The normal density of gold is \(\rho\) and its bulk modulus is \(\mathrm{K}\). The increase in density of a…
The normal density of gold is \(\rho\) and its bulk modulus is \(\mathrm{K}\). The increase in density of a lump of gold, when a pressure \(\mathrm{P}\) is applied uniformly on all sides is
\(\frac{\rho \mathrm{P}}{\mathrm{K}}\)
\(\frac{\mathrm{P}}{\mathrm{\rho K}}\)
\(\frac{\mathrm{K}}{\mathrm{\rho P}}\)
\(\frac{\rho K}{\mathrm{P}}\)
Solution
density \(\mathrm{d}=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\mathrm{v}}=\mathrm{mv}^{-1} \Rightarrow \mathrm{dav}^{-1}\)
\(\Rightarrow\) relative change in density
will be \(\frac{\Delta \mathrm{d}}{\mathrm{d}}=\frac{\Delta \mathrm{v}}{\mathrm{v}}\)
only magnitude relation \(\frac{\Delta \mathrm{d}}{\mathrm{d}}=\frac{\Delta \mathrm{v}}{\mathrm{v}}\)
bulk modulus \(\mathrm{k}\) is given as
\(\mathrm{p}=\mathrm{k} \frac{\Delta \mathrm{v}}{\mathrm{v}} ; \mathrm{p}=\) applied pressure
\(\Rightarrow \frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{p}}{\mathrm{k}}\)
so change in density (fractional)
will be \(\frac{\Delta \mathrm{d}}{\mathrm{d}}=\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{p}}{\mathrm{k}}\)
\(\Rightarrow \Delta \mathrm{d}=\mathrm{d} \times \frac{\mathrm{p}}{\mathrm{k}}\) but \(\mathrm{d}=\rho\) (given)
So \(\Delta \mathrm{d}=\frac{\rho\mathrm{p}}{\mathrm{k}}\)