$A$ body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1\, cm$. If the angular velocity is doubled,the elongation in the spring is $5\, cm$. The original length of the spring is ......... $cm$

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 $g=9.8 \,ms^{-2}$):

To the free end of a spring hanging from a rigid support, a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly, the amount of maximum stretching is: (force constant, $k$)

$A$ 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:

When the acceleration of the $5\ kg$ block is $2\ m/s^2$ as shown in the figure,what will be the acceleration of the $15\ kg$ block? (in $m/s^2$)

$A$ bead of mass $m = 100 \,g$ is attached to one end of a spring of natural length $L$ and spring constant $k = \frac{(\sqrt{3}+1) mg}{L}$. The other end of the spring is fixed at point $A$ on a smooth vertical ring of radius $R$. The bead is at point $B$ such that the spring makes an angle of $30^{\circ}$ with the horizontal diameter. The normal reaction at $B$ just after it is released to move is (take $g = 9.8 \,ms^{-2}$): (in $\,N$)