Test: Fluid Properties - 1 - Mechanical Engineering MCQ

Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w / v
Thus, unit of is N / m³.

The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it unitless.

Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density.

• Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e.,
  [p] = [m]/[v] = [m] /[L³] = [ML^-3]
• The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it dimensionless.
• Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus,
  [v] = [V]/[m] = [L³]/[M] = [M^-1L³].
• Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,

Viscosity of the fluid µ = 0.44 kg/m·s 

Density of fluid ρ = 88 Kg/m³

The force per unit area (magnitude) required to maintain the bottom plate stationary will be 

Hence, the correct answer is 26.4.

In a fluid, the following forces are present

Gravity force (F_g) due to gravity

Pressure force F_p due to pressure of fluid

Viscous force F_ν due to viscosity

Tension force F_s due to surface tension

Turbulent force F_t due to turbulence.

F_c due to compressibility.

∴ F_net = F_g + F_p + F_ν + F_s + F_t + F_c

Viscosity:

Viscosity is the property of a fluid which determines its resistance to shearing stresses or which opposes the relative motion between the different layers.

Cause of Viscosity:

It is due to cohesion and molecular momentum exchange between fluid layers.

Unit of viscosity:

S.I. Unit: Pa.s or N.s/m²

C.G.S Unit of viscosity is Poise = dyne-sec/cm²

Important Points:

Surface tension is the elastic tendency of a fluid surface which makes it acquire the least surface area possible. It occurs at the interface of two liquids, due to the intermolecular force of cohesion.

Compressibility is the ability of fluid to change its volume under pressure.

Diameter of shaft (d) = 40 mm = 0.04 m

Length of shaft (L) = 40 mm = 0.04 m

Angular speed of shaft (ω)= 20 rad/s

Viscosity of fluid (μ) = 20 × 10−3 Pa-s

Clearance (dy) = 0.02 mm = 2 × 10⁻⁵ m

Velocity at periphery of shaft = u

= 0.04 N.m

Hence, the correct option is (a).

Fluid dynamics is the study of forces responsible for the motion of fluids.

The specific weight of the fluid depends upon gravitational acceleration and mass density of the fluid.

The weight of the liquid per unit volume under standard temperature and pressure conditions is defined as the specific weight of that liquid.

ϒ=W/V

where,

W = weight of the fluid

V = volume of the fluid

We know that,

W = m×g

where,

m = mass of the fluid

g = acceleration due to gravity

ϒ=m×g/V

Mass density of the fluid,ρ =(mass of the fluid)/(volume of the fluid)=m/V

ϒ=(m/V) x g

ϒ=ρ x g

Therefore, it can be concluded that the specific weight of the fluid depends upon gravitational acceleration and the mass density of the fluid.

Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2
will be represented by v1 < v2.

Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w / V
Thus, γ = 6.5 ⁄10^-3 N ⁄ m³ = 6.5 kN/m³.

For Newtonian fluid, shear stress (τ)

Shear stress (τ) α shear strain rate
Hence, the correct option is (c).

Let the tension in the string is T.
Surface area of contact (A) = L × L = L²
Free body diagram:

From diagram →

T = mg

Drag force = Shear force = T

T = (Shear stress) × Contact area surface (A)

Hence, the correct option is (c)