Q. If two point charges, +Q and -Q, are separated by a distance d, what is the magnitude of the electric field (E) at the midpoint?
-
A.
E = 0
-
B.
E = k * Q / (d/2)^2
-
C.
E = k * Q / d^2
-
D.
E = k * Q / (d^2/4)
Solution
At the midpoint, the electric fields due to both charges cancel each other out, resulting in E = 0.
Correct Answer:
A
— E = 0
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Q. If two resistors, R1 and R2, are in series, what is the equivalent resistance (R_eq)?
-
A.
R_eq = R1 + R2
-
B.
R_eq = R1 * R2
-
C.
R_eq = R1 / R2
-
D.
R_eq = R1 - R2
Solution
The equivalent resistance of resistors in series is given by R_eq = R1 + R2.
Correct Answer:
A
— R_eq = R1 + R2
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Q. In a Carnot engine operating between 300 K and 600 K, what is the maximum efficiency?
-
A.
50%
-
B.
60%
-
C.
33.33%
-
D.
100%
Solution
Efficiency = 1 - (T_cold/T_hot) = 1 - (300/600) = 0.5 or 50%.
Correct Answer:
B
— 60%
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Q. In a Carnot engine operating between 500 K and 300 K, what is the maximum efficiency?
-
A.
40%
-
B.
60%
-
C.
50%
-
D.
20%
Solution
Efficiency = 1 - (Tc/Th) = 1 - (300/500) = 0.4 or 40%.
Correct Answer:
B
— 60%
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Q. In a Carnot engine, if the hot reservoir is at 500 K and the cold reservoir is at 300 K, what is the maximum efficiency?
-
A.
40%
-
B.
60%
-
C.
50%
-
D.
20%
Solution
The maximum efficiency of a Carnot engine is given by 1 - (T_c/T_h), which is 1 - (300/500) = 0.4 or 40%.
Correct Answer:
B
— 60%
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Q. In a circuit with a 12 V battery and a 4 Ω resistor, what is the current flowing through the circuit?
-
A.
3 A
-
B.
2 A
-
C.
4 A
-
D.
1 A
Solution
Using Ohm's law, I = V/R = 12 V / 4 Ω = 3 A.
Correct Answer:
B
— 2 A
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Q. In a circuit with a 12 V battery and a resistor of 4 Ω, what is the current flowing through the circuit?
-
A.
3 A
-
B.
4 A
-
C.
2 A
-
D.
6 A
Solution
Using Ohm's Law: I = V/R = 12 V / 4 Ω = 3 A.
Correct Answer:
A
— 3 A
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Q. In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?
Solution
Total resistance = 4Ω + 8Ω = 12Ω. Current I = V / R = 12V / 12Ω = 1A. Voltage drop across 8Ω = I * R = 1A * 8Ω = 8V.
Correct Answer:
B
— 6V
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Q. In a circuit with a 24V battery and a total resistance of 6Ω, what is the total current flowing through the circuit?
Solution
Using Ohm's Law, I = V / R = 24V / 6Ω = 4A.
Correct Answer:
A
— 4A
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Q. In a circuit with a 24V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?
-
A.
8V
-
B.
12V
-
C.
16V
-
D.
20V
Solution
Total resistance R = 4Ω + 8Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop across 8Ω = I * R = 2A * 8Ω = 16V.
Correct Answer:
B
— 12V
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Q. In a circuit with a 24V battery and two resistors in series (8Ω and 4Ω), what is the voltage drop across the 4Ω resistor?
-
A.
6 V
-
B.
8 V
-
C.
12 V
-
D.
4 V
Solution
Total resistance = 8Ω + 4Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop across 4Ω = I * R = 2A * 4Ω = 8 V.
Correct Answer:
A
— 6 V
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Q. In a circuit with a 24V power supply and two resistors of 8Ω and 4Ω in series, what is the voltage drop across the 4Ω resistor?
-
A.
4.8 V
-
B.
8 V
-
C.
12 V
-
D.
16 V
Solution
Total resistance R = 8Ω + 4Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop V_R = I * R = 2A * 4Ω = 8 V.
Correct Answer:
C
— 12 V
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Q. In a discharging RC circuit, what is the expression for the voltage across the capacitor at time t?
-
A.
V = V0*e^(-t/RC)
-
B.
V = V0*(1 - e^(-t/RC))
-
C.
V = V0*t
-
D.
V = V0/t
Solution
In a discharging RC circuit, the voltage across the capacitor at time t is given by V = V0*e^(-t/RC), where V0 is the initial voltage.
Correct Answer:
A
— V = V0*e^(-t/RC)
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Q. In a double-slit experiment, what causes the interference pattern?
-
A.
Reflection
-
B.
Refraction
-
C.
Diffraction
-
D.
Interference
Solution
The interference pattern is caused by the superposition of light waves from the two slits.
Correct Answer:
D
— Interference
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Q. In a frictionless environment, if a 4 kg object is pushed with a force of 16 N, what will be its acceleration?
-
A.
2 m/s²
-
B.
4 m/s²
-
C.
6 m/s²
-
D.
8 m/s²
Solution
Acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
B
— 4 m/s²
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Q. In a heat engine, if the input heat is 1000 J and the work done is 300 J, what is the efficiency of the engine?
-
A.
30%
-
B.
70%
-
C.
50%
-
D.
10%
Solution
Efficiency is calculated as η = (Work output / Heat input) * 100%. Here, η = (300 J / 1000 J) * 100% = 30%.
Correct Answer:
A
— 30%
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Q. In a heat engine, if the input heat is 800 J and the output work is 300 J, what is the efficiency?
-
A.
0.375
-
B.
0.5
-
C.
0.25
-
D.
0.625
Solution
Efficiency = Work output / Heat input = 300 J / 800 J = 0.375 or 37.5%.
Correct Answer:
A
— 0.375
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Q. In a heat engine, what is the purpose of the heat reservoir?
-
A.
To absorb heat
-
B.
To release heat
-
C.
To convert heat to work
-
D.
To increase efficiency
Solution
The heat reservoir serves as a source of thermal energy that the heat engine absorbs heat from to perform work.
Correct Answer:
A
— To absorb heat
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Q. In a heat engine, what is the purpose of the heat sink?
-
A.
To absorb heat from the engine
-
B.
To release heat to the environment
-
C.
To increase efficiency
-
D.
To store energy
Solution
The heat sink in a heat engine is used to release excess heat to the environment, allowing the engine to operate efficiently.
Correct Answer:
B
— To release heat to the environment
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Q. In a heat engine, what is the work done equal to?
-
A.
Heat input minus heat output
-
B.
Heat output minus heat input
-
C.
Total heat energy
-
D.
Change in internal energy
Solution
The work done by a heat engine is equal to the heat input minus the heat output.
Correct Answer:
A
— Heat input minus heat output
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Q. In a p-n junction diode, what does the 'p' stand for?
-
A.
Positive charge carriers
-
B.
Photons
-
C.
Protons
-
D.
Potential difference
Solution
The 'p' in p-n junction diode stands for positive charge carriers, which are holes created by the absence of electrons.
Correct Answer:
A
— Positive charge carriers
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Q. In a p-n junction diode, what type of charge carriers are predominant in the p-type region?
-
A.
Electrons
-
B.
Holes
-
C.
Ions
-
D.
Neutrons
Solution
In the p-type region of a p-n junction diode, holes are the predominant charge carriers, created by the addition of acceptor impurities.
Correct Answer:
B
— Holes
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Q. In a parallel circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the total current supplied by the battery?
-
A.
2 A
-
B.
3 A
-
C.
4 A
-
D.
1 A
Solution
Total resistance, R_eq = 1/(1/4 + 1/6) = 2.4 Ω. Total current, I = V/R_eq = 12 V / 2.4 Ω = 5 A.
Correct Answer:
B
— 3 A
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Q. In a parallel circuit with a 12 V battery, what is the voltage across each branch?
-
A.
12 V
-
B.
6 V
-
C.
24 V
-
D.
0 V
Solution
In a parallel circuit, the voltage across each branch is equal to the source voltage, which is 12 V.
Correct Answer:
A
— 12 V
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Q. In a parallel circuit with a 12 V supply, what is the voltage across each resistor?
-
A.
12 V
-
B.
6 V
-
C.
24 V
-
D.
0 V
Solution
In a parallel circuit, the voltage across each component is equal to the supply voltage, so it is 12 V.
Correct Answer:
A
— 12 V
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Q. In a parallel circuit with a 12V battery and two resistors (4Ω and 12Ω), what is the total current supplied by the battery?
-
A.
1 A
-
B.
2 A
-
C.
3 A
-
D.
4 A
Solution
Total current I = V/R_total. R_total = 1/(1/4 + 1/12) = 3Ω. I = 12V / 3Ω = 4 A.
Correct Answer:
C
— 3 A
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Q. In a parallel circuit with a total voltage of 12 V, what is the voltage across each resistor?
-
A.
6 V
-
B.
12 V
-
C.
24 V
-
D.
0 V
Solution
In a parallel circuit, the voltage across each resistor is equal to the total voltage, so it is 12 V.
Correct Answer:
B
— 12 V
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Q. In a parallel circuit with three resistors of 2Ω, 3Ω, and 6Ω, what is the total resistance?
Solution
1/Rtotal = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1Ω.
Correct Answer:
A
— 1Ω
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Q. In a parallel circuit with two resistors of 6 ohms and 12 ohms, what is the equivalent resistance?
-
A.
4 ohms
-
B.
8 ohms
-
C.
2 ohms
-
D.
3 ohms
Solution
For parallel resistors, 1/R_eq = 1/R1 + 1/R2. Here, 1/R_eq = 1/6 + 1/12 = 2/12 + 1/12 = 3/12, so R_eq = 4 ohms.
Correct Answer:
A
— 4 ohms
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Q. In a parallel circuit with two resistors of 6Ω and 12Ω, what is the total resistance?
Solution
1/R_eq = 1/R1 + 1/R2 = 1/6 + 1/12 = 1/4. Therefore, R_eq = 4Ω.
Correct Answer:
A
— 4Ω
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