Question

A cube of wood floating in water supports a $200g$ mass resting on the center of its top face. When the mass is removed the cube rises $2cm$. Determine the volume of the cube?

Answer 1

Hint: First we will assume the length of the edge of the cube. Then we will have to find the volume of water displaced by the $200g$ object. Then from the statement we can say that the mass of the $200g$ object is equal to the mass of the displaced water. Now using mass, volume and density relation we can find the edge of the cube. Once we get the edge we can find the volume of the cube.

Complete step by step answer:
As per the problem there is a cube of wood floating in water supporting a $200g$ mass resting on the center of its top face. When the mass is removed the cube rises $2cm$.
We need to find the volume of the cube.
Let us assume that the edge length of the cube is $xcm$.
As in the problem it is saying when mass is removed the cube rises $2cm$ so we can say the volume of water displaced by $200g$ object is,
${V_{WD}} = lbh$
Where,
${V_{WD}}$ is the volume of water displaced.
The length and breadth are the edge of the cube and the height is $2cm$.
Now on putting the respective values we will get,
${V_{WD}} = x \times x \times 2c{m^3}$
$ \Rightarrow {V_{WD}} = 2{x^2}c{m^3}$
Now we can say that the mass of the $200g$ object is equal to the mass of displaced water.
Mathematically we can write,
Mass of $200g$ object = Mass of water displaced
${M_{Object}} = {M_{WD}} \ldots \ldots \left( 1 \right)$
We know,
Mass of the object be ${M_{Object}} = 200g$.
Mass of water displaced can be represented as,
${M_{WD}} = {V_{WD}} \times \rho $
Where,
${M_{WD}}$ is the mass of water displaced.
${V_{WD}}$ is the volume of water displaced.
$\rho $ is the density of water which is a constant term equal to $1gc{m^{ - 3}}$.
Now putting the know vales we will get,
${M_{WD}} = 2{x^2}c{m^3} \times 1gc{m^{ - 3}} = 2{x^2}g$
Now putting the known values in equation $\left( 1 \right)$ we will get,
$200g = 2{x^2}g$
Cancelling and rearranging we will get,
$x = 10cm\left( {edge} \right)$
We know the volume of the cube as,
${V_{Cube}} = {\left( {edge} \right)^3} = {\left( {10cm} \right)^3}$
Hence the volume of the cube is $1000c{m^3}$.

Note: Here we have used the density of water as $1gc{m^{ - 3}}$ which is an approximate value the actual vale $0.997gc{m^{ - 3}}$ or $997kg{m^{ - 3}}$. Note that this density of water is measured in $25^\circ $ Celsius which is the room temperature. Remember that density of water is directly proportional to mass which is that it increases with increase in mass while it is inversely proportional to volume which is that it decreases with increase in volume.