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 cyclist is moving at a speed of 15 km/h. If a car is moving in the same direction at 30 km/h, what is the relative speed of the car with respect to the cyclist?
A.
15 km/h
B.
30 km/h
C.
45 km/h
D.
0 km/h
Solution
Relative speed = speed of car - speed of cyclist = 30 - 15 = 15 km/h.
Q. A cyclist is moving in a circular track of radius 30 m with a speed of 15 m/s. What is the net force acting on the cyclist if the mass of the cyclist is 60 kg?
A.
180 N
B.
120 N
C.
90 N
D.
60 N
Solution
Centripetal force F = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.
Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the angular velocity of the cyclist?
Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist towards the center of the track?
A.
50 N
B.
75 N
C.
100 N
D.
125 N
Solution
Centripetal force (F_c) = mv²/r. Assuming mass m = 100 kg, F_c = (100 kg)(15 m/s)² / (30 m) = 75 N.
Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist if the mass of the cyclist is 60 kg?
A.
180 N
B.
120 N
C.
90 N
D.
60 N
Solution
Centripetal force F_c = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.
Q. A cyclist is negotiating a circular turn of radius 30 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent slipping?
A.
0.25
B.
0.5
C.
0.75
D.
1
Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15²)/(30*9.8) = 0.25.
Q. A cyclist is pedaling at a constant speed and exerts a power of 200 W. If the cyclist increases their power output to 400 W, what happens to their speed assuming no other forces act?
A.
Speed remains the same
B.
Speed doubles
C.
Speed increases by 41%
D.
Speed increases by 100%
Solution
Power is proportional to the cube of the speed in cycling. If power doubles, speed increases by a factor of (2)^(1/3) which is approximately 1.26, or about a 41% increase.
Q. A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
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 cylinder, I = (1/2)mR^2 and ω = v/R. Solving gives v = √(3gh).
Q. A cylinder rolls down a hill. If it has a radius R and rolls without slipping, what is the relationship between its linear velocity v and its angular velocity ω?
A.
v = Rω
B.
v = 2Rω
C.
v = ω/R
D.
v = R^2ω
Solution
For rolling without slipping, the relationship is v = Rω.
Q. A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
A.
√(gh)
B.
√(2gh)
C.
√(3gh)
D.
√(4gh)
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 cylinder, I = 1/2 mr^2, leading to v = √(2gh).
Q. A cylindrical conductor of radius R carries a uniform charge per unit length λ. What is the electric field at a distance r from the axis of the cylinder (r > R)?
A.
0
B.
λ/(2πε₀r)
C.
λ/(2πε₀R)
D.
λ/(4πε₀r²)
Solution
For a point outside the cylinder, the electric field is given by E = λ/(2πε₀r).
Q. A cylindrical Gaussian surface encloses a charge Q. If the height of the cylinder is doubled while keeping the radius constant, what happens to the electric flux through the curved surface?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
Solution
The electric flux through the curved surface is proportional to the charge enclosed, which remains constant, so the flux through the curved surface doubles if the height is doubled.
Q. A cylindrical Gaussian surface encloses a charge Q. If the radius of the cylinder is r and its height is h, what is the electric flux through the curved surface?
A.
Q/ε₀
B.
Q/(2ε₀)
C.
Q/(4ε₀)
D.
0
Solution
The electric flux through the curved surface of a cylinder is given by Φ = Q_enc/ε₀, where Q_enc = Q.
