A brass rod of length `50 cm` and diameter `3.0 cm` is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at `250^(@)C`, if the original length are at `40.0^(@)C`?
(Coefficient of linear expansion of brass `=2.0 xx 10^(-5)//^(@)C, steel = 1.2 xx 10^(-5)//^(@)C`

Options

A. `0.27 cm`
B. `0.34 cm`
C. `0.21 cm`
D. `0.18 cm`

Explanation

To find the change in length of the combined brass and steel rods when the temperature changes from \(40^\circ C\) to \(250^\circ C\), we will use the formula for linear expansion:

\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]

where:
- \(\Delta L\) is the change in length,
- \(L_0\) is the original length,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step 1: Calculate the change in temperature (\(\Delta T\))
\[ \Delta T = 250^\circ C - 40^\circ C = 210^\circ C \]

### Step 2: Calculate the change in length for the brass rod
Given:
- Length of brass rod, \(L_{brass} = 50 \, \text{cm}\)
- Coefficient of linear expansion for brass, \(\alpha_{brass} = 2.0 \times 10^{-5} \, \text{°C}^{-1}\)

Using the formula:
\[ \Delta L_{brass} = L_{brass} \cdot \alpha_{brass} \cdot \Delta T \]
\[ \Delta L_{brass} = 50 \, \text{cm} \cdot (2.0 \times 10^{-5} \, \text{°C}^{-1}) \cdot (210 \, \text{°C}) \]
\[ \Delta L_{brass} = 50 \cdot 2.0 \times 10^{-5} \cdot 210 \]
\[ \Delta L_{brass} = 0.21 \, \text{cm} \]

### Step 3: Calculate the change in length for the steel rod
Given:
- Length of steel rod, \(L_{steel} = 50 \, \text{cm}\)
- Coefficient of linear expansion for steel, \(\alpha_{steel} = 1.2 \times 10^{-5} \, \text{°C}^{-1}\)

Using the formula:
\[ \Delta L_{steel} = L_{steel} \cdot \alpha_{steel} \cdot \Delta T \]
\[ \Delta L_{steel} = 50 \, \text{cm} \cdot (1.2 \times 10^{-5} \, \text{°C}^{-1}) \cdot (210 \, \text{°C}) \]
\[ \Delta L_{steel} = 50 \cdot 1.2 \times 10^{-5} \cdot 210 \]
\[ \Delta L_{steel} = 0.126 \, \text{cm} \]

### Step 4: Calculate the total change in length of the combined rod
\[ \Delta L_{net} = \Delta L_{brass} + \Delta L_{steel} \]
\[ \Delta L_{net} = 0.21 \, \text{cm} + 0.126 \, \text{cm} \]
\[ \Delta L_{net} = 0.336 \, \text{cm} \]

### Final Answer
The total change in length of the combined rod at \(250^\circ C\) is approximately \(0.34 \, \text{cm}\).

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