Q. What is the electric potential energy of a charge of 1 C at a point where the electric potential is 10 V?
A.
10 J
B.
1 J
C.
0 J
D.
100 J
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Solution
Electric potential energy U = q * V = 1 C * 10 V = 10 J.
Correct Answer:
A
— 10 J
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Q. What is the electric potential energy of a charge of 1 C placed in an electric field of 10 N/C at a distance of 2 m?
A.
20 J
B.
10 J
C.
5 J
D.
2 J
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Solution
Electric potential energy U = q * V = q * E * d = 1 C * 10 N/C * 2 m = 20 J.
Correct Answer:
A
— 20 J
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Q. What is the electric potential energy of a charge of 1 μC placed in an electric potential of 200 V?
A.
0.2 mJ
B.
0.1 mJ
C.
0.4 mJ
D.
0.5 mJ
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Solution
Electric potential energy U = q * V = 1 × 10^-6 C * 200 V = 0.2 mJ.
Correct Answer:
A
— 0.2 mJ
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Q. What is the electric potential energy of a system of two charges +q and -q separated by a distance r?
A.
0
B.
kq²/r
C.
-kq²/r
D.
kq/r
Show solution
Solution
The electric potential energy U = k(q1*q2)/r = k(+q*-q)/r = -kq²/r.
Correct Answer:
C
— -kq²/r
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Q. What is the electric potential energy of a system of two charges of +2 μC and -2 μC separated by 0.5 m?
A.
-72 J
B.
72 J
C.
0 J
D.
-36 J
Show solution
Solution
U = k * (q1 * q2) / r = (9 × 10^9 N m²/C²) * (2 × 10^-6 C * -2 × 10^-6 C) / 0.5 m = -72 J.
Correct Answer:
D
— -36 J
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Q. What is the electric potential energy of a system of two point charges Q1 and Q2 separated by a distance r?
A.
kQ1Q2/r
B.
kQ1Q2/2r
C.
kQ1Q2/r²
D.
kQ1Q2
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Solution
The electric potential energy U of a system of two point charges is given by U = kQ1Q2/r.
Correct Answer:
A
— kQ1Q2/r
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Q. What is the electron configuration of the element with atomic number 26?
A.
1s2 2s2 2p6 3s2 3p6 4s2 3d6
B.
1s2 2s2 2p6 3s2 3p6 4s2 3d5
C.
1s2 2s2 2p6 3s2 3p6 4s2 3d7
D.
1s2 2s2 2p6 3s2 3p6 4s2 3d8
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Solution
The electron configuration of iron (Fe), which has an atomic number of 26, is 1s2 2s2 2p6 3s2 3p6 4s2 3d6.
Correct Answer:
A
— 1s2 2s2 2p6 3s2 3p6 4s2 3d6
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Q. What is the energy band gap of silicon at room temperature?
A.
0.1 eV
B.
1.1 eV
C.
1.5 eV
D.
2.0 eV
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Solution
The energy band gap of silicon at room temperature is approximately 1.1 eV.
Correct Answer:
B
— 1.1 eV
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Q. What is the energy difference between the n=1 and n=2 levels in a hydrogen atom?
A.
10.2 eV
B.
13.6 eV
C.
1.89 eV
D.
3.4 eV
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Solution
The energy difference between n=1 and n=2 in hydrogen is 10.2 eV.
Correct Answer:
A
— 10.2 eV
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Q. What is the energy of a photon emitted during the transition from n=3 to n=2 in a hydrogen atom?
A.
10.2 eV
B.
1.89 eV
C.
12.1 eV
D.
3.4 eV
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Solution
The energy of the photon can be calculated using the Rydberg formula. The transition from n=3 to n=2 emits a photon of energy approximately 1.89 eV.
Correct Answer:
B
— 1.89 eV
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Q. What is the energy of a photon with a frequency of 5 x 10^14 Hz?
A.
3.1 x 10^-19 J
B.
2.5 x 10^-19 J
C.
4.0 x 10^-19 J
D.
6.6 x 10^-19 J
Show solution
Solution
The energy of a photon is given by E = hf. Using h = 6.63 x 10^-34 J.s, E = 6.63 x 10^-34 * 5 x 10^14 = 3.31 x 10^-19 J.
Correct Answer:
A
— 3.1 x 10^-19 J
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Q. What is the energy of a photon with a frequency of 5 x 10^14 Hz? (h = 6.63 x 10^-34 J.s)
A.
3.31 x 10^-19 J
B.
1.32 x 10^-19 J
C.
2.65 x 10^-19 J
D.
4.98 x 10^-19 J
Show solution
Solution
The energy of a photon is given by E = hν. Substituting the values gives E = 6.63 x 10^-34 J.s * 5 x 10^14 Hz = 3.31 x 10^-19 J.
Correct Answer:
A
— 3.31 x 10^-19 J
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Q. What is the energy of a photon with a frequency of 6 x 10^14 Hz?
A.
3.31 x 10^-19 J
B.
1.24 x 10^-19 J
C.
4.14 x 10^-19 J
D.
2.00 x 10^-19 J
Show solution
Solution
The energy of a photon is given by E = h * f, where h = 6.63 x 10^-34 J·s. Thus, E = 6.63 x 10^-34 * 6 x 10^14 = 4.14 x 10^-19 J.
Correct Answer:
C
— 4.14 x 10^-19 J
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Q. What is the energy of a photon with a wavelength of 500 nm?
A.
3.98 eV
B.
2.48 eV
C.
1.24 eV
D.
0.62 eV
Show solution
Solution
The energy of a photon is given by E = hc/λ. Using h = 6.626 x 10^-34 J·s and c = 3 x 10^8 m/s, E = (6.626 x 10^-34)(3 x 10^8)/(500 x 10^-9) = 3.98 eV.
Correct Answer:
A
— 3.98 eV
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Q. What is the energy of the ground state of a hydrogen atom?
A.
-13.6 eV
B.
-3.4 eV
C.
0 eV
D.
13.6 eV
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Solution
The energy of the ground state of a hydrogen atom is -13.6 eV.
Correct Answer:
A
— -13.6 eV
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Q. What is the energy stored in a capacitor of capacitance 10μF charged to a potential difference of 50V?
A.
0.0125 J
B.
0.025 J
C.
0.05 J
D.
0.1 J
Show solution
Solution
U = 1/2 * C * V² = 1/2 * (10 × 10^-6 F) * (50 V)² = 0.0125 J.
Correct Answer:
B
— 0.025 J
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Q. What is the energy stored in a capacitor of capacitance C charged to a voltage V?
A.
1/2 CV
B.
CV
C.
1/2 C/V
D.
C/V
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Solution
The energy (U) stored in a capacitor is given by the formula U = 1/2 CV².
Correct Answer:
A
— 1/2 CV
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Q. What is the energy stored in a capacitor with capacitance C charged to a voltage V?
A.
1/2 CV²
B.
CV
C.
1/2 V²/C
D.
C²V
Show solution
Solution
The energy (U) stored in a capacitor is given by the formula U = 1/2 CV².
Correct Answer:
A
— 1/2 CV²
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Q. What is the energy stored in a capacitor with capacitance C charged to voltage V?
A.
1/2 CV
B.
CV
C.
1/2 C/V
D.
C/V
Show solution
Solution
The energy (U) stored in a capacitor is given by the formula U = 1/2 CV².
Correct Answer:
A
— 1/2 CV
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Q. What is the entropy change for a reversible isothermal expansion of an ideal gas?
A.
nR ln(Vf/Vi)
B.
0
C.
nR(Tf - Ti)
D.
nC ln(Vf/Vi)
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Solution
The entropy change for a reversible isothermal expansion of an ideal gas is ΔS = nR ln(Vf/Vi).
Correct Answer:
A
— nR ln(Vf/Vi)
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Q. What is the entropy change for a reversible isothermal process?
A.
Zero
B.
nR ln(Vf/Vi)
C.
nR(Tf - Ti)
D.
nR ln(Tf/Ti)
Show solution
Solution
The entropy change for a reversible isothermal process is ΔS = nR ln(Vf/Vi).
Correct Answer:
B
— nR ln(Vf/Vi)
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Q. What is the entropy change for a reversible process?
A.
Always positive
B.
Always negative
C.
Can be zero
D.
Depends on the path taken
Show solution
Solution
For a reversible process, the entropy change can be zero if the process is isothermal and reversible.
Correct Answer:
C
— Can be zero
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Q. What is the equation for the displacement of a damped harmonic oscillator?
A.
x(t) = A e^(-bt) cos(ωt)
B.
x(t) = A e^(bt) cos(ωt)
C.
x(t) = A cos(ωt)
D.
x(t) = A e^(-bt) sin(ωt)
Show solution
Solution
The displacement of a damped harmonic oscillator is given by x(t) = A e^(-bt) cos(ωt), where b is the damping coefficient.
Correct Answer:
A
— x(t) = A e^(-bt) cos(ωt)
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Q. What is the equation of motion for a damped harmonic oscillator?
A.
m d²x/dt² + b dx/dt + kx = 0
B.
m d²x/dt² + kx = 0
C.
m d²x/dt² + b dx/dt = 0
D.
m d²x/dt² + b dx/dt + kx = F(t)
Show solution
Solution
The equation of motion for a damped harmonic oscillator is m d²x/dt² + b dx/dt + kx = 0.
Correct Answer:
A
— m d²x/dt² + b dx/dt + kx = 0
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Q. What is the equation of motion for a simple harmonic oscillator with amplitude A and angular frequency ω?
A.
x(t) = A cos(ωt)
B.
x(t) = A sin(ωt)
C.
x(t) = A e^(ωt)
D.
x(t) = A ωt
Show solution
Solution
The equation of motion for SHM is x(t) = A cos(ωt) or x(t) = A sin(ωt).
Correct Answer:
A
— x(t) = A cos(ωt)
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Q. What is the equivalent of 1 liter in cubic meters?
A.
0.001 m³
B.
0.01 m³
C.
0.1 m³
D.
1 m³
Show solution
Solution
1 liter is equal to 0.001 cubic meters.
Correct Answer:
A
— 0.001 m³
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Q. What is the equivalent resistance of a 10Ω and a 5Ω resistor connected in parallel?
A.
3.33Ω
B.
7.5Ω
C.
15Ω
D.
5Ω
Show solution
Solution
R_eq = 1/(1/10 + 1/5) = 3.33Ω.
Correct Answer:
A
— 3.33Ω
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Q. What is the equivalent resistance of a 6Ω and a 3Ω resistor connected in parallel?
A.
2Ω
B.
4Ω
C.
1.5Ω
D.
9Ω
Show solution
Solution
R_eq = 1/(1/6 + 1/3) = 2Ω.
Correct Answer:
B
— 4Ω
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Q. What is the equivalent resistance of a circuit with a 10 ohm and a 5 ohm resistor in series?
A.
15 ohms
B.
5 ohms
C.
10 ohms
D.
20 ohms
Show solution
Solution
For resistors in series, R_eq = R1 + R2 = 10 ohms + 5 ohms = 15 ohms.
Correct Answer:
A
— 15 ohms
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Q. What is the equivalent resistance of a circuit with two 10Ω resistors in series and one 5Ω resistor in parallel with them?
A.
5Ω
B.
10Ω
C.
15Ω
D.
20Ω
Show solution
Solution
R_series = 10Ω + 10Ω = 20Ω. R_parallel = 1/(1/20 + 1/5) = 4Ω. Therefore, R_eq = 4Ω.
Correct Answer:
C
— 15Ω
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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!
