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 man is standing 40 meters away from a building. If the angle of elevation to the top of the building is 30 degrees, what is the height of the building?
Q. A man is standing 50 meters away from a vertical pole. If he looks up at an angle of elevation of 60 degrees to the top of the pole, what is the height of the pole?
A.
25 m
B.
30 m
C.
35 m
D.
40 m
Solution
Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.
Q. A man is standing on a hill 80 meters high. If he looks at a point on the ground at an angle of depression of 45 degrees, how far is the point from the base of the hill?
Q. A man is standing on the ground and looking at the top of a 15 m high pole. If he is 20 m away from the base of the pole, what is the angle of elevation?
A.
36.87 degrees
B.
45 degrees
C.
60 degrees
D.
30 degrees
Solution
Using tan(θ) = height/distance, we have tan(θ) = 15/20. Therefore, θ = tan⁻¹(0.75) which is approximately 36.87 degrees.
Q. A man is standing on the ground and looking at the top of a 40 m high building. If the angle of elevation is 60 degrees, how far is he from the building?
A.
20 m
B.
40 m
C.
20√3 m
D.
40√3 m
Solution
Using tan(60°) = height/distance, we have distance = height/tan(60°) = 40/√3 = 20√3 m.
Q. A man is standing on the ground and looking at the top of a building. If the angle of elevation is 45 degrees and he is 10 meters away from the building, what is the height of the building?
Q. A man is standing on the ground and looking at the top of a tree. If the angle of elevation is 60 degrees and he is 10 meters away from the base of the tree, what is the height of the tree?
A.
5√3 m
B.
10√3 m
C.
15√3 m
D.
20√3 m
Solution
Using tan(60°) = height/10, we have √3 = height/10. Therefore, height = 10√3 m.
Q. A man is standing on the ground and observes the top of a building at an angle of elevation of 60 degrees. If he is 50 m away from the building, what is the height of the building?
A.
25 m
B.
43.3 m
C.
50 m
D.
86.6 m
Solution
Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.
Q. A man is standing on the ground and observes the top of a tree at an angle of elevation of 45 degrees. If he is 10 meters away from the tree, what is the height of the tree?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
Solution
Using tan(45°) = height/10, we have 1 = height/10. Therefore, height = 10 m.
Q. A mass attached to a spring oscillates with a damping coefficient of 0.3 kg/s. If the mass is 1 kg and the spring constant is 4 N/m, what is the damping ratio?
A.
0.1
B.
0.3
C.
0.5
D.
0.75
Solution
Damping ratio (ζ) = c / (2√(mk)) = 0.3 / (2√(1*4)) = 0.3 / 4 = 0.075.
Q. A mass is measured as 15.0 kg with an uncertainty of ±0.3 kg. If this mass is used to calculate the force (F = ma) with an acceleration of 9.8 m/s², what is the uncertainty in the force?
A.
0.3 N
B.
2.94 N
C.
0.5 N
D.
1.5 N
Solution
Uncertainty in force = a * (uncertainty in mass) = 9.8 * 0.3 = 2.94 N.
Q. A mass is measured as 5.0 kg with an uncertainty of ±0.1 kg. If this mass is used to calculate weight (W = mg), what is the uncertainty in weight if g = 9.8 m/s²?
A.
±0.2 N
B.
±0.5 N
C.
±0.1 N
D.
±0.4 N
Solution
Uncertainty in weight = g * (uncertainty in mass) = 9.8 * 0.1 = ±0.98 N, rounded to ±1 N.
Q. A mass m is attached to a spring of spring constant k. If the mass is displaced by a distance x from its equilibrium position, what is the restoring force acting on the mass?
A.
kx
B.
-kx
C.
mg
D.
-mg
Solution
The restoring force in simple harmonic motion is given by Hooke's law, which states that the force is proportional to the displacement and acts in the opposite direction. Therefore, the restoring force is -kx.
Q. A mass m is attached to a spring of spring constant k. If the mass is displaced from its equilibrium position and released, what is the time period of the oscillation?
A.
2π√(m/k)
B.
2π√(k/m)
C.
π√(m/k)
D.
π√(k/m)
Solution
The time period T of a mass-spring system in simple harmonic motion is given by T = 2π√(m/k).
Q. A mass m is attached to a string and is whirled in a horizontal circle. If the radius of the circle is halved, what happens to the tension in the string if the speed remains constant?
A.
It doubles
B.
It remains the same
C.
It halves
D.
It quadruples
Solution
Tension T = mv²/r. If r is halved, T doubles for constant speed.
Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, the tension in the string is T. What is the expression for T?
A.
T = mg
B.
T = mg - mv²/r
C.
T = mg + mv²/r
D.
T = mv²/r
Solution
At the highest point, T + mg = mv²/r, thus T = mg - mv²/r.
Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the minimum speed required to keep the mass in circular motion?
A.
√(g*r)
B.
g*r
C.
2g*r
D.
g/2
Solution
At the highest point, the centripetal force is provided by the weight. Minimum speed = √(g*r).
Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the condition for the mass to just complete the circular motion?
A.
Tension = 0
B.
Tension = mg
C.
Tension = 2mg
D.
Tension = mg/2
Solution
At the highest point, the centripetal force is provided by the weight of the mass, so T + mg = mv²/r. For T = 0, mg = mv²/r.
