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 capillary tube is dipped in water. What is the shape of the water surface inside the tube?
A.
Flat
B.
Concave
C.
Convex
D.
Irregular
Solution
The water surface inside the capillary tube is concave due to the adhesive forces between water and the tube material being stronger than the cohesive forces among water molecules.
Q. A capillary tube is dipped into water. How high will the water rise in the tube if the radius is 1 mm?
A.
2.5 cm
B.
5 cm
C.
10 cm
D.
15 cm
Solution
Using the capillary rise formula, h = (2γcosθ)/(ρgr), where γ is surface tension, θ is contact angle, ρ is density, g is acceleration due to gravity, and r is radius.
Q. A capillary tube is dipped into water. The height to which water rises in the tube is determined by:
A.
Surface tension and density of the liquid
B.
Only surface tension
C.
Only density of the liquid
D.
Viscosity of the liquid
Solution
The height of the liquid column in a capillary tube is determined by both surface tension and the density of the liquid, as described by the capillary rise formula.
Correct Answer:
A
— Surface tension and density of the liquid
Q. A capillary tube is dipped into water. The water rises in the tube due to which of the following?
A.
Surface tension and adhesion
B.
Surface tension and cohesion
C.
Only adhesion
D.
Only cohesion
Solution
The rise of water in a capillary tube is due to both surface tension (which pulls the liquid up) and adhesion (the attraction between water molecules and the tube's surface).
Q. A capillary tube of radius 0.5 mm is dipped in water. What is the height of the water column raised in the tube? (Surface tension = 0.072 N/m, density of water = 1000 kg/m³)
A.
0.5 m
B.
0.1 m
C.
0.2 m
D.
0.3 m
Solution
Using the formula h = 2γ/(ρgr), h = 2 × 0.072 N/m / (1000 kg/m³ × 9.81 m/s² × 0.0005 m) = 0.2 m.
Q. A car is moving at 80 km/h and a motorcycle is moving at 100 km/h in the same direction. What is the relative speed of the motorcycle with respect to the car?
A.
20 km/h
B.
180 km/h
C.
100 km/h
D.
80 km/h
Solution
Relative speed = Speed of motorcycle - Speed of car = 100 km/h - 80 km/h = 20 km/h.
Q. A car is moving at 80 km/h and a motorcycle is moving at 60 km/h in the same direction. What is the relative speed of the motorcycle with respect to the car?
A.
20 km/h
B.
60 km/h
C.
80 km/h
D.
140 km/h
Solution
Relative speed = Speed of motorcycle - Speed of car = 60 km/h - 80 km/h = -20 km/h (20 km/h behind).
Q. A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
A.
0.1
B.
0.2
C.
0.3
D.
0.4
Solution
The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.
Q. A car is negotiating a curve of radius 100 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent the car from skidding?
A.
0.15
B.
0.25
C.
0.30
D.
0.35
Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15 m/s)² / (100 m * 9.8 m/s²) ≈ 0.23.
