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
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
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
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
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
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
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
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
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
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
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
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
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
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B.
It doubles
-
C.
It halves
-
D.
It quadruples
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
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
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)
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
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C.
μ₀ times the total current enclosed
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D.
The electric field times the area
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 calculated?
-
A.
Sum of individual fluxes
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B.
Product of charges
-
C.
Sum of enclosed charges divided by ε₀
-
D.
Average of charges
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 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
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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
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 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
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B.
Multiply the individual moments
-
C.
Subtract the individual moments
-
D.
Divide the individual moments
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
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
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
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C.
μ₀I/4πR
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D.
μ₀I/πR
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
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
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
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 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
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 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
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 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
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
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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