Question

A uniform block of height 0.390m, floats in water of density 1000kg/m3. The top of the block is 0.0473m above the surface of the water. What is the density of the block?
density of the block = 878.72 kg/m^3

Ethyl alcohol with a density of 791kg/m3 is poured slowly over the block until the block is totally immersed. Assume that the two liquids do not mix. How much does the depth to which the block immersed in the water change? For convention use positive as meaning more of the block is immersed in the water and negative as less.

Answer 1

hint:
the volume of alcohol displaced by the block, is:
\[V = S\left( {h - {h_0}} \right)\]
where S is the base area of the block h is the height of the block and h_0 is the eight from the surface of the alcohol

Answer 2

now using the principle of Archimede, we can write:
\[\Large {\delta _{block}}Shg = {\delta _{alcohol}}S\left( {h - {h_0}} \right)\]
where the \delta_block is the density of the block, and 1delat_alcohol is the density of alcohol
What is h_0?

Answer 3

oops... I have made a typo error, here is the right formula:
\[\Large {\delta _{block}}Shg = {\delta _{alcohol}}S\left( {h - {h_0}} \right)g\]
where g is, of course, the gravity, namely:
g=981 cm/sec