What must be the lengths of steel and copper rods at `0^(@)C` for the difference in their lengths to be 10 cm at any common temperature?
`(alpha_(steel)=1.2xx10^(-5).^(@)K^(-1)and alpha_("copper")=1.8xx10^(-5).^(@)K^(-1))`

Options

A. 30 cm for steel and 20 cm for copper
B. 20 cm for steel and 40 cm for copper
C. 40 cm for steel and 30 cm for copper
D. 30 cm for steel and 40 cm for copper

Answer

To find the lengths of steel and copper rods at \(0^\circ C\) such that the difference in their lengths is \(10 \, cm\) at any common temperature, we can follow these steps:

### Step 1: Understand the formula for linear expansion
The linear expansion of a material can be described by the formula:
\[ L = L_0 (1 + \alpha \Delta T) \]
where:
- \(L\) is the final length after expansion,
- \(L_0\) is the initial length,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step 2: Set up the equations for both materials
Let \(L_s\) be the length of the steel rod and \(L_c\) be the length of the copper rod at \(0^\circ C\). The lengths at a common temperature \(T\) will be:
\[ L_s = L_{0s} (1 + \alpha_s \Delta T) \]
\[ L_c = L_{0c} (1 + \alpha_c \Delta T) \]
where:
- \(\alpha_s = 1.2 \times 10^{-5} \, K^{-1}\) (for steel),
- \(\alpha_c = 1.8 \times 10^{-5} \, K^{-1}\) (for copper).

### Step 3: Express the difference in lengths
The problem states that the difference in lengths at any common temperature is \(10 \, cm\):
\[ |L_s - L_c| = 10 \, cm \]

### Step 4: Substitute the equations into the difference
Substituting the expressions for \(L_s\) and \(L_c\):
\[ |L_{0s} (1 + \alpha_s \Delta T) - L_{0c} (1 + \alpha_c \Delta T)| = 10 \, cm \]

### Step 5: Simplify the equation
This can be rewritten as:
\[ |L_{0s} + L_{0s} \alpha_s \Delta T - L_{0c} - L_{0c} \alpha_c \Delta T| = 10 \, cm \]
Factoring out common terms:
\[ |L_{0s} - L_{0c} + (L_{0s} \alpha_s - L_{0c} \alpha_c) \Delta T| = 10 \, cm \]

### Step 6: Set up the ratio of lengths
Since the difference must be constant, we can set:
\[ L_{0s} \alpha_s = L_{0c} \alpha_c \]
This gives us the ratio of the initial lengths:
\[ \frac{L_{0s}}{L_{0c}} = \frac{\alpha_c}{\alpha_s} \]

### Step 7: Substitute the values of \(\alpha\)
Substituting the values:
\[ \frac{L_{0s}}{L_{0c}} = \frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}} = \frac{1.8}{1.2} = \frac{3}{2} \]

### Step 8: Conclusion
Let \(L_{0c} = 2x\) and \(L_{0s} = 3x\). The difference in lengths can be expressed as:
\[ L_{0s} - L_{0c} = 3x - 2x = x \]
Setting this equal to \(10 \, cm\):
\[ x = 10 \, cm \implies L_{0s} = 30 \, cm \quad \text{and} \quad L_{0c} = 20 \, cm \]

Thus, the lengths of the steel and copper rods at \(0^\circ C\) must be \(30 \, cm\) and \(20 \, cm\) respectively.