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.
Q. A solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one will have a greater translational speed when they reach the ground?
A.
Solid sphere
B.
Hollow sphere
C.
Both will have the same speed
D.
Depends on the mass
Solution
The solid sphere will have a greater translational speed because it has a smaller moment of inertia.
Q. A solid sphere of mass M and radius R is rolling without slipping on a horizontal surface. What is the expression for its total angular momentum about its center of mass?
A.
(2/5)MR^2ω
B.
MR^2ω
C.
MR^2
D.
0
Solution
Total angular momentum L = Iω, where I = (2/5)MR^2 for a solid sphere.
Q. A solid sphere of mass m and radius r rolls without slipping down an inclined plane of angle θ. What is the acceleration of the center of mass of the sphere?
A.
g sin(θ)
B.
g sin(θ)/2
C.
g sin(θ)/3
D.
g sin(θ)/4
Solution
The acceleration of the center of mass of a rolling object is given by a = g sin(θ) / (1 + k^2/r^2). For a solid sphere, k^2/r^2 = 2/5, thus a = g sin(θ) / (1 + 2/5) = g sin(θ) / (7/5) = (5/7)g sin(θ).
Q. A solid sphere of radius R rolls without slipping down an inclined plane of angle θ. What is the acceleration of the center of mass of the sphere?
A.
g sin(θ)
B.
g sin(θ)/2
C.
g sin(θ)/3
D.
g sin(θ)/4
Solution
The acceleration of the center of mass of a solid sphere rolling down an incline is given by a = g sin(θ) / (1 + (2/5)) = g sin(θ) / (7/5) = (5/7) g sin(θ).
Q. A solid sphere rolls down a hill without slipping. If the height of the hill is h, what is the speed of the sphere at the bottom of the hill?
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(5gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid sphere, I = (2/5)mr^2 and ω = v/r. Solving gives v = √(2gh).
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.
1:2
B.
2:3
C.
1:3
D.
1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:3.
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom of the incline?
A.
1:2
B.
2:3
C.
1:3
D.
1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:5, which simplifies to 2:3.
Q. A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.
1:2
B.
2:3
C.
1:1
D.
1:3
Solution
For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
Q. A solution has a density of 1.2 g/mL and contains 30 g of solute. What is the molarity if the molar mass of the solute is 60 g/mol?
A.
0.5 M
B.
1 M
C.
2 M
D.
1.5 M
Solution
Volume of solution = mass / density = 30 g / 1.2 g/mL = 25 mL = 0.025 L. Moles of solute = 30 g / 60 g/mol = 0.5 moles. Molarity = 0.5 moles / 0.025 L = 20 M.
Q. A solution is prepared by dissolving 50 g of glucose (C6H12O6) in 250 g of water. What is the mass percent of glucose in the solution? (Molar mass of glucose = 180 g/mol)
A.
20%
B.
15%
C.
25%
D.
10%
Solution
Mass percent = (mass of solute / (mass of solute + mass of solvent)) × 100 = (50 g / (50 g + 250 g)) × 100 = 20%.
Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. What is the concentration in terms of molarity? (Molar mass of NaCl = 58.5 g/mol)
A.
1 M
B.
2 M
C.
0.5 M
D.
0.25 M
Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.
Q. A solution is prepared by dissolving 58.5 g of NaCl in enough water to make 1 L of solution. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)
A.
1 M
B.
2 M
C.
0.5 M
D.
0.1 M
Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.
