Question

A cube of wood supporting a $200\text{g}$ mass just floats in water. When the mass is removed, the cube rises by $2\text{cm}$. What is the size 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 $10\text{cm}$

Let $l$ be the side of the cube, $h$ be the height of the cube above water and $\rho$ be the 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

$\therefore \text{Weight of cube + weight of mass= weight of liquid displaced}$

$\Rightarrow l^3\rho g+(200\times g)={\rho }_w\times l^2(l-h)g$

Given, ${\rho }_w=1{\text{g/cm}}^3$

$\Rightarrow l^3\rho +200=l^2(l-h).........\left(1\right)$

After the removal of $200\text{g}$ mass, the cube rises by $2\text{cm}$.

Volume of cube in water$=l^2\times (l-(h+2\left)\right)$

For flotation,

${\rho }_w\times l^2\times (l-(h+2\left)\right)g=l^3\rho g$

$\Rightarrow l^2\times (l-(h+2\left)\right)=l^3\rho .........\left(2\right)$

Substituting the value of $l^3\rho$ from equation $\left(2\right)$ in equation $\left(1\right)$, we get

$l^2\times (l-(h+2\left)\right)+200=l^2(l-h)$

$l^3-l^{2}h-2l^2+200=l^3-l^{2}h$

$\Rightarrow 2l^2=200$

$\Rightarrow l=10\text{cm}$

Hence, option $\left(\text{A}\right)$ is correct.

$\text{Alternative Solution:-}$ If we put $200\text{g}$ mass back, then $2\text{cm}$ of the block will get submerged and this will increase the buoyant force, which is sufficient to balance the $200\text{g}$ mass. Increase in buoyant force = weight of $200\text{g}$ 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 $=2l^{2}g\times 1$ Increase in buoyant force = $2l^{2}g\times 1$ Weight of $200\text{g}$ mass = $200\times g$ $\therefore 2l^{2}g\times 1$= $200\times g$ $\Rightarrow l=10\text{cm}$ Hence, option $\left(\text{A}\right)$ is correct.