
$\mathrm{v}_2=\text { volume immersed in oil. }$
$\mathrm{v}_1 \rho_{\mathrm{w}} \mathrm{g}+\mathrm{v}_2 \rho_{\mathrm{o}} \mathrm{g}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \rho_{\mathrm{c}} \mathrm{g}$
$\mathrm{v}_1+\frac{\mathrm{v}_2 \rho_0}{\rho_w}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \frac{\rho_{\mathrm{c}}}{\rho_w}$
$=\mathrm{v}_1+0.8 \mathrm{v}_2=0.9 \mathrm{v}_1+0.9 \mathrm{v}_2$
$=0.1 \mathrm{v}_1=0.1 \mathrm{v}_2$
$\mathrm{v}_1: \mathrm{v}_2=1: 1$
(given atmospheric pressure $P_{A}=1.01 \times 10^{5}\,Pa$, density of water $\rho_{ w }=1000\,kg / m ^{3}$ and gravitational acceleration $g=10\,m / s ^{2}$ )
