Q. At what temperature will the RMS speed of a gas be 1000 m/s if its molar mass is 0.044 kg/mol? (R = 8.314 J/(mol K))
A.
500 K
B.
600 K
C.
700 K
D.
800 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), we solve for T: T = (v_rms^2 * M) / (3R) = (1000^2 * 0.044) / (3 * 8.314) = 700 K.
Correct Answer:
C
— 700 K
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Q. At what temperature will the RMS speed of a gas be 300 m/s if its molar mass is 28 g/mol?
A.
300 K
B.
600 K
C.
900 K
D.
1200 K
Show solution
Solution
Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T. Setting v_rms = 300 m/s and M = 28 g/mol, we find T = (M * v_rms^2)/(3R) = 600 K.
Correct Answer:
B
— 600 K
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Q. At what temperature will the RMS speed of a gas be 500 m/s if its molar mass is 0.02 kg/mol? (2000)
A.
250 K
B.
500 K
C.
1000 K
D.
2000 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), rearranging gives T = (v_rms^2 * M) / (3R). Substituting values gives T = 500 K.
Correct Answer:
B
— 500 K
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Q. At what temperature will the RMS speed of a gas be 600 m/s if its molar mass is 0.02 kg/mol?
A.
300 K
B.
600 K
C.
900 K
D.
1200 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Plugging in values gives T = (600^2 * 0.02) / (3 * 8.314) = 900 K.
Correct Answer:
C
— 900 K
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Q. Calculate the moment of inertia of a hollow sphere of mass M and radius R about an axis through its center.
A.
2/5 MR^2
B.
3/5 MR^2
C.
2/3 MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a hollow sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer:
B
— 3/5 MR^2
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Q. Calculate the RMS speed of a gas with molar mass 0.028 kg/mol at 300 K. (R = 8.314 J/(mol K))
A.
500 m/s
B.
600 m/s
C.
700 m/s
D.
800 m/s
Show solution
Solution
Using v_rms = sqrt(3RT/M), we find v_rms = sqrt(3 * 8.314 * 300 / 0.028) = 600 m/s.
Correct Answer:
B
— 600 m/s
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Q. Convert 5 kilometers to meters.
A.
500
B.
5000
C.
50
D.
5
Show solution
Solution
5 kilometers is equal to 5000 meters.
Correct Answer:
B
— 5000
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Q. Determine the moment of inertia of a solid sphere of mass M and radius R about an axis through its center.
A.
2/5 MR^2
B.
3/5 MR^2
C.
4/5 MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer:
A
— 2/5 MR^2
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Q. During a phase change, the temperature of a substance:
A.
Increases
B.
Decreases
C.
Remains constant
D.
Varies unpredictably
Show solution
Solution
During a phase change, the temperature of a substance remains constant while the substance absorbs or releases heat.
Correct Answer:
C
— Remains constant
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Q. During an isochoric process, the volume of the gas:
A.
Increases
B.
Decreases
C.
Remains constant
D.
Varies with temperature
Show solution
Solution
In an isochoric process, the volume of the gas remains constant.
Correct Answer:
C
— Remains constant
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Q. During an isochoric process, the volume of the system:
A.
Increases
B.
Decreases
C.
Remains constant
D.
Varies with temperature
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Solution
In an isochoric process, the volume of the system remains constant.
Correct Answer:
C
— Remains constant
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Q. During an isothermal expansion of an ideal gas, what happens to the internal energy?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Depends on the amount of gas
Show solution
Solution
In an isothermal process for an ideal gas, the internal energy remains constant because the temperature does not change.
Correct Answer:
C
— Remains constant
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Q. For a charged plane sheet, if the surface charge density is doubled, what happens to the electric field?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
Show solution
Solution
The electric field due to a charged plane sheet is directly proportional to the surface charge density. Therefore, if σ is doubled, E also doubles.
Correct Answer:
B
— It doubles
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Q. For a charged sphere, what happens to the electric field inside the sphere as the radius increases?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
Show solution
Solution
The electric field inside a uniformly charged sphere is zero.
Correct Answer:
D
— Becomes zero
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Q. For a charged spherical conductor, what happens to the electric field inside the conductor when it is charged?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
Show solution
Solution
The electric field inside a charged conductor in electrostatic equilibrium is zero.
Correct Answer:
D
— Becomes zero
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Q. For a circular loop of radius R carrying a current I, what is the magnetic field at the center of the loop?
A.
B = μ₀I/(2R)
B.
B = μ₀I/(4R)
C.
B = μ₀I/(πR)
D.
B = μ₀I/(2πR)
Show solution
Solution
The magnetic field at the center of a circular loop of radius R carrying current I is given by B = μ₀I/(2πR).
Correct Answer:
D
— B = μ₀I/(2πR)
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Q. For a closed loop of wire carrying current, what does the line integral of the magnetic field equal?
A.
Zero
B.
The product of current and resistance
C.
μ₀ times the total current enclosed
D.
The electric field times the area
Show solution
Solution
According to Ampere's Law, the line integral of the magnetic field around a closed loop equals μ₀ times the total current enclosed by the loop.
Correct Answer:
C
— μ₀ times the total current enclosed
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Q. For a closed surface enclosing multiple charges, how is the total electric flux related to the enclosed charges?
A.
It is proportional to the sum of the charges
B.
It is inversely proportional to the sum of the charges
C.
It is independent of the charges
D.
It is proportional to the square of the charges
Show solution
Solution
According to Gauss's law, the total electric flux through a closed surface is proportional to the total charge enclosed.
Correct Answer:
A
— It is proportional to the sum of the charges
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Q. For a closed surface enclosing multiple charges, how is the total electric flux calculated?
A.
Sum of individual fluxes
B.
Product of charges
C.
Sum of enclosed charges divided by ε₀
D.
Average of charges
Show solution
Solution
The total electric flux through a closed surface is given by Φ = ΣQ_enc/ε₀, where Q_enc is the total charge enclosed.
Correct Answer:
C
— Sum of enclosed charges divided by ε₀
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Q. For a composite body made of a solid cylinder and a solid sphere, how do you calculate the total moment of inertia about the same axis?
A.
Add the individual moments
B.
Multiply the individual moments
C.
Subtract the individual moments
D.
Divide the individual moments
Show solution
Solution
The total moment of inertia of a composite body about the same axis is the sum of the individual moments of inertia.
Correct Answer:
A
— Add the individual moments
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Q. For a composite body made of two solid cylinders of mass M1 and M2 and radius R, what is the total moment of inertia about the same axis?
A.
I1 + I2
B.
I1 - I2
C.
I1 * I2
D.
I1 / I2
Show solution
Solution
The total moment of inertia of a composite body is the sum of the individual moments of inertia: I_total = I1 + I2.
Correct Answer:
A
— I1 + I2
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Q. For a current-carrying loop, what is the magnetic field at the center if the radius is halved?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
Show solution
Solution
The magnetic field at the center of a loop is inversely proportional to the radius. If the radius is halved, the magnetic field quadruples.
Correct Answer:
C
— It quadruples
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Q. For a cylindrical conductor of radius R carrying current I, what is the magnetic field at a point outside the conductor?
A.
0
B.
μ₀I/2πR
C.
μ₀I/4πR
D.
μ₀I/πR
Show solution
Solution
Using Ampere's Law, B = μ₀I/2πR for points outside the cylindrical conductor.
Correct Answer:
B
— μ₀I/2πR
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Q. For a cylindrical conductor of radius R carrying current I, what is the magnetic field at a point outside the cylinder?
A.
0
B.
μ₀I/2πr
C.
μ₀I/4πr
D.
μ₀I/πr
Show solution
Solution
For points outside the cylinder, B = μ₀I/2πr.
Correct Answer:
B
— μ₀I/2πr
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Q. For a damped oscillator, what is the relationship between the natural frequency and the damped frequency?
A.
Damped frequency is greater
B.
Damped frequency is equal
C.
Damped frequency is less
D.
No relationship
Show solution
Solution
The damped frequency is less than the natural frequency due to the effect of damping.
Correct Answer:
C
— Damped frequency is less
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Q. For a diffraction grating with 500 lines per mm, what is the angle of the first order maximum for light of wavelength 600 nm?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
15 degrees
Show solution
Solution
Using the grating equation d sin θ = nλ, where d = 1/500000 m, n = 1, and λ = 600 x 10^-9 m, we find θ ≈ 30 degrees.
Correct Answer:
A
— 30 degrees
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Q. For a diffraction pattern produced by a single slit, how does the width of the central maximum change if the slit width is halved?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Show solution
Solution
If the slit width is halved, the width of the central maximum increases because the angle for the first minimum increases.
Correct Answer:
A
— Increases
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Q. For a diffraction pattern produced by a single slit, how does the width of the central maximum compare to the other maxima?
A.
Wider than all other maxima
B.
Narrower than all other maxima
C.
Equal to all other maxima
D.
None of the above
Show solution
Solution
The central maximum in a single-slit diffraction pattern is wider than all other maxima.
Correct Answer:
A
— Wider than all other maxima
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Q. For a gas at 300 K, if the RMS speed is 500 m/s, what will be the RMS speed at 600 K?
A.
500 m/s
B.
707 m/s
C.
1000 m/s
D.
250 m/s
Show solution
Solution
RMS speed is proportional to the square root of temperature, so v_rms at 600 K = 500 * sqrt(600/300) = 707 m/s.
Correct Answer:
B
— 707 m/s
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Q. For a gas at 300 K, what is the RMS speed if the molar mass is 0.028 kg/mol?
A.
500 m/s
B.
600 m/s
C.
700 m/s
D.
800 m/s
Show solution
Solution
Using v_rms = sqrt(3RT/M), we calculate v_rms = sqrt(3 * 8.314 * 300 / 0.028) which gives approximately 600 m/s.
Correct Answer:
B
— 600 m/s
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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!