$A$ spring of force constant $15 \ N/m$ is cut into two pieces. If the ratio of their lengths is $1 : 3$, then the force constant of the smaller piece is . . . . . . $N/m$.

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 $g = 10 \, m/s^2$)

$A$ block of $3 \; kg$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in the figure. If spring $A$ is cut from the lower point at $t = 0$,find the acceleration of the block in $m/s^2$ at $t = 0$.

One-fourth length of a spring of force constant $K$ is cut away. The force constant of the remaining spring will be

$A$ spring has a spring constant $k$ and original length $l$. If it is cut into pieces in the ratio $\alpha : \beta : \gamma$,find the spring constant of each piece in terms of the original spring constant $k$ (where $\alpha, \beta$,and $\gamma$ are integers).

Two springs $A$ and $B$ are fixed at the top and are stretched by $8 \,cm$ and $16 \,cm$ respectively, when loads of $20 \,N$ and $10 \,N$ are suspended at the lower ends. The ratio of the spring constants of the springs $A$ and $B$ is (in $: 1$)