Q. What is the voltage across a 10Ω resistor if a current of 3A flows through it?
A.
30 V
B.
15 V
C.
10 V
D.
20 V
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Solution
Using Ohm's law, V = I * R = 3A * 10Ω = 30 V.
Correct Answer:
A
— 30 V
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Q. What is the voltage across a 10Ω resistor if the current flowing through it is 3A?
A.
30V
B.
10V
C.
15V
D.
20V
Show solution
Solution
Using Ohm's law, V = I * R = 3A * 10Ω = 30V.
Correct Answer:
A
— 30V
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Q. What is the voltage across a 5 Ω resistor carrying a current of 3 A?
A.
15 V
B.
10 V
C.
5 V
D.
20 V
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Solution
Using Ohm's law, V = I * R = 3 A * 5 Ω = 15 V.
Correct Answer:
A
— 15 V
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Q. What is the voltage across a capacitor (V) after one time constant in an RC charging circuit?
A.
V = V0(1 - e^(-t/τ))
B.
V = V0 * e^(-t/τ)
C.
V = V0(1 + e^(-t/τ))
D.
V = V0 * t/τ
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Solution
The voltage across a charging capacitor after one time constant is V = V0(1 - e^(-t/τ)).
Correct Answer:
A
— V = V0(1 - e^(-t/τ))
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Q. What is the voltage across a capacitor after 3 time constants if it was charged to 12 V?
A.
12 V
B.
8 V
C.
4 V
D.
0 V
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Solution
After 3 time constants, a capacitor is considered fully charged, so the voltage remains at 12 V.
Correct Answer:
A
— 12 V
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Q. What is the voltage across a capacitor after 3 time constants in an RC circuit with a supply voltage of 12 V?
A.
12 V
B.
8 V
C.
4 V
D.
0 V
Show solution
Solution
Voltage across capacitor V(t) = V0 * (1 - e^(-t/τ)). After 3τ, V(3τ) = 12 V * (1 - e^(-3)) ≈ 12 V * (1 - 0.05) ≈ 11.4 V.
Correct Answer:
B
— 8 V
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Q. What is the voltage across a capacitor after one time constant in an RC charging circuit?
A.
63.2% of the supply voltage
B.
50% of the supply voltage
C.
100% of the supply voltage
D.
36.8% of the supply voltage
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Solution
After one time constant, the voltage across the capacitor reaches approximately 63.2% of the supply voltage.
Correct Answer:
A
— 63.2% of the supply voltage
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Q. What is the voltage across the capacitor after one time constant in an RC charging circuit?
A.
V(1 - e^(-1))
B.
V(1 + e^(-1))
C.
V/e
D.
V
Show solution
Solution
After one time constant, the voltage across the capacitor is V(1 - e^(-1)), where V is the applied voltage.
Correct Answer:
A
— V(1 - e^(-1))
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Q. What is the voltage across the capacitor in an RC circuit after one time constant?
A.
0.63V
B.
0.37V
C.
1.0V
D.
0.5V
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Solution
After one time constant (τ), the voltage across the capacitor reaches approximately 63% of the applied voltage V.
Correct Answer:
A
— 0.63V
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Q. What is the work done against gravity when lifting a 5 kg object to a height of 2 m?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Work = mgh = 5 kg × 9.8 m/s² × 2 m = 98 J.
Correct Answer:
B
— 20 J
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Q. What is the work done by a constant force F acting on a particle that moves a distance d in the direction of the force?
A.
F * d
B.
F / d
C.
F + d
D.
F - d
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Solution
The work done by a constant force is calculated as the product of the force and the distance moved in the direction of the force, W = F * d.
Correct Answer:
A
— F * d
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Q. What is the work done by a constant force on a rigid body moving in the direction of the force?
A.
Force × Distance
B.
Force × Distance × cos(θ)
C.
Force × Distance × sin(θ)
D.
Force / Distance
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Solution
The work done is calculated as W = Fd cos(θ), where θ is the angle between the force and the direction of motion.
Correct Answer:
B
— Force × Distance × cos(θ)
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Q. What is the work done by a constant force?
A.
W = Fd cos(θ)
B.
W = Fd sin(θ)
C.
W = F + d
D.
W = F/d
Show solution
Solution
The work done by a constant force is calculated using the formula W = Fd cos(θ), where θ is the angle between the force and the direction of motion.
Correct Answer:
A
— W = Fd cos(θ)
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Q. What is the work done by a force of 15 N acting at an angle of 60 degrees to the direction of motion over a distance of 4 m?
A.
30 J
B.
60 J
C.
120 J
D.
15 J
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Solution
Work = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
B
— 60 J
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Q. What is the work done by a gas that expands isothermally from volume V1 to V2 at temperature T?
A.
nRT ln(V2/V1)
B.
nRT (V2 - V1)
C.
P(V2 - V1)
D.
0
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Solution
For isothermal expansion, the work done is given by W = nRT ln(V2/V1).
Correct Answer:
A
— nRT ln(V2/V1)
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Q. What is the work done by a torque of 15 N·m that rotates an object through an angle of 60 degrees?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
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Solution
Work done W = τθ, where θ = 60 degrees = π/3 radians. Thus, W = 15 N·m * π/3 = 15 * 1.047 = 15.71 J.
Correct Answer:
B
— 30 J
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Q. What is the work done by a torque of 5 N·m that rotates an object through an angle of 90 degrees?
A.
0.5 J
B.
1 J
C.
2.5 J
D.
5 J
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Solution
Work done by torque is W = τθ, where θ = π/2 rad. Thus, W = 5 N·m * (π/2) = 5 * 1.57 = 7.85 J.
Correct Answer:
C
— 2.5 J
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Q. What is the work done when a force of 10 N moves an object 3 meters in the direction of the force?
A.
30 J
B.
10 J
C.
3 J
D.
0 J
Show solution
Solution
Work done is calculated as W = F * d * cos(θ). Here, θ = 0 degrees, so W = 10 N * 3 m * cos(0) = 30 J.
Correct Answer:
A
— 30 J
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Q. What is the work done when a force of 15 N moves an object 3 meters in the direction of the force?
A.
45 J
B.
30 J
C.
15 J
D.
60 J
Show solution
Solution
Work done W = F × d = 15 N × 3 m = 45 J.
Correct Answer:
A
— 45 J
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Q. What is the work done when a force of 50 N moves an object 3 meters?
A.
150 J
B.
100 J
C.
200 J
D.
50 J
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Solution
Work W = F * d = 50 N * 3 m = 150 J.
Correct Answer:
A
— 150 J
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Q. What is the work done when a force of 50 N moves an object 4 meters in the direction of the force?
A.
100 J
B.
200 J
C.
150 J
D.
50 J
Show solution
Solution
Work W = F * d = 50 N * 4 m = 200 J.
Correct Answer:
B
— 200 J
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Q. What is the work function of a material in the context of the photoelectric effect?
A.
The energy required to remove an electron from the material
B.
The energy of the incident photon
C.
The energy of the emitted electron
D.
The total energy of the material
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Solution
The work function is the minimum energy required to remove an electron from the surface of a material.
Correct Answer:
A
— The energy required to remove an electron from the material
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Q. What phenomenon does the photoelectric effect demonstrate?
A.
Light can behave as both a wave and a particle.
B.
Electrons can be emitted from a material when it absorbs light.
C.
Energy is conserved in all processes.
D.
Mass can be converted into energy.
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Solution
The photoelectric effect shows that when light of sufficient frequency strikes a material, it can cause the emission of electrons, demonstrating the particle-like properties of light.
Correct Answer:
B
— Electrons can be emitted from a material when it absorbs light.
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Q. What phenomenon explains the discrete lines in atomic spectra?
A.
Blackbody radiation
B.
Quantum transitions between energy levels
C.
Thermal expansion
D.
Photoelectric effect
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Solution
The discrete lines in atomic spectra are explained by quantum transitions between energy levels, where electrons absorb or emit photons.
Correct Answer:
B
— Quantum transitions between energy levels
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Q. What phenomenon explains the discrete lines observed in atomic spectra?
A.
Blackbody radiation
B.
Quantum transitions between energy levels
C.
Thermal expansion
D.
Photoelectric effect
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Solution
The discrete lines in atomic spectra are due to quantum transitions between energy levels of electrons in an atom.
Correct Answer:
B
— Quantum transitions between energy levels
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Q. What phenomenon explains the emission of electrons from a material when light shines on it?
A.
Photoelectric effect
B.
Compton scattering
C.
Rayleigh scattering
D.
Blackbody radiation
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Solution
The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light of sufficient frequency.
Correct Answer:
A
— Photoelectric effect
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Q. What phenomenon explains the emission of electrons from a metal surface when light shines on it?
A.
Photoelectric effect
B.
Compton scattering
C.
Electron diffraction
D.
Thermal emission
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Solution
The photoelectric effect occurs when light of sufficient frequency strikes a metal surface, causing the emission of electrons due to the absorption of energy from the photons.
Correct Answer:
A
— Photoelectric effect
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Q. What phenomenon occurs when light passes through a narrow slit?
A.
Reflection.
B.
Refraction.
C.
Diffraction.
D.
Interference.
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Solution
Diffraction occurs when light passes through a narrow slit, causing it to spread out and create a pattern of light and dark regions.
Correct Answer:
C
— Diffraction.
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Q. What phenomenon occurs when two waves meet and combine to form a new wave?
A.
Diffraction
B.
Interference
C.
Refraction
D.
Reflection
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Solution
The phenomenon that occurs when two waves meet and combine to form a new wave is called interference.
Correct Answer:
B
— Interference
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Q. What type of energy is associated with a rolling object due to its motion?
A.
Kinetic energy
B.
Potential energy
C.
Thermal energy
D.
Elastic energy
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Solution
A rolling object possesses kinetic energy due to both its translational and rotational motion.
Correct Answer:
A
— Kinetic energy
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