A brass rod and lead rod each 80 cm long at 0°C are clamped together at one end with their free ends coinciding. The separation of free ends of the rods if the system is placed in a steam bath is (αbᵣₐₛₛ =18 x 10⁻⁶ /°C and αₗₑₐd = 28 x 10⁻⁶ /°C )

To find the separation between the free ends of the brass and lead rods when heated, we can analyze the linear thermal expansion of each rod individually.

The formula for linear thermal expansion is given by:
$\Delta L=L·\alpha ·\Delta T$
where:

• L is the initial length of the rods (80 cm at 0°C).
• α is the coefficient of linear expansion.
• ΔT is the change in temperature.

When the system is placed in a steam bath, the temperature rises from 0°C to 100°C. Thus, the change in temperature is:
$\Delta T=100\text{°C}-0\text{°C}=100\text{°C}$

The separation between their free ends after expansion will be the difference between their individual changes in length:
$\text{Separation}=\Delta L_{\text{lead}}$
Substituting the expansion formula, we get:
$\text{Separation}=L·({\alpha }_{\text{lead}}-{\alpha }_{\text{brass}})·\Delta T$

Now, let's plug in the given values:

• $L=80\text{cm}$
• ${\alpha }_{\text{lead}}=28\times {10}^{-6}{\text{/°C}}^{}$
• ${\alpha }_{\text{brass}}=18\times {10}^{-6}{\text{/°C}}^{}$
• $\Delta T=100\text{°C}$

Calculating the separation:
$\text{Separation}=80\text{cm}\times (28\times {10}^{-6}-18\times {10}^{-6}){\text{/°C}}^{\times }$
$\text{Separation}=80\times (10\times {10}^{-6})\times 100\text{cm}$
$\text{Separation}=80\times {10}^{-3}\text{cm}$
$\text{Separation}=0.08\text{cm}$

To convert this result into millimeters, recall that $1\text{cm}=10\text{mm}$:
$\text{Separation}=0.08\times 10\text{mm}=0.8\text{mm}$

Therefore, the separation of the free ends of the rods is 0.8 mm.