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Work, Energy & Fluid Dynamics (Physics) Solved Questions & Notes (2026) - Apex Rankers

Natural & Physical Sciences > Physics > Work, Energy & Fluid Dynamics

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Work, Energy & Fluid Dynamics

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Q. 1 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
How much work is done by a centripetal force on a satellite orbiting Earth in a circular orbit?
A
Zero work, because the force is always perpendicular to the displacement vector
✓ Correct
B
Positive work equal to G*M*m/r
C
Negative work equal to kinetic energy
D
Infinite work over infinite revolutions
💡 Step-by-Step Explanation & Concept Rationale
Work W = F * d * cos(theta). For circular motion, theta = 90 deg, so cos(90) = 0, meaning no work is done.
Q. 2 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
What fundamental law of physics is represented by Bernoulli's Principle in fluid dynamics?
A
The Law of Conservation of Energy
✓ Correct
B
The Law of Conservation of Momentum
C
The Law of Conservation of Mass
D
The Second Law of Thermodynamics
💡 Step-by-Step Explanation & Concept Rationale
Bernoulli's equation (P + 1/2 rho*v² + rho*g*h = constant) is a direct formulation of conservation of mechanical energy for steady, incompressible, non-viscous fluid flow.
Q. 3 Physics (Secondary & College Level)
Difficulty: HARD (1 Mark)
What happens to the terminal velocity of a small spherical raindrop falling through air if its radius is doubled (Stokes' Law)?
A
It increases by a factor of 4 (quadruples)
✓ Correct
B
It doubles
C
It remains unchanged
D
It increases by a factor of 8
💡 Step-by-Step Explanation & Concept Rationale
Terminal velocity v_t is proportional to r² (from equating weight proportional to r³ with viscous drag 6*pi*eta*r*v). Doubling radius increases v_t by 2² = 4.
Q. 4 Physics (Secondary & College Level)
Difficulty: HARD (1 Mark)
If the momentum of a moving body is increased by 50%, what is the percentage increase in its kinetic energy?
A
100%
B
125%
✓ Correct
C
50%
D
225%
💡 Step-by-Step Explanation & Concept Rationale
KE is proportional to p^2. If p becomes 1.5p, KE becomes (1.5)^2 = 2.25 times, representing a 125% increase.
Q. 5 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
A projectile has maximum height H and horizontal range R. If R = 4H, the angle of projection is:
A
75 degrees
B
60 degrees
C
30 degrees
D
45 degrees
✓ Correct
💡 Step-by-Step Explanation & Concept Rationale
Since R = 4H / tan(theta), if R = 4H then tan(theta) = 1, so theta = 45 degrees.
Q. 6 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #14: 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. 7 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #20: 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. 8 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #26: 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. 9 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #32: 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. 10 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
The equation of continuity (A1 * v1 = A2 * v2) expresses the conservation of:
A
Energy
B
Mass
✓ Correct
C
Momentum
D
Pressure
💡 Step-by-Step Explanation & Concept Rationale
The equation of continuity is a direct mathematical expression of the conservation of mass in fluid dynamics.
Q. 11 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #9: 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. 12 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #15: 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. 13 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #21: 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. 14 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #27: 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. 15 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
The maximum theoretical efficiency of a heat engine operating between temperatures Th and Tc is given by Carnot as:
A
1 + (Tc / Th)
B
(Th - Tc) / Tc
C
Tc / Th
D
1 - (Tc / Th)
✓ Correct
💡 Step-by-Step Explanation & Concept Rationale
Carnot efficiency eta = 1 - (T_cold / T_hot) where temperatures are in Kelvin.
Q. 16 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #9: 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. 17 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #15: 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. 18 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #21: 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. 19 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #27: 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. 20 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
When light enters from a rarer medium into a denser medium, which of its properties remains completely unchanged?
A
Wavelength
B
Frequency
✓ Correct
C
Velocity
D
Amplitude
💡 Step-by-Step Explanation & Concept Rationale
Frequency is determined by the source of radiation and remains constant regardless of the optical medium.
Q. 21 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #9: 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. 22 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #15: 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. 23 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #21: 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. 24 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #27: 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. 25 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #33: 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).
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