$A$ spring whose unstretched length is $\ell$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $\ell_1$ and $\ell_2$ where $\ell_1 = n\ell_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants $k_1$ and $k_2$ will be:

$A$ spring of force constant $k$ is cut into three equal parts. The force constant of each part would be

$A$ uniform spring of force constant $k$ is cut into two pieces,the lengths of which are in the ratio $1 : 2$. The ratio of the force constants of the shorter and the longer pieces is

The length of a spring is $l$ and its force constant is $k$. When a weight $W$ is suspended from it,its length increases by $x$. If the spring is cut into two equal parts and put in parallel and the same weight $W$ is suspended from them,then the extension will be

Two blocks with masses $100 \text{ g}$ and $200 \text{ g}$ are attached to the ends of springs $A$ and $B$ as shown in the figure. The energy stored in spring $A$ is $E$. The energy stored in spring $B$,when the spring constants $k_{A}$ and $k_{B}$ of springs $A$ and $B$ respectively satisfy the relation $4k_{A} = 3k_{B}$,is:

If a spring is extended to length $l,$ then according to Hooke's law,