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Step-by-Step Solution
Step 1: Recall the Definition of Bulk Modulus
The bulk modulus of elasticity, denoted by $K$, is defined (in magnitude) as
$K = -\,\frac{\Delta P}{\frac{\Delta V}{V}}$,
where $\Delta P$ is the change in external pressure, $V$ is the initial volume, and $\Delta V$ is the change in volume.
Step 2: Express Volume Change in Terms of Density
Since density $\rho$ is defined by
$\rho = \frac{M}{V}$,
a change in volume can be related to a change in density. If $\rho$ changes by a small amount $\Delta \rho$, then we can write:
\[
\frac{-\,\Delta \rho}{\rho} = \frac{\Delta V}{V}.
\]
The negative sign appears because an increase in density corresponds to a decrease in volume and vice versa.
Step 3: Reformulate Bulk Modulus in Terms of Density Change
Using
\(
\frac{\Delta V}{V} = -\,\frac{\Delta \rho}{\rho},
\)
we substitute into the expression for $K$:
\[
K = -\,\frac{\Delta P}{\frac{\Delta V}{V}} = -\,\frac{\Delta P}{\left(-\,\frac{\Delta \rho}{\rho}\right)} = \frac{\rho\,\Delta P}{\Delta \rho}.
\]
Step 4: Solve for the Change in Density, $\Delta \rho$
Rearranging the above equation,
\[
\Delta \rho = \frac{\rho \,\Delta P}{K}.
\]
Step 5: Substitute the Given Pressure $P$
Since the applied pressure is $P$ (instead of $\Delta P$), we have:
\[
\Delta \rho = \frac{\rho \,P}{K}.
\]
This is the increase in the density of the material under the applied uniform pressure $P$.
Final Answer
$\Delta \rho = \dfrac{\rho\,P}{K}\,.$
