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Waves, Oscillations & Optics (Physics) Solved Questions & Notes (2026) - Apex Rankers

Natural & Physical Sciences > Physics > Waves, Oscillations & Optics

36 Total Solved Questions
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Waves, Oscillations & Optics

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Q. 1 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
What is the time period T of a simple pendulum of length L in a gravitational field g for small angular displacements?
A
T = 2 * pi * sqrt(L / g)
✓ Correct
B
T = 2 * pi * sqrt(g / L)
C
T = pi * sqrt(L * g)
D
T = 2 * pi * (L / g)
💡 Step-by-Step Explanation & Concept Rationale
The period of a simple pendulum is T = 2*pi*sqrt(L/g), independent of the mass of the bob.
Q. 2 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
When light travels from an optically denser medium (refractive index n1) into a rarer medium (n2) at an angle of incidence greater than the critical angle, what phenomenon occurs?
A
Total Internal Reflection
✓ Correct
B
Polarization by Scattering
C
Diffraction
D
Refraction towards the normal
💡 Step-by-Step Explanation & Concept Rationale
Total internal reflection occurs when light strikes a boundary with a rarer medium at an angle exceeding critical angle theta_c = arcsin(n2/n1).
Q. 3 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In Young's Double-Slit Experiment, what is the fringe width (fringe spacing beta) on a screen at distance D for light of wavelength lambda with slit separation d?
A
beta = (lambda * D) / d
✓ Correct
B
beta = (lambda * d) / D
C
beta = (d * D) / lambda
D
beta = lambda / (d * D)
💡 Step-by-Step Explanation & Concept Rationale
Fringe spacing beta = lambda * D / d.
Q. 4 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
When a body moves with uniform velocity in a circle, its acceleration is directed:
A
Zero
B
Tangentially forward
C
Away from the center
D
Towards the center of the circle
✓ Correct
💡 Step-by-Step Explanation & Concept Rationale
Centripetal acceleration is always directed radially inward towards the center of curvature.
Q. 5 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
Escape velocity from the surface of Earth is approximately equal to:
A
7.9 km/s
B
11.2 km/s
✓ Correct
C
9.8 km/s
D
15.0 km/s
💡 Step-by-Step Explanation & Concept Rationale
v_escape = sqrt(2*g*R) = sqrt(2 * 9.8 * 6.4x10^6) ≈ 11.2 km/s.
Q. 6 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #16: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both stop immediately
B
Both masses stick together and move with half velocity
C
Both bounce back with equal speeds
D
The incident mass stops and target mass moves with original velocity
✓ Correct
💡 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 #22: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both masses stick together and move with half velocity
B
The incident mass stops and target mass moves with original velocity
✓ Correct
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. 8 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #28: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both stop immediately
B
Both masses stick together and move with half velocity
C
Both bounce back with equal speeds
D
The incident mass stops and target mass moves with original velocity
✓ Correct
💡 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 #34: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both masses stick together and move with half velocity
B
The incident mass stops and target mass moves with original velocity
✓ Correct
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. 10 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
Surface tension of a liquid decreases when:
A
Pressure is reduced
B
Temperature is decreased
C
Impurity is added
D
Temperature is increased
✓ Correct
💡 Step-by-Step Explanation & Concept Rationale
As temperature increases, molecular thermal kinetic energy increases, weakening intermolecular cohesive forces and decreasing surface tension.
Q. 11 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #11: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Mach Number
B
Reynolds Number (Re)
✓ Correct
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. 12 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #17: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Prandtl Number
B
Mach Number
C
Froude Number
D
Reynolds Number (Re)
✓ Correct
💡 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 #23: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Mach Number
B
Reynolds Number (Re)
✓ Correct
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. 14 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #29: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Prandtl Number
B
Mach Number
C
Froude Number
D
Reynolds Number (Re)
✓ Correct
💡 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: EASY (1 Mark)
The entropy of an isolated system undergoing an irreversible natural process always:
A
Decreases
B
Increases
✓ Correct
C
Remains constant
D
Fluctuates to zero
💡 Step-by-Step Explanation & Concept Rationale
According to the Second Law of Thermodynamics, the total entropy of an isolated system always increases over time.
Q. 16 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #11: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
2.00
B
1.40 (7/5)
C
1.33 (4/3)
D
1.67 (5/3)
✓ Correct
💡 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 #17: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.40 (7/5)
B
1.67 (5/3)
✓ Correct
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. 18 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #23: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
2.00
B
1.40 (7/5)
C
1.33 (4/3)
D
1.67 (5/3)
✓ Correct
💡 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 #29: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.40 (7/5)
B
1.67 (5/3)
✓ Correct
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. 20 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
The total internal reflection occurs only when light travels from:
A
Any medium at 45 degrees
B
Rarer to denser medium at normal incidence
C
Denser to rarer at angle less than critical angle
D
Denser medium to rarer medium at an angle greater than critical angle
✓ Correct
💡 Step-by-Step Explanation & Concept Rationale
TIR requires light to travel from optically denser to rarer medium with angle of incidence exceeding the critical angle.
Q. 21 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #11: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(g / L)
B
2 * pi * sqrt(L / g)
✓ Correct
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. 22 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #17: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * (L / g)^2
B
2 * pi * sqrt(g / L)
C
pi * sqrt(L * g)
D
2 * pi * sqrt(L / g)
✓ Correct
💡 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 #23: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(g / L)
B
2 * pi * sqrt(L / g)
✓ Correct
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. 24 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #29: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * (L / g)^2
B
2 * pi * sqrt(g / L)
C
pi * sqrt(L * g)
D
2 * pi * sqrt(L / g)
✓ Correct
💡 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)
In wave theory #35: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(g / L)
B
2 * pi * sqrt(L / g)
✓ Correct
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).
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