Question

A cube of wood supporting $200g$ mass just floats in water. When mass is removed cube rises by $2cm$. What is side of cube in $cm$? (Density of water = 1000 $\frac{kg}{m^3}$)

Answer 1

Let $M$ be the mass of wooden cube and $a$ be the side of the cube.
The body is in equilibrium in both cases. So lets equate the forces in both cases.
Case I:
$0.2g+Mg={\rho }_w{V}_{1}g......\left(1\right)$

Case II:
$Mg={\rho }_w{V}_{2}g.......\left(2\right)$

Here, $V_1=V_2+a^2.\left(0.02\right)\Rightarrow V_1-V_2=a^2.\left(0.02\right)$

Now, by subtracting $\left(2\right)$ from $\left(1\right)$, we get
$0.2g+Mg-Mg={\rho }_w(V_1-V_2)g$
$0.2g={\rho }_{w}g\left(\mathrm{0.02.}{a}^2\right)$
$\Rightarrow a^2=\frac{0.2}{0.02\times {\rho }_w}=\frac{10}{1000}=0.01m^2\Rightarrow a=0.1m\Rightarrow a=10cm$