GATE Past Year Questions: Fluid Statics | Fluid Mechanics for Mechanical Engineering PDF Download

For floating body Buoyancy force acts through the center of buoyancy which is C.G for displaced volume.

Given data
l = 5 m
θ = 30°
ρ = 1000 kg/m3
h = 2.5 sin 30° = 1.25m

F = ρgAh = 1000 x 9.81 x 5 x 1.25 = 61312.5N

Applying Bernoulli’s equation at exit, we get

We know Z₁ = Z₂, V₁ = 0 & P₂ = P_atm
Hence it reduce to

But P₁ = ρ₁gh₁ + ρ₂gh₂ + ρ₃gh₃
Upon substituting, we get

For stable floating body metacentre should be above centre of gravity.
For unstable floating body, metacentre should be below centre of gravity.
For neutral equilibrium, both metacentre and centre of gravity should coincide.

Bulk modules of elasticity,
 

K = 2000 MPa

P = 5 + 1 + 1.01 = 7.01 bar

We know that

 [∵ v = ω × r]

Area of circular ring = 2πrdr
Force on elementry ring
= Intensity of pressure × Area of ring

∴ Total force on the top of the cylinder

Thrust at the bottom of the cylinder
= Weight of water in cylinder + Total force on the top of cylinder

Let T = Tension in string
T₁ = Mg = Buoyancy force
T + ρ_sH.A.g = ρ.h A.g
∴ T = g.A(ρh - ρs.H)

F_x = 2ρghrw

P_A = 50 cm Hg vacuum = p_H

P_atm = 760 mm Hg

Let P_A be the absolute pressure at 'A'
Pressure at D = Pressure at E
or P_A + P_CD = P_atm + P_AF + P_FE

P_{A= 771.2 mm Hg}

Gauge pressure = (δm - δw)gh
= (13600 - 1000) × 9.81 × 10 × 10-3 = 1236 Pa

Height of water column over plate is H.
F = P × A = ρgHA

Fall in larger limb = Rise in smaller limb
A₁ Δh = A₂h

Total hydrostatic pressure force, F = ρghA
F = 1000 × 9.81 × 8.5 × 0.785 = 65458 N
or
F = 65.45 KN