A steel rod is 4.00cm in diameter at `30^(@)C` A brass ring has an interior diameter of 3.992cm at `30^(@)` in order that the ring just slides onto the steel rod the common temperature of the two should be nearly
`(alpha_(steel)`=`11xx10^(-6)(^(@)C)` and `alpha_(brass)=19xx10^(-6)(^(@)C)`

Options

A. `200^(@)C`
B. `350^(@)C`
C. `280^(@)C`
D. `300^(@)C`

Answer

To solve the problem, we need to find the common temperature at which a brass ring can slide onto a steel rod. We will use the concept of linear expansion for both materials.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Diameter of the steel rod at 30°C, \( D_s = 4.00 \, \text{cm} \)
- Interior diameter of the brass ring at 30°C, \( D_r = 3.992 \, \text{cm} \)
- Coefficient of linear expansion for steel, \( \alpha_s = 11 \times 10^{-6} \, \text{°C}^{-1} \)
- Coefficient of linear expansion for brass, \( \alpha_r = 19 \times 10^{-6} \, \text{°C}^{-1} \)

2. **Set Up the Linear Expansion Formula:**
The formula for linear expansion is given by:
\[ L_f = L_0 (1 + \alpha \Delta T) \]
where \( L_f \) is the final length, \( L_0 \) is the initial length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

3. **Express the Final Diameters:**
- For the steel rod:
\[ D_s' = D_s (1 + \alpha_s \Delta T) \]
- For the brass ring:
\[ D_r' = D_r (1 + \alpha_r \Delta T) \]

4. **Set the Final Diameters Equal:**
For the ring to slide onto the rod, the final diameter of the steel rod must equal the final diameter of the brass ring:
\[ D_s (1 + \alpha_s \Delta T) = D_r (1 + \alpha_r \Delta T) \]

5. **Substituting the Values:**
Substituting the values of \( D_s \) and \( D_r \):
\[ 4.00 (1 + 11 \times 10^{-6} \Delta T) = 3.992 (1 + 19 \times 10^{-6} \Delta T) \]

6. **Expanding the Equation:**
Expanding both sides gives:
\[ 4.00 + 4.00 \cdot 11 \times 10^{-6} \Delta T = 3.992 + 3.992 \cdot 19 \times 10^{-6} \Delta T \]

7. **Rearranging the Equation:**
Rearranging the equation to isolate \( \Delta T \):
\[ 4.00 - 3.992 = 3.992 \cdot 19 \times 10^{-6} \Delta T - 4.00 \cdot 11 \times 10^{-6} \Delta T \]
\[ 0.008 = (3.992 \cdot 19 \times 10^{-6} - 4.00 \cdot 11 \times 10^{-6}) \Delta T \]

8. **Calculating the Coefficients:**
Calculate the coefficients:
\[ 3.992 \cdot 19 \approx 75.848 \quad \text{and} \quad 4.00 \cdot 11 \approx 44 \]
Thus,
\[ 0.008 = (75.848 - 44) \times 10^{-6} \Delta T \]
\[ 0.008 = 31.848 \times 10^{-6} \Delta T \]

9. **Solving for \( \Delta T \):**
\[ \Delta T = \frac{0.008}{31.848 \times 10^{-6}} \approx 251.2 \, \text{°C} \]

10. **Finding the Final Temperature:**
Since the initial temperature is 30°C:
\[ T_f = 30 + \Delta T \approx 30 + 251.2 \approx 281.2 \, \text{°C} \]

### Final Answer:
The common temperature at which the brass ring just slides onto the steel rod is approximately **281.2°C**.