Question

The normal density of gold is $\rho$ and its bulk modulus is $K$ .The Increase in the density of a lump of gold, when pressure $P$ is applied uniformly on all sides is
A. ${\displaystyle \frac{\rho P}{K}}$
B. ${\displaystyle \frac{\rho K}{P}}$
C. ${\displaystyle \frac{P}{\rho K}}$
D. ${\displaystyle \frac{K}{\rho P}}$

Answer 1

Hint: We first write the bulk modulus formula, then we know in bulk modules there is no change in mass, applying this statement in the problem we get a relation between volume and density of the gold. Use this relation in the bulk modulates formula.Now after rearrangement we will get the increase in the density of the lump gold.

Formula used:
$K=-{\displaystyle \frac{dP}{{\displaystyle \frac{dV}{V}}}}$
Where, Bulk modulus = $K$, Pressure applied = $dP$ and Change in its volume =${\displaystyle \frac{dV}{V}}$.
$\rho ={\displaystyle \frac{m}{V}}$
Where, Density = $\rho$, Mass = $m$ and Volume = $V$.

Complete step by step answer:
As given in the problem, there is a lump of gold whose normal density is $\rho$ , bulk modulus is $K$ and the pressure applied is $P$.Due to this applied pressure there is an increase in the density of the lump of gold.Bulk modulus is defined as the pressure that we applied divided by the fractional change in its volume due to application of this pressure.Mathematically we can write,
$K=-{\displaystyle \frac{dP}{{\displaystyle \frac{dV}{V}}}}$

Now by applying this bulk formula we will get,
$K=-{\displaystyle \frac{dP}{{\displaystyle \frac{dV}{V}}}}$
$\Rightarrow K=-V{\displaystyle \frac{dP}{dV}}\dots \dots \left(1\right)$
We can define density as mass divided by volume of the lump of gold.
Mathematically,
$\rho ={\displaystyle \frac{m}{V}}$
$\Rightarrow \rho =m{V}^{-1}$
Now differentiating the above expression we will get,
$\mathrm{\Delta }\rho =V^{-1}\mathrm{\Delta }m-m{V}^{-2}\mathrm{\Delta }V$
Divind both side by $\rho$ we will get,
${\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}={\displaystyle \frac{V^{-1}\mathrm{\Delta }m}{\rho }}-{\displaystyle \frac{m{V}^{-2}\mathrm{\Delta }V}{\rho }}$

Putting $\rho$ value in RHS we will get,
${\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}={\displaystyle \frac{V^{-1}\mathrm{\Delta }m}{m{V}^{-1}}}-{\displaystyle \frac{m{V}^{-2}\mathrm{\Delta }V}{m{V}^{-1}}}$
Cancelling the common terms we will get,
${\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}={\displaystyle \frac{\mathrm{\Delta }m}{m}}-{\displaystyle \frac{\mathrm{\Delta }V}{V}}$
We know that in bulk modulus change in mass is equal to zero.
Hence the above equation will become,
${\displaystyle \frac{\mathrm{\Delta }\rho }{\rho }}=-{\displaystyle \frac{\mathrm{\Delta }V}{V}}$
Small change in the term,
${\displaystyle \frac{d\rho }{\rho }}=-{\displaystyle \frac{dV}{V}}\dots \dots \left(2\right)$

Replacing equation $\left(1\right)$ with equation $\left(2\right)$ we will get,
$K=-V{\displaystyle \frac{dP}{dV}}$
$\Rightarrow K=-{\displaystyle \frac{dP}{{\displaystyle \frac{dV}{V}}}}$
By putting,
$K={\displaystyle \frac{dP}{{\displaystyle \frac{d\rho }{\rho }}}}$
$\Rightarrow K=\rho {\displaystyle \frac{dP}{d\rho }}$
As given in the problem here $dP=P$
Now replacing the value in the above equation we will get,
$K=\rho {\displaystyle \frac{P}{d\rho }}$
Rearranging the above equation we will get,
$d\rho =\rho {\displaystyle \frac{P}{K}}$
Where, change in density = $d\rho$

Therefore the correct option is $\left(A\right)$.

Note: Be careful while differentiating the density term. Differentiate the RHS term into two different parts: first differentiate with respect to mass by keeping volume constant and then again differentiate with respect to volume by keeping mass constant. This eases the calcul;ations in the problem.