Answer: 1.83 cm below the interface
Explanation:
There are three forces acting on the block of wood here:
- The weight of the block, downward, of magnitude
[tex]W=\rho_b V g[/tex]
where
[tex]\rho_b = 961 kg/m^3[/tex] is the density of the block (wood)
V is the total volume of the block
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
- The buoyant force on the part of block immersed in the water, upward, of magnitude
[tex]B_w = \rho_w V_w g[/tex]
where
[tex]\rho_w = 1000 kg/m^3[/tex] is the water density
[tex]V_w[/tex] is the fraction of block immersed in the water
- The buoyant force on the part of block in the oil, upward, of magnitude
[tex]B_o = \rho_o V_o g[/tex]
where
[tex]\rho_o=923 kg/m^3[/tex] is the oil density
[tex]V_o[/tex] is the part of block immersed in the oil
Since the block is in equilibrium, we have
[tex]W=B_w+B_o\\\rho_b V g = \rho_w V_w g + \rho_o V_o g\\\rho_b V = \rho_w V_w + \rho_o V_o[/tex]
We also know that
[tex]V=V_o+V_w[/tex]
Since the block is completely immersed in the two liquids; so we can re-arrange the equations and find:
[tex]\rho_b (V_o+V_w) = \rho_w V_w + \rho_o V_o\\\rho_b V_o+\rho_b V_w = \rho_w V_w + \rho_o V_o\\V_o(\rho_b-\rho_o)=V_w(\rho_w-\rho_b)\\\frac{V_w}{V_o}=\frac{\rho_b-\rho_o}{\rho_w-\rho_b}=\frac{961-923}{1000-961}=0.974[/tex]
This means that the fraction of volume in water is 0.974 times the fraction of volume in oil.
We also know that the volume of the block in each medium is proportional to the height of the block in each medium, so
[tex]\frac{h_w}{h_o}=0.974[/tex] (1)
We also know that the total height of the block is
[tex]h=h_w+h_o = 3.72 cm[/tex]
so
[tex]h_o = 3.72-h_w[/tex]
And substituting into (1), we find
[tex]h_w = 0.974 h_o = 0.974(3.72-h_w)=3.62-0.974h_w\\h_w=\frac{3.62}{1+0.974}=1.83 cm[/tex]
