Ablock of wood floats in fresh water with 0.663 of its volume v submerged and in oil with 0.913 v submerged. find the density of (a) the wood and (b) the oil.

(a). 663kg/m³.

(b). 726.18kg/m³

Explanation:

(a).  Let V be the volume of the block of wood,

Then the submerged volume in fresh water is

V₍s₎=0.663V

Where V₍s₎= submerged volume in water.

Since the block is floating, according to Archimedes principle,

The weight of the displaced fresh water is equal to the weight of block.

∴D₍w₎V₍s₎ = DV (1)

Where D₍w₎ = Density of fresh water, D = Density of Block of wood

Since V₍s₎=0.663V, and D₍w₎=1000kg/m³.

We substitute the value of V₍s₎ and D₍w₎ in equation(1) above.

⇒ 1000×0.663V = D × V

∴ D × V = 1000 × 0.663V

Dividing both side of the equation by V

∴ D = 663kg/m³.

(b). Let V be the volume of the wood block. and V₍o₎ is the submerged volume in oil.

V₍o₎ = 0.913V

Since the wood block is floating, according to Archimedes principle,

The weight of the displaced oil is equal to the weight of the wood block.

IF D₍o₎ = Density of  oil.

D₍o₎V₍o₎  = DV(2)

D=663kg/m³, and V₍o₎ =0.913V,

Substituting the value of D and V₍o₎  in equation(2)

D₍o₎ × 0.913V = 663 × V

Dividing both side of the equation by the coefficient of D₍o₎

∴D₍o₎ × 0.913V/0.913V = 663V/0.913V

∴ D₍o₎ = 663/0.913

   D₍o₎ = 726.177

  D₍o₎ ≈ 726.18kg/m³

we need to know the time interval during which the force acts. impulse of force = momentum change