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. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 1700 Hz, what is the wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
A
— 0.2 m
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Q. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength?
A.
2 m
B.
1 m
C.
0.5 m
D.
0.2 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 170 Hz = 2 m.
Correct Answer:
A
— 2 m
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Q. A sound wave travels through a medium with a speed of 340 m/s and has a frequency of 1700 Hz. What is its wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength = Speed / Frequency = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
B
— 0.5 m
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Q. A sound wave travels through air at a speed of 340 m/s. If the frequency of the sound is 1700 Hz, what is the wavelength?
A.
0.2 m
B.
0.5 m
C.
1 m
D.
2 m
Show solution
Solution
Using the formula v = fλ, we can rearrange to find λ = v/f. Thus, λ = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer:
A
— 0.2 m
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Q. A sound wave travels through water at a speed of 1500 m/s. If the frequency of the sound wave is 300 Hz, what is the wavelength?
A.
2 m
B.
3 m
C.
4 m
D.
5 m
Show solution
Solution
Wavelength λ = v/f = 1500 m/s / 300 Hz = 5 m.
Correct Answer:
A
— 2 m
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Q. A speed is measured as 20 m/s with an uncertainty of ±0.5 m/s. If this speed is used to calculate kinetic energy, what is the percentage error in kinetic energy?
A.
5%
B.
2.5%
C.
1%
D.
10%
Show solution
Solution
K.E. = 0.5 * m * v², percentage error = 2 * (Δv/v) = 2 * (0.5/20) = 0.05 or 5%.
Correct Answer:
B
— 2.5%
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Q. A speed is measured as 20 m/s with an uncertainty of ±0.5 m/s. What is the absolute error in the speed measurement?
A.
0.5 m/s
B.
0.25 m/s
C.
1 m/s
D.
0.1 m/s
Show solution
Solution
The absolute error is given directly as ±0.5 m/s.
Correct Answer:
A
— 0.5 m/s
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Q. A speed is recorded as 60 km/h with an error of 2 km/h. What is the percentage error?
A.
3.33%
B.
2.5%
C.
1.67%
D.
4%
Show solution
Solution
Percentage error = (Absolute error / Measured value) * 100 = (2 / 60) * 100 = 3.33%
Correct Answer:
A
— 3.33%
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Q. A speed of 30 m/s is measured with an uncertainty of ±0.5 m/s. What is the total uncertainty if this speed is used to calculate kinetic energy?
A.
0.25 J
B.
0.5 J
C.
1 J
D.
2 J
Show solution
Solution
Kinetic energy = 0.5 * m * v²; uncertainty in KE = 2 * v * uncertainty in v = 2 * 30 * 0.5 = 30 J.
Correct Answer:
C
— 1 J
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Q. A sphere rolls down a ramp of height h. What is the total mechanical energy at the top?
A.
mgh
B.
1/2 mv^2
C.
mgh + 1/2 mv^2
D.
0
Show solution
Solution
The total mechanical energy at the top is purely potential energy, which is mgh.
Correct Answer:
A
— mgh
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Q. A sphere rolls down a ramp. If the height of the ramp is h, what is the speed of the sphere at the bottom assuming no energy loss?
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(gh)
Show solution
Solution
Using conservation of energy, the potential energy at height h converts to kinetic energy at the bottom, giving speed √(2gh).
Correct Answer:
A
— √(2gh)
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Q. A sphere rolls on a flat surface with a speed v. What is the kinetic energy of the sphere?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/5)mv^2
C.
(1/2)mv^2 + (2/5)mv^2
D.
(1/2)mv^2 + (3/5)mv^2
Show solution
Solution
The total kinetic energy of a rolling sphere is the sum of translational and rotational kinetic energy, which is (1/2)mv^2 + (2/5)mv^2.
Correct Answer:
C
— (1/2)mv^2 + (2/5)mv^2
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Q. A sphere rolls without slipping on a flat surface. If it has a radius R and rolls with a speed v, what is its angular speed?
A.
v/R
B.
2v/R
C.
v/2R
D.
v²/R
Show solution
Solution
The angular speed ω of a rolling sphere is given by ω = v/R.
Correct Answer:
A
— v/R
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Q. A spherical conductor has a charge Q. What is the electric potential inside the conductor?
A.
0
B.
Q/(4πε₀r)
C.
Q/(4πε₀R)
D.
Constant throughout
Show solution
Solution
The electric potential inside a charged spherical conductor is constant and equal to the potential on its surface, which is Q/(4πε₀R).
Correct Answer:
D
— Constant throughout
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Q. A spherical conductor has a radius R and carries a charge Q. What is the electric potential on its surface?
A.
kQ/R
B.
kQ/2R
C.
0
D.
kQ/R²
Show solution
Solution
The electric potential V on the surface of a charged spherical conductor is given by V = kQ/R.
Correct Answer:
A
— kQ/R
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Q. A spherical Gaussian surface of radius R encloses a charge Q. What is the electric field at a distance 2R from the center?
A.
Q/4πε₀R²
B.
Q/4πε₀(2R)²
C.
0
D.
Q/ε₀(2R)²
Show solution
Solution
The electric field outside a spherical charge distribution is given by E = Q/4πε₀r². At 2R, it becomes Q/4πε₀(2R)².
Correct Answer:
B
— Q/4πε₀(2R)²
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Q. A spherical shell of radius R carries a total charge Q. What is the electric field at a point outside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
Q/(4πε₀R)
D.
Q/(4πε₀R³)
Show solution
Solution
For a spherical shell, the electric field outside the shell behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).
Correct Answer:
B
— Q/(4πε₀R²)
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Q. A spherical shell of radius R carries a uniform charge Q. What is the electric field inside the shell?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
D.
Q/(4πε₀)
Show solution
Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer:
B
— 0
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Q. A spherical shell of radius R carries a uniform surface charge density σ. What is the electric field inside the shell?
A.
0
B.
σ/ε₀
C.
σ/2ε₀
D.
σ/4ε₀
Show solution
Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer:
A
— 0
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Q. A spring obeys Hooke's law. If the spring constant is doubled, what happens to the elongation for the same applied force?
A.
Elongation doubles
B.
Elongation halves
C.
Elongation remains the same
D.
Elongation quadruples
Show solution
Solution
According to Hooke's law, elongation is inversely proportional to the spring constant; thus, if the spring constant is doubled, the elongation halves.
Correct Answer:
B
— Elongation halves
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Q. A spring stretches 5 cm when a load of 10 N is applied. What is the spring constant?
A.
200 N/m
B.
100 N/m
C.
50 N/m
D.
25 N/m
Show solution
Solution
Using Hooke's law, k = F/x = 10 N / 0.05 m = 200 N/m.
Correct Answer:
B
— 100 N/m
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Q. A spring with a spring constant of 200 N/m is compressed by 0.1 m. What is the potential energy stored in the spring?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
Show solution
Solution
Potential Energy = 0.5 × k × x² = 0.5 × 200 N/m × (0.1 m)² = 1 J.
Correct Answer:
A
— 1 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.1 m. What is the work done in compressing the spring?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
Show solution
Solution
Work done = 0.5 × k × x^2 = 0.5 × 200 N/m × (0.1 m)^2 = 1 J.
Correct Answer:
A
— 1 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.5 m. What is the potential energy stored in the spring?
A.
25 J
B.
50 J
C.
100 J
D.
200 J
Show solution
Solution
Potential Energy in spring = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.
Correct Answer:
B
— 50 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.5 m. What is the work done in compressing the spring?
A.
25 J
B.
50 J
C.
100 J
D.
200 J
Show solution
Solution
Work done = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.
Correct Answer:
B
— 50 J
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Q. A stone is dropped from a height of 20 m. How long does it take to reach the ground?
A.
2 s
B.
1 s
C.
3 s
D.
4 s
Show solution
Solution
Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.
Correct Answer:
A
— 2 s
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Q. A stone is dropped from a height of 20 m. How long will it take to reach the ground?
A.
2 s
B.
1 s
C.
3 s
D.
4 s
Show solution
Solution
Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.
Correct Answer:
A
— 2 s
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Q. A stone is dropped from a height of 45 m. How far will it have fallen after 2 seconds?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
Show solution
Solution
Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(2²) = 19.6 m.
Correct Answer:
C
— 30 m
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Q. A stone is dropped from a height of 45 m. How far will it have fallen after 3 seconds?
A.
10.5 m
B.
20 m
C.
30 m
D.
40.5 m
Show solution
Solution
Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(3²) = 44.1 m.
Correct Answer:
D
— 40.5 m
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Q. A stone is dropped from a height of 45 m. How far will it travel horizontally if it is thrown horizontally with a speed of 10 m/s?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
Show solution
Solution
Time to fall (t) = √(2h/g) = √(2*45/9.8) ≈ 3 s; horizontal distance = v*t = 10*3 = 30 m.
Correct Answer:
B
— 30 m
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