The normal density of gold is p and its bulk modulus is K The increase in density of a lump of gold when a pressure P applied uniformly on all side is

$\mathrm{K}=\frac{\mathrm{P}}{(\mathrm{\Delta V}/\mathrm{V})}\text{ or }\frac{\mathrm{\Delta V}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{K}}$………..(1)
Further.$\mathrm{\rho }=\frac{\mathrm{M}}{\mathrm{V}}\text{ and }{\mathrm{\rho }}^{\mathrm{\prime }}=\frac{\mathrm{M}}{\mathrm{V}-\mathrm{\Delta V}}$
$\therefore \frac{{\mathrm{\rho }}^{\mathrm{\prime }}}{\mathrm{\rho }}=\frac{\mathrm{V}}{\mathrm{V}-\mathrm{\Delta V}}=\frac{1}{\{1-(\mathrm{\Delta V}/\mathrm{V}\left)\right\}}$
$={\left(1-\frac{\mathrm{\Delta V}}{\mathrm{V}}\right)}^{-1}=\left(1+\frac{\mathrm{\Delta V}}{\mathrm{V}}\right)$………..(2)
From eqs. (1) and (2), we get
$\frac{{\mathrm{\rho }}^{\mathrm{\prime }}}{\mathrm{\rho }}=\left(1+\frac{\mathrm{P}}{\mathrm{K}}\right)\text{ or }\frac{{\mathrm{\rho }}^{\mathrm{\prime }}}{\mathrm{\rho }}-1=\frac{\mathrm{P}}{\mathrm{K}}$
or $\frac{{\mathrm{\rho }}^{\mathrm{\prime }}-\mathrm{\rho }}{\mathrm{\rho }}=\frac{\mathrm{\rho }}{\mathrm{K}}\text{ or }{\mathrm{\rho }}^{\mathrm{\prime }}-\mathrm{\rho }=\frac{\mathrm{\rho P}}{\mathrm{K}}$