(a) Here, $\alpha(steel) = \ 1.1 \times 10^{- 5}℃^{- 1}$
$\alpha(copper) = 1.7 \times 10^{- 5}℃^{- 1}$
$\frac{l_{0}(s)}{l_{0}(c)} = \frac{\alpha(c)}{\alpha(s)} = \ \frac{1.7\ \times 10^{- 5}}{1.1 \times 10^{- 5}} = 1.545$
$\therefore\ l_{0}(s) = 1.545l_{0}(c)$
Also, $l_{0}(s) - l_{0}(c) = 5$
0.545$l_{0}(c) = 5\ $
$l_{0}(c) = \frac{5}{0.545} = 9.17\ cm$
And $l_{0}(s) = 1.545 \times 9.17\ cm\ 14.17\ cm$