Step-by-step

Step 1 of 2

(a)

The weight of the block is,

${W_{{\rm{block}}}} = {m_{{\rm{block}}}}g$ …… (1)

Here, ${m_{{\rm{block}}}}$ is the mass of the block and $g$ is the acceleration due to gravity.

The weight of the water is,

${W_{{\rm{water}}}} = {m_{{\rm{water}}}}g$ …… (2)

Here, ${m_{{\rm{water}}}}$ is the mass of water.

Since the block of wood floats when submerged in water. Therefore, the weight of the water displaced by the block will be equal to the weight of the block.

Therefore,

${m_{{\rm{block}}}}g = {m_{{\rm{water}}}}g$ …… (3)

Substitute ${\rho _{{\rm{block}}}}{V_{{\rm{block}}}}$ for ${m_{{\rm{block}}}}$ and ${\rho _{{\rm{water}}}}{V_{{\rm{water}}}}$ for ${m_{{\rm{water}}}}$ in the equation (3).

${\rho _{{\rm{block}}}}{V_{{\rm{block}}}}g = {\rho _{{\rm{water}}}}{V_{{\rm{water}}}}g$

The above equation is modified as,

${\rho _{{\rm{block}}}} = {\rho _{{\rm{water}}}}\left( {\frac{{{V_{{\rm{water}}}}}}{{{V_{{\rm{block}}}}}}} \right)$ …… (4)

Substitute $1000{\rm{ kg/}}{{\rm{m}}^3}$ for ${\rho _{{\rm{water}}}}$ and $\frac{2}{3}{V_{{\rm{block}}}}$ for ${V_{{\rm{water}}}}$ in the equation (4).

$\begin{array}{c}\\{\rho _{{\rm{block}}}} = {\rho _{{\rm{water}}}}\left( {\frac{{{V_{{\rm{water}}}}}}{{{V_{{\rm{block}}}}}}} \right)\\\\ = 1000{\rm{ kg/}}{{\rm{m}}^3}\left( {\frac{{\frac{2}{3}{V_{{\rm{block}}}}}}{{{V_{{\rm{block}}}}}}} \right)\\\\ = 666.66{\rm{ kg/}}{{\rm{m}}^3}\\\end{array}$

Part a

The density of the block of wood is $666.66{\rm{ kg/}}{{\rm{m}}^3}$

Explanation | Common mistakes | Hint for next step

When the object floats, the weight of the object is equal to the weight of the fluid displaced by the object. Hence, the buoyant force acting on the object will be equal to the weight of the object.

Step 2 of 2

(b)

The block of wood floats when submerged in oil. Therefore, the weight of the oil displaced by the block will be equal to the weight of the block.

${\rho _{{\rm{block}}}}{V_{{\rm{block}}}}g = {\rho _{{\rm{oil}}}}{V_{{\rm{oil}}}}g$

The above equation is modified as,

${\rho _{{\rm{oil}}}} = {\rho _{{\rm{block}}}}\left( {\frac{{{V_{{\rm{block}}}}}}{{{V_{{\rm{oil}}}}}}} \right)$ …… (5)

Here, ${V_{{\rm{oil}}}}$ is the volume of the oil.

Substitute ${\rm{666}}{\rm{.66 kg/}}{{\rm{m}}^3}$ for ${\rho _{{\rm{block}}}}$ and $0.9{V_{{\rm{block}}}}$ for ${V_{{\rm{oil}}}}$ in the equation (5).

$\begin{array}{c}\\{\rho _{{\rm{oil}}}} = {\rho _{{\rm{block}}}}\left( {\frac{{{V_{{\rm{block}}}}}}{{{V_{{\rm{oil}}}}}}} \right)\\\\ = {\rm{666}}{\rm{.66 kg/}}{{\rm{m}}^3}\left( {\frac{{{V_{{\rm{block}}}}}}{{0.9{V_{{\rm{block}}}}}}} \right)\\\\ = 740.73{\rm{ kg/}}{{\rm{m}}^3}\\\end{array}$

Part b

The density of oil is $740.73{\rm{ kg/}}{{\rm{m}}^3}$

Explanation | Common mistakes

The density of the oil is less than the density of the block and therefore the block will be able to float when submerged in the oil.

Answer

Part a

The density of the block of wood is $666.66{\rm{ kg/}}{{\rm{m}}^3}$

Part b

The density of oil is $740.73{\rm{ kg/}}{{\rm{m}}^3}$