The length of s steel rod exceeds that of a brass rod by 5 cm. If the difference in their lengths remains same at all temperature, then the length of brass rod will be: (`alpha` for iron and brass are `12xx10^(-6)//^(@)C` and `18xx10^(-6)//^(@)C`, respectively)

Options

A. 15 cm
B. 5 cm
C. 10 cm
D. 2 cm

Explanation

To solve the problem, we need to use the concept of linear thermal expansion. The change in length of a rod due to temperature change can be expressed as:

\[ L' = L(1 + \alpha \Delta T) \]

Where:
- \(L'\) is the new length after temperature change,
- \(L\) is the original length,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

Given:
- The length of the steel rod exceeds that of the brass rod by 5 cm, which means:
\[ L_s - L_b = 5 \text{ cm} \]
Where \(L_s\) is the length of the steel rod and \(L_b\) is the length of the brass rod.

### Step 1: Write the equations for the lengths after temperature change
After a temperature change \(\Delta T\), the lengths of the steel and brass rods can be expressed as:
\[ L_s' = L_s(1 + \alpha_s \Delta T) \]
\[ L_b' = L_b(1 + \alpha_b \Delta T) \]

### Step 2: Set up the equation based on the constant difference
According to the problem, the difference in lengths remains constant at all temperatures:
\[ L_s' - L_b' = 5 \text{ cm} \]

Substituting the expressions for \(L_s'\) and \(L_b'\):
\[ L_s(1 + \alpha_s \Delta T) - L_b(1 + \alpha_b \Delta T) = 5 \]

### Step 3: Rearranging the equation
Expanding the equation gives:
\[ L_s + L_s \alpha_s \Delta T - L_b - L_b \alpha_b \Delta T = 5 \]
Rearranging yields:
\[ (L_s - L_b) + (L_s \alpha_s - L_b \alpha_b) \Delta T = 5 \]

### Step 4: Substitute the known difference
Since \(L_s - L_b = 5\):
\[ 5 + (L_s \alpha_s - L_b \alpha_b) \Delta T = 5 \]
Subtracting 5 from both sides results in:
\[ (L_s \alpha_s - L_b \alpha_b) \Delta T = 0 \]

### Step 5: Analyze the equation
For the above equation to hold true for any \(\Delta T\), the term in parentheses must equal zero:
\[ L_s \alpha_s = L_b \alpha_b \]

### Step 6: Substitute the values of \(\alpha\)
Given:
- \(\alpha_s = 12 \times 10^{-6} \, \text{°C}^{-1}\) (for steel)
- \(\alpha_b = 18 \times 10^{-6} \, \text{°C}^{-1}\) (for brass)

Substituting these values into the equation:
\[ L_s (12 \times 10^{-6}) = L_b (18 \times 10^{-6}) \]
Cancelling out \(10^{-6}\) gives:
\[ L_s \cdot 12 = L_b \cdot 18 \]

### Step 7: Express \(L_s\) in terms of \(L_b\)
Rearranging gives:
\[ \frac{L_s}{L_b} = \frac{18}{12} = \frac{3}{2} \]
This means:
\[ L_s = \frac{3}{2} L_b \]

### Step 8: Substitute back into the original difference equation
Now substitute \(L_s\) back into the difference equation:
\[ \frac{3}{2} L_b - L_b = 5 \]
This simplifies to:
\[ \frac{1}{2} L_b = 5 \]
Thus:
\[ L_b = 10 \text{ cm} \]

### Final Answer
The length of the brass rod is \(10 \text{ cm}\).

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