If the Bulk modulus of lead is `8.0xx10^(9)N//m^(2)` and the initial density of the lead is `11.4g//"cc"`, then under the pressure of `2.0xx10^(8)N//m^(2)`, the density of the lead is

Options

A. 11.3 g/cc
B. 11.5 g/cc
C. 11.6 g/cc
D. 11.7 g/cc

Answer

To find the new density of lead under the given pressure, we can use the formula that relates the change in density to the change in pressure and the bulk modulus. The formula is:

\[ \Delta \rho = \frac{P}{B} \cdot \rho_0 \]

Where:
- \(\Delta \rho\) is the change in density,
- \(P\) is the applied pressure,
- \(B\) is the bulk modulus,
- \(\rho_0\) is the initial density.

### Step-by-Step Solution:

1. **Identify the given values**:
- Bulk modulus of lead, \(B = 8.0 \times 10^9 \, \text{N/m}^2\)
- Initial density of lead, \(\rho_0 = 11.4 \, \text{g/cc} = 11.4 \times 10^3 \, \text{kg/m}^3\) (since \(1 \, \text{g/cc} = 1000 \, \text{kg/m}^3\))
- Applied pressure, \(P = 2.0 \times 10^8 \, \text{N/m}^2\)

2. **Calculate the change in density**:
- Using the formula for change in density:
\[ \Delta \rho = \frac{P}{B} \cdot \rho_0 \]
- Substitute the values:
\[ \Delta \rho = \frac{2.0 \times 10^8}{8.0 \times 10^9} \cdot (11.4 \times 10^3) \]

3. **Calculate the fraction**:
- Calculate \(\frac{P}{B}\):
\[ \frac{2.0 \times 10^8}{8.0 \times 10^9} = \frac{1}{40} = 0.025 \]

4. **Calculate the change in density**:
- Now, calculate \(\Delta \rho\):
\[ \Delta \rho = 0.025 \cdot (11.4 \times 10^3) = 0.025 \cdot 11400 = 285 \]

5. **Calculate the new density**:
- The new density \(\rho\) is given by:
\[ \rho = \rho_0 + \Delta \rho \]
- Substitute the values:
\[ \rho = 11.4 \times 10^3 + 285 = 11400 + 285 = 11685 \, \text{kg/m}^3 \]

6. **Convert back to g/cc**:
- Convert the new density back to g/cc:
\[ \rho = \frac{11685}{1000} = 11.685 \, \text{g/cc} \]

### Final Answer:
The density of lead under the pressure of \(2.0 \times 10^8 \, \text{N/m}^2\) is approximately \(11.69 \, \text{g/cc}\).