Question

A block of wood floats in a liquid of density $0.8\,g\,c{m^{ - 3}}$ with one fourth of its volume submerged. In oil the block floats with 60% of its volume submerged. Find the density of (a) wood and (b) oil

Answer 1

Calculate the weight of the wood that is submerged in the liquid. Then use the law of floatation or Archimedes principle to determine the density of wood. The Buoyant force is equal to the weight of the floating object.

Formula used:
The density of the object of mass m is,
$\rho = \dfrac{m}{V}$
Here, V is the volume of the object.

Complete step by step answer:
(a) We assume the volume of wood is V, then we have given that one fourth of the volume of wood is submerged in the liquid. Therefore, the volume of wood submerged in the liquid is $\dfrac{V}{4}$.
We also assume the density of wood is ${\rho _w}$ and density of the given liquid is ${\rho _l}$.

We can express the weight of wood using the relation between mass, density and volume as follows,
${W_w} = {\rho _w}Vg$

Here, g is the acceleration due to gravity.

Also, we can express the weight of the liquid displaced by the wood is,
${W_l} = \dfrac{{{\rho _l}Vg}}{4}$

Now, we can use Archimedes principle to calculate the buoyant force acting on the wood. According to Archimedes principle, the buoyant force acting on the object submerged in the liquid is equal to weight of the liquid displaced by the object. Therefore,
${\rho _w}Vg = \dfrac{{{\rho _l}Vg}}{4}$
${\rho _w} = \dfrac{{{\rho _l}}}{4}$

Substitute $0.8\,g\,c{m^{ - 3}}$ for ${\rho _l}$ in the above equation.
${\rho _w} = \dfrac{{0.8\,g\,c{m^{ - 3}}}}{4}$
$\Rightarrow {\rho _w} = 0.2\,g\,c{m^{ - 3}}$

Therefore, the density of wood is $0.2\,g\,c{m^{ - 3}}$.

(b) We have given that the volume of wood submerged in the oil is 60%. Therefore, the volume of submerged wood is, $\dfrac{{3V}}{5}$.

Now, according to Archimedes principle, the buoyant force on the wood is, $\dfrac{{{\rho _{oil}}3Vg}}{5}$. Here, ${\rho _{oil}}$ is the density of oil.

We have, the weight of wood is equal to buoyant force. Therefore,
${\rho _w}Vg = \dfrac{{{\rho _{oil}}3Vg}}{5}$
$\Rightarrow {\rho _{oil}} = \dfrac{{5{\rho _w}}}{3}$

Substitute $0.2\,g\,c{m^{ - 3}}$ for ${\rho _w}$ in the above equation.
${\rho _{oil}} = \dfrac{{5\left( {0.2} \right)}}{3}$
$\Rightarrow {\rho _{oil}} = 0.33\,g\,c{m^{ - 3}}$

Therefore, the density of oil is $0.33\,g\,c{m^{ - 3}}$.

Note:
Students should always use Archimedes' principle when the question is based on an object submerged or floating in the liquid. To cross check the solution, students should always check whether the density of a floating or submerged object is less than the liquid or not. The density of the floating object is always less than the density of liquid.