Question

A cubical block of wood 10 cm on a side & rho wood= 0.5 g/cm^3 floats in a jar of water. Oil of density 0.8 g/cm^3 is poured into the water until the top of the oil layer is 4 cm below the top of the block. How deep is the oil?

Answer 1

Answer on Question #72396-Physics-Mechanics-Relativity

A cubical block of wood 10 cm on a side & rho wood= 0.5 g/cm^3 floats in a jar of water. Oil of density 0.8 g/cm^3 is poured into the water until the top of the oil layer is 4 cm below the top of the block. How deep is the oil?

Solution

Side of cubical block is

$L=0.1\ m$

Density of wood is

$d_{1}=500\frac{kg}{m^{3}}$

Density of oil is

$d_{2}=800\frac{kg}{m^{3}}$

The top of the oil layer is 0.04 m below the top of the block.

Area of face of wooden block is

$A=0.01\ m^{2}$

Volume of wooden block is

$V=0.001\ m^{3}.$

Weight of wooden block is

$W_{b}=0.1A\cdot 500g=50Ag.$

Suppose depth of oil layer is x.

Volume of oil displaced is

$V_{o}=xA$

Weight of oil displaced is

$W_{o}=800\ xAg$

Volume of water displaced is

$V_{w}=(0.1-0.04-x)A=(0.06-x)A$

Weight of water displaced is

$W_{w}=1000(0.06-x)\ Ag$

From law of floatation:

$$W_o + W_w = W_b$$ $$800 \, xAg + 1000(0.06 - x)Ag = 50Ag$$ $$800x + 1000(0.06 - x) = 50$$ $$x = 0.05 \, m = 5 \, cm.$$

Answer: $5 \, cm$.

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