$A$ force of $20\,dyne$ applied to the end of a spring increases its length by $1\,mm$. What will be the force constant of the spring?

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 three equal parts. The force constant of each part would be

$A$ spring of force constant $k$ is cut into two pieces such that one piece is three times the length of the other. The longer piece will have a force constant of

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:

$A$ spring of force constant $15 \ N/m$ is cut into two pieces. If the ratio of their lengths is $1 : 3$, then the force constant of the smaller piece is . . . . . . $N/m$.