A copper bar of length `L` and area of cross section `A` is placed in a chamber at atmospheric pressure. If the chamber is evacuated, the percentage change in its volume will be (compressibility of copper is `8 xx 10^(-12) m^(2)//N` and `1 atm = 10^(5) N//m)`

Options

A. `8xx10^(-7)`
B. `8xx10^(-5)`
C. `1.25xx10^(-4)`
D. `1.25xx10^(-5)`

Explanation

To solve the problem of determining the percentage change in volume of a copper bar when the chamber is evacuated, we can follow these steps:

### Step 1: Understand the given data
We have:
- Length of the copper bar = \( L \)
- Area of cross-section = \( A \)
- Compressibility of copper, \( \beta = 8 \times 10^{-12} \, \text{m}^2/\text{N} \)
- Atmospheric pressure, \( P = 1 \, \text{atm} = 10^5 \, \text{N/m}^2 \)

### Step 2: Use the definition of compressibility
The compressibility \( \beta \) is defined as:
\[ \beta = -\frac{1}{V} \frac{\Delta V}{\Delta P} \]
Where:
- \( \Delta V \) is the change in volume
- \( \Delta P \) is the change in pressure
- \( V \) is the original volume

### Step 3: Calculate the change in pressure
When the chamber is evacuated, the pressure changes from atmospheric pressure to zero. Therefore, the change in pressure is:
\[ \Delta P = P - 0 = 10^5 \, \text{N/m}^2 \]

### Step 4: Relate the change in volume to compressibility
Rearranging the formula for compressibility gives:
\[ \Delta V = -\beta V \Delta P \]

### Step 5: Calculate the original volume
The original volume \( V \) of the copper bar can be calculated as:
\[ V = A \times L \]

### Step 6: Substitute values into the change in volume equation
Substituting \( V \) into the equation for \( \Delta V \):
\[ \Delta V = -\beta (A \times L) \Delta P \]
Substituting \( \Delta P \):
\[ \Delta V = -\beta (A \times L) (10^5) \]

### Step 7: Calculate the percentage change in volume
The percentage change in volume is given by:
\[ \text{Percentage change} = \left( \frac{\Delta V}{V} \right) \times 100 \]
Substituting for \( \Delta V \) and \( V \):
\[ \text{Percentage change} = \left( \frac{-\beta (A \times L) (10^5)}{A \times L} \right) \times 100 \]
This simplifies to:
\[ \text{Percentage change} = -\beta (10^5) \times 100 \]

### Step 8: Substitute the value of compressibility
Now substituting the value of compressibility:
\[ \text{Percentage change} = - (8 \times 10^{-12}) (10^5) \times 100 \]
Calculating this gives:
\[ \text{Percentage change} = - (8 \times 10^{-7}) \times 100 = -8 \times 10^{-5} \% \]

### Final Result
The percentage change in the volume of the copper bar when the chamber is evacuated is:
\[ \text{Percentage change} = -8 \times 10^{-5} \% \]