Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of `2.0xx 10^5 N m^-2` from the following data. Density of kerosene `= 800 kg m^-3` and speed of sound in kerosene `= 1330 m s^-1`

Explanation

To find the change in volume of 1.0 liter of kerosene when subjected to an extra pressure of \(2.0 \times 10^5 \, \text{N/m}^2\), we can follow these steps:

### Step 1: Convert Volume to Cubic Meters
1. The volume of kerosene is given as 1.0 liter.
2. Convert liters to cubic meters:
\[ V = 1.0 \, \text{liters} = 1.0 \times 10^{-3} \, \text{m}^3 \]

### Step 2: Identify Given Values
1. The extra pressure (\(\Delta P\)) is given as:
\[ \Delta P = 2.0 \times 10^5 \, \text{N/m}^2 \]
2. The density (\(\rho\)) of kerosene is:
\[ \rho = 800 \, \text{kg/m}^3 \]
3. The speed of sound in kerosene (\(v\)) is:
\[ v = 1330 \, \text{m/s} \]

### Step 3: Calculate Bulk Modulus
1. The bulk modulus (\(K\)) can be calculated using the formula:
\[ K = \rho v^2 \]
2. Substitute the values:
\[ K = 800 \, \text{kg/m}^3 \times (1330 \, \text{m/s})^2 \]

### Step 4: Calculate Change in Volume
1. The change in volume (\(\Delta V\)) can be calculated using the formula derived from the bulk modulus:
\[ \Delta V = \frac{V \Delta P}{K} \]
2. Substitute the values:
\[ \Delta V = \frac{1.0 \times 10^{-3} \, \text{m}^3 \times 2.0 \times 10^5 \, \text{N/m}^2}{K} \]

### Step 5: Perform the Calculations
1. Calculate \(K\):
\[ K = 800 \times (1330)^2 = 800 \times 1768900 = 1415120000 \, \text{N/m}^2 \]
2. Now substitute \(K\) back into the change in volume equation:
\[ \Delta V = \frac{1.0 \times 10^{-3} \times 2.0 \times 10^5}{1415120000} \]
3. Calculate \(\Delta V\):
\[ \Delta V = \frac{2.0 \times 10^{-3}}{1415120000} \approx 0.14 \times 10^{-6} \, \text{m}^3 \]

### Step 6: Convert to Centimeter Cubes
1. Convert \(\Delta V\) from cubic meters to cubic centimeters:
\[ \Delta V \approx 0.14 \, \text{cm}^3 \]

### Final Answer
The change in volume of 1.0 liter of kerosene when subjected to an extra pressure of \(2.0 \times 10^5 \, \text{N/m}^2\) is approximately \(0.14 \, \text{cm}^3\).

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