Question

A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aqueous solution, the block of material floats with 0.800 in. below the surface. Estimate the specific gravities of the block and the solution.

Answer 1

Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  $\frac{1.60}{1.60+0.400}$  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = $\frac{1.60+0.400}{0.800}$ × 0.8

specific gravity of  solution  = 2