At `40^(@)C`, a brass rod has a length 50 cm and a diameter 3.0 mm. it is joined to a steel rod of the same length and diameter at the same temperature. What is the change in the length of the composite rod when it is heated to `240^(@)C`? The coefficient of liear expansion of brass and steel are `2.0xx10^(-5).^(@)C^(-1)` and `1.2xx10^(-5).^(@)C^(-1)` respectively:

Options

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

Answer

To solve the problem, we need to calculate the change in length of both the brass rod and the steel rod when the temperature is increased from \(40^\circ C\) to \(240^\circ C\). 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
First, we calculate the change in temperature (\(\Delta T\)):
\[ \Delta T = T_2 - T_1 = 240^\circ C - 40^\circ C = 200^\circ C \]

### Step 2: Calculate the change in length of the brass rod
Next, we calculate the change in length of the brass rod (\(\Delta L_b\)):
- The original length of the brass rod \(L_b = 50 \, \text{cm}\)
- The coefficient of linear expansion for brass \(\alpha_b = 2.0 \times 10^{-5} \, ^\circ C^{-1}\)

Using the formula:
\[ \Delta L_b = L_b \cdot \alpha_b \cdot \Delta T \]
Substituting the values:
\[ \Delta L_b = 50 \, \text{cm} \cdot (2.0 \times 10^{-5} \, ^\circ C^{-1}) \cdot (200 \, ^\circ C) \]
\[ \Delta L_b = 50 \cdot 2.0 \times 10^{-5} \cdot 200 \]
\[ \Delta L_b = 50 \cdot 4.0 \times 10^{-3} = 0.2 \, \text{cm} \]

### Step 3: Calculate the change in length of the steel rod
Now, we calculate the change in length of the steel rod (\(\Delta L_s\)):
- The original length of the steel rod \(L_s = 50 \, \text{cm}\)
- The coefficient of linear expansion for steel \(\alpha_s = 1.2 \times 10^{-5} \, ^\circ C^{-1}\)

Using the formula:
\[ \Delta L_s = L_s \cdot \alpha_s \cdot \Delta T \]
Substituting the values:
\[ \Delta L_s = 50 \, \text{cm} \cdot (1.2 \times 10^{-5} \, ^\circ C^{-1}) \cdot (200 \, ^\circ C) \]
\[ \Delta L_s = 50 \cdot 1.2 \times 10^{-5} \cdot 200 \]
\[ \Delta L_s = 50 \cdot 2.4 \times 10^{-3} = 0.12 \, \text{cm} \]

### Step 4: Calculate the total change in length of the composite rod
Finally, we add the changes in length of both rods to find the total change in length of the composite rod:
\[ \Delta L_{\text{total}} = \Delta L_b + \Delta L_s \]
\[ \Delta L_{\text{total}} = 0.2 \, \text{cm} + 0.12 \, \text{cm} = 0.32 \, \text{cm} \]

### Final Answer
The change in the length of the composite rod when heated to \(240^\circ C\) is \(0.32 \, \text{cm}\).

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