A copper collar is to fit tightly about a steel shaft that has a diameter of 6 cm at `20^@C`. The inside diameter of the copper collar at the temperature is `5.98cm`
Q. To what temperature must the copper collar be raised to that it will just slip on the steel shaft, assuming the steel shaft remains at `20^@C` ? `(alpha_("copper")=17xx10^(-6)//K`)

Options

A. `324^@C`
B. `21.7^@C`
C. `217^@C`
D. `32.4^@C`

Explanation

To solve the problem, we need to determine the temperature to which the copper collar must be raised so that it can fit over the steel shaft. We will use the concept of linear expansion and the formula for change in length due to temperature change.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Diameter of the steel shaft, \(D_s = 6 \, \text{cm}\)
- Inside diameter of the copper collar, \(D_c = 5.98 \, \text{cm}\)
- Initial temperature, \(T_0 = 20 \, ^\circ C\)
- Coefficient of linear expansion for copper, \(\alpha_{\text{copper}} = 17 \times 10^{-6} \, \text{K}^{-1}\)

2. **Convert Diameters to Radii**:
- Radius of the steel shaft, \(R_s = \frac{D_s}{2} = \frac{6 \, \text{cm}}{2} = 3 \, \text{cm}\)
- Inside radius of the copper collar, \(R_c = \frac{D_c}{2} = \frac{5.98 \, \text{cm}}{2} = 2.99 \, \text{cm}\)

3. **Use the Linear Expansion Formula**:
The change in length (or diameter in this case) due to temperature change can be expressed as:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
where \(L_0\) is the original length (or diameter), \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature.

4. **Set Up the Equations**:
For the steel shaft (which remains at \(20 \, ^\circ C\)):
\[ D_s = D_{s0} \cdot (1 + \alpha_s \cdot (T_s - T_0)) \]
For the copper collar:
\[ D_c = D_{c0} \cdot (1 + \alpha_c \cdot (T_c - T_0)) \]
Here, \(D_{s0} = 6 \, \text{cm}\) and \(D_{c0} = 5.98 \, \text{cm}\).

5. **Equate the Diameters**:
At the temperature \(T_c\) where the collar will just slip over the shaft:
\[ D_s = D_c \]
Thus, we have:
\[ 6 = 5.98 \cdot (1 + 17 \times 10^{-6} \cdot (T_c - 20)) \]

6. **Solve for \(T_c\)**:
Rearranging the equation:
\[ 6 = 5.98 + 5.98 \cdot 17 \times 10^{-6} \cdot (T_c - 20) \]
\[ 6 - 5.98 = 5.98 \cdot 17 \times 10^{-6} \cdot (T_c - 20) \]
\[ 0.02 = 5.98 \cdot 17 \times 10^{-6} \cdot (T_c - 20) \]
\[ T_c - 20 = \frac{0.02}{5.98 \cdot 17 \times 10^{-6}} \]
\[ T_c - 20 = \frac{0.02}{1.0156 \times 10^{-4}} \approx 196.1 \]
\[ T_c \approx 196.1 + 20 \approx 216.1 \, ^\circ C \]

7. **Final Result**:
The temperature to which the copper collar must be raised is approximately \(T_c \approx 216.1 \, ^\circ C\).