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Step-by-Step Detailed Solution
Step 1: Understanding the Initial Condition (Block in Water)
When the wooden block is floating in water such that
4/5 of its volume is submerged, the buoyant force
(from water) balances the weight of the block.
Mathematically,
$V_b\,\rho_b\,g = V_s\,\rho_w\,g$,
where
$V_b$ = total volume of the block,
$V_s$ = submerged volume,
$\rho_b$ = density of the block, and
$\rho_w$ = density of water.
Simplifying, we get
$\frac{V_s}{V_b} = \frac{\rho_b}{\rho_w} = \frac{4}{5}\,.$
This implies
$\rho_b = \frac{4}{5}\rho_w$.
Step 2: Analyzing the Final Situation (Block Partly in Water and Partly in Oil)
After pouring the oil, the block is found to be just under
the surface of oil with half of its volume in water and half
in oil. Let $\rho_o$ be the density of oil. The new buoyant
force is now contributed by both water and oil.
Hence, total buoyant force
$= \left(\frac{V_b}{2}\right)\rho_w\,g + \left(\frac{V_b}{2}\right)\rho_o\,g$,
and this must equal the weight of the block, $V_b\,\rho_b\,g$.
Therefore,
$V_b\,\rho_b\,g = \left(\frac{V_b}{2}\right)\rho_o\,g + \left(\frac{V_b}{2}\right)\rho_w\,g\,.$
Step 3: Deriving the Relation Between Oil's Density and Water's Density
Cancel $V_b$ and $g$ from both sides:
$\rho_b = \frac{1}{2}(\rho_o + \rho_w)$.
Since from Step 1 we have
$\rho_b = \frac{4}{5}\,\rho_w$,
substituting this value of $\rho_b$ into the above equation gives:
$\frac{4}{5}\rho_w = \frac{1}{2}(\rho_o + \rho_w)$
$\Rightarrow \frac{8}{5}\rho_w = \rho_o + \rho_w$
$\Rightarrow \rho_o = \frac{8}{5}\rho_w - \rho_w = \frac{3}{5}\rho_w$.
Thus, the density of the oil relative to that of water is
$\frac{\rho_o}{\rho_w} = \frac{3}{5} = 0.6$.
Step 4: Final Answer
The density of the oil (relative to the density of water) is 0.6.
