Question

A wooden cube just floats inside water with a $200\text{gm}$ mass placed on it. When the mass is removed, the cube floats with its top surface $2\text{cm}$ above the water level. What is the side of the cube? [Take $g=10{\text{m/s}}^2$]

• A. $6\text{cm}$
• B. $8\text{cm}$
• C. $10\text{cm}$
• D. $12\text{cm}$

Answer 1

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

Let the side of the wooden cube be $L\text{cm}$.
In initial case, the wooden cube was floating completely submerged just below the water surface.


When the mass of $200\text{gm}$ is removed, the cube comes out $2\text{cm}$ above the water surface, which means that weight of the mass was balanced by the weight of the liquid displaced corresponding to the volume of the exposed part of cube.
Volume of the exposed part of the cube,
$V=(L^2\times 2){\text{cm}}^3=2L^2\times {10}^{-6}{\text{m}}^3$
Weight of fluid displaced by the exposed part of the cube
$W_f=V\times {\rho }_f\times g$
Weight of $200\text{gm}$ mass is,
$W=mg=200\times {10}^{-3}\times 10\text{N}$

Hence we can write,
$\text{Weight of mass}=\text{Weight of fluid displaced}$
Or $W=W_f$
$\Rightarrow 200\times {10}^{-3}\times 10=V\times {\rho }_f\times g$
$\because {\rho }_f=1000{\text{kg/m}}^3$
$\Rightarrow 200\times {10}^{-2}=2L^2\times {10}^{-6}\times 1000\times 10$
$\Rightarrow L^2=100$
$\therefore L=10\text{cm}$ is the required side of cube.