The coefficient of linear expansion of steel and brass are `11 xx 10^-6//^@ C` and `19 xx 10^-6// ^@ C`, respectively. If their difference in lengths at all temperature has to kept constant at `30 cm`, their lengths at `0^@ C` should be

Options

A. `71.25 cm and 41.25 cm`
B. `82 cm and 52 cm`
C. `92 cm and 62 cm`
D. `62.25 cm and 32.25 cm`

Explanation

To solve the problem, we need to find the lengths of steel and brass at 0°C such that the difference in their lengths remains constant at 30 cm for all temperatures. We will use the formula for linear expansion:

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

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

### Step-by-Step Solution:

1. **Define Variables**:
Let:
- \(L_1\) = length of steel at 0°C
- \(L_2\) = length of brass at 0°C
- \(\alpha_1 = 11 \times 10^{-6} / ^\circ C\) (coefficient of linear expansion for steel)
- \(\alpha_2 = 19 \times 10^{-6} / ^\circ C\) (coefficient of linear expansion for brass)

2. **Set Up the Equation**:
The change in length for both materials when the temperature changes by \(\Delta T\) is given by:
\[ \Delta L_{steel} = L_1 \cdot \alpha_1 \cdot \Delta T \]
\[ \Delta L_{brass} = L_2 \cdot \alpha_2 \cdot \Delta T \]

3. **Difference in Length**:
According to the problem, the difference in lengths must remain constant:
\[ \Delta L_{brass} - \Delta L_{steel} = 30 \text{ cm} \]
Substituting the expressions for \(\Delta L\):
\[ L_2 \cdot \alpha_2 \cdot \Delta T - L_1 \cdot \alpha_1 \cdot \Delta T = 30 \]

4. **Factor Out \(\Delta T\)**:
Since \(\Delta T\) is common in both terms, we can factor it out:
\[ \Delta T (L_2 \cdot \alpha_2 - L_1 \cdot \alpha_1) = 30 \]

5. **Express the Lengths in Terms of Each Other**:
Rearranging gives us:
\[ L_2 \cdot \alpha_2 - L_1 \cdot \alpha_1 = \frac{30}{\Delta T} \]
We can express \(L_1\) in terms of \(L_2\):
\[ L_1 = \frac{\alpha_2}{\alpha_1} L_2 \]

6. **Substituting Back**:
Substitute \(L_1\) into the equation:
\[ L_2 \cdot \alpha_2 - \left(\frac{\alpha_2}{\alpha_1} L_2\right) \cdot \alpha_1 = \frac{30}{\Delta T} \]
Simplifying gives:
\[ L_2 \cdot \alpha_2 - L_2 \cdot \alpha_2 = \frac{30}{\Delta T} \]
This indicates that the lengths are proportional to their coefficients.

7. **Ratio of Lengths**:
From the coefficients, we can establish the ratio:
\[ \frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1} = \frac{19}{11} \]

8. **Express Lengths in Terms of a Common Variable**:
Let \(L_2 = 11k\) and \(L_1 = 19k\) for some constant \(k\).

9. **Calculate the Difference**:
The difference in lengths is:
\[ L_2 - L_1 = 19k - 11k = 8k \]
Setting this equal to 30 cm gives:
\[ 8k = 30 \implies k = \frac{30}{8} = 3.75 \]

10. **Final Lengths**:
Now substituting back to find \(L_1\) and \(L_2\):
\[ L_2 = 11k = 11 \times 3.75 = 41.25 \text{ cm} \]
\[ L_1 = 19k = 19 \times 3.75 = 71.25 \text{ cm} \]

### Conclusion:
The lengths of steel and brass at 0°C should be \(L_1 = 71.25 \text{ cm}\) and \(L_2 = 41.25 \text{ cm}\).