A spring of force constant k is cut into length of ratio 1 : 2 : 3. They are connected in series and the new force constant is ${\mathrm{k}}^{\text{'}}$.Then they are connected in parallel and force constant is ${\mathrm{k}}^{\text{'}\text{'}}$. Then ${\mathrm{k}}^{\text{'}}$ : ${\mathrm{k}}^{\text{'}\text{'}}$ is :

As we know spring constant $k \propto \frac{1}{l}$

Length of the spring segments = $\frac{l}{6}, \frac{l}{3}, \frac{l}{2}$

So spring constants for spring segments will be ${\mathrm{k}}_1 = 6\mathrm{k}, {\mathrm{k}}_2 = 3\mathrm{k}, {\mathrm{k}}_3 = 2\mathrm{k}$

So in parallel combination,

${\mathrm{k}}^{\text{'}\text{'}} = {\mathrm{k}}_1+{\mathrm{k}}_2+{\mathrm{k}}_3 = 11\mathrm{k}$

In series combination, $\frac{1}{{\mathrm{k}}^{\text{'}}} = \frac{1}{6\mathrm{k}}+\frac{1}{3\mathrm{k}}+\frac{1}{2\mathrm{k}}=\frac{1+2+3}{6\mathrm{k}}=\frac{1}{\mathrm{k}}$

Hence ${\mathrm{k}}^{\text{'}} : {\mathrm{k}}^{\text{'}\text{'}} = 1 : 11$