Answer:
(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
