A steel rod is 3 cm in diameter at `-10.00^@C` . A brass ring has an interior diameter of 2.992 cm at `-10.00^@C` . At what common temperature will the ring just slide onto the rod ?

Answer

To find the common temperature at which the brass ring will just slide onto the steel rod, we need to set the final diameters of both the steel rod and the brass ring equal to each other. Here are the steps to solve the problem:

### Step 1: Identify the initial conditions
- The initial diameter of the steel rod at -10.00°C is 3 cm.
- The initial interior diameter of the brass ring at -10.00°C is 2.992 cm.

### Step 2: Write the formula for thermal expansion
The formula for the final diameter after thermal expansion is given by:
\[ D_f = D_i \times (1 + \alpha \times \Delta T) \]
where:
- \(D_f\) = final diameter
- \(D_i\) = initial diameter
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature

### Step 3: Set up the equations for both materials
1. **For the steel rod**:
\[ D_{f, \text{steel}} = 3 \, \text{cm} \times \left(1 + (11 \times 10^{-6}) \times (T - (-10))\right) \]
Simplifying this gives:
\[ D_{f, \text{steel}} = 3 \, \text{cm} \times \left(1 + (11 \times 10^{-6}) \times (T + 10)\right) \]

2. **For the brass ring**:
\[ D_{f, \text{brass}} = 2.992 \, \text{cm} \times \left(1 + (19 \times 10^{-6}) \times (T - (-10))\right) \]
Simplifying this gives:
\[ D_{f, \text{brass}} = 2.992 \, \text{cm} \times \left(1 + (19 \times 10^{-6}) \times (T + 10)\right) \]

### Step 4: Set the final diameters equal to each other
To find the temperature at which the ring just slides onto the rod, we set the two final diameters equal:
\[ 3 \times \left(1 + (11 \times 10^{-6}) \times (T + 10)\right) = 2.992 \times \left(1 + (19 \times 10^{-6}) \times (T + 10)\right) \]

### Step 5: Solve for \(T\)
Expanding both sides:
\[ 3 + 3 \times (11 \times 10^{-6}) \times (T + 10) = 2.992 + 2.992 \times (19 \times 10^{-6}) \times (T + 10) \]

Rearranging terms and simplifying:
\[ 3 + 3 \times (11 \times 10^{-6}) \times (T + 10) - 2.992 = 2.992 \times (19 \times 10^{-6}) \times (T + 10) \]

Let’s denote \(x = T + 10\):
\[ 3 + 3 \times (11 \times 10^{-6}) \times x - 2.992 = 2.992 \times (19 \times 10^{-6}) \times x \]

Now, solving for \(x\) will yield the value of \(T\).

### Step 6: Calculate the temperature
After performing the calculations, we find:
\[ T = 325 \, °C \]

### Final Answer
The common temperature at which the brass ring will just slide onto the steel rod is **325°C**.

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