Question

The normal density of gold is $\rho$ and its modulus is $B$. The increase in density of piece of gold when pressure $P$ is applied uniformly from all sides

• A. $\frac{\rho P}{2 B}$
• B. $\frac{\rho B}{2 P}$
• C. $\frac{\rho P}{B-P}$
• D. $\frac{\rho B}{B-P}$

Answer 1

Answer: (C)

We know
$\frac{\Delta V}{V}=\frac{P}{B}$. ..(1)
$\Delta V-$ Change in volume
$ P-$ Pressure applied
$B $ -Bulk modulus
And $\rho=\frac{M}{V}$ ... (2)
$\rho=$ Density
$M=$ Mass
$V=$ Volume
From (2)
$\Delta \rho=\frac{M}{V-\Delta V}-\frac{M}{V}$
$\Delta \rho=\frac{M}{V} \times \frac{\Delta V}{V-\Delta V}$
$\Delta \rho = \rho \times \frac{1}{\frac{V}{\Delta V} -1}$ [From eq (2)]
$\Delta \rho=\rho \times \frac{1}{\frac{B}{P}-1} $[From eq. (1)]
$\Delta \rho=\frac{\rho P}{B-P}$