Q. What is the impedance of a series RLC circuit at resonance?
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Solution
At resonance, the impedance of a series RLC circuit is equal to the resistance R.
Correct Answer:
A
— R
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Q. What is the induced EMF in a loop of wire if the magnetic field through the loop changes at a rate of dB/dt?
A.
-A(dB/dt)
B.
A(dB/dt)
C.
-L(dB/dt)
D.
L(dB/dt)
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Solution
The induced EMF (ε) is given by Faraday's law as ε = -A(dB/dt), where A is the area of the loop.
Correct Answer:
A
— -A(dB/dt)
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Q. What is the induced EMF in a loop of wire if the magnetic flux through it changes at a rate of dΦ/dt?
A.
-dΦ/dt
B.
dΦ/dt
C.
Φ
D.
0
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Solution
According to Faraday's law of electromagnetic induction, the induced EMF is given by ε = -dΦ/dt.
Correct Answer:
A
— -dΦ/dt
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Q. What is the induced EMF in a loop of wire when the magnetic field changes at a rate of 5 T/s?
A.
0 V
B.
5 V
C.
10 V
D.
15 V
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Solution
The induced EMF (ε) is given by Faraday's law of electromagnetic induction: ε = -dΦ/dt. If the change in magnetic field is 5 T/s, then ε = 5 V.
Correct Answer:
B
— 5 V
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Q. What is the induced EMF in a loop of wire when the magnetic field through it changes at a rate of 5 T/s?
A.
0 V
B.
5 V
C.
10 V
D.
15 V
Show solution
Solution
The induced EMF (ε) is given by Faraday's law of electromagnetic induction: ε = -dΦ/dt. If the rate of change of magnetic field is 5 T/s, then ε = 5 V.
Correct Answer:
B
— 5 V
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Q. What is the induced EMF in a loop of wire when the magnetic field through it changes at a rate of dB/dt?
A.
-dB/dt
B.
dB/dt
C.
μ₀dB/dt
D.
0
Show solution
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is given by ε = -dΦ/dt, where Φ is the magnetic flux.
Correct Answer:
A
— -dB/dt
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Q. What is the integral form of Ampere's Law?
A.
∮B·dl = μ₀I_enclosed
B.
∮E·dl = -dΦ/dt
C.
∮F·dl = m*a
D.
∮V·dl = Q/C
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Solution
The integral form of Ampere's Law states that the line integral of the magnetic field B around a closed loop is equal to μ₀ times the enclosed current I.
Correct Answer:
A
— ∮B·dl = μ₀I_enclosed
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Q. What is the latent heat of fusion for ice?
A.
334 J/g
B.
2260 J/g
C.
4190 J/g
D.
1000 J/g
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Solution
The latent heat of fusion for ice is approximately 334 J/g.
Correct Answer:
A
— 334 J/g
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Q. What is the latent heat of fusion?
A.
Heat required to change a solid to a liquid
B.
Heat required to change a liquid to a gas
C.
Heat required to change a gas to a solid
D.
Heat required to change a liquid to a solid
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Solution
Latent heat of fusion is the heat required to change a solid into a liquid at its melting point.
Correct Answer:
A
— Heat required to change a solid to a liquid
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Q. What is the magnetic field at a distance r from an infinitely long straight wire carrying current I?
A.
μ₀I/2πr
B.
μ₀I/4πr
C.
μ₀I/πr
D.
0
Show solution
Solution
Using Ampere's Law, B = μ₀I/2πr for an infinitely long straight wire.
Correct Answer:
A
— μ₀I/2πr
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Q. What is the magnetic field at a point due to a long straight current-carrying conductor using Biot-Savart Law?
A.
μ₀I/(2πr)
B.
μ₀I/(4πr²)
C.
μ₀I/(2r)
D.
μ₀I/(4r)
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Solution
The magnetic field B at a distance r from a long straight conductor carrying current I is given by B = μ₀I/(2πr) according to Biot-Savart Law.
Correct Answer:
A
— μ₀I/(2πr)
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Q. What is the magnetic field at a point on the axis of a circular loop of radius R carrying current I?
A.
μ₀I/(2R)
B.
μ₀I/(4R)
C.
μ₀I/(2R²)
D.
μ₀I/(2√2R)
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Solution
The magnetic field on the axis of a circular loop is given by B = (μ₀I/(2R)) * (1/(1 + (z/R)²)^(3/2)) where z is the distance along the axis.
Correct Answer:
D
— μ₀I/(2√2R)
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Q. What is the magnetic field at a point on the axis of a circular loop of radius R carrying a current I at a distance x from the center?
A.
(μ₀I)/(2R) * (R²/(R²+x²)^(3/2))
B.
(μ₀I)/(4R) * (R²/(R²+x²)^(3/2))
C.
(μ₀I)/(2R) * (1/(R²+x²)^(3/2))
D.
(μ₀I)/(4R) * (1/(R²+x²)^(3/2))
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Solution
The magnetic field on the axis of a circular loop is given by B = (μ₀I)/(2R) * (R²/(R²+x²)^(3/2)).
Correct Answer:
A
— (μ₀I)/(2R) * (R²/(R²+x²)^(3/2))
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Q. What is the magnetic field at the center of a circular loop of radius R carrying current I?
A.
μ₀I/2R
B.
μ₀I/R
C.
μ₀I/4R
D.
μ₀I/πR
Show solution
Solution
The magnetic field at the center of a circular loop is given by B = μ₀I/2R.
Correct Answer:
A
— μ₀I/2R
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Q. What is the magnetic field at the center of a circular loop of radius R carrying a current I?
A.
μ₀I/(2R)
B.
μ₀I/(4R)
C.
μ₀I/(R)
D.
μ₀I/(8R)
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Solution
The magnetic field at the center of a circular loop is given by B = μ₀I/(2R).
Correct Answer:
A
— μ₀I/(2R)
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Q. What is the magnetic field at the center of a square loop of side a carrying current I?
A.
μ₀I/4a
B.
μ₀I/2a
C.
μ₀I/a
D.
μ₀I/8a
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Solution
The magnetic field at the center of a square loop is B = μ₀I/4a.
Correct Answer:
A
— μ₀I/4a
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Q. What is the magnetic field at the midpoint of a wire carrying current I in opposite directions?
A.
Zero
B.
μ₀I/2
C.
μ₀I
D.
Depends on distance
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Solution
At the midpoint, the magnetic fields due to the two currents cancel each other out, resulting in zero net magnetic field.
Correct Answer:
A
— Zero
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Q. What is the magnetic field due to a circular loop of radius R carrying current I at a point on its axis at a distance x from the center?
A.
μ₀I/(2R)
B.
μ₀I/(2(x² + R²)^(3/2))
C.
μ₀I/(4πR)
D.
μ₀I/(x² + R²)
Show solution
Solution
The magnetic field at a point on the axis of a circular loop at a distance x from the center is given by B = μ₀I/(2(x² + R²)^(3/2)).
Correct Answer:
B
— μ₀I/(2(x² + R²)^(3/2))
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Q. What is the magnetic field due to a current I flowing through a straight wire at a distance d?
A.
μ₀I/(2πd)
B.
μ₀I/(4πd²)
C.
μ₀I/(d)
D.
μ₀I/(2d)
Show solution
Solution
The magnetic field B at a distance d from a straight wire carrying current I is given by B = μ₀I/(2πd).
Correct Answer:
A
— μ₀I/(2πd)
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Q. What is the magnetic field due to a current loop at a point on its axis?
A.
μ₀I/2R
B.
μ₀I/4R
C.
μ₀I/2R²
D.
μ₀I/4R²
Show solution
Solution
Using Ampere's Law, B = μ₀I/2R² at a point on the axis of a current loop.
Correct Answer:
C
— μ₀I/2R²
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Q. What is the magnetic field due to a long straight wire carrying current I at a distance r from the wire?
A.
μ₀I/2πr
B.
μ₀I/4πr
C.
μ₀I/πr
D.
μ₀I/8πr
Show solution
Solution
The magnetic field due to a long straight wire is given by B = (μ₀I)/(2πr).
Correct Answer:
A
— μ₀I/2πr
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Q. What is the magnetic field due to a magnetic dipole at a point along its axial line?
A.
(μ₀/4π) * (2m/r³)
B.
(μ₀/4π) * (m/r³)
C.
(μ₀/4π) * (m/r²)
D.
(μ₀/4π) * (m/r⁴)
Show solution
Solution
The magnetic field due to a magnetic dipole at a point along its axial line is given by B = (μ₀/4π) * (2m/r³).
Correct Answer:
A
— (μ₀/4π) * (2m/r³)
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Q. What is the magnetic field due to a solenoid of length L, carrying n turns per unit length and current I?
A.
μ₀nI
B.
μ₀nI/L
C.
μ₀nI/2
D.
μ₀nI/L²
Show solution
Solution
The magnetic field inside a long solenoid is B = μ₀nI.
Correct Answer:
A
— μ₀nI
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Q. What is the magnetic field due to a straight conductor at a point 1 meter away carrying a current of 5 A?
A.
0.01 T
B.
0.02 T
C.
0.03 T
D.
0.04 T
Show solution
Solution
Using Biot-Savart Law, B = μ₀I/(2πr) = (4π × 10^-7 Tm/A)(5 A)/(2π(1 m)) = 0.01 T.
Correct Answer:
B
— 0.02 T
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Q. What is the magnetic field due to a straight wire at a distance of 0.5 m carrying a current of 10 A?
A.
0.4 μT
B.
0.2 μT
C.
0.1 μT
D.
0.8 μT
Show solution
Solution
Using B = μ₀I/(2πr), substituting μ₀ = 4π × 10⁻⁷ Tm/A, I = 10 A, and r = 0.5 m gives B = 0.4 μT.
Correct Answer:
A
— 0.4 μT
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Q. What is the magnetic field inside a hollow conductor carrying current?
A.
Zero
B.
Uniform
C.
Varies with distance
D.
Depends on the current
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Solution
According to Ampere's law, the magnetic field inside a hollow conductor carrying current is zero.
Correct Answer:
A
— Zero
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Q. What is the magnetic field inside a hollow cylindrical shell carrying current I?
A.
0
B.
μ₀I/2πR
C.
μ₀I/4πR
D.
μ₀I/πR
Show solution
Solution
Inside a hollow cylindrical shell, the magnetic field is zero.
Correct Answer:
A
— 0
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Q. What is the magnetic field inside a long solenoid carrying current I?
A.
Zero
B.
μ₀nI
C.
μ₀I/n
D.
μ₀I/2
Show solution
Solution
The magnetic field inside a long solenoid is given by B = μ₀nI, where n is the number of turns per unit length.
Correct Answer:
B
— μ₀nI
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Q. What is the magnetic field inside a long solenoid carrying current?
A.
Zero
B.
Uniform and parallel to the axis
C.
Varies with distance
D.
Depends on the current only
Show solution
Solution
The magnetic field inside a long solenoid is uniform and parallel to the axis of the solenoid.
Correct Answer:
B
— Uniform and parallel to the axis
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Q. What is the magnetic field inside a long solenoid with n turns per unit length carrying a current I?
A.
μ₀nI
B.
μ₀I/n
C.
μ₀I/2n
D.
μ₀I/4n
Show solution
Solution
The magnetic field inside a long solenoid is given by B = μ₀nI.
Correct Answer:
A
— μ₀nI
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Showing 3841 to 3870 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!
