For metal-1,
\(a_1 = \frac{Δl}{lΔt} = \frac{0.10}{50.0 \times 100}\)
\(a_1 = 2.0 \times 10^{-5 }/ °C\)
For metal -2,
\(a_2\) = \(\frac{Δl}{lΔt}\) = \(\frac{0.08}{80.0 \times 100}\)
\(a_2\) = \(1.0 \times 10^{-5} / °C\)
Let the lengths of metal - 1 and metal - 2 in the third rod at 0°C be l1 and l2 respectively.
Then l1 + l2 = 50.0 ....(1)
When this rod is heated to 50°C, then