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 a given pressure using the bulk modulus, we can follow these steps:

### Step 1: Understand the formula
The formula for the new density (\( \rho' \)) under pressure is given by:

\[ \rho' = \frac{\rho}{1 - \frac{P}{B}} \]

where:
- \( \rho \) = initial density
- \( P \) = applied pressure
- \( B \) = bulk modulus

### Step 2: Identify the given values
From the question, we have:
- Bulk modulus of lead, \( B = 8.0 \times 10^9 \, \text{N/m}^2 \)
- Initial density of lead, \( \rho = 11.4 \, \text{g/cc} \)
- Applied pressure, \( P = 2.0 \times 10^8 \, \text{N/m}^2 \)

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula:

\[ \rho' = \frac{11.4}{1 - \frac{2.0 \times 10^8}{8.0 \times 10^9}} \]

### Step 4: Calculate the fraction
First, calculate the fraction \( \frac{P}{B} \):

\[ \frac{P}{B} = \frac{2.0 \times 10^8}{8.0 \times 10^9} = \frac{2}{8} \times \frac{10^8}{10^9} = \frac{1}{4} \times \frac{1}{10} = 0.025 \]

### Step 5: Substitute back into the equation
Now substitute this value back into the density equation:

\[ \rho' = \frac{11.4}{1 - 0.025} = \frac{11.4}{0.975} \]

### Step 6: Perform the division
Now, calculate \( \rho' \):

\[ \rho' \approx \frac{11.4}{0.975} \approx 11.692 \]

### Step 7: Round the result
Rounding \( 11.692 \) to one decimal place gives:

\[ \rho' \approx 11.7 \, \text{g/cc} \]

### Final Answer
The new density of lead under the given pressure is approximately \( 11.7 \, \text{g/cc} \).

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