Let's solve the problem step by step: ### Given: - Initial volume, \( V_1 = 1005 \ \text{cm}^3 \) - Final volume, \( V_2 = 1000 \ \text{cm}^3 \) - Initial pressure, \( P_1 = 1 \ \text{MN/m}^2 = 1 \times 10^6 \ \text{N/m}^2 \) - Final pressure, \( P_2 = 2 \ \text{MN/m}^2 = 2 \times 10^6 \ \text{N/m}^2 \) ### Bulk Modulus Formula: The bulk modulus \( K \) is given by: \[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \] Where: - \( \Delta P = P_2 - P_1 \) - \( \Delta V = V_2 - V_1 \) - \( V \) is the original volume (use \( V_1 \) for small changes) ### Calculate the changes: #### 1. Change in pressure: \[ \Delta P = P_2 - P_1 = (2 - 1) \times 10^6 = 1 \times 10^6 \ \text{N/m}^2 \] #### 2. Change in volume: \[ \Delta V = V_2 - V_1 = 1000 - 1005 = -5 \ \text{cm}^3 \] #### 3. Volumetric strain: \[ \frac{\Delta V}{V_1} = \frac{-5}{1005} \] ### Substitute into the formula: \[ K = -\frac{1 \times 10^6}{\frac{-5}{1005}} \] \[ K = -\frac{1 \times 10^6}{-0.004975} \] \[ K = \frac{1 \times 10^6}{0.004975} \] \[ K \approx 201,005,025 \ \text{N/m}^2 \] \[ K \approx 201 \ \text{MPa} \] ### **Final Answer:** The correct answer is: \[ \boxed{201 \ \text{MPa}} \]