At `50^(@)C` , a brass rod has a length 50 cm and a diameter 2 mm . It is joined to a steel rod of the same length and diameter at the same temperature . The change in the length of the composite rod when it is heated to `250^(@)C` is (Coefficient of linear expansion of brass = `2.0 xx 10^(-5)"^(@) C^(-1)` , coefficient of linear expansion of steel = `1.2 xx 10^(-5) "^(@) C^(-1)`)

Options

A. `0.28` cm
B. `0.30` cm
C. `0.32` cm
D. `0.34` cm

Explanation

To solve the problem of finding the change in length of the composite rod when heated from \(50^\circ C\) to \(250^\circ C\), we will calculate the change in length for both the brass and steel rods separately, and then sum these changes.

### Step 1: Calculate the change in length of the brass rod

The formula for the change in length due to thermal expansion is given by:

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

Where:
- \(\Delta L\) = change in length
- \(\alpha\) = coefficient of linear expansion
- \(L_0\) = original length
- \(\Delta T\) = change in temperature

For the brass rod:
- \(\alpha_B = 2.0 \times 10^{-5} \, ^\circ C^{-1}\)
- \(L_{0B} = 50 \, \text{cm} = 0.50 \, \text{m}\)
- \(\Delta T = 250^\circ C - 50^\circ C = 200^\circ C\)

Substituting the values:

\[ \Delta L_B = (2.0 \times 10^{-5}) \times (0.50) \times (200) \]

Calculating this gives:

\[ \Delta L_B = (2.0 \times 10^{-5}) \times (0.50) \times (200) = 0.002 \, \text{m} = 0.2 \, \text{cm} \]

### Step 2: Calculate the change in length of the steel rod

Using the same formula for the steel rod:

For the steel rod:
- \(\alpha_S = 1.2 \times 10^{-5} \, ^\circ C^{-1}\)
- \(L_{0S} = 50 \, \text{cm} = 0.50 \, \text{m}\)
- \(\Delta T = 250^\circ C - 50^\circ C = 200^\circ C\)

Substituting the values:

\[ \Delta L_S = (1.2 \times 10^{-5}) \times (0.50) \times (200) \]

Calculating this gives:

\[ \Delta L_S = (1.2 \times 10^{-5}) \times (0.50) \times (200) = 0.0012 \, \text{m} = 0.12 \, \text{cm} \]

### Step 3: Calculate the total change in length of the composite rod

Now, we sum the changes in length of both rods:

\[ \Delta L_{\text{total}} = \Delta L_B + \Delta L_S \]

Substituting the values:

\[ \Delta L_{\text{total}} = 0.2 \, \text{cm} + 0.12 \, \text{cm} = 0.32 \, \text{cm} \]

### Final Answer

The change in the length of the composite rod when heated to \(250^\circ C\) is:

\[ \Delta L_{\text{total}} = 0.32 \, \text{cm} \]

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