From question. liron − lbrass = 10 cm = constant at all temperatures. Let l0 be length at temperature 0ºC and length l after change in temperature.
\(l^o_{iron}\)− \(l^o_{brass}\) = 10 cm at all temperature
∴ \(l^o_{iron}\) (1 + αiron∆t) − \(l^o_{brass}\)(1 + αbrass∆t) = 10 cm
\(l^o_{iron}\)\(\alpha^o_{iron}\) = \(l^0_{brass} \)αbrass
∴ \(\frac{l^o_{iron}}{l^o_{brass}}=\frac{1.8\times10^{-5}}{1.2\times10^{-5}}=\frac{1.8}{1.2}=\frac{3}{2}\)
Then,
\(l^o_{iron}\)- \(l^0_{brass} \) = 10
3x - 2x = 10
x = 10
Length of iron rod = 3 × 10 = 30 cm
Length of brass rod = 2 × 10 = 20 cm
The difference between is 10cm.