A cylindrical steel plug is inserted into a circular hole of diameter 2.60 cm in a brass plate. When the plug and the plates are at a temperature of `20^(@)C,` the diameter of the plug is 0.010 cm smaller than that of the hole. The temperature at which the plug will just fit in it is
`("Given",alpha_(steel)=11xx10^(-6)C^(-1)and alpha_(brass)=19xx10^(-6)C^(-1)`

Options

A. `-48^(@)C`
B. `-20^(@)C`
C. `-10^(@)C`
D. `-458^(@)C`

Explanation

To solve the problem, we need to determine the temperature at which a cylindrical steel plug will fit perfectly into a circular hole in a brass plate. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Diameter of the hole in brass plate, \( D_{hole} = 2.60 \, \text{cm} \)
- Diameter of the plug at \( 20^\circ C \), \( D_{plug} = D_{hole} - 0.01 \, \text{cm} = 2.60 \, \text{cm} - 0.01 \, \text{cm} = 2.59 \, \text{cm} \)
- Coefficient of linear expansion for steel, \( \alpha_{steel} = 11 \times 10^{-6} \, \text{C}^{-1} \)
- Coefficient of linear expansion for brass, \( \alpha_{brass} = 19 \times 10^{-6} \, \text{C}^{-1} \)

### Step 2: Set up the equation for thermal expansion
The diameter of the hole in the brass plate and the diameter of the steel plug will change with temperature. The final diameters can be expressed as:

\[ D_{final, brass} = D_{hole} + \Delta D_{brass} \]
\[ D_{final, steel} = D_{plug} + \Delta D_{steel} \]

Where:
- \(\Delta D_{brass} = D_{hole} \cdot \alpha_{brass} \cdot \Delta T\)
- \(\Delta D_{steel} = D_{plug} \cdot \alpha_{steel} \cdot \Delta T\)

### Step 3: Equate the final diameters
For the plug to just fit into the hole, we set the final diameters equal:

\[ D_{hole} + D_{hole} \cdot \alpha_{brass} \cdot \Delta T = D_{plug} + D_{plug} \cdot \alpha_{steel} \cdot \Delta T \]

### Step 4: Substitute the known values
Substituting the values we have:

\[ 2.60 + 2.60 \cdot (19 \times 10^{-6}) \cdot \Delta T = 2.59 + 2.59 \cdot (11 \times 10^{-6}) \cdot \Delta T \]

### Step 5: Simplify the equation
This simplifies to:

\[ 2.60 + 4.94 \times 10^{-6} \cdot \Delta T = 2.59 + 2.849 \times 10^{-6} \cdot \Delta T \]

### Step 6: Rearrange the equation
Rearranging gives:

\[ 2.60 - 2.59 = (2.849 \times 10^{-6} - 4.94 \times 10^{-6}) \cdot \Delta T \]

\[ 0.01 = -2.091 \times 10^{-6} \cdot \Delta T \]

### Step 7: Solve for \(\Delta T\)
Now, solving for \(\Delta T\):

\[ \Delta T = \frac{0.01}{-2.091 \times 10^{-6}} \approx -4780.4 \, \text{C} \]

### Step 8: Calculate the final temperature
The final temperature \( T_f \) is given by:

\[ T_f = 20 + \Delta T = 20 - 4780.4 \approx -4758.4 \, \text{C} \]

### Final Answer
The temperature at which the plug will just fit in the hole is approximately \( -4758.4 \, \text{C} \).

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