An iron rod and another of brass, both at `27^@C` differ in length by `10^-3`m. The coefficient of linear expansion for iron is `1.1xx10^(-5)//^(@)C` and for brass is `1.9xx10^(-5)//^(@)C`. The temperature at which both these rods will have the same length is

Options

A. `0^@C`
B. `152^@C`
C. `175^@C`
D. Data is insufficient

Explanation

To find the temperature at which the lengths of the iron rod and brass rod will be equal, we can use the formula for linear expansion:

\[ L = L_0 (1 + \alpha \Delta T) \]

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

Given:
- The initial temperature \( T_0 = 27^\circ C \).
- The difference in length between the iron rod and brass rod at \( 27^\circ C \) is \( 10^{-3} \) m.
- Coefficient of linear expansion for iron, \( \alpha_{iron} = 1.1 \times 10^{-5} / ^\circ C \).
- Coefficient of linear expansion for brass, \( \alpha_{brass} = 1.9 \times 10^{-5} / ^\circ C \).

### Step 1: Write the equations for lengths of both rods at temperature \( T \).

For the iron rod:
\[ L_{iron} = L_0 (1 + \alpha_{iron} (T - 27)) \]

For the brass rod:
\[ L_{brass} = L_0 (1 + \alpha_{brass} (T - 27)) \]

### Step 2: Set up the equation for the difference in lengths.

At \( T = 27^\circ C \), the difference in lengths is given as \( 10^{-3} \) m. Therefore, we can write:
\[ L_{brass} - L_{iron} = 10^{-3} \]

Substituting the expressions for \( L_{iron} \) and \( L_{brass} \):
\[ L_0 (1 + \alpha_{brass} (T - 27)) - L_0 (1 + \alpha_{iron} (T - 27)) = 10^{-3} \]

### Step 3: Simplify the equation.

Factoring out \( L_0 \):
\[ L_0 \left( \alpha_{brass} (T - 27) - \alpha_{iron} (T - 27) \right) = 10^{-3} \]

This simplifies to:
\[ L_0 (\alpha_{brass} - \alpha_{iron})(T - 27) = 10^{-3} \]

### Step 4: Substitute the values of \( \alpha_{brass} \) and \( \alpha_{iron} \).

Substituting the coefficients:
\[ L_0 (1.9 \times 10^{-5} - 1.1 \times 10^{-5})(T - 27) = 10^{-3} \]
\[ L_0 (0.8 \times 10^{-5})(T - 27) = 10^{-3} \]

### Step 5: Solve for \( T \).

Rearranging gives:
\[ T - 27 = \frac{10^{-3}}{L_0 (0.8 \times 10^{-5})} \]
\[ T - 27 = \frac{10^{-3}}{0.8 \times 10^{-5} L_0} \]

Now we need to find \( L_0 \). Since \( L_0 \) is not provided, we can assume it is a positive value that will yield a valid temperature.

### Step 6: Calculate \( T \).

Assuming \( L_0 \) is a reasonable length (for example, 1 m or 1000 mm), we can calculate:
\[ T - 27 = \frac{10^{-3}}{0.8 \times 10^{-5} \times 1} \]
\[ T - 27 = \frac{10^{-3}}{0.8 \times 10^{-5}} \]
\[ T - 27 = 12.5 \]
\[ T = 27 + 12.5 = 39.5^\circ C \]

### Conclusion

The temperature at which both the iron rod and brass rod will have the same length is approximately \( 39.5^\circ C \).