Question

A cube of wood supporting a $200\text{gms}$ mass just floats in water. When the mass is removed, the cube rises by $2\text{cm}$. What is the size (side length) of the cube?
Density of water = $1{\text{g/cm}}^3$.
Assume initial and final situations to be under equilibrium.

• A. $10\text{cm}$
  
• B. $20\text{cm}$
• C. $30\text{cm}$
• D. $40\text{cm}$

Answer 1

The correct option is A $10\text{cm}$


If, $l$ =side of cube, $h$= height of cube above water and $\rho$ =density of wood.
Mass of the cube $=l^3\rho$
Volume of cube in water $=l^2(l-h)$
Volume of the displaced water $=l^2(l-h)$

As the cube is floating,
Weight of cube + Weight of wood =weight of liquid displaced
or $l^3\rho +200=l^2(1-h)....\left(i\right)$

After the removal of $200\text{gm}$ mass, the cube rises by $2\text{cm}$
$l^2\times l-(h+2)$
Volume of cube in water
or $l^2\times l-(h+2)=l^3\rho ...\left(ii\right)$

substituting the value of $l^3\rho$ from equation $\left(ii\right)$, in equation $\left(i\right)$, we get,
$l^2\times l-(h+2)+200=l^2(1-h)$
$l^3-l^{2}h-2l^2+200=l^3-l^2$
$2l^2=200$
$\Rightarrow l=10\text{cm}$

Alternative method:–
If we put $200\text{gm}$ mass back then $2\text{cm}$ of block will get submerged and this will increase the buoyant force, which is sufficient to balance the weight of $200\text{gm}$ mass.

Increase in buoyant force $=$ Weight of $200\text{gm}$ mass.
Let the length of the block be $l\text{cm}$ and density of water $=$ $1{\text{g/cm}}^3$.

Considering everything in $CGS$ units,

Weight of Liquid displaced by $2\text{cm}$ of the block $=$ $2l^{2}g\left(1\right)$

Increase in buoyant force $=2l^{2}g\left(1\right)$

Weight of 200gm mass $=200g$

$2l^{2}g\left(1\right)=200g$

$l=10\text{cm}$

Hence, $\left(A\right)$ is the correct answer.

Why this Question? Buoyant force is equal to the weight of the fluid displaced by the body.