Hydrostatic Force MCQ Quiz in தமிழ் - Objective Question with Answer for Hydrostatic Force - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Hydrostatic pressure at any depth from the water surface is given by –

\(P = {\gamma _L}h = {\rho _L}gh\)

Where, γ_L = unit weight of liquid (N/m³),

ρ_L = density of liquid (Kg/m³),

g = acceleration due to gravity (m/sec²)

h = depth of location measured from water surface (m) and

P = Hydrostatic pressure (N/m²)

It should also be noted that the Pressure Gauges always reads the excess pressure from the surrounding.

Calculation:

Given,

\(g = 10\;m/se{c^2},\;{P_A} = 50kPa\;at\;5m\;and\;{P_B} = 100\;kPa\;at\;10m\)

\({P_A} = 50\; \times 1000\;N/{m^2} = \;{\rho _L} \times 10 \times 5 \Rightarrow {\rho _L} = 1000\;kg/{m^3}\)

\({P_B} = 100\; \times 1000\;N/{m^2} = \;{\rho _L} \times 10 \times 10 \Rightarrow {\rho _L} = 1000\;kg/{m^3}\)

∴ Density of liquid is 1000 kg/m³

Concept:

For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:

A = Total area of an inclined surface

h̅ = Depth of centre of gravity of inclined area from a free surface

h* = Distance of centre of pressure from the free surface of a liquid

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} +\bar h \)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A ̅ h}} +\bar h\)

Calculation:

Diameter of circular plate (D) = 1.5 m

h̅ = h_CG = 2 m

Position of centre of pressure is given by,

\({h_{CP}} = {h_{CG}} + \frac{{{I_{xx}}}}{{A{h_{CG}}}}\)

\({I_{xx}} = \frac{{\pi {D^4}}}{{64}},A = \frac{{\pi {D^2}}}{4}\)

\(\frac{I}{A} = \frac{{\pi {D^4}}}{{64}} \times \frac{4}{{\pi {D^2}}} = \frac{{{D^2}}}{{16}}\)

\(h_{CP}= {h_{CG}} + \frac{{{D^2}}}{{16{h_{CG}}}}\)

\(h_{cp}= 2 + \frac{{{{\left( {1.5} \right)}^2}}}{{\left( {16} \right)\left( 2 \right)}}= 2.07031~ m\)

h_CP = 2.07 m

The density of seawater (ρ_sw) > density of river water (ρ_rw)

For the same ship, balancing the buoyant force for both conditions,

\({{\rm{\rho }}_{{\rm{sw}}}}{\rm{ \times g \times }}{{\rm{h}}_{{\rm{sw}}}\times A}{\rm{ = }}{{\rm{\rho }}_{{\rm{rw}}}}{\rm{ \times g \times }}{{\rm{h}}_{{\rm{rw}}}\times A}\)

\({\rho _{{\rm{rw}}}} < {\rho _{{\rm{sw}}}} \Rightarrow {{\rm{h}}_{{\rm{sw}}}}{\rm{ < }}{{\rm{h}}_{{\rm{rw}}}},\;\)

where h is the depth of ship into the water

As the depth of ship in sea water decreases, that means the height of ship (rises) above water increases.

Explanation:

Hydrostatic pressure, P = ρgh

Hydrostatic force, F = ρgh̅A

Where h̅ is the depth of the centroid of the area A of the surface

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A\bar h}} + \bar h\)

Since the pressure increases with the depth, the centre of pressure P must lie below the centroid of the area of the surface.

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A\bar h}} + \bar h\)

Consider a surface is rectangular, with dimension b × d, and its one end is placed immediately beneath the free level of the liquid.

The centroid is located at d/2 below the free level of the liquid.

h̅ = d/2, IG = bd3/12

\({h^*} = \frac{{\frac{{b{d^3}}}{{12}}}}{{bd \times \frac{d}{2}}} + \frac{d}{2} = \frac{2}{3}d\)

The centre of pressure for a plane vertical rectangular surface lies at a depth of two-third the height of the immersed surface.

Concept:

Hydrostatic pressure is given by,

P = ρ g H

Where, ρ = density, g = gravitational acceleration, H = pressure head

Solution:

1. N/mm² = 10⁶ Pa
2. Kgf/cm² = 9.81/10^-4 Pa
3. Millibar = 10² Pa
4. mm of mercury = 10^-3 × 13600 × 9.81 Pa = 133.416 Pa

Hydrostatic pressure (P) at any depth (h) for a fluid of density (ρ) is given by

\(P = \rho \times g \times h\)

Therefore hydrostatic pressure for incompressible fluid varies linearly with the depth.

Important Point:

Hydrostatic pressure for incompressible fluid varies linearly with the depth.

Hydrostatic pressure for compressible fluid varies exponentially with the depth.

Concept:

 The depth of the center of pressure is given by

\({h_{cp}} = \bar h + \frac{{{I_{CG}}}}{{A\bar h}}\)

Calculation:

Given:

D = 1.6 m, 

\({h_{cp}} = \frac{{1.6}}{2} + \frac{{\left( {\frac{\pi }{{64}}} \right){{\left( {1.6} \right)}^4}}}{{\left( {\frac{\pi }{4}} \right){{\left( {1.6} \right)}^2} \; \times\; 0.8}}\)

∴ h_cp = 1 m

\({\bar h = 1.5 + \frac{{4.5}}{2} = 3.75\;m}\)

\({I_G}_{xx} = \frac{{1 \times {{4.5}^3}}}{{12}}\)= 7.593 m⁴  , Width b = 1 m

θ = 90°

A = 1 × 4.5 m²

\({\bar h_{CP}} = 3.75 + \frac{{7.593 \times 1}}{{3.75 \times 4.5}}\)

Concept:

• According to Pascal’s Law, the pressure or intensity of pressure at a point in a static fluid is equal in all directions.
• For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:

A = Total area of an inclined surface, h̅ = Depth of centre of gravity of inclined area from a free surface, h* = Distance of centre of pressure from the free surface of a liquid

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A \bar h}} + \bar h\)

The following facts can be concluded from the above equations:

• Center of pressure lies below the centroid because for any plane surface, the factor \(\frac{I_G}{A \bar h}\) is always positive
• Deeper the surface is lowered into the liquid (i.e. greater is the value of h̅), closer comes the centre of pressure to the centroid of the area
• Depth of the centre of pressure is independent of the specific weight of the liquid and is consequently same for all liquids

Centre of the pressure of rectangular plate:

Consider a rectangular plate of width b and height h is immersed in water.

h̅ = Depth of centre of gravity of rectangular plate from free surface = h/2

Area = bh² Moment of inertia through centre of gravity. I_G = bh³/12

h* = Distance of centre of pressure from the free surface of a liquid = \({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h=\frac{{\frac{{bh^3}}{{12}}}}{{bh~\times~\frac{{h}}{{2}}}}+\frac{{h}}{{2}}=\frac{{2h}}{{3}}\)

The distance of centre of pressure from base = h - 2h/3 = h/3

Concept:

• According to Pascal’s Law, the pressure or intensity of pressure at a point in a static fluid is equal in all directions.
• For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:

A = Total area of an inclined surface, h̅ = Depth of centre of gravity of inclined area from a free surface, h* = Distance of centre of pressure from the free surface of a liquid

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}}}{{A \bar h}} + \bar h\)

Centre of the pressure of rectangular plate:

Consider a rectangular plate of width b and height h is immersed in water.

h̅ = Depth of centre of gravity of rectangular plate from free surface = h/2

Area = bh2 Moment of inertia through centre of gravity. IG = bh3/12

h* = Distance of centre of pressure from the free surface of a liquid = \({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)

For vertical plane surface: θ = 90°

\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h=\frac{{\frac{{bh^3}}{{12}}}}{{bh~×~\frac{{h}}{{2}}}}+\frac{{h}}{{2}}=\frac{{2h}}{{3}}\)

Calculation:

Centre of the pressure of rectangular plate = 2h/3 = (2 × 75)/ 3 = 50 mm.