There is some change in length when a `33000 N` tensile force is applied on a steel rod of area of cross-section `10^(-3)m^(2)`. The change in temperature of the steel rod when heated is
`(Y = 3 xx 10^(11)N // m^(2) , alpha = 1.1 xx 10^(-5 //@)C)` a) 20 degree C b) 15 degree C c) 10 degree C d) 0 degree C

Options

A. `20^@C`
B. `15^@C`
C. `10^@C`
D. `0^@C`

Answer

To solve the problem, we need to find the change in temperature (ΔT) of a steel rod when a tensile force is applied. We will use the relationship between stress, strain, Young's modulus, and the coefficient of linear expansion.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Force (F) = 33,000 N
- Area of cross-section (A) = \(10^{-3} \, m^2\)
- Young's modulus (Y) = \(3 \times 10^{11} \, N/m^2\)
- Coefficient of linear expansion (α) = \(1.1 \times 10^{-5} /°C\)

2. **Calculate Stress (σ):**
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{33,000 \, N}{10^{-3} \, m^2} = 33,000,000 \, N/m^2 = 3.3 \times 10^7 \, N/m^2 \]

3. **Relate Young's Modulus to Stress and Strain:**
The formula for Young's modulus (Y) is:
\[ Y = \frac{\sigma}{\text{strain}} \]
Strain (ε) is given by:
\[ \text{strain} = \alpha \Delta T \]
Thus, we can write:
\[ Y = \frac{\sigma}{\alpha \Delta T} \]

4. **Rearranging to Find ΔT:**
Rearranging the equation gives:
\[ \Delta T = \frac{\sigma}{Y \alpha} \]

5. **Substituting the Values:**
\[ \Delta T = \frac{3.3 \times 10^7 \, N/m^2}{(3 \times 10^{11} \, N/m^2)(1.1 \times 10^{-5}/°C)} \]

6. **Calculating ΔT:**
\[ \Delta T = \frac{3.3 \times 10^7}{3.3 \times 10^6} = 10 \, °C \]

### Final Answer:
The change in temperature (ΔT) is **10°C**.