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 point charge of +Q is placed at the center of a spherical shell of radius R with surface charge density σ. What is the electric field inside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
σ/ε₀
D.
Q/(4πε₀R)
Solution
According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
Q. A potentiometer is used to compare two emf sources. If the first source gives a balance length of 60cm and the second gives 90cm, what is the ratio of their emfs?
A.
2:3
B.
3:2
C.
1:1
D.
4:5
Solution
The ratio of emfs is equal to the ratio of the balance lengths, so it is 60cm:90cm = 2:3.
Q. A potentiometer is used to compare two EMFs. If the known EMF is 6V and the length of the wire is 120 cm, what is the potential gradient if the length of the wire is used to balance an unknown EMF of 4V?
A.
0.05 V/cm
B.
0.03 V/cm
C.
0.04 V/cm
D.
0.02 V/cm
Solution
The potential gradient is calculated as (6V / 120 cm) = 0.05 V/cm. For the unknown EMF of 4V, the length used would be (4V / 0.05 V/cm) = 80 cm.
Q. A potentiometer wire has a uniform cross-section and a potential difference of 12V across it. If the length of the wire is 6m, what is the potential gradient?
Q. A potentiometer wire has a uniform cross-section and a total length of 10 m. If a potential difference of 5 V is applied across it, what is the potential gradient?
A.
0.5 V/m
B.
1 V/m
C.
2 V/m
D.
5 V/m
Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Q. A projectile is launched at an angle of 30° with an initial speed of 40 m/s. What is the maximum height reached by the projectile?
A.
80 m
B.
60 m
C.
40 m
D.
20 m
Solution
Using the formula H = (u² * sin²θ) / (2g), where u = 40 m/s, θ = 30°, and g = 9.8 m/s², we find H = (40² * (1/4)) / (2*9.8) = 40.82 m, approximately 60 m.
