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

Concept:

The Triangle centroid lies at a depth of 2h/3 from the apex considering h is the height of triangle.

Using symmetry center of pressure will lie on the same axis as G.

Depth of COP is given as

$h_{cp}=\overline{x}+\frac{I_{CG}}{A\overline{x}}$

Calculation:

Given:

G = centroid, Center of pressure acts at a distance of h_cp

From the figure,

$\overline{x}=\frac{1}{3}\times \frac{\sqrt{3}a}{2}=\frac{\sqrt{3}a}{6}m$

$I_{CG}=\frac{b{h}^3}{36}=\frac{a}{36}\times (\frac{\sqrt{3}}{2}a{)}^3=\frac{3\sqrt{3}{a}^4}{288}{m}^4$

$A=\frac{1}{2}\times a\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{4}{a}^2{m}^2$

$\therefore h_{cp}=\frac{\sqrt{3}a}{6}+\frac{(3\sqrt{3}{a}^4/288)}{(\frac{\sqrt{3}{a}^2}{4})\times (\frac{\sqrt{3}a}{6})}=\frac{\sqrt{3}a}{4}m$

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

$h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}$

For vertical plane surface: θ = 90°

$h^{\ast }=\frac{I_G}{A̅h}+\overline{h}$

Calculation:

Diameter of circular plate (D) = 1.5 m

h̅ = h_CG = 2 m

Position of centre of pressure is given by,

$h_{CP}=h_{CG}+\frac{I_{xx}}{A{h}_{CG}}$

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

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

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

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

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,

${\rho }_{\mathrm{s}\mathrm{w}}\times \mathrm{g}\times {\mathrm{h}}_{\mathrm{s}\mathrm{w}}\times A={\rho }_{\mathrm{r}\mathrm{w}}\times \mathrm{g}\times {\mathrm{h}}_{\mathrm{r}\mathrm{w}}\times A$

${\rho }_{\mathrm{r}\mathrm{w}}<{\rho }_{\mathrm{s}\mathrm{w}}\Rightarrow {\mathrm{h}}_{\mathrm{s}\mathrm{w}}<{\mathrm{h}}_{\mathrm{r}\mathrm{w}},$

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

$h^{\ast }=\frac{I_G{\sin }^2\theta }{A\overline{h}}+\overline{h}$

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°

$h^{\ast }=\frac{I_G}{A\overline{h}}+\overline{h}$

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

$h^{\ast }=\frac{\frac{b{d}^3}{12}}{bd\times \frac{d}{2}}+\frac{d}{2}=\frac{2}{3}d$

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

$h_{cp}=\overline{h}+\frac{I_{CG}}{A\overline{h}}$

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

$\overline{h}=1.5+\frac{4.5}{2}=3.75m$

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

θ = 90°

A = 1 × 4.5 m²

${\overline{h}}_{CP}=3.75+\frac{7.593\times 1}{3.75\times 4.5}$

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

$h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}$

For vertical plane surface: θ = 90°

$h^{\ast }=\frac{I_G}{A\overline{h}}+\overline{h}$

The following facts can be concluded from the above equations:

• Center of pressure lies below the centroid because for any plane surface, the factor $\frac{I_G}{A\overline{h}}$ 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 = $h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}$

For vertical plane surface: θ = 90°

$h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}=\frac{\frac{b{h}^3}{12}}{bh\times \frac{h}{2}}+\frac{h}{2}=\frac{2h}{3}$

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

$h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}$

For vertical plane surface: θ = 90°

$h^{\ast }=\frac{I_G}{A\overline{h}}+\overline{h}$

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 = $h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}$

For vertical plane surface: θ = 90°

$h^{\ast }=\frac{I_G{\sin }^2\theta }{A̅h}+\overline{h}=\frac{\frac{b{h}^3}{12}}{bh\times \frac{h}{2}}+\frac{h}{2}=\frac{2h}{3}$

Calculation:

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