A spring of force constant $$k$$ is cut into lengths of ratio $$1:2:3.$$  They are connected in series and the new force constant is $$k'.$$  If they are connected in parallel and force constant is $$k'',$$  then 1 $$k':k''$$  is

Solution :
When the spring is cut into pieces, they will have the new force constant. The spring is divided into $$1:2:3$$   ratio.
When the pieces are connected in series, the resultant force constant
$$\eqalign{ & \frac{1}{{v'}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}} \cr & \frac{1}{{v'}} = \frac{1}{x} + \frac{1}{{2x}} + \frac{1}{{3x}} \cr & v' = \frac{{6x}}{{11}} \cr} $$
In parallel, the net force constant
$$K'' = x + 2x + 3x = 6x = 11K$$
The required ratio $$\frac{K}{{K''}} = \frac{{\frac{{6x}}{{11}}}}{{6x}} = 1:11$$