Q. For a thin circular ring of mass M and radius R, what is the moment of inertia about an axis perpendicular to its plane through its center?
A.
MR^2
B.
1/2 MR^2
C.
2/3 MR^2
D.
1/3 MR^2
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Solution
The moment of inertia of a thin circular ring about an axis through its center and perpendicular to its plane is I = MR^2.
Correct Answer: A — MR^2
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Q. For a toroidal solenoid with N turns and radius R carrying current I, what is the magnetic field inside the toroid?
A.
μ₀NI/2πR
B.
μ₀NI/R
C.
μ₀NI/4πR
D.
μ₀NI/2R
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Solution
The magnetic field inside a toroid is given by B = μ₀NI/2πR.
Correct Answer: B — μ₀NI/R
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Q. For a uniformly charged sphere of radius R and total charge Q, what is the electric field at a distance r from the center where r > R?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀r)
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Solution
For r > R, the electric field behaves as if all the charge were concentrated at the center, given by E = Q/(4πε₀r²).
Correct Answer: A — Q/(4πε₀r²)
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Q. For an ideal gas, if the temperature is increased, what happens to the RMS speed?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Depends on the gas
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Solution
The RMS speed increases with temperature as v_rms = sqrt(3RT/M) shows that it is directly proportional to the square root of temperature T.
Correct Answer: A — Increases
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Q. For an ideal gas, if the volume is halved while keeping the temperature constant, what happens to the pressure?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
According to Boyle's law, for a given mass of gas at constant temperature, the pressure is inversely proportional to the volume. Halving the volume will double the pressure.
Correct Answer: B — It doubles
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Q. For an ideal gas, the equation of state is given by:
A.
PV = nRT
B.
PV = NkT
C.
PV = mRT
D.
PV = kT
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Solution
The equation of state for an ideal gas is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
Correct Answer: A — PV = nRT
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Q. For an ideal gas, which equation relates pressure, volume, and temperature?
A.
PV = nRT
B.
PV = nR
C.
PV = RT
D.
PV = nT
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Solution
The ideal gas law is given by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
Correct Answer: A — PV = nRT
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Q. For an ideal gas, which of the following equations is correct?
A.
PV = nRT
B.
PV = nR/T
C.
PV = nT/R
D.
PV = nRT^2
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Solution
The ideal gas law is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
Correct Answer: A — PV = nRT
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Q. For an ideal gas, which of the following is true at constant temperature?
A.
PV = nRT
B.
P1V1 = P2V2
C.
P/T = constant
D.
V/T = constant
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Solution
At constant temperature, Boyle's law states that the product of pressure and volume (PV) is constant, hence P1V1 = P2V2.
Correct Answer: B — P1V1 = P2V2
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Q. For an infinite plane sheet of charge with surface charge density σ, what is the electric field at any point?
A.
σ/2ε₀
B.
σ/ε₀
C.
0
D.
σ/4πε₀
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Solution
The electric field due to an infinite plane sheet of charge is constant and given by E = σ/2ε₀ on either side of the sheet.
Correct Answer: A — σ/2ε₀
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Q. For an infinite plane sheet of charge with surface charge density σ, what is the electric field at a point near the sheet?
A.
σ/2ε₀
B.
σ/ε₀
C.
0
D.
σ/4πε₀
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Solution
Using Gauss's law, the electric field due to an infinite plane sheet of charge is E = σ/2ε₀ on either side of the sheet.
Correct Answer: A — σ/2ε₀
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Q. For destructive interference to occur in a thin film, the path difference must be equal to:
A.
nλ/2 (n is an integer)
B.
nλ (n is an integer)
C.
λ/4
D.
λ/2
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Solution
Destructive interference occurs when the path difference is an odd multiple of λ/2 (i.e., (2n+1)λ/2).
Correct Answer: A — nλ/2 (n is an integer)
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Q. How does the addition of a surfactant affect the surface tension of water?
A.
Increases surface tension
B.
Decreases surface tension
C.
No effect on surface tension
D.
Surface tension becomes negative
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Solution
Surfactants lower the surface tension of water by disrupting the cohesive forces between water molecules.
Correct Answer: B — Decreases surface tension
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Q. How does the addition of salt to water affect its surface tension?
A.
Increases surface tension
B.
Decreases surface tension
C.
No effect on surface tension
D.
Salt has no effect on water
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Solution
The addition of salt to water can increase the surface tension due to the ionic interactions that enhance the cohesive forces among water molecules.
Correct Answer: A — Increases surface tension
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Q. How does the addition of soap to water affect its surface tension?
A.
Increases surface tension
B.
Decreases surface tension
C.
No effect on surface tension
D.
Changes surface tension unpredictably
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Solution
The addition of soap decreases the surface tension of water by disrupting the cohesive forces between water molecules.
Correct Answer: B — Decreases surface tension
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Q. How does the gravitational field strength vary with distance from a point mass?
A.
It increases linearly.
B.
It decreases with the square of the distance.
C.
It remains constant.
D.
It decreases linearly.
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Solution
The gravitational field strength decreases with the square of the distance from the point mass.
Correct Answer: B — It decreases with the square of the distance.
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Q. How does the gravitational force between two objects change if both masses are doubled?
A.
It becomes four times stronger
B.
It becomes twice as strong
C.
It remains the same
D.
It becomes half as strong
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Solution
Doubling both masses results in a fourfold increase in gravitational force, as it is directly proportional to the product of the masses.
Correct Answer: A — It becomes four times stronger
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Q. How does the gravitational force between two objects change if the distance between them is halved?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
According to Newton's law of gravitation, if the distance is halved, the force increases by a factor of 4 (F ∝ 1/r²).
Correct Answer: B — It quadruples
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Q. How does the gravitational force between two objects change if the mass of one object is tripled?
A.
It triples
B.
It doubles
C.
It remains the same
D.
It increases by a factor of nine
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Solution
Gravitational force is directly proportional to the product of the masses. Tripling one mass triples the force.
Correct Answer: A — It triples
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Q. How does the gravitational potential change as you move away from a planet?
A.
It increases.
B.
It decreases.
C.
It remains constant.
D.
It oscillates.
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Solution
The gravitational potential decreases as you move away from a planet.
Correct Answer: B — It decreases.
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Q. How does the gravitational potential energy of a system of two masses change as they move closer together?
A.
It increases.
B.
It decreases.
C.
It remains constant.
D.
It becomes zero.
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Solution
The gravitational potential energy decreases as the two masses move closer together.
Correct Answer: B — It decreases.
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Q. How does the gravitational potential energy of an object change when it is lifted to a height 'h' above the ground?
A.
It decreases.
B.
It increases.
C.
It remains the same.
D.
It becomes zero.
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Solution
The gravitational potential energy increases as the object is lifted to a height 'h' above the ground.
Correct Answer: B — It increases.
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Q. How many centimeters are there in a meter?
A.
10
B.
100
C.
1000
D.
10000
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Solution
There are 100 centimeters in a meter.
Correct Answer: B — 100
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Q. How many millimeters are there in a meter?
A.
10
B.
100
C.
1000
D.
10000
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Solution
There are 1000 millimeters in a meter.
Correct Answer: C — 1000
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Q. How many seconds are there in an hour?
A.
60
B.
3600
C.
600
D.
120
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Solution
There are 3600 seconds in an hour (60 seconds x 60 minutes).
Correct Answer: B — 3600
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Q. How many significant figures are in the measurement 0.004560?
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Solution
The measurement 0.004560 has 4 significant figures (the leading zeros are not counted).
Correct Answer: B — 4
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Q. How many significant figures are in the measurement 0.05060?
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Solution
Leading zeros do not count, but the trailing zero after the decimal does. Therefore, 0.05060 has 4 significant figures.
Correct Answer: B — 4
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Q. How many significant figures are in the number 0.000500?
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Solution
The significant figures are 5, 0, and the trailing zero counts, totaling 3.
Correct Answer: B — 3
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Q. How many significant figures are in the number 0.004560?
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Solution
The number 0.004560 has 4 significant figures (the leading zeros are not counted).
Correct Answer: B — 4
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Q. How many significant figures are in the number 0.00456?
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Solution
Leading zeros do not count as significant figures. The significant figures in 0.00456 are 4, 5, and 6, which totals to 3 significant figures.
Correct Answer: B — 3
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