Question

A rectangular block floats in pure water with 0.5 in, above the surface and 1.5 in below the surface. When placed in an aqueous solution, the block of material floats with 1 in. below the surface. Estimate the specific gravities of the block and the solution. (Suggestion: Call the horizontal cross-sectional area of the block A. A should cancel in your calculations.)

Answer 1

Explanation:

We will calculate the specific gravity of the block as follows.

    Specific gravity of block = $\frac{\text{block vol below}}{\text{total block vol } \times \text{specific gravity of water}}$

                   = $\frac{1.5}{(1.5 + 0.5) \times 1}$

                   = 0.75

And, the specific gravity of the solution is as follows.

 Specific gravity of solution = $\frac{\text{total block vol}}{\text{block vol below} \times \text{specific gravity of block}}$

            = $\frac{2}{1 \times 0.75}$

            = 1.5

As the given block is rectangular, and its volume is directly proportional to its height and A is not needed in these calculations.

Also, we can note here that the term specific gravity is no longer approved and has been replaced by the term relative density.