A cubical block of iron, 5 cm on each side is floating in mercury taken in a vessel. Then the height of the block above mercury level. $\left({\rho }_{\text{Hg}}=13.\text{6}{\text{g/cm}}^{\text{3}}\text{,}{\rho }_{\text{Fe}}=7.2{\text{g/cm}}^{\text{3}}\right)$  

Given that a cubical block of Iron has side s = 5 cm.

Volume of Iron $={\text{s}}^3=5^3{\text{cm}}^3=125\text{cc}$

It is said to be floating in mercury present in a vessel.

So weight of Iron cube = Weight of mercury displaced

${\rho }_{\text{Iron}}\times {\text{V}}_{\text{Iron}\text{cube}}\times \text{g}={\text{V}}_{\text{Mercury}\text{displaced}}\times {\rho }_{\text{mercury}}\times \text{g}$

${\rho }_{\text{Iron}}=7.2\text{g/cc,}{\rho }_{\text{mercury}}=13.6\text{g/cc}$

$\Rightarrow 7.2\times 125={\text{V}}_{\text{Mercury}\text{displaced}}\times 13.6$

${\text{V}}_{\text{above the}\text{water}}=300\text{cc}$                     

${\text{V}}_{\text{immersed}}=66.2\text{cc}$

Height of block immersed × base area = 66.2

${\text{V}}_{\text{immersed}}\times 5^2=66.2$

$\therefore {\text{h}}_{\text{immersed}}=\frac{66.2}{25}=2.65\text{cm}$

Then height of block present above surface of mercury$=5-2.65=2.35\text{cm}$