A cube of wood supporting 200g mass just floats in water the mass is removed, cube raises by 2cm, the side of cube is

Given that A cube is supporting 200 g of mass

Just floating in water as shown

M = 200 g = 0.2 kg

Let side of cube = s

Mass of cube $\text{M}=\text{d}\times \text{v}$

$\text{M}={\text{ds}}^3$

We know weight of body = weight of liquid displaced.

$\text{Vdg}+0.\text{2g}={\text{Vd}}_{\text{2}}\text{g}$

${\text{S}}^{\text{3}}\text{d}+0.\text{2}={\text{s}}^{\text{3}}\text{1}000$

$0.\text{2}={\text{s}}^{\text{3}}\left(\text{1}000-\text{d}\right)\mathrm{.............}\left(\text{1}\right)$

${\text{m}}_{\text{wood}}{\text{=V}}_{\text{b}}{\text{d}}_{\text{w}}g$

${\text{s}}^{\text{3} }\text{d}={\text{s}}^{\text{2}} \left(\text{s}-0.0\text{2}\right){\text{d}}_{w }$  

$\Rightarrow \text{sd}={\text{sd}}_{\text{w}}-0.02{\text{d}}_{\text{w}}$                    $\left(\because {\text{d}}_{\text{w}}=1000{\text{kg/m}}^{\text{3}}\right)$

$\Rightarrow \text{sd}=\text{s}\left(\text{1}000\right)-\text{2}0$

$\text{2}0=\text{S}\left(\text{1}000-\text{d}\right)\mathrm{............}\left(\text{2}\right)$

Divide eqn (2) with (1) $\Rightarrow \frac{20}{0.2}=\frac{\text{s}\left(1000-\text{d}\right)}{{\text{s}}^{\text{3}}\left(1000-\text{d}\right)}$

$\Rightarrow \frac{20}{0.2}=\frac{s}{s^3}$

$\therefore s^2=\frac{1}{100}$

$\Rightarrow \text{s}=\text{1}{0}^{-\text{1} }\text{=1}0\text{cm}$