Answer
Density of the oil $=7.4\times10^{2}$ $kg/m^{3}$
Work Step by Step
First consider the case when it's floating in water.
Since the block is floating and it is undergoes no net acceleration, the weight of block $(W)$ is equal to the buoyant force $(F_{b})$ by water.
This can be written as:
$W=F_{b}$ . . . . . . . . . . . . . . . . . . . . . . .(1)
Since two-thirds of the volume $V$ is under water, buoyant force is given by:
$F_{b}=\frac{2}{3}V \rho_{w}g$
Substituting the known values and setting $F_{b}$ as $\frac{2}{3}V \rho_{w}g$ in the previous equation and solving gives:
$mg = \frac{2}{3}V \rho_{w} g$
$m=\frac{2}{3}V\times1000$ $kg/m^{3}$ . . . . . . . . . . . . . . . . . . (2)
Here the equation is in terms of mass and volume but we need that in terms of density of the block.
So writing mass in terms of volume and density gives:
$mass=density\times volume$
$m=\rho_{block}\times V$
Substituting the value of $m$ in equation (2) and solving gives:
$\rho_{block}\times V=\frac{2}{3}V\times1000$ $kg/m^{3}$
$\rho_{block}= \frac{2}{3}\times1000$ $kg/m^{3}= 666.66$ $kg/m^{3}=6.6666\times10^{2}$ $kg/m^{3}$
$\rho_{block}\approx 6.7\times10^{2}$ $kg/m^{3}$
So the density of the wooden block is $6.7\times10^{2}$ $kg/m^{3}$
Similarly, when it floats in oil with $\frac{90}{100}V$ submerged :
$W=F_{b}$ . . . . . . . . . . . . . . . . . . . . . . .(3)
and the magnitude of buoyant force is given by:
$F_{b}=\frac{90}{100}V \rho_{oil}$ $g$
Substituting the known values and setting $F_{b}$ as $\frac{90}{100}VĪ_{oil}$ $g$ in the equation (3) and solving gives:
$mg=\frac{90}{100}V\times \rho_{oil}$ ${g}$ . . . . . . . . . . . . . . . . . . . . . . .(4)
Writing the mass of block in terms of volume and density gives:
$mass=density\times volume$
$m=\rho_{block}\times V$
Substituting this value of $m$ in Equation (4) and setting $\rho_{block} =6.7\times10^{2}$$kg/m^{3}$ gives:
$\rho_{block}\times V \times g=\frac{90}{100}V\times \rho_{oil}$ ${g}$
$(6.7\times 10^{2}$ $kg/m^{3})\times V \times g=\frac{90}{100}V \times \rho_{oil}\times g$
$6.7\times10^{2}$ $kg/m^{3}=\frac{9}{10}\rho_{oil}$
$\rho_{oil}=6.7\times10^{2}$ $kg/m^{3}\times\frac{10}{9}$
$\rho_{oil}=744.4$ $kg/m^{3}$
$\rho_{oil}=7.4\times10^{2}$ $kg/m^{3}$
So the density of oil $=7.4\times10^{2}$ $kg/m^{3}$
