Find the density of the metal under a pressure of 20,000 N `cm^(-2)`. Given density of the metal=11 g `cm^(-3)`, bulk modulus of the metal `=8xx10^(9) Nm^(-2)`

Answer

To find the density of the metal under a pressure of 20,000 N/cm², we will follow these steps:

### Step 1: Convert the pressure to SI units
The given pressure is 20,000 N/cm². We need to convert this to N/m² (Pascals).

\[ 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \]
\[ P = 20,000 \text{ N/cm}^2 = 20,000 \times 10^4 \text{ N/m}^2 = 2 \times 10^8 \text{ N/m}^2 \]

### Step 2: Write the formula for bulk modulus
The bulk modulus \( K \) is defined as:

\[ K = \frac{\text{Stress}}{\text{Volumetric Strain}} = \frac{P}{\frac{\Delta V}{V}} \]

Where:
- \( P \) is the pressure applied,
- \( \Delta V \) is the change in volume,
- \( V \) is the original volume.

### Step 3: Calculate the volumetric strain
From the definition of bulk modulus, we can rearrange the formula to find the volumetric strain:

\[ \frac{\Delta V}{V} = \frac{P}{K} \]

Substituting the values:

\[ K = 8 \times 10^9 \text{ N/m}^2 \]
\[ \frac{\Delta V}{V} = \frac{2 \times 10^8}{8 \times 10^9} = \frac{1}{40} \]

### Step 4: Find the change in volume
The change in volume \( \Delta V \) can be expressed as:

\[ \Delta V = \frac{V}{40} \]

### Step 5: Calculate the final volume
The final volume \( V' \) after the pressure is applied can be calculated as:

\[ V' = V - \Delta V = V - \frac{V}{40} = V \left(1 - \frac{1}{40}\right) = V \left(\frac{39}{40}\right) \]

### Step 6: Relate the mass and density
Since the mass of the metal remains constant, we can relate the initial and final states using the formula:

\[ \text{Mass} = \text{Initial Volume} \times \text{Initial Density} = \text{Final Volume} \times \text{Final Density} \]

Let \( \rho \) be the initial density (11 g/cm³), then:

\[ V \cdot \rho = V' \cdot \rho' \]

Substituting \( V' \):

\[ V \cdot \rho = \left(\frac{39}{40} V\right) \cdot \rho' \]

### Step 7: Solve for the final density
Dividing both sides by \( V \):

\[ \rho = \frac{39}{40} \cdot \rho' \]

Rearranging gives:

\[ \rho' = \frac{40}{39} \cdot \rho \]

Substituting the initial density:

\[ \rho' = \frac{40}{39} \cdot 11 \text{ g/cm}^3 \]

Calculating this:

\[ \rho' = \frac{440}{39} \approx 11.28 \text{ g/cm}^3 \]

### Final Answer
The density of the metal under a pressure of 20,000 N/cm² is approximately **11.28 g/cm³**.

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