Artificial diamond is made by subjected carbon in the form of graphite to a pressure of `1.55 xx 10^(10) N//m^(2)` at a high temperature. Assuming that the natrual diamond formed at similar high presssure within the earth, calcualte its original volume if the mass of the diamond before cutting was 150g. The density of diamond = 3000 `kg//m^(2) `.Bulk modulus `= 6.2 xx 10^(111) N//m^(2) `.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass of the diamond from grams to kilograms.
Given:
- Mass of diamond = 150 g

To convert grams to kilograms:
\[ \text{Mass in kg} = \frac{\text{Mass in grams}}{1000} = \frac{150}{1000} = 0.15 \text{ kg} \]

### Step 2: Calculate the volume of the diamond using its density.
Given:
- Density of diamond = 3000 kg/m³

Using the formula for volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{0.15 \text{ kg}}{3000 \text{ kg/m}^3} \]
\[ \text{Volume} = 5 \times 10^{-5} \text{ m}^3 \]

### Step 3: Calculate the change in volume due to pressure using the bulk modulus.
Given:
- Pressure (P) = \(1.55 \times 10^{10} \text{ N/m}^2\)
- Bulk Modulus (K) = \(6.2 \times 10^{11} \text{ N/m}^2\)

The change in volume (\(\Delta V\)) can be calculated using the formula:
\[ \Delta V = -\frac{P \cdot V}{K} \]
Substituting the values:
\[ \Delta V = -\frac{(1.55 \times 10^{10}) \cdot (5 \times 10^{-5})}{6.2 \times 10^{11}} \]
Calculating:
\[ \Delta V = -\frac{7.75 \times 10^{5}}{6.2 \times 10^{11}} \approx -1.25 \times 10^{-6} \text{ m}^3 \]

### Step 4: Calculate the original volume of the diamond.
The original volume (\(V_0\)) is the sum of the volume calculated in Step 2 and the change in volume calculated in Step 3:
\[ V_0 = V + \Delta V \]
Substituting the values:
\[ V_0 = 5 \times 10^{-5} + (-1.25 \times 10^{-6}) \]
\[ V_0 = 5 \times 10^{-5} - 1.25 \times 10^{-6} = 5.125 \times 10^{-5} \text{ m}^3 \]

### Final Answer:
The original volume of the diamond is:
\[ V_0 \approx 5.125 \times 10^{-5} \text{ m}^3 \]

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