Major Competitive Exams
Q. What is the effect on the total current in a circuit if an additional resistor is added in parallel to an existing resistor?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Adding a resistor in parallel decreases the total resistance, which increases the total current according to Ohm's law.
Correct Answer: A — Increases
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Q. What is the effect on the total resistance in a circuit when more resistors are added in series?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Adding more resistors in series increases the total resistance, as total resistance is the sum of individual resistances.
Correct Answer: A — Increases
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Q. What is the effect on the total resistance in a series circuit if one resistor is removed? (2023)
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Removing a resistor from a series circuit decreases the total resistance since the sum of the remaining resistances is less.
Correct Answer: B — Decreases
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Q. What is the effect on the total resistance when a resistor is added in parallel to an existing resistor?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Doubles
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Solution
Adding a resistor in parallel decreases the total resistance of the circuit.
Correct Answer: B — Decreases
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Q. What is the effect on the total resistance when more resistors are added in parallel?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Depends on the resistors
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Solution
Adding resistors in parallel decreases the total resistance.
Correct Answer: B — Decreases
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Q. What is the effect on the total resistance when more resistors are added in parallel to a circuit?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Doubles
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Solution
Adding more resistors in parallel decreases the total resistance of the circuit.
Correct Answer: B — Decreases
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Q. What is the effect on the total resistance when more resistors are added in series?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Depends on the resistors
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Solution
Total resistance increases when resistors are added in series.
Correct Answer: A — Increases
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Q. What is the effect on the total resistance when more resistors are added in series to a circuit?
A.
It decreases
B.
It increases
C.
It remains the same
D.
It becomes zero
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Solution
In series, the total resistance increases as more resistors are added.
Correct Answer: B — It increases
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Q. What is the efficiency of a Carnot engine operating between 300 K and 600 K? (2022)
A.
0.5
B.
0.33
C.
0.25
D.
0.67
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Solution
Efficiency = 1 - (T_c/T_h) = 1 - (300/600) = 0.5.
Correct Answer: A — 0.5
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Q. What is the efficiency of a Carnot engine operating between 500 K and 300 K? (2022)
A.
0.4
B.
0.5
C.
0.6
D.
0.7
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Solution
Efficiency = 1 - (T2/T1) = 1 - (300/500) = 0.4 or 40%.
Correct Answer: C — 0.6
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Q. What is the efficiency of a Carnot engine operating between a hot reservoir at 600 K and a cold reservoir at 300 K?
A.
0.5
B.
0.33
C.
0.25
D.
0.75
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Solution
Efficiency (η) = 1 - (T_c/T_h) = 1 - (300/600) = 0.5 or 50%.
Correct Answer: B — 0.33
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Q. What is the efficiency of a Carnot engine operating between temperatures of 500 K and 300 K?
A.
0.4
B.
0.5
C.
0.6
D.
0.7
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Solution
Efficiency (η) = 1 - (T2/T1) = 1 - (300/500) = 0.4 or 40%.
Correct Answer: C — 0.6
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Q. What is the efficiency of a Carnot engine operating between temperatures T1 and T2? (2019) 2019
A.
(T1 - T2) / T1
B.
T2 / T1
C.
(T1 - T2) / T2
D.
1 - (T2 / T1)
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Solution
The efficiency (η) of a Carnot engine is given by η = 1 - (T2 / T1), where T1 is the higher temperature and T2 is the lower temperature.
Correct Answer: D — 1 - (T2 / T1)
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Q. What is the efficiency of a Carnot engine operating between temperatures T1 and T2?
A.
(T1 - T2) / T1
B.
T2 / T1
C.
(T1 - T2) / T2
D.
1 - (T2 / T1)
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Solution
The efficiency of a Carnot engine is given by the formula: efficiency = 1 - (T2 / T1), where T1 is the higher temperature and T2 is the lower temperature.
Correct Answer: D — 1 - (T2 / T1)
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Q. What is the electric field at a distance d from an infinitely long charged wire with linear charge density λ?
A.
λ/(2πε₀d)
B.
λ/(4πε₀d²)
C.
λ/(2πε₀d²)
D.
0
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Solution
The electric field due to an infinitely long charged wire is given by E = λ/(2πε₀d).
Correct Answer: A — λ/(2πε₀d)
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Q. What is the electric field at a distance of 1m from a charge of +1μC?
A.
9 × 10^9 N/C
B.
9 × 10^6 N/C
C.
9 × 10^3 N/C
D.
9 × 10^12 N/C
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Solution
Electric field E = k * q / r^2 = (9 × 10^9) * (1 × 10^-6) / (1)^2 = 9 × 10^3 N/C.
Correct Answer: B — 9 × 10^6 N/C
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Q. What is the electric field at a distance r from a uniformly charged disk of radius R and surface charge density σ?
A.
σ/(2ε₀)
B.
σ/(4ε₀)
C.
σ/(2ε₀) * (1 - r/√(R² + r²))
D.
Zero
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Solution
The electric field at a distance r from a uniformly charged disk is given by E = σ/(2ε₀) * (1 - r/√(R² + r²)).
Correct Answer: C — σ/(2ε₀) * (1 - r/√(R² + r²))
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Q. What is the electric field at a distance r from a uniformly charged sphere of radius R and total charge Q, when r > R?
A.
Q/(4πε₀r²)
B.
Q/(4πε₀R²)
C.
Zero
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, thus E = Q/(4πε₀r²).
Correct Answer: A — Q/(4πε₀r²)
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Q. What is the electric field at a distance r from an infinitely long line charge with linear charge density λ?
A.
λ/2πε₀r
B.
λ/4πε₀r²
C.
λ/ε₀r
D.
λ/2ε₀r²
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Solution
The electric field due to an infinite line charge is given by E = λ/2πε₀r.
Correct Answer: A — λ/2πε₀r
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Q. What is the electric field at a distance r from an infinitely long line of charge with linear charge density λ?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
λ/(2πε₀r²)
D.
0
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Solution
The electric field due to an infinitely long line of charge is given by E = λ/(2πε₀r), directed radially outward from the line.
Correct Answer: A — λ/(2πε₀r)
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Q. What is the electric field at a point due to a positive charge?
A.
Directed towards the charge
B.
Directed away from the charge
C.
Zero
D.
Depends on the distance from the charge
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Solution
The electric field due to a positive charge is directed away from the charge.
Correct Answer: B — Directed away from the charge
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Q. What is the electric field at a point due to a uniformly charged infinite plane sheet? (2020)
A.
0
B.
σ/2ε₀
C.
σ/ε₀
D.
σ/4ε₀
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Solution
The electric field due to an infinite plane sheet is E = σ/2ε₀ on both sides, but the total field is σ/ε₀.
Correct Answer: C — σ/ε₀
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Q. What is the electric field at a point just outside a charged conductor?
A.
0
B.
σ/ε₀
C.
σ/2ε₀
D.
σ/4ε₀
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Solution
The electric field just outside a charged conductor is given by E = σ/ε₀, where σ is the surface charge density.
Correct Answer: B — σ/ε₀
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Q. What is the electric field at a point midway between two equal and opposite charges?
A.
Zero
B.
Maximum
C.
Minimum
D.
Depends on distance
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Solution
The electric fields due to both charges cancel each other out at the midpoint, resulting in zero electric field.
Correct Answer: A — Zero
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Q. What is the electric field at a point on the axis of a dipole at a distance d from the center of the dipole?
A.
0
B.
p/(4πε₀d²)
C.
p/(2πε₀d²)
D.
p/(4πε₀d³)
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Solution
The electric field along the axis of a dipole at a distance d is given by E = (1/(4πε₀)) * (2p/d³), where p is the dipole moment.
Correct Answer: D — p/(4πε₀d³)
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Q. What is the electric field at a point outside a uniformly charged sphere of radius R and total charge Q?
A.
0
B.
Q/(4πε₀R²)
C.
Q/(4πε₀R)
D.
Q/(2πε₀R²)
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Solution
For a point outside a uniformly charged sphere, the electric field behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).
Correct Answer: B — Q/(4πε₀R²)
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Q. What is the electric field due to a point charge at a distance 'r' from it? (2021)
A.
k * q / r^2
B.
k * q / r
C.
k * q * r
D.
k * q * r^2
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Solution
The electric field (E) due to a point charge (q) at a distance (r) is given by E = k * q / r^2, where k is Coulomb's constant.
Correct Answer: A — k * q / r^2
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Q. What is the electric field due to a point charge at a distance r?
A.
k * q / r^2
B.
k * q / r
C.
k * q * r
D.
k * q * r^2
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Solution
The electric field E due to a point charge q at a distance r is given by E = k * q / r^2, where k is Coulomb's constant.
Correct Answer: A — k * q / r^2
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Q. What is the electric field due to a point charge of +10μC at a distance of 0.2m?
A.
22500 N/C
B.
45000 N/C
C.
50000 N/C
D.
75000 N/C
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Solution
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (0.2 m)² = 225000 N/C.
Correct Answer: C — 50000 N/C
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Q. What is the electric field due to a point charge of +1μC at a distance of 0.1m?
A.
9000 N/C
B.
900 N/C
C.
90 N/C
D.
9 N/C
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Solution
Electric field E = k * |q| / r^2 = (9 × 10^9) * (1 × 10^-6) / (0.1)^2 = 9000 N/C.
Correct Answer: A — 9000 N/C
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