The force constant of a spring is $K$. If half of the spring is cut off,what will be the force constant of the remaining spring?

Two massless springs of spring constant $K_1$ and $K_2$ are connected in series,suspended vertically,and a certain mass is attached to the free end. If $e_1$ and $e_2$ are their respective extensions and $f$ is the stretching force,the total extension produced is:

$A$ spring of length $l$ has a force constant $k$. When a weight $W$ is attached to it,the extension produced is $x$. If the spring is cut into two equal parts and these parts are connected in parallel to support the same weight $W$,what will be the new extension?

$A$ 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 $k'$. Then they are connected in parallel and the force constant is $k''$. Then $k':k''$ is

Two springs $A$ and $B$ having spring constants $K_{A}$ and $K_{B}$ $(K_{A} = 2 K_{B})$ are stretched by applying forces of equal magnitude. If the energy stored in spring $A$ is $E_{A}$,then the energy stored in spring $B$ will be:

Two masses of $10\, kg$ and $20\, kg$ respectively are connected by a massless spring as shown in the figure. $A$ force of $200\, N$ acts on the $20\, kg$ mass. At the instant when the $10\, kg$ mass has an acceleration of $12\, m\, s^{-2}$,the acceleration of the $20\, kg$ mass is ...... $m\, s^{-2}$.