A wooden block floating in a bucket of water has 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.
- A
0.5
- B
0.8
- C
0.7
- D
0.6
$\begin{array}{l}V\sigma g = \dfrac{4}{5}v{\rho _\omega }g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(1)\\V\sigma g = \dfrac{v}{2}{\rho _\omega }g + \dfrac{v}{2}{\rho _0}g\\ \Rightarrow \,\left( {\dfrac{{{\rho _\omega }}}{2} + \dfrac{{{\rho _{oil}}}}{2}} \right) = \dfrac{4}{5}{\rho _\omega }\\ \Rightarrow \,\dfrac{{{\rho _{oil}}}}{2} = {\rho _\omega }\left( {\dfrac{4}{5} - \dfrac{1}{2}} \right) = \dfrac{3}{{10}}{\rho _\omega }\\ \Rightarrow {\rho _{oil}} = \dfrac{3}{5}{\rho _\omega } = 0.6{\rho _\omega }\end{array}$
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