a
(a)
Let the natural length of wire be $=1$
When only $M_1$ hanging
Using $\Delta l=\frac{F L}{A Y}$
$\left(l_1-l \right)=\frac{M_1 g \cdot l}{A Y \ldots(1)}$
When both $M_1, M_2$ hanging
$\left(l_2-l\right)=\frac{\left(M_1+M_2\right) g \cdot l}{A Y} \ldots(2)$
Dividing $(1)$ by $(2)$
$\frac{l_1-l}{l_2-l}=\frac{M_1}{M_1+M_2}$
Solving this we get
$I=\frac{M_1}{M_2}\left(l_1-l_2\right)+I_1$