If a compressive force of `3.0xx10^(4)N` is exerted on the end of a 20 cm long bone of cross-sectional area `3.6cm^(2)` (i) will the bone break and (ii) if not by how much length does it shorten? Given compresive strength of bone `=7.7xx10^(8)N//m^(2)` and young's modulus of bone `=1.5xx10^(10)N//m^(2)`

Answer

To solve the problem, we need to determine if the bone will break under the applied compressive force and, if it does not break, calculate how much it shortens.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Compressive force, \( F = 3.0 \times 10^4 \, \text{N} \)
- Length of the bone, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Cross-sectional area, \( A = 3.6 \, \text{cm}^2 = 3.6 \times 10^{-4} \, \text{m}^2 \)
- Compressive strength of bone, \( \sigma_{\text{strength}} = 7.7 \times 10^8 \, \text{N/m}^2 \)
- Young's modulus of bone, \( Y = 1.5 \times 10^{10} \, \text{N/m}^2 \)

2. **Calculate the Applied Stress:**
The applied stress \( \sigma_{\text{applied}} \) can be calculated using the formula:
\[ \sigma_{\text{applied}} = \frac{F}{A} \]
Substituting the values:
\[ \sigma_{\text{applied}} = \frac{3.0 \times 10^4 \, \text{N}}{3.6 \times 10^{-4} \, \text{m}^2} = 8.33 \times 10^7 \, \text{N/m}^2 \]

3. **Check if the Bone Will Break:**
Compare the applied stress with the compressive strength of the bone:
\[ \sigma_{\text{applied}} = 8.33 \times 10^7 \, \text{N/m}^2 < \sigma_{\text{strength}} = 7.7 \times 10^8 \, \text{N/m}^2 \]
Since the applied stress is less than the compressive strength, the bone will **not break**.

4. **Calculate the Change in Length:**
To find the change in length \( \Delta L \), we use the formula derived from Young's modulus:
\[ Y = \frac{\sigma_{\text{applied}}}{\frac{\Delta L}{L}} \]
Rearranging gives:
\[ \Delta L = \frac{\sigma_{\text{applied}} \cdot L}{Y} \]
Substituting the known values:
\[ \Delta L = \frac{8.33 \times 10^7 \, \text{N/m}^2 \cdot 0.2 \, \text{m}}{1.5 \times 10^{10} \, \text{N/m}^2} \]
Calculating this:
\[ \Delta L = \frac{1.666 \times 10^7}{1.5 \times 10^{10}} = 1.11 \times 10^{-3} \, \text{m} = 1.11 \, \text{mm} \]

### Final Answers:
(i) The bone will **not break**.
(ii) The bone will shorten by approximately **1.11 mm**.