JEE Main MCQ & Objective Questions
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. If a fluid has a viscosity of 0.5 Pa·s, what does this indicate about its flow characteristics?
A.
It flows easily
B.
It is very thick
C.
It is a gas
D.
It is a low-density fluid
Show solution
Solution
A viscosity of 0.5 Pa·s indicates that the fluid is relatively thick and flows less easily compared to fluids with lower viscosity.
Correct Answer:
B
— It is very thick
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Q. If a force of 10 N is applied at an angle of 30 degrees to the horizontal, what is the horizontal component of the force?
A.
5 N
B.
8.66 N
C.
10 N
D.
0 N
Show solution
Solution
The horizontal component is calculated as 10 N * cos(30°) = 8.66 N.
Correct Answer:
B
— 8.66 N
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Q. If a force of 10 N is applied to move an object 5 m in the direction of the force, what is the work done?
A.
50 J
B.
30 J
C.
20 J
D.
10 J
Show solution
Solution
Work Done (W) = Force * Distance = 10 N * 5 m = 50 J
Correct Answer:
A
— 50 J
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Q. If a force of 10 N is applied to move an object 5 m, what is the work done?
A.
50 J
B.
25 J
C.
10 J
D.
5 J
Show solution
Solution
Work Done (W) = Force * Distance = 10 N * 5 m = 50 J
Correct Answer:
A
— 50 J
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Q. If a force of 12 N is applied at an angle of 30 degrees to a lever arm of 1 m, what is the torque about the pivot?
A.
6 Nm
B.
10 Nm
C.
12 Nm
D.
15 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 12 N × 1 m × sin(30°) = 12 N × 1 m × 0.5 = 6 Nm.
Correct Answer:
A
— 6 Nm
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Q. If a force of 12 N is applied at an angle of 30 degrees to the horizontal while moving an object 3 m, what is the work done by the force?
A.
18 J
B.
24 J
C.
30 J
D.
36 J
Show solution
Solution
Work done = F × d × cos(θ) = 12 N × 3 m × cos(30°) = 12 N × 3 m × (√3/2) = 18 J.
Correct Answer:
B
— 24 J
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Q. If a force of 12 N is applied to a 4 kg object, what is the resulting acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we find a = F/m = 12 N / 4 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. If a force of 15 N acts on an object and moves it 4 m in the direction of the force, what is the work done?
A.
30 J
B.
60 J
C.
75 J
D.
90 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
B
— 60 J
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Q. If a force of 15 N is applied at a distance of 0.4 m from the pivot, what is the torque?
A.
6 Nm
B.
12 Nm
C.
15 Nm
D.
20 Nm
Show solution
Solution
Torque = Force × Distance = 15 N × 0.4 m = 6 Nm.
Correct Answer:
B
— 12 Nm
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Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1.5 m, what is the torque about the pivot?
A.
3.75 Nm
B.
7.5 Nm
C.
11.25 Nm
D.
12.99 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1.5 m × sin(30°) = 15 × 1.5 × 0.5 = 11.25 Nm.
Correct Answer:
B
— 7.5 Nm
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Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1 m, what is the torque about the pivot?
A.
7.5 Nm
B.
15 Nm
C.
12.99 Nm
D.
10 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1 m × sin(30°) = 15 N × 1 m × 0.5 = 7.5 Nm.
Correct Answer:
C
— 12.99 Nm
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Q. If a force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done by the force?
A.
30 J
B.
60 J
C.
120 J
D.
180 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 120 J
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Q. If a force of 15 N is applied at an angle of 60° to the horizontal while moving an object 4 m, what is the work done by the force?
A.
30 J
B.
60 J
C.
120 J
D.
180 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 120 J
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Q. If a force of 15 N is applied to a mass of 3 kg, what is the net force acting on the mass if there is a frictional force of 5 N opposing the motion?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
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Q. If a force of 15 N is applied to a mass of 3 kg, what is the net force acting on the mass if it is also experiencing a frictional force of 5 N?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
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Q. If a force of 15 N is applied to move an object 3 m in the direction of the force, what is the work done?
A.
45 J
B.
30 J
C.
15 J
D.
60 J
Show solution
Solution
Work done = Force × Distance = 15 N × 3 m = 45 J.
Correct Answer:
A
— 45 J
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Q. If a force of 15 N is applied to move an object 4 m in the direction of the force, what is the work done?
A.
30 J
B.
60 J
C.
45 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
B
— 60 J
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Q. If a force of 30 N is applied to a mass of 10 kg, what is the net force acting on the mass if there is a frictional force of 10 N opposing the motion?
A.
20 N
B.
30 N
C.
40 N
D.
10 N
Show solution
Solution
Net force = applied force - frictional force = 30 N - 10 N = 20 N.
Correct Answer:
A
— 20 N
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Q. If a force of 5 N acts on an object and moves it 4 m in the direction of the force, what is the work done?
A.
10 J
B.
15 J
C.
20 J
D.
25 J
Show solution
Solution
Work done = Force × Distance = 5 N × 4 m = 20 J.
Correct Answer:
C
— 20 J
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Q. If a force of 5 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done by the force?
A.
10 J
B.
20 J
C.
5 J
D.
15 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 5 N × 4 m × cos(60°) = 5 N × 4 m × 0.5 = 10 J.
Correct Answer:
A
— 10 J
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Q. If a force of 5 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m, what is the work done by the force?
A.
7.5 J
B.
15 J
C.
12.99 J
D.
10 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 5 N × 3 m × cos(60°) = 5 N × 3 m × 0.5 = 7.5 J.
Correct Answer:
C
— 12.99 J
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Q. If a forced oscillator is driven at a frequency much lower than its natural frequency, what happens to the amplitude?
A.
Increases significantly
B.
Decreases
C.
Remains constant
D.
Fluctuates
Show solution
Solution
At frequencies much lower than the natural frequency, the amplitude of the forced oscillator increases significantly.
Correct Answer:
B
— Decreases
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Q. If a galvanometer shows a deflection when connected to a potentiometer, what does it indicate?
A.
The circuit is open.
B.
The potential difference is zero.
C.
The potential difference is equal to the reference voltage.
D.
The current is flowing through the galvanometer.
Show solution
Solution
A deflection in the galvanometer indicates that the potential difference across the galvanometer is equal to the reference voltage.
Correct Answer:
C
— The potential difference is equal to the reference voltage.
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Q. If a gas expands against a constant external pressure, the work done by the gas is given by:
A.
W = P_ext * ΔV
B.
W = ΔU + Q
C.
W = Q - ΔU
D.
W = P_ext / ΔV
Show solution
Solution
The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.
Correct Answer:
A
— W = P_ext * ΔV
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Q. If a gas expands and does 150 J of work while absorbing 100 J of heat, what is the change in internal energy?
A.
-50 J
B.
50 J
C.
250 J
D.
100 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.
Correct Answer:
A
— -50 J
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Q. If a gas expands and does 50 J of work while absorbing 30 J of heat, what is the change in internal energy?
A.
-20 J
B.
20 J
C.
80 J
D.
30 J
Show solution
Solution
Using the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 30 J - 50 J = -20 J.
Correct Answer:
B
— 20 J
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Q. If a gas occupies 10 L at 1 atm, what will be its volume at 2 atm if the temperature remains constant?
A.
5 L
B.
10 L
C.
20 L
D.
15 L
Show solution
Solution
Using Boyle's Law (P1V1 = P2V2), if P1 = 1 atm, V1 = 10 L, and P2 = 2 atm, then V2 = (P1V1)/P2 = (1 atm * 10 L) / 2 atm = 5 L.
Correct Answer:
A
— 5 L
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Q. If a gas occupies a volume of 10 L at 1 atm, what will be its volume at 2 atm if the temperature remains constant?
A.
5 L
B.
10 L
C.
20 L
D.
15 L
Show solution
Solution
Using Boyle's Law (P1V1 = P2V2), if the pressure doubles from 1 atm to 2 atm, the volume will halve from 10 L to 5 L.
Correct Answer:
A
— 5 L
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Q. If a hollow cylinder and a solid cylinder of the same mass and radius roll down the same incline, which one reaches the bottom first?
A.
Hollow cylinder
B.
Solid cylinder
C.
Both reach at the same time
D.
Depends on the angle of incline
Show solution
Solution
The solid cylinder reaches the bottom first because it has a smaller moment of inertia compared to the hollow cylinder.
Correct Answer:
B
— Solid cylinder
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Q. If a hollow cylinder rolls down an incline, how does its acceleration compare to that of a solid cylinder?
A.
Hollow cylinder accelerates faster
B.
Solid cylinder accelerates faster
C.
Both accelerate equally
D.
Depends on the angle of incline
Show solution
Solution
The solid cylinder has a lower moment of inertia compared to the hollow cylinder, thus it accelerates faster down the incline.
Correct Answer:
B
— Solid cylinder accelerates faster
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