension

Q. A family has 2 children. What is the probability that both children are boys if it is known that at least one is a boy?

A. 1/2
B. 1/3
C. 1/4
D. 1/5

Solution

The possible combinations of children are BB, BG, GB, GG. Given that at least one is a boy, we can eliminate GG, leaving us with BB, BG, GB. Out of these 3 combinations, only 1 is BB. Therefore, the probability is 1/3.

Correct Answer: B — 1/3

Q. A family has 2 children. What is the probability that both children are boys, given that at least one is a boy?

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

Solution

The possible combinations are BB, BG, GB. Given at least one is a boy, the combinations are BB, BG, GB. The probability of both being boys is 1/3.

Correct Answer: A — 1/3

Q. A family has 2 children. What is the probability that both children are boys?

A. 1/4
B. 1/2
C. 1/3
D. 1/5

Solution

The possible combinations of children are BB, BG, GB, GG. Out of these, only 1 combination is both boys (BB). Thus, the probability is 1/4.

Correct Answer: A — 1/4

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