Question

A rectangular block floats in pure water with 0.5 inch above the surface and 1.5 inches below the surface. When placed in an aqueous solution, the block of material floats with 1 inch below the surface. Estimate the density of the block and the solution

Answer 1

750 kg/m³

1500 kg/m³

Step-by-step explanation:

SG = Specific gravity of water

Specific gravity of block

`\frac{\text{Block volume below}}{\text{Total block volume}}* SG=(1.5)/(1.5+0.5)1=0.75`

Density of block

`\rho_b=0.75* 1000\\\Rightarrow \rho_b=750\ kg/m^3`

Density of block is 750 kg/m³

Specific gravity of solution

`\frac{\text{Total block volume}}{\text{Block volume below}}* SG=(2)/(1)0.75=1.5`

Density of solution

`\rho_s=1.5* 1000\\\Rightarrow \rho_s=1500\ kg/m^3`

Density of block is 1500 kg/m³