The volume of a solid is `6 xx 10^(-3) m^(3)` under 2 atm pressure. Find the change in volume when subjected to a pressure of 102 atm. Bulk modulus of the material `= 10^(11) Nm^(-2)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Initial volume \( V_0 = 6 \times 10^{-3} \, \text{m}^3 \)
- Initial pressure \( P_1 = 2 \, \text{atm} \)
- Final pressure \( P_2 = 102 \, \text{atm} \)
- Bulk modulus \( K = 10^{11} \, \text{N/m}^2 \)

### Step 2: Calculate the change in pressure
The change in pressure \( \Delta P \) can be calculated using the formula:
\[ \Delta P = P_2 - P_1 \]
Substituting the values:
\[ \Delta P = 102 \, \text{atm} - 2 \, \text{atm} = 100 \, \text{atm} \]

### Step 3: Convert the change in pressure to SI units
1 atm is equivalent to \( 1.013 \times 10^5 \, \text{N/m}^2 \). Therefore, we convert \( \Delta P \):
\[ \Delta P = 100 \, \text{atm} \times 1.013 \times 10^5 \, \text{N/m}^2/\text{atm} = 1.013 \times 10^7 \, \text{N/m}^2 \]

### Step 4: Use the bulk modulus formula to find the change in volume
The formula relating change in pressure, bulk modulus, and change in volume is:
\[ \Delta P = -K \frac{\Delta V}{V_0} \]
Rearranging this formula to solve for \( \Delta V \):
\[ \Delta V = -\frac{\Delta P \cdot V_0}{K} \]
Substituting the known values:
\[ \Delta V = -\frac{1.013 \times 10^7 \, \text{N/m}^2 \times 6 \times 10^{-3} \, \text{m}^3}{10^{11} \, \text{N/m}^2} \]

### Step 5: Calculate \( \Delta V \)
Calculating the above expression:
\[ \Delta V = -\frac{1.013 \times 10^7 \times 6 \times 10^{-3}}{10^{11}} = -\frac{6.078 \times 10^4}{10^{11}} = -6.078 \times 10^{-7} \, \text{m}^3 \]

### Conclusion
The change in volume when subjected to a pressure of 102 atm is:
\[ \Delta V \approx -6.08 \times 10^{-7} \, \text{m}^3 \]