Answer to Question #157422 in Mechanics | Relativity for Femi

Question

A block of wood weighs 65g in air. A lead sinker weighs 70g in water. The sinker is attached to the wood and both together weighs 65g in water. Find the relative density of the wood.

Answer 1

By the definition of the buoyant force, we have:

$$\rho Vg=mg.$$

For the block of wood, we get:

$$\rho_{wood}V_{wood}=65,$$ $$V_{wood}=\dfrac{65}{\rho_{wood}}.$$

For the lead sinker, we get:

$$(\rho_{lead}-\rho_{water})V_{lead}=70,$$ $$V_{lead}=\dfrac{70}{(\rho_{lead}-\rho_{water})}.$$

For the sinker attached to the wood and both submerged in water, we get:

$$\rho_{lead}V_{lead}+\rho_{wood}V_{wood}-\rho_{water}(V_{lead}+V_{wood})=65.$$

Substituting, $V_{lead}$ and $V_{wood}$ into the previous equation, we get:

$\rho_{lead}\dfrac{70}{(\rho_{lead}-\rho_{water})}+\rho_{wood}\dfrac{65}{\rho_{wood}}-\rho_{water}(\dfrac{70}{(\rho_{lead}-\rho_{water})}+\dfrac{65}{\rho_{wood}})=65.$

After simplification, we get:

$$70-65(\dfrac{\rho_{water}}{\rho_{wood}}-1)=65,$$ $$\dfrac{\rho_{water}}{\rho_{wood}}-1=\dfrac{5}{65},$$ $$\dfrac{\rho_{water}}{\rho_{wood}}=\dfrac{14}{13}.$$

Finally, we can find the relative density of the wood:

$$\dfrac{\rho_{wood}}{\rho_{water}}=\dfrac{13}{14}=0.928$$

Answer:

$\dfrac{\rho_{wood}}{\rho_{water}}=0.928$