Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm³. If the mass of the other is 48 g, its density in g / cm³ is

Apparent weight  $=\mathrm{V}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}=\frac{\mathrm{m}}{\mathrm{\rho }}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}$
where  m =  mass of the body, 
$\rho =$  density of the body  
$\sigma =$  density of water 
If two bodies are in equilibrium then their apparent weight must be equal.
$\frac{{\mathrm{m}}_1}{{\mathrm{\rho }}_1}({\mathrm{\rho }}_1-\mathrm{\sigma })=\frac{{\mathrm{m}}_2}{{\mathrm{\rho }}_2}({\mathrm{\rho }}_2-\mathrm{\sigma })$    
$\Rightarrow \frac{36}{9}(9-1)=\frac{48}{{\rho }_2}\left({\rho }_2-1\right)$  
By solving we get  ${\mathrm{\rho }}_2=3$.