Question

Oil having a density of 930 kg/m^3 floats on

Answer 1

Answer:

0.0268 m

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.

Sum of forces in the y direction:

∑F = ma

B₁ + B₂ − mg = 0

ρ₁V₁g + ρ₂V₂g − mg = 0

ρ₁V₁ + ρ₂V₂ = m

ρ₁V₁ + ρ₂V₂ = ρV

ρ₁Ah₁ + ρ₂Ah₂ = ρAh

ρ₁h₁ + ρ₂h₂ = ρh

(930 kg/m³)h₁ + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

Since the block is fully submerged, h₁ + h₂ = 4.93 cm.

(930 kg/m³) (4.93 cm − h₂) + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

h₂ = 2.68 cm

h₂ = 0.0268 m