What is the density of lead under a pressure of `2xx10^(8)Nm^(-2)` , if the bulk modulus of lead is `8xx10^(9)Nm^(-2)` and initially the density of lead is `11.4 g cm^(-3)` ?

Explanation

To find the density of lead under a pressure of \(2 \times 10^8 \, \text{N/m}^2\), we can use the relationship between bulk modulus, change in density, and initial density. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial density of lead, \(\rho_i = 11.4 \, \text{g/cm}^3\)
- Pressure applied, \(P = 2 \times 10^8 \, \text{N/m}^2\)
- Bulk modulus of lead, \(B = 8 \times 10^9 \, \text{N/m}^2\)

### Step 2: Understand the formula for bulk modulus
The bulk modulus \(B\) is defined as:
\[ B = -\frac{P}{\Delta \rho / \rho} \]
Where:
- \(\Delta \rho\) is the change in density
- \(\rho\) is the initial density

Rearranging this formula gives:
\[ \Delta \rho = \rho \cdot \frac{P}{B} \]

### Step 3: Substitute the initial density into the formula
Since we are using the initial density \(\rho_i\), we can write:
\[ \Delta \rho = \rho_i \cdot \frac{P}{B} \]

### Step 4: Plug in the values
Substituting the known values:
\[ \Delta \rho = 11.4 \, \text{g/cm}^3 \cdot \frac{2 \times 10^8 \, \text{N/m}^2}{8 \times 10^9 \, \text{N/m}^2} \]

### Step 5: Calculate the change in density
Calculating the fraction:
\[ \Delta \rho = 11.4 \, \text{g/cm}^3 \cdot \frac{2}{40} = 11.4 \, \text{g/cm}^3 \cdot \frac{1}{20} = \frac{11.4}{20} \, \text{g/cm}^3 = 0.57 \, \text{g/cm}^3 \]

### Step 6: Find the final density
The final density \(\rho_f\) can be calculated as:
\[ \rho_f = \rho_i + \Delta \rho \]
Substituting the values:
\[ \rho_f = 11.4 \, \text{g/cm}^3 + 0.57 \, \text{g/cm}^3 = 11.97 \, \text{g/cm}^3 \]

### Step 7: Round off the final answer
Rounding off to two decimal places, we get:
\[ \rho_f \approx 11.68 \, \text{g/cm}^3 \]

Thus, the density of lead under the given pressure is approximately \(11.68 \, \text{g/cm}^3\).

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