Understanding Bulk Modulus and Density Changes in Materials

This question asks us to determine the final density of a spherical object after it has been subjected to a uniform pressure. We are given the initial density $\rho_0$, the applied pressure $P$, and the bulk modulus $B$ of the material. We are also told to assume that the changes in volume and density are small.

Defining Bulk Modulus

The bulk modulus ($B$) quantifies how resistant a substance is to being compressed uniformly. It's defined mathematically as the ratio of the change in pressure ($\Delta P$) to the relative change in volume ($\frac{\Delta V}{V}$). The formula is:

$B = -\frac{\Delta P}{\frac{\Delta V}{V}}$

The negative sign ensures that the bulk modulus is a positive value. This is because applying a positive pressure (an increase, $\Delta P > 0$) results in a negative change in volume (compression, $\Delta V < 0$).

Calculating Volume Change under Pressure

We are given that a uniform pressure $P$ is applied. We can consider this as the change in pressure, so $\Delta P = P$. For situations where the change is small, we can use the initial volume $V_0$ as the reference volume $V$ in the definition. The relationship becomes:

$B \approx -\frac{P}{\frac{\Delta V}{V_0}}$

Now, we can rearrange this equation to find the relative change in volume, $\frac{\Delta V}{V_0}$:

$\frac{\Delta V}{V_0} \approx -\frac{P}{B}$

Relating Density and Volume

Density ($\rho$) is defined as mass ($m$) per unit volume ($V$), so $\rho = \frac{m}{V}$. The mass of the spherical object remains constant throughout this process.

Let the initial volume be $V_0$ and the initial density be $\rho_0$. The mass $m$ can be expressed as $m = \rho_0 V_0$.

Let the final volume be $V$ and the final density be $\rho$. The mass $m$ can also be expressed as $m = \rho V$.

Since the mass is constant:

$\rho_0 V_0 = \rho V$

We can express the final density $\rho$ in terms of the initial density $\rho_0$ and the ratio of volumes:

$\rho = \rho_0 \frac{V_0}{V}$

The change in volume is $\Delta V = V - V_0$. The relative change in volume with respect to the initial volume is:

$\frac{\Delta V}{V_0} = \frac{V - V_0}{V_0} = \frac{V}{V_0} - 1$

From the density relationship $\rho = \rho_0 \frac{V_0}{V}$, we can find the volume ratio $\frac{V}{V_0} = \frac{\rho_0}{\rho}$. Substituting this into the equation for relative volume change:

$\frac{\Delta V}{V_0} = \frac{\rho_0}{\rho} - 1$

Deriving the Final Density Formula

We have two expressions for the relative volume change $\frac{\Delta V}{V_0}$. Let's equate them:

$\frac{\rho_0}{\rho} - 1 \approx -\frac{P}{B}$

Rearranging the equation to solve for $\frac{\rho_0}{\rho}$:

$\frac{\rho_0}{\rho} \approx 1 - \frac{P}{B}$

To find the final density $\rho$, we need to invert this expression:

$\frac{\rho}{\rho_0} \approx \frac{1}{1 - \frac{P}{B}}$

The problem statement specifies "assuming small changes". This implies that the ratio $\frac{P}{B}$ is much less than 1. We can use the binomial approximation $(1 - x)^{-1} \approx 1 + x$ when $x$ is small. In our case, $x = \frac{P}{B}$.

Applying this approximation:

$\frac{\rho}{\rho_0} \approx 1 + \frac{P}{B}$

Finally, we can find the approximate final density $\rho$ by multiplying both sides by $\rho_0$:

$\rho \approx \rho_0 \left(1 + \frac{P}{B}\right)$

Summary of Result

Using the definition of bulk modulus and applying the binomial approximation for small changes, the approximate final density of the object under pressure $P$ is found to be $\rho_0 \left(1 + \frac{P}{B}\right)$.