The diameter of a brass rod is 4 mm and Young's modulus of brass is `9 xx 10^(10) N//m^(2)`. The force required to stretch by 0.1 % of its length is

Options

A. `360 pi N`
B. `36 pi N`
C. `144 pi xx 10^(3)N`
D. `36 pi xx 10^(5) N`

Explanation

To find the force required to stretch a brass rod by 0.1% of its length, we can follow these steps:

### Step 1: Understand the parameters given
- Diameter of the brass rod, \( D = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- Young's modulus of brass, \( Y = 9 \times 10^{10} \, \text{N/m}^2 \)
- The extension (change in length) is \( 0.1\% \) of its original length \( L \).

### Step 2: Calculate the change in length
The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = 0.1\% \times L = \frac{0.1}{100} \times L = 0.001L \]

### Step 3: Calculate the strain
Strain \( \epsilon \) is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L}{L} = \frac{0.001L}{L} = 0.001 \]

### Step 4: Calculate the stress
Stress \( \sigma \) is given by the formula:
\[ \sigma = Y \cdot \epsilon \]
Substituting the values we have:
\[ \sigma = 9 \times 10^{10} \, \text{N/m}^2 \times 0.001 = 9 \times 10^{7} \, \text{N/m}^2 \]

### Step 5: Calculate the cross-sectional area of the rod
The radius \( r \) of the rod is half of the diameter:
\[ r = \frac{D}{2} = \frac{4 \times 10^{-3}}{2} = 2 \times 10^{-3} \, \text{m} \]
The cross-sectional area \( A \) is given by:
\[ A = \pi r^2 = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) = 4\pi \times 10^{-6} \, \text{m}^2 \]

### Step 6: Calculate the force
Using the relation between stress, force, and area:
\[ \sigma = \frac{F}{A} \implies F = \sigma \cdot A \]
Substituting the values:
\[ F = (9 \times 10^{7} \, \text{N/m}^2) \cdot (4\pi \times 10^{-6} \, \text{m}^2) \]
Calculating this gives:
\[ F = 36\pi \times 10^{1} \, \text{N} = 360\pi \, \text{N} \]

### Final Answer
The force required to stretch the brass rod by 0.1% of its length is:
\[ F \approx 360\pi \, \text{N} \]