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A wooden block floating in a bucket of water has $\frac{4}{5}$ of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is
$0.5$
$0.7$
$0.6$
$0.8$
Solution
$In\,{1^{st\,}}\,situation$
${V_b}{\rho _b}g = {V_s}{\rho _w}g$
$\frac{{{V_s}}}{{{V_b}}} = \frac{{{\rho _b}}}{{{\rho _w}}} = \frac{4}{5}$ $…\left( i \right)$
Here $V_b$ is volume of block
$V_s$ is submerged volume of block
${\rho _b}$ is density of block
${\rho _w}$ id density of water $\& $ Let ${\rho _0}$ is density of oil
Finally in equilibrium condition
${V_b}{\rho _b}g = \frac{{{V_b}}}{2}{\rho _0}g + \frac{{{V_b}}}{2}{\rho _w}g$
$2{\rho _b} = {\rho _0} + {\rho _w}$
$ \Rightarrow \frac{{{\rho _0}}}{{{\rho _w}}} = \frac{3}{5} = 0.6$
