Two springs of force constants $K_1$ and $K_2$ are loaded with weights $W_1$ and $W_2$ respectively. Assume that the length of each spring is increased by the same amount. If $K_1 = 2 K_2$, then the ratio $\frac{W_2}{W_1}$ is

If a spring is extended to length $l,$ then according to Hooke's law,

$A$ spring of force constant $k$ is cut into three equal pieces. If these three pieces are connected in parallel,the force constant of the combination will be

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

$A$ spring of force constant $k$ is cut into two equal halves. The force constant of each half is

When a $1.0\,kg$ mass hangs attached to a spring of length $50\,cm$,the spring stretches by $2\,cm$. The mass is pulled down until the length of the spring becomes $60\,cm$. What is the amount of elastic energy stored in the spring in this condition,if $g = 10\,m/s^2$? (in $Joule$)