The bulk modulus of a metal is `8 xx 10^(9) Nm^(-2)` and its density is 11 g `cm^(-2)`. The density of this metal under a pressure of 20000 N `cm^(-2)` will be (in g `cm^(-3)`)

Options

A. `(440)/(39)`
B. `(431)/(39)`
C. `(451)/(39)`
D. `(40)/(39)`

Explanation

To solve the problem, we need to find the new density of the metal under a pressure of 20000 N/cm², given its bulk modulus and initial density. Let's break down the solution step by step.

### Step 1: Understand the given data
- **Bulk Modulus (B)**: \( 8 \times 10^9 \, \text{N/m}^2 \)
- **Initial Density (\( \rho_1 \))**: \( 11 \, \text{g/cm}^3 \)
- **Pressure (\( P \))**: \( 20000 \, \text{N/cm}^2 \)

### Step 2: Convert pressure to N/m²
Since the bulk modulus is given in N/m², we need to convert the pressure from N/cm² to N/m².

\[ P = 20000 \, \text{N/cm}^2 = 20000 \times 10^4 \, \text{N/m}^2 = 2 \times 10^8 \, \text{N/m}^2 \]

### Step 3: Use the bulk modulus formula
The bulk modulus is defined as:

\[ B = -\frac{P}{\frac{\Delta V}{V}} \]

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

Rearranging gives:

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

### Step 4: Relate volume change to density
The initial volume \( V_1 \) can be expressed in terms of density and mass:

\[ V_1 = \frac{m}{\rho_1} \]

The final volume \( V_2 \) after applying pressure will be:

\[ V_2 = V_1 - \Delta V \]

Substituting \( \Delta V \):

\[ V_2 = V_1 - \frac{P V_1}{B} = V_1 \left(1 - \frac{P}{B}\right) \]

### Step 5: Calculate the change in volume
Substituting \( P \) and \( B \):

\[ V_2 = V_1 \left(1 - \frac{2 \times 10^8}{8 \times 10^9}\right) \]

Calculating the fraction:

\[ \frac{2 \times 10^8}{8 \times 10^9} = \frac{1}{40} \]

Thus,

\[ V_2 = V_1 \left(1 - \frac{1}{40}\right) = V_1 \left(\frac{39}{40}\right) \]

### Step 6: Relate the new density to the new volume
The new density \( \rho_2 \) can be expressed as:

\[ \rho_2 = \frac{m}{V_2} \]

Since mass \( m = \rho_1 V_1 \):

\[ \rho_2 = \frac{\rho_1 V_1}{V_2} = \frac{\rho_1 V_1}{V_1 \left(\frac{39}{40}\right)} = \frac{\rho_1 \cdot 40}{39} \]

### Step 7: Substitute the initial density
Substituting \( \rho_1 = 11 \, \text{g/cm}^3 \):

\[ \rho_2 = \frac{11 \cdot 40}{39} = \frac{440}{39} \, \text{g/cm}^3 \]

### Final Answer
Thus, the density of the metal under a pressure of 20000 N/cm² is:

\[ \rho_2 \approx 11.28 \, \text{g/cm}^3 \]