Answer
7.6 cm
Work Step by Step
By considering the equilibrium of the system, we can write.
Buoyant force + Buoyant force (from oil) = Weight of the cylinder
(from water)
$F_{Bw}+F_{Bo}=mg$
By using the principle of Archimedes, we can write,
$V_{w}\rho_{w}g+V_{o}\rho_{o}g=mg$ ; Let's plug known values into this equation
$V_{w}\times1000\space kg/m^{3}+V_{o}\times 750\space kg/m^{3}=7\space kg$
$V_{w}+0.75V_{o}=7\times10^{-3}m^{3}-(1)$
From the figure, we can get,
$V_{w}+V_{o}=8.48\times10^{-3}m^{3}-(2)$
(2)-(1)=>
$0.275V_{o}=1.48\times10^{-3}m^{3}$
$V_{o}=5.38\times10^{-3}m^{3}$
We can write,
$V_{o}=\pi r^{2}h=>h=\frac{V_{o}}{\pi r^{1}}=\frac{5.38\times10^{-3}m^{3}}{\pi (0.15\space m)^{2}}=7.6\space cm$
Thus, the height of the cylinder that is in the oil = 7.6 cm
