Introduction & Properties of Fluid Level- 1 - GATE ME Mechanics Free MCQ

Assume unit dimension perpendicular to the plane of paper.

W = FT

(wh × 1)ρg = (σ × 1) × cos θ The weight of the fluid that rises up in the slit is balanced by the vertical component of surface tension force.

⇒ h = 2σ cos θ wρg For a tube with diameter w We know h = 4σ cos θ ρgw Clearly the capillary rise in slit is 2 times less than tube.

By Ostwal-de Waal model

τ = ( du / dy )ⁿ + B

μapp = A ( du / dy )^n-1

Given τ = τ0 + C ( du / dy )^0.5

∴ μapp = C ( du / dy )^-0.5

⇒

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As the plate starts sliding down because of its weight component, the oil between the plate and plane shall start flowing and viscous stress will act on the plate. With acceleration the velocity of plate and thus deformation rate of oil will increase. After the initial period of acceleration the viscous force will balance the weight component and plate will achieve equilibrium moving with constant velocity.

This final constant velocity is called terminal velocity. ∴ Gravitational force = Viscous Force

⇒ mg cos α = μ × VT / ε × A

⇒ given mg = W; we get VT = ε W cos α μA

Height of water column added

If this line is moved by a distance dy perpendicular to its length.

The area generated = dx × dy

The work required to increase the

Area = Fsurface × dy = σdxdy = σdA

i. e The work required = σ × (Increase of area) So , !rea of 1mm dia drop

= 4π (0.5 × 10^-3)² = 3.14 × 10^-6m

Volume of each droplet = 4 / 3 πr³ / 64 = πR³ / 48

∴ Radius of each droplet

r = [ 3 / 4π (πR3 / 48)^1/3

r = 0.125 mm

Surface area of all 64 drops = 64 × 4 × π (0.125 × 10^-3)2 = 1.25 × 10−5m² Increase in surface area = 1.25 × 10−5 − 3.14 × 10^-6 = 9.36 × 10^-6

The work required

= σ × ΔA = 0.073 × 9.36 × 10^-6

= 6.83 × 10^-7J

= 0.68 × 10^-6J

From Newton’s law of viscosity,

τ = μ du / dy

Given, u = 3 / 4y − y2

⇒ du / dy = 3 / 4− 2y

∴ τ0.3 = μ ( 3 / 4 − 2(0.3))

= 0.15 μ

τ0.2 = μ (3 / 4− 2(0.2))

= 0.35 μ

According to question,

τ0.3 = k τ0.2

⇒ 0.15 μ = k × 0.35 μ

⇒ k = 3 / 7

Therefore option (C) is correct.

Specific gravity(S) of a liquid is defined as the ratio of the density of the liquid(pl) to that of water(pw).

Thus, S = 0:66.

wh π(r02 − ri2) = σ cos θ × 2π(r0 + ri)

∴ h = σ cos θ × 2π(r0 + ri) / wπ(r₀² − r₁²)

= 2σ cos θ / w(r0 − ri)

Specific gravity(S) of a liquid is defined as the ratio of the density of the liquid(pl) to that of water(pw).

Thus, S = 0.66.

Shear stress τ = μ du / dy = μ u / h

Shear force = shear stress × area

= μ u / h × 2π r dr

Viscous torque on the element,

dT = μ u / h (2π r dr)r

Since the shaft rotates with angular velocity ω radians per second, the peripheral velocity u = ω r dT = μ ωr h (2π r dr)r = 2π μ ωr3dr h

Total viscous torque required to overcome the shearing force can be obtained by integrating from r = 0 to r = R.

Total torque T = 2π μ ω / h ∫_R⁰ r³ dr

= 1 / 2 π μ ω / h R⁴

During the test, initial volume of container (when it is empty) is

V1 = 0.5 m3

The volume increases by 6%, thus final volume is

V2 = V1 + 6% of V1 = 0.5 × 1.06

= 0.53 m3

Let the water pumped into the container have a volume of V3 when it is outside (i.e. under atmospheric conditions). But when it is pumped into the container it’s final volume will be V2 and pressure equal to 300 bar.

∴ for water

K = dp / (−dV / V)

i. e. −dV / V = dp / K

⇒ −(V2 − V3) / V3 = dp / K

⇒ V3 − V2 / V3 = dp / K

⇒ V3 (1 − dp / K) = V2

⇒ V3 = V2 / 1 −dp / K

V3 = 0.53 / 1 −300×10⁵ / 2×10⁹

⇒ V3 = 0.538

Mass of water = ρ × V3 = 1000 × 0.538 = 538 kg

Kinematic viscosity

ν = 3.5 stokes = 3.5 cm²/s = 3.5 × 10−4 m²/s

Mass density ρ = 0.85 × 1000 = 850 kg/m³

∴ Dynamic viscosity μ = ρ ν

= 850 × (3.5 × 10^-4)

= 0.2975 N s/m²

Thickness of oil film

=(20.016 − 20) / 2 × 10^-2

= 0.00008 m

Shear stress τ = μ du / dy = μ V / t

= 0.2975 × 0.15 / 0.00008 m

= 557.81 N/m²

Frictional resistance

= shear stress × area

= 557.81 × (π × 0.20 × 3.2)

= 1121 N = 1.12 kN

The slope of τ vs ϕ̇ is equal to apparent viscosity of fluid.

Consider average slope (average apparent viscosity)

1. Data points ② and ①

μapp = 1.4 − 0 / 2 − 0 = 0.7

2. Data points ③ and ②

μapp = 2.6 − 1.4 / 3 − 2 = 1.2

3. Data points ④ and ③

μapp = 4 − 2.6 / 4 − 3 = 1.4

Thus as τ increases μapp also increases i.e. the fluid is shear thickening fluid (dilatant).