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 10 kg object is subjected to a net force of 50 N, what is its acceleration?
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
15 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 50 N / 10 kg = 5 m/s².
Correct Answer:
B
— 5 m/s²
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Q. If a 10 kg object is subjected to a net force of 50 N, what will be its acceleration?
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 50 N / 10 kg = 5 m/s².
Correct Answer:
B
— 5 m/s²
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Q. If a 10 ohm resistor is connected to a 20V battery, what is the energy consumed in 5 seconds?
A.
20 J
B.
40 J
C.
10 J
D.
50 J
Show solution
Solution
Power P = V^2 / R = 20^2 / 10 = 40 W. Energy = Power * time = 40 W * 5 s = 200 J.
Correct Answer:
B
— 40 J
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Q. If a 10 ohm resistor is connected to a 30V battery, what is the energy consumed in 5 seconds?
A.
15 J
B.
30 J
C.
75 J
D.
150 J
Show solution
Solution
Power P = V^2 / R = (30V)^2 / 10 ohms = 90 W. Energy = Power * Time = 90 W * 5 s = 450 J.
Correct Answer:
C
— 75 J
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Q. If a 1000 kg car accelerates from rest to a speed of 20 m/s in 10 seconds, what is the average power output of the car's engine?
A.
2000 W
B.
4000 W
C.
6000 W
D.
8000 W
Show solution
Solution
First, calculate the work done: W = 0.5 * m * v^2 = 0.5 * 1000 kg * (20 m/s)^2 = 200000 J. Then, power is P = W/t = 200000 J / 10 s = 20000 W.
Correct Answer:
C
— 6000 W
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Q. If a 1000 W heater is used for 3 hours, how much energy does it consume in kilowatt-hours?
A.
3 kWh
B.
2 kWh
C.
1 kWh
D.
0.5 kWh
Show solution
Solution
Energy consumed in kilowatt-hours is calculated as E = P (kW) * t (h). Here, P = 1000 W = 1 kW and t = 3 hours. Thus, E = 1 kW * 3 h = 3 kWh.
Correct Answer:
A
— 3 kWh
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Q. If a 1000 W heater runs for 2 hours, how much energy does it consume?
A.
7200 J
B.
14400 J
C.
20000 J
D.
3600000 J
Show solution
Solution
Energy = Power × Time = 1000 W × (2 × 3600 s) = 7200000 J.
Correct Answer:
D
— 3600000 J
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Q. If a 10Ω resistor is connected to a 20V battery, what is the power dissipated by the resistor?
A.
20W
B.
40W
C.
100W
D.
200W
Show solution
Solution
Power (P) can be calculated using P = V^2/R. Here, P = 20V^2 / 10Ω = 400 / 10 = 40W.
Correct Answer:
B
— 40W
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Q. If a 10Ω resistor is connected to a 20V source, what is the energy consumed in 5 seconds?
A.
20J
B.
40J
C.
50J
D.
100J
Show solution
Solution
Power P = V^2 / R = 20V^2 / 10Ω = 40W. Energy = Power * time = 40W * 5s = 200J.
Correct Answer:
B
— 40J
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Q. If a 12 kg object is moving with a constant velocity, what can be said about the net force acting on it?
A.
It is zero
B.
It is equal to its weight
C.
It is equal to the applied force
D.
It is increasing
Show solution
Solution
An object moving with constant velocity has a net force of zero according to Newton's first law.
Correct Answer:
A
— It is zero
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Q. If a 12 kg object is pushed with a force of 48 N, what is the acceleration of the object?
A.
2 m/s²
B.
4 m/s²
C.
6 m/s²
D.
8 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 48 N / 12 kg = 4 m/s².
Correct Answer:
B
— 4 m/s²
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Q. If a 12V battery is connected across a 4 ohm resistor, what is the current flowing through the resistor?
A.
2 A
B.
3 A
C.
4 A
D.
6 A
Show solution
Solution
Using Ohm's law, I = V/R = 12V / 4Ω = 3 A.
Correct Answer:
A
— 2 A
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Q. If a 12V battery is connected across a 4 ohm resistor, what is the power dissipated in the resistor?
A.
12 W
B.
24 W
C.
36 W
D.
48 W
Show solution
Solution
Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 144 / 4 = 36 W.
Correct Answer:
B
— 24 W
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Q. If a 12V battery is connected across a 4Ω resistor, what is the current flowing through the resistor?
Show solution
Solution
Using Ohm's law, I = V/R = 12V / 4Ω = 3A.
Correct Answer:
B
— 2A
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Q. If a 12V battery is connected to a 4 ohm resistor, what is the power dissipated by the resistor?
A.
12 W
B.
24 W
C.
36 W
D.
48 W
Show solution
Solution
Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 36 W.
Correct Answer:
B
— 24 W
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Q. If a 2 kg object is acted upon by a net force of 6 N, what is the object's 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 = 6 N / 2 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using energy conservation: PE_initial = KE_final; m * g * h = 1/2 * m * v^2; v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s
Correct Answer:
A
— 10 m/s
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Q. If a 2 kg object is dropped from a height of 5 m, what is its velocity just before it hits the ground? (g = 10 m/s²)
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using energy conservation, v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s
Correct Answer:
A
— 10 m/s
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Q. If a 2 kg object is moving with a velocity of 3 m/s and a force of 6 N is applied in the opposite direction, what will be its final velocity after 2 seconds?
A.
0 m/s
B.
1 m/s
C.
2 m/s
D.
3 m/s
Show solution
Solution
Net force = -6 N, acceleration = F/m = -6 N / 2 kg = -3 m/s². Final velocity = initial velocity + at = 3 m/s + (-3 m/s² * 2 s) = 0 m/s.
Correct Answer:
B
— 1 m/s
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Q. If a 2 kg object is pulled with a force of 10 N and experiences a frictional force of 4 N, what is its acceleration?
A.
3 m/s²
B.
5 m/s²
C.
2 m/s²
D.
1 m/s²
Show solution
Solution
Net force = applied force - friction = 10 N - 4 N = 6 N. Acceleration a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer:
A
— 3 m/s²
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Q. If a 2 kg object is subjected to a net force of 6 N, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. If a 3 kg object is in free fall, what is the force acting on it due to gravity?
A.
3 N
B.
9 N
C.
30 N
D.
None of the above
Show solution
Solution
Force due to gravity F = mg = 3 kg * 9.8 m/s² = 29.4 N, approximately 9 N for simplification.
Correct Answer:
B
— 9 N
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by friction?
A.
-24 J
B.
-48 J
C.
0 J
D.
24 J
Show solution
Solution
Initial kinetic energy = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer:
B
— -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction?
A.
-24 J
B.
-48 J
C.
-12 J
D.
-36 J
Show solution
Solution
Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer:
B
— -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction force?
A.
-24 J
B.
-48 J
C.
-72 J
D.
-96 J
Show solution
Solution
Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer:
B
— -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s, what is the total mechanical energy if it is at a height of 2 m?
A.
30 J
B.
40 J
C.
50 J
D.
60 J
Show solution
Solution
Total mechanical energy = K.E + P.E = 0.5 × 3 kg × (4 m/s)² + 3 kg × 9.8 m/s² × 2 m = 24 J + 58.8 J = 82.8 J.
Correct Answer:
C
— 50 J
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Q. If a 3 kg object is moving with a velocity of 4 m/s, what is its momentum?
A.
12 kg m/s
B.
6 kg m/s
C.
8 kg m/s
D.
10 kg m/s
Show solution
Solution
Momentum (p) = m * v = 3 kg * 4 m/s = 12 kg m/s
Correct Answer:
A
— 12 kg m/s
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Q. If a 4 kg object is acted upon by a net force of 16 N, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
C
— 4 m/s²
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Q. If a 4 kg object is at rest and a net force of 8 N is applied, what will be its velocity after 2 seconds?
A.
4 m/s
B.
2 m/s
C.
1 m/s
D.
0 m/s
Show solution
Solution
Using F = ma, a = F/m = 8 N / 4 kg = 2 m/s². Velocity after 2 seconds = a * t = 2 m/s² * 2 s = 4 m/s.
Correct Answer:
B
— 2 m/s
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Q. If a 4 kg object is moving with a constant velocity, what can be said about the net force acting on it?
A.
It is zero
B.
It is equal to its weight
C.
It is equal to the applied force
D.
It is maximum
Show solution
Solution
According to Newton's first law, if an object is moving with constant velocity, the net force acting on it is zero.
Correct Answer:
A
— It is zero
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