Consider three cases, same spring is attached with 2 kg, 3 kg and I kg blocks as shown in figure. lf ${\mathrm{x}}_1, {\mathrm{x}}_2, {\mathrm{x}}_3$ be the extensions in the spring in the three cases, then

 

lf ${\mathrm{T}}_1, {\mathrm{T}}_2 , {\mathrm{T}}_3$, are the tensions in the strings in the three cases, we have

$\mathrm{In} \mathrm{case} \mathrm{I} : {\mathrm{T}}_1 = \frac{2{\mathrm{m}}_1{\mathrm{m}}_2\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2} = \frac{2\times 2\times 2\mathrm{g}}{(2+2)} = 20 \mathrm{N}$

In case II : ${\mathrm{T}}_2 = \frac{2\times 3\times 2\mathrm{g}}{(3+2)} = 24 \mathrm{N}$

In case III : ${\mathrm{T}}_3 = \frac{2\times 1\times 2\mathrm{g}}{(1+2)} = \frac{40}{3}\mathrm{N}$

At T = Kx

or $\mathrm{x} \propto \mathrm{T} \mathrm{here} {\mathrm{T}}_2 > {\mathrm{T}}_1 > {\mathrm{T}}_3$

Hence, ${\mathrm{x}}_2 > {\mathrm{x}}_1 > {\mathrm{x}}_3$