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:

$A$ spring of force constant $10\, N/m$ has an initial stretch $0.20\, m$. In changing the stretch to $0.25\, m$,the increase in potential energy is about.....$joule$.

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 . . . . . . $ms^{-1}$.

Two springs of spring constants $K$ and $2K$ are stretched by the same force. If $W_{1}$ and $W_{2}$ are the energies stored in them respectively,then:

$A$ spring of force constant $800\, N/m$ has an extension of $5\, cm$. The work done in extending it from $5\, cm$ to $15\, cm$ is ............. $J$.

When a spring is stretched by $2 \, cm$,it stores $100 \, J$ of energy. If it is stretched further by $2 \, cm$,the stored energy will be increased by ............. $J$.