$A$ 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?

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:

The system shown in the figure is in equilibrium and at rest. The spring and string are massless. Now, the string is cut. The acceleration of mass $2m$ and $m$ just after the string is cut 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 shown,the $10 \, kg$ mass has an acceleration of $12 \, m/s^2$ towards the right. The acceleration of the $20 \, kg$ mass at this instant is ........ $m/s^2$.

Two identical springs have the same force constant $73.5 \,Nm^{-1}$. The elongation produced in each spring in the three cases shown in Figure-$1$,Figure-$2$,and Figure-$3$ are (given $g=9.8 \,ms^{-2}$):

When a body is suspended from a spring and released slowly,the extension in the spring is $x$. If the body is released suddenly,the extension in the spring is: