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 specific gravities of the block and the solution. (Suggestion: Call the horizontal crosssectional area of the block A. A should cancel in your calculations.)

Answer 1

Answer:

- the specific gravity of the block is 0.75

- the specific gravity of the solution is 1.5

Explanation:

Given the data in the question;

first we find the specific gravity of a block SGB

SGB = ( block vol below / total block vol ) × the specific gravity of water

we substitute

SG$_{BLOCK$ = ( 1.5 / (1.5 + 0.5 ) ) × 1

SG$_{BLOCK$ = ( 1.5 / (1.5 + 0.5 ) ) × 1

SG$_{BLOCK$ = (1.5 / 2) × 1

SG$_{BLOCK$ = 0.75

Therefore, the specific gravity of the block is 0.75

specific gravity of solution SG$_{SOLUTION$

SG$_{SOLUTION$ = (total block vol / block below ) × SG$_{BLOCK$

we substitute

SG$_{SOLUTION$ = ( 2 / 1 ) × 0.75

SG$_{SOLUTION$ = 2  × 0.75

SG$_{SOLUTION$ = 1.5

Therefore, the specific gravity of the solution is 1.5