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Electricity, Magnetism & AC Circuits (Physics) Solved Questions & Notes (2026) - Apex Rankers

Natural & Physical Sciences > Physics > Electricity, Magnetism & AC Circuits

36 Total Solved Questions
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Electricity, Magnetism & AC Circuits

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Q. 1 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
What is Coulomb's law for the electrostatic force between two point charges q1 and q2 separated by distance r in vacuum?
A
F = (1 / (4 * pi * epsilon_0)) * (q1 * q2 / r²)
✓ Correct
B
F = (1 / (4 * pi * epsilon_0)) * (q1 * q2 / r)
C
F = epsilon_0 * (q1 * q2 / r²)
D
F = (q1 + q2) / r²
💡 Step-by-Step Explanation & Concept Rationale
Coulomb's Law states force is directly proportional to the product of charges and inversely proportional to the square of distance.
Q. 2 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
What is the equivalent capacitance when two capacitors C1 and C2 are connected in parallel?
A
C_eq = C1 + C2
✓ Correct
B
C_eq = (C1 * C2) / (C1 + C2)
C
C_eq = 1/C1 + 1/C2
D
C_eq = sqrt(C1 * C2)
💡 Step-by-Step Explanation & Concept Rationale
In parallel, capacitors add directly: C_eq = C1 + C2.
Q. 3 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
According to Faraday's Law of Electromagnetic Induction and Lenz's Law, what is the induced electromotive force (EMF)?
A
EMF = - N * (dPhi / dt)
✓ Correct
B
EMF = + N * (dPhi / dt)
C
EMF = N * I * R
D
EMF = B * A * t
💡 Step-by-Step Explanation & Concept Rationale
Faraday-Lenz Law states induced EMF is proportional to the negative rate of change of magnetic flux through the circuit.
Q. 4 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
The rate of change of angular momentum is equal to:
A
Applied Torque
✓ Correct
B
Centripetal Force
C
Linear Momentum
D
Rotational Kinetic Energy
💡 Step-by-Step Explanation & Concept Rationale
According to Newton's second law in rotational motion, tau = dL/dt (Torque equals rate of change of angular momentum).
Q. 5 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #11: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both bounce back with equal speeds
B
Both masses stick together and move with half velocity
C
The incident mass stops and target mass moves with original velocity
✓ Correct
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 6 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #17: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
The incident mass stops and target mass moves with original velocity
✓ Correct
B
Both masses stick together and move with half velocity
C
Both bounce back with equal speeds
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 7 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #23: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both bounce back with equal speeds
B
Both masses stick together and move with half velocity
C
The incident mass stops and target mass moves with original velocity
✓ Correct
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 8 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #29: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
The incident mass stops and target mass moves with original velocity
✓ Correct
B
Both masses stick together and move with half velocity
C
Both bounce back with equal speeds
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 9 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #35: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both bounce back with equal speeds
B
Both masses stick together and move with half velocity
C
The incident mass stops and target mass moves with original velocity
✓ Correct
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 10 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #6: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Reynolds Number (Re)
✓ Correct
B
Mach Number
C
Froude Number
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 11 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #12: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Froude Number
B
Mach Number
C
Reynolds Number (Re)
✓ Correct
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 12 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #18: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Reynolds Number (Re)
✓ Correct
B
Mach Number
C
Froude Number
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 13 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #24: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Froude Number
B
Mach Number
C
Reynolds Number (Re)
✓ Correct
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 14 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #30: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Reynolds Number (Re)
✓ Correct
B
Mach Number
C
Froude Number
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 15 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #6: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.33 (4/3)
B
1.40 (7/5)
C
1.67 (5/3)
✓ Correct
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 16 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #12: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.67 (5/3)
✓ Correct
B
1.40 (7/5)
C
1.33 (4/3)
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 17 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #18: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.33 (4/3)
B
1.40 (7/5)
C
1.67 (5/3)
✓ Correct
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 18 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #24: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.67 (5/3)
✓ Correct
B
1.40 (7/5)
C
1.33 (4/3)
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 19 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #30: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.33 (4/3)
B
1.40 (7/5)
C
1.67 (5/3)
✓ Correct
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 20 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #6: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(L / g)
✓ Correct
B
2 * pi * sqrt(g / L)
C
pi * sqrt(L * g)
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 21 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #12: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
pi * sqrt(L * g)
B
2 * pi * sqrt(g / L)
C
2 * pi * sqrt(L / g)
✓ Correct
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 22 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #18: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(L / g)
✓ Correct
B
2 * pi * sqrt(g / L)
C
pi * sqrt(L * g)
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 23 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #24: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
pi * sqrt(L * g)
B
2 * pi * sqrt(g / L)
C
2 * pi * sqrt(L / g)
✓ Correct
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 24 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #30: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(L / g)
✓ Correct
B
2 * pi * sqrt(g / L)
C
pi * sqrt(L * g)
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 25 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
Coulomb's law states that the electrostatic force between two point charges is inversely proportional to:
A
Product of their charges
B
Distance between them (r)
C
Square of the distance between them (r^2)
✓ Correct
D
Permittivity of free space
💡 Step-by-Step Explanation & Concept Rationale
Coulomb's Law: F = (1 / 4*pi*epsilon_0) * (q1 * q2 / r^2).
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