The force constants of two springs are $K_1$ and $K_2$. Both are stretched until their elastic potential energies are equal. If the stretching forces applied are $F_1$ and $F_2$,then the ratio $F_1:F_2$ is:

One end of a spring of spring constant $80 \ Nm^{-1}$ and unstretched length of $30 \ cm$ is fixed at point $A$ and the other end of the spring is fitted with a smooth ring of mass $300 \ g$ as shown in the figure. The ring is allowed to slide on a horizontal rod fixed at a height of $40 \ cm$. Initially the spring makes an angle of $60^{\circ}$ with the vertical and the system of spring and ring is released from rest. The speed of the ring when the spring becomes vertical is . . . . . . $ms^{-1}$.

$A$ spring $40 \, mm$ long is stretched by the application of a force. If $10 \, N$ force is required to stretch the spring by $1 \, mm$,then the work done in stretching the spring by $40 \, mm$ is ............. $J$.

The potential energy of a certain spring when stretched through a distance $S$ is $10 \, J$. The amount of work (in $J$) that must be done on this spring to stretch it through an additional distance $S$ will be

Two springs have spring constants $k_{A}$ and $k_{B}$ such that $k_{A} > k_{B}$. If both springs are stretched by the same extension $x$,the work required will be:

$A$ bungee jumper is jumping with the help of an elastic ideal rope (force constant $K$). The jumper steps off the bridge and falls from rest towards the river below. He does not hit the water. The mass of the jumper is $m$,and the natural length of the rope is $l$. Gravity is $g$. Assuming everything is ideal,choose the incorrect option.