Explanation

(b) $T_{1} = K\left( l - l_{1} \right)$
$T_{2} = K(l - l_{2})$
So, $\frac{T_{1}}{T_{2}} = \ \frac{l - l_{1}}{\left( l - l_{2} \right)}\ $
$\therefore\ \ T_{1}l - T_{1}l_{2} = T_{2}l - T_{2}l_{1}$
$\left( T_{1} - T_{2} \right)l = T_{1}l_{2} - T_{2}l_{1}$
$l = \ \frac{T_{1}l_{2} - T_{2}l_{1}}{(T_{1} - T_{2})}\ $
$l = (5a - 4b)$ …….(i)
$k = \frac{1}{b - a\ }$ …….(ii)
So, length of wire when tension is 9 N
$9 = kl^{'}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (l^{'} =$ change in length)
9 = $\frac{1}{(b - a)}\ \times l^{'} \Rightarrow l^{'} = \ 9b - 9a$
Hence, final length = $l + l^{'}$
= $5a - 4a + 9a - 9a$
$l_{0} = 5b - 4a$