determine the lengths of an iron rod and a copper ruler at 0°C if the difference in their lengths at 50°C and450°C is and equal to 2cm

Let the lengths of the iron rod and the copper ruler at 0°C be \( L_i \) and \( L_c \) respectively. The thermal expansion equation is given by:

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

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

Given that the difference in their lengths at 50°C and 450°C is equal to 2 cm, we can write the expansion equations for both temperatures:

1. **At 50°C**
- Length of iron rod:
\[
L_{i,50} = L_i (1 + \alpha_i \times 50)
\]
- Length of copper ruler:
\[
L_{c,50} = L_c (1 + \alpha_c \times 50)
\]
- Difference:
\[
(L_c (1 + \alpha_c \times 50)) - (L_i (1 + \alpha_i \times 50)) = d_1
\]

2. **At 450°C**
- Length of iron rod:
\[
L_{i,450} = L_i (1 + \alpha_i \times 450)
\]
- Length of copper ruler:
\[
L_{c,450} = L_c (1 + \alpha_c \times 450)
\]
- Difference:
\[
(L_c (1 + \alpha_c \times 450)) - (L_i (1 + \alpha_i \times 450)) = d_2
\]

Since the difference in their lengths at 450°C and 50°C is given as 2 cm:

\[
d_2 - d_1 = 2
\]

Substituting the values:

\[
[L_c (1 + \alpha_c \times 450) - L_i (1 + \alpha_i \times 450)] - [L_c (1 + \alpha_c \times 50) - L_i (1 + \alpha_i \times 50)] = 2
\]

\[
L_c (\alpha_c \times 400) - L_i (\alpha_i \times 400) = 2
\]

\[
400 (L_c \alpha_c - L_i \alpha_i) = 2
\]

\[
L_c \alpha_c - L_i \alpha_i = \frac{1}{200}
\]

Given:
- Coefficient of linear expansion for iron: \( \alpha_i = 1.2 \times 10^{-5} \) per °C
- Coefficient of linear expansion for copper: \( \alpha_c = 1.7 \times 10^{-5} \) per °C

\[
L_c (1.7 \times 10^{-5}) - L_i (1.2 \times 10^{-5}) = \frac{1}{200}
\]

Multiplying by \( 10^5 \):

\[
1.7 L_c - 1.2 L_i = 5
\]

For simplicity, assume \( L_i = x \) and \( L_c = y \), giving:

\[
1.7y - 1.2x = 5
\]

We need another equation to solve for \( x \) and \( y \). Assuming the difference in their initial lengths is small, we use an approximate relation:

\[
y = x + d
\]

Using typical rod and ruler sizes, assume \( d \approx 10 \) cm. Solving:

\[
1.7(x + 10) - 1.2x = 5
\]

\[
1.7x + 17 - 1.2x = 5
\]

\[
0.5x = -12
\]

\[
x = 24 \text{ cm}
\]

\[
y = 34 \text{ cm}
\]

Thus, the iron rod is **24 cm**, and the copper ruler is **34 cm** at 0°C.