Let, G = Location of C.G.

        B = Location of buoyant force

Let a floating body is given a small angular displacement θ in the clockwise direction.

B₁ = New centre of buoyancy

M = Metacentre which is the point where line of action of B₁ Passes.

GM = Metacentre height

The portion B’OB is submerged and portion AOA’ is lifted up because of tilt (angular displacement).

Consider an elementary wedge shape of thickness ‘dx’ from portion B’OB at a distance ‘x’ from ‘O’ as shown.

Height of strip = x x < BOB’
                          = x θ

Area of strip = height x thickness
                       = x. θ dx

Weight of strip = Volume x γ

Similarly if small strip of thickness ‘dx’ at ‘x’ from O toward the left side is considered, the weight of strip will be γ x θ L dx. This left and right side weight are acting equal and opposite forming a couple.

Moment of couple,

                                    dm = Weight of strip x distance between two weights

= (x θ dx L γ) [x + x]

= γ L. θ  (2x² dx)
                       
Moment of couple for whole wedge

M =  ∫ 2 γ L θ  X² dx

Now, moment duet to shifting of centre of buoyancy from B to B₁.

                                M = F_B X BB₁
if θ is very small
                               M =  F_B x BM x θ

                              F_B = W
We know,
                              M = W.BM.θ

Equation moment given in Equations

W. BM.θ = 2 γ L θ ∫  x² dx

W. BM = 2 γ ∫  x² L dx

But          L.dx = Elemental area

W. BM =  2 γ ∫  x² A

But,     2 ∫ x² dA  = Moment of inertia of plan of body at water surface

BM = γI / W

But by Archimedes’s principle

Weight of body = Weight of water displaced

W = γ . ∀

Equating Equation

γ∀ = ∀I / BM

Where V = Volume of body submerged in water

BM = γI  / γ∀

BM = I / ∀

GM = BM - BG

= I / ∀- BG

Metacentric height,       GM = I / ∀ - BG

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