Question

A wooden block floating in seawater has two thirds of its volume submerged. When the block is placed in mineral oil, \(80.0 \%\) of its volume is submerged. Find the density of (a) the wooden block and (b) the mineral oil.

Short answer

#tag_title#Step 2: Apply the principle of buoyancy to the block in seawater#tag_content# The buoyant force acting on the wooden block when submerged in seawater is equal to the weight of the seawater displaced by the volume submerged. Thus, \[F_B={V_{submerged_{sw}}}\times {\rho_{sw}}\times g= \frac{2}{3}V \times {\rho_{sw}}\times g\] The weight of the wooden block is \[W={m_W}\times g=V\times {\rho_W}\times g\] For the block to be in equilibrium (neither sinking nor floating), the buoyant force must equal the weight: \[\frac{2}{3}V \times {\rho_{sw}}\times g = V\times {\rho_W}\times g\] #tag_title#Step 3: Solve for the density of the wooden block#tag_content# We can find the density of the wooden block by dividing both sides of the equation from step 2 by Vg, and we get: \[\frac{2}{3}{\rho_{sw}}={\rho_W}\] Now, we plug in the given density of seawater, ${\rho_{sw}}=1030 \frac{kg}{m^3}$, and solve for ${\rho_W}$: \[\rho_W=\frac{2}{3}\times1030 = 686 \frac{kg}{m^3}\] #tag_title#Step 4: Apply the principle of buoyancy to the block in mineral oil#tag_content# Similar to step 2, the buoyant force acting on the wooden block when submerged in mineral oil is equal to the weight of the mineral oil displaced by the volume submerged. Thus, \[F_B={V_{submerged_{mo}}}\times {\rho_{mo}}\times g= 0.8V \times {\rho_{mo}}\times g\] Since the weight of the wooden block remains the same, we again have the buoyant force equaling the weight: \[0.8V \times {\rho_{mo}}\times g = V\times {\rho_W}\times g\] #tag_title#Step 5: Solve for the density of mineral oil#tag_content# We can find the density of the mineral oil by dividing both sides of the equation from step 4 by 0.8Vg, and we get: \[\rho_{mo}= \frac{\rho_W}{0.8}\] Now, we plug in the density of the wooden block, ${\rho_W}=686 \frac{kg}{m^3}$, and solve for ${\rho_{mo}}$: \[{\rho_{mo}}= \frac{686}{0.8}= 857.5 \frac{kg}{m^3}\] The density of the wooden block is $686 \frac{kg}{m^3}$, and the density of the mineral oil is $857.5 \frac{kg}{m^3}$.

Step-by-step solution

Write down the given volume percentages submerged in seawater and mineral oil

In seawater: Volume submerged, \(V_{submerged_{sw}}= \frac{2}{3}V\) In mineral oil: Volume submerged, \(V_{submerged_{mo}}= 0.8V\)