The density of a metal at a normal pressure is $\rho$. Its density when it is subjected to an excess pressure p is $\rho$'. If B is the bulk modulus of the metal, the ratio $\rho \text{'}/\rho$ is

Let V be the volume of the metal at normal pressure. 

From B =$-\frac{p}{∆V/V}$ , the decrease in the volume when the metal is subjected to an excess pressure p is given by 

          $\left|∆V\right|=\frac{pV}{B}$

$\therefore$    New volume of metal is V' = V$-\frac{pV}{B}$ = V$\left(1-\frac{p}{B}\right)$

Now mass of the metal is m = $\rho$V. Therefore, its new density is 

           $\rho \text{'}=\frac{m}{V\text{'}}=\frac{\rho V}{V\left(1-{\displaystyle \frac{p}{B}}\right)}=\frac{\rho }{1-{\displaystyle \frac{p}{B}}}$

or      $\frac{\rho \text{'}}{\rho }=\frac{1}{1-p/B}$