Question

The normal density of gold is $\rho$ and its bulk modulus is $K$. The increase in density of a lump of gold, when a pressure $P$ is applied uniformly on all sides is

• A. $\frac{\rho P}{K}$
• B. $\frac{\rho K}{P}$
• C. $\frac{P}{\rho K}$
• D. $\frac{K}{\rho P}$

Answer 1

The correct option is A $\frac{\rho P}{K}$

density $d=\frac{mass}{volume}=\frac{m}{v}=m{v}^{-1}\Rightarrow d\alpha v^{-1}$

$\Rightarrow$ relative change in density

will be $\frac{\Delta d}{d}=\frac{\Delta v}{v}$

only magnitude relation $\frac{\Delta d}{d}=\frac{\Delta v}{v}$

bulk modulus k is given as

$p=k\frac{\Delta v}{v};p=$ applied pressure

$\Rightarrow \frac{\Delta v}{v}=\frac{p}{k}$

so change in density (fractional)

will be $\frac{\Delta d}{d}=\frac{\Delta v}{v}=\frac{p}{k}$

$\Rightarrow \Delta d=d\times \frac{p}{k}$ but $d=p$ (given)

So $\Delta d=\frac{pp}{k}$