In the setup shown,a $200\, N$ block is supported in equilibrium with the help of strings and a spring. At point $O$,the strings are knotted. The extension in the spring is $4\, cm$. The force constant of the spring is closest to ............ $N/m$ $[g = 10\, m/s^2]$.

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:

$A$ simple spring has length '$l$' and force constant '$K$'. It is cut into two springs of length '$l_1$' and '$l_2$' such that $l_1 = n l_2$ ($n$ is an integer). The force constant of the spring of length '$l_1$' is:

Initially,a spring is at its natural length. $A$ block of mass $2 \, kg$ is attached to the lower end,and a block of mass $0.25 \, kg$ is at the top. If the system is released from the position where the spring is at its natural length,find the maximum force exerted by the system on the floor? (in $N$,take $g = 10 \, m/s^2$)

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:

Assertion $(A)$: When a spring is cut into two equal parts,the spring constant of each piece is twice that of the original spring. Reason $(R)$: Spring constant is inversely proportional to the length of the spring.