Q. What is the electric field due to a point charge at a distance r?
A.
k * q / r^2
B.
k * q / r
C.
k * q * r
D.
k * q * r^2
Show solution
Solution
The electric field E due to a point charge q at a distance r is given by E = k * q / r^2, where k is Coulomb's constant.
Correct Answer:
A
— k * q / r^2
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Q. What is the electric field due to a point charge of +10μC at a distance of 0.2m?
A.
22500 N/C
B.
45000 N/C
C.
50000 N/C
D.
75000 N/C
Show solution
Solution
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (0.2 m)² = 225000 N/C.
Correct Answer:
C
— 50000 N/C
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Q. What is the electric field due to a point charge of +1μC at a distance of 0.1m?
A.
9000 N/C
B.
900 N/C
C.
90 N/C
D.
9 N/C
Show solution
Solution
Electric field E = k * |q| / r^2 = (9 × 10^9) * (1 × 10^-6) / (0.1)^2 = 9000 N/C.
Correct Answer:
A
— 9000 N/C
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Q. What is the electric field due to a point charge of +1μC at a distance of 1m?
A.
9 × 10^9 N/C
B.
1 × 10^6 N/C
C.
9 × 10^6 N/C
D.
1 × 10^9 N/C
Show solution
Solution
Electric field E = k * |q| / r^2 = (9 × 10^9) * (1 × 10^-6) / (1)^2 = 9 × 10^6 N/C.
Correct Answer:
C
— 9 × 10^6 N/C
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Q. What is the electric field due to a point charge of +4μC at a distance of 0.1m?
A.
36000 N/C
B.
40000 N/C
C.
44000 N/C
D.
48000 N/C
Show solution
Solution
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (4 × 10^-6 C) / (0.1 m)² = 36000 N/C.
Correct Answer:
B
— 40000 N/C
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Q. What is the electric field due to a point charge of +4μC at a distance of 0.2m?
A.
4500 N/C
B.
9000 N/C
C.
18000 N/C
D.
36000 N/C
Show solution
Solution
E = k * |q| / r^2 = (9 × 10^9) * (4 × 10^-6) / (0.2)^2 = 9000 N/C.
Correct Answer:
B
— 9000 N/C
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Q. What is the electric field due to a point charge of +5μC at a distance of 0.1 m?
A.
4500 N/C
B.
5000 N/C
C.
5500 N/C
D.
6000 N/C
Show solution
Solution
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (0.1 m)² = 4500 N/C.
Correct Answer:
B
— 5000 N/C
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Q. What is the electric field due to a point charge of +5μC at a distance of 0.1m?
A.
4500 N/C
B.
5000 N/C
C.
45000 N/C
D.
50000 N/C
Show solution
Solution
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (0.1 m)² = 45000 N/C.
Correct Answer:
C
— 45000 N/C
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Q. What is the electric field due to a point charge of +5μC at a distance of 0.2m?
A.
11250 N/C
B.
4500 N/C
C.
2250 N/C
D.
5625 N/C
Show solution
Solution
E = k * |q| / r^2 = (9 × 10^9) * (5 × 10^-6) / (0.2)^2 = 11250 N/C.
Correct Answer:
A
— 11250 N/C
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Q. What is the electric field due to a point charge of +5μC at a distance of 0.3m? (2000)
A.
1500 N/C
B.
5000 N/C
C.
1000 N/C
D.
2000 N/C
Show solution
Solution
E = k * |q| / r^2 = (9 × 10^9) * (5 × 10^-6) / (0.3)^2 = 5000 N/C.
Correct Answer:
B
— 5000 N/C
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Q. What is the electric field due to a uniformly charged infinite plane sheet with surface charge density σ?
A.
σ/2ε₀
B.
σ/ε₀
C.
2σ/ε₀
D.
0
Show solution
Solution
The electric field due to an infinite plane sheet is E = σ/2ε₀ on both sides, thus total E = σ/ε₀.
Correct Answer:
C
— 2σ/ε₀
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Q. What is the electric field due to a uniformly charged infinite plane sheet?
A.
0
B.
σ/2ε₀
C.
σ/ε₀
D.
σ/4ε₀
Show solution
Solution
According to Gauss's law, the electric field due to an infinite plane sheet with surface charge density σ is E = σ/2ε₀, directed away from the sheet.
Correct Answer:
B
— σ/2ε₀
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Q. What is the electric field due to a uniformly charged line of charge with linear charge density λ at a distance r from the line?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
2λ/(πε₀r)
D.
λ/(ε₀r)
Show solution
Solution
Using Gauss's law, the electric field due to a uniformly charged line of charge is E = λ/(2πε₀r).
Correct Answer:
A
— λ/(2πε₀r)
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Q. What is the electric field due to an infinite plane sheet of charge with surface charge density σ?
A.
σ/2ε₀
B.
σ/ε₀
C.
σ/4ε₀
D.
0
Show solution
Solution
The electric field due to an infinite plane sheet of charge is given by E = σ/2ε₀, directed away from the sheet if the charge is positive.
Correct Answer:
A
— σ/2ε₀
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Q. What is the electric field inside a charged conductor in electrostatic equilibrium?
A.
Zero
B.
Constant
C.
Varies with distance
D.
Depends on charge density
Show solution
Solution
Inside a charged conductor in electrostatic equilibrium, the electric field is zero.
Correct Answer:
A
— Zero
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Q. What is the electric field inside a uniformly charged hollow sphere?
A.
Zero
B.
Uniform and equal to the surface field
C.
Varies linearly with distance from the center
D.
Depends on the charge outside the sphere
Show solution
Solution
According to Gauss's law, the electric field inside a uniformly charged hollow sphere is zero.
Correct Answer:
A
— Zero
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Q. What is the electric field inside a uniformly charged spherical shell?
A.
Zero
B.
Uniform
C.
Varies linearly
D.
Depends on the charge outside
Show solution
Solution
The electric field inside a uniformly charged spherical shell is zero.
Correct Answer:
A
— Zero
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Q. What is the electric field outside a uniformly charged sphere of radius R with total charge Q?
A.
0
B.
Q/(4πε₀R²)
C.
Q/(4πε₀R)
D.
Q/(2πε₀R²)
Show solution
Solution
For a uniformly charged sphere, outside the sphere, the electric field behaves as if all the charge were concentrated at the center, E = Q/(4πε₀R²).
Correct Answer:
B
— Q/(4πε₀R²)
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Q. What is the electric flux through a closed surface surrounding a charge of -3Q?
A.
-3Q/ε₀
B.
3Q/ε₀
C.
0
D.
-6Q/ε₀
Show solution
Solution
According to Gauss's law, the electric flux through a closed surface is Φ = Q_enc/ε₀. Here, Q_enc = -3Q, so Φ = -3Q/ε₀.
Correct Answer:
A
— -3Q/ε₀
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Q. What is the electric flux through a closed surface surrounding a charge Q?
A.
0
B.
Q/ε₀
C.
Q/2ε₀
D.
Q/4ε₀
Show solution
Solution
According to Gauss's law, the electric flux Φ through a closed surface is given by Φ = Q/ε₀.
Correct Answer:
B
— Q/ε₀
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Q. What is the electric flux through a closed surface that encloses no charge?
A.
0
B.
Q/ε₀
C.
Q
D.
4πQ/ε₀
Show solution
Solution
According to Gauss's law, if there is no charge enclosed, the electric flux through the surface is zero.
Correct Answer:
A
— 0
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Q. What is the electric potential at a distance of 3 m from a charge of 10 μC?
A.
3000 V
B.
9000 V
C.
6000 V
D.
1000 V
Show solution
Solution
V = k * q / r = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (3 m) = 30000 V / 3 = 9000 V.
Correct Answer:
B
— 9000 V
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Q. What is the electric potential at a distance of 4 m from a charge of 8 μC? (2000)
A.
4500 V
B.
1800 V
C.
2000 V
D.
None of the above
Show solution
Solution
Electric potential V = k * q / r = (9 × 10^9 N m²/C²) * (8 × 10^-6 C) / (4 m) = 1800 V.
Correct Answer:
B
— 1800 V
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Q. What is the electric potential at a point 0.3m away from a charge of +4μC?
A.
12000 V
B.
3000 V
C.
6000 V
D.
9000 V
Show solution
Solution
V = k * q / r = (9 × 10^9) * (4 × 10^-6) / 0.3 = 12000 V.
Correct Answer:
A
— 12000 V
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Q. What is the electric potential at a point due to a point charge?
A.
kQ/r
B.
Q/(4πε₀r²)
C.
kQ/(4πε₀r)
D.
Q/(4πr²)
Show solution
Solution
The electric potential V at a distance r from a point charge Q is given by V = kQ/r, where k is Coulomb's constant.
Correct Answer:
A
— kQ/r
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Q. What is the electric potential at a point due to a positive point charge?
A.
Zero
B.
Positive
C.
Negative
D.
Depends on distance
Show solution
Solution
The electric potential due to a positive point charge is positive and decreases with distance from the charge.
Correct Answer:
B
— Positive
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Q. What is the electric potential at a point in space where the electric field is zero?
A.
Zero
B.
Positive
C.
Negative
D.
Undefined
Show solution
Solution
The electric potential can be positive or negative; it is not necessarily zero when the electric field is zero.
Correct Answer:
B
— Positive
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Q. What is the electric potential due to a dipole at a point along the axial line at a distance 'r' from the center of the dipole?
A.
0
B.
k * p / r²
C.
k * p / r
D.
k * p / 2r
Show solution
Solution
The potential V = k * p / r, where p is the dipole moment.
Correct Answer:
C
— k * p / r
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Q. What is the electric potential due to a point charge at a distance r?
A.
k * q / r
B.
k * q / r^2
C.
k * q * r
D.
k * q * r^2
Show solution
Solution
The electric potential V due to a point charge q at a distance r is given by V = k * q / r.
Correct Answer:
A
— k * q / r
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Q. What is the electric potential due to a point charge of 5 μC at a distance of 2 m?
A.
0 V
B.
2250 V
C.
1125 V
D.
4500 V
Show solution
Solution
Electric potential V = k * q / r = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m) = 2250 V.
Correct Answer:
B
— 2250 V
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Showing 3631 to 3660 of 5000 (167 Pages)
Physics Syllabus (JEE Main) MCQ & Objective Questions
The Physics Syllabus for JEE Main is crucial for students aiming to excel in their exams. Understanding this syllabus not only helps in grasping fundamental concepts but also enhances problem-solving skills through practice. Engaging with MCQs and objective questions is essential for effective exam preparation, as it allows students to identify important questions and strengthen their knowledge base.
What You Will Practise Here
Mechanics: Laws of Motion, Work, Energy, and Power
Thermodynamics: Laws of Thermodynamics, Heat Transfer
Waves and Oscillations: Simple Harmonic Motion, Wave Properties
Electromagnetism: Electric Fields, Magnetic Fields, and Circuits
Optics: Reflection, Refraction, and Optical Instruments
Modern Physics: Quantum Theory, Atomic Models, and Nuclear Physics
Fluid Mechanics: Properties of Fluids, Bernoulli's Principle
Exam Relevance
The Physics Syllabus (JEE Main) is integral to various examinations, including CBSE, State Boards, and competitive exams like NEET and JEE. Questions often focus on conceptual understanding and application of theories. Common patterns include numerical problems, conceptual MCQs, and assertion-reason type questions, which test both knowledge and analytical skills.
Common Mistakes Students Make
Misinterpreting the question stem, leading to incorrect answers.
Neglecting units and dimensions in calculations.
Overlooking the significance of diagrams in understanding concepts.
Confusing similar concepts, such as velocity and acceleration.
Failing to apply formulas correctly in different contexts.
FAQs
Question: What are the key topics in the Physics Syllabus for JEE Main?Answer: Key topics include Mechanics, Thermodynamics, Waves, Electromagnetism, Optics, Modern Physics, and Fluid Mechanics.
Question: How can I improve my performance in Physics MCQs?Answer: Regular practice of MCQs, understanding concepts deeply, and revising important formulas can significantly enhance your performance.
Start solving practice MCQs today to test your understanding of the Physics Syllabus (JEE Main). This will not only boost your confidence but also prepare you effectively for your upcoming exams. Remember, consistent practice is the key to success!
