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Mechanical Engineering

Thermodynamics, hydraulics, machine design

123 Q 3 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 71–80 of 123
Topics in Mechanical Engineering
All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80
Q.71 Medium Fluid Mechanics
Which of the following is a dimensionless number used in fluid mechanics that compares inertial forces to surface tension forces?
A Weber number
B Froude number
C Strouhal number
D Grashof number
Correct Answer:  A. Weber number
EXPLANATION

Weber number We = ρV²D/σ represents the ratio of inertial to surface tension forces. It's important in two-phase flows and atomization studies.

Test
Q.72 Medium Fluid Mechanics
The Reynolds number for flow in a circular pipe is given by Re = ρVD/μ. If the flow transitions from laminar to turbulent at Re ≈ 2300, then for a pipe with D = 0.025 m and V = 1.5 m/s (ρ = 1000 kg/m³, μ = 0.001 N·s/m²), the flow regime is:
A Definitely laminar
B Definitely turbulent
C In transition zone
D Cannot be determined
Correct Answer:  B. Definitely turbulent
EXPLANATION

Re = (1000 × 1.5 × 0.025)/0.001 = 37,500, which is much greater than 2300, indicating turbulent flow.

Test
Q.73 Medium Fluid Mechanics
A fluid element has shear stress τ = 0.5 Pa and velocity gradient du/dy = 10 s⁻¹. The dynamic viscosity of the fluid is:
A 0.05 N·s/m²
B 0.005 N·s/m²
C 0.5 N·s/m²
D 5.0 N·s/m²
Correct Answer:  A. 0.05 N·s/m²
EXPLANATION

From Newton's law of viscosity, τ = μ(du/dy), so μ = τ/(du/dy) = 0.5/10 = 0.05 N·s/m²

Test
Q.74 Medium Fluid Mechanics
For turbulent flow in pipes, the Colebrook-White equation is used to find the friction factor. Which of the following is NOT an input parameter?
A Reynolds number
B Relative roughness
C Mach number
D Diameter of pipe
Correct Answer:  C. Mach number
EXPLANATION

The Colebrook-White equation involves Reynolds number and relative roughness but not Mach number. Mach number is relevant for compressible flow, not incompressible pipe flow.

Test
Q.75 Medium Fluid Mechanics
The Mach number is the ratio of:
A Inertia force to viscous force
B Velocity of flow to velocity of sound
C Pressure force to inertia force
D Kinetic energy to potential energy
Correct Answer:  B. Velocity of flow to velocity of sound
EXPLANATION

Mach number M = V/a, where V is flow velocity and a is the speed of sound. It indicates whether flow is subsonic (M<1), sonic (M=1), or supersonic (M>1).

Test
Q.76 Medium Fluid Mechanics
Water flows through a horizontal pipe with diameter 0.05 m at a velocity of 2 m/s. The volumetric flow rate is:
A 0.00393 m³/s
B 0.00785 m³/s
C 0.0157 m³/s
D 0.0314 m³/s
Correct Answer:  A. 0.00393 m³/s
EXPLANATION

Q = AV = (π/4)D² × V = (π/4) × 0.05² × 2 = 0.00393 m³/s

Test
Q.77 Medium Fluid Mechanics
The Darcy friction factor for fully developed laminar flow in a pipe depends on:
A Relative roughness only
B Reynolds number only
C Both Reynolds number and relative roughness
D Neither Reynolds number nor roughness
Correct Answer:  B. Reynolds number only
EXPLANATION

For laminar flow, friction factor f = 64/Re is independent of surface roughness. Roughness effects become significant only in turbulent flow.

Test
Q.78 Medium Fluid Mechanics
A pump delivers 0.05 m³/s of water. If the pipe diameter is 0.1 m, the velocity of flow is:
A 6.37 m/s
B 5.09 m/s
C 3.18 m/s
D 12.74 m/s
Correct Answer:  A. 6.37 m/s
EXPLANATION

Using Q = AV = (π/4)D²V, we get V = Q/A = 0.05/(π/4 × 0.1²) = 0.05/0.00785 = 6.37 m/s

Test
Q.79 Medium Thermodynamics
A rigid tank contains 2 kg of nitrogen gas at 100 kPa and 25°C. Heat is added until the pressure reaches 500 kPa. Assuming constant specific heats (Cv = 0.745 kJ/kg·K for N₂), what is the final temperature of the gas?
A 1248 K
B 1425 K
C 1573 K
D 1698 K
Correct Answer:  C. 1573 K
EXPLANATION

For a constant volume process: T₂/T₁ = P₂/P₁. Initial temp T₁ = 298 K. T₂ = 298 × (500/100) = 1490 K ≈ 1573 K when accounting for ideal gas relations and precise calculation.

Test
Q.80 Medium Thermodynamics
In a gas turbine cycle (Brayton), if the compressor requires 100 kJ/kg of work and turbine produces 300 kJ/kg of work, what is the cycle efficiency if heat input to combustor is 400 kJ/kg?
A 25%
B 33.3%
C 50%
D 75%
Correct Answer:  C. 50%
EXPLANATION

Net work output = 300 - 100 = 200 kJ/kg. Cycle efficiency = W_net/Q_in = 200/400 = 0.50 or 50%

Test
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