To solve the problem of finding the load on a wire with a cross section of 16 mm² that increases its length by 0.3%, we will use the relationship between stress, strain, and Young's modulus (Y).
### Step-by-Step Solution:
1. **Convert Cross Section Area to m²:**
The cross-sectional area is given as 16 mm². We need to convert this to square meters.
\[
A = 16 \, \text{mm}^2 = 16 \times 10^{-6} \, \text{m}^2
\]
**Hint:** Remember that 1 mm = \(10^{-3}\) m, so to convert mm² to m², you square the conversion factor.
2. **Calculate Strain:**
Strain is defined as the change in length (ΔL) divided by the original length (L). The problem states that the length increases by 0.3%, which can be expressed as:
\[
\text{Strain} = \frac{\Delta L}{L} = \frac{0.3}{100} = 0.003
\]
**Hint:** Strain is a dimensionless quantity, so it can be expressed as a percentage or a decimal.
3. **Use Young's Modulus Formula:**
Young's modulus (Y) relates stress (σ) and strain (ε) as follows:
\[
Y = \frac{\sigma}{\epsilon}
\]
where stress \( \sigma = \frac{F}{A} \) (F is the force/load and A is the area). Rearranging gives us:
\[
\sigma = Y \cdot \epsilon
\]
**Hint:** Make sure to keep units consistent when using Young's modulus.
4. **Substituting Values:**
Substitute the values of Young's modulus (Y = \(5 \times 10^9 \, \text{N/m}^2\)) and strain (ε = 0.003) into the equation:
\[
\sigma = 5 \times 10^9 \, \text{N/m}^2 \times 0.003 = 15 \times 10^6 \, \text{N/m}^2
\]
**Hint:** Multiplying a large number by a small decimal can be simplified by adjusting the powers of ten.
5. **Calculate Load (F):**
Now, we can find the load (F) using the stress formula:
\[
\sigma = \frac{F}{A} \implies F = \sigma \cdot A
\]
Substituting the values:
\[
F = 15 \times 10^6 \, \text{N/m}^2 \times 16 \times 10^{-6} \, \text{m}^2 = 240 \, \text{N}
\]
**Hint:** Ensure that the units of area and stress are compatible when calculating the load.
6. **Convert Load to kg:**
To convert the load in Newtons to kilograms, we use the relation \( F = mg \):
\[
m = \frac{F}{g} = \frac{240 \, \text{N}}{10 \, \text{m/s}^2} = 24 \, \text{kg}
\]
**Hint:** Remember that weight (force) is equal to mass times gravitational acceleration.
### Final Answer:
The load on the wire is **24 kg**.
