The difference in length between two rods \(\mathrm{A}\) and \(\mathrm{B}\) is \(60 \mathrm{~cm}\) at all temperatures. If \(\alpha_{\mathrm{A}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}\) and \(\beta_{\mathrm{B}}=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}\), the lengths of the two rods are

Given: \(\Delta l=60 \mathrm{~cm}, \alpha_{\mathrm{A}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}\),
\(\alpha_{\mathrm{B}}=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}\)
\(\Delta l\) is constant at all temperatures.
We know \(\Delta l=l \alpha \Delta \mathrm{t}\)
Let the length of the rods at a temperature \(0^{\circ} \mathrm{C}\) be \(l_{\mathrm{A}}\) and, \(l_{\mathrm{B}}\)
\(\therefore \quad\) At temperature \(\mathrm{t}^{\circ} \mathrm{C}\)
\(\begin{array}{ll}
& l_{\mathrm{A}} a_{\mathrm{A}} \mathrm{t}_{\mathrm{A}}=l_{\mathrm{B}} a_{\mathrm{B}} \mathrm{t}_{\mathrm{B}} \\
& l_{\mathrm{A}}(18) \times 10^{-6}=l_{\mathrm{B}}(27) \times 10^{-6}.. (i) \\
& \Delta l=l_{\mathrm{A}}-l_{\mathrm{B}} \\
\Delta l & =\frac{3}{2} l_{\mathrm{B}}-l_{\mathrm{B}} .. from (i)\\
& \Delta l=\frac{1}{2} l_{\mathrm{B}} \\
\therefore \quad & l_{\mathrm{B}}=2 \Delta l \\
\therefore \quad & l_{\mathrm{B}}=2 \times 60=120 \mathrm{~cm} \\
\therefore \quad & l_{\mathrm{A}}=\frac{3}{2} \times 120=180 \mathrm{~cm}
\end{array}\)