Explanation

(a) Let the original unstretched length be $l.$
$Y = \frac{Stress}{Strain} = \frac{T/A}{\mathrm{\Delta}l/l} = \frac{T}{A} \times \frac{l}{\mathrm{\Delta}l}$
Now, $Y = \frac{4}{A}$ $\frac{l}{\left( l_{1} - l \right)} = \frac{6}{A}\ \frac{l}{\left( l_{2} - l \right)} = \frac{9}{A}\ \frac{l}{\left( l_{3} - l \right)}$
$\therefore\ \ $ $4\left( l_{3} - l \right) = 9\left( l_{1} - l \right)$
$\Longrightarrow 4l_{3} + 5l = 9l_{1}\ \ \ldots(i)$
Again, $6\left( l_{3} - l \right) = 9\left( l_{2} - l \right)$
⟹$\ \ 2l_{3} + l = 3l_{2}\ \ \ \ \ldots(ii)$
Solving Eqs. (i) and (ii), we obtain
$l_{3} = (2.5l_{2} - 1.5l_{1})$