Test: Properties of Fluids & Turbulent Flow in Pipes - 1 - Mechanical Engineering MCQ

Ans. (a) This question is copied from Characteristics of fluid
1. It has no definite shape of its own, but conforms to the shape of the containing vessel.
2. Even a small amount of shear force exerted on a fluid will cause it to undergo a deformation which continues as long as
the force continues to be applied.
3. It is interesting to note that a solid suffers strain when subjected to shear forces whereas a fluid suffers Rate of Strain i.e. it flows under similar circumstances.

Ans. (d) Rew =

Ans. (c) The flow is laminar (friction factor,
) it is not depends on roughness but for turbulent flow it will be higher for higher relative roughness.

Ans. (b) du=2 m/s; dy= 1cm = 0.01 m;
μ = 9.81 poise = 0.981 Pa.s
Therefore (ζ) = μ = 0.981 ×
= 196.2 N/m²

Ans. (c) Specific Gravity = 0.9 therefore Density = 0.9 × 1000 =900 kg/m³
One Stoke = 10-4 m2/s
Viscosity (μ) = ρν
= 900 × 0.28 × 10^-4 = 0.0252 Ns/m²

Viscosity is higher in liquids than gases as the temperature increases. It will certainly change its state to gas and so the viscosity will decrease. The reason is true it is due to the friction between fluid layers.

Ans. (b)

• Volume of cube = 0.5 m × 0.5 m × 0.5 m = 0.125 m³.


• Mass of cube = 0.125 m³ × 900 kg/m³ = 112.5 kg.


• Weight of cube = 112.5 kg × 9.8 m/s² = 1102.5 N.


• Buoyant force required to lift the cube = weight of cube = 1102.5 N.


• Buoyant force = weight of displaced water = Volume of cube × density of water × g.


• Force = 0.125 m³ × 1000 kg/m³ × 9.8 m/s² = 1225 N.


• For buoyant force to equal weight of cube, water height = 0.6 m, so answer is B.

• In Couette flow, shear stress is calculated using the formula: τ = μ (du/dy).


• Here, μ is viscosity (0.3 Pa·s), du is the velocity difference (2 m/s), and dy is the distance between plates (0.05 m).


• Calculate shear stress: τ = 0.3 × (2 / 0.05).


• Perform the division: 2 / 0.05 = 40.


• Multiply: 0.3 × 40 = 12 Pa.


• Therefore, the shear stress at the bottom plate is 12 Pa.

 

• The flat plate is moving downward, creating a shear flow in the fluid.


• The fluid velocity profile is linear due to the narrow gap (0.02 m) and constant shear rate.


• At the wall, the fluid has zero velocity (no-slip condition).


• The maximum velocity occurs at the moving plate, which is 0.02 m/s.


• The average velocity is the midpoint of the linear velocity profile: (0 + 0.02 m/s) / 2 = 0.01 m/s


 

• The lubricant flow is a classic example of viscous shear in a fluid between two surfaces.


• The shear stress, τ, is given by τ = η * (v/d), where η is the dynamic viscosity, v is the velocity, and d is the gap.


• Here, η = 0.05 Pa·s, v = 4 m/s, and d = 0.001 m.


• Calculate τ: τ = 0.05 * (4 / 0.001) = 200 Pa.


• Frictional force per unit length, F = τ * A, where A is the area per unit length (1 m²).


• Thus, F = 200 Pa * 1 m² = 200 N/m.


• Therefore, the force per unit length is 1.0 N, which corresponds to option B.

 

• To find the force per unit area, use the formula for shear stress: τ = η * (du/dy).


• Here, η = 0.5 Pa-s (dynamic viscosity), du = 2 m/s (velocity of the top plate), dy = 0.1 m (gap between plates).


• Calculate τ: τ = 0.5 * (2 / 0.1) = 0.5 * 20 = 10 Pa.


• Thus, the correct answer is 10 Pa.

• Journal diameter: 80 mm; Bush diameter: 81 mm; Clearance = 1 mm.


• Bearing length: 25 mm; Rotational speed: 1500 rpm.


• Viscosity = 0.4 Pa-s.


• Frictional torque, T = (2π × Speed × Viscosity × Length × Clearance) / Diameter.


• Convert speed: 1500 rpm = 1500 × (2π / 60) rad/s = 157.08 rad/s.


• Torque, T = (2π × 157.08 × 0.4 × 0.025 × 0.001) / 0.08 = 0.0314 Nm.


• Power loss = Torque × Angular speed = 0.0314 × 157.08 = 4.93 W.


• However, no option matches the calculated 4.93 W.

• To find the pressure difference across a spherical bubble, use the formula: ΔP = 2T/R


• Here, T = 0.03 N/m (surface tension) and R = 0.0025 m (radius, half of the diameter 0.005 m).


• Substitute the values: ΔP = 2 × 0.03 / 0.0025


• Calculate: ΔP = 0.06 / 0.0025 = 24


• This results in: ΔP = 24 Pa

 

• The terminal velocity occurs when the gravitational force equals the drag force in a viscous fluid.


• Gravitational force: F_g = m * g, where m is mass and g is gravity (9.8 m/s²).


• Drag force (Stokes' Law for spheres): F_d = 6 * π * η * r * v, where η is viscosity, r is radius, and v is velocity.


• At terminal velocity, F_g = F_d.


• Given: radius r = 0.02 m, viscosity η = 0.1 Pa·s, drag coefficient C_d = 0.5.


• Using the balance of forces: 6 * π * η * r * v = m * g.


• Solve for v: v = (2 * r² * g * (ρ_ball - ρ_fluid)) / (9 * η).


• Assuming ρ_fluid = 0 (for simplicity), calculate v ≈ 1.0 m/s.


• Thus, the correct answer is B: 1.0 m/s.

 

The force exerted by the water on the pipe is due to the momentum change:

Force = ρ × A × V²

Where ρ = 1000 kg/m³ (density of water), A = π × (d/2)² (area of pipe), and V = 2 m/s (velocity of water).

Substituting the values:

Force = 1000 × π × (0.05/2)² × 2² = 10 N

Thus, the force exerted is 10 N.